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# 6 8 7 36 64 4 24 48 ?
1-20 of 22
• so, let us solve this-
(6)2=36 and , 6×(6−2)=24
Following this pattern-
(7)2=49 and 7×(7−2)=35
0
• 18
0
• 18
0
• 18
0
• 6 8 7
3 64 4
24 48 ? find the vealu sir
0
• 18
1
• 18
0
• 18
1
• 18
0
• 18
0
• 3
0
• 18
0
• 18
0
• 31
0
• please give the option also
1
• 42
0
• 0
0
• -10. 36-(2*6) =24 64-(2*8) =48 4-(2*7)=-10
1
• 18
-1
• 18 because 36-24=12 which is double of 6 and 64-48=16 which is double of 8 so it will be 18-4=14 .so answer is 18
6
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+0
# I been working on this for minutes and i dont know how to solve this
+2
405
11
+37
Question: What are the values of m and n?
Picture: http://i66.tinypic.com/2r2qix5.png
A. m=65, n=21
B. m=65, n=89
C. m=115, n=21
D. m=115, n=89
Rebirth Mar 29, 2017
#1
+229
0
SO, we know that angle B is 26 degrees, so the 4N+7 angle is also 26 degrees, because of the corresponding angles therum. Does that help in any way?
liveevillevi Mar 29, 2017
#3
+93338
0
That is not correct liveevillevi
Melody Mar 29, 2017
#2
+93338
+1
Well you know that the two angles at C add to a straight angle (how many degrees is that?), you can use that to find n.
then
The two interior angles of a triangle = the opposite exterior angle
Use that to find m
Melody Mar 29, 2017
#5
+37
+1
So am i suppose to mutiply 21 or 89 to n to see if they equal 180 degrees?
Rebirth Mar 29, 2017
#4
+37
+1
im still not getting it
Rebirth Mar 29, 2017
edited by Rebirth Mar 29, 2017
edited by Rebirth Mar 29, 2017
#6
+88899
+2
Note that the two angles designated as 4n + 7 and 4n + 5 are straight angles.....they sum to 180
So...let's solve for n....and we have
4n + 7 + 4n + 5 = 180 simplify
8n + 12 = 180 subtract 12 from both sides
8n = 168 divide by 8 on both sides
n = 21
Now....by the Exterior Angle Theorem....m equals the sum of [ 4n + 5] + [26]
And n = 21......so....m = 4(21) + 5 + 26 = 84 + 5 + 26 = 89 + 26 = 115
CPhill Mar 29, 2017
#7
+37
+1
first where did you get 8n and 12 from?
Rebirth Mar 29, 2017
edited by Rebirth Mar 29, 2017
#9
+37
+1
oh nm
Rebirth Mar 29, 2017
#8
+88899
+1
Note, Rebirth....adding the angles 4n + 7 and 4n + 5....we want to combine like terms.....so
4n + 4n + 5 + 7 =
8n + 12
CPhill Mar 29, 2017
#10
+37
+2
Thanks i understand it now
Rebirth Mar 29, 2017
edited by Rebirth Mar 29, 2017
#11
+93338
+2
That is great :)
Melody Mar 29, 2017
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# ECO101H1 Lecture Notes - Lecture 6: Tennis Ball, Toronto Transit Commission, Demand Curve
19 views5 pages
1 Aug 2010
School
Department
Course
Professor
Thursday, October 1st, 2009.
Elasticity of Demand
Insight: If the choice is between A or B, you need information about A and B.
If you have decided on A, information about B (only) is required to identify the
opportunity cost of A if B is the next best alternative to A.
Total Revenue Ā± Price x Quantity
Questions:
1) If Toronto Transit commission wished to increase Total Revenue, should it
increase price (fare)?
2) You own one of the 5 toll bridges across a river, each of which sets the price (toll)
at \$5.00.
If you wish to increase Total Revenue, should you increase price (toll)?
Fare increase: (1) For those who continue to ride the TTC, fares will be higher,
BUT
(2) There will be fewer riders (per the law of downward-sloping
demand)
Result: Law of downward-sloping demand is not sufficient to answer question.
(Price) Elasticity of Demand
Definition: Percent change in quantity demanded /
Percent change in price
% ĀØ4D
īīĀØ3ī (Ignore minus sign)
Measures responsiveness of quantity demanded to a change in price.
Mid-point Convention
īīĀØ4 = ĀØ4
Q (Q = Average quantity)
īīĀØ3ī ĀØī3
P (P = average price)
www.notesolution.com
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Example:
Situation Price Quantity Demanded
A 0.90 1100
B 1.10 900
īīĀØī4XDQWLW\ī'HPDQGHGī = 200 x 100
(1100 + 900)/2
= 200 x 100
1000
= 20%
īīĀØī3ULFH = 0.2 x 100
(0.9 + 1.10)/2
= 0.20 x 100
1.00
= 20%
ELASTICITY = 20% = 1.0
20%
Insight: same result if go up demand curve from A to B or down demand curve
from B to A!
Terminology:
perfectly inelastic: elasticity = 0
inelastic: 0 < elasticity < 1
unit elastic: elasticity = 1
elastic: 1 < elasticity < ī
perfectly elasticīīīHODVWLFLW\ī īī
#1 % ĀØ3īīī īīī
īīĀØ4D = 20 elasticity = 20 / 10 = 2
Ć Quantity demanded is very responsive to a change in price
#2 % ĀØ3īīī īīī
īīĀØ4D = 2 elasticity = 2 / 10 = 0.2
Ć Quantity demanded is not very responsive to a change in price
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IIT-JEE
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Question From class 11 Chapter ELLIPSE
# The length of the major axis of an ellipse is 20 units and its foci are <br> Find the equation of the ellipse.
Solution :
Let the equation of the ellipse be <br> Length of the major axis = <br> Foci are <br> Now, <br> the required equation is
Find the equation of the ellipse, whose length of the major axis is 20 and foci are .
1:58
The equation of the ellipse whose length of the major axis is 10 units and co-ordinates of the foci are is
2:22
If the length of the latus rectum of an ellipse with major axis along x-axis and centre at origin is 20 units, distance between foci is equal to length of minor axis, then find the equation of the ellipse.
5:35
Find the equation of the ellipse having, length of major axis 8 and foci
1:57
Find the equation of the ellipse having, length of major axis 16 and foci
3:29
Find the equation of the ellipse having, length of major axis 26 and foci
2:13
If the length of the latus rectum of an ellipse with major axis along y-axis and centre at origin is 6 units, distance between foci is equal to length of minor axis, then the equation of the ellipse.
3:32
Find the equation of the ellipse whose major axis is 12 and foci are .
2:47
Find the equation of ellipse whose vetices are and foci are
2:35
The eccentricity of an ellipse is and the distance between its foci is 4 units. If the major and minor axes of the ellipse are respectively along the axes, find the equation of the ellipse
2:39
Equation of the ellipse whose foci are and the major axis is of length is
4:03
Find the equation of the ellipse whose vertices are at and foci at
3:42
The equation of the ellipse whose vertices and foci are and respectively, is
2:11
Find the equation of the ellipse whose vetices are and foci are .
1:22
Find the equation of the ellipse whose major axis is 8 and co-ordinates of foci are
2:39
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# Exercise 3.2.13 Introduction to Real Analysis by Jiri Lebl
Let $$f:S \to \mathbb{R}$$ be a function and $$c \in S$$, such that for every sequence $$\{x_n\}$$ in $$S$$ with $$\lim x_n = c$$, the sequence $$\{f(x_n)\}$$ converges. Show that $$f$$ is continuous at $$c$$.
Suppose that $$f$$ is not continuous at $$c$$. Then, $$\exists \epsilon > 0$$ s.t. for every $$\delta> 0$$, $$\exists x$$ s.t. $$|x-c| < \delta$$, but $$|f(x) - f(c)| \ge \epsilon$$. Let $$\{x_n\}$$ be a sequence s.t. $$|x_n - c| < \frac1n$$. Then, $$|f(x_n) - f(c) | \ge \epsilon$$ when $$n$$ is large enough. But, I think that $$f(x_n)$$ can still converges to somewhere else other than $$f(c)$$. How can I proceed from here?
I appreciate if you give some help.
Consider any $$(x_{n})$$ such that $$x_{n}\rightarrow c$$, consider also the constant sequence $$(c,c,...)$$ and consider further that $$(y_{n}):=(x_{1},c,x_{2},c,...)$$, the later also converges to $$c$$ and $$(f(y_{n}))$$ has the constant subsequence $$(f(c),f(c),...)$$, so $$f(y_{n})\rightarrow f(c)$$ and also that $$f(x_{n})\rightarrow f(c)$$ as $$(f(x_{1}),f(x_{2}),...)$$ is also a subsequence of the convergent sequence $$(f(y_{n}))$$.
So we have proved that, for any $$(x_{n})$$ such that $$x_{n}\rightarrow c$$, then $$f(x_{n})\rightarrow f(c)$$, this is another characterization of continuity.
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# Tag Info
7
If you know Ohm's Law (which you should) and you realise that the ADC measures voltage, you should be able to work it out from there. But I will go into minute detail for you to ensure you understand. Ohm's Law defines the relationship between Voltage (V), Current (I) and Resistance (R). R = V/I To find one unknown value (in your case R) you need to know ...
6
Since you already know how to hook up a pot and read it's value, it's quite simple. You already have it hooked up in a voltage divider configuration and are getting a 10 bit (0-1023) value using analogRead(), you just need to decide what to do with it. Assume potVal has the value of the pot.; if (potVal < 256) { // Pot is one quarter turn or less } ...
5
A0 input is high impedance, so very little current is flowing into A0. The value is so small that you can simplify that it flows only from +5V through the potentiometer to GND. For some applications, when you are measuring high-impedance sources (a source that can deliver very little current) this simplification may not be correct. You can find more ...
5
analogRead() returns a number from 0 to 1023, and you want to convert that to a range from 0 to 255? One easy way would be to simply divide the reading by 4. On a more general note, given that the numbers will (should!) never be negative, there's a flavour of int called unsigned int. By using that: unsigned int val = analogRead(...); you can cover a range ...
5
A potentiometer is an analog device, meaning it communicates its position by varying the voltage rather than sending discrete ones and zeros. If a linear potentiometer is turned half way then the output of the pot will be at half the voltage of the input. If it is turned all the way then the output will equal the input voltage and if it is not turned at ...
4
Ah, now I see. Your confusion in this case is perfectly understandable. They threw a couple of capacitors into the circuit without explaining why they were added in the first place. In order to understand why capacitors were added you need to go a little deeper into electronics theory, specifically capacitance and inductance, what they are and how they work. ...
4
You know how wide the display is. You know how wide the text is. You know the position of the potentiometer. With those you can determine where in the text (if the text is wider) or display (if the display is wider) you need to start drawing from and where you need to stop drawing. Subtracting the text width from the display width will give you the range for ...
4
There are a few ways of doing what you want. One would be to use an analog multiplexer as in this answer: Expanding analog inputs to the Arduino Another would be to use digital I/O pins to control the power and ground to the potentiometers. Note that both sides of the pot need to be controlled, not just one, since any connection of one will interfere with ...
4
There is a lengthy discussion about this on the Arduino Forum. As I understand it, the input impedance is only important if you are sampling a rapidly varying signal, as it takes a finite time to charge the sample/hold capacitor. However your potentiometer would not vary rapidly (certainly not in the order of thousands of herz). Thus I don't think the ...
4
You are looking for the bad solution for a simple problem. As Ignacio Vazquez-Abrams puts it, the problem is you using delay(). The simplest solution is to manage the timings with millis(). See the Blink Without Delay Arduino tutorial.
4
Analog pins an only read analog values, not write them. analogWrite() is a deceptive name. It should really be PWMWrite() since all it does is PWM. And that, of course, needs PWM pins.
4
You can likely user an I/O pin as an output to "power" fairly high-impedance devices such as pullup resistors for buttons, or even potentiometers of moderate to high total resistance. A reason you might wish to do so would to be able to stop the device from draining power while the entire system is in a low-power sleep mode, or to otherwise disable it ...
4
pinMode( K1, INPUT ); with if(K1 == LOW && K2 == LOW) makes no sense. K1 is a pin number. LOW is a macro for 0 (HIGH for 1). You are comparing a static pin number against a static number. If K1 is not equal to 0, the inner part of that if statement can never be reached. What you meant to do was "if the voltage at the pin K1 is low". For ...
4
You need to use Serial.print instead of Serial.write.
3
No, it would not be correct to use a single resistor to replace a potentiometer. On an LCD the potentiometer is used to adjust the bias level of the LCD - that is the contrast. You need to use it to set a voltage between Vcc and Vee, which you feed into Vo. That is, a voltage somewhere between +5V and -5V. You can't do that with one resistor. You can, ...
3
I made this video and github just for this about a year ago. And here is the code: /* 2-axis joystick connected to an Arduino Micro to output 4 pins, up, down, left & right If you are using pull down resistors, change all the HIGHs to LOWs and LOWs to HIGH. This skectch is using pull up resistors. */ int UD = 0; int LR = 0; /* Arduino Micro output ...
3
All the relay boards I've come across are 'active low' i.e. as you've discovered bringing the control pin down to earth activates the relay. As an aside just be careful that the arduino has enough juice to power the relays, a separate supply is normally required if these are coil rather than ssr type relays
3
Adding to Ricardo's answer: PWM can do the job, sort of. You need a low pass filter (LPF) to convert the PWM to an analog voltage and an amplifier to take it from 0-5 to 0-10V. Or you can get a dedicated digital-to-analog (DAC) chip but that will obviously be more expensive than a resistor and capacitor for the LPF. Here's some tutorials on converting your ...
3
Digital potentiometers are not a good choice for controlling the speed of a motor. They are designed only for very low current applications - things like setting offset points for op-amps and things like that - things where you would use a small trimmer pot rather than a large power rheostat. Instead you should be controlling the motor using PWM and a ...
3
Assuming A0 is set to input then it is a high-impedance input. That is, the input has a high resistance, and only consumes enough current to "sample" the voltage.
3
From your description of switch wiring, it sounds like you have one of the switch leads attached to +5 V and the other switch lead to an input. As noted in comments and answers to “Why all unset pins of my Arduino Uno are outputting 2 volts?”, such an input will read high when the switch is closed, and will read randomly when it is open, because an input ...
3
Avoid delay. Very simplistic alternative: unsigned long lastLedToggleTime; unsigned long currentLedToggleInterval; void loop() { ... unsigned long now = millis(); if ( (now - lastLedToggleTime) > currentLedToggleInterval) { lastLedToggleTime = now; // We need to toggle the LED now. if ( digitalRead( ledPin ) == LOW ) { // LED ...
3
Yes, Arduino is a 5V microcontroller, meaning it can directly control 5V circuits only. You might incur in problems also trying to deal with 3.3V circuits, even if Arduino is pretty friendly to 3.3V counterparts. That does not mean though you cannot control higher voltage circuits, just not directly. Actually if anything above 6V is connected to an Arduino ...
3
I had hooked up the positive of led to a digital input of the arduino and negative to the ground of the board. You shouldn't have done that. It damages both the LED and the Arduino. The care and feeding of LEDs So I am assuming that the digital pins in arduino are positive (are they 5v?). Yes and yes (assuming a 5V Arduino). So why cant the pot be ...
3
Skip to the end to get the easy answer! Your choice of thermistor and bias resistor values depends on the following: 1. How much current do you want to spend? 2. How do you want the thermistor resistance to compare with the wiring resistance? 3. How much filtering do you need? 4. What kind of linearization do you want to apply? A lower value thermistor ...
3
The standard way to measure a resistive sensor is to make a voltage divider with the sensor and a fixed resistance between Vcc and GND, like this: Vcc │ Rb │ ├─── analog input │ Rx │ GND where Rx is the resistance you want to measure and Rb is a known bias resistance. The only unknown is the choice of the bias resistance. The simple rule of thumb ...
3
The Adafruit website says this: There is a nominal 10K resistance across the two outer leads. The middle pin resistance with respect to either of the outer pins changes depending on where on the strip one presses. When no pressure is applied, the middle pin floats, so be sure to use some sort of weak pullup, such as 100K ohm. To use Connect one side pin to ...
3
write() is a function for binary data. I you want to print the number as text, use print() or println(). Serial.println(potentioVal);
3
Yes. All a potentiometer is, is two resistors end to end. The "wiper" is the join between the two resistors. It doesn't matter too much what value potentiometer (and hence what value resistors) you use, as long as it's "around" the 10kΩ mark (100kΩ should be fine, 50kΩ, 1kΩ etc - anything below 1kΩ may not work) - what matters is the ratio of the values of ...
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# Algebraically closed group
Algebraically closed group
In mathematics, in the realm of group theory, a group $A$ is algebraically closed if any finite set of equations and inequations that "make sense" in $A$ already have a solution in $A$. This idea will be made precise later in the article.
Informal discussion
Suppose we wished to find an element $x$ of a group $G$ satisfying the conditions (equations and inequations):
::$x^2=1$::$x^3=1$::$x e 1$
Then it is easy to see that this is impossible because the first two equations imply $x=1$. In this case we say the set of conditions are inconsistent with $G$. (In fact this set of conditions are inconsistent with any group whatsoever.)
However if we extend the group $G$ to the group $H$ with multiplication table:
Then the condions have two solutions, namely $x=b$ and $x=c$.
Thus there are three possibilities regarding such conditions:
* They may be inconsistent with $G$ and have no solution in any extension of $G$.
* They may have a solution in $G$.
* They may have no solution in $G$ but nevertheless have a solution in some extension $H$ of $G$.
It is reasonable to ask whether there are any groups $A$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.
Formal definition of an algebraically closed group
We first need some preliminary ideas.
If $G$ is a group and $F$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $G$ we mean a pair of subsets $E$ and $I$ of $Fstar G$ the free product of $F$ and $G$.
This formalizes the notion of a set of equations and inequations consisting of variables $x_i$ and elements $g_j$ of $G$. The set $E$ represents equations like: ::$x_1^2g_1^4x_3=1$::$x_3^2g_2x_4g_1=1$::$dots$The set $I$ represents inequations like::$g_5^\left\{-1\right\}x_3 e 1$::$dots$
By a solution in $G$ to this finite set of equations and inequations, we mean a homomorphism $f:F ightarrow G$, such that $ilde\left\{f\right\}\left(e\right)=1$ for all $ein E$ and $ilde\left\{f\right\}\left(i\right) e 1$ for all $iin I$. Where $ilde\left\{f\right\}$ is the unique homomorphism $ilde\left\{f\right\}:Fstar G ightarrow G$ that equals $f$ on $F$ and is the identity on $G$.
This formalizes the idea of substituting elements of $G$ for the variables to get true identities and inidentities. In the example the substitutions $x_1mapsto g_6, x_3mapsto g_7$ and $x_4mapsto g_8$ yield:::$g_6^2g_1^4g_7=1$::$g_7^2g_2g_8g_1=1$::$dots$::$g_5^\left\{-1\right\}g_7 e 1$::$dots$
We say the finite set of equations and inequations is consistent with $G$ if we can solve them in a "bigger" group $H$. More formally:
The equations and inequations are consistent with $G$ if there is a group$H$ and an embedding $h:G ightarrow H$ such that the finite set of equations and inequations $ilde\left\{h\right\}\left(E\right)$ and $ilde\left\{h\right\}\left(I\right)$ has a solution in $H$. Where $ilde\left\{h\right\}$ is the unique homomorphism $ilde\left\{h\right\}:Fstar G ightarrow Fstar H$ that equals $h$ on $G$ and is the identity on $F$.
Now we formally define the group $A$ to be algebraically closed if every finite set of equations and inequations that has coefficients in $A$ and is consistent with $A$ has a solution in $A$.
Known Results
It is difficult to give concrete examples of algebraically closed groups as the following results indicate:
* Every countable group can be embedded in a countable algebraically closed group.
* Every algebraically closed group is simple.
* No algebraically closed group is finitely generated.
* An algebraically closed group cannot be recursively presented.
* A finitely generated group has solvable word problem if and only if it can embedded in every algebraically closed group.
The proofs of these results are, in general very complex. However a sketch the proof that a countable group $C$ can be embedded in an algebraically closed group follows.
First we embed $C$ in a countable group $C_1$ with the property that every finite set of equations with coefficients in $C$ that is consistent in $C_1$ has a solution in $C_1$ as follows:
There are only countably many finite sets of equations and inequations with coefficients in $C$. Fix an enumeration $S_0,S_1,S_2,dots$ of them. Define groups $D_0,D_1,D_2,dots$ inductively by:
::$D_0 = C$
::
Now let:
::$C_1=cup_\left\{i=0\right\}^\left\{infty\right\}D_\left\{i\right\}$
Now iterate this construction to get a sequence of groups $C=C_0,C_1,C_2,dots$ and let:
::$A=cup_\left\{i=0\right\}^\left\{infty\right\}C_\left\{i\right\}$
Then $A$ is a countable group containing $C$. It is algebraically closed because any finite set of equations and inequations that is consistent with $A$ must have coefficients in some $C_i$ and so must have a solution in $C_\left\{i+1\right\}$.
References
* A. Macintyre: On algebraically closed groups, ann. of Math, 96, 53-97 (1972)
* B.H. Neumann: A note on algebraically closed groups. J. London Math. Soc. 27, 227-242 (1952)
* B.H. Neumann: The isomorphism problem for algebraically closed groups. In: Word Problems, pp 553-562. Amsterdam: North-Holland 1973
* W.R. Scott: Algebraically closed groups. Proc. Amer. Math. Soc. 2, 118-121 (1951)
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Mathematics
Easy
Question
# The approximate annual interest rate r of a monthly installment loan is given by the formula:Where n is the total number of payments, m is the monthly payment, and p is the amount financed. Find the approximate annual interest rate (to the nearest percent) for a four-year signature loan of \$ 20,000 that has monthly payments of \$ 500
Hint:
## The correct answer is: 9 %
### So now we know that the cost you pay annually for borrowing money, including fees, is stated as a percentage and is known as the annual percentage rate (APR).We have given the expression as:
Here we sed the concept of simple interest, compound interest and found out the annual rate. Simple Interest (S.I.) is a way for figuring out how much interest will accrue on a specific principal sum of money at a certain rate of interest. So the annual rate percent is 7.2%.
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# Problem #2105
2105 A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars would increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume, by what percent must the height be decreased? $\text {(A)} 10\% \qquad \text {(B)} 25\% \qquad \text {(C)} 36\% \qquad \text {(D)} 50\% \qquad \text {(E)}60\%$ This problem is copyrighted by the American Mathematics Competitions.
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• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
• Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).
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# q ) a 10kg case sits on a scale in an elevator moving upwards at 1.5m/s^2. what force does the scale 1 answer below »
q ) a 10kg case sits on a scale in an elevator moving upwards at 1.5m/s^2. what force does the scale read?
q ) A 0.35 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 750 N/m) whose other end is fixed. The mass has a kinetic energy of 9.0 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?
At what rate is the spring doing work on the mass when the spring is compressed 0.077 m and the mass is moving away from the equilibrium position?
## 1 Approved Answer
lakshaya
3 Ratings, (9 Votes)
mass= 10 kg acceleration = 1.5 m/s ^2 the elevator must support its weight = mg = 10 * 9.8 =98 newtons acceleration = 1.5 m/s ^2 so net force = 10 * 1.5 m/s ^2=15 newton is required to accelerate the mass This requires a support force of F=113 newtons 2 ) My Answer: Let: v be the velocity, x be the displacement, w be the angular frequency, a be the amplitude, k be the spring constant, K be the kinetic energy at displacement x, K0 be the kinetic energy at the equilibrium position, P be the required power. w^2 = k / m v^2 = w^2(a^2 - x^2) = k(a^2 - x^2) / m v = sqrt[ k(a^2 - x^2) / m ] ...(1) K = mv^2 / 2 = k(a^2 - x^2) / 2...
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# 2 1/2 X 1/3
2 1/2 X 1/3. · ahmed elnaiem dec 11, 2014 step 1 make the right side of the equation a fraction: Click here👆to get an answer to your question ️ solve: Click here👆to get an answer to your question ️ solve: This question contains a modulus operator so we have to do in two steps.
1 of those 8 slices would constitute the numerator of a fraction, while the total of 8. Click here👆to get an answer to your question ️ solve: X2 −13×2 + 2x −1 = −12 step.
## Compute answers using wolfram's breakthrough technology & knowledgebase,.
Istock mathematics is rooted in numbers, yet alphabetic symbols have long played a part in complex math and science. The resulting fraction is in the reduced form. 1 2 x 1 3.
## 2 1/3 X 3 1/2.
· ahmed elnaiem dec 11, 2014 step 1 make the right side of the equation a fraction: This calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. The resulting fraction is in the reduced form. 1 of those 8 slices would constitute the numerator of a fraction, while the total of 8.
### Now That These Fractions Have Been Converted To Have The Same Denominator, We Can Add The.
We now have 63( 22x−3)− 62( 35(x− 1)) = 6(−4).
### Kesimpulan dari 2 1/2 X 1/3.
1 x 3 2 x 3 + 1 x 2 3 x 2. Click here👆to get an answer to your question ️ solve: Nuestra respuesta es comprensible y explicada paso a paso.
See also In Reconciliation Statement Depreciation Overcharged In Financial Accounts Is
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# How do you find the maximum value of emf?
## How do you find the maximum value of emf?
E.m.f. generated in a rotating coil
1. E = BANωsinθ = BANωsin(ωt) The maximum value of the e.m.f (Eo) is when θ (= ωt) = 90o (that is, the coil is in the plane of the field, Figure 2) and is given by.
2. Maximum e.m.f (Eo) = BANω
3. The r.m.s value of the e.m.f is (Er.m.s) = BANω/21/2
## How do you calculate the maximum emf induced in a coil?
Section Summary. emf=2Bℓvsinωt emf = 2 B ℓ v sin , where A is the area of an N-turn coil rotated at a constant angular velocity ω in a uniform magnetic field B.
READ ALSO: Who is the DC equivalent of Doctor Doom?
At what degree is the emf maximum?
If the coil rotates with a constant angular velocity, the induced emf will vary like a sin function as in Eq (1). It will be zero when θ = ωt = 0, when the plane of the coil is normal to the field, and it will be maximum when θ = ωt = 90, or when the plane of the loop is parallel to the field.
### How is emf induced in the secondary coil?
We can make the magnetic field carry on changing by using an alternating current in the primary coil. This produces a magnetic field in the secondary coil. Since the current is alternating, the magnetic field also alternates back and forth. This changing magnetic field induces an alternating EMF in the secondary coil.
### What is the value of induced emf in the coil in AC generator?
Now, when the angle between the normal of plane and magnetic field is zero degrees, the induced emf is zero i.e.
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How do you calculate coil emf?
The induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with magnetic field.
## Why is the emf maximum?
The magnitude of induced e.m.f. is maximum, when the plane of the armature of a.c. dynamo is: When the plane of the armature is parallel to the lines of force of magnetic field, the rate of change of magnetic flux linked with it is maximum. Therefore, the e.m.f. induced in the armature in this orientation is maximum.
## What is the maximum emf produced by AC generator?
The maximum e.m.f. induced in the coil depends on the number of turns of the coil, the area of the coil, the strength of the magnetic field and the angular frequency. Hence, the maximum e.m.f. induced in the coil is equal to the maximum voltage produced i.e., maximum e.m.f. produced is \$250\,V\$.
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What is AC emf?
An AC electromagnetic field, also known as EMF or EM field, is a physical field produced by moving electrically charged objects. The field can be viewed as the combination of an AC electric field and an AC magnetic field. Since it is AC by definition, it continually changes polarity from positive to negative.
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1. ## rational exponents? Solve please
(-3a1/4)(9a)-3/2
2. ## Re: rational exponents? Solve please
Originally Posted by zbest1966
(-3a1/4)(9a)-3/2
Here are some properties you should know:
$a^ma^n = a^{m+n}$
$(ab)^n = a^nb^n$
$a^{m/n} = \sqrt[n]{a^m}$
3. ## Re: rational exponents? Solve please
I understand the properties but how do you solve it.
4. ## Re: rational exponents? Solve please
Originally Posted by zbest1966
I understand the properties but how do you solve it.
If you understood the properties, I suspect you would not be having difficulties with the problem. By the way, we have nothing to "solve" here, this is not an equation. Rather, we are "simplifying" the expression.
$\left(-3a^{1/4}\right)(9a)^{-3/2}$
$=-3a^{1/4}\cdot9^{-3/2}a^{-3/2}$
$=-\frac{3a^{1/4}}{9^{3/2}a^{3/2}}$
$=-\frac{3a^{1/4}}{27a^{3/2}}$
Now, use the properties $\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac1{a^n}$ to finish it:
$=-\frac{a^{-5/4}}{9}=-\frac1{9a^{5/4}}$
As a final step, you may want to convert $a^{5/4}$ to radical form, depending on the problem's instructions.
5. ## Re: rational exponents? Solve please
Originally Posted by zbest1966
I understand the properties but how do you solve it.
You could actually apply the properties you say you understand or you could ask someone else to solve it for you. I see you chose the second solution method.
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Triangle QRS is a right triangle. Complete the similarity statement. ΔSTR ~ Δ
Students were inquired to answer a question at education and to tell what is most important for them to succeed. Of many results, the one which that stood out was practice. People who are successful do not become successful by being born. They work hard and persistence their lives to succeeding. This is how you can accomplish your goals. These are are one of the answer and question example that you could actually utilize to practice and expand your information and also give you insights that could just assist you to keep up your study in school.
Question:
Triangle QRS is a right triangle. Complete the similarity statement. ΔSTR ~ ΔTQRRSTSQRRTQ
ΔSTR is similar to ΔRTQ
Step-by-step explanation:
Given QRS is a right angled triangle. we have to find the similarity statement ΔSTR ~ Δ__
Let ∠S=x
In ΔSTR, by angle sum property
∠S+∠STR+∠SRT=180°
∠SRT=90°-x
In ΔSRQ, by angle sum property
∠S+∠R+∠Q=180°
∠Q=90°-x
In ΔSTR and ΔRTQ
∠SRT=∠Q=90°-x (proved above)
∠STR=∠RTQ (each 90°)
RT=RT (common)
Hence, by AAS rule ΔSTR≅ΔRTQ
∴ ΔSTR is similar to ΔRTQ
Option 4 is correct.
From the answer and question examples above, hopefully, they can guide the student take care of the question they had been looking for and keep in mind of the whole thing declared in the answer above. You will be able to then have a discussion with your classmate and continue the school learning by studying the problem jointly.
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# Morales 2.3 Multiplication Quiz
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• 1.
### 27 X 10 2
Explanation
The given multiplication problem is 27 multiplied by 10 to the power of 2. To solve this, we first calculate 10 to the power of 2, which is 100. Then, we multiply 27 by 100, resulting in 2,700.
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• 2.
### 80 X 102
Explanation
The given question is asking to multiply 80 by 102. The product of these two numbers is 8,160, not 8,000. Therefore, the given answer of 8,000 is incorrect.
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• 3.
### 306 X 102
Explanation
The given question is asking for the product of 306 and 102. To find the answer, we multiply 306 by 102, which equals 30,600. Therefore, the correct answer is 30,600.
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• 4.
### 145 X 102
Explanation
The given question asks for the product of 145 and 102. The correct answer is 14,500, which is obtained by multiplying 145 by 102.
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• 5.
### 520 X 102
Explanation
The given question asks for the product of 520 and 102. When multiplying these two numbers, we can multiply 520 by 2, which gives us 1040, and then multiply 520 by 100, which gives us 52,000. Therefore, the correct answer is 52,000.
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• 6.
### 60 X 103
Explanation
The given question is asking for the product of 60 multiplied by 103. The correct answer is 60,000. This is because when you multiply 60 by 103, you get a result of 6,180.
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• 7.
### 48 X 103
Explanation
The given question is asking for the product of 48 and 103. To find the answer, we multiply 48 by 103, which equals 4,944. Therefore, the correct answer is 4,944.
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• 8.
### 143 X 103
Explanation
The given question asks for the product of 143 and 103. To find the product, we multiply the two numbers together. When we multiply 143 by 103, we get 14,729. Therefore, the correct answer is 14,729.
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• 9.
### 630 X 103
Explanation
The given question asks for the product of 630 and 103. When multiplying these two numbers, we get 65,190. However, the correct answer provided is 630,000. It seems that there may be an error in the answer given, as it does not match the result of the multiplication.
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• 10.
### 35 X 105
Explanation
The correct answer is obtained by multiplying 35 by 105, which equals 3,675.
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• 11.
### 198 X 108
Explanation
The correct answer is obtained by multiplying 198 by 108. The product of these two numbers is 19,800, which can be represented as 19,800,000,000 in scientific notation.
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• 12.
### 256 X 102
Explanation
The given question is asking for the product of 256 and 102. To find the answer, we multiply 256 by 102, which gives us 25,600.
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# Prove that the square of any positive integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$.
Given: Statement "Square of any positive integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$".
To prove: Here we have to prove the given statement.
Solution:
According to Euclid's lemma,
If $a$ and $b$ are two positive integers;
• $a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.
If $b\ =\ 3$, then;
• $a\ =\ 3q\ +\ r$, where $0\ \underline{< }\ r\ <\ 3$.
• So, $r\ =\ 0,\ 1,\ 2$
When, $r\ =\ 0$:
$a\ =\ 3q$
Squaring on both sides, we get:
$a^2\ = (3q)^2$
$a^2\ = 9q^2$
$a^2\ = 3(3q^2)$
$a^2\ = 3m$, where $m\ =\ 3q^2$
When, $r\ =\ 1$:
$a\ =\ 3q\ +\ 1$
Squaring on both sides, we get:
$a^2\ = (3q\ +\ 1)^2$
$a^2\ = 9q^2\ +\ 6q\ + 1$
$a^2\ = 3(3q^2\ +\ 2q)\ +\ 1$
$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 2q$
When, $r\ =\ 2$:
$a\ =\ 3q\ +\ 2$
Squaring on both sides, we get:
$a^2\ = (3q\ +\ 2)^2$
$a^2\ = 9q^2\ +\ 12q\ + 4$
$a^2\ = 9q^2\ +\ 12q\ + 3\ +\ 1$
$a^2\ = 3(3q^2\ +\ 4q\ +\ 1)\ +\ 1$
$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 4q\ +\ 1$
Hence, the square of any positive integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$.
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# Algebra Functions
## Functions Are Functional
Jennifer Ledwith
A function is a set of data that has one distinct output for each input, but functions are much more than x and y living together on a Cartesian plane in a graph paper galaxy. They have applications in real life. Functions describe the relationship between an input and its output.
Graphs, charts, equations, and verbal descriptions – these depict functions.
Here are other words for input and output:
### Input:
• x-value
• independent variable
• independent quantity
• domain
• cause
### Output:
• y-value
• dependent variable
• dependent quantity
• range
• effect
As you have probably surmised, you control the input. Once the input travels through the function, you get your output. OK, I’m a big math nerd, but I will leave Algebra land to explain.
Imagine that you’re running late for math tutoring. You’re speeding 60 miles per hour in a 20 mile per hour school zone. A cop pulls you over and writes you a \$300 speeding ticket. Let’s break down this real world cause and effect situation.
Input: You chose to exceed the speed limit by 40 miles per hour.
Output: You received a speeding ticket.
You see, you can control how fast you’re going, but the consequence, or the output, is up to the cop.
### Algebra That’s Useful During March Madness
Let me give you another example. I have a perfect function that determines basketball scores. (Shhh…It’ll be our secret.)
y = 15x + 10
x is the number of hours per day that a basketball team practices.
y is the number of points scored in the game.
For a breakdown of the results in a table, refer to the Basketbal Scores Table.
For a depiction of the results in a graph, look at the Basketball Scores Graph in More Images.
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even and odd perfect numbers existence
Thank for my previous post. Also, thank you so much for this site (m.s.e)
1) If odd perfect numbers there, those numbers can be expressible $12k + 1$ or $324k + 81$ or $468k + 117$. If yes, please discuss, how far I am correct.
2) If $K$ = $(4^n - 2^n)$/2 is perfect, when $k = 1^3 + 3^3 + ...$
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You might want to add how you came up with these results ... – Hagen von Eitzen Feb 3 '13 at 8:56
@HagenvonEitzen! sinece even perfect numbers cannot wirtien in 12k + 1 and so on,.. – Jiha Feb 3 '13 at 9:19
Well, even perfect numbers surely cannot be written as $12k+1$. What was your reasoning about odd perfect numbers? – Hagen von Eitzen Feb 3 '13 at 9:24
The formula $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$ is well-known and leads to $$1^3+3^3+\ldots+(2m-1)^3=\sum_{k=1}^{2m}k^3-\sum_{k=1}^{m}(2k)^3\\=\frac{(2m)^2(2m+1)^2}{4}-8\cdot\frac{m^2(m+1)^2}{4}\\=m^2(2m^2-1).$$ Since it is well-known that even perfect numbers $N$ are of the form $N=2^{p-1}(2^p-1)$ with $p=2n+1$ an odd prime and $2^p-1$ a Mersenne prime, letting $m=2^n$, you find that indeed $$N=\frac{4^n-2^n}{2}=m^2(2m^2-1)=1^3+3^3+\ldots+(2m-1)^3.$$
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# GEOM 1 | Lesson 1 | Try This (Modifying Rectangles) Solutions
Draw a series of shapes starting with a rectangle to show how you can modify the rectangle to create a parallelogram, triangle, and trapezoid, and describe how the area of each is related to each other as the shape changes.
In the first figure the area of the rectangle and the area of the parallelogram are the same, note that they have the same base and height. The area of the triangle is half the area of the rectangle. The area of the trapezoid is also equal to the area of the rectangle. In this case the areas of the rectangle, the parallelogram and the trapezoid all have the same area. The triangles in both the rectangle and the parallelogram would also have the same area.
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• # question_answer Direction: Study the following Graph carefully and answer the questions given below: Total Number of Arts, Science and Commerce Students in Various Colleges Different Colleges In college F, the students taking Commerce is what percent of the total number of students taking Arts, Science and Commerce in all ? A) 45 B) 55 C) 59 D) 25
(d) In College F, number of students taking Arts, Science and Commerce = (40+50+30) thousand =120 thousand $\therefore$ Required percentage = $\frac{30}{120}\times 100$ = 25
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# The London Marathon took place on 25 April this year.
## Presentation on theme: "The London Marathon took place on 25 April this year."— Presentation transcript:
The London Marathon took place on 25 April this year.
Last year, a new course record was set by Samuel Wanjiru. He completed the 26 miles and 385 yards in two hours five minutes and nine seconds. For how far could the fastest runner in your class keep up with him? Could you keep up for 100m? 400m? A mile? 5 miles?
In 2005, Jeremy Clarkson raced a car against an amateur marathon runner around the course. Given that Clarkson says that the average speed of a car in London is 10.6mph who should win? What other factors might affect the result? Who do you think will win? Click here to watch the video
Up2d8 maths London Marathon Teacher Notes
London Marathon Introduction: The first London Marathon took place on 29 March 1981, with 6 625 runners completing the course. The American Dick Beardsley and Norwegian Inge Simonsen won the men’s event dramatically, holding hands as they crossed the finish line in a dead heat. The marathon has since grown enormously with many runners raising money for charity. The sponsors claim that the London Marathon is the biggest fundraising event in the world. This resource uses the context of the London Marathon to pose two questions for students to work on: firstly, students are asked to consider how long the fastest runner in your class could keep up with a marathon runner; and then they are asked to predict whether running or driving is a faster way to travel around London. Content objectives: This context provides the opportunity for teachers and students to explore a number of objectives. Some that may be addressed are: interpret results involving uncertainty and prediction interpret graphs and diagrams and make inferences to support or cast doubt on initial conjecture interpret and use compound measures, including from other subjects and real life. Process objectives: These will depend on the amount of freedom you allow your class with the activity. It might be worth considering how you’re going to deliver the activity and highlighting the processes that this will allow on the diagram below:
Assessment: You may wish to consider how you will assess the task and how you will record your assessment. This could include developing the assessment criteria with your class. You might choose to focus on the content objectives or on the process objectives. You might decide that this activity lends itself to comment only marking or to student self-assessment. If you use the APP model of assessment then you might use this activity to help you in building a picture of your students’ understanding Assessment criteria to focus on might be: use their own strategies within mathematics and in applying mathematics to practical contexts (Using and Applying Mathematics level 4) draw simple conclusions of their own and give an explanation of their reasoning (Using and Applying Mathematics level 5) solve problems and carry through substantial tasks by breaking them into smaller, more manageable tasks, using a range of efficient techniques, methods and resources, including ICT; give solutions to an appropriate degree of accuracy (Using and Applying Mathematics level 6) Probing questions: You may wish to introduce some points into the discussion which might include: what was Wanjiru’s average speed? where do you think a marathon runner is going at their slowest? At their fastest? how much slower is an amateur runner likely to be than a world record holder? what factors influenced Clarkson’s drive that were not ‘average’? if the car vs runner were to happen again, who would you predict would win?
You will need: The PowerPoint presentation (and some idea about how fast someone in your class can run. If impractical then, if your school has a sports day, maybe the records from this or a conversation with the PE dept about expected times). There are three slides: The first slide sets the scene The final slide sets the second task, asking students to construct and interpret a mathematical model to predict whether a marathon runner or a car is faster around the marathon route. The second slide sets the first activity, asking students to estimate for how long they could keep up with Olympic Champion Samuel Wanjiru
Download ppt "The London Marathon took place on 25 April this year."
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```Question 251005
{{{log((n+2)) + log((8)) = log((n^2 + 7n +10))}}}<br>
Solving equations where the variable is in the argument of one or more logarithms usually involves transforming the equation into one fo the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)<br>
Since your equation has logarithms on both sides of the equation already, we will aim for the second form. We just have to find a way to combine the two logarithms on the left into a single logarithm. Fortunately we have a property of logarithms, {{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}, which allows us to do exactly what we want. Using this property on your equation gives us:
{{{log((n+2)*8) = log((n^2 + 7n +10))}}}
which simplifies to:
{{{log((8n+16)) = log((n^2 + 7n +10))}}}<br>
We now have the desired (second) form. The next step uses the idea that if the logarithm of n+2 is the same as the logarithm of {{{n^2 +7b +10}}} then
n+2 must be the same as {{{n^2 +7b +10}}}:
{{{8n+16 = n^2 + 7n +10}}}
The variable is now out of the argument of any logarithms. This is why we use the desired forms. Equations in those forms can be rewritten without logarithms. (BTW: If you use the first form, you rewrite it in exponential form.)<br>
We now have a "normal" quadratic equation to solve. So we'll get one side equal to zero (by subtracting 8n and 16 from ecah side):
{{{0 = n^2 - n - 6}}}
Next we factor or use the Quadratic Formula. This factors pretty easily:
{{{0 = (n-3)(n+2)}}}
By the Zero Product Property this product can be zero only if one of the factors is zero. So:
{{{n-3 = 0}}} or {{{n+2 = 0}}}
Solving these we get:
{{{n = 3}}} or {{{n = -2}}}<br>
With logarithmic equations it is important to check your answers. We must reject any solutions which make an argument of a logarithm negative or zero. Always use the original equation to check your answers.
{{{log((n+2)) + log((8)) = log((n^2 + 7n +10))}}}
Checking x = 3:
{{{log(((3)+2)) + log((8)) = log(((3)^2 + 7(3) +10))}}}
{{{log((5)) + log((8)) = log((9 + 7(3) +10))}}}
{{{log((5)) + log((8)) = log((9 + 21 +10))}}}
{{{log((5)) + log((8)) = log((40))}}}
As you can see, all the arguments of the logarithms are positive. So we have no reason to reject x = 3. To finish the check we can use the property from before to combine the logarithms on the left side:
{{{log((5*8)) = log((40))}}}
{{{log((40)) = log((40))}}} Check!<br>
Checking x = -2:
{{{log(((-2)+2)) + log((8)) = log(((-2)^2 + 7(-2) +10))}}}
{{{log((0)) + log((8)) = log((4 + (-14) +10))}}}
Already we have a problem. We have an argument to a logarithm that is zero. So we must reject this solution. (We only have to find one argument to a logarithm that is zero or negative to reject a solution. It makes no difference that the logarithm on the right ends up with a zero argument. Nor does it make any difference that log(8) is OK.)<br>
So the only solution to your equation is x = 3.
```
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A loop within another loop is known as a(n) ____ loop
A loop within another loop is known as a nested loop.
Question
Updated 3/23/2017 10:54:41 AM
Rating
3
A loop within another loop is known as a nested loop.
Confirmed by jeifunk [3/23/2017 11:03:06 AM]
Questions asked by the same visitor
(x + 5)(x - 5)
Weegy: Simplifying 4(x + 3) + -1(x + -2) = 0 Reorder the terms: 4(3 + x) + -1(x + -2) = 0 (3 * 4 + x * 4) + -1(x + -2) = 0 (12 + 4x) + -1(x + -2) = 0 Reorder the terms: 12 + 4x + -1(-2 + x) = 0 12 + 4x + (-2 * -1 + x * -1) = 0 12 + 4x + (2 + [ -1x) = 0 Reorder the terms: 12 + 2 + 4x + -1x = 0 Combine like terms: 12 + 2 = 14 14 + 4x + -1x = 0 Combine like terms: 4x + -1x = 3x 14 + 3x = 0 Solving 14 + 3x = 0 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-14' to each side of the equation. 14 + -14 + 3x = 0 + -14 Combine like terms: 14 + -14 = 0 0 + 3x = 0 + -14 3x = 0 + -14 Combine like terms: 0 + -14 = -14 3x = -14 Divide each side by '3'. x = -4.666666667 Simplifying x = -4.666666667 ] (More)
Question
Updated 8/13/2014 8:19:28 AM
(x + 5)(x - 5)
= x^2 - 5x + 5x - 25;
= x^2 - 25
Confirmed by andrewpallarca [8/13/2014 8:28:01 AM]
27,191,362
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Posted 1 Mar 2011
# Kruskal Algorithm
, 5 Jul 2012
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Implementation of Kruskal Algorithm in C#
## Introduction
According to Wikipedia:"Kruskal's algorithm is an algorithm in graph theory that finds a minimum spanning tree for a connectedweighted graph. This means it finds a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is minimized. If the graph is not connected, then it finds a minimum spanning forest (a minimum spanning tree for each connected component). Kruskal's algorithm is an example of a greedy algorithm."
In short, Kruskal's algorithm is used to connect all nodes in a graph, using the least cost possible.
#### Example
A cable TV company is laying a cable in a new neighborhood. An internet cafe is connecting all PCs via network.
## Using the Demo
Click anywhere to plot the vertices. Hold down ctrl and select two vertices to create an edge. A popup window appears to enter edge cost. Having finished plotting the graph, click Solve.
## Using the Code
`IList<Edge> Solve(IList<Edge> graph, out int totalCost); `
How the Graph is formed with GDI is not covered as it is irrelevant.
## Classes
Typically, our graph consists of two components, Vertices, and Edges connecting these vertices.
Each Edge is marked by a value or weight, which is the Cost of connecting the two vertices.
### Vertex
Holds:
• Vertex Name (which must be unique within Graph) and its Drawing Point
• Rank and Root (we'll get to those later)
```using System;
using System.Drawing;
namespace Kruskal
{
public class Vertex
{
#region Members
private int name;
#endregion
#region Public Properties
public int Name
{
get
{
return name;
}
}
public int Rank { get; set; }
public Vertex Root { get; set; }
public Point Position { get; set; }
#endregion
#region Constructor
public Vertex(int name, Point position)
{
this.name = name;
this.Rank = 0;
this.Root = this;
this.Position = position;
}
#endregion
#region Methods
internal Vertex GetRoot()
{
if (this.Root != this)// am I my own parent ? (am i the root ?)
{
this.Root = this.Root.GetRoot();// No? then get my parent
}
return this.Root;
}
internal static void Join(Vertex root1, Vertex root2)
{
if (root2.Rank < root1.Rank)//is the rank of Root2 less than that of Root1 ?
{
root2.Root = root1;//yes! then Root1 is the parent of Root2 (since it has the higher rank)
}
else //rank of Root2 is greater than or equal to that of Root1
{
root1.Root = root2;//make Root2 the parent
if (root1.Rank == root2.Rank)//both ranks are equal ?
{
root2.Rank++;//increment Root2, we need to reach a single root for the whole tree
}
}
}
#endregion
}
} ```
### Edge
Holds two Vertices, Cost of connection between them, and CostDrawing Point.
Note that it implements `IComparable`, we'll need it to sort Edges by Cost later on.
```using System;
using System.Drawing;
namespace Kruskal
{
public class Edge : IComparable
{
#region Members
private Vertex v1, v2;
private int cost;
private Point stringPosition;
#endregion
#region Public Properties
public Vertex V1
{
get
{
return v1;
}
}
public Vertex V2
{
get
{
return v2;
}
}
public int Cost
{
get
{
return cost;
}
}
public Point StringPosition
{
get
{
return stringPosition;
}
}
#endregion
#region Constructor
public Edge(Vertex v1, Vertex v2, int cost, Point stringPosition)
{
this.v1 = v1;
this.v2 = v2;
this.cost = cost;
this.stringPosition = stringPosition;
}
#endregion
#region IComparable Members
public int CompareTo(object obj)
{
Edge e = (Edge)obj;
return this.cost.CompareTo(e.cost);
}
#endregion
}
} ```
## Algorithm Implementation
### Sorting Edges
Edges are sorted in ascending order according to Cost using Quick Sort.
### Making Sets
Initially, every Vertex is its own Root and has rank zero.
```public Vertex(int name, Point position)
{
this.name = name;
this.Rank = 0;
this.Root = this;
this.Position = position;
}
```
## Why This Step?
We need it to ensure that, on adding our Vertices, we are not forming a loop.
Consider this example:
The Edge BD was not considered, because B,D are already connected through B,A,D.
Thus, for every Edge we examine, we must inspect that its two Vertices belong to different sets (trees).
## How To Find Out If Two Vertices Are In Different Sets?
Using the recursive function `GetRoot()`.
```internal Vertex GetRoot()
{
if (this.Root != this)// am I my own parent ? (am i the root ?)
{
this.Root = this.Root.GetRoot();// No? then get my parent
}
return this.Root;
} ```
If roots are indeed different, (each Vertex is in a different set), `Join()` the two Vertices.
```internal static void Join(Vertex root1, Vertex root2)
{
if (root2.Rank < root1.Rank)//is the rank of Root2 less than that of Root1 ?
{
root2.Root = root1;//yes! then Root1 is the parent of Root2 (since it has the higher rank)
}
else //rank of Root2 is greater than or equal to that of Root1
{
root1.Root = root2;//make Root2 the parent
if (root1.Rank == root2.Rank)//both ranks are equal ?
{
root2.Rank++;//increment Root2, we need to reach a single root for the whole tree
}
}
} ```
## Conclusion
Hope I delivered a clear explanation.
## License
This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)
## About the Author
Software Developer Australia
Enthusiastic programmer/researcher, passionate to learn new technologies, interested in problem solving, data structures, algorithms, AI, machine learning and nlp.
Amateur guitarist/ keyboardist, squash player.
## Comments and Discussions
FirstPrev Next
KRUSKAL ALGORITM EASY IMPLENTATION For A BEginer Learner Member 104879102-Jan-14 10:08 Member 10487910 2-Jan-14 10:08
Great article, Go 5! The Manoj Kumar21-Nov-13 8:13 The Manoj Kumar 21-Nov-13 8:13
Re: Great article, Go 5! Omar Gameel Salem21-Nov-13 21:44 Omar Gameel Salem 21-Nov-13 21:44
This is a masterpiece ThirstyMind13-Oct-13 6:21 ThirstyMind 13-Oct-13 6:21
Re: This is a masterpiece Omar Gameel Salem13-Oct-13 6:34 Omar Gameel Salem 13-Oct-13 6:34
aaa Member 102258431-Oct-13 20:35 Member 10225843 1-Oct-13 20:35
What compiler you use?? thanks :) Member 100400969-May-13 23:09 Member 10040096 9-May-13 23:09
Re: What compiler you use?? thanks :) Omar Gameel Salem10-May-13 2:53 Omar Gameel Salem 10-May-13 2:53
My vote of 5 Kenneth Haugland10-Aug-12 23:00 Kenneth Haugland 10-Aug-12 23:00
My vote of 5 Abinash Bishoyi6-Apr-12 3:05 Abinash Bishoyi 6-Apr-12 3:05
My vote of 5 manoj kumar choubey27-Mar-12 23:44 manoj kumar choubey 27-Mar-12 23:44
My Vote of 5 RaviRanjankr30-Dec-11 4:58 RaviRanjankr 30-Dec-11 4:58
Re: My Vote of 5 mobilat30-Dec-11 6:39 mobilat 30-Dec-11 6:39
Quick sort mobilat28-Dec-11 22:48 mobilat 28-Dec-11 22:48
My vote of 5 al13n9-Oct-11 21:42 al13n 9-Oct-11 21:42
Some responses to unjustified criticisms [modified] AndyUk0611-Sep-11 4:02 AndyUk06 11-Sep-11 4:02
Re: Some responses to unjustified criticisms Omar Gamil11-Sep-11 4:09 Omar Gamil 11-Sep-11 4:09
Re: Some responses to unjustified criticisms AndyUk0611-Sep-11 6:52 AndyUk06 11-Sep-11 6:52
My vote of 5 AndyUk069-Sep-11 1:01 AndyUk06 9-Sep-11 1:01
My vote of 5 naser abu khas2-Apr-11 14:24 naser abu khas 2-Apr-11 14:24
My vote of 5 paryj9-Mar-11 18:36 paryj 9-Mar-11 18:36
My vote of 5 EyalSch28-Mar-11 4:57 EyalSch2 8-Mar-11 4:57
My vote of 1 JV99992-Mar-11 4:16 JV9999 2-Mar-11 4:16
My vote of 1 Petr Pechovic2-Mar-11 0:57 Petr Pechovic 2-Mar-11 0:57
Re: My vote of 1 Omar Gamil2-Mar-11 1:54 Omar Gamil 2-Mar-11 1:54
Response intended to Dave also: "you provide no detailed discussion of the concepts behind the algorithm" That's absolutely right, no excuses "what it's used for" have you read the article, you'd have noticed--> In short, Kruskal algorithm is used to connect all nodes in a graph, using the least cost possible "examples of its use" Again, if you bothered to look at the article--> Example: A cable TV company is laying cable in a new neighborhood. An internet Cafe is connecting all PC's via network. and finally "how the code works" i explained all steps except sorting, which is known to any computer science first grader, and the article is aimed to intermediate level anyway. to sum up, i dont think you or dave even read the article at all, and giving such low rank for someone who took the time and effort to try to help others is unprofessional
Re: My vote of 1 Petr Pechovic2-Mar-11 7:58 Petr Pechovic 2-Mar-11 7:58
Re: My vote of 1 Omar Gamil2-Mar-11 21:33 Omar Gamil 2-Mar-11 21:33
Re: My vote of 1 Dave Kreskowiak2-Mar-11 9:05 Dave Kreskowiak 2-Mar-11 9:05
Re: My vote of 1 M i s t e r L i s t e r2-Mar-11 12:36 M i s t e r L i s t e r 2-Mar-11 12:36
My vote of 1 snortle2-Mar-11 0:14 snortle 2-Mar-11 0:14
My vote of 2 Adrian Pasik1-Mar-11 22:45 Adrian Pasik 1-Mar-11 22:45
Re: My vote of 2 Omar Gamil1-Mar-11 23:28 Omar Gamil 1-Mar-11 23:28
My vote of 5 TV Mogul1-Mar-11 10:17 TV Mogul 1-Mar-11 10:17
Re: My vote of 5 Omar Gamil1-Mar-11 21:58 Omar Gamil 1-Mar-11 21:58
[My vote of 1] Not good a article... Dave Kreskowiak1-Mar-11 8:39 Dave Kreskowiak 1-Mar-11 8:39
Last Visit: 31-Dec-99 18:00 Last Update: 23-Aug-17 6:48 Refresh 1
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800score.comhttp://www.800score.com/forum/ GMAT Ratio (Data Sufficiency)http://www.800score.com/forum/viewtopic.php?f=3&t=2651 Page 1 of 1
Author: questioner [ Fri Feb 17, 2012 7:21 pm ] Post subject: GMAT Ratio (Data Sufficiency) A certain science class has a total of 35 students. If the ratio of girls to boys in the class is 4 : 3, how many more girls than boys are in the class?A. 1B. 5C. 7D. 15E. 20(B) If there are 4 girls for every 3 boys, then the class is 4/7 girls and 3/7 boys. The number of girls in the class is 4/7 × 35 = 20, and the number of boys is 3/7 × 35 = 15. The question asks how many more girls there are than boys: 20 – 15 = 5 more girls than boys.The correct answer is choice (B).----------I do not understand why the explanation can use ratio 3/7 and 4/7. Because I read an information, sound likes if girls is A, boys is B => A/B = 4/3. So I can solve with two conditions such as A/B = 4/3 and A + B = 35. Hence, B = 15, A = 20.
Author: Gennadiy [ Fri Feb 17, 2012 7:36 pm ] Post subject: Re: t.1, qt.2: ratio, fractions Quote:because I read an information, sound likes if girls is A, boys is B => A/B = 4/3.You use the classical ratio approach. A : B = 4 : 3. That's completely fine and works here well.Note, that you get the same result. A – B = 20 – 15 = 5 more girls than boys.Quote:I do not understand why the explanation can use ratio 3/7 and 4/7.How do we get 3/7 and 4/7 so fast, knowing the ratio 3 : 4?The ratio 3 : 4 gives us:boys = 3x, girls = 4x. The total number of students is 3x + 4x = 7xSo boys make 3x/7x = 3/7. Girls make 4x/7x = 4/7.So if we know that some total set is composed of just A and B in ratio a : b, then A makes a/(a + b) and B makes b/(a + b) of the whole.Furthermore, it is true for any number of subset objects. Just make sure they all compose the whole set. For example,If an alloy is composed of three elements in ratio 2 : 3 : 4, thenelement 1 is 2/9 of the alloyelement 2 is 3/9 of the alloyelement 3 is 4/9 of the alloy
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# Steps To Solving Math Word Problems
She loves creating learning resources for students and teachers.She authors Teacher Trap, a blog aimed at sharing her challenges, successes, and insights as a teacher.
She loves creating learning resources for students and teachers.She authors Teacher Trap, a blog aimed at sharing her challenges, successes, and insights as a teacher.
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Ten of her students are great at word problems involving addition, and only 7 seem to understand subtraction word problems.
Five of her students are bored with the easy problems.
Thirteen students are still struggling with basic math facts and 3 have trouble reading the word problems at all. Issue #2: Flexible Thinking Many kids are taught to solve word problems methodically, with a prescriptive step-by-step plan using key words that don’t always work.
Plans are great, but not when students use them as a crutch rather than a tool.
And it’s not as if somehow when you’re studying you can avoid math word problems. But before we try to break it down, it’s best to just try to figure out what this problem is about – generally speaking. You won’t solve it until you are at least familiar with the situation.
Use these simple steps to solve every math word problem with ease (well – as much ease as you can have when solving math problems. There is an interesting difference between math word problems and simply solving an equation: math word problems don’t give you the equation. So much of math is about solving equations properly.
When I was in elementary school, this was actually a pretty reliable strategy!
But today, kids are asked to solve much more complex problems, often with tricky wording or intentional distractors.
Here’s an example showing student work on two similar problems about monkeys.
## Comments Steps To Solving Math Word Problems
• ###### Solving Word Problems
Solving Word Problems. “Word Problems scare me!” “Word Problems make me nervous!” “Word Problems? I just skip them!” Don't Worry! With just 4 easy steps.…
• ###### Simple Steps to Solve Math Word Problems Udemy
Learn how to reduce stress and anxiety by implementing a step by step approach to solve almost any math word problem.…
• ###### Steps to Student Success With Word Problems Using ST Math
Conceptual understanding opens the door to higher order thinking and complex problem solving. In order to be successful with word problems, students must.…
• ###### Step word problem truffles video Khan Academy
Sal solves a two-step word problem by drawing a picture and creating an equation. CCSS Math 3. Practice Represent 2-step word problems with equations.…
• ###### How to Solve Algebra Problems Step-By-Step - ThoughtCo
Feb 20, 2019. Use the Following Steps to Solve Word Problems Identify the. Apply Math to Real Life With These Proportions Word Problems. Aerial view of.…
• ###### Secret to Solving Math Word Problems. Hint It's Not about.
Nov 12, 2017. How to Use Reading Skills for Solving Math Word Problems. Provide students. they can do it. Find a printable checklist of these steps here.…
• ###### Two-step equation word problem garden video Khan.
Here's a nifty word problem in which we find the dimensions of a garden given only the perimeter. Let's create an equation to solve. CCSS Math 7B.4, 7B.4a. About. Practice Two-step equations word problems · Next lesson.…
• ###### Strategies to Conquer Math Word Problems - Laura Candler
Many kids are taught to solve word problems methodically, with a prescriptive step-by-step plan using key words that don't always work. Plans are great, but not.…
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# An Otto cycle has a compression ratio of 8. If 250 kJ of work is extracted from the cycle, the heat rejected by the cycle is
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1. 500 kJ
2. 442.69 kJ
3. 331.4 kJ
4. 192.69 kJ
Option 4 : 192.69 kJ
Free
CT 3: Building Materials
2480
10 Questions 20 Marks 12 Mins
## Detailed Solution
Explanation:
According to the given data compression ratio of the Otto cycle r = 8
Work extracted from the cycle (W) = 250 KJ
Let the heat input into the cycle is Q1 and heat rejected from the cycle is Q2
The efficiency of the Otto cycle ηotto = 1 $$- \frac{1}{{{{\rm{r}}^{\gamma - 1}}}}$$ $$\frac{{\rm{W}}}{{{\rm{W}} + {{\rm{Q}}_2}}}$$
So, 1$$- \frac{1}{{{8^{0.4}}}}$$ = 0.5647 = $$\frac{{250}}{{250 + {{\rm{Q}}_2}}}$$
Or, $$\frac{{250}}{{250 + {{\rm{Q}}_2}}}$$ = 0.5647
Or, 250 + Q2 = $$\frac{{250}}{{0.5647}}$$ = 442.7129
∴ Q2 = 442.7129-250 = 192.7129 kJ
Important Points
[How to solve this question without a calculator?
1$$- \frac{1}{{{8^{0.4}}}}$$ is just a little greater than 1 $$- \frac{1}{{{8^{0.333}}}}$$or greater than 1$$- \frac{1}{{{8^{\left( {\frac{1}{3}} \right)}}}}$$ or greater than 0.5
So, $$\frac{{250}}{{250 + {{\rm{Q}}_2}}}$$ is just a little greater than 0.5
Or, $$\frac{{250}}{{250 + {{\rm{Q}}_2}}}$$ > 0.5
Or, 250 + Q2 < 500
Or,Q2 < 250
Hence the only option is an option (4)
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Published by Pearson
# Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 344: 50
#### Answer
$\left( \dfrac{1}{6}a+m \right)\left( \dfrac{1}{6}a-m \right)$
#### Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, then the factored form of the given expression, $\dfrac{1}{36}a^2-m^2$, is \begin{array}{l} \left( \dfrac{1}{6}a+m \right)\left( \dfrac{1}{6}a-m \right) .\end{array}
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# warm up write down objective and homework in agenda lay out homework (none) homework (systems of...
Post on 22-Dec-2015
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• Slide 1
• Warm Up Write down objective and homework in agenda Lay out homework (none) Homework (Systems of Equations graphing) WELCOME BACK! Get a calculator!
• Slide 2
• Warm Up Graph the following lines
• Slide 3
• Vocabulary system of linear equations: two or more linear equations graphed in the same coordinate plane solution of a system of linear equations: any ordered pair in a system that makes all the equations true no solution: when two lines are parallel, there are no points of intersection infinitely many solutions: when the graphs of the systems of equations are the same line
• Slide 4
• What is a system of equations? A system of equations is when you have two or more equations using the same variables. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair. When graphing, you will encounter three possibilities.
• Slide 5
• A SYSTEM of equations A system can have one solution, infinitely many solutions, or no solutions. One solution means the lines intersect No solutions means the lines never touch (parallel) Many solutions means the lines are the same
• Slide 6
• Intersecting Lines The point where the lines intersect is your solution. The solution of this graph is (1, 2) (1,2)
• Slide 7
• Parallel Lines These lines never intersect! Since the lines never cross, there is NO SOLUTION! Parallel lines have the same slope with different y- intercepts.
• Slide 8
• Coinciding Lines These lines are the same! Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS! Coinciding lines have the same slope and y-intercepts.
• Slide 9
• Solving a system of equations by graphing. There are 3 steps to solving a system using a graph. Step 1: Graph both equations. Step 2: Do the graphs intersect? Step 3: Check your solution. Graph using slope and y intercept or x- and y-intercepts. Be sure to use a ruler and graph paper! This is the solution! LABEL the solution! Substitute the x and y values into both equations to verify the point is a solution to both equations.
• Slide 10
• What is the solution of the system graphed below? 1.(2, -2) 2.(-2, 2) 3.No solution 4.Infinitely many solutions
• Slide 11
• Graph the equations. 2x + y = 4 (0, 4) and (2, 0) x - y = 2 (0, -2) and (2, 0) Where do the lines intersect? (2, 0) 2x + y = 4 x y = 2
• Slide 12
• Check your answer! To check your answer, plug the point back into both equations. 2x + y = 4 2(2) + (0) = 4 x - y = 2 (2) (0) = 2
• Slide 13
• Graph the equations. y = 2x 3 m = 2 and b = -3 y = 2x + 1 m = 2 and b = 1 Where do the lines intersect? No solution! Notice that the slopes are the same with different y- intercepts. If you recognize this early, you dont have to graph them!
• Slide 14
• Example Graph the following equations on the same graph and find the solution to the system of equations: y = 2x -7 y = 1 a) What solution did you find? (4, 1) b) To check: Plug your solution(s) into each equation to see if it works. c) Consider the point (5, 3). Does it work in the 1 st equation? The 2nd? Is this a solution to the system? First but not second, so not a system!
• Slide 15
• Example Graph and find the solution: y = 3 x - 6 4 y = 3x + 1 4 What is the solution to this system? NO solution Explain why you came to this conclusion.
• Slide 16
• Example Graph and find the solution: 2x + y = 6 y = -2x + 6 What is the solution to this system? Coinciding lines! All solutions Explain why you came to this conclusion.
• Slide 17
• You Try! Graph the following equations on the same graph and find the solution to the system of equations: y = -2x + 3 y = 1/2x - 2 a) What solution did you find? (2, -1) b) To check: Plug your solution(s) into each equation to see if it works.
• Slide 18
• You Try! Graph the following equations on the same graph and find the solution to the system of equations: y = 3x + 5 y = 3x - 2 a) What solution did you find? NO Solution! b) To check: Plug your solution(s) into each equation to see if it works.
• Slide 19
• Graph and find the solution: 2x + y = 6 y = -2x + 6 What is the solution to this system? Infinite Solutions Explain why you came to this conclusion. They are coinciding lines, which means they intersect at every point; therefore there is infinite solutions
• Slide 20
• 1.Graph the following equations on the same graph and find the solution to the system of equations: y = 2x 7 y = x + 2 a)What solution did you find? (9,11) b ) To check: Plug your solution(s) into each equation to see if it works. c)Consider the point (5, 3). Does it work in the 1 st equation? the 2 nd ? Is this a solution to the system? It works in the 1 st, but not 2 nd ; therefore its NOT a solution.
• Slide 21
• Graph and find the solution: y = x 6 y = x + 1 What is the solution to this system? No solution! Explain why you came to this conclusion. They are parallel and will never intersect
• Slide 22
• What is the solution of this system? 3x y = 8 2y = 6x -16 1.(3, 1) 2.(4, 4) 3.No solution 4.Infinitely many solutions
• Slide 23
• Summarize the number of possible solutions to a system of two equations in two variables and explain how each possibility could occur. (There are THREE)
• Slide 24
• Extra Resources http://www.phschool.com/webcodes10/index.cfm?fuseaction=home.gotoWebCode&wcpref ix=ata&wcsuffix=0701 http://www.phschool.com/webcodes10/index.cfm?fuseaction=home.gotoWebCode&wcpref ix=ata&wcsuffix=0701 http://www.regentsprep.org/Regents/math/A LGEBRA/AE3/PracGr.htm http://www.regentsprep.org/Regents/math/A LGEBRA/AE3/PracGr.htm http://www.regentsprep.org/Regents/math/A LGEBRA/AE3/GrSys.htm http://www.regentsprep.org/Regents/math/A LGEBRA/AE3/GrSys.htm http://www.quia.com/cz/43456.html?AP_ran d=1117418363 http://www.quia.com/cz/43456.html?AP_ran d=1117418363
• Slide 25
• Steps to Graphing Systems on Calc. Step 1: Press y =, clear out old equations and enter new Step 2: Press GRAPH Step 3:Find the intersection of the 2 equations by pressing 2nd CALC (over TRACE) 5: intersect. Press ENTER 3 times to find the intersection
• Slide 26
• Solving systems on graphing calculator Example ALWAYS put in slope intercept form FIRST Step 1: Press y = and clear any old equations. Enter: y = x + 6 y = 2x + 4
• Slide 27
• Step 2 Step 2: Press GRAPH (If the intersection is off the graph, press ZOOM; arrow down until you see 0: ZoomFit and hit enter or you can adjust the window.) If you cant see the intersection on your screen, the calculator wont find it!!!
• Slide 28
• Step 3: Find the intersection of the 2 equations by pressing 2nd CALC (over TRACE) 5: intersect. Press ENTER 3 times to find the intersection The intersection is (2, 8). *Check the point of intersection by substituting the x- and y-values into both equations
• Slide 29
• Examples Find the intersection of each system of equations by using a graphing calculator. Check your solutions. (Hint: Sometimes you have to solve for y first.) 1. y = 4x 1 2. y =3x 4 y = x + 2 y=5x12 Answers: 1. (-1, 3) 2. (4, 8)
• Slide 30
• Practice No Solution (5, 5) 5
• Slide 31
• Practice 3. 2x4y=8 4. x + y = 6 x y = 4 x 5y =0 3. (4, 0) 4. (-5, -1)
• Slide 32
• You Try! Hint: Sometimes you have to solve for y first.
• Slide 33
• Answers 1. (-1, 3) 2. (4, 8) 3. (4, 0) 4. (-5, -1)
• Slide 34
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# Definition of integers
We’ve already learned what natural numbers are and how to perform simple mathematical operations with them, such as addition, subtraction, multiplication and division. We represent the set of natural numbers with $\mathbb{N} = \{ 1, 2, 3, 4 \ldots\}$. These are the numbers we use when we’re counting objects. We can also add zero to the set of natural numbers to represent the absence of quantity (when there is nothing there). Then we mark the set with $\mathbb{N_{0}} = \{ 0, 1, 2, 3, 4 \ldots \}.$
Before we explain what integers are, we would like you to observe the following example:
The air temperature today at noon was 39.2°F and by the evening the air temperature declined by 42.8°F. What was the air temperature in the evening?
We know what natural numbers are and we know how to add or subtract natural numbers. Can we solve example above using just the knowledge we have on natural numbers?
Let’s rewrite the problem using mathematical language:
$39.2-42.8=$
We don’t know how to subtract natural numbers when the subtrahend (second number) is larger than the minuend (first number) so we need to expand the set of natural numbers. To easily understand how we do that, let’s put first few natural numbers on the number line:
All of these numbers are sorted in order from smallest to the greatest. Now, let’s try to do some subtraction with two arbitrarily chosen natural numbers. For example, we know that $\ 4 – 3 = 1$. But what happens if we want to subtract natural numbers when the subtrahend (second number) is larger than the minuend (first number)? Can we do that? Are there numbers that are smaller than zero? Where can we find those numbers on our number line? Well, yes, we can subtract any two arbitrarily chosen natural numbers. Also, yes, there are numbers that are smaller than zero, and, to answer the last question, look at the picture below.
Take a look at this sequence. We have a zero in the middle. Now look at the first number to the right and first one to the left of zero – those are 1 and -1. If we look at the second number to the right and the second one to the left, those are 2 and -2. The third ones are 3 and -3. We can notice the pattern and, to be mathematically correct, we can come up with the following definition: Numbers that are found on opposite sides from zero and are equidistant (or equally distant) with respect to zero are called the opposite numbers or additive inverses.
The set that contains both positive and negative numbers is called the set of whole numbers or the set of integers and we mark the set with: $\mathbb{Z} = \{ \ldots-4, -3, -2, -1, 0, 1, 2, 3, 4 \ldots\}$. It consists of natural numbers and their additive inverses – numbers that, when added to their natural counterparts, result in zero (like -4 and 4, or -16 and 16, etc.). This set is also unbounded from both ends, which means that the smallest and largest integer do not exist! There is always a number that will be smaller than the previous smallest number, and a number that will be larger than the previous largest number.
Every operation that can be done with natural numbers can be done as easily with integers. Addition, subtraction, multiplication, division – you name it! To make things easier and explain them a bit better, we divided these topics into separate lessons, which you can access via the main menu or by clicking on their names a couple of rows above.
That’s it for this lesson! Feel free to explore the website a bit and learn how integers can be used in mathematical operations.
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### Volume of square base pyramid
Assignment Help Mathematics
##### Reference no: EM13134706
Given a structure shaped like a pyramid with a square base measuring 9 feet on a side and a height of 8 feet, what is the volume of the tent?
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A number a is divisible by the number b if a div b has actually a remainder that zero (0). Because that example, 15 divided by 3 is exactly 5 which indicates that that remainder is zero. Us then say the 15 is divisible through 3.
You are watching: What numbers are divisible by 6
In our various other lesson, we questioned the divisibility rules because that 7, 11, and 12. This time, we will certainly cover the divisibility rule or tests for2, 3, 4, 5, 6, 9, and 10. Believe me, girlfriend will be able to learn them very quickly because you may not recognize that you currently have a straightforward and intuitive understanding of it. For instance, it is obvious that all also numbers space divisible through 2. The is pretty lot the divisibility dominion for 2. The goal of this divisibility rules lesson is to formalize what you currently know.
Divisibility rules help us to determine if a number is divisible by another without going v the actual department process such as the long department method. If the number in inquiry are numerically small enough, we might not have to use the rule to test because that divisibility. However, fornumbers whose values are large enough, we want to have actually some rule to serve as “shortcuts” to assist us figure out if lock are undoubtedly divisible by each other.
A number is divisible by2if the last digit of the number is 0, 2, 4, 6, or 8.
Example 1: Is the number 246 divisible by 2?
Solution: because the last digit that the number 246 end in 6, that way it is divisible through 2.
Example 2: which of the numbers 100, 514, 309, and 768 are divisible by 2?
Solution: If we research all four numbers, only the number 309 doesn’t end with 0, 2, 4, 6, or 8. We deserve to conclude the all the numbers above except 309 space divisible through 2.
A number is divisible by 3 if the amount of the digits of the number is divisible through 3.
Example 1: Is the number 111 divisible by 3?
Solution: Let’s include the number of the number 111. We have 1 + 1 + 1 = 3. Since the amount of the number is divisible by the 3, as such the number 111 is likewise divisible by 3.
Example 2: Which among the two numbers 522 and 713 is divisible through 3?
Solution: The amount of the digits of 522 (5+2+2=9) is 9 which is divisible through 3. That provides 522 divisible by 3. However, the number 713 has 11 as the sum of its digits which is plainly not divisible through 3 therefore 713 is not divisible by 3. Therefore, only 522 is divisible by 3.
A number is divisible by 4 if the last two digits the the number are divisible by 4.
Example 1: What is the just number in the set below is divisible by 4?
945, 736, 118, 429
Solution: Observe the last 2 digits the the 4 numbers in the set. Notice that 736 is the only number within the last 2 digits (36) is divisible by 4. We deserve to conclude the 736 is the only number in the set that is divisible by 4.
Example 2: True or False. The number 5,554 is divisible by 4.
Solution: The last two digits of the number 5,554 is 54 i beg your pardon is not divisible through 4. That method the provided number is no divisible by 4 therefore the price is false.
A number is divisible by 5 if the critical digit that the number is 0 or 5.
Example 1: multiple Choice. I beg your pardon number is divisible through 5.
A) 68
B) 71
C) 20
D) 44
Solution: In order for a number to it is in divisible by 5, the critical digit that the number need to be either 0 or 5. Going over the choices, only the number 20 is divisible through 5 for this reason the price is choice C.
Example 2: pick all the number that space divisible through 5.
A) 27
B) 105
C)556
D) 343
E) 600
Solution: Both 105 and also 600 are divisible by 5 since they either end in 0 or 5. Thus, options B and also E are the correct answers.
A number is divisible by 6 if the number is divisible by both 2 and 3.
Example 1: Is the number 255 divisible by 6?
Solution: For the number 255 to it is in divisible by 6, it need to divisible by 2 and also 3. Let’s check very first if the is divisible through 2. Note that 255 is not an even number (any number finishing in 0, 2, 4, 6, or 8) which makes it no divisible 2. There’s no need to inspect further. We deserve to now conclude the this is not divisible by 6. The answer is NO.
Example 2: Is the number 4,608 divisible through 6?
Solution: A number is an even number so the is divisible by 2. Now inspect if the is divisible through 3. Let’s do that by adding all the digits of 4,608 i m sorry is 4 + 6+ 0 + 8 = 18. Obviously, the sum of the digits is divisible by 3 since 18 ÷ 3 = 6. Because the number 4,608 is both divisible by 2 and also 3 climate it must likewise be divisible by 6. The prize is YES.
A number is divisible by 9 if the amount of the number is divisible by 9.
Example 1: Is the number 1,764 divisible by 9?
Solution: For a number to be divisible through 9, the amount of its number must additionally be divisible by 9. Because that the number 1,764 we obtain 1 + 7 + 6 + 4 = 18. Due to the fact that the amount of the number is 18 and also is divisible by 9 because of this 1,764 should be divisible through 9.
Example 2: choose all the number that space divisible by 9.
A) 7,065
B) 3,512
C) 8,874
D) 22,778
E) 48,069
Solution: Let’s include the digits of each number and check if its amount is divisible by 9.
For 7,065, 7 + 0 + 6 + 5 = 18 i beg your pardon is divisible by 9.For 3,512, 3 + 5 + 1 + 2 = 11 which is NOT divisible through 9.For 8,874, 8 + 8 + 7 + 4 = 27 i beg your pardon is divisible by 9.For 22,778, 2 + 2 + 7 + 7 + 8 = 26 i m sorry is NOT divisible by 9.For 48,069, 4 + 8 + 0 + 6 + 9 = 27 which is divisible by 9.
Therefore, choices A, C, and also E room divisible by 9.
A number is divisible by 10 if the critical digit that the number is 0.
See more: Andy Biersack As A Young Kid I Was Definitely What You, Andy Biersack
The number 20, 40, 50, 170, and also 990 room all divisible by 10 because their last digit is zero, 0. Top top the other hand, 21, 34, 127, and also 468 are not divisible through 10 due to the fact that they don’t end with zero.
You might likewise be interested in:
Divisibility Rules because that 7, 11, and 12
MATH SUBJECTSIntroductory AlgebraIntermediate AlgebraAdvanced AlgebraAlgebra indigenous ProblemsGeometryIntro come Number TheoryBasic math Proofs
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# Fluid mechanics of water in a pipe
Hello All,
I've frequented this forum for various purposes and this is the first time I've decided to post! Hopefully I can find the help or conformation I'm looking for. I may not have all the necessary info, but I feel this is basic enough that I should be able to come up with an appropriate answer.
This is a basic setup where I have water being pumped through steel piping and I'm trying to find the exit pressure and velocity of the water. Here is what I know for sure.
Pipe Diameter, d = .151in or .003835m
Pipe Length, l = 72in or 1.8288m
Pressure from pump, P = 100psi or 689400Pa
Temp, T = 140°F or 60°C
Density, ρ = 983.2 kg/m³
Viscosity, μ = 0.467*10-³ N*s/m²
Roughness steel, ε = .000025 m
1. To get the velocity of the water at the starting point I use Bernoulli's equation for dynamic pressure, q=1/2*ρ*v². Solving for v, I get v=37.45 m/s.
2. I calculate my Reynolds Number and relative roughness of the steel pipe, then use the Moody chart to determine the Friction Factor.
Re = (ρ*v*d)/μ = (983.2*37.45*.003835)/(.467*10-³) = 301,978
RR = ε/d = (.000025)/(.003835) = .00652
From the Moody chart: f = .031
3. Calculating the pressure drop due to the friction in the pipe (I feel this is my problem)
ΔP= (ρ*v²*f*l)/(2*d) = (983.2*37.45²*.031*1.8288)/(2*.003835) = 10192437.5 Pa
From there I would use Bernoulli's equation for the flow at the beginning of the pipe, P + 1/2*ρ*v² + ρ*g*h (last part is negligible for me), set equal to an equation for the exit flow which incorporates the friction.
I feel as though I am making an incorrect assumption (or 2 or 3...) which is leading to such a large drop in pressure. Areas I feel I could have gone astray; calculating the initial velocity using the dynamic pressure equation, assuming that friction will even have an effect here, or even that the pressure of the pump is way too high for such a small diameter pipe (and my calculations are somewhat correct).
If there is anyone out there that can comment on this and perhaps guide me in the right direction I'd greatly appreciate it.
Andy Resnick
Yikes... I was going to use Poiseuille flow, but your Re >>2000, so that's no good.
The exit is at atmospheric pressure...does the pipe let water freely squirt out, or is it held at some other pressure? That gives you your pressure drop across the length of the pipe.
100 psi into a 1/10 inch diameter tube... yikes.
Yea, I'm way past laminar flow with the current setup. I'm looking into more details right now, but I'm pretty sure there is not a resovoir at the end of the tubing and there is not any kind of nozzle (which would probably shoot off like a bullet anyways).
After looking at it for a bit I too started to think that the 100 psi was just way too much. Thank you for your reply, it had been some time since I dealt with fluid mechanics and I needed some one else's opinion on this.
Before I keep digging into this topic, can anyone confirm that the way in which I used the dynamic pressure equation, q = 1/2*ρ*v², was in fact correct? I thought it to be since the only initial pressure in the system is from the pump, but I'd really like to know that I do infact have the basics correct.
Your velocity is probably closer to 9 meters per second. Reynolds number is about 36.000. See thumbnail.
Bob S
#### Attachments
• Pressure_drop.jpg
17.4 KB · Views: 370
Bob,
I was wondering how you came to the conclusion that the initial velocity would be around 9 m/s. Assuming everything is correct, pressure drop = 108 psi and pressure of water at the exit = 7.45 psi, I can plug those numbers into the bernoulli equation, which should look like this:
1/2*ρ*v1² + P1 = 1/2*ρ*v2² + P2
and with the numbers:
1/2*983*(9)² + 792810(pascals) = 1/2*983*(v2)² + 51705(pascals)
solving for v2 = 39.86 m/s²
I know that a drop in pressure leads to increased velocity typically where the flow path has decreased in area, but I'm not positive that the same applies when the drop in pressure is due to head loss or the friction in the pipe.
Also, I'm still looking for confirmation on how to calculate an initial velocity with the given data. Need to be sure I have the basic of basics down before I'm satisfied with the other results. Once again, any help is appreciated! Cheers!
After going through everything again, I realize I'm completely off base in my last post. That program Bob used gave me my output velocity via the flow rate. If thats the case, then my exit velocity is roughly 0.06 m/s. To me that seems much too low. I am wandering off the path somewhere and making poor assumptions.
Still interested in how to determine my initial velocity though.
Q_Goest
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8th Grade Math - Statistics and Probability
Common Core Math Practice
This quiz illustrates the range of difficulty expected by the following Common Core State (CCS) standards:
Question Common Core Standard # 1 8.SP.1 2 8.SP.2 3 8.SP.3 4 8.SP.4 5 8.SP.4 6 8.SP.4
Question 1: The scatter plot in the figure below shows the GPA of the students of a class versus their commute time. Which of the following statements is true?
Question 2: All students of a school were asked what their preferred sport is and the results were recorded below.
Favorite Sport by Gender
Boys25412490
Girls284240110
Totals538364200
Which of the following statements are correct?
Questions 3 and 4: The graph below shows the diameters of a number of white oak trees versus their ages.
The growth factor of white oak trees is approximatively:
Assuming a linear growth rate, what is the diameter of a 130 years old white oak tree?
Questions 5 and 6: A study investigates the relationship between learning piano and the mathematics level of fifth grade students, by testing them with a mathematics quiz and recording the results in the table below.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Student Studies Piano Yes No Yes No No Yes Yes No No Yes Yes No No No Yes No No No Yes No Mathematics Quiz Grade A C B D B A A C A D A A B C A D C B A B
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Divide both sides by 3:
`log(2x - 1) = 1/3`
So
`10^(1"/"3)=2x-1`
`x=1/2(10^(1"/"3)+1)`
`=1.577217345`
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Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm,
Question:
In a ΔABC, if a = 3 cm, b = 5 cm and c = 7 cm, find cos A, cos B, cos C.
Solution:
Given: a = 3 cm, b = 5 cm and c = 7 cm
Need to find: cos A, cos B, cos C
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{5^{2}+7^{2}-3^{2}}{2.5 .7}=\frac{65}{70}=\frac{13}{14}$
$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}=\frac{7^{2}+3^{2}-5^{2}}{2.7 .3}=\frac{33}{42}$
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{3^{2}+5^{2}-7^{2}}{2.3 .5}=\frac{-15}{30}=-\frac{1}{2}$
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You are on page 1of 30
MATH 211 Winter 2013
Lecture Notes
(Adapted by permission of K. Seyarth)
Sections 2.6
Sections 2.6 Page 1/1
2.6 Matrix Transformations
Sections 2.6 Page 2/1
Linear Transformations
Denition
A transformation T : R
n
R
m
is a linear transformation if and only if,
for all X, Y R
n
and all scalars a,
T1. T(X + Y) = T(X) + T(Y) (preservation of addition)
T2. T(aX) = aT(X) (preservation of scalar multiplication)
As a consequence of T2, for any linear transformation T,
T(0X) = 0T(X), implying T(0) = 0,
and
T((1)X) = (1)T(X), implying T(X) = T(X),
i.e., T preserves the zero vector and T preserves the negative of a vector.
Sections 2.6 Page 3/1
Furthermore, if X
1
, X
2
, . . . , X
k
are vectors in R
n
and Y is a linear
combination of X
1
, X
2
, . . . , X
k
, i.e.,
Y = a
1
X
1
+ a
2
X
2
+ + a
k
X
k
for some a
1
, a
2
, . . . , a
k
R, then T1 and T2 used repeatedly give us
T(Y) = T(a
1
X
1
+ a
2
X
2
+ + a
k
X
k
)
= a
1
T(X
1
) + a
2
T(X
2
) + + a
k
T(X
k
),
i.e.,T preserves linear combinations.
Sections 2.6 Page 4/1
Example
Let T : R
4
R
3
be a linear transformation such that
T
_
_
1
1
0
2
_
_
=
_
_
2
3
1
_
_
and T
_
_
0
1
1
1
_
_
=
_
_
5
0
1
_
_
. Find T
_
_
1
3
2
4
_
_
.
The only way it is possible to solve this problem is if
_
_
1
3
2
4
_
_
is a linear combination of
_
_
1
1
0
2
_
_
and
_
_
0
1
1
1
_
_
,
i.e., if there exist a, b R so that
_
_
1
3
2
4
_
_
= a
_
_
1
1
0
2
_
_
+ b
_
_
0
1
1
1
_
_
Sections 2.6 Page 5/1
Example (continued)
Solve the system of four equations in two variables:
_
_
1 0 1
1 1 3
0 1 2
2 1 4
_
_
_
_
1 0 1
0 1 2
0 0 0
0 0 0
_
_
Thus a = 1, b = 2, and
_
_
1
3
2
4
_
_
=
_
_
1
1
0
2
_
_
2
_
_
0
1
1
1
_
_
.
Sections 2.6 Page 6/1
Example (continued)
It follows that
T
_
_
1
3
2
4
_
_
= T
_
_
_
_
_
_
1
1
0
2
_
_
2
_
_
0
1
1
1
_
_
_
_
_
_
= T
_
_
1
1
0
2
_
_
2T
_
_
0
1
1
1
_
_
=
_
_
2
3
1
_
_
2
_
_
5
0
1
_
_
=
_
_
8
3
3
_
_
Sections 2.6 Page 7/1
Example (2.6 Example 2)
Every matrix transformation is a linear transformation.
Proof.
Suppose T : R
n
R
m
is a matrix transformation induced by the m n
matrix A, i.e., T(X) = AX for each X R
n
.
Let X, Y R
n
and let a R. Then
T(X + Y) = A(X + Y) = AX + AY = T(X) + T(Y),
proving that T preserves addition. Also,
T(aX) = A(aX) = a(AX) = aT(X),
proving that T preserves scalar multiplication.
Since T1 and T2 are satised, T is, in fact, a linear transformation.
Sections 2.6 Page 8/1
It turns out that the converse of this statement is also true, i.e., every
linear transformation from R
n
to R
m
is a matrix transformation.
Theorem (2.6 Theorem 2)
Let T : R
n
R
m
be a transformation.
1
T is linear if and only if T is a matrix transformation.
2
If T is linear, then T is induced by the unique matrix
A =
_
T(E
1
) T(E
2
) T(E
n
)
,
where E
j
is the j
th
column of I
n
.
Why does this work for nding A?
Sections 2.6 Page 9/1
The uniqueness in Theorem 2 guarantees that there is exactly one matrix
for any linear transformation, so it makes sense to say the matrix of a
linear transformation.
Sections 2.6 Page 10/1
Examples
Q
0
: R
2
R
2
is reection in the x-axis, i.e.,
Q
0
_
x
y
_
=
_
x
y
_
.
We saw earlier that Q
0
is induced by the matrix A =
_
1 0
0 1
_
, so
Q
0
is a linear transformation.
By Theorem 2, Q
0
is induced by the matrix
_
Q
0
(E
1
) Q
0
(E
2
)
=
_
Q
0
_
1
0
_
Q
0
_
0
1
_ _
=
_
1 0
0 1
_
= A.
Sections 2.6 Page 11/1
Examples (continued)
R
2
: R
2
R
2
is (counterclockwise) rotation about the origin through
an angle of
2
, and
R
2
_
x
y
_
=
_
y
x
_
.
We saw earlier that R
2
is induced by the matrix B =
_
0 1
1 0
_
, so
R
2
is a linear transformation.
By Theorem 2, R
2
is induced by the matrix
_
R
2
(E
1
) R
2
(E
2
)
=
_
R
2
_
1
0
_
R
2
_
0
1
_ _
=
_
0 1
1 0
_
= B.
Sections 2.6 Page 12/1
Example (2.6 Example 4)
Q
1
: R
2
R
2
is reection in the line y = x, i.e.,
Q
1
_
a
b
_
=
_
b
a
_
.
Notice that
_
b
a
_
=
_
0 1
1 0
_ _
a
b
_
,
so Q
1
is a matrix transformation, and hence a linear transformation.
Again, illustrating Theorem 2, Q
1
is induced by the matrix
_
Q
1
(E
1
) Q
1
(E
2
)
=
_
Q
1
_
1
0
_
Q
1
_
0
1
_ _
=
_
0 1
1 0
_
.
Sections 2.6 Page 13/1
Example
Let T : R
2
R
3
be a transformation dened by
T
_
x
y
_
=
_
_
2x
y
x + 2y
_
_
.
Show that T is a linear transformation.
Solution. If T were a linear transformation, then T would be induced by
the matrix
A =
_
T(E
1
) T(E
2
)
=
_
T
_
1
0
_
T
_
0
1
_ _
=
_
_
2 0
0 1
1 2
_
_
.
Sections 2.6 Page 14/1
Example
Let T : R
2
R
3
be a transformation dened by
T
_
x
y
_
=
_
_
2x
y
x + 2y
_
_
.
Show that T is a linear transformation.
Solution. If T were a linear transformation, then T would be induced by
the matrix
A =
_
T(E
1
) T(E
2
)
=
_
T
_
1
0
_
T
_
0
1
_ _
=
_
_
2 0
0 1
1 2
_
_
.
Sections 2.6 Page 15/1
Example (continued)
Since
A
_
x
y
_
=
_
_
2 0
0 1
1 2
_
_
_
x
y
_
=
_
_
2x
y
x + 2y
_
_
= T
_
x
y
_
,
T is a matrix transformation, and therefore a linear transformation.
Sections 2.6 Page 16/1
Example
Let T : R
2
R
2
be a transformation dened by
T
_
x
y
_
=
_
xy
x + y
_
.
Is T is a linear transformation? Explain.
Solution. If T were a linear transformation, then T would be induced by
the matrix
A =
_
T(E
1
) T(E
2
)
=
_
T
_
1
0
_
T
_
0
1
_ _
=
_
0 0
1 1
_
.
Now
A
_
x
y
_
=
_
0 0
1 1
_ _
x
y
_
=
_
0
x + y
_
.
Sections 2.6 Page 17/1
Example (continued)
In particular,
A
_
1
1
_
=
_
0 0
1 1
_ _
1
1
_
=
_
0
2
_
,
while
T
_
1
1
_
=
_
1
2
_
.
Since A
_
1
1
_
= T
_
1
1
_
, T is not a linear transformation.
There is an alternate way to show that T is not a linear transformation.
Sections 2.6 Page 18/1
Example (continued)
Notice that
_
1
1
_
=
_
1
0
_
+
_
0
1
_
, and
T
_
1
1
_
=
_
1
2
_
, T
_
1
0
_
=
_
0
1
_
, T
_
0
1
_
=
_
0
1
_
.
From this we see that
T
_
1
1
_
= T
_
1
0
_
+ T
_
0
1
_
,
i.e., T does not preserve addition, and so T is not a linear transformation.
Sections 2.6 Page 19/1
Denition (Vector scalar multiplication)
Let X R
2
and let k R. Then kX is the vector in R
2
that is |k| times
the length of X; kX points the same directions as X if k > 0, and
opposite to X if k < 0.
Let X, Y R
2
, and consider the parallelogram dened by 0, X and Y.
The vector X + Y corresponds to the fourth vertex of this parallelogram.
Sections 2.6 Page 20/1
Rotation in R
2
Let R
: R
2
R
2
denote counterclockwise rotation about the origin
through and angle of , and let X, Y R
2
and k R. Then
R
(kX) = kR
(X),
and
R
(X + Y) = R
(X) + R
(Y).
This means that R
is a linear transformation, and hence a matrix
transformation.
Therefore, by our earlier theorem, we can nd the matrix that induces R
by simply nding R
(E
1
) and R
(E
2
).
Sections 2.6 Page 21/1
R
(E
1
) = R
_
1
0
_
=
_
cos
sin
_
,
and
R
(E
2
) = R
_
0
1
_
=
_
sin
cos
_
Theorem (2.6 Theorem 4)
The transformation R
: R
2
R
2
is a linear transformation, and is
induced by the matrix
_
cos sin
sin cos
_
.
Sections 2.6 Page 22/1
Reection in R
2
Let Q
m
: R
2
R
2
denote reection in the line y = mx, and let
X, Y R
2
and k R. Then
Q
m
(kX) = kQ
m
(X),
and
Q
m
(X + Y) = Q
m
(X) + Q
m
(Y).
This means that Q
m
is a linear transformation and hence a matrix
transformation.
Therefore, by our earlier theorem, we can nd the matrix that induces Q
m
by simply nding Q
m
(E
1
) and Q
m
(E
2
).
Sections 2.6 Page 23/1
However, an easier way to obtain the matrix for Q
m
is to use the following
observation:
Q
m
= R
Q
0
R
,
where is the angle between y = mx and the positive x axis.
Using the standard trigonometric identities:
cos =
1
1 + m
2
and sin =
m
1 + m
2
,
the matrix for Q
m
can be found by computing the product
_
cos sin
sin cos
_ _
1 0
0 1
_ _
cos() sin()
sin() cos()
_
Sections 2.6 Page 24/1
Theorem (2.6 Theorem 5)
The transformation Q
m
: R
2
R
2
, denoting reection in the line y = mx,
is a linear transformation and is induced by the matrix
1
1 + m
2
_
1 m
2
2m
2m m
2
1
_
.
For reection in the x-axis, m = 0, and the theorem yields the expected
matrix
_
1 0
0 1
_
.
However, the y-axis has undened slope, so the theorem does not apply.
We use Q
Y
to denote reection in the y-axis, and as weve already seen,
Q
Y
is induced by the matrix
_
1 0
0 1
_
,
and thus is a linear transformation.
Sections 2.6 Page 25/1
Example
Find the rotation or reection that equals reection in the x-axis followed
by rotation through an angle of
2
.
We must nd the matrix for the transformation R
2
Q
0
.
Q
0
is induced by A =
_
1 0
0 1
_
, and R
2
is induced by
B =
_
cos
2
sin
2
sin
2
cos
2
_
=
_
0 1
1 0
_
R
2
Q
0
is induced by
BA =
_
0 1
1 0
_ _
1 0
0 1
_
=
_
0 1
1 0
_
.
Sections 2.6 Page 26/1
Example (continued)
Notice that BA =
_
0 1
1 0
_
is a reection matrix.
How do we know this?
Now, since 1 m
2
= 0, we know that m = 1 or m = 1. But
2m
1+m
2
= 1 > 0, so m > 0, implying m = 1.
Therefore,
R
2
Q
0
= Q
1
,
reection in the line y = x.
Sections 2.6 Page 27/1
In general,
The composite of two reections is a rotation.
The composite of two rotations is a rotation.
The composite of a reection and a rotation is a reection.
Sections 2.6 Page 28/1
Example
Find the rotation or reection that equals reection in the line y = x
followed by reection in the y-axis.
We must nd the matrix for the transformation Q
Y
Q
1
.
Q
1
is induced by
A =
1
2
_
0 2
2 0
_
=
_
0 1
1 0
_
,
and Q
Y
is induced by
B =
_
1 0
0 1
_
.
Therefore, Q
Y
Q
1
is induced by BA.
Sections 2.6 Page 29/1
Example (continued)
BA =
_
1 0
0 1
_ _
0 1
1 0
_
=
_
0 1
1 0
_
.
What transformation does BA induce?
Rotation through an angle such that
cos = 0 and sin = 1.
Therefore, Q
Y
Q
1
= R
2
= R3
2
.
Sections 2.6 Page 30/1
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|
# Mathematica : Numerically Integrate function with undefined variables?
• Mathematica
• Hepth
In summary, the conversation is about a large function that needs to be integrated over two variables, x and y, both ranging from 0 to 1. The function is expressed as a sum of terms, each of which can be separated into a function of x and y. The G functions are complicated and take a long time to integrate algebraically, but can be quickly numerically integrated. The question is whether there is a way to numerically integrate the entire equation without expanding it and dividing out the functions A[q] and B[q] for each term. The suggested solution is to substitute temporary variables Aq and Bq for A[q] and B[q], use CoefficientList to extract the parts of the equation that are multiplied by
Hepth
Gold Member
I have a LARGE function that I must integrate over 2 variables, {x,y} both from 0 to 1.
The Function looks like :
F[x_,y_] = A[q] G[x,y] + B[q] G2[x,y] + (...)
Where (...) MAY be 100's of terms long, if not thousands. But the fact is each term is separable, in SOME way to a function of x and y (and no other units or unknown variables).
Each G function is quite complicated as well, and so far takes over 3 minutes each to integrate algebraically, though can be numerically integrated in under a second or two. They're all well behaved in the integration region.
Is there a way, other than expand F, and dividing out A[q] etc for each, doing the NIntegrate, then multiplying it again, to numerically integrate this equation so it ends up being:
result = (0.14325) A[q] + (-4.5602134) B[q] + (...)
Can I tell NIntegrate to integrate around variables?
I know something like NIntegrate[ a x^2,{x,0,1}] will throw an error due to an undefined "a" inserted.
## 1. How do I numerically integrate a function in Mathematica with undefined variables?
To numerically integrate a function with undefined variables in Mathematica, you need to first assign values to those variables using the "Set" function. For example, if your function is f(x,y) and you want to integrate from x=0 to x=1 and y=2, you would use the command "Set[x,0]; Set[y,2]; NIntegrate[f[x,y],{x,0,1}]. This will evaluate the integral numerically with the given values for x and y.
## 2. Can I use functions with undefined variables in Mathematica's numerical integration?
Yes, you can use functions with undefined variables in Mathematica's numerical integration. However, you will need to assign values to those variables before integrating, as mentioned in the previous answer. If you do not assign values, Mathematica will not be able to evaluate the integral.
## 3. How accurate is Mathematica's numerical integration with undefined variables?
The accuracy of Mathematica's numerical integration with undefined variables will depend on the accuracy of the assigned values for those variables. If the values are precise, the integration will be accurate. However, if the values are only approximate, the integration may also be approximate.
## 4. Can I use symbolic variables in Mathematica's numerical integration?
Yes, you can use symbolic variables in Mathematica's numerical integration. However, you will need to use the "Set" function to assign values to those variables before integrating, as mentioned in the first answer. You can also use the "Assuming" function to specify assumptions about the symbolic variables.
## 5. How can I check the accuracy of the numerical integration in Mathematica?
To check the accuracy of the numerical integration in Mathematica, you can use the "AccuracyGoal" and "PrecisionGoal" options in the "NIntegrate" function. These options allow you to specify the desired level of accuracy for the integration and can be adjusted to achieve a more accurate result.
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|
## Differential equation study guide for exam (formula sheet)
Partial Differential Equation Toolbox User's Guide. 5/26/2018 · Differential equation study guide for exam (formula sheet) 1. Author: Daniel Albert May 25, 2018 www.dennisusa.com 1 Differential Equation and Linear Algebra Exam Review (The Cheat Sheet) Chapter 1: First order linear ODE • Differential equation: A …, A first-order differential equation is said to be homogeneous if it can be written in the form dy dx = F ( y x) Such an equation can be solved by using the change of variables: v = y x. which transforms the equation into one that is separable. To discover more on this type of equations, check this complete guide on Homogeneous Differential.
### Partial Differential Equation Toolbox User's Guide
1. Solving Differential Equations intmath.com. Di erential Equations Study Guide1 First Order Equations General Form of ODE: dy dx = f(x;y(1) ) Indicial Equation: r2 +(p(0) 1)r+q(46) (0) = 0 First Solution: y 1 = (x x 0)r 1 X1 k=0 (47) a k(x x k)k Where r 1 is the larger real root if both roots of (46) are real or either root if …, Study Guide Block 2: Ordinary Differential Equations Quiz 4. The curve C satisfies the differential equation (where C doesn't intersect the line x = -1). Find the equation of C. 5. The curve C satisfies the differential equation y" -3xy' -3y = 0. It passes through the point (0,l) and has its slope equal to 0 at that point..
Solutions of Differential Equations. As expected for a second‐order differential equation, the general solution contains two parameters ( c 0 and c 1), which will be determined by the initial conditions. Since y(0) = 2, it is clear that c 0 = 2, and then, since y′(0) = 3, the … The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial
Solving a differential equation. From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it. The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial
The Partial Differential Equation (PDE) Toolbox provides a powerful and flexible environment for the study and solution of partial differential equations in two space dimensions and time. The equations are discretized by the Finite Element Method (FEM). The objectives of … Choose an ODE Solver Ordinary Differential Equations. An ordinary differential equation (ODE) contains one or more derivatives of a dependent variable, y, with respect to a single independent variable, t, usually referred to as time.The notation used here for representing derivatives of y with respect to t is y ' for a first derivative, y ' ' for a second derivative, and so on.
5/26/2018 · Differential equation study guide for exam (formula sheet) 1. Author: Daniel Albert May 25, 2018 www.dennisusa.com 1 Differential Equation and Linear Algebra Exam Review (The Cheat Sheet) Chapter 1: First order linear ODE • Differential equation: A … A Differential Equation is a n equation with a function and one or more of its derivatives: Example: an equation with the function y and its derivative dy dx . Solving. We solve it when we discover the function y (or set of functions y). There are many "tricks" to solving Differential Equations (if they can be solved!).
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Differential Equations Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Try this differential equation to see what I mean: dx dy y x 2 2 2 =-How about it? One of the first things that should occur to you is that this isn’t a linear differential equation, so the techniques in the first part of this chap-ter won’t help. However, you’ll probably notice that you can write this equa-tion …
The result would be a differential equation of the order. Conversley if we are given a differential equation of the order we can, in general, obtain an equivalent relationship containing no derivatives but n arbitrary constants. This relationship is called "The General Solution" A … A first-order differential equation is said to be homogeneous if it can be written in the form dy dx = F ( y x) Such an equation can be solved by using the change of variables: v = y x. which transforms the equation into one that is separable. To discover more on this type of equations, check this complete guide on Homogeneous Differential
Diп¬Ђerential Equations Study Sheet Chesnes. Study Guide Block 2: Ordinary Differential Equations Quiz 4. The curve C satisfies the differential equation (where C doesn't intersect the line x = -1). Find the equation of C. 5. The curve C satisfies the differential equation y" -3xy' -3y = 0. It passes through the point (0,l) and has its slope equal to 0 at that point., 4/5/2019 · Differential Equations Here are my notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations..
### Differential Equations—Wolfram Language Documentation
Second Order Linear Nonhomogeneous Differential Equations. A differential equation is a mathematical equation that relates some function with its derivatives.In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two., A Differential Equation is a n equation with a function and one or more of its derivatives: Example: an equation with the function y and its derivative dy dx . Solving. We solve it when we discover the function y (or set of functions y). There are many "tricks" to solving Differential Equations (if they can be solved!)..
### Solving Systems of Linear Differential Equations by
Solving Systems of Linear Differential Equations by. Partial Differential Equation Toolbox Product Description Solve partial differential equations using finite element analysis Partial Differential Equation Toolbox provides functions for solving partial differential equations (PDEs) in 2-D, 3-D, and time using finite element analysis. https://en.wikipedia.org/wiki/Degree_of_a_differential_equation From Differential Equations For Dummies. By Steven Holzner . To confidently solve differential equations, you need to understand how the equations are classified by order, how to distinguish between linear, separable, and exact equations, and how to identify homogenous ….
An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations. Note this is the dialog that defines the equation, not the dialog used for each fit. • Fitting a differential equation requires more calculations, so it takes noticeably more time that fitting the usual kind of equation. • It is only possible to define Y'. It is not possible to use differential equations to define intermediate variables.
Homogeneous Differential Equations This guide helps you to identify and solve homogeneous first order ordinary differential equations. Introduction A differential equation (or DE) is any equation which contains derivatives, see study guide: Basics of Differential Equations. To … Try this differential equation to see what I mean: dx dy y x 2 2 2 =-How about it? One of the first things that should occur to you is that this isn’t a linear differential equation, so the techniques in the first part of this chap-ter won’t help. However, you’ll probably notice that you can write this equa-tion …
An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations. Note this is the dialog that defines the equation, not the dialog used for each fit. • Fitting a differential equation requires more calculations, so it takes noticeably more time that fitting the usual kind of equation. • It is only possible to define Y'. It is not possible to use differential equations to define intermediate variables.
Note this is the dialog that defines the equation, not the dialog used for each fit. • Fitting a differential equation requires more calculations, so it takes noticeably more time that fitting the usual kind of equation. • It is only possible to define Y'. It is not possible to use differential equations to define intermediate variables. Homework Help in Differential Equations from CliffsNotes! Need help with your homework and tests in Differential Equations and Calculus? These articles can hel
A differential equation is a mathematical equation that relates some function with its derivatives.In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from
The equation is written as a system of two first-order ordinary differential equations (ODEs). These equations are evaluated for different values of the parameter μ.For faster integration, you should choose an appropriate solver based on the value of μ.. For μ = 1, any of the MATLAB ODE solvers can solve the van der Pol equation efficiently.The ode45 solver is one such example. 4/5/2019 · Differential Equations Here are my notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations.
From Differential Equations For Dummies. By Steven Holzner . To confidently solve differential equations, you need to understand how the equations are classified by order, how to distinguish between linear, separable, and exact equations, and how to identify homogenous … An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.
## Differential Equations—Wolfram Language Documentation
Diп¬Ђerential Equations Study Sheet Chesnes. Homework Help in Differential Equations from CliffsNotes! Need help with your homework and tests in Differential Equations and Calculus? These articles can hel, 5/26/2018 · Differential equation study guide for exam (formula sheet) 1. Author: Daniel Albert May 25, 2018 www.dennisusa.com 1 Differential Equation and Linear Algebra Exam Review (The Cheat Sheet) Chapter 1: First order linear ODE • Differential equation: A ….
### Differential Equations MATLAB & Simulink Example
Homogeneous differential equations Portal - UEA. The equation is written as a system of two first-order ordinary differential equations (ODEs). These equations are evaluated for different values of the parameter μ.For faster integration, you should choose an appropriate solver based on the value of μ.. For μ = 1, any of the MATLAB ODE solvers can solve the van der Pol equation efficiently.The ode45 solver is one such example., Separable Differential Equation. Sanjay is a microbiologist, and he's trying to come up with a mathematical model to describe the population growth of a certain type of bacteria..
4/5/2019 · Differential Equations Here are my notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations. Solving a differential equation. From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it.
2/3/2016 · Learn to Solve Ordinary Differential Equations. Math and Science In this lesson the student will learn what a differential equation is and how to solve them. Intro to 2nd Order The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial
Separable Differential Equation. Sanjay is a microbiologist, and he's trying to come up with a mathematical model to describe the population growth of a certain type of bacteria. Di erential Equations Study Guide1 First Order Equations General Form of ODE: dy dx = f(x;y(1) ) Indicial Equation: r2 +(p(0) 1)r+q(46) (0) = 0 First Solution: y 1 = (x x 0)r 1 X1 k=0 (47) a k(x x k)k Where r 1 is the larger real root if both roots of (46) are real or either root if …
Note. By default, the required order of the first two arguments of func are in the opposite order of the arguments in the system definition function used by the scipy.integrate.ode class and the function scipy.integrate.solve_ivp.To use a function with the signature func(t, y,...), … Try this differential equation to see what I mean: dx dy y x 2 2 2 =-How about it? One of the first things that should occur to you is that this isn’t a linear differential equation, so the techniques in the first part of this chap-ter won’t help. However, you’ll probably notice that you can write this equa-tion …
4/5/2019 · Differential Equations Here are my notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations. Free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Differential Equations Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.
5/26/2018 · Differential equation study guide for exam (formula sheet) 1. Author: Daniel Albert May 25, 2018 www.dennisusa.com 1 Differential Equation and Linear Algebra Exam Review (The Cheat Sheet) Chapter 1: First order linear ODE • Differential equation: A … The equation is written as a system of two first-order ordinary differential equations (ODEs). These equations are evaluated for different values of the parameter μ.For faster integration, you should choose an appropriate solver based on the value of μ.. For μ = 1, any of the MATLAB ODE solvers can solve the van der Pol equation efficiently.The ode45 solver is one such example.
Note this is the dialog that defines the equation, not the dialog used for each fit. • Fitting a differential equation requires more calculations, so it takes noticeably more time that fitting the usual kind of equation. • It is only possible to define Y'. It is not possible to use differential equations to define intermediate variables. The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial
Differential equations introductionWriting a differential equationWorked example: linear solution to differential equation. Practice. Verify solutions to differential equations economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education A differential equation (de) is an equation involving a function and its deriva-tives. Differential equations are called partial differential equations (pde) or or-dinary differential equations (ode) according to whether or not they contain partial derivatives. The order of a differential equation is the highest order derivative occurring.
Second Order Linear Homogeneous Differential Equations with Constant Coefficients For the most part, we will only learn how to solve second order linear equation with constant coefficients (that is, when p(t) and q(t) are constants). Since a homogeneous equation is easier to solve compares to its Equation type. Choose whether you are entering an ordinary explicit equation (that defines Y as a function of X and parameters), an implicit equation (Y appears on both sides of the equals sign) or a differential equation (which defines the derivative of Y with respect to X). Name. Enter a name you will recognize, so you can choose the equation
8/9/2014 · Will edit with respect to usefulness when I've completed the class but wanted to outline the topics contained in this Quick Study for Differential Equations (DEQ going forward) because it was hard for me to get the full picture with just pieces of the picture. 8/9/2014 · Will edit with respect to usefulness when I've completed the class but wanted to outline the topics contained in this Quick Study for Differential Equations (DEQ going forward) because it was hard for me to get the full picture with just pieces of the picture.
The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial 8/9/2014 · Will edit with respect to usefulness when I've completed the class but wanted to outline the topics contained in this Quick Study for Differential Equations (DEQ going forward) because it was hard for me to get the full picture with just pieces of the picture.
Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial
### Second Order Linear Nonhomogeneous Differential Equations
Choose an ODE Solver MATLAB & Simulink. Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from, Separable Differential Equation. Sanjay is a microbiologist, and he's trying to come up with a mathematical model to describe the population growth of a certain type of bacteria..
### GraphPad Prism 8 Curve Fitting Guide How to Enter a new
1. Solving Differential Equations intmath.com. Solving a differential equation. From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it. https://en.wikipedia.org/wiki/Exact_differential_equation A differential equation (de) is an equation involving a function and its deriva-tives. Differential equations are called partial differential equations (pde) or or-dinary differential equations (ode) according to whether or not they contain partial derivatives. The order of a differential equation is the highest order derivative occurring..
From Differential Equations For Dummies. By Steven Holzner . To confidently solve differential equations, you need to understand how the equations are classified by order, how to distinguish between linear, separable, and exact equations, and how to identify homogenous … Solving a differential equation. From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it.
Automatically selecting between hundreds of powerful and in many cases original algorithms, the Wolfram Language provides both numerical and symbolic solving of differential equations (ODEs, PDEs, DAEs, DDEs,). With equations conveniently specified symbolically, the Wolfram Language uses both its rich set of special functions and its unique symbolic interpolating functions to represent Automatically selecting between hundreds of powerful and in many cases original algorithms, the Wolfram Language provides both numerical and symbolic solving of differential equations (ODEs, PDEs, DAEs, DDEs,). With equations conveniently specified symbolically, the Wolfram Language uses both its rich set of special functions and its unique symbolic interpolating functions to represent
The basic equation addressed by the software is the PDE expressed in Ω, which we shall refer to as theelliptic equation, regardless of whether its coefficients and boundary conditions make the PDE problem elliptic in the mathematical sense. Analogously, we shall use the terms parabolic equation and hyperbolic equation for equations with spatial 2/3/2016 · Learn to Solve Ordinary Differential Equations. Math and Science In this lesson the student will learn what a differential equation is and how to solve them. Intro to 2nd Order
Di erential Equations Study Guide1 First Order Equations General Form of ODE: dy dx = f(x;y(1) ) Indicial Equation: r2 +(p(0) 1)r+q(46) (0) = 0 First Solution: y 1 = (x x 0)r 1 X1 k=0 (47) a k(x x k)k Where r 1 is the larger real root if both roots of (46) are real or either root if … Solving a differential equation. From the above examples, we can see that solving a DE means finding an equation with no derivatives that satisfies the given DE. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of DE we are dealing with before we attempt to solve it.
• The order of a differential equation is the highest deriviative in the equation. • A differential equation is linear if it is linear in parameters such that the coefficients on each deriviative of y term is a function of the independent variable (t). • Solutions: Explicit → Written as a … Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from
Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from 25) LaPlace Equation Nothing to enter for this and for the other PDE’s = Partial Differential Equations. Their solutions are simply displayed here.
2/3/2016 · Learn to Solve Ordinary Differential Equations. Math and Science In this lesson the student will learn what a differential equation is and how to solve them. Intro to 2nd Order A Differential Equation is a n equation with a function and one or more of its derivatives: Example: an equation with the function y and its derivative dy dx . Solving. We solve it when we discover the function y (or set of functions y). There are many "tricks" to solving Differential Equations (if they can be solved!).
From Differential Equations For Dummies. By Steven Holzner . To confidently solve differential equations, you need to understand how the equations are classified by order, how to distinguish between linear, separable, and exact equations, and how to identify homogenous … 2/3/2016 · Learn to Solve Ordinary Differential Equations. Math and Science In this lesson the student will learn what a differential equation is and how to solve them. Intro to 2nd Order
A differential equation (de) is an equation involving a function and its deriva-tives. Differential equations are called partial differential equations (pde) or or-dinary differential equations (ode) according to whether or not they contain partial derivatives. The order of a differential equation is the highest order derivative occurring. Homework Help in Differential Equations from CliffsNotes! Need help with your homework and tests in Differential Equations and Calculus? These articles can hel
Study Guide Block 2: Ordinary Differential Equations Quiz 4. The curve C satisfies the differential equation (where C doesn't intersect the line x = -1). Find the equation of C. 5. The curve C satisfies the differential equation y" -3xy' -3y = 0. It passes through the point (0,l) and has its slope equal to 0 at that point. Solving linear differential equations may seem tough, but there's a tried and tested way to do it! We'll explore solving such equations and how this relates to the technique of elimination from
Homogeneous Differential Equations This guide helps you to identify and solve homogeneous first order ordinary differential equations. Introduction A differential equation (or DE) is any equation which contains derivatives, see study guide: Basics of Differential Equations. To … 4/5/2019 · Differential Equations Here are my notes for my differential equations course that I teach here at Lamar University. Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to learn how to solve differential equations or needing a refresher on differential equations.
Note this is the dialog that defines the equation, not the dialog used for each fit. • Fitting a differential equation requires more calculations, so it takes noticeably more time that fitting the usual kind of equation. • It is only possible to define Y'. It is not possible to use differential equations to define intermediate variables. Solutions of Differential Equations. As expected for a second‐order differential equation, the general solution contains two parameters ( c 0 and c 1), which will be determined by the initial conditions. Since y(0) = 2, it is clear that c 0 = 2, and then, since y′(0) = 3, the …
5/26/2018 · Differential equation study guide for exam (formula sheet) 1. Author: Daniel Albert May 25, 2018 www.dennisusa.com 1 Differential Equation and Linear Algebra Exam Review (The Cheat Sheet) Chapter 1: First order linear ODE • Differential equation: A … 7/29/2012 · Free ebook http://tinyurl.com/EngMathYT Easy way of remembering how to solve ANY differential equation of first order in calculus courses. The secret involve...
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Stat Trek
Teach yourself statistics
Teach yourself statistics
# Experimental Design for ANOVA
There is a close relationship between experimental design and statistical analysis. The way that an experiment is designed determines the types of analyses that can be appropriately conducted.
In this lesson, we review aspects of experimental design that a researcher must understand in order to properly interpret experimental data with analysis of variance.
## What Is an Experiment?
An experiment is a procedure carried out to investigate cause-and-effect relationships. For example, the experimenter may manipulate one or more variables (independent variables) to assess the effect on another variable (the dependent variable).
Conclusions are reached on the basis of data. If the dependent variable is unaffected by changes in independent variables, we conclude that there is no causal relationship between the dependent variable and the independent variables. On the other hand, if the dependent variable is affected, we conclude that a causal relationship exists.
## Experimenter Control
One of the features that distinguish a true experiment from other types of studies is experimenter control of the independent variable(s).
In a true experiment, an experimenter controls the level of the independent variable administered to each subject. For example, dosage level could be an independent variable in a true experiment; because an experimenter can manipulate the dosage administered to any subject.
## What is a Quasi-Experiment?
A quasi-experiment is a study that lacks a critical feature of a true experiment. Quasi-experiments can provide insights into cause-and-effect relationships; but evidence from a quasi-experiment is not as persuasive as evidence from a true experiment. True experiments are the gold standard for causal analysis.
A study that used gender or IQ as an independent variable would be an example of a quasi-experiment, because the study lacks experimenter control over the independent variable; that is, an experimenter cannot manipulate the gender or IQ of a subject.
As we discuss experimental design in the context of a tutorial on analysis of variance, it is important to point out that experimenter control is a requirement for a true experiment; but it is not a requirement for analysis of variance. Analysis of variance can be used with true experiments and with quasi-experiments that lack only experimenter control over the independent variable.
Note: Henceforth in this tutorial, when we refer to an experiment, we will be referring to a true experiment or to a quasi-experiment that is almost a true experiment, in the sense that it lacks only experimenter control over the independent variable.
## What Is Experimental Design?
The term experimental design refers to a plan for conducting an experiment in such a way that research results will be valid and easy to interpret. This plan includes three interrelated activities:
• Write statistical hypotheses.
• Collect data.
• Analyze data.
Let's look in a little more detail at these three activities.
## Statistical Hypotheses
A statistical hypothesis is an assumption about the value of a population parameter. There are two types of statistical hypotheses:
• Null hypothesis: The null hypothesis is the statement subjected to a statistical test in an experiment. It is denoted by H0. For example, consider the following null hypothesis:
H0: μi = μj
Here, μi is the population mean for group i, and μj is the population mean for group j. This hypothesis makes the assumption that population means in groups i and j are equal.
• Alternative hypothesis: The alternative hypothesis is the hypothesis that is tenable if the null hypothesis is rejected. It is denoted by H1 or Ha. For example, consider the following alternative hypothesis:
H1: μi ≠ μj
This hypothesis makes the assumption that population means in groups i and j are not equal.
The null hypothesis and the alternative hypothesis are written to be mutually exclusive. If one is true, the other is not.
Experiments rely on sample data to test the null hypothesis. If experimental results, based on sample statistics, are consistent with the null hypothesis, the null hypothesis cannot be rejected; otherwise, the null hypothesis is rejected in favor of the alternative hypothesis.
## Data Collection
The data collection phase of experimental design is all about methodology - how to run the experiment to produce valid, relevant statistics that can be used to test a null hypothesis.
### Identify Variables
Every experiment exists to examine a cause-and-effect relationship. With respect to the relationship under investigation, an experimental design needs to account for three types of variables:
• Dependent variable. The dependent variable is the outcome being measured, the effect in a cause-and-effect relationship.
• Independent variables. An independent variable is a variable that is thought to be a possible cause in a cause-and-effect relationship.
• Extraneous variables. An extraneous variable is any other variable that could affect the dependent variable, but is not explicitly included in the experiment.
Note: The independent variables that are explicitly included in an experiment are also called factors.
### Define Treatment Groups
In an experiment, treatment groups are built around factors, each group defined by a unique combination of factor levels.
For example, suppose that a drug company wants to test a new cholesterol medication. The dependent variable is total cholesterol level. One independent variable is dosage. And, since some drugs affect men and women differently, the researchers include an second independent variable - gender.
This experiment has two factors - dosage and gender. The dosage factor has three levels (0 mg, 50 mg, and 100 mg), and the gender factor has two levels (male and female). Given this combination of factors and levels, we can define six unique treatment groups, as shown below:
Gender Dose
0 mg 50 mg 100 mg
Male Group 1 Group 2 Group 3
Female Group 4 Group 5 Group 6
Note: The experiment described above is an example of a quasi-experiment, because the gender factor cannot be manipulated by the experimenter.
### Select Factor Levels
A factor in an experiment can be described by the way in which factor levels are chosen for inclusion in the experiment:
• Fixed factor. The experiment includes all factor levels about which inferences are to be made.
• Random factor. The experiment includes a random sample of levels from a much bigger population of factor levels.
Experiments can be described by the presence or absence of fixed or random factors:
• Fixed-effects model. All of the factors in the experiment are fixed.
• Random-effects model. All of the factors in the experiment are random.
• Mixed model. At least one factor in the experiment is fixed, and at least one factor is random.
The use of fixed factors versus random factors has implications for how experimental results are interpreted. With a fixed factor, results apply only to factor levels that are explicitly included in the experiment. With a random factor, results apply to every factor level from the population.
For example, consider the blood pressure experiment described above. Suppose the experimenter only wanted to test the effect of three particular dosage levels - 0 mg, 50 mg, and 100 mg. He would include those dosage levels in the experiment, and any research conclusions would apply to only those particular dosage levels. This would be an example of a fixed-effects model.
On the other hand, suppose the experimenter wanted to test the effect of any dosage level. Since it is not practical to test every dosage level, the experimenter might choose three dosage levels at random from the population of possible dosage levels. Any research conclusions would apply not only to the selected dosage levels, but also to other dosage levels that were not included explicitly in the experiment. This would be an example of a random-effects model.
### Select Experimental Units
The experimental unit is the entity that provides values for the dependent variable. Depending on the needs of the study, an experimental unit may be a person, animal, plant, product - anything. For example, in the cholesterol study described above, researchers measured cholesterol level (the dependent variable) of people; so the experimental units were people.
Note: When the experimental units are people, they are often referred to as subjects. Some researchers prefer the term participant, because subject has a connotation that the person is subservient.
If time and money were no object, you would include the entire population of experimental units in your experiment. In the real world, where there is never enough time or money, you will usually select a sample of experimental units from the population.
Ultimately, you want to use sample data to make inferences about population parameters. With that in mind, it is best practice to draw a random sample of experimental units from the population. This provides a defensible, statistical basis for generalizing from sample findings to the larger population.
Finally, it is important to consider sample size. The larger the sample, the greater the statistical power; and the more confidence you can have in your results.
### Assign Experimental Units to Treatments
Having selected a sample of experimental units, we need to assign each unit to one or more treatment groups. Here are two ways that you might assign experimental units to groups:
• Independent groups design. Each experimental unit is randomly assigned to one, and only one, treatment group. This is also known as a between-subjects design.
• Repeated measures design. Experimental units are assigned to more than one treatment group. This is also known as a within-subjects design.
### Control for Extraneous Variables
Extraneous variables can mask effects of independent variables. Therefore, a good experimental design controls potential effects of extraneous variables. Here are a few strategies for controlling extraneous variables:
• Randomization Assign subjects randomly to treatment groups. This tends to distribute effects of extraneous variables evenly across groups.
• Repeated measures design. To control for individual differences between subjects (age, attitude, religion, etc.), assign each subject to multiple treatments. This strategy is called using subjects as their own control.
• Counterbalancing. In repeated measures designs, randomize or reverse the order of treatments among subjects to control for order effects (e.g., fatigue, practice).
As we describe specific experimental designs in upcoming lessons, we will point out the strategies that are used with each design to control the confounding effects of extraneous variables.
## Data Analysis
Researchers follow a formal process to determine whether to reject a null hypothesis, based on sample data. This process, called hypothesis testing, consists of five steps:
• Formulate hypotheses. This involves stating the null and alternative hypotheses. Because the hypotheses are mutually exclusive, if one is true, the other must be false.
• Choose the test statistic. This involves specifying the statistic that will be used to assess the validity of the null hypothesis. Typically, in analysis of variance studies, researchers compute a F ratio to test hypotheses.
• Compute a P-value, based on sample data. Suppose the observed test statistic is equal to S. The P-value is the probability that the experiment would yield a test statistic as extreme as S, assuming the null hypothesis is true.
• Choose a significance level. The significance level, denoted by α, is the probability of rejecting the null hypothesis when it is really true. Researchers often choose a significance level of 0.05 or 0.01.
• Test the null hypothesis. If the P-value is smaller than the significance level, we reject the null hypothesis; if it is larger, we fail to reject.
A good experimental design includes a precise plan for data analysis. Before the first data point is collected, a researcher should know how experimental data will be processed to accept or reject the null hypotheses.
Problem 1
In a well-designed experiment, which of the following statements is true?
I. The null hypothesis and the alternative hypothesis are mutually exclusive.
II. The null hypothesis is subjected to statistical test.
III. The alternative hypothesis is subjected to statistical test.
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
Solution
The correct answer is (D). The null hypothesis and the alternative hypothesis are mutually exclusive; if one is true, the other must be false. Only the null hypothesis is subjected to statistical test. When the null hypothesis is accepted, the alternative hypothesis is rejected. The alternative hypothesis is not tested explicitly.
Problem 2
In a true experiment, each subject is assigned to only one treatment group. What type of design is this?
(A) Independent groups design
(B) Repeated measures design
(C) Within-subjects design
(D) None of the above
(E) All of the above
Solution
The correct answer is (A). In an independent groups design, each experimental unit is assigned to one treatment group. In the other two designs, each experimental unit is assigned to more than one treatment group.
Problem 3
In a true experiment, which of the following does the experimenter control?
(A) How to manipulate independent variables.
(B) How to assign subjects to treatment conditions.
(C) How to control for extraneous variables.
(D) None of the above
(E) All of the above
Solution
The correct answer is (E). The experimenter chooses factors and factor levels for the experiment, assigns experimental units to treatment groups (often through a random process), and implements strategies (randomization, counterbalancing, etc.) to control the influence of extraneous variables.
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# Sales Price Using Net Profit Per Hour, Part 2
Originally published: 08.01.06 by Ruth King
Last month, I described how to calculate your selling price based on net profit per hour calculations. This month, I'll give you examples of how to better estimate your selling price to the customer.
Whether you calculate your prices on a gross margin basis or net profit per hour basis, you must have accurate labor costs. Both require that you estimate the number of hours to complete the job. If you miss the hours, you will miss the estimate. Then, either your job will be at a loss or wildly profitable.
Assumptions:
Hourly cost is \$40 per hour (including burden)
Overhead cost is \$25 per hour
Desired profit is \$30 per hour
Gross margin is 35 percent
Exercise #1:
You estimate a job will take 16 hours. The estimated cost for equipment and materials is \$1,000. What is the selling price to the customer?
Using the gross margin method:
Total labor cost = 16 hours x \$40 per hour = \$640
Total equipment/material cost = \$1,000
Total direct cost = \$1,640
Selling price
to the customer = \$1,640 ÷ .65 = \$2,523
The total direct cost (\$1,640) is divided by 1 - .35 or .65 (the gross margin).
Using the net profit per hour method:
Total labor cost = 16 hours x \$40 per hour = \$640
Total equipment/material cost = \$1,000
Total overhead cost = 16 hours x \$25 per hour = \$400
Total profit = 16 hours x \$30 per hour = \$480
Selling price to the customer = \$2,520
The results for Exercise #1 are typical for cases where the labor cost and material cost are close. Let’s look at what happens when you have a high materials cost situation (Exercise #2) and a high labor cost situation (Exercise #3).
Exercise #2:
You estimate a job will take 16 hours. The estimated equipment and materials is \$2,000. What is the selling price to the customer?
Using the gross margin method:
Total labor cost = 16 hours x \$40 per hour = \$640
Total equipment/material cost = \$2,000
Total direct cost = \$2,640
Selling price to the customer = \$2,640 ÷ .65 = \$4,062
The total direct cost (\$2,640) is divided by 1 - .35 or .65 (the gross margin).
Estimated profit: \$4,062 (selling price) – 2,640 (direct cost) – 400 (overhead cost) = \$1,022
Using the net profit per hour method:
Total labor cost = 16 hours x \$40 per hour = \$640
Total equipment/material cost = \$2,000
Total overhead cost = 16 hours x \$25 per hour = \$400
Total profit = 16 hours x \$30 per hour = \$480
Selling price to the customer = \$3,520
In this case, the person estimating on a net profit per hour basis would win the job. The contractor bidding on a gross margin basis would say that he couldn’t do the job for that price. He wouldn’t realize that his estimated profit was \$1,022. To meet the other contractor’s price, he would have to lower his gross margin to 25 percent, which he thinks would be an unprofitable job when in actuality it wouldn’t be.
Exercise #3:
You estimate a job will take 32 hours. The estimated equipment and materials is \$1,000.What is the selling price to the customer?
Using the gross margin method:
Total labor cost = 32 hours x \$40 per hour = \$1,280
Total equipment/material cost = \$1,000
Total direct cost = \$2,280
Selling price to the customer = \$2,280 ÷ .65 = \$3,508
The total direct cost (\$2,280) is divided by 1 - .35 or .65 (the gross margin).
Estimated profit: 3,508 (selling price) – 2,280 (total direct cost) – 800 (overhead cost) = \$428
Using the net profit per hour method:
Total labor cost = 32 hours x \$40 per hour = \$1,280
Total equipment/material cost = \$1,000
Total overhead cost = 32 hours x \$25 per hour = \$800
Total profit = 32 hours x \$30 per hour = \$960
Selling price to the customer = \$4,040
In Exercise #3, by using the gross margin method you’ll only make a \$428 profit on the job. The contractor using the net profit per hour method keeps his profit per hour consistent at \$30 per hour or he can choose to lower his net profit per hour to meet the competition.
The high labor jobs are dangerous to use gross margin to calculate sales price. Let’s take Exercise #4:
For Exercise #4, let’s say that Exercise #3’s job actually took 40 hours to complete instead of 32. What happened?
Total labor cost increased to \$1,600.
Material cost stayed the same at \$1,000.
Total overhead cost increased to \$1,000.
Using the gross margin calculation the profit is:
\$3,508 (selling price) – 2,600 (revised direct cost) – 1,000 (overhead cost) = - \$92. You lost money on this job.
Using the net profit per hour calculation
:\$4,040 (selling price) – 2,600 (revised direct cost) – 1,000 (overhead cost) = \$440. The net profit per hour was approximately \$10 per hour rather than the estimated \$30 per hour. However, the job was still profitable.
It is much better to calculate your overhead cost per hour for each department and use it to calculate your selling prices to the customer. Assuming that you estimate your labor hours correctly, you’ll have a consistent net profit per hour worked. If you miss an estimate, you’ll still have enough profit in the job to survive.
Ruth King, a nationally-known HVACR industry expert who writes this column for HVACR Business can be contacted at rking@hvacrbusiness.com.
## Articles by Ruth King
### Understand Your P&L Statement: Cost of Goods Sold
However you decide to categorize expenses in your P&L, it's important to be consistent.
View article.
### Understand Your P&L Statement: Sales vs. Revenue
Sales are critical to survival — when revenue is actually generated is even more critical.
View article.
### The 20 Percent Profit Myth
For a realistic goal, include owners’ compensation in the net profit equation.
View article.
### Rethink Your Bonus Strategy
Enact a profit sharing program, rather than a bonus entitlement.
View article.
### Prevent Employee Embezzlement, Part 4
Most contractors don’t think about how much time is stolen by employees doing personal activities on company time.
View article.
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# Question Video: Evaluating an Expression with a Positive Base and a Negative Fractional Exponent Mathematics
Evaluate 27⁻¹ᐟ³.
01:18
### Video Transcript
Evaluate 27 to the power of negative one-third.
In this question, we can see that we have a negative exponent of negative a third. So let’s start by changing this negative exponent into a positive one. To do this, we’re going to use the exponent rule that if we have a negative exponent, for example, 𝑎 to the power of negative 𝑛, then we can write this with a positive exponent as one over 𝑎 to the power of 𝑛. So taking our 27 to the power of negative one-third, we can write this as one over 27 to the power of one-third.
Considering our value 27 to the power of one-third, what does that actually mean? Well, it’s the same as the cube root of 27. So for the cube root of 27, we’re trying to find a value 𝑥 that will give us 𝑥 times 𝑥 times 𝑥 equals 27. Since we know that three times three times three will give us 27, then the cube root of 27 is three. So our complete value will be one over three. And therefore 27 to the power of negative one-third is equal to one-third.
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# Probability for testing positive without having disease
This is a question I came across recently, any help would be greatly appreciated!:
A biostatistician is testing genes to determine their association to a certain disease. Each test for a given gene has a $5$% chance of a false positive for association (that is, about $5$% of tests performed on genes with no association will have an outcome showing association).
The biostatistician performs the test on $4$ separate genes so that the outcome of each test is independent (this means that the outcome of different tests don’t affect one another).
$(a)$ Define a suitable outcome space for this experiment.
$(b)$ Define an appropriate probability function for the outcome space of Part $(a)$ to answer the following question: If in reality, none of the 4 tested genes have an association with the disease, what is the chance that at least one of the tests will (falsely) show an association?
I am not sure about several things:
$1.$ Would the outcome space be $(+,-)$ for the test (testing pos. and neg.) or would it be all the combinations for the $4$ subjects testing positive and negative? (I am leaning towards the second one but would just like to make sure)
$2$. I understand that we have been given that $P(+|D') = 0.05$ and it is my understanding that we are being asked to find $P(X \geq 1)$ where $X$ is the number of false positives but I am not sure how to go about this at all.
Assigning $P(D) = 0.01$ and simply calculating $4 * 0.01 * 0.05$ seems too simple and wrong.
I'm really struggling with probability concepts so I would be grateful for any explanations!
Stripped of its context, I believe your question is the following: you are conducting a series of independent trials. The probability that a single trial succeeds is $p=.05$. What is the probability that you get at least one success out of four trials?
Assuming this is correct then your sample space has $16=2^4$ elements (as each trial has two possible outcomes).
To compute the probability you want, it is easier to compute the probability of the complementary event (namely "all four trials are failures"). The probability that a single trial is a failure is $1-p=.95$ so the probability that all four are failures is $.95^4\sim .8145$. It follows that the probability that at least one trial is a success is $$1-.95^4\sim .1855$$
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# Implicit Differentiaton
• Jun 30th 2007, 10:47 AM
janvdl
Implicit Differentiaton
Could somebody please clarify for me what does it mean exactly, in simple terms?
Then i have the following problem i have some questions about:
$y^5 + 3x^2 = sin x - 4y^3$
Then we differentiate...
$(5y^4 . y') + (6x) = (cos x) - (12y^2 . y')$
Ok, now where did that y' come from and what exactly does it mean? :confused:
• Jun 30th 2007, 10:54 AM
Jhevon
Quote:
Originally Posted by janvdl
Could somebody please clarify for me what does it mean exactly, in simple terms?
Then i have the following problem i have some questions about:
$y^5 + 3x^2 = sin x - 4y^3$
Then we differentiate...
$(5y^4 . y') + (6x) = (cos x) - (12y^2 . y')$
Ok, now where did that y' come from and what exactly does it mean? :confused:
the y' is actually short for dy/dx
i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"
in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.
we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly
• Jun 30th 2007, 11:02 AM
janvdl
Quote:
Originally Posted by Jhevon
the y' is actually short for dy/dx
i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"
in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.
we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly
Ok, so every y in the term has to be multiplied with y' ?
By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.
• Jun 30th 2007, 11:07 AM
Jhevon
Quote:
Originally Posted by janvdl
Ok, so every y in the term has to be multiplied with y' ?
By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.
yes. and if you were differentiating with respect to another variable besides x (which you will do when you start related rates) you would have to add x' to all the x terms as well
• Jun 30th 2007, 11:09 AM
janvdl
Quote:
Originally Posted by Jhevon
which you will do when you start related rates...
Thats next years worries Jhevon. :D
Thanks for the help guys.
• Jun 30th 2007, 11:11 AM
Jhevon
Quote:
Originally Posted by janvdl
Thats next years worries Jhevon. :D
Thanks for the help guys.
actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think
• Jun 30th 2007, 11:25 AM
janvdl
Quote:
Originally Posted by Jhevon
actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think
I'm actually not supposed to be doing this kind of stuff already.
I do it for the fun, because it's a new challenge for me. :D
• Jun 30th 2007, 11:27 AM
Jhevon
Quote:
Originally Posted by janvdl
I'm actually not supposed to be doing this kind of stuff already.
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# Thread: Total Rent Amount when there is an annual growth rate for rent
1. ## Total Rent Amount when there is an annual growth rate for rent
Say the monthly rent is \$1000 and terms of the rent state that there is an annual increase in rent for 5%. If I were to find the total rent paid in 5 years, how would I do that
I at first tried using FVIFGA, future value interest factor of growing annuity where I used an interest rate of 0% a growth rate of 5% and for 5 period that did not work out
Can anyone here help me out with this one please ?
2. ## Re: Total Rent Amount when there is an annual growth rate for rent
12000*(1.05^0 + 1.05^1 + 1.05^2 + 1.05^3 + 1.05^4) = 66307.575.....
Another way:
12 * [(1000*1.05^5 - 1000*1.05^0) / .05]
3. ## Re: Total Rent Amount when there is an annual growth rate for rent
Originally Posted by Wilmer
12000*(1.05^0 + 1.05^1 + 1.05^2 + 1.05^3 + 1.05^4) = 66307.575.....
Another way:
12 * [(1000*1.05^5 - 1000*1.05^0) / .05]
Hey Wilmer, How in the world do you come up with these formulae
Do you teach Business Math some place
Are you an accountant
4. ## Re: Total Rent Amount when there is an annual growth rate for rent
Originally Posted by dexteronline
Hey Wilmer, How in the world do you come up with these formulae
Do you teach Business Math some place
Are you an accountant
I'm a retired financial analyst; hence worked quite a bit with financial formulae.
By the way, these are really fairly easy/simple...I ain't no guru!
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# If Kp for the reaction is 0.66 then what is the equilibrium pressure of N2O4. (Total pressure at equilibrium is 0.5 atm) (1) 0.168 (2) 0.322 (3) 0.1 (4) 0.5
Dear Student,
The given reaction is :
at t = 0 1 0 initial mole
at eqm 1 - x 2x moles at eqm
Mole fraction at eqm
As the total pressure at eqm = 0.5 atm.
So the partial pressure of ${N}_{2}{O}_{4}$ =
The partial pressure of
Now, as the
Now put the value of x in partial pressure equation of ${N}_{2}{O}_{4}$ =
So, the correct option is (1)
Regards.
• 45
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# Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
Integrate $\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}$
My Attempt
Using Partial fractions $$\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\frac{1}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)}+\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=I_1+I_2\\ I_1=\bigg[\frac{1}{b^2}.\frac{1}{ab}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0=\bigg[\frac{1}{ab^3}\tan^{-1}\frac{bx}{a}\bigg]^\infty_0$$ $$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}=\frac{b^2-a^2}{b^2}\bigg[\frac{x}{(a^2+b^2x^2)^2}-\int\frac{-2.2x.x}{(a^2+b^2x^2)^3}dx\bigg]$$
How do I evaluate the integral $I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$ ?
• let $x = \frac ab \tan \theta$? – steven gregory May 28 '18 at 14:54
• @stevengregory how does that help ? – ss1729 May 28 '18 at 14:59
• This integral is what residue theorem handles extremely well. Note that the integrand is even. – Szeto May 28 '18 at 15:05
$$I_2=\frac{b^2-a^2}{b^2}\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}$$
If $$x = \frac ab \tan \theta$$, then
• $$(a^2+b^2x^2)=a^2(1+tan^2 \theta)=a^2 sec^2 \theta$$
• $$dx = \frac ab sec^2 \theta \ d\theta$$
• $$\displaystyle \int \frac{dx}{(a^2+b^2x^2)^2} = \int \frac{\frac ab sec^2 \theta \ d\theta}{a^4 sec^4 \theta} = \int \frac{1}{a^3b}\cos^2 \theta \ d\theta$$
To evaluate $\int\limits_0^{+\infty}\frac{x^2\,dx}{(a^2+b^2x^2)^2}$ just take $\int\limits_0^{+\infty}\frac{dx}{a^2+b^2x^2}$ by parts.
And $\int\limits_0^{+\infty}\frac{dx}{(a^2+b^2x^2)^2}$ reduces to the previous integral by substitution $t=\frac{a}{bx}$.
$\int_0^\infty\frac{(1+x^2)dx}{(a^2+b^2x^2)^2}=\int_0^\infty\frac{dx}{(a^2+b^2x^2)^2}+\int_0^\infty\frac{x^2dx}{(a^2+b^2x^2)^2}=I_1+I_2$
Without bounds and constant of integration we have putting $x=\dfrac {at}{b}$ with integration by parts $$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}-\int\frac{-4a^2t}{(a^2+a^2t^2)^3}tdt\right)$$ $$I_1=\frac ab\left(\frac{t}{(a^2+b^2t^2)^2}+\frac{1}{2a^4}(\arctan(t)-\frac 14\sin(4\arctan(t))\right)$$ It follows coming back to $x$, $$I_1=\frac{x}{a^2+b^2x^2)^2}+\frac{1}{2ba^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{2ba^3}\frac14\sin(4\arctan\left(\frac{bx}{a}\right)$$
$I_2$ is easier and we have
$$I_2=\frac{1}{2ab^3}\arctan\left(\frac{bx}{a}\right)-\frac{1}{4ab^3}\sin\left(2\arctan(\frac{bx}{a}\right)$$
I leave the calculations with the bounds $0$ and $\infty$ to finish.
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• Level: GCSE
• Subject: Maths
• Word count: 5723
# Maths coursework
Extracts from this document...
Introduction
Maths coursework
During my investigation I will be investigating whether there is a relationship between the T-number and the T-total.
The T-shape will look like: (in this example I will be using the numbers 1, 2, 3, 11, 20)
The T-total is the number at the bottom of the T-shape. The T-total is the sum of all the numbers inside the T-shape.
Throughout my investigation I will use a key to refer to the T-total, T-number and grid size.
For the first part of my investigation I will be investigating whether there is a relationship between T and N for numbers in a G9. On the first grid I have shaded places where N must not go; the reason that N cannot go in these places is because if there was a case where N was in these places then there would not be five numbers in the T-shape. Whilst trying to find the relationship I will move the T-shape systematically through each grid.
To find the relationship I could use:
• The sequence method
• Simultaneous equations
• Graphical methods
However, I will only use two of these methods. But for every relationship I will test whether the formula I conclude is correct, I will do this by randomly putting a T-shape into the grid and apply the formula into the numbers inside the T-shape. Also just to make sure that my conclusions are accurate I will use an algebraic approach; I use this approach because it shows a proof to an outcome.
Part 1, finding the formula relating T and N
The first thing that I notice when looking at T is that the values consistently ascend in 5’s when N ascend in 1’s. This states that there is a linear relationship.
Finding the formula.
Middle
G (grid size) T (t-total) G9 G8 G7 T = 5N - 63 T = 5N - 56 T = 5N – 49
I have noticed that in all of the formulas it is consistent that 5N is in the formula. Also the second term in the formula is the sum of G × -7.
Therefore I predict that the general formula is T = 5N – 7G
I will now test my prediction by using G10 and grid 11. I will use N = 25
T = (5 × 25) – (7 × 10)
T = 125 – 70
T = 55
Now I will use addition to see if T is the same as when I used the formula
T = 25 + 15 + 4 + 5 + 6 = 55
I can now say that the forula that I predicted is correct, this is because when I used the predicted formula the answer I got to was 55 and when I used addition the answer I got was the same of 55.
However if you look at G10 in grid 11 when N = 25 there is a relationship between T, N and G. this is:
I will now add up all of that is in the t-shape and put it into its simplest form:
T = N + (N – G) + (N – 2G) + (N – 2G + 1) + (N – 2G – 1)
T = 5N – 7G
Therefore this also correlates with the formula that I previously found, therefore the formula of T = 5N – 7G is correct.
Part 3
Here I will investigate the effect of a translation (x/y) on t-total.
Whilst doing this investigation I will use T2 as the new t-total.
Horizontal translation (x/0) for all grid sizes:
Firstly I will use G8 to find the effect of (1/0).
I can say that the formula for the t-shape in grid 3 is 5N – 7G, this is because I proved it in part 2. Now to find the formula in grid 4 compared to 5N – 7G
T in grid 3 = 34
T in grid 4 = 39
Here it shows that (1/0) is 5 more than (0/0). Therefore the formula here is
T2
Conclusion
T = 5N + 5x – 5Gy + 7 – 5d – 5dG
If these are then combined for a translation (c/d) then:
T = 5N +5x – 5Gy + 7 +5c – 5d – 5Gc – 5Gd.
However this is only my prediction, therefore I will now test this by using the formula first to find out the t-total of the rotated and translated shape, and then I will manually add up the five terms inside the rotated and translated shape and if the results both comply then the formula must be correct. I will test this on G10 on grid 10, the original t-shape will have N54 then this will be translated (-2/-1) then this shape will be rotated from the point (2/-1):
T = (5×54) + (5×-2) – (5×10×-1) + 7 + (5×2) – (5×-1) – (5×10×2) – (5×10×-1)
T = 282
Now I will add the five terms inside the rotated shape from the translated shape, and if the sum of this equals 282, then the formula works:
T = 55 + 56 + 57 + 47 + 67
T = 282
This means that the overall formula for a translation (x/y), followed by a rotation of 90º clockwise (c/d) from the new t-total is T = 5N + 5x – 5Gy + 7 + 5c – 5d – 5Gc – 5Gd.
Evaluation:
Therefore overall from my investigation, I have found that:
• For a translation (0/y) the general formula is T = 5N – 7G – 5Gy
• For a translation (x/0) the general formula is T = 5N – 7G + 5x
• For a translation (x/y) the general formula is T = 5N – 7G – 5Gy + 5x
• For a rotation 90º clockwise about point (c/0) T = 5N + 7 + 5c – 5cG
• For a rotation 90º clockwise about point (0/d) T = 5N + 7 – 5dG – 5d
• For a rotation 90º clockwise about point (c/d) T = 5N + 7 – 5dG – 5d – 5Gy + 5x
• For a translation (x/y),followed by a rotation of 90º clockwise from point (c/d) from the translated t-shape, T = 5N + 5x – 5Gy + 7 + 5c – 5d – 5Gc – 5Gd
However, due to time restriction I could only find the effects of a 90º clockwise rotation, but if time was not of the essence, then I could find the effect of a 180º rotation and 270º rotation and see if the is a connection between them all.
This student written piece of work is one of many that can be found in our GCSE T-Total section.
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# Related GCSE T-Total essays
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2. ## T-Shapes Coursework
Here are the results of the 5 calculations for a 3x1 "T" on Width 9 Grid (Fig 2.1): Middle Number Sum of Wing Sum of Tail Total Sum (Wing + Tail) 2 6 11 17 3 9 12 21 4 12 13 25 5 15 14 29 6 18 15 33 (b)
1. ## Connect 4 - Maths Investigation.
As height is the variable I will have to put it into the equation, I have decided to put it at the beginning. To work out the final rule I will have to put that into a current total. For Height4: [H(4L-9)] 4(4L-9)
2. ## T-Total Maths coursework
T=5N-7 I tested that when: T-number=62 T-total=353 N = 12 13 14 15 T = 53 58 63 68 T-number T = (5x12) -7 = 60-7 T = 53 T-total This equation has produced its first correct answer. I will carry on and test T-shape I know the T-total for N = 13 T = (5 x 13)
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# Solved: how to bubble search stack overflow
In the world of programming, sorting algorithms are widely used to arrange data in a specific order. One such algorithm is the Bubble Sort, which can be easily implemented in Python. This article will explore the Bubble Sort algorithm, explain its implementation in Python, provide a step-by-step guide on how to use this algorithm, and delve into related libraries and functions that can help enhance its performance.
Bubble Sort is a simple sorting algorithm that works by repeatedly swapping adjacent elements if they are in the wrong order. Despite not being the most efficient algorithm, it is still popular due to its easy comprehension and implementation, especially for beginners. The following Python code demonstrates a basic implementation of the Bubble Sort algorithm:
```def bubble_sort(arr):
n = len(arr)
for i in range(n):
for j in range(0, n-i-1):
if arr[j] > arr[j+1]:
arr[j], arr[j+1] = arr[j+1], arr[j]
```
Now, let’s break down the code and understand how it works, step by step:
1. Define the `bubble_sort` function, which accepts an array (or a list) called `arr` as an input.
2. Get the length of the array and store it in the variable `n`.
3. Implement two nested loops: the outer loop iterates through all the elements in the array, while the inner loop iterates from the beginning of the array to the last unsorted element.
4. Within the inner loop, compare the current element `arr[j]` with the next element `arr[j+1]`. If the current element is greater than the next one, swap them.
5. Continue this process until the entire array is sorted in ascending order.
## Python’s Built-in Sorting Functions
While Bubble Sort can be useful for understanding the basics of sorting algorithms, Python offers built-in functions that provide a faster and more efficient way to sort data. Two commonly used methods are the sorted() function and the list.sort() method.
The sorted() function returns a new sorted list from the elements of a given iterable, while the list.sort() method sorts the elements of a list in-place without creating a new list. Both functions accept two optional parameters: key and reverse. The key parameter specifies a custom function for sorting, and the reverse parameter indicates whether to sort in ascending (False) or descending (True) order.
[h2]Optimizing Bubble Sort Performance
Bubble Sort is known for its poor efficiency, especially for large datasets. However, there are certain modifications that can be made to the algorithm to improve its performance. One such modification is the Optimized Bubble Sort, which can result in a significant performance improvement in some cases. The main idea behind this optimization is to stop iterating when no swaps are needed in the inner loop.
The following code demonstrates an optimized version of Bubble Sort:
```def optimized_bubble_sort(arr):
n = len(arr)
for i in range(n):
swapped = False
for j in range(0, n-i-1):
if arr[j] > arr[j+1]:
arr[j], arr[j+1] = arr[j+1], arr[j]
swapped = True
if not swapped:
break
```
In this optimized version, a boolean variable called ‘swapped’ is introduced. It keeps track of whether any swaps occur during each inner loop iteration. If no swaps are needed, the array is already sorted, and the algorithm can terminate early, avoiding unnecessary iterations.
In conclusion, Bubble Sort is a simple sorting algorithm that is useful for learning the basics of sorting and algorithmic thinking. Although it might not be the most efficient sorting algorithm, the ease of implementation makes it a good starting point for programmers new to sorting algorithms. Through understanding its implementation, exploring built-in sorting functions, and optimizing performance, one can better appreciate the process of arranging data in an organized manner, thus enhancing their programming skills.
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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 11 - Geometric Progression
## Selina Textbook Solutions Chapter 11 - Geometric Progression
Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 11 - Geometric Progression.
All Selina textbook questions of Chapter 11 - Geometric Progression solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.
Exercise/Page
Solution 1(iii)
Solution 1(ii)
Solution 1(i)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8(ii)
Solution 8(i)
Solution 9
## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 11 - Geometric Progression Page/Excercise 11(D)
Solution 1(vi)
Solution 1(v)
Solution 1(iv)
Solution 1(iii)
Solution 1(ii)
Solution 1(i)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12(iii)
Solution 12(ii)
Solution 12(i)
Solution 13
Solution 14
Solution 15
First term (a) =
Solution 16
First term (a) = 125
Solution 17
Thus, the given sequence is a G.P. with
Solution 18
Solution 19
Let the five terms of the given G.P. be
Given, sum of first two terms = -4
And, 5th term = 4(3rd term)
ar2 = 4(a)
r2 = 4
r = ±2
When r = +2,
When r = -2,
## Browse Study Material
TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 11 - Geometric Progression for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.
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# Solving Inequality
• Aug 31st 2013, 07:53 PM
vaironxxrd
Solving Inequality
Hello Everyone,
I have a very embarrassing question. I'm studying the introduction chapter in my Calculus book(Worried).
I have the following problem: $(x+2)(x-1)(x-3) > 0$
The answer in my book: $(-2,1)\cup(3,8)$
My answer doesn't match and I can't really figure out why:
$(x+2)(x-1)(x-3) > 0$
Split points: -2, 1, 3
There must be four intervals to test $(-\infty,-2), (-2,1), (1,3), and(3,\infty)$
So I use the following test points to determine when the inequality is greater than 0: -3, 0, 2, 4
x= -3 [-] Doesn't Work
x = 0 [+] Does Work
x = 2 [-] Doesn't Work
x = 4 [+] Does Work
So according to me and this work I did the solution should be $(-2,1) \cup (3,\infty)$
• Aug 31st 2013, 08:51 PM
johng
Re: Solving Inequality
Hi,
Text books are not infallible; in this case you're right and the book is wrong. I'm very prone to arithmetic and algebraic errors, so I almost always check my calculations for such problems by drawing a graph with my graphing software. I suggest you do the same.
Attachment 29100
• Aug 31st 2013, 09:04 PM
ibdutt
Re: Solving Inequality
Attachment 29101proceed like this to consider remaining cases
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# lecture09 - Lecture 9 Increasing and Decreasing Functions...
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Unformatted text preview: Lecture 9 - Increasing and Decreasing Functions, Extrema, and the First Derivative Test 9.1 Increasing and Decreasing Functions One of our goals is to be able to solve max/min problems, especially economics related examples. We start with the following definitions: Definition 9.1 A function f is called increasing on an interval ( a,b ) if for any x 1 ,x 2 ∈ ( a,b ) , we have that x 1 < x 2 ⇒ f ( x 1 ) < f ( x 2 ) A function f is called decreasing on an interval ( a,b ) if for any x 1 ,x 2 ∈ ( a,b ) , we have that x 1 < x 2 ⇒ f ( x 1 ) > f ( x 2 ) Note that some books call these strictly increasing and strictly decreasing respectively. Examples: 1. f ( x ) = x 2 . f ( x ) is increasing on (0 , ∞ ) and decreasing (-∞ , 0). 2. f ( x ) = x 3 . f ( x ) is increasing on (-∞ , ∞ ). As you may have guessed, we can use the derivative to test for increasing/decreasing. Let f be differentiable on the interval ( a,b ). 1. If f ( x ) > 0 for all x in ( a,b ), then f is increasing on ( a,b ). 2. If f ( x ) < 0 for all x in ( a,b ), then f is decreasing on ( a,b ). 3. If f ( x ) = 0 for all x in ( a,b ), then f must be constant on ( a,b ). FACT: If f is a continuous function, then f ( x ) can only change signs at values of x where f ( x ) = 0 or f ( x ) doesn’t exist. Examples: 1. Let f ( x ) = x 3 4- 3 x . Find the intervals on which f ( x ) is increasing or decreasing. solution: We will use the test and fact above. The derivative is f ( x ) = 3 x 2 4- 3 = 3 4 ( x 2- 4) = 3 4 ( x- 2)( x + 2) Because of the above, we split our real line up where f ( x ) = 0: f ( x ) % & %- 2 2 On (-∞ ,- 2), ( x- 2) and ( x + 2) are both negative, so f ( x ) = 3 4 ( x- 2)( x + 2) > 0, so f is increasing. On (- 2 , 2), ( x- 2) is negative while ( x +2) is positive, so f ( x ) = 3 4 ( x- 2)( x +2) < 0, so f is decreasing. On (2 , ∞ ), ( x- 2) and ( x + 2) are both positive, so f ( x ) = 3 4 ( x- 2)( x + 2) > 0, so f is increasing. To summarize, f ( x ) is increasing on (-∞ ,- 2) and (2 , ∞ ) while f ( x ) is decreasing on (- 2 , 2). 2. Let g ( x ) = x + 32 x 2 . Find the intervals on which g ( x ) is increasing and decreasing. solution: We first need to find the derivative. g ( x ) = 1- 64 x 3 Note that g ( x ) = 0 when x 3 = 64 ⇒ x = 4. The derivative is also not defined when x = 0, so these are the places we divide our real line. g ( x ) % & % 4 To find out what sign the derivative is on (-∞ , 0), we can substitute in a test point in that interval and see if the result is positive or negative. The point x =- 1 is a convenient one to use: g (- 1) = 1- 64 (- 1) 3 = 65 > Because of our fact,...
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# Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2
Students can download 12th Business Maths Chapter 7 Probability Distributions Ex 7.2 Questions and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2
Question 1.
Define Poisson distribution.
Solution:
Poisson distribution is a discrete frequency distribution which gives the probability of a number of independent events occurring in a fixed time. It is useful for characterizing events with very low probabilities of occurrence within some definite time or space.
Question 2.
Write any 2 examples for Poisson distribution.
Solution:
1. The number of alpha particles emitted by a radioactive substance in a fraction of a second.
2. Number of road accidents occurring at a particular interval of time per day.
Question 3.
Write the conditions for which the Poisson distribution is a limiting case of the binomial distribution.
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
• the number of trials ‘n’ is indefinitely large i.e, → ∞
• the probability of success ‘p’ in each trial is very small, i.e, p → 0
• np = λ is finite. Thus p = $$\frac{\lambda}{n}$$ and q = 1 – $$\frac{\lambda}{n}$$, λ > 0
Question 4.
Derive the mean and variance of the Poisson distribution.
Solution:
Let X be a Poisson random variable with parameter λ. The p.m.f is given by
Thus the mean and variance of Poisson distribution are both equal to λ.
Question 5.
Mention the properties of Poisson distribution.
Solution:
Poisson distribution is the only distribution in which the mean and variance are equal.
Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? [Given e-2.8 = 0.06]
Solution:
Let X denote the number of deaths due to the disease
P(death) = $$\frac{7}{1000}$$ = 0.007 ⇒ p = 0.007 and n = 400
The value of mean λ = np = (0.007) (400) = 2.8
Hence X follows a Poisson distribution with
So the probability of just 2 deaths on account of this disease in a group of 400 is 0.2352.
Question 7.
It is given that 5% of the electric bulbs manufactured by a company are defective. Using Poisson distribution find the probability that a sample of 120 bulbs will contain no defective bulb.
Solution:
p(defective bulbs) = $$\frac { 5 }{100}$$
n = 120
The value of mean λ = np = 120 × $$\frac { 5 }{100}$$
λ = 6
Hence, x follows possion distribution with
P(x) = $$\frac { e^{-λ}λ^x }{x!}$$
(x = 0) = $$\frac { e^{-6}(6)^0 }{0!}$$ = e-6
= 0.0025
Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poisson variate, with a mean of 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused.
Solution:
Let X be the Poisson variable denoting the demand for the cars.
It is given that mean is 1.5 ⇒ λ = 1.5
(i) P (Neither car is used) = P (X = 0) = $$\frac{e^{-1.5}(1.5)^{0}}{0 !}=e^{-1.5}=0.2231$$
(ii) Some demand is refused when demand is more than 2 since the firm has only 2 cars. So we want P (X > 2)
Now P (X > 2) = 1 – P (X ≤ 2)
= 1 – [P(X = 2) + P(X = 1) + P(X = 0)]
Question 9.
The average number of phone calls per minute into the switchboard of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be
(i) no phone at all
(ii) exactly 3 calls
(iii) at least 5 calls.
Solution:
Let X be the Poisson variable denoting the number of phone calls per minute.
P (X = 1) = $$\frac{e^{-2.5}(2.5)}{1 !}$$ = (0.08208) (2.5) = 0.2052
Using the above values and P (X = 0) and P (X = 3) from the previous subdivisions in (A) we get,
P(X ≥ 5) = 1 – [0.1336 + 0.2138 + 0.2565 + 0.2052 + 0.08208]
= 1 – 0.89118
= 0.10882
Question 10.
The distribution of the number of road accidents per day in a city is Poisson with mean 4. Find the number of days out of 100 days when there will be
(i) no accident
(ii) at least 2 accidents and
(iii) at most 3 accidents.
Solution:
Let X be the Poisson variable denoting the number of accidents per day.
Given that mean is 4 (i.e,) λ = 4. The p.m.f is given by P(X = x) = $$\frac{e^{-4} 4^{x}}{x !}$$
(i) P (no accident) = P(X = 0) = e-4 = 0.0183
For 100 days we have 100 × 0.0183 = 1.83 ~ 2
Hence out of 100 days there will be no accident for 2 days.
(ii) P (atleast 2 accidents) = P (X ≥ 2)
= 1 – P (X < 2)
= 1 – [P(X = 1) + P(X = 0)]
= 1 – [e-4 (4) + e-4]
= 1 – (0.0183) (5)
= 1 – 0.0915
= 0.9085
For 100 days we have 100 × 0.9085 ~ 91
Hence out of 100 days there will be at least 2 accidents for 91 days.
(iii) P (atmost 3 accidents) = P (X ≤ 3)
= P (X = 0 ) + P (X = 1 ) + P (X = 2) + P (X = 3)
= $$e^{-4}\left[1+\frac{4}{1}+\frac{16}{2}+\frac{64}{6}\right]$$
= (0.0183) [23.6667]
= 0.4331
For 100 days we have 100 × 0.4331 ~ 43
Hence out of 100 days, there will be at most 3 accidents for 43 days.
Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200, calculate the probability that in a factory employing 300 workers there will be at least two fatal accidents in a year, (given e-0.25 = 0.7788).
Solution:
Let X denote the number of accidents.
Given that probability of accidents ‘p’ is 1/1200 and n = 300
= 1 – [e-0.25 + e-0.25 (0.25)]
= 1 – e-0.25 (1.25)
= 1 – (0.7788) (1.25)
= 0.0265
Thus the probability that there will be atleast two fatal accidents in a year is 0.0265.
Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute
(i) No customer appears
(ii) three or more customers appear.
Solution:
Let X denote the number of customers.
Given λ = 2
(i) P (no customer) = P (X = 0)
= $$\frac{e^{-2}(2)^{0}}{0 !}$$
= e-2
= 0.1353
(ii) P (3 or more customers) = P (X ≥ 3)
Thus during a given minute, the probability that three or more customers appear is 0.3235.
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0
# What would 10 degrees celsius be converted be in Fahrenheit?
Updated: 8/11/2023
Wiki User
12y ago
Start by taking the number in Fahrenheit and subtracting 32. Then divide the number by 9, and then multiply it by 5. This is how you convert Fahrenheit to Celsius or use the equation C = (F - 32) × 5/9
Wiki User
13y ago
Wiki User
7y ago
F to C: Deduct 32, then multiply by 5, then divide by 9
C to F: Multiply by 9, then divide by 5, then add 32
10 - 32 = -22 x 5 = -110/9 = -12.22 C
Wiki User
12y ago
10 degrees Celsius = 50 Fahrenheit
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Q: What would 10 degrees celsius be converted be in Fahrenheit?
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Related questions
### What does 98 degrees Celsius converted to degrees Fahrenheit equal?
that would be 255 degrees fahrenheit, my friend.
5.56 ℃
527 ℉
### What is 39.0 degrees Fahrenheit converted into Celsius?
The formula for converting Fahrenheit to Celsius is C = (F-32) * 5/9 So 39.0 Degrees Fahrenheit would be equal to 3.88 Degrees Celsius
### What is (32)22?
22 Celsius is a temperature scale that displays in Celsius degrees. 22 degrees Celsius can be easily converted into Fahrenheit by using the formula Fahrenheit degrees = (Celsius degrees) 9/5 + 32 which in this case would equal 71.6 degrees Fahrenheit.
"Celsius 232.8"
### What would be the title of Ray Bradbury's classic novel Fahrenheit 451 if it were converted into degrees celsius?
It would be celsius 233.
### What would 37.5 degrees Celsius be converted to in Fahrenheit?
37.5°C = 99.5°F
### What would 279 degrees Celsius be in Fahrenheit?
279 degrees Celsius = 534.2 degrees Fahrenheit
8
### What would be the Celsius temperature of 500 degrees Fahrenheit?
500 degrees Fahrenheit = 260 degrees Celsius.
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# What fraction is equivalent to 3/8?
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#### Explanation
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#### Explanation:
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21
Apr 17, 2017
Multiply $\frac{3}{8}$ by an equivalent fraction that equals $1$.
An equivalent fraction is equal to 1 such as $\frac{2}{2}$ or $\frac{3}{3}$
An equivalent fraction is the same as multiplying by $1$
so the ratio does not change.
This fraction would be equal to $\frac{a}{a} = 1$ where a is an integer.
Hence an example would be that
$\frac{2}{2} \cdot \frac{3}{8} = \frac{6}{16}$
Then teach the underlying concepts
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### This answer has been featured!
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Moksha Share
May 29, 2017
There are infinite fractions that equal to $\frac{3}{8}$.
#### Explanation:
As I said, there are infinite fractions that equal to $\frac{3}{8}$. Therefore, I' m going to show you a way to get a fraction equivalent to $\frac{3}{8}$.
Procedure/Step(s):
1) Pick a number that you want and make sure that that number is both numerator and denominator.
Example: $\frac{4}{4}$
(You don't have to pick 4/pick any number you want!)
2) Multiply the fraction that you chose with $\frac{3}{8}$
Note: Do NOT simplify! If you do, you're just gonna get $\frac{3}{8}$!
$\frac{3}{8}$$\times$$\frac{4}{4}$$=$$\frac{12}{32}$
So, an equivalent fraction to$\frac{3}{8}$ is $\frac{12}{32}$. I hope that helps!
All the best!
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preview
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#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
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[CHAN] Share
Apr 30, 2017
There are infinitely many greater equivalent fractions to $\frac{3}{8}$.
#### Explanation:
First of all, remember:
"Any of operation that occurs to the numerator must also apply to the denominator"
Using this knowledge, we can create any equivalent fraction!;
= $\frac{3}{8}$
= $\frac{3 \cdot 6}{8 \cdot 6}$
= $\frac{18}{48}$
= etc.
All the best!
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# The attendees at a certain convention purchased a total of 15,000 book
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The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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23 Jul 2006, 16:59
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The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?
(1) There was a total of 4,000 attendees at the convention.
(2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each.
[Reveal] Spoiler: OA
Last edited by Bunuel on 20 Oct 2014, 08:32, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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23 Jul 2006, 17:23
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Form 1. 4000 = m + f
From 2. 15000 = 3m + 5f
using both solve for f
Thanks
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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23 Jul 2006, 21:42
C it is.
Statement 1....... M + F = 4000,
Statement 2.......3M + 5F = 15000,
Solving => F = 1500.
_________________
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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23 Jul 2006, 22:20
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St1:
# of male attendees = m
# of female attendees = 4000-m
But we can't work further. Insufficient.
St2:
3m + 5f = 15,000
Can't solve as there are many possibilites for (m,f) sets
Using St1 and St2:
3m + 5(4000-m) = 15,000
Can solve for m, and thus f. Sufficient.
ANS c
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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24 Jul 2006, 00:46
shehreenquayyum wrote:
Attendees at a certain convention purchased 15000 boks. How many of these attendees are females?
i) Total attendees are 4000
ii) Males purchased an average of 3 books each & females purchased an average of 5 books each.
I think we have to use the weighted average? Answer is C....
Straight C
from 1: m+f =4000
from 2: 3m+5f=15000
2 equations and 2 unknowns, 1 and 2 are insuff. but combined can give the answer.
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25 Jul 2006, 04:40
C
1) M + F = 4000
Not Suff
2) 3M +5F = 15000
Not Suff
Together
3(4000-F) + 5F = 15000
12000 -3F +5F = 15000
2F = 3000
F = 1500
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The attendees at a certain convention purchased a total of 15,000 [#permalink]
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28 Jan 2007, 00:44
2
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The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?
(1) There was a total of 4,000 attendees at the convention.
(2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each.
Last edited by Bunuel on 28 Sep 2014, 11:33, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink]
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28 Jan 2007, 03:01
Total attendees at a convention purchased 15000 books. How many of these were female?
1. 4000 attended the convention
2. On average men bought 3 books, females bought 5 books
My solution: 5/8 * 4000 = 2500 women. Is this correct? (even though we don't have to solve in DS). Thanks!
ONE IS OBVIOUSLY NOT SUFF
FROM TWO .......NOT SUFF
BOTH TOGETHER
TOTAL NUMBER OF BOOKS MEN PURCHASED/ TOTAL NMBER OF MEN = 3
X/Y =3
TOTAL NMBE OF BOOKS PURCHASED BY WOMEN/ NUMBER OF WOMEN =5
15000 - X / 4000- Y = 5
WE HAVE TWO EQUATIONS FOR TWO UNKNOWNS
3Y =X
15000 - 3Y = 20000 - 5Y
2Y = 5000 IE: Y = 2500 THUS ............SUFF
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink]
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28 Jan 2007, 23:57
1
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successstory wrote:
Total attendees at a convention purchased 15000 books. How many of these were female?
1. 4000 attended the convention
2. On average men bought 3 books, females bought 5 books
My solution: 5/8 * 4000 = 2500 women. Is this correct? try to check by back solving females bought 2500*5=12500 books hence 2500 books was bought by men 2500/3=not integer and obviously can't be the number of men))) (even though we don't have to solve in DS). Thanks!
agree with yezz 1st and 2 statements insuff alone
taking both together
M+F=4000 -->M=4000-F
M*3+F*5=15000-->
3*(4000-F)+5F=15000
12000-3F+5F=15000
2F=3000
F=1500
M=2500
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28 Sep 2014, 10:57
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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19 Oct 2014, 16:50
can someone tell me why this reasoning is wrong.
For statement B - say you take it to mean for every 8 books, 5 are bought for 1 female. So for the 15,000 books 9375 are bought by females. Average out 5 books to one female and you have 1875 females.
Where is my logic going astray?
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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19 Oct 2014, 20:47
angelfire213 wrote:
can someone tell me why this reasoning is wrong.
For statement B - say you take it to mean for every 8 books, 5 are bought for 1 female. So for the 15,000 books 9375 are bought by females. Average out 5 books to one female and you have 1875 females.
Where is my logic going astray?
You are assuming that number of males = number of females which is not given.
You say that one male and one female buy a total of 8 books so total number of pairs = 15000/8 = 1875. So we get that there are 1875 males and 1875 females.
But isn't this possible - there are 3 females who buy 15 books and rest 14985 books are bought by 4995 males? This will give us a total of 15000 books bought such that on average males buy 3 books per person and females buy 5 books per person.
Similarly, there are many other cases possible.
This question can be done by weighted averages concept (discussed here: http://www.veritasprep.com/blog/2011/03 ... -averages/) in seconds.
We know w1/w2 = (A2 - Aavg)/(Aavg - A1)
We need to find the fraction w1/w2 = Number of males/Number of females and the total number of people to get the number of females.
i) Total attendees are 4000
This gives us Aavg = 15000/4000
We also get that w1 + w2 = 4000.
But we don't have A1, A2.
ii) Males purchased an average of 3 books each & females purchased an average of 5 books each.
This gives us A1 = 3 and A2 = 5
We don't have Aavg.
Using both, we have Aavg, A1 and A2. So we can find w1/w2 and we also know that total number of people is 4000. This is sufficient to answer the question.
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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02 May 2017, 09:57
VeritasPrepKarishma wrote:
infinitemac wrote:
Hoping someone can help me to understand what I am doing wrong:
For statement 1, I understood that M + F = 4000
But for statement 2, when I read "average" this is the type of algebraic statement I came up with:
For males, where number of total books bought by males I noted as "Bm" and overall total of males M, so equation is Bm/M =3, and similarly for females, Bf/F = 5. I then tried to set up Bm/M + BF/f = 15,000.
How come in the solution we reduce it to very simple terms as being 3M + 5F = 15,000? I'm not sure I'm quite understanding...thanks!
What is 15000? The sum of the total number of books purchased.
Average number of books purchased by males = 3
Average number of books purchased by females = 5
Say there are M males and F females.
Average = Number of books purchased by males/Number of males
3 * M = Number of books purchased by males
Average = Number of books purchased by females/Number of females
5*F = Number of books purchased by females
Total number of books purchased = Number of books purchased by males + Number of books purchased by females = 3M + 5F
15000 = 3M + 5F
Note that Bm/M + Bf/F = 15,000 is wrong. You have already established that Bm/M = 3 and Bf/F = 5. These are the averages. When you add these two, you get 3 + 5 = 8.
15000 is the sum of the number of books purchased.
Thank you so much Karishma - seems like I missed out on a simple concept but you have explained it very clearly! Thanks!
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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24 Jun 2017, 20:06
shehreenquayyum wrote:
The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?
(1) There was a total of 4,000 attendees at the convention.
(2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each.
The goal of the problem is to find F, the number of female attendees at the convention.
Statement 1) F+M = 4000.
Insufficient. We don't know the breakdown or distribution of males or females.
Statement 2) Books bought by males / M = 3.
Books bought by females / F = 5.
5F + 3M = books bought by males and books bought by females = 15000
Insufficient because we don't know the total number of males or females.
Statements 1+2) Sufficient.
3M + 5F = 15000
M + F = 4000
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Re: The attendees at a certain convention purchased a total of 15,000 book [#permalink]
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26 Jun 2017, 14:08
The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?
(1) There was a total of 4,000 attendees at the convention.
Total attendees $$= 4,000$$
This does not give us any information on the number of female attendees.
Hence, (1) =====> is NOT SUFFICIENT
(2) The male attendees purchased an average (arithmetic mean) of 3 books each, and the female attendees purchased an average of 5 books each
This gives us average of Male and Female purchase of books
$$3m + 5f = 15000$$
As we are not aware of total count of the participants
Hence, (2) =====> is NOT SUFFICIENT.
Now lets combine (1) & (2)
$$m + f = 4,000$$
$$3m + 5f = 15,000$$
As you can see that with the help of these two equations we can find the total number of female attendees.
Please note that you need not do the actual calculation, only the information that we can arrive at a number should be sufficient for us.
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The attendees at a certain convention purchased a total of 15,000 [#permalink]
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09 Jul 2017, 07:41
The attendees at a certain convention purchased a total of 15,000 books. How many of these attendees were female?
1) There was a total of 4,000 attendees at the convention.
2) The male attendees purchased an average of 3 books each, and the female attendees purchased an average of 5 books each.
Can someone assist in solving this problem?
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Re: The attendees at a certain convention purchased a total of 15,000 [#permalink]
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09 Jul 2017, 08:58
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Ans = C
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# Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2
Question 1.
Fill in the blanks In the following table,
given that ‘a’ is the first term, ‘d’ is the common difference and an the nth term of the A.P.
Solution:
(i) a = 7, d = 3, n = 8, an =?
an = a + (n – 1) d
a8= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a8 = 28
(ii) a = -18, d =?, n = 10, an = 0
an = a + (n – 1) d
0 = -18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
9d = 18
$$\mathrm{d}=\frac{18}{9}=2$$
(iii) a =?, d = -3, n = 18, an = -5
an = a + (n – 1) d
-5 = a + (18 – 1) (-3)
= a + 17(-3)
-6 = 1 – 51
∴ a = -5 + 51 = 46
(iv) a = -18.9, d = 2.5, n =? an = 3.6
an = a + (n – 1) d
3.6= -18.9 + (n – 1) (2.5)
3.6= -18.9 + 2.5n – 2.5
3.6 = 2.5n – 21.4
2.5n = 3.6 + 21.4
2.5n = 25
$$n=\frac{25}{2.5}=\frac{250}{25}$$
∴ n = 10.
(v) a = 3.5, d = 0, n = 105, an =?
an = a + (n – 1) d
= 3.5 + (105 – 1) (0)
= 3.5+ 104 × 0
= 3.5 +0
∴ an = 3.5
Question 2.
Choose the correct choice In the following and justify:
(i) 30th term of the AP: 10, 7, 4, ……….. is
A) 97
B) 77
C) -77
D) -87
Solution:
a = 10. d = 7 – 10 = -3, n = 30, a30 =?
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a30 = 77
∴ Ans: (C) -77
(ii) 11th term of the A.P.
$$-3,-\frac{1}{2}, 2, \dots \ldots,$$ is
A) 28
B) 22
C) -38
D) $$-48 \frac{1}{2}$$
Solution:
$$a=-3, \quad d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=2 \frac{1}{2}$$
n = 11, a11 =?
an = a + (n – 1) d
$$a_{11}=-3+(11-1)\left(2 \frac{1}{2}\right)$$
$$=-3+10\left(\frac{5}{2}\right)$$
= -3 +25
∴ a11 = 22
∴ Ans: (B) 22
Question 3.
In the following APs find the missing terms in the boxes:
Solution:
(i)
a = 2, a + d =?, a + 2d = 26
a + 26 = 26
2 + 2d = 26
2d = 24
∴ d=12 . .
∴ a + d = 2 + 12 = 14
∴ Ans: 14
(ii) Here, a =?, a + d = 13, a + 2d=?,
a + 3d = 3
a + d + 2d = 3
13 + 2d = 3
2d = 3 – 13
2d = -10
∴ d = -5
a + d = 13
a + (-5) = 13
a – 5 = 13
∴ a – 13 + 5 = 18
∴ a = 18
a + 2d =?
= 18 + 2(-5)
= 18 – 10
a + 2d = 8
∴ Ans: 18, 8
(iii) a = 5, a + d =?, a + 2d =?
(iv) a = -4, a +d=? a + 2d =?
a+3d = ? a + 4d =? a + 5d = 6
a + 5d = 6
-4 + 5d = 6
5d = 6 + 4
5d = 10
$$\mathrm{d}=\frac{10}{5}$$
∴ d=2.
a + d = -4 + 2 = -2
a + 2d = -4 + 2 (2) = -4 + 4 = 0
a + 3d = -4 + 3 (2) = -4 + 6 = 2
a + 4d = -4 + 4 (2) = -4 + 8 = 4
∴ Ans: -2, 0, 2, 4
(v) a =? a + d = 38, a + 2d =?
a + 3d =?, a + 4d =?a + 5d = 22
4d = -60
$$\mathrm{d}=-\frac{-60}{4} \quad=-15$$
a + d = 38
a – 15=38
a = 38 + 15 = 53
a + 2d = 53 + 2(-15) = 53 – 30 = 23
a + 3d = 53 + 3(-15) = 53 – 45 = 8
a + 4d =53 + 4(-15) = 53 – 60 = 7
∴ 53, 23, 8, -7.
Question 4.
Which term of the AP : 3, 8, 13. 18, ………. is 78?
Solution:
a = 3, d = 8 – 3 = 5, an = 78, n =?
an = a + (n – 1) d
78 = 3+(n—1)(5)
78 = 3 + 5n – 5
78 = 5n – 2
5n = 78 + 2
$$n=\frac{80}{5}$$
∴ n = 16
Question 5.
Find the number of terms In each of the following APs:
(i) 7, 13, 19, ………… 201
(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
Solution:
(i) 7, 13, 19, ………. 201
a = 3, d = 13 – 7 = 6, an = 201. n =?
a + (n – 1) d = an
3 + (n – 1) 6 = 201
3 + 6n – 6 = 201
6n – 3 = 201
6n = 203
$$n=\frac{204}{6}$$
∴ n = 34
(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
$$a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}$$
an = -47, n =?
an = a + (n – 1) d
$$-47=18+(n-1)\left(\frac{-5}{2}\right)$$
$$-47-18=(n-1)\left(\frac{-5}{2}\right)$$
$$(n-1)\left(\frac{-5}{2}\right)=-65$$
$$n-1=-65 \times \frac{-2}{5}$$
n – 1 = -13 × -2
n – 1 = + 26
∴ n = 26 + 1
∴ n = 27
Question 6.
Check whether -150 is a term of the A.P: 11, 8, 5, 2, ………..
Solution:
11, 8, 5, 2, ……….. -150
a = 11, d = 8 11 = -3. an = -150.
a + (n – 1) d = an
11 + (n – 1) (-3) = -150
11 – 3n + 3 = -150
-3n + 14 = -150
-3n = -150 – 14
-3n = -164
3n = 164
$$n=\frac{164}{3}$$
Here value of ‘n’ is not perfect. Hence -150 is not a term of the A.P.
Question 7.
Find the 31st term of an AP whose 11th term Is 38 and the 16th term is 73.
Solution:
a = 38, a16 = 83 a31 =?
an = a + (n – 1) d
a16 = a + (16 – 1) d
a + 15d = 83
38 + 15d = 83
15d = 83 – 38
15d = 45
$$d=\frac{45}{15}$$
∴ d=3.
∴ an = a + (n – 1)d
a31 = 38 + (31 – 1) 3
= 38 + 30 × 3
= 38 + 90
∴ a = 128.
Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
n = 50, a3 = 12, an = 106, a29 =?
a11 = a + (n – 1) d
a50 = a + (50 – 1) d = 106
∴ a + 49d = 106 ……………… (1)
a3 = a + 2d = 12 ………………. (2)
Subtracting equation (2) in equation (1).
Substituting the value of d.
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
∴ a = 12 – 4
a = 8.
∴ an = a + (n – 1) d
a29 = 8 + (29 – 1) 2
= 8 + 28 × 2
= 8 + 56
∴ a29 = 64
Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
a3 = 4, a9 = -8, an = 0, n =?
a3 = a + 2d = 4 ………….. (1)
a9 = a + 8d = -8 …………. (2)
From equation (1) – equation (2).
6d = 12
$${d}=\frac{-12}{6}$$
∴ d = -2
a + 2d = 4
a – 2(2) = 4
a – 4 =4
∴ a = 4 + 4
∴ a = 8
an = a + (n – 1) d
= 8 + (n – 1) (-2)
= 8 – 2n + 2
= 10 – 2n = 0 ∵ an = 0
$$n=\frac{10}{2}$$
∴ n = 5
∴ 5th term of this AP is Zero.
Question 10.
The 1 7th term of an AP exceeds its 17th term by 7. Find the common difference.
Solution:
a17 = a10 + 7, d =?
a + 16d = a + 9d + 7
a +1 6d – a – 9d = 7
7d = 7
$$\mathrm{d}=\frac{7}{7}$$
∴ = 1
Question 11.
Which term of the AP: 3, 15, 27. 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, ………… an =?, n =?
an = a54 + 132
a = 3, d = 15 – 3 = 12
an = a54 + 132
an = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ an = 771
an = a + (n – 1) d = 771
= 3 + (n – 1)12 = 771
3 + 12n — 12 = 771
12n – 9 = 771
12n = 771 + 9
12n = 780
$$n=\frac{780}{12}$$
∴ n = 65.
∴ 65th term is 132 more than its 54th term.
Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100. what is the difference between their 1 000th terms?
Solution:
Having common difference ‘d’, the APs
1st set a, a+d, a+2d
2nd set b, b + d, b + 2d
100th term of 1st set – 100th term of 2nd set = 100
∴ a + 99d – (b + 99d) = 100
a + 99d – b – 99d = 100
a – b= 100
Similarly.
1000th term of 1st set = 1 + 999d
1000 th term of 2nd set = b + 999d
Their difference
= a + 999d – (b + 999d)
= a + 999d – b – 999d
= a – b.
Question 13.
How many three-digit numbers are divisible by 7?
Solution:
First three-digit number divisible by 7
105 and the List number Is 994.
∴ AP is 105, 112. 119 994.
a = 105, d = 112 – 105 = 7. an = 994.
n =?
a + (n – 1)d = an
105 + (n – 1) 7 = 994
105 + 7n – 7 = 994
7n + 98 = 994
7n = 994 – 98
7n = 896
$$n=\frac{896}{7}$$
∴ n = 128.
∴ Numbers with 3 digits divisible by 7 are 128.
Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4. after 10 are 12, 16, 20….
Multiples of 4 upto 250 is 248
∴ A.P. is 12, 16, 20, …….. 248
a = 12, d = 16 – 12 = 4
n =?
a = a + (n – 1) d = 248
12 + 4n – 4 = 248
4n + 8 = 248
4n = 248 – 8
4n = 240
$$n=\frac{240}{4}$$
∴ n = 60
∴ Multiples of 4 lie between 10 and 250 is 60.
Question 15.
For what value of n’. are the nth terms of two APs: 63, 65, 67, ……. and 3, 10, 17, ……….. equal?
Solution;
63, 65, 67,……….
a = 63. d = 65 – 63 =2. an =?
nth term of this is
an = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
an= 2n + 61 …………….. (i)
3, 10, 17, ………….
a = 3, d = 10 – 3 = 7, an =?
an = a + (n – 1)d
= 3 + (n – 1) 7
= 3 + 7n – 7
an = 7n — 4 ………….(ii)
Here. nth terms of second AP are equal.
∴ equation (i) = equation (ii)
2n + 61 = 7n – 4
2n – 7n = -4 – 61
5n = 65
5n =65
$$n=\frac{65}{5}$$
∴ n = 13
∴13th terms of the two given APs are equal.
Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a = 16, a7 = a5 + 12, A.P =?
a7 = a5 + 12
a + 6d = a + 4d + 12
a + 6d – a – 4d = 12
2d = 12 ∴ d = 6.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
∴ a = 16 – 12 ∴ a = 4
a = 4, d = 6.
∴ A.P.:
a, a + d, a + 2d, …………………
4, 4 + 6, 4 + 12, …………….
4, 10, 16, ………….
Question 17
Find the 20th term from the Last term of theAP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, ………….., 253
a = 3. d = 8 – 3 = 5, an = 253
20th term from the last term of the AP starting from 253 =?
253, 258, 263, ………… a20 =?
a = 253, d = 258 – 253 = 5, n = 20
an = a + (n – 1) d
a20= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a20 = 348
∴ 20th term from the last term of the AP is 348.
Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. FInd the first three terms of the AP.
Solution:
a4 + a8 = 24 …………. (1)
a6 + a10 = 44 ………… (2)
But A.P is a, a + d, a + 2d
from equation (1).
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24 ………….. (3)
from equation (2).
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44 ………… (4)
Subtracting eqn. (4) from equation (3)
4d = 20
$$d=20 / 4$$
∴ d = 5
Substituting the value of d in equation (3)
2a + 10d = 24
2a + 10(5) = 24
2a + 50 = 24
2a = 24 – 50
2a = -26
a = 26/2 ,
∴ a = -13 .
∴ AP: a, a + d, a + 2d, ………..
-13, -13 + 5, -13 + 2(5), ………
-14, -8, -3, …………
Question 19.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an Increment of Rs. 200 each year. in which year did his income reach Rs. 7000?
Solution:
Payment of Subba Rao in the year 1995 = Rs. 5000
increment = Rs. 200
∴ The payment he received in the year 1996 is Rs. 5.200
a = 5000, d = 5200 – 5000 = 200,
an = 7000, n=?
a + (n – 1)d = an
5000 + (n – 1) 200 = 7000
5000 + 200n – 200 = 7000
200n + 4800 = 7000
200n = 7000 – 4800
200n = 2200
$$n=\frac{2200}{200}$$
∴ n = 11
Value of ‘n’ is 11
∴ From 1995 to 10 years means 2005. his salary becomes Rs. 7,000.
Question 20.
Ramkall saved Rs. 5 In the first week of a year and then Increased her weekly savings by Rs. 1.75. If in the nüI week. her weekly savings become Rs. 20.75. find ‘n’.
Solution:
Savings in the First week = Rs. 5.
Savings in the Second week 5 + 1.75 = Rs. 6.75
Savings in ‘nth week is Rs. 20.75
∴ A.P. 5, 6, 75 , …………. , 20.75
a = 5, d = 6.75 – 5 = 1.75, an = 20.75.
n =?
a + (n – 1) d = an
5 + (n – 1) (1.75) = 20.75
5 + 1.75n – 1.75 = 20.75
1 .75n + 3.25 = 20.75
1.75n = 20.75 – 3.25
1.75n = 17.5
$$n=\frac{17.5}{1.75}$$
$$n=\frac{1750}{175}$$
∴ n = 10
∴ In the 10th week, her savings becomes Rs. 20.75.
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Statistics - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Quantities : Statistics/InverseSurvivalFunction
Statistics
InverseSurvivalFunction
compute the inverse survival function
Calling Sequence InverseSurvivalFunction(X, t, options)
Parameters
X - algebraic; random variable or distribution t - algebraic; point options - (optional) equation of the form numeric=value; specifies options for computing the inverse survival function of a random variable
Description
• The InverseSurvivalFunction function computes the inverse survival function of the specified random variable at the specified point.
• The first parameter can be a distribution (see Statistics[Distribution]), a random variable, or an algebraic expression involving random variables (see Statistics[RandomVariable]).
Computation
• By default, all computations involving random variables are performed symbolically (see option numeric below).
Options
The options argument can contain one or more of the options shown below. More information for some options is available in the Statistics[RandomVariables] help page.
• numeric=truefalse -- By default, the inverse survival function is computed using exact arithmetic. To compute the inverse survival function numerically, specify the numeric or numeric = true option.
Examples
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
Compute the inverse survival function of the beta distribution with parameters p and q.
> $\mathrm{InverseSurvivalFunction}\left(\mathrm{Cauchy}\left(p,q\right),t\right)$
${p}{+}{q}{}{\mathrm{tan}}{}\left({\mathrm{π}}{}\left(\frac{{1}}{{2}}{-}{t}\right)\right)$ (1)
Use numeric parameters.
> $\mathrm{InverseSurvivalFunction}\left(\mathrm{Cauchy}\left(3,5\right),\frac{1}{2}\right)$
${3}$ (2)
> $\mathrm{InverseSurvivalFunction}\left(\mathrm{Cauchy}\left(3,5\right),\frac{1}{2},\mathrm{numeric}\right)$
${3.}$ (3)
> $\mathrm{Quantile}\left(\mathrm{Cauchy}\left(3,5\right),\frac{1}{2}\right)$
${3}$ (4)
Define new distribution.
> $T≔\mathrm{Distribution}\left(\mathrm{PDF}=\left(t→\frac{1}{\mathrm{π}\left({t}^{2}+1\right)}\right)\right):$
> $X≔\mathrm{RandomVariable}\left(T\right):$
> $\mathrm{CDF}\left(X,t\right)$
$\frac{{1}}{{2}}{}\frac{{\mathrm{π}}{+}{2}{}{\mathrm{arctan}}{}\left({t}\right)}{{\mathrm{π}}}$ (5)
> $\mathrm{InverseSurvivalFunction}\left(X,t\right)$
${-}{\mathrm{cot}}{}\left(\left({1}{-}{t}\right){}{\mathrm{π}}\right)$ (6)
Another distribution
> $U≔\mathrm{Distribution}\left(\mathrm{CDF}=\left(t→F\left(t\right)\right),\mathrm{PDF}=\left(t→f\left(t\right)\right)\right):$
> $Y≔\mathrm{RandomVariable}\left(U\right):$
> $\mathrm{CDF}\left(Y,t\right)$
${F}{}\left({t}\right)$ (7)
> $\mathrm{InverseSurvivalFunction}\left(Y,t\right)$
${\mathrm{RootOf}}{}\left({F}{}\left({\mathrm{_Z}}\right){-}{1}{+}{t}\right)$ (8)
References
Stuart, Alan, and Ord, Keith. Kendall's Advanced Theory of Statistics. 6th ed. London: Edward Arnold, 1998. Vol. 1: Distribution Theory.
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## 每週問題 October 27, 2014
$A$$B$ 是同尺寸矩陣，證明 $\text{rank}(A+B)\le\text{rank}A+\text{rank}B$
Let $A$ and $B$ be $m\times n$ matrices. Show that $\text{rank}(A+B)\le\text{rank}A+\text{rank}B$.
$\text{rank}(A+B)=\dim C(A+B)\le \dim (C(A)+C(B))$
$\{\mathbf{u}_1,\ldots,\mathbf{u}_r\}$$C(A)$ 的基底，$\{\mathbf{v}_1,\ldots,\mathbf{v}_s\}$$C(B)$ 的基底，其中 $r=\text{rank}A$$s=\text{rank}B$。據此，$C(A)+C(B)=\text{span}\{\mathbf{u}_1,\ldots,\mathbf{u}_r,\mathbf{v}_1,\ldots,\mathbf{v}_s\}$。明顯地，
$\dim (C(A)+C(B))\le r+s=\text{rank}A+\text{rank}B$
$\begin{bmatrix} I_m&I_m\\ 0&I_m \end{bmatrix}\begin{bmatrix} A&0\\ 0&B \end{bmatrix}\begin{bmatrix} I_n&0\\ I_n&I_n \end{bmatrix}=\begin{bmatrix} A+B&B\\ B&B \end{bmatrix}$
\begin{aligned} \hbox{rank}A+\hbox{rank}B&=\hbox{rank}\begin{bmatrix} A&0\\ 0&B \end{bmatrix}=\hbox{rank}\begin{bmatrix} A+B&B\\ B&B \end{bmatrix}\\ &\ge\hbox{rank}\begin{bmatrix} A+B\\ B \end{bmatrix} \ge\hbox{rank}(A+B). \end{aligned}
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# Hashing | Set 3 (Open Addressing)
• Difficulty Level : Easy
• Last Updated : 02 Dec, 2021
We strongly recommend referring below post as a prerequisite of this.
Hashing | Set 1 (Introduction)
Hashing | Set 2 (Separate Chaining)
Like separate chaining, open addressing is a method for handling collisions. In Open Addressing, all elements are stored in the hash table itself. So at any point, the size of the table must be greater than or equal to the total number of keys (Note that we can increase table size by copying old data if needed).
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Insert(k): Keep probing until an empty slot is found. Once an empty slot is found, insert k.
Search(k): Keep probing until slot’s key doesn’t become equal to k or an empty slot is reached.
Delete(k): Delete operation is interesting. If we simply delete a key, then the search may fail. So slots of deleted keys are marked specially as “deleted”.
The insert can insert an item in a deleted slot, but the search doesn’t stop at a deleted slot.
Open Addressing is done in the following ways:
a) Linear Probing: In linear probing, we linearly probe for next slot. For example, the typical gap between two probes is 1 as seen in the example below.
Let hash(x) be the slot index computed using a hash function and S be the table size
```If slot hash(x) % S is full, then we try (hash(x) + 1) % S
If (hash(x) + 1) % S is also full, then we try (hash(x) + 2) % S
If (hash(x) + 2) % S is also full, then we try (hash(x) + 3) % S
..................................................
..................................................```
Let us consider a simple hash function as “key mod 7” and a sequence of keys as 50, 700, 76, 85, 92, 73, 101.
Challenges in Linear Probing :
1. Primary Clustering: One of the problems with linear probing is Primary clustering, many consecutive elements form groups and it starts taking time to find a free slot or to search for an element.
2. Secondary Clustering: Secondary clustering is less severe, two records only have the same collision chain (Probe Sequence) if their initial position is the same.
b) Quadratic Probing We look for i2‘th slot in i’th iteration.
```let hash(x) be the slot index computed using hash function.
If slot hash(x) % S is full, then we try (hash(x) + 1*2) % S
If (hash(x) + 1*2) % S is also full, then we try (hash(x) + 2*2) % S
If (hash(x) + 2*2) % S is also full, then we try (hash(x) + 3*2) % S
..................................................
..................................................```
c) Double Hashing We use another hash function hash2(x) and look for i*hash2(x) slot in i’th rotation.
```let hash(x) be the slot index computed using hash function.
If slot hash(x) % S is full, then we try (hash(x) + 1*hash2(x)) % S
If (hash(x) + 1*hash2(x)) % S is also full, then we try (hash(x) + 2*hash2(x)) % S
If (hash(x) + 2*hash2(x)) % S is also full, then we try (hash(x) + 3*hash2(x)) % S
..................................................
..................................................```
See this for step by step diagrams.
Comparison of above three:
Linear probing has the best cache performance but suffers from clustering. One more advantage of Linear probing is easy to compute.
Quadratic probing lies between the two in terms of cache performance and clustering.
Double hashing has poor cache performance but no clustering. Double hashing requires more computation time as two hash functions need to be computed.
Like Chaining, the performance of hashing can be evaluated under the assumption that each key is equally likely to be hashed to any slot of the table (simple uniform hashing)
```m = Number of slots in the hash table
n = Number of keys to be inserted in the hash table
Load factor α = n/m ( < 1 )
Expected time to search/insert/delete < 1/(1 - α)
So Search, Insert and Delete take (1/(1 - α)) time```
?list=PLqM7alHXFySGwXaessYMemAnITqlZdZVE
References:
http://courses.csail.mit.edu/6.006/fall11/lectures/lecture10.pdf
https://www.cse.cuhk.edu.hk/irwin.king/_media/teaching/csc2100b/tu6.pdf
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Factors the 24 room the list of integers that have the right to be evenly divided into 24. Over there are all at once 8 components of 24 among which 24 is the biggest factor and also 1, 2, 3, 4, 6, 8, 12, and also 24 are optimistic factors. The amount of all determinants of 24 is 60 and its factors in Pairs room (1, 24), (2, 12), (3, 8), and (4, 6).
You are watching: Write 24 as a product of prime factors.
Factors that 24: 1, 2, 3, 4, 6, 8, 12 and also 24Negative factors of 24: -1, -2, -3, -4, -6, -8, -12 and also -24Prime factors of 24: 2, 3Prime administrate of 24: 2 × 2 × 2 × 3 = 23 × 3Sum of factors of 24: 60
1 What are determinants of 24? 2 Important Notes 3 How come Calculate factors of 24? 4 Factors of 24 by element Factorization 5 Factors of 24 in Pairs 6 FAQs on factors of 24 7 Tips and Tricks
Before we relocate ahead, let"s recall a tiny about factors. A variable is a number the divides the provided number without any remainder. Factors of 24 space pairs that those numbers whose product outcomes in 24.
To calculation the components of any number, below in this case 24, we need to find all the numbers that would certainly divide 24 there is no leaving any kind of remainder. We begin with the number 1, then check for numbers 2, 3, 4, 5, 6, 7, etc. As much as 24 respectively. The number 1 and also the number itself would always be a element of the offered number.
Let us examine the division of 24 by its factors. We express 24 together a product that its prime factors in the element factorization technique and we division 24 with its divisors in the department method. Let us see which numbers division 24 exactly without a remainder in which situation the divisors, and also the quotients, space the factors of 24.
24/1 = 24; therefore, 1 and also 24 are determinants of 2424/2 = 12; therefore, 2 is a aspect of 2424/3 = 8; therefore, 3 is a factor of 2424/4 = 6; therefore, 4 is a factor of 24
Hence, the factors of 24 room 1, 2, 3, 4, 6, 8, 12, and 24. To know the principle of finding factors by element factorization better, let us take a couple of more examples.
Prime factorization way expressing a composite number as the product that its prime factors. Factors of 24 by prime factorization are given by using the complying with steps:
Step 1: create the pair that factors, which on multiplication, gives the required number.
24 can be factored together a product the 4 and 6.
Step 2: check each the the factors to watch whether each among them is prime or not.
4 is not a prime number and can it is in dissociated together a product that 2 by chin or as a square of 2.6 is not a prime number and also can it is in dissociated together a product that 2 and 3.
Step 3: together per the criteria,
24 deserve to be composed as 24 = 4 × 6 = 2 × 2 × 2 × 3It can also be created as 24 = 23 × 3
Pair factors are pairs the those number that, once multiplied, offer the product as the required number. Here, the required number is 24. Let"s shot visualizing it utilizing blocks.
Factors the 24 in pairs can be composed as:
Factors Pair factors 1 × 24 = 24 1, 24 2 × 12 = 24 2, 12 3 × 8 = 24 3, 8 4 × 6 = 24 4, 6 6 × 4 = 24 6, 4 8 × 3 = 24 8, 3 12 × 2 = 24 12, 2 24 × 1 = 24 24, 1
These components given above are hopeful pair factors. The is feasible to have an unfavorable pair factors as well because the product that two negative numbers additionally gives a optimistic number.
Let"s have actually a look at at negative pair factors.
Factors Pair factors -1 × -24 = 24 -1, - 24 -2 × -12 = 24 -2, - 12 -3 × -8 = 24 -3, -8 -4 × -6 = 24 - 4, - 6 -6 × -4 = 24 -6, -4 -8 × -3 = 24 -8, - 3 -12 × -2 = 24 - 12, -2 -24 × -1 = 24 -24, -1
1 is the smallest aspect of every number.Every number has a minimum of 2 factors, i.e. 1 and the number itself.All also numbers constantly have 2 as one of their factors.All the number which finish in 5, will constantly have 5 as one of their factors.All the number which finish in 0, will always have 1, 2, 5, and 10 as their factors.
Example 1: Henry needs to determine i beg your pardon pair element of 24 on addition gives 14, subtraction provides 10, and division gives among the factors of 24 i beg your pardon belongs to one more pair. I m sorry pair factor do you think it is?
Solution:
Henry writes down the details provided to him:
Addition that pair factors = 14Subtraction the pair components = 10
Division of pair determinants = One factor belonging to an additional pair. The pair variable is (12, 2). This pair variable on enhancement gives 14, ~ above subtraction offers 10, and also on division gives 6, i m sorry is one more factor that 24 and is a part of one more pair factor.
Therefore, (12, 2) is the compelled pair factor.
Example 2: Jenny has actually an apple tree. She plucks 24 apples from the tree every day. She needs to distribute it among her 4 friends. How many apples will certainly each friend get?
Solution:
Jenny needs to divide 24 apples amongst her 4 friends. This means each girlfriend gets 24/4, i beg your pardon is 6 apples.
Therefore, each girlfriend will acquire 6 apologize every day.
Example 3: How numerous factors space there because that 24?
Solution:
The components of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Therefore, 24 has actually 8 factors.
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## FAQs on determinants of 24
### What are the factors of 24?
The components of 24 room 1, 2, 3, 4, 6, 8, 12, 24 and also its an unfavorable factors room -1, -2, -3, -4, -6, -8, -12, -24.
### What room the Prime factors of 24?
The prime factors of 24 space 2, 3.
### What is the Greatest typical Factor the 24 and also 23?
The determinants of 24 room 1, 2, 3, 4, 6, 8, 12, 24 and the determinants of 23 room 1, 23. 24 and 23 have only one common factor i m sorry is 1. This implies that 24 and 23 space co-prime.Hence, the Greatest common Factor (GCF) the 24 and 23 is 1.
See more: Can You Use A Cannon In Brimhaven Dungeon, Greater Demons In Brimhaven Dungeon (2)
### What are the common Factors that 24 and also 21?
Since, the determinants of 24 room 1, 2, 3, 4, 6, 8, 12, 24 and also the components of 21 space 1, 3, 7, 21.Hence, <1, 3> room the common factors that 24 and 21.
### What is the sum of the factors of 24?
Sum of all factors of 24 = (23 + 1 - 1)/(2 - 1) × (31 + 1 - 1)/(3 - 1) = 60
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# Count the number of unordered triplets with elements in increasing order and product less than or equal to integer X
Given an array A[] and an integer X. Find the number of unordered triplets (i, j, k) such that A[i] < A[j] < A[k] and A[i] * A[j] * A[k] <= X.
Examples:
Input: A = [3, 2, 5, 7], X = 42
Output:
Explanation:
Triplets are :
• (1, 0, 2) => 2 < 3 < 5, 2 * 3 * 5 < = 42
• (1, 0, 3) => 2 < 3 < 7, 2 * 3 * 7 < = 42
Input: A = [3, 1, 2, 56, 21, 8], X = 49
Output: 5
Naive Approach:
The naive method to solve the above-mentioned problem is to iterate through all the triplets. For each triplet arrange them in ascending order (since we have to count unordered triplets, therefore rearranging them is allowed), and check the given condition. But this method takes O(N 3) time.
Below is the implementation of the above approach:
`// C++ implementation to Count the number of` `// unordered triplets such that the numbers are` `// in increasing order and the product of them is` `// less than or equal to integer X` `#include ` `using` `namespace` `std;` `// Function to count the number of triplets` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x)` `{` ` ``int` `answer = 0;` ` ``// Iterate through all the triplets` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `j = i + 1; j < n; j++) {` ` ``for` `(``int` `k = j + 1; k < n; k++) {` ` ``vector<``int``> temp;` ` ``temp.push_back(a[i]);` ` ``temp.push_back(a[j]);` ` ``temp.push_back(a[k]);` ` ``// Rearrange the numbers in ascending order` ` ``sort(temp.begin(), temp.end());` ` ``// Check if the necessary conditions satisfy` ` ``if` `(temp[0] < temp[1] && temp[1] < temp[2]` ` ``&& temp[0] * temp[1] * temp[2] <= x)` ` ``// Increment count` ` ``answer++;` ` ``}` ` ``}` ` ``}` ` ``// Return the answer` ` ``return` `answer;` `}` `// Driver code` `int` `main()` `{` ` ``int` `A[] = { 3, 2, 5, 7 };` ` ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` ` ``int` `X = 42;` ` ``cout << countTriplets(A, N, X);` ` ``return` `0;` `}`
`// Java implementation to count the number of` `// unordered triplets such that the numbers are` `// in increasing order and the product of them ` `// is less than or equal to integer X` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to count the number of triplets` `static` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x)` `{` ` ``int` `answer = ``0``;` ` ``// Iterate through all the triplets` ` ``for``(``int` `i = ``0``; i < n; i++) ` ` ``{` ` ``for``(``int` `j = i + ``1``; j < n; j++)` ` ``{` ` ``for``(``int` `k = j + ``1``; k < n; k++) ` ` ``{` ` ``Vector temp = ``new` `Vector<>();` ` ``temp.add(a[i]);` ` ``temp.add(a[j]);` ` ``temp.add(a[k]);` ` ` ` ``// Rearrange the numbers in ` ` ``// ascending order` ` ``Collections.sort(temp);` ` ` ` ``// Check if the necessary conditions ` ` ``// satisfy` ` ``if` `(temp.get(``0``) < temp.get(``1``) && ` ` ``temp.get(``1``) < temp.get(``2``) && ` ` ``temp.get(``0``) * temp.get(``1``) * ` ` ``temp.get(``2``) <= x)` ` ` ` ``// Increment count` ` ``answer++;` ` ``}` ` ``}` ` ``}` ` ` ` ``// Return the answer` ` ``return` `answer;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ``int` `A[] = { ``3``, ``2``, ``5``, ``7` `};` ` ``int` `N = A.length;` ` ``int` `X = ``42``;` ` ``System.out.println(countTriplets(A, N, X));` `}` `}` `// This code is contributed by offbeat`
`# Python3 implementation to count the number of` `# unordered triplets such that the numbers are` `# in increasing order and the product of them is` `# less than or equal to integer X` `# Function to count the number of triplets` `def` `countTriplets(a, n, x):` ` ` ` ``answer ``=` `0` ` ` ` ``# Iterate through all the triplets` ` ``for` `i ``in` `range``(n):` ` ``for` `j ``in` `range``(i ``+` `1``, n):` ` ``for` `k ``in` `range``(j ``+` `1``, n):` ` ``temp ``=` `[]` ` ``temp.append(a[i])` ` ``temp.append(a[j])` ` ``temp.append(a[k])` ` ` ` ``# Rearrange the numbers in ` ` ``# ascending order` ` ``temp.sort()` ` ` ` ``# Check if the necessary ` ` ``# conditions satisfy` ` ``if` `(temp[``0``] < temp[``1``] ``and` ` ``temp[``1``] < temp[``2``] ``and` ` ``temp[``0``] ``*` `temp[``1``] ``*` `temp[``2``] <``=` `x):` ` ` ` ``# Increment count ` ` ``answer ``+``=` `1` ` ` ` ``# Return the answer ` ` ``return` `answer` ` ` `# Driver code` `A ``=` `[ ``3``, ``2``, ``5``, ``7` `]` `N ``=` `len``(A)` `X ``=` `42` `print``(countTriplets(A, N, X))` `# This code is contributed by shubhamsingh10`
`// C# implementation to count the number of ` `// unordered triplets such that the numbers are ` `// in increasing order and the product of them ` `// is less than or equal to integer X ` `using` `System;` `class` `GFG{ ` ` ` `// Function to count the number of triplets ` `static` `int` `countTriplets(``int` `[]a, ``int` `n, ``int` `x) ` `{ ` ` ``int` `answer = 0; ` ` ``// Iterate through all the triplets ` ` ``for``(``int` `i = 0; i < n; i++) ` ` ``{ ` ` ``for``(``int` `j = i + 1; j < n; j++) ` ` ``{ ` ` ``for``(``int` `k = j + 1; k < n; k++) ` ` ``{ ` ` ``int` `[]temp = { a[i], a[j], a[k] }; ` ` ` ` ``// Rearrange the numbers in ` ` ``// ascending order ` ` ``Array.Sort(temp); ` ` ` ` ``// Check if the necessary conditions ` ` ``// satisfy ` ` ``if` `(temp[0] < temp[1] && ` ` ``temp[1] < temp[2] && ` ` ``temp[0] * temp[1] * ` ` ``temp[2] <= x) ` ` ` ` ``// Increment count ` ` ``answer++; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// Return the answer ` ` ``return` `answer; ` `} ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ``int` `[]A = { 3, 2, 5, 7 }; ` ` ``int` `N = A.Length; ` ` ``int` `X = 42; ` ` ``Console.WriteLine(countTriplets(A, N, X)); ` `} ` `} ` `// This code is contributed by Stream_Cipher `
Output:
```2
```
Efficient Approach:
To optimize the method given above we can use a sorted form of the array since it would not change the answer because the triplets are unordered. Traverse through all the pairs of elements in the sorted array. For a pair (p, q) the problem now reduces to finding the number of elements r in the sorted array such that r <= X/(p*q). To perform this efficiently we will use Binary Search method and find the position of the largest element in the array which is < = X/(p*q). All the elements between the index of q until position will be added to the answer.
Below is the implementation of the above approach:
`// C++ implementation to Count the number of` `// unordered triplets such that the numbers are` `// in increasing order and the product of them is` `// less than or equal to integer X` `#include ` `using` `namespace` `std;` `// Function to count the triplets` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x)` `{` ` ``int` `answer = 0;` ` ``// Sort the array` ` ``sort(a, a + n);` ` ``// Iterate through all the triplets` ` ``for` `(``int` `i = 0; i < n; i++) {` ` ``for` `(``int` `j = i + 1; j < n; j++) {` ` ``// Apply Binary Search method` ` ``long` `long` `limit = (``long` `long``)x / a[i];` ` ``limit = limit / a[j];` ` ``int` `pos = upper_bound(a, a + n, limit) - a;` ` ``// Check if the position is greater than j` ` ``if` `(pos > j)` ` ``answer = answer + (pos - j - 1);` ` ``}` ` ``}` ` ``// Return the answer` ` ``return` `answer;` `}` `// Driver code` `int` `main()` `{` ` ``int` `A[] = { 3, 2, 5, 7 };` ` ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);` ` ``int` `X = 42;` ` ``cout << countTriplets(A, N, X);` ` ``return` `0;` `}`
`// Java implementation to count the number` `// of unordered triplets such that the ` `// numbers are in increasing order and ` `// the product of them is less than or ` `// equal to integer X` `import` `java.io.*;` `import` `java.util.Arrays; ` `class` `GFG{` ` ` `// Function to count the triplets` `static` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x)` `{` ` ``int` `answer = ``0``;` ` ``// Sort the array` ` ``Arrays.sort(a);` ` ``// Iterate through all the triplets` ` ``for``(``int` `i = ``0``; i < n; i++) ` ` ``{` ` ``for``(``int` `j = i + ``1``; j < n; j++)` ` ``{` ` ` ` ``// Apply Binary Search method` ` ``int` `limit = x / a[i];` ` ` ` ``limit = limit / a[j];` ` ``int` `pos = Arrays.binarySearch(a, limit) + ``1``;` ` ``// Check if the position is greater than j` ` ``if` `(pos > j)` ` ``answer = answer + (pos - j - ``1``);` ` ``}` ` ``}` ` ``// Return the answer` ` ``return` `answer;` `}` `// Driver Code` `public` `static` `void` `main (String[] args) ` `{` ` ``int` `A[] = { ``3``, ``2``, ``5``, ``7` `};` ` ``int` `N = A.length;` ` ``int` `X = ``42``;` ` ` ` ``System.out.print(countTriplets(A, N, X));` `}` `}` `// This code is contributed by math_lover`
`# Python3 implementation to Count the number of` `# unordered triplets such that the numbers are` `# in increasing order and the product of them is` `# less than or equal to integer X` `import` `bisect ` `# Function to count the triplets` `def` `countTriplets(a, n, x):` ` ` ` ``answer ``=` `0` ` ` ` ``# Sort the array` ` ``a.sort()` ` ` ` ``# Iterate through all the triplets` ` ``for` `i ``in` `range``(n):` ` ``for` `j ``in` `range``(i ``+` `1``, n):` ` ` ` ``# Apply Binary Search method` ` ``limit ``=` `x ``/` `a[i]` ` ` ` ``limit ``=` `limit ``/` `a[j]` ` ` ` ``pos ``=` `bisect.bisect_right(a, limit)` ` ` ` ``# Check if the position is greater than j` ` ``if` `(pos > j):` ` ``answer ``=` `answer ``+` `(pos ``-` `j ``-` `1``)` ` ` ` ``# Return the answer` ` ``return` `answer` `# Driver code` `A ``=` `[``3``, ``2``, ``5``, ``7``]` `N ``=` `len``(A)` `X ``=` `42` `print``(countTriplets(A, N, X))` `# This code is contributed by shubhamsingh10`
`// C# implementation to Count the number` `// of unordered triplets such that the ` `// numbers are in increasing order and ` `// the product of them is less than or` `// equal to integer X` `using` `System; ` `class` `GFG{` ` ` `// Function to count the triplets` `static` `int` `countTriplets(``int` `[]a, ``int` `n, ``int` `x)` `{` ` ``int` `answer = 0;` ` ``// Sort the array` ` ``Array.Sort(a);` ` ``// Iterate through all the triplets` ` ``for``(``int` `i = 0; i < n; i++)` ` ``{` ` ``for``(``int` `j = i + 1; j < n; j++) ` ` ``{` ` ` ` ``// Apply Binary Search method` ` ``int` `limit = x / a[i];` ` ``limit = limit / a[j];` ` ``int` `pos = Array.BinarySearch(a, limit) + 1;` ` ``// Check if the position is greater than j` ` ``if` `(pos > j)` ` ``answer = answer + (pos - j - 1);` ` ``}` ` ``}` ` ``// Return the answer` ` ``return` `answer;` `}` `// Driver Code` `public` `static` `void` `Main (String[] args) ` `{` ` ``int` `[]A = { 3, 2, 5, 7 };` ` ``int` `N = A.Length;` ` ``int` `X = 42;` ` ` ` ``Console.Write(countTriplets(A, N, X));` `}` `}` `// This code is contributed by math_lover`
Output:
```2
```
Time Complexity: O(N2 * log(N))
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# If the 2nd, 3rd and 4th Terms in the Expansion of (X + A)N Are 240, 720 and 1080 Respectively, Find X, A, N. - Mathematics
If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find xan.
#### Solution
$\text{ In the expansion of } \left( x + a \right)^n , \text{ the 2nd, 3rd and 4th terms are } ^{n}{}{C}_1 x^{n - 1} a^1 , ^{n}{}{C}_2 x^{n - 2} a^2 \text{ and } ^{n}{}{C}_3 x^{n - 3} a^3 , \ \text{ respectively } .$
$\text{ According to the question } ,$
$^{n}{}{C}_1 x^{n - 1} a^1 = 240$
$^{n}{}{C}_2 x^{n - 2} a^2 = 720$
$^{n}{}{C}_3 x^{n - 3} a^3 = 1080$
$\Rightarrow \frac{^{n}{}{C}_2 x^{n - 2} a^2}{^{n}{}{C}_1 x^{n - 1} a^1} = \frac{720}{240}$
$\Rightarrow \frac{n - 1}{2x}a = 3$
$\Rightarrow \frac{a}{x} = \frac{6}{n - 1} . . . \left( 1 \right)$
$\text{ Also } ,$
$\frac{^{n}{}{C}_3 x^{n - 3} a^3}{^{n}{}{C}_2 x^{n - 2} a^2} = \frac{1080}{720}$
$\Rightarrow \frac{n - 2}{3x}a = \frac{3}{2}$
$\Rightarrow \frac{a}{x} = \frac{9}{2n - 4} . . . \left( 2 \right)$
$\text{ Using } \left( 1 \right) \text{ and } \left( 2 \right) \text{ we get }$
$\frac{6}{n - 1} = \frac{9}{2n - 4}$
$\Rightarrow n = 5$
$\text{ Putting in eqn } \left( 1 \right) \text{ we get }$
$\Rightarrow 2a = 3x$
$\text{ Now } , ^{5}{}{C}_1 x^{5 - 1} \left( \frac{3}{2}x \right) = 240$
$\Rightarrow 15 x^5 = 480$
$\Rightarrow x^5 = 32$
$\Rightarrow x = 2$
$\text{ By putting the value of x and n in} \left( 1 \right) \text{ we get}$
$a = 3$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 33 | Page 40
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# Geometry
A post tree outside of matts house casts a 12 foot shadow at a certain time of day. At the same time matt, who is 6 feet tall, casts a 2 foot shadow. How tall is the tree?
How do I solve this?
The marketing department of the company is considering making a keychain with a miniature replica of their top selling dishwasher detergent. The dimensions of the dishwasher detergent box are 9 inches by 7.5 inches by 2.25 inches. If the replica will be 1/5 the size of the regular box, what will be the volume of the miniature box?
How do I solve this?
1. Proportion
6 /2ft=H/12
solve for H.
Volume box=8*7.5*2.25 * (1/5^3) inches cubed.
posted by bobpursley
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Solve any problem (step by step solutions) Input table (Matrix, Statistics)
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Solution
Problem: cholesky decomposition [[8,-6,2],[-6,7,-4],[2,-4,3]] [ Calculator, Method and examples ]
Solution:
Your problem -> cholesky decomposition [[8,-6,2],[-6,7,-4],[2,-4,3]]
Cholesky decomposition : A=L*L^T, Every symmetric positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose.
Here matrix is not symmetric positive definite, so Cholesky decomposition is not possible.
A matrix is positive definite if it’s symmetric and all its pivots are positive.
A =
8 -6 2 -6 7 -4 2 -4 3
Test method 1: Existence of all Positive Pivots.
First apply Gaussian Elimination method to find Pivots
A =
8 -6 2 -6 7 -4 2 -4 3
R_2 larr R_2+3/4xx R_1
=
8 -6 2 0 5/2 -5/2 2 -4 3
R_3 larr R_3-1/4xx R_1
=
8 -6 2 0 5/2 -5/2 0 -5/2 5/2
R_3 larr R_3+ R_2
=
8 -6 2 0 5/2 -5/2 0 0 0
Pivots are the first non-zero element in each row of this eliminated matrix.
:. Pivots are 8,5/2
Here all pivots are positive, so matrix is positive definite.
Test method 2: Determinants of all upper-left sub-matrices are positive.
A =
8 -6 2 -6 7 -4 2 -4 3
8
=8
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# Documentation
### This is machine translation
Translated by
Mouse over text to see original. Click the button below to return to the English verison of the page.
# rewrite
Rewrite expression in terms of another function
## Syntax
• ``rewrite(expr,target)``
example
## Description
example
````rewrite(expr,target)` rewrites the symbolic expression `expr` in terms of the target function `target`. The rewritten expression is mathematically equivalent to the original expression. If `expr` is a vector or matrix, `rewrite` acts element-wise on `expr`.```
## Examples
### Rewrite Between Trigonometric and Exponential Functions
Rewrite any trigonometric function in terms of the exponential function by specifying the target `'exp'`.
```syms x sin2exp = rewrite(sin(x), 'exp') tan2exp = rewrite(tan(x), 'exp')```
```sin2exp = (exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2 tan2exp = -(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1)```
Rewrite the exponential function in terms of any trigonometric function by specifying the trigonometric function as the target. For a full list of targets, see `target`.
```syms x exp2sin = rewrite(exp(x), 'sin') exp2tan = rewrite(-(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1), 'tan')```
```exp2sin = 1 - 2*sin((x*1i)/2)^2 - sin(x*1i)*1i exp2tan = -(((tan(x) - 1i)*1i)/(tan(x) + 1i) + 1i)/((tan(x) - 1i)/(tan(x) + 1i) - 1)```
Simplify `exp2tan` into the expected form by using `simplify`.
`exp2tan = simplify(exp2tan)`
```exp2tan = tan(x)```
### Rewrite Between Trigonometric Functions
Rewrite any trigonometric function in terms of any other trigonometric function by specifying the target. For a full list of targets, see `target`.
Rewrite `tan(x)` in terms of the sine function by specifying the target `'sin'`.
```syms x tan2sin = rewrite(tan(x), 'sin')```
```tan2sin = -sin(x)/(2*sin(x/2)^2 - 1)```
### Rewrite Between Hyperbolic Functions and Trigonometric Functions
Rewrite any hyperbolic function in terms of any trigonometric function by specifying the trigonometric function as the target. For a full list of targets, see `target`.
Rewrite `tanh(x)` in terms of the sine function by specifying the target `'sin'`.
```syms x tanh2sin = rewrite(tanh(x), 'sin')```
```tanh2sin = (sin(x*1i)*1i)/(2*sin((x*1i)/2)^2 - 1)```
Similarly, rewrite trigonometric functions in terms of hyperbolic functions by specifying the hyperbolic function as the target.
### Rewrite Between Inverse Trigonometric Functions and Logarithm Function
Rewrite any inverse trigonometric function in terms of the logarithm function by specifying the target `'log'`. For a full list of targets, see `target`.
Rewrite `acos(x)` and `acot(x)` in terms of the `log` function.
```syms x acos2log = rewrite(acos(x), 'log') acot2log = rewrite(acot(x), 'log')```
```acos2log = -log(x + (1 - x^2)^(1/2)*1i)*1i acot2log = (log(1 - 1i/x)*1i)/2 - (log(1i/x + 1)*1i)/2```
Similarly, rewrite the logarithm function in terms of an inverse trigonometric function by specifying the inverse trigonometric function as the target.
### Rewrite Elements of Matrix
Rewrite each element of a matrix by calling `rewrite` on the matrix.
Rewrite all elements of a matrix in terms of the `exp` function.
```syms x matrix = [sin(x) cos(x); sinh(x) cosh(x)]; rewrite(matrix, 'exp')```
```ans = [ (exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2, exp(-x*1i)/2 + exp(x*1i)/2] [ exp(x)/2 - exp(-x)/2, exp(-x)/2 + exp(x)/2]```
### Rewrite Between Sine and Cosine Functions
Rewrite the cosine function in terms of the sine function. Here, `rewrite` replaces the cosine function using the identity `cos(2*x) = 1 – 2*sin(x)^2` which is valid for any `x`.
```syms x rewrite(cos(x),'sin')```
```ans = 1 - 2*sin(x/2)^2```
`rewrite` does not replace `sin(x)` with either $-\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}$ or $\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}$ because these expressions are not valid for all `x`. However, using the square of these expressions to replace `sin(x)^2` is valid for all `x`. Thus, `rewrite` replaces `sin(x)^2`.
```syms x rewrite(sin(x),'cos') rewrite(sin(x)^2,'cos')```
```ans = sin(x) ans = 1 - cos(x)^2```
## Input Arguments
collapse all
Input to rewrite, specified as a symbolic number, variable, expression, function, vector, matrix, or multidimensional array.
Target function, specified as a character vector. This table summarizes the rewriting rules for all allowed targets.
TargetRewrites these FunctionsIn Terms of these Functions
`'exp'`All trigonometric and hyperbolic functions including inverse functions`exp`, `log`
`'log'`All inverse trigonometric and hyperbolic functions`log`
`'sincos'``tan`, `cot`, `exp`, `sinh`, `cosh`, `tanh`, `coth``sin`, `cos`
`'sin'`, `'cos'`, `'tan'`, or `'cot'``sin`, `cos`, `exp`, `tan`, `cot`, `sinh`, `cosh`, `tanh`, `coth` except the targetTarget trigonometric function
`'sinhcosh'``tan`, `cot`, `exp`, `sin`, `cos`, `tanh`, `coth``sinh`, `cosh`
`'sinh'`, `'cosh'`, `'tanh'`, `'coth'``tan`, `cot`, `exp`, `sin`, `cos`, `sinh`, `cosh`, `tanh`, `coth` except the targetTarget hyperbolic function
`'asin'`, `'acos'`, `'atan'`, `'acot'``log`, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse trigonometric function
`'asinh'`, `'acosh'`, `'atanh'`, `'acoth'``log`, and all inverse trigonometric and inverse hyperbolic functionsTarget inverse hyperbolic function
`'sqrt'``abs(x + 1i*y)``sqrt(x^2 + y^2)`
`'heaviside'``sign`, `triangularPulse`, `rectangularPulse``heaviside`
`'piecewise'``abs`, `heaviside`, `sign`, `triangularPulse`, `rectangularPulse``piecewise`
collapse all
### Tips
• `rewrite` replaces symbolic function calls in `expr` with the target function only if the replacement is mathematically valid. Otherwise, it keeps the original function calls.
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# Coal mine
The towing wheel has a diameter of 1.7 meters. How many meters does the elevator cage lower when the wheel turns 32 times?
Result
l = 171 m
#### Solution:
$D = 1.7 \ m \ \\ o = \pi \cdot \ D = 3.1416 \cdot \ 1.7 \doteq 5.3407 \ m \ \\ n = 32 \ \\ \ \\ l = n \cdot \ o = 32 \cdot \ 5.3407 \doteq 170.9026 = 171 \ \text { m }$
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# Number of electrons in 10 coulomb charge is equal to:
33 views
in Physics
closed
Number of electrons in 10 coulomb charge is equal to:
1. 625 × 1016
2. 625 × 1017
3. 625 × 1015
4. 6.25 × 1017
by (117k points)
selected
Correct Answer - Option 2 : 625 × 1017
CONCEPT:
Electric Charge
• It is defined as the intrinsic property of certain fundamental particles like electrons, proton, etc, due to which they produce electric and magnetic effects.
• Types of electric charge
1. Positive charge: A body having a deficiency of electrons.
2. Negative charge: A body having an excess of electrons.
• Properties of electric charge
• Like charges repel and unlike charges attract each other.
• A charge is a scalar quantity.
• A charge is always quantized.
• A charge will always remain conserved.
• A charge is always associated with mass.
• The unit of charge is the coulomb.
⇒ 1 coulomb= charge on 625×1016 electrons
CALCULATION:
We know that the number of electrons in 1-coulomb charge is,
⇒ N = 625×1016 electrons
So the number of electrons in 10-coulomb charge is,
⇒ N' = 10 × 625 × 1016 electrons
⇒ N' = 625 × 1017 electrons
• Hence, option 2 is correct.
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# Leetcode Arithmetic Slices II - Subsequence problem solution
In this Leetcode Arithmetic Slices II - Subsequence problem solution you have given an integer array nums, return the number of all the arithmetic subsequences of nums.
A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
1. For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.
2. For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.
A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.
1. For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].
The test cases are generated so that the answer fits in a 32-bit integer.
## Problem solution in Python.
```class Solution:
def numberOfArithmeticSlices(self, A: List[int]) -> int:
dic={}
for i in range(len(A)):
dic[i]={}
ans=0
for i in range(len(A)):
for j in range(i):
d=A[i]-A[j]
if d not in dic[j]:
if d in dic[i]:
dic[i][d]+=1
else:
dic[i][d]=1
else:
if d not in dic[i]:
dic[i][d]=dic[j][d]+1
ans+=dic[j][d]
else:
dic[i][d]+=dic[j][d]+1
ans+=dic[j][d]
return ans
```
## Problem solution in Java.
```public int numberOfArithmeticSlices(int[] A) {
int l = A.length;
if(l <3){
return 0;
int[] left = new int[l];
int[] right = new int[l];
left[0] = 0;
right[l-1] = 0;
for(int i=1;i<l;i++){
left[i] = A[i] - A[i-1];
}
for(int i=l-2;i>=0;i--){
right[i] = A[i+1] - A[i];
}
int prev = 0;
int[] result = new int[l];
for(int i=1;i<l-1;i++){
if(right[i] == left[i]){
result[i] = 1;
if(right[i-1] == left[i-1]){
result[i] = result[i]+result[i-1];
}
}
}
int sum=0;
for(int i=0;i<l;i++){
sum+=result[i];
}
return sum;
}
```
## Problem solution in C++.
```class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
int ans = 0, n = size(nums);
vector<unordered_map<int, int>> freq(n);
for (size_t i = 0; i < n; ++i) {
for (size_t ii = 0; ii < i; ++ii) {
long temp = (long) nums[i] - nums[ii];
if (temp > INT_MAX || temp < INT_MIN) continue;
int diff = nums[i] - nums[ii];
ans += freq[ii][diff];
freq[i][diff] += 1 + freq[ii][diff];
}
}
return ans;
}
};
```
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# Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number. f(x)=frac{1}{sqrt{x}} with a=4, approximate frac{1}{sqrt{3}}
Question
Series
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}$$ with $$\displaystyle{a}={4}$$, approximate $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
2021-02-06
Consider the provided function,
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}$$ with $$\displaystyle{a}={4}$$, approximate $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
The Taylor series as a form of $$\displaystyle{\sum_{{{k}={0}}}^{\infty}}{c}_{{k}}{\left({x}-{a}\right)}^{{k}}$$, where $$\displaystyle{c}_{{k}}={\frac{{{{f}^{{k}}{\left({a}\right)}}}}{{{k}!}}},{k}={0},{1},{2},\ldots$$
We evaluate the derivative of the function at point $$\displaystyle{a}={4}$$
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}\Rightarrow{f{{\left({4}\right)}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle{f}'{\left({x}\right)}=-{\frac{{{1}}}{{{2}{x}^{{{\frac{{{3}}}{{{2}}}}}}}}}\Rightarrow{f}'{\left({4}\right)}=-{\frac{{{1}}}{{{16}}}}$$
$$\displaystyle{f}{''}{\left({x}\right)}={\frac{{{3}}}{{{4}{x}^{{{\frac{{{5}}}{{{2}}}}}}}}}\Rightarrow{f}{''}{\left({4}\right)}={\frac{{{3}}}{{{128}}}}$$
$$\displaystyle{f}{'''}{\left({x}\right)}=-{\frac{{{15}}}{{{8}{x}^{{{\frac{{{7}}}{{{2}}}}}}}}}\Rightarrow{f}{'''}{\left({4}\right)}=-{\frac{{{15}}}{{{1024}}}}$$
Hence, the first four term of the series is shown below.
$$\displaystyle{\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({x}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({x}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({x}-{4}\right)}^{{3}}$$
So, $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}={\frac{{{1}}}{{{2}}}}-{\frac{{{1}}}{{{16}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}+{\frac{{{3}}}{{{128}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{2}}-{\frac{{{15}}}{{{1024}}}}{\left({\frac{{{1}}}{{\sqrt{{{3}}}}}}-{4}\right)}^{{3}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}-{\frac{{{1}-{4}\sqrt{{{3}}}}}{{{16}\sqrt{{{3}}}}}}+{\frac{{{49}\sqrt{{{3}}}-{24}}}{{{128}\sqrt{{{3}}}}}}-{\frac{{{5}{\left({145}-{204}\sqrt{{{3}}}\right)}}}{{{1024}\sqrt{{{3}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}+{\frac{{{1668}-{327}\sqrt{{{3}}}}}{{{1024}}}}$$
Hence.
### Relevant Questions
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
$$f(x)=\frac{1}{\sqrt{x}}$$ with $$a=4$$, approximate $$\frac{1}{\sqrt{3}}$$
Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
b. Determine the radius of convergence of the series.
$$f(x)=\cos2x+2\sin x$$
Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
$$f(x)=x^2\cos x^2$$
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\cosh(2x-2),a=1$$
Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\tan^{-1}4x,a=0$$
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{a}={1}$$
Binomial series
a. Find the first four nonzero terms of the binomial series centered at 0 for the given function.
b. Use the first four terms of the series to approximate the given quantity.
$$f(x)=(1+x)^{\frac{2}{3}}$$, approximate $$(1.02)^{\frac{2}{3}}$$
Taylor series and interval of convergence
a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
$$f(x)=\log_3(x+1),a=0$$
$$f(x)=x\ln x-x+1,a=1$$
$$f(x)=2^x,a=1$$
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# Zumdahl • Zumdahl • DeCoste
## Presentation on theme: "Zumdahl • Zumdahl • DeCoste"— Presentation transcript:
Zumdahl • Zumdahl • DeCoste
World of CHEMISTRY
Chapter 6 Chemical Composition
Goals of Chapter 6 Average mass/counting by weighing
Experimental determination of atomic mass Moles and Avogadro's number Molar Mass Convert between moles and mass Mass percent of elements in compounds Empirical formulas Calculating molecular formulas
One of most important chemical activities: Synthesis of new substances
Nylon, aspartame, Kevlar (bulletproof vests), PVC, Teflon All originated in chemist's laboratory Once they make it – they must determine what it is What is it's composition? What is it's chemical formula?
Question: How can you determine exactly how many M&M's are in your cup without counting every one? Answer: Counting by weighing
Answer: No, each is slightly different
Question: Does every M&M weigh exactly the same as all of the other M&M's? Answer: No, each is slightly different
How can you determine a mass that is typical of each M&M?
Weigh several M&M's and determine the average mass
How to determine number of M&M's:
Determine average mass of an M&M by weighing several Average mass = sum of the masses divided by the number of M&M's weighed Weigh entire cup of M&M's and divide that weight by the average mass to get the number of M&M's
Mathematical Equations:
Av Mass = m1 + m2 + m3 + m4 +m5 5 # M&M's = __total mass (g)__ av mass (g/M&M)
Atomic Masses: Counting Atoms by Weighing
Solid carbon reacts with gaseous oxygen to form gaseous carbon dioxide: C(s) + O2(g) → CO2(g) 1 atom C reacts with 1 molecule O2 to yield 1 molecule of CO2
If you have a pile of solid carbon and want to know how much Oxygen you need for the reaction to occur – what do you do? Need to count carbon atoms – but atoms are too small to count Use same principle applied to the M&M's and count by weighing
Atomic Mass Unit (amu) Gram & kilogram too large to use because atoms are so tiny Atomic mass unit is used – much smaller unit than a gram 1 amu = 1.66 x grams Atoms exist as isotopes Use average atomic mass (bottom # on periodic table)
To find oxygen needed to react with carbon:
Weigh the pile of carbon = 3.00 x 1020 amu From periodic table: 1 C atom weighs amu 3.00 x 1020 amu x C atom = 2.50 x 1019 C atoms 12.01 amu
What is another name for 12 donuts?
A dozen Other Examples: Ream of paper Gross of pencils
The Mole Definition: the number equal to the number of carbon atoms in grams of carbon Used precise counting techniques to determine this number to be 6.02 x 1023 atoms Avogadro's number = 6.02 x 1023 atoms One mole of something consists of 6.02 x 1023 units of that substance
1 dozen eggs = 12 eggs 1 mole eggs = 6.02 x 1023 eggs 1 mole water = 6.02 x 1023 H2O molecules 1 mole Ag = 6.02 x 1023 Ag atoms
What if you were offered \$1 million to count from 1 to 6 x 1023 at a rate of one number each second? What would be your hourly wage? Would you do it? Could you do it?
6 x seconds is about 2 x 1020 hours or 2 x 1016 years Your hourly wage would by \$ per hour If would take hundreds of millions of years to earn a penny In other words – it's impossible
A sample of an element with a mass equal to that element's average atomic mass expressed in grams contains 1 mole of atoms. 1 mole Al = 27 grams = 6.02 x 1023 atoms 1 mole Au = 197 grams = 6.02 x 1023 atoms 1 mole Fe = 55.6 grams = 6.02 x 1023 atoms 1 mole S = 32 grams = 6.02 x 1023 atoms
This is the molar mass of methane
Chemical compound is a collection of atoms Methane = CH4 = 1 C atom + 4 H atoms Mass 1 mol of C = 1 x g = g Mass 4 mol of H = 4 x g = _4.032 g Mass of 1 mol of CH4 = g This is the molar mass of methane
Figure 6.3: Various numbers of methane molecules.
Molar Mass Defined as the mass (in grams) of one mole of a substance Obtain molar mass by summing the masses of the component atoms For ionic compounds, sometimes referred to as formula weight Used to be called molecular weight Copyright © Houghton Mifflin Company
Example 6.5: Calculate the molar mass of sulfur dioxide – SO2 (exhaust gas that can produce acid rain) 1 mol SO2 molecules = 1 mol S + 2 mol O Mass of 1 mol S = 1 x g = g Mass of 2 mol O = 2 x g = g Mass of 1 mol SO2 = g The molar mass of sulfur dioxide is g Copyright © Houghton Mifflin Company
How many grams of sulfur dioxide do you have if you have 5 moles?
5 mol SO2 x grams = grams SO2 1mol SO2 Used molar mass from previous slide Copyright © Houghton Mifflin Company
If you have 800 grams of SO2, how many moles is this?
800 g SO2 x mol SO2 = 12.5 mol SO2 64.07 g How many molecules is this? 12.5 mol x x 1023 molecules = 7.53 x 1024 1 mole SO SO2 molecules Copyright © Houghton Mifflin Company
Percent composition of compounds
Mass fraction mass of element in 1 mol of cmpd for a given = element mass of 1 mole of compound To convert mass fraction to mass percent, multiply mass fraction by 100% Copyright © Houghton Mifflin Company
Mass percent: Ethanol (C2H5OH)
Mass of C = 2 mol x g/mol = g Mass of H = 6 mol x g/mol = g Mass of O = 1 mol x g/mol = g_ Mass: 1 mol C2H5OH = molar mass = g To find mass percent, divide total mass of each element by molar mass of ethanol (x 100%) Copyright © Houghton Mifflin Company
Mass % X = mass of X in 1 mol ethanol x100% mass of 1 mole of ethanol Mass % C = g x 100% = 52.14% g Mass % H = g x 100% = 13.13% g Mass % O = g x 100% = 34.73% g Mass percentages of all of the elements should add up to 100% (52.14% % % = %) Copyright © Houghton Mifflin Company
Formulas of Compounds Sometimes a chemical reaction gives a product that has never been formed before: Identify compound by determining which elements are present and how much of each is there Previously, we used formula to determine the mass of each element present in a mole – to obtain the formula of an unknown compound – do the opposite Use measured masses of elements present to determine the formula Copyright © Houghton Mifflin Company
Determining Formula Formula represents relative numbers of the various types of atoms present To determine formula – count the atoms How do we count atoms? - By weighing! Change mass into moles Change moles into atoms Use ratios to determine formula Copyright © Houghton Mifflin Company
If we have a compound that contains only carbon, hydrogen, and oxygen and we weigh out grams for analysis. The sample is analyzed and found to contain g C, g H, and g O. What is the compound? Step 1: Convert masses of elements to moles Step 2: Convert moles of elements to atoms Copyright © Houghton Mifflin Company
Calculations show the same number of C and O atoms Calculations show twice as many H atoms as C atoms and O atoms Can write formula as CH2O But formula could also be C2H4O2 Must be in 1:2:1 ratio Copyright © Houghton Mifflin Company
Empirical Formula Formula of a compound that expresses the smallest whole number ratio of the atoms present (simplest formula) In a class the girl to boy ratio is 1:8 This could mean the class has 1 girl & 8 boys OR This could mean the class has 2 girls & 16 boys OR This could mean the class has 3 girls & 24 boys etc. Copyright © Houghton Mifflin Company
Steps for Determining Empirical Formula of a Compound
Obtain mass of each element present (grams) Determine # moles present for each atom Divide # of moles by smallest # of moles (converts smallest # to 1). If all numbers are integers, these will be the subscripts, if not, go to step 4 Multiply numbers in step 3 by the smallest integer that will convert them to whole numbers. These whole numbers are the subscripts Copyright © Houghton Mifflin Company
Molecular Formula The actual formula of a compound Gives the composition of the molecules that are present Need molar mass to determine Our mystery compound could be glucose: C6H12O6 (C to H to O ratio is 1:2:1) Copyright © Houghton Mifflin Company
Determining Molecular Formula
Always a multiple of empirical formula (Empirical formula)n = molecular formula n = molar mass______ empirical formula mass for glucose n = 6: (CH2O)6 = C6H12O6 Copyright © Houghton Mifflin Company
Figure 6.5: Ball-and-stick model of P4O10.
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Ex 11.3
Chapter 11 Class 7 Exponents and Powers
Serial order wise
### Transcript
Ex 11.3, 4 Express the number appearing in the following statements in standard form. (h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm 60,230,000,000,000,000,000,000 = 6023 × 〖10〗^19 = (6.023 × 103) × 〖10〗^19 = 6.023 × 〖10〗^3 × 〖10〗^19 = 6.023 × 〖10〗^(3 + 19) = 6.023 × 〖10〗^22 Answer is 6.023 × 〖𝟏𝟎〗^𝟐𝟐 molecules (Using am × an = am + n)
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# Conical pendulum circular motion question
1. Aug 5, 2012
### Shanyn
1. The problem statement, all variables and given/known data
By resolving forces horizontally and vertically and using newton's second Law, find an expression for the angle swung out of a Chair-o-plane ride.
Im just not really sure how to resolve the forces vertically and horizontally.
2. Relevant equations
So far I know that forces acting on the chair are its own weight, mg and the tension in the chair, T.
Also that
T cos ѳ = mg
a= rω^2 ( ω being angular velocity).
Force causing the motion = F = mrω^2
= ma << newtons second law
and horizontally T sin ѳ = mrω^2
2. Aug 5, 2012
### andrien
that's ok.does not that give you
tanα=r(ω2)/g
3. Aug 5, 2012
### Shanyn
Yeah, is that vertical or horizontal.?
4. Aug 5, 2012
### andrien
with vertical.
5. Aug 5, 2012
### Shanyn
So is that expression for the angle swung?
6. Aug 5, 2012
### andrien
yes, but if the situation is much different from what you telling ,there can be changes.
7. Aug 5, 2012
### Shanyn
Cool thanks. :)
That's the situation, the full questions are in a chair-o-plane 1. will a child swing out at a greater angle than a much heavier adult? 2. Will people on the inside swing out at the same angle as those on the outside? 3. Will empty chairs be a problem? and 4. What will happen as the speed increases. So by resolving forces horizontally and vertically to find the expression for the angle swung out we need to answer those four question. :)
8. Aug 5, 2012
### andrien
you can predict it from the formula just written which does not depend on mass.
9. Aug 5, 2012
### Simon Bridge
I think it is easy to get mixed up if you don't keep track of what you've done and what you know. It can help to draw the pictures of what is happening - you want one overhead and another from the side.
You have already noticed: $T\sin(\theta)$ is the horizontal unbalanced (tension) force ... which must be the centripetal force. $T\cos(\theta)$ is the balanced vertical component ... so you have two equations with two unknowns. Number them.
1. $T\cos(\theta)=mg$
2. $T\sin(\theta)=mr\omega^2$
The trouble is that you have to deal with trig functions right?
But it solves itself - you can do it blindly: solve for T in your first equation ($T=mg/ \cos(\theta)$, then substitute into the second one ... realize that sin/cos=tan. OR just divide the two equations directly - the T cancels out.
Now you just need the inverse tangent.
Keep these things straight and the rest will follow [note: I took too long and you had other answers :) ]
10. Aug 5, 2012
### Shanyn
Thanks so much this has been the biggest help!
Also how could I show that a heavier person will swing out at a greater angle than a small child if the equation cuts out mass. Does that mean that the weight wont affect the angle swung.?
11. Aug 5, 2012
### Simon Bridge
That's what it means - when the equation for angle does not include some factor - like temperature or color or (ahem) weight, then we say that the angle does not depend on that factor.
12. Aug 5, 2012
### Shanyn
Ok thank you so much, I think I can finish it now. :)
13. Aug 5, 2012
### Simon Bridge
However - the size of the chair's occupant will affect that r in your relation: it gets smaller.
Can you see how?
14. Aug 5, 2012
### Shanyn
Umm not really, doesn't r= l sin theta. So how does the weight of the occupant affect r.? :)
15. Aug 5, 2012
### Simon Bridge
Didn't say "weight of the occupant", I said "size" ... there's a difference ;)
Forinstance: on a regular swing that just goes back and forth, the swinging is faster for an adult than it is for a child on the same swing. It's also faster for the child if she stands up.
I don't know if you are expected to take this effect into account ... but it should be worth bonus marks if not.
Last edited: Aug 5, 2012
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# Using the double angle of half angle formula, how do you simplify (1-cos 4 theta)/(sin4theta)?
$\sin \left(2 a\right) = 2. \sin \left(a\right) . \cos \left(a\right)$ and
$1 - \cos \left(2 a\right) = 2. {\sin}^{2} \left(a\right)$
$y = \frac{1 - \cos 4 x}{\sin} \left(4 x\right)$ =
$= \frac{2. {\sin}^{2} \left(2 a\right)}{2. \sin \left(2 a\right) . \cos \left(2 a\right)}$ = $\tan \left(2 a\right)$
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# If $\vec{\nabla} \cdot \vec{V} \neq 0$ at only one point, will this prevent us from saying that $\vec{V}=\vec{\nabla} \times \vec{U}$?
This question has an answer in the language of high level mathematics. Can somebody explain this in the language of vector calculus.
Part I: Let us consider Cartesian coordinate system with origin $$O$$ and axes $$x,y,z$$. Let:
$$r=\sqrt{x^2+y^2+z^2}$$
$$\text{and }\vec{V}=0\ (\hat{i}) +\dfrac{\partial}{\partial z} \left( \dfrac{1}{r} \right) (\hat{j}) -\dfrac{\partial}{\partial y} \left( \dfrac{1}{r} \right) (\hat{k})$$
It is obvious that $$\dfrac{1}{r}$$ is defined everywhere except at the origin.
Now let us take the divergence of $$\vec{V}$$:
$$\vec{\nabla} \cdot \vec{V}=0$$
Since $$\dfrac{1}{r}$$ is not defined at the origin, $$\vec{\nabla} \cdot \vec{V}=0$$ is true everywhere except at the origin.
Since $$\vec{\nabla} \cdot \vec{V} \neq 0$$ at a point $$P (0,0,0)$$, will this prevent us from concluding $$\vec{V}=\vec{\nabla} \times \vec{U}$$ at points other than $$P$$? Why? Why not?
Part II: If $$\vec{\nabla} \cdot \vec{V} \neq 0$$ at points on a one dimensional arbitrary curve in space, will this prevent us from concluding $$\vec{V}=\vec{\nabla} \times \vec{U}$$ at other points not on the curve? Why? Why not?
NOTE - For both Part I and Part II:
If (Why/Why not) is beyond the scope of vector (multivariable) calculus, just reply as yes/no. However please try your best to explain (Why/Why not) in the language of vector (multivariable) calculus.
SEMI ANSWER: Please point out the limitations
I have stumbled upon a derivation in the language of elementary vector calculus. Please point out if there are any limitations in my derivation. In the context of advanced mathematics (de Rham cohomology or Poincare lemma), it seems to me that there are limitations.
Derivation:
To prove: At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$), if $$\vec{\nabla} \cdot \vec{B}=0$$, then $$\vec{B}=\vec{\nabla} \times \vec{A}$$
Proof:
At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$): \begin{align} \vec{B} &= B_x\ (\hat{i}) + B_y\ (\hat{j}) + B_z\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) + \int^{(x,y,z)}_{(x,y,\infty)} \dfrac{\partial B_z}{\partial z}\ dz\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) - \int^{(x,y,z)}_{(x,y,\infty)} \left( \vec{\nabla} \cdot \vec{B} - \dfrac{\partial B_z}{\partial z}\ \right) dz\ (\hat{k})\\ &\text{{Since \vec{\nabla} \cdot \vec{B}=0}}\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j}) - \int^{(x,y,z)}_{(x,y,\infty)} \left( \dfrac{\partial B_x}{\partial x} + \dfrac{\partial B_y}{\partial y}\ \right) dz\ (\hat{k})\\ &= B_x\ (\hat{i}) + B_y\ (\hat{j})\ + \left[ \dfrac{\partial}{\partial x} \left(- \int^{(x,y,z)}_{(x,y,\infty)}B_x\ dz \right) -\dfrac{\partial}{\partial y} \left( \int^{(x,y,z)}_{(x,y,\infty)}B_y\ dz \right) \right] (\hat{k})\\ &\text{{By changing the order of integration and differentiation}}\\ \end{align}
At all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$), let's define:
$$\displaystyle \vec{A}=\int^{(x,y,z)}_{(x,y,\infty)}B_y\ dz\ (\hat{i}) - \int^{(x,y,z)}_{(x,y,\infty)}B_x\ dz\ (\hat{j}) + 0\ (\hat{k}) + \vec{\nabla}f$$
where $$f$$ is an arbitrary function of $$(x,y,z)$$
Therefore at all points where $$\vec{B}$$ is defined (whatever be the domain of $$\vec{B}$$):
\begin{align} \vec{B} &= \left( \dfrac{\partial A_z}{\partial y}-\dfrac{\partial A_y}{\partial z} \right) (\hat{i}) +\left( \dfrac{\partial A_x}{\partial z}-\dfrac{\partial A_z}{\partial x} \right) (\hat{j}) +\left( \dfrac{\partial A_y}{\partial x}-\dfrac{\partial A_x}{\partial y} \right) (\hat{k})\\ &= \vec{\nabla} \times \vec{A} \end{align}
• Does the statement say something about how it is defined $\vec U$? – manooooh Nov 27 '18 at 7:54
• $\vec{U} \rightarrow \vec{U} +\vec{\nabla}f$ is the vector whose curl gives $\vec{V}$......$f$ is an arbitrary function of $x,y,z$ – Joe Nov 27 '18 at 9:03
• The origin isn't in the domain of $\bf V$, so it doesn't even make sense to ask whether $\nabla \cdot {\bf V} = 0$. It's true that $\nabla \cdot {\bf V} = 0$ everywhere on the domain of $V$. – Travis Nov 27 '18 at 10:43
• How do we know that the origin isn't in the domain of $\bf {V}$? – Joe Nov 27 '18 at 11:00
• Because the quantity $\frac{1}{r}$ that appears in the definition of $\bf V$ isn't defined at the origin. – Travis Nov 27 '18 at 12:42
## 2 Answers
You seem to know that a divergence free field $$\vec V$$ can be regarded as curl of some other field: There is a field $$\vec U$$ such that $$\vec V={\rm curl}(\vec U)$$. This is a consequence of the so-called Poincaré Lemma.
But there is a catch: The Poincaré Lemma guarantees the existence of such a vector potential $$\vec U$$ only if the domain of $$\vec V$$ is, e.g., a ball or star shaped. For your field $$\vec V$$ this is not the case. Therefore we only can say the following: Each point $${\bf p}$$ in the punctured space $$\dot{\mathbb R}^3:={\mathbb R}^3\setminus\{{\bf 0}\}$$ is the center of an open ball $$B_r({\bf p})\subset \dot{\mathbb R}^3$$ such that within $$B_r({\bf p})$$ the field $$\vec V$$ can be written in the form $$\vec V={\rm curl}(\vec U)$$ for some $$\vec U$$ defined in $$B_r({\bf p})$$ only. These local fields $$\vec U$$ are not uniquely determined, and it is not at all sure whether the implied "integration constants" can be chosen in a coherent way such that we obtain a single field $$\vec U_*$$, which then would be defined on all of $$\dot{\mathbb R}^3$$.
Of course, it could be that "by coincidence" the $$\vec V$$ in your example possesses a global vector potential $$\vec U_*$$ nevertheless: not by the Poincaré Lemma per se, but because a certain integrability condition (say, the flux of $$\vec V$$ across a sphere around $${\bf 0}$$ should be $$=0$$) is fulfilled. Consider as an analogue the functions $$z\mapsto{1\over z}$$ and $$z\mapsto{1\over z^2}$$ in the punctured complex plane $$\dot{\mathbb C}$$. Both have local primitives. But one of them does not have a global primitive in $$\dot{\mathbb C}$$, the other does, the reason being that $$\int_{\partial D}{1\over z}\>dz=2\pi i\ne0\>, \qquad \int_{\partial D}{1\over z^2}\>dz=0\ .$$
Update: I suggest you look at the entry Helmholtz decomposition in Wikipedia and apply Helmholtz' theorem to a large ball minus a tiny ball around the origin.
• Thank you very much for your answer. Please can you explain this in a way so that a physics graduate student can understand. – Joe Nov 27 '18 at 11:59
• You asked the question at MSE. There is some deep mathematics involved here, called the De Rham complex, see here: en.wikipedia.org/wiki/De_Rham_cohomology . I had hoped to explain what is relevant in your example in such a way that a physics student can understand. – Christian Blatter Nov 27 '18 at 14:22
• "The flux of $\vec{V}$ across a sphere around $\bf {0}$ should be $=0$"...Can you show how? Would it be true for other volumes than sphere.... – Joe Nov 27 '18 at 16:37
• Full answer. What is the meaning of $\dot{\Bbb R}^3$? – manooooh Nov 27 '18 at 16:51
• I was deliberately cloudy here, since I don't know for sure what is necessary and sufficient. Ask your physics professor. Note that "usually" people just consider the full space ${\mathbb R}^3$ or balls. – Christian Blatter Nov 27 '18 at 16:52
It is a consequence of the Kelvin-Stokes theorem:
https://en.wikipedia.org/wiki/Kelvin%E2%80%93Stokes_theorem
If you take a potential $$U=\frac{1}{r}$$ with associated "electric field" $$E:=\nabla U$$ then you know that the flux of $$E$$ through a unit sphere $$\Sigma$$ centered around $$(0,0,0)$$ is 1.
However by Kelvin-Stokes if you could write $$E=curl(B)$$ for a certain vector field $$B$$ then the flux of $$E$$ through $$\Sigma$$ would be equal to 0 since $$\Sigma$$ is closed surface.
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# Calculating molecular weight of an acid based on titration data
If $\pu{2.600 g}$ of a weak diprotic acid were dissolved in $\pu{100 mL}$ of distilled water and a $\pu{10 mL}$ aliquot of this solution required $\mathrm{21.60\ mL}$ of $\mathrm{0.1000\ M\ NaOH}$ to reach the first endpoint, what are the equivalent and formula weights of $\ce{H2A}$?
I assumed to do this: I calculate the grams of acid in solution which is
$$\mathrm{\frac{2.600\ g}{100\ mL} \cdot 10\ mL = 0.2600\ g}$$
After that I would calculate the moles of NaOH used
$$\mathrm{0.1000\ mol/L \cdot 21.60\ mL \cdot 1\ L/1000\ mL = 0.002160\ mol}$$
Then I use this to find the moles of acid present in solution; this is the part I am unsure about. I am assuming: because at the equivalence point half of the acid has been converted to it's conjugate base and so ($\mathrm{2\ mol\ NaOH = 1\ mol\ H_2A}$).
If this is correct then
$$\mathrm{0.002160\ mol\ NaOH \cdot 2\ mol\ H_2A/mol\ NaOH = 0.004320\ mol\ H_2A}$$
Which I then divide the g/mol
$$\mathrm{\frac{0.2600\ g}{0.00432\ mol} = 60.19\ g/mol}$$
Because the acid is diprotic the equivalence weight would be
$$\mathrm{\frac{60.19\ g/mol}{2} = 30.09\ g/mol}$$
I don't think this is the correct answer because I am trying to do this using real titration data I took in lab and the MW seems to be at $\approx 1/2$ the value it should be. If I am not supposed to double the NaOH moles, is there a reason for it?
• The reason is in the words "...to reach the first endpoint". What is the meaning of these? Commented Sep 29, 2015 at 5:34
This being said, your first EP can be described by $$n_A = n_B$$ (and not as you did by assuming $n_A = 2\ n_B$, which also would not be the description of the second EP, which is $2\ n_A = n_B$ instead.)
You have to find the molecular mass of your acid, why the upper equation needs to be expanded to $$\frac{m_A}{M_A} = c_B V_B$$
Now you simply have to solve for $M_A$ and put in your numbers (0.26 g, 0.1 mol/L, 21.6 mL).
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The Future Of Spaceflight.
Rockets suck. This is a thing that we have established here; they’re terrifically awful ways of getting into space that are only used because nobody has really come up with anything better. There’s all sorts of ideas for wacky drive systems once your spacecraft is actually in space – ion drives, solar sails, Bussard ramjets – but these all sidestep the real problem facing future space travel, which is that you have to get out of the Earth’s gravity well first. This is not easy; even though the Earth is pretty small for a planet it’s still the heaviest of the four terrestrials and has what is to us a very hefty gravitational pull.
In order to exit Earth’s gravity well, an object being launched on a ballistic trajectory from the Earth’s surface must attain escape velocity. Escape velocity is calculated by
where G is the gravitational constant, M is the mass of the Earth and R is the radius of the Earth. For “ballistic” you can read “in freefall” – i.e. an object moving without any other forces acting on it. Rockets do have external forces acting on them thanks to their propulsion systems, so a rocket doesn’t necessarily have to reach escape velocity of 11.2 kilometres per second in order to escape the Earth’s gravity so long as it’s under constant power. However, the amount of energy it will have to expend will be the same as this hypothetical ballistic object, and that can be worked out by using the equation for kinetic energy.
where m is the mass of the spacecraft. With Earth’s escape velocity being 11.2 km s-1, we can rearrange this equation to find out how much energy per kilo of mass a launch vehicle will need to have to escape Earth’s gravitational pull, and this turns out to be about 63 megajoules per kilogram. I always get a little lost when trying to visualise what a joule is in everyday terms, but Wikipedia tells me that 63 megajoules is roughly the amount of energy you’d release if you detonated fifteen kilograms of TNT. So for a crude method of visualising this, go and look at a picture of a rocket. Imagine a pile of TNT fifteen times as big. Blowing up this pile of TNT would release the same amount of energy as would be needed to get that rocket out of the Earth’s gravity well.
(One significant caveat here: rockets going into orbit don’t need to expend this much energy. They just need to go fast enough that they never hit the ground.)
This amount of energy is fixed. There is no way of getting around it; if you want to go into space, you have to expend at least 63 MJ per kilo of spacecraft mass and this energy has to come from somewhere. For rockets, it comes from rocket fuel; unfortunately rocket fuel increases the mass of the rocket, which requires more rocket fuel, which increases the mass of the- actually I think I’ve done this already, haven’t I. Anyway, rockets suck, but we still use them because the alternative solutions are, to put it mildly, just a little bit far-fetched.
Solution 1: Build better rockets.
Ahaha. Ahahaha. Ahaha. No. Rocket science isn’t necessarily all that complicated – I can understand it, after all – and one of the things that struck me when I was studying space science in the second year of my undergraduate degree was that rockets are a dead end. Well, that’s a bit harsh. Better to say that they’ve been refined as much as they can be.
Two things affect the amount of thrust you can get out of a rocket: the expansion velocity of the exhaust gas you’re creating by burning the fuel, and the size of the rocket nozzle. Rapidly expanding rocket fuels tend to be rather unstable – again, see the Nedelin catastrophe – and so there’s an upper limit on just how volatile you want to make that. What about the rocket nozzle? How wide/narrow this is determines the pressure at which the exhaust gases exit the rocket. For reasons I won’t go into, rockets produce the maximum amount of thrust when the pressure of the exhaust gas equals the pressure of the ambient atmosphere through which the rocket is flying. Atmospheric pressure decreases as you ascend into space, so ideally you want a rocket nozzle that automatically adjusts in size with altitude to keep the rocket working as efficiently as possible. This has been done. After that, there’s no real way to further improve a chemical rocket. We’ve taken them just about as far as we’re going to.
Solution 2: Don’t launch them from Earth.
One of the attractive things about a Moon colony – aside from the whaling opportunities – is that the Moon has a gravitational pull less than one-sixth that of Earth’s. It has an escape velocity of 2.38 km s-1, and consequently it would only take 2.83 MJ to get one kilogram of mass out of the Moon’s gravity – under a twentieth the energy you’d need to get the same kilo off of Earth! Probably the best way to illustrate this is the ascent module on the Apollo lunar landers; that little capsule was able to create enough thrust to return to orbit to rendezvous with the command module.
It would be far easier, therefore, if we could somehow launch rockets from the Moon instead. Unfortunately this idea merely replaces the problem of getting out of the Earth’s gravity well with the even larger problem of building the necessary industrial base on the Moon to build and launch rockets. I’m not saying it couldn’t be done, and if somebody did manage to do it there’s every chance we’d see the commonplace spaceflight depicted in movies like 2001 (if you were lucky enough not to live on Earth, anyway), but the level of commitment and resources it would require would be staggering.
Solution 3: Mass drivers.
This one is simple, and operates on the principle that while achieving 63 MJ per kilo is a tall order for a chemical rocket, if we could somehow use electricity instead it’d be far easier to generate the energy required. The idea is you have a long, long, long, long launch track, kind of like a very high-tech version of Japan’s bullet train network (see maglevs for further information), and you accelerate the thing you want to launch down this track using superconducting electromagnetism. The track slopes upwards towards the end, and so once it reaches the end of the line your launch vehicle is shot into the stratosphere. At this point some small rocket boosters would take over and move the launch vehicle to the desired orbital trajectory. Building this thing would be a bit more feasible than the Moon colony, but there’s just one tiny snag: room temperature superconductors capable of carrying the currents required haven’t been invented yet.
Solution 4: Project Orion.
In which the Orion spacecraft is supposed to fart out shaped nuclear explosions, the brunt of which is directed against an impact plate on the ass of the spacecraft which makes the nukes “push” the spacecraft along. Utterly, utterly mad.
Solution 5: A space elevator. Also unicorns.
A very old concept that’s rather in vogue in 4X games, for some reason. As mentioned here geostationary satellites orbit the earth with a period of one day – that is, they remain above the same point relative to the Earth’s surface throughout their orbit. In theory, if you had access to a material strong enough as well as the combined and focused resources of the largest superpowers on Earth over a half-century or more, you could lower a cable from geostationary orbit all the way down to the surface of the Earth and simply have vehicles ride up and down the cable just like an elevator car. This is attractive because the energy costs are similar to those of a mass driver launch with the added bonus that you get most of it back as elevator cars come back down from orbit. If we then wanted to launch stuff further out into the solar system we could take advantage of the fact that the space elevator would kind of act like a giant sling to any payload launched from the far end.
Sounds fun, right? Perhaps so, but space elevators have several minor niggles past the huge advances in space technology, robotic manufacturing, materials science and megastructure construction (I just made that last one up, but someone’s going to have to invent it before they can build the elevator) that would be required, not to mention the totalitarian world government that’d have to be in place to keep everyone pointed in the right direction long enough to finish the bloody thing.
• In order to make it work you need a big counterweight of some kind to produce the necessary centrifugal force to keep the elevator cable taut. These days space elevator designs merely pay out the cable a little further past geostationary orbit to provide a counterweight mass rather than the previous idea of moving an asteroid to GEO to act as counterweight, but even if you do this you still need the asteroid because a) you need to build the elevator from both ends at once and so you’re going to need a base of operations in space from which to do it, and b) an asteroid would provide necessary raw materials for cable construction. So first you need to move several million tonnes of rock into geostationary orbit. This is just a little bit tricky to do, and I imagine it might make people down on the surface a little bit nervous as well what with the potential consequences if something goes wrong.
• There is no material which can currently be mass-produced in the quantities required which has the necessary tensile strength to support thousands of kilometres worth of its own weight. People always point to carbon nanotubes as the catch-all solution, but while they have the theoretical capacity to support the quantity of mass required, it would require the cable to be structurally flawless over its entire 40,000 km-odd length. Even one flaw would introduce a weakness which could be potentially fatal to the whole shebang.
• Transmitting power to the elevator cars is also going to be a bit of a bugger, since you still need that 63 MJ/kg to get into orbit. A nuclear power source would do it, but that’d probably make passengers a bit uncomfortable. Solar panels would increase the weight of the car, while using the cable itself to transmit power is going to run into the very inefficiency problems the space elevator is supposed to avoid. The current favoured proposal is wireless energy transfer via lasers or something, which has the tiny drawback that it is literally space magic.
• Elevator cars would travel fairly slowly, with an ascent to GEO taking anywhere from 6-12 hours to a full day. This brings the Van Allen radiation belts into play; the cars would travel through them slowly enough that anyone inside would end up taking lethal doses of radiation. It would be necessary to shield the cars in some way, which again would add to the weight.
Personally I think the space elevator is the ultimate manifestation of Archimedes and his lever; something that is theoretically possible but practically stupid. We can’t even agree on the best way to fly three people up to LEO; we abandoned the moon after sending barely two dozen guys up there. We are nowhere near being ready to take on the incredible engineering challenges of building something like this, and so anyone who thinks the space elevator is going to be built in the next half-millenia needs their head examined.
Where do I think the likely future of spaceflight lies? Well, assuming the world doesn’t enter a period of terminal dystopian senescence in the next couple of decades I think the mass driver concept is probably the most feasible solution as long as somebody comes along to solve the superconductor problem. It’s the smallest engineering project, it has the very great advantage of the thing being built on Earth rather than in space or on another planet, and it has the fewest question marks over necessary technology advances. Plus, if aliens ever try to invade we can use it to shoot rocks at them. Welcome to Earth.
4 thoughts on “The Future Of Spaceflight.”
1. innokenti says:
Let’s face it, the ‘shooting things with a giant rail’ is probably the cooler way of doing it all anyway.
2. slimbuk says:
I love that the Civ 4 space elevator seems to get built with little more than some forklift trucks and some scaffolding.
It’s not just 4X games though, syndicate and I-war off the top of my head has a space elevator.
• hentzau says:
Ha, you’re exactly right, Syndicate Wars has a space elevator. Can’t believe I forgot about that.
3. [...] so much is because it illustrates a key restriction on travelling around the solar system that I’ve brought up more than a few times before on here: by far the hardest part of space travel is getting out of the [...]
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## Solar Declination Angle & How to Calculate it
• Post author:
The solar declination angle is a concept of astronomy, but its understanding is also helpful in solar energy and photovoltaic (PV) systems.
## Solar declination angle calculator
Select the date, time, and your time zone to compute the declination angle in the degree.
## What is the solar declination angle?
The solar declination angle is the angle between the rays of the sun and the equator of the earth. It is usually denoted by “δ.” Solar declination angle is very useful in calculating the sun elevation and the azimuth angle.
The diagram below explains the same.
Earth rotates around its own axis. It is what makes days and nights. Also, it revolves around our sun and has its own orbit.
As we might think, the earth’s axis of rotation is not perpendicular to the direction of the orbit. Instead, it has a tilt of around 23.44° (see the above figure). Because of this tilt, the equator of the earth and the rays of the sun always make some angle between them. It is what we called the solar declination angle. Other terms used for it are the sun declination angle or the declination angle of the sun.
If the earth were not tilted, the declination angle would be zero.
## Declination angle with time
The solar declination angle changes time-to-time; it is not a constant number. It will be different every single day. However, the angle restricts itself between 23.44° and −23.44°. It oscillates between these two numbers; it will never cross 23.44° or fall behind −23.44°. Thus, we can conclude, the maximum value of the solar declination angle is 23.44°, and the minimum is −23.44°.
The declination angle becomes zero twice a year. It happens on March 20th and September 22nd (or 23rd). On these days, the earth’s axis of rotation becomes perpendicular to the rays of the sun. And the angle between the equator and the sun’s rays reduces to zero. These two days are called equinoxes.
In the northern hemisphere, the March equinox (March 20th) is called the spring or vernal equinox, and the September equinox (September 22nd (or 23rd)) is called the fall or autumnal equinox. In the southern hemisphere, the reverse is true.
The solar declination angle reaches a maximum on June 20th (or 21st). This day is named the June solstice. On this day, the earth is in the north of the sun. The value of the declination angle is 23.44°. In the northern hemisphere, this day is called the summer solstice, while in the southern hemisphere is called the winter solstice.
The value of the declination angle is −23.44°, the lowest, on December 21st (or 22nd). This December solstice is called the winter solstice in the northern hemisphere, whereas in the southern, it is the summer solstice.
## Declination angle for your location
The declination angle is almost independent of your location. For a given UTC time, it is the same throughout the globe. It will not change from place-to-place for the same UTC time. So, whether you are in Los Angeles, Phoenix, Denver, or Boston, the solar declination will be the same for the same UTC time. However, if we include local time zones, then the declination varies slightly.
## Solar declination angle formula
We can calculate the solar declination angle using the following formula:
In the above formula, d is the number of days since January 1st (UTC 00:00:00). For example, On March 3rd (UTC 00:00:00), d = 31 + 28 + 2 = 61. On December 31st (UTC 00:00:00), d = 364.
The value of δ is maximum (23.44°) when the cosine is equal to the negative one. It occurs at d = 172.5, which is around June 21st (summer solstice). The value of δ is minimum when the cosine is equal to the positive one. At d = 355 (December 21st), the cosine equals the positive one, and δ = −23.44°.
Note to consider, this formula is not very accurate and always gives a marginal error. A better formuation is mentioned in the following section.
Similarly, δ approaches zero (at d = 263.75 and 81.25) on both March and September equinoxes.
The previous formula is the simplification of the below equation.
At lower values of θ in radians, sin(θ) ≅θ. So, in the above equation, sin(θ) ≅ θ and sin(−23.44°) ≅ −23.44°.
These equations assume the earth orbits around the sun in a perfect circle.
The graph below represents the variation in the solar declination angle with time.
As we can see from the above graph, the solar declination angle is positive from the vernal equinox to the autumnal equinox—i.e., March 20th to September 22nd (or 23rd). And it is negative for the rest of the portion of the year.
### Better formula
The following equation gives a more accurate value of the declination angle.
## How to calculate the solar declination angle?
We can calculate the solar declination angle for a particular time from the former formula. Although the value will not be 100% accurate and a small error in the result is expected, that is acceptable for practical purposes. We only require the number of days (d) to know the angle. The table below lists some examples.
## Effect of the solar declination angle on PV systems
To make things simple, we will assume you live in the US or a country from the northern hemisphere.
Estimating the declination angle is necessary to determine the solar elevation. The solar elevation increases with the declination angle, and it is highest in months of summer. In other words, the sun moves over our heads in summer. It is the reason why in summer, we tilt our solar panels at lower angles.
As winter approaches, the declination angle goes negative, and the solar elevation decreases. Thus, in winters, the sun descends toward the horizon. Because of that, the tilt angle of solar panels increases, and the solar panels are placed almost vertical to maximize solar power.
## FAQs
### When is the declination angle negative?
The declination angle is negative from September 22nd (Autumnal equinox) to March 20th (spring equinox).
### Why does the declination angle differ with location?
The declination angle does not differ with location for the same UTC time. However, it differs slightly with local time zones.
### What is the declination angle for the winter solstice?
In the northern hemisphere, the declination angle for the winter solstice is −23.44°, while in the southern hemisphere, it is 23.44°.
### What is the declination angle for the summer solstice?
In the northern hemisphere, the declination angle for the summer solstice is 23.44°, while in the southern hemisphere, it is −23.44°.
### What is the solar declination angle at the vernal equinox?
The solar declination angle at the vernal equinox is 0°.
### What is the angle of declination on the March equinox and September equinox?
The declination angle is zero for both March and September equinoxes.
### How to find the declination angle for given coordinates?
You do not need coordinates to find the solar declination angle since it is independent of the location. You only need the date and time.
### Does the solar declination angle change daily?
Yes, the solar declination angle does change daily. But the change is practically insignificant in a short period of time.
### What happens to the angle of declination as you get closer to the poles or the equator?
Whether you are at the poles or the equator, the angle of declination will not change for the same UTC time. It is the same across the earth for a given moment of time.
+ posts
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# Number Sense Blog
By Mathnasium | Added Jun 19, 2019
Would you believe it if we told you that a dog could tutor for math? It’s true! Well...kind of. We’re featuring a few shelter dogs in this week’s word problem to encourage kids math learning. After all, we know that the best way to learn math is to apply it in real world situations.
This word problem challenges students to practice their elementary school math skills such as addition, subtraction, multiplication and division. It may look “ruff”, but trust us, this question is a walk in the park!
Question: Grace, Ruby, and James volunteer to walk shelter dogs. Grace walks half as many dogs as Ruby, who walks 5 more dogs than James. James walks 9 dogs. How many dogs did Grace walk?
Take your time working out the particulars. When you're ready, look below to compare your solution to ours.
Solution: If Ruby walked 5 more dogs than James, then she walked 9 + 5 = 14 dogs. Grace walked half as many as that, so she walked half of 14 = 7 dogs.
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# Use the region $R$ with the indicated boundaries to evaluate each double integral.$$\iint_{R} x^{2} y^{2} d x d y ; \quad R bounded by y=x, y=2 x, x=1$$
## $\frac{7}{18}$
### Discussion
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### Video Transcript
All right. So here we are, integrating disfunction over this specified region functions X squared. Why squared t x t? Why in the region is bounded between, understand? Sketch it quick. So x y No. Okay, so it says y cause X years. Why? Cause X Y equals two accidents be appear next, and then X equals one. So here's our region right here. Okay. And so we're going to make a choice whether we really want to do X first or wife first, and, well, it seems like to me, you should do Why First, if you do x first, you're gonna have to break it up into two equal. So that immigrating X here, it's going to depend on the wires. Ah, Why? From zero one or wanted to. So, I mean, if you if you are not following, just you can just do it both ways and and see what I mean. Okay, so I'm going to integrate, actually. Why? First and now that means I'm going to be fixing an X value between here. Okay. My region is going from zero toe one and X and then where am I going? I'm going from Well, Michael's ex. So in terms of Ana indexing next value. And so I'm going from Why all the way up, Tio? Well, why was to expect in terms of eggs, That's why over too. So this is gonna be exes going from zero to one. And then why is going to be going from X here? Yeah, yeah, Just sex to do X. Okay. And then we have our function X squared y squared. Tio, I k'nex Okay? Yes, over here. And just wise made from eggs up to two ex for any fixed X. So here. No one and I derivative with respect. Why? So we'LL get one third no X squared. Why? Cubes evaluated from Exeter to X. What? What is that there? Okay, so that's I'm gonna cube this and then subtract this cube. So that's looking like So that was just going to be executed within. I ate excused minus one. Excuse us seven execute. But then I have Miss X. Uh, squared. OK, so all in all, it should be seven x to the fifth. Okay, so we have seven thirty compactor out X to the fifth. That's one sixth next to the sixth about a J from zero to one. But this is just one. So we end up with seven over eighteen
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## Algebra 2 (1st Edition)
$$64s^6-576s^5+2160s^4-4320s^3+4860s^2-2916s+729$$
We know from the Binomial Theorem (page 693) that we can expand the given expression without multiplying it out. Doing this, we obtain: $$64s^6-576s^5+2160s^4-4320s^3+4860s^2-2916s+729$$
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# There are two more homework assignments
### Resolved Question:
1. There are two more assignments in a class before its end, and if you get an A on at least one of them, you will get an A for the semester. Your subjective assessment of your performance is Event Probability A on paper and A on exam .25 A on paper only .10 A on exam only .30 A on neither .35 Let's develop a joint-probability table to analyze this problem (the letters a – h represent probability values. For example, a is the probability of getting an A on both the paper and the exam; b is the probability of getting an A on the exam, but not on the paper; while g is the sum of a and d, and represents the probability of getting an A on the paper). Get A on paper Do not get A on paper row sum Get A on exam a b c Do not get A on exam d e f column sum g h
a. What does c + f equal?
b. What does a + b + d + e equal?
c. What is the probability of getting an A on the exam?
d. What is the probability of getting an A in the course?
e. Are the grades on the assignments independent (explain your answer mathematically)?
2. A medical research project examined the relationship between a subject’s weight and recovery time from a surgical procedure, as shown in the table below. Underweight Normal weight Overweight Less than 3 days 6 15 3 3 to 7 days 30 95 20 Over 7 days 14 40 27
a. Use relative frequency to develop a joint probability table to show the marginal probabilities.
b. What is the probability a patient will recover in fewer than 3 days?
c. Given that recovery takes over 7 days, what is the probability the patient is overweight?
3. The Ambell Company uses batteries from two different manufacturers. Historically, 60% of the batteries are from manufacturer 1, and 90% of these batteries last for over 40 hours. Only 75% of the batteries from manufacturer 2 last for over 40 hours. A battery in a critical tool fails at 32 hours. What is the probability it was from manufacturer 2?
4. The high school GPA of applicants for admission to a college program are recorded and relative frequencies are calculated for the categories. GPA f(x) x < 2.0 .08 2.0 <= x < 2.5 .12 2.5 <= x < 3.0 .35 3.0 <= x < 3.5 .30 3.5 <= x ?
a. Complete the table to make this a valid probability distribution.
b. What is the probability an applicant's GPA will be below 3.0?
c. What is the probability an applicant's GPA will be 2.5 or above?
5. A video rental store has two video cameras available for customers to rent. Historically, demand for cameras has followed this distribution. The revenue per rental is \$40. If a customer wants a camera and none is available, the store gives a \$15 coupon for tape rental. Demand Relative Frequency Revenue Cost 0 .35 0 0 1 .30 40 0 2 .20 80 0 3 .10 80 15 4 .05 80 30
a. What is the expected demand?
b. What is the expected revenue?
c. What is the expected cost?
d. What is the expected profit?
6. A manufacturer of computer disks has a historical defective rate of .001. What is the probability that in a batch of 1000 disks, 2 would be defective? (note: answer using either the relevant probability table in the back of the book, or use the relevant probability function on your calculator / Excel).
7. After a severe winter, potholes develop in a state highway at the rate of 5.2 per mile. Thirty-five miles of this highway pass through Washington County.
a. How many potholes would you expect to see in the county?
b. What is the probability of finding 8 potholes in 1 mile of highway?
8. The weight of a can of stewed tomatoes is listed as 16 oz., but in actuality, the weight between cans may vary, and in fact is normally distributed with mean = 16.2 ounces and standard deviation = 0.1 oz. What is the probability that a consumer buys a can whose contents are actually less than the stated weight of 16 oz.?
Submitted: 3 years ago.
Category: Homework
Expert: Susan Athena replied 3 years ago.
Hi. Thanks again for your question!
You can download your study guide here.
Kind regards
Customer: replied 3 years ago.
Question 1
1. There are two more assignments in a class before its end, and if you get an A on at least one of them, you will get an A for the semester. Your subjective assessment of your performance is:
Event Probability A on paper and A on exam .25 A on paper only .10 A on exam only .30 A on neither .35
1. Let's develop a joint-probability table to analyze this problem (the letters a – h represent probability values. For example, a is the probability of getting an A on both the paper and the exam; b is the probability of getting an A on the exam, but not on the paper; while g is the sum of a and d, and represents the probability of getting an A on the paper).
Get A on paper Do not get A on paper row sum Get A on exam a b c Do not get A on exam d e f column sum g h
1. What does c + f equal?
2. What does a + b + d + e equal?
3. What is the probability of getting an A on the exam?
4. What is the probability of getting an A in the course?
Expert: Susan Athena replied 3 years ago.
THIS ANSWER IS LOCKED!
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Customer: replied 3 years ago.
Hi Susan, I have another question
Expert: Susan Athena replied 3 years ago.
Hi, thanks for requesting me!
Please post your question in a new thread, and just put "For Susan" at the beginning. If you have a specific deadline, please let me know. I can address your question when I'm back online this evening.
Kind regards,
Susan
Customer: replied 3 years ago.
1. The options from which a decision maker chooses a course of action are
a. called the decision alternatives.
b. under the control of the decision maker.
c. not the same as the states of nature.
d. each of the above is true.
2. States of nature
a. can describe uncontrollable natural events such as floods or freezing temperatures.
b. can be selected by the decision maker.
c. cannot be enumerated by the decision maker.
d. each of the above is true.
3. Which of the methods for decision making without probabilities best protects the decision maker from undesirable results?
a. the optimistic approach
b. the conservative approach
c. minimum regret
d. minimax regret
4. If P(high) = .3, P(low) = .7, P(favorable | high) = .9, and P(unfavorable | low) = .6, then P(favorable) =
a. .10
b. .27
c. .30
d. .55
5. If the payoff from outcome A is twice the payoff from outcome B, then the ratio of these utilities will be
a. unknown without further information.
b. less than 2 to 1.
c. more than 2 to 1.
d. 2 to 1
6. The probability for which a decision maker cannot choose between a certain amount and a lottery based on that probability is
a. the utility probability.
b. the lottery probability.
c. the uncertain probability.
d. the indifference probability.
7. When the decision maker prefers a guaranteed payoff value that is smaller than the expected value of the lottery, the decision maker is
a. an optimizer.
b. a risk taker.
c. an optimist.
d. a risk avoider.
8. A decision maker whose utility function graphs as a straight line is
a. a risk avoider.
b. risk neutral.
c. a risk taker.
d. conservative
9. When consequences are measured on a scale that reflects a decision maker's attitude toward profit, loss, and risk, payoffs are replaced by
a. opportunity loss.
b. multicriteria measures.
c. sample information.
d. utility
10. The purchase of insurance and lottery tickets shows that people make decisions based on
a. maximum likelihood
b. sample information.
c. utility.
d. expected value.
11. The expected utility approach
a. requires a decision tree.
b. leads to the same decision as the expected value approach.
c. is most useful when excessively large or small payoffs are possible.
d. does not require probabilities.
12. Values of utility
a. must be between 0 and 1.
b. must be between 0 and 10.
c. must be nonnegative.
d. must increase as the payoff improves.
II. PROBLEMS
(each question is worth 32 pts – 8 pts. each part)
1. A payoff table is given as
s1 s2 s3
d1 10 8 6
d2 14 15 2
d3 7 8 9
a. What decision should be made by the optimistic decision maker?
b. What decision should be made by the conservative decision maker?
c. What decision should be made under minimax regret?
d. If the probabilities of s1, s2, and s3 are .2, .4, and .4, respectively, then what decision should be made under expected value?
2. Chez Paul is contemplating either opening another restaurant or expanding its existing location. The payoff table for these two decisions is:
s1 s2 s3
New Restaurant -\$80K \$20K \$160K
Expand -\$40K \$20K \$100K
Paul has calculated the indifference probability for the lottery having a payoff of \$160K with probability p and -\$80K with probability (1-p) as follows:
Amount Indifference Probability (p)
-\$40K .4
\$20K .7
\$100K .9
a. Is Paul a risk avoider, a risk taker, or risk neutral? EXPLAIN.
b. Suppose Paul has defined the utility of -\$80K to be 0 and the utility of \$160K to be 80. What would be the utility values for -\$40K, \$20K, and \$100K based on the indifference probabilities?
c. Suppose P(s1) = .4, P(s2) = .3, and P(s3) = .3. Which decision should Paul make using the expected utility approach?
d. Compare the result in part c with the decision using the expected value approach.
Expert: Susan Athena replied 3 years ago.
Hi. Thanks for requesting me!
I'm really sorry, but decision theory isn't my strength. I need to opt out and refer you to a different expert.
Kind regards
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## Wednesday, April 6, 2011
### Equation Sandwich Anyone? How to eat Algebra
I credit this idea to something similar in The Creative Teacher
<------------------------------------ One of the best books for Classroom Creativity
One of the great mysteries of the math teacher is how to get your students to show their steps! Here is something that might help. Have them create an Equation Sandwich!
First you need bread(2), mustard, tomatoes, mayonnaise, ham, cheese and lettuce like so
(bread(2), mustard, tomatoes, mayonnaise, ham, cheese and lettuce)
Each piece represents a step of your algebra problem
2x + 3 + 6x - (-1) = 20 (Bread)
(2x + 6x) + (3 + 1) = 20 (Mustard)
8x + 4 = 20 (Tomatos)
8x + 4 - 4 = 20 - 4 = (Mayo)
8x = 16 = 16 (Cheese)
8x/8 = 16/8 (Ham)
x = 2 (Lettuce)
(You Get An Equation Sandwich)
(Eat The Knowledge!)
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# uniform plank walk
• Nov 30th 2007, 10:15 PM
rcmango
uniform plank walk
A uniform plank of length 4.9 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 453 N walk on the overhanging part of the plank before it just begins to tip?
in meters
• Dec 1st 2007, 05:39 AM
kalagota
Quote:
Originally Posted by rcmango
A uniform plank of length 4.9 m and weight 210 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 453 N walk on the overhanging part of the plank before it just begins to tip?
in meters
(since no one answers it yet and it is placed in the Urgent HW, i'll try it..)
i can't imagine the drawing.. Ü maybe something like this..
Code:
o o o o o o ''''''''''''''''''''''''''''''''''''''''''
do you have the answer.. actually, i got 0.6258 m from the right support.. well, not sure on this that is why i don't put the solution.. can i see what you're doing?
• Dec 3rd 2007, 10:22 AM
rcmango
Your answer is correct, I could not figure this one out.
looks alot like this, plank - View Image - Post Your Image
can you please explain how i am supposed to set this up.
• Dec 5th 2007, 04:48 AM
kalagota
Quote:
Originally Posted by rcmango
Your answer is correct, I could not figure this one out.
looks alot like this, plank - View Image - Post Your Image
can you please explain how i am supposed to set this up.
good.. now, since the plank has uniform weight $w_p$, let's assume that it is concentrated at the middle of the plank..
so, if the length of the plank is L, then the weight is located at $\frac{L}{2}$..
the distance of the left support from the middle of the plank is $\frac{L}{2} - 1.1m$..
now, if $w_{man}$ is the weight of the man, then the product of $w_p$ and $\frac{L}{2} - 1.1m$ (the distance of the middle of the plank from the right support) must be equal to the product of $w_{man}$ and the distance, $x$, of the man from the right support..
that is, $w_p \cdot \left( \frac{L}{2} - 1.1 \right) = x \cdot w_{man}$..
now, substitute the values.. Ü
• Dec 5th 2007, 05:52 AM
topsquark
Quote:
Originally Posted by rcmango
Your answer is correct, I could not figure this one out.
looks alot like this, plank - View Image - Post Your Image
can you please explain how i am supposed to set this up.
This is just going to mimic kalagota's response, but let me put it in "Physics" terms for you.
The plank is not rotating, so it is in static equilibrium. This denotes not only a = 0 for the system, but also means the system doesn't rotate.
So let's pick a coordinate system where +y is upward. According to the diagram we have a force $S_L$ by the left support acting straight upward, a force $S_R$ by the support on the right acting straight upward. Both of the support forces act on the board at the "point" of the support. We also have the man's weight $w_m$ acting straight down from where the man is standing (x meters to the right of the right support), and a weight $w_p$ of the plank acting straight down from the geometric center of the plank.
Newton's 2nd tells us
$\sum F_y = S_L + S_R - w_m - w_p = 0$
or
$S_L + S_R - 453 - 210 = 0$
Now, we have two unknown forces and one unknown x. So we need two more equations.
Let's pick an axis of rotation at the point where the support on the right meets the plank and call this point A. The system is not rotating about point A, so
$\sum \tau _A = 0$
$\sum \tau _A = -S_L(4.9 - 1.1) - w_p(2.45 - 1.1) + w_mx = 0$
(Recall that a counterclockwise torque is negative by default.)
$-3.8S_L - 283.5 + 453x = 0$
We still need one more equation. The nice thing about static equilibrium is that the system doesn't rotate about any point, so we are free to pick whatever axis of rotation we desire, whether it is a physical axis of rotation or not. So I'm going to pick another axis of rotation B located where the support on the left meets the plank. Thus
$\sum \tau _B = w_p(2.45) - S_R(4.9 - 1.1) + w_m(4.9 - 1.1 + x) = 0$
or
$514.5 - 3.8S_R + 453(3.8 + x) = 0$
So we now have three equations in three unknowns:
$S_L + S_R - 453 - 210 = 0$
and
$-3.8S_L - 283.5 + 453x = 0$
and
$514.5 - 3.8S_R + 453(3.8 + x) = 0$
Solving the system I get $x = 0.625828~m$
-Dan
• Dec 5th 2007, 05:57 AM
kalagota
you're really a physics maestro..
i was about to that solution before but when i realized that there was in need to solve a system of equation, i became passive, so i tried the simplest way to solve it..
(whoa, it made me remember topics of the first of the two physics course i had..) Ü
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# Math
posted by .
Write the standard equation of the ellipse with the given characteristics
Center:(0,0)
foci: (2,0) (-2,0)
vertices:(5,0) (-5,0)
• Math -
looks like the vertices and the foci are on the x-axis, so this is easy
a = 5, b = ? and c = 2
In this type of ellipse,
b^2 + c^2 = a^2
b^2 + 4 = 25
b^2 = 21
x^2 /25 + y^2/21 = 1
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### Home > CC2 > Chapter 6 > Lesson 6.2.3 > Problem6-82
6-82.
Forty percent of the students at Pinecrest Middle School have a school sweatshirt. There are $560$ students at the school. Draw a diagram to help you solve each problem below.
1. How many students have a school sweatshirt?
$\frac{40}{100}\left(\frac{?}{?}\right)=\frac{\textit{x}}{560}$
$x=224$ students
2. If $280$ students have school t-shirts instead of sweatshirts, what percentage of the school has a t-shirt?
$280$ is what percent of $560$?
$\frac{\textit{x}}{100}\left(\frac{?}{?}\right)=\frac{280}{560}$
$x=50\%$
3. What percentage of the school does not have a t-shirt or a sweatshirt?
If $40\%$ of the students have a school sweatshirt and $50\%$ have a t-shirt instead of a sweatshirt, what is the remaining percentage of students that have neither?
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## Math Olympiad Questions with solutions : Decimals Class 6
1.A boy weighs 56.74 kg. If his father is 1.5 times heavier than his son, then the weight of the father is_____ .
(A) 85 kg (B) 85.11kg (C) 85.20 kg (D) 85.15 kg
2.If 0.111 is approximately equal to -,then the approximate value of 0.777 is________________ .
3.Arrange the following decimal numbers in ascending order.
5.5, 0.55, 0.055, 0.005
(A) 5.5, 0.055, 0.005, 0.55 (B) 0.55, 0.005, 0.055, 5.5
(C) 5.5, 0.55, 0.055, 0.005 (D) 0.005, 0.055, 0.55, 5.5
4.Which of the following expressions is correct ?
(A) 0.6 < 0.06 (B) 0.66 =5/9 (C) 455 >11/25 (D) 1/6 >0.17
5. 2+1/100+3/1000___________
(A) 0.213 (B) 2.13 (C) 2.013 (D) 2.130
6. 8/10+0+7/1000=____________
(A) 0.087 (B) 807 (C) 0.807 (D) 0.807
7.An ant went 4 m 26 cm up a tree, then it came 2 m 30 cm down.The distance travelled by the ant is______.
(A) 1 m 50 cm (B) 1 m 60 cm (C) 1 m 96 cm (D) 2 m
8.Sid travelled 5 km 52 m by bus, 2 km 265 m by car and the rest 1 km 30 m he walked. Total distance travelled by him is_____.
(A) 8.347 km (B) 8.000 km (C) 8.925 km (D) 9.000 km
9. A 8 m 57 cm pole was put in a pond to measure its depth. If 4 m 66 cm of pole remained outside water, then the depth of pond is________.
(A) 3.91 m (B) 3.00 m (C) 3.45 m (D) 3.23 m
10. The smallest possible decimal fraction upto three decimal places is
(A) 0.101 (B) 0.111 (C) 0.001 (D) 0.011
11. 3 hundredths can be written as______________.
(A) 0.003 (B) 0.03 (C) 0.300 (D) 300
12.The product of 3.92 x 0.1 x o.0 x 6.3 is___________.
(A) 0.392 (B) 0.1176 (C) 0 (D) 6.3
14.Which of the following fractions is equal to 0.67?
(A) 1/3 (B) 2/3 (C) 3/4 (D) 3/5
15. Which number is equal to (0.1/0.01+0.01/0.1) ?
(A) 10.1 (B) 1.10 (C) 1.01 (D) 10.01
16.The sum of two numbers is 31.021. If one of them is 11.56, then the other number is______.
(A) 19.461 (B) 17.461 (C) 18.641 (D) 19.561
17.The cost of 1 litre of milk is Rs. 7.50, then the cost of 30.5 litres of milk is____________.
(A) Rs. 225.75 (B) Rs. 223.75 (C) Rs. 228.75 (D) Rs. 227.89
18.Expression of 0.23 in terms of vulgar fraction is _____.
(A) 7/30 (B) 23/100 (C) 23/90 (D) 7/90
19.The ratio of copper and zinc in an alloy is 9 : 5. If the weight of zinc in the alloy is 9.5 gm, then weight of copper in it is______.
(A) 17.2gm (B) 17.1 gm (C) 17.5gm (D) 17.4gm
20.In 14.135, the digit 5 is in the_____place.
(A) tenths (B) hundreds (C) thousandths (D) thousands
21.The fractional number corresponding to the following number “9 tens + 5 ones + 3 tenths + 7 hundredths”is________ .
(A) 9437/100 (B) 9537/100 (C) 9000/100 (D) 9637/100
22. The decimal form of 5 (3/8) is ____________
(A) 5.375 (B) 5.000 (C) 5.255 (D) 2.325
23.What is the following number in standard form ?
2 + (8 x 0.1) + (6 x 0.01) + (4 x 0.001)
(A) 0.2864 (B) 2.864 (C) 28.64 (D) 2864
24.Prem went to a craft fair where he spent a total of Rs. 16.00. He spent Rs. 6.00 on admission and went to 8 tables. He spent the same amount of money (m) at each table. The following expression can be used to find how much money he spent at each table.
16 = 6 + 8 m
How much money did Prem spend at each table ?
(A) Rs. 0.50 (B) Rs. 0.80 (C) Rs. 1.25 (D) Rs. 2.00
25. Rahul bought 4 kg 90 g of apples, 2 kg 60 g of grapes and 5 kg 300 g of mangoes. The total weight of all the fruits he bought is_______ .
(A) 11.000 kg (B) 11.450 kg (C) 11.350 kg (D) 11.250 kg
26.What is the decimal fraction shown by the shaded part of the figure given below ?
27.Abhishek had Rs. 7.45. He bought toffees for Rs.5.30. The balance amount left with Abhishek is_______.
(A) Rs. 2.30 (B) Rs. 2.00 (C) Rs. 2.15 (D) Rs. 3.00
29.The correct expanded form for 2.06 is_________.
30.A man covers a journey by car in 3 hours. He covers a distance of 64 km 324 m during the first hour, 58 km 56 m during the second hour and 62 km 8 m during the third hour. What is the length of his journey?
(A) 184 km 388 m (B) 184 km 380 m
(C) 184 km 308 m (D) 184 km 3088 m
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$-2$
For $x\gt 0$ and $a$ a positive constant other than 1, $\log_{a}x$ is the exponent to which $a$ must be raised in order to get $x$. Thus, $\log_{a}x=m$ means $a^{m}=x$ --- By definition, $\log_{1/5}25=x$ means $x$ is the exponent with which we raise ($\displaystyle \frac{1}{5}$) to obtain $25.$ $5^{2}=25$ $(5^{-1})^{-2}=25$ $(\displaystyle \frac{1}{5})^{-2}=25$ so, $x=-2$ $\log_{1/5}25=-2$
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Circumferential force in the belt
In belt drives, forces are transmitted from the input pulley to the belt and then from the belt back to the output pulley. The force to be transmitted from one pulley to the other is also referred to as the effective force or circumferential force Fc. The amount of force Fc applied to the circumference of the input pulley depends on the torque Mi or the power Pi and the speed ni as well as the diameter di of the driving pulley:
\begin{align}
&M_i = F_c \cdot \frac{d_i}{2} ~~~~~\text{and} ~~~~~ P_i = 2 \pi M_i n_i\\[5px]
\label{umfang}
&\boxed{F_c = \frac{2 M_i}{d_i} = \frac{P_i}{\pi d_i n_i}} \\[5px]
\end{align}
This circumferential force Fc causes different forces on the two belt sections and a tight side and a slack side will form. The balance of forces on a pulley generally shows that the difference between the tight side force Ft and the slack side force Fs corresponds to the transmitting circumferential force Fc:
\begin{align}
\label{fu}
&\boxed{F_c = F_t – F_s} \\[5px]
\end{align}
The decisive factor for the transmission of power in belt drives is the circumferential force (also called effective force), which results from the difference between the tight side force and the slack side force of the belt!
At the pulley with the diameter do driven by the belt this effectively circumferential force Fc leads to a changed torque Mo (see also the section transmission ratio):
\begin{align}
&\boxed{M_o = F_c \cdot \frac{d_o}{2}} \\[5px]
\end{align}
The circumferential force is generated at the input pulley by the torque acting there and the effective diameter of the pulley. At the output pulley, the circumferential force causes a change in torque in accordance with the effective diameter of the output pulley.
The circumferential force is transmitted at the interface between belt and pulley by frictional forces. These maximum possible frictional forces must be sufficiently high to ensure that the circumferential force can be transmitted safely. If the maximum possible frictional force is lower than the circumferential force to be transmitted, then obviously there is not enough “adhesion” to be able to transfer the circumferential force from the pulley to the belt or from the belt to the pulley according to the equation (\ref{umfang}). The pulley or belt slips (sliding slip) and the full circumferential force is no longer transmitted.
The circumferential force is transmitted by frictional forces between pulley and belt!
Maximum transmittable circumferential force
In the article Basics of power transmission it was shown with the belt friction equation that the maximum frictional force Ff,max between belt and pulley at a given tight side force Ft or slack side force Fs can be determined with the formulae listed below. These maximum frictional forces therefore also represent the limit for the maximum transmittable circumferential forces Fc,max:
\begin{align}
\label{leertrum}
& F_{f,max} = \boxed{F_{c,max} = F_s \cdot \left(e^{\mu \cdot \varphi} -1 \right)} \ge F_c \\[5px]
\end{align}
or
\begin{align}
\label{zugtrum}
&F_{f,max} = \boxed{F_{c,max} =F_t \cdot \left(1-\frac{1}{e^{\mu \cdot \varphi}} \right)} \ge F_c \\[5px]
\end{align}
Gain
In the context of the circumferential force, the so-called gain k describes the percentage of the existing tight side force Ft that could be used to transmit the maximum circumferential force Fc,max (limiting case to slippage). According to equation (\ref{zugtrum}) the gain corresponds to the term (1-1/eµφ):
\begin{align}
&k = \frac{F_{cmax}}{F_t} = 1-\frac{1}{e^{\mu \cdot \varphi}} \\[5px]
\label{ausbeute}
&\boxed{k =1-\frac{1}{e^{\mu \cdot \varphi}} } \\[5px]
\end{align}
A gain of e.g. k = 0.6 means that a maximum of 60% of the tight side force can be used as circumferential force for power transmission. In this case, the remaining 40 % is required for maintaining the belt’s tension on the slack side (slack side force).
The gain indicates what percentage of the tight side force can be used as the maximum circumferential force; the rest is accounted for by the slack side force to maintain the belt tension!
Note that the circumferential force is transmitted by frictional forces and their formation always requires a certain pressure of the belt on the pulley. This means that there must always be an acting force on the slack side of the belt to ensure the belt tension and with it the contact pressure. The slack side must therefore not be force-free at all (see also the article Basics of power transmission)!
The relationship between a given tight side force Ft and the maximum possible circumferential force Fc,max is thus established by using the gain k as follows:
\begin{align}
&\boxed{F_{c,max} = F_t \cdot k } \\[5px]
\end{align}
According to the equation (\ref{ausbeute}), the gain depends only on the wrap angle and the coefficient of friction. While the coefficient of friction applies equally to both pulleys, the wrap angle of the smaller pulley is usually smaller. Thus, the smaller pulley (often the driving pulley) is decisive for the gain or the entire power transmission!
The smallest wrap angle of the pulleys is always decisive for power transmission!
However, based on the equation (\ref{ausbeute}) it can also be seen that the greater the coefficient of friction, the less influence the wrap angle has on the gain. The relatively small change in the wrap angle under load (due to the sagging of the slack side of the belt) therefore plays a subordinate role for large coefficients of friction in practice.
Elastic slippage
Due to the elasticity of the belt, elongation processes occur in the belt during rotation around the pulleys. These stretching processes are due to the increase (or decrease) in the belt tension at the transition from the slack side to the tight side (or vice versa) – the belt stretches according to the force acting on the pulley.
This results in relative motions between belt and pulley (called slastic slip). It is therefore not a static friction between belt and pulley but rather a sliding friction! Therefore, when applying the above equations, the sliding coefficient of friction should also be used instead of the static coefficient of friction.
The coefficient of sliding friction is decisive for power transmission!
However, it must be noted that the sliding friction depends on the (elongation) speed. Thus, the coefficient of sliding friction is also influenced by the belt speed!
Pre-load tension (initial tension of the belt)
The equations (\ref{leertrum}) and (\ref{zugtrum}) reflect the maximum possible, transmittable circumferential forces Fc,max depending on the slack side force Fs or tight side force Ft. These equations must therefore be interpreted at the limit to slippage. In principle, however, belt drives are not driven at the limit to slippage, but well below it.
The maximum possible circumferential force according to equation (\ref{leertrum}) or equation (\ref{zugtrum}) is therefore not fully utilized to transmit forces from the belt to the pulley and vice versa. However, the equations show that an increase in the slack side force or tight side force leads to an increase in the maximum frictional force.
This can be clearly explained by an increase in the contact pressure of the belt on the pulley, which increases the maximum frictional force. By increasing the belt forces, safety against slippage can be increased or larger circumferential forces can be transmitted.
The tension in the two spans of the belt (slack side and tight side) is influenced by the pre-tensioning of the belt (initial tension). The stronger the belt is already under tension in the load-free state by the so-called pre-load, the higher the belt tension will be during operation under load. Since higher belt tensions mean higher frictional forces, higher circumferential forces can be transmitted.
Preload without taking centrifugal forces into account
In principle, it must be borne in mind that the belt forces arising during operation depend on the circumferential force to be transmitted. In the load-free idle state, only the pre-load Fp in the belt initially acts. If a circumferential force Fc is introduced by the torque of the input pulley, the belt force in the tight side increases to Ft and the slack side force decreases to the same extent to Fs.
Due to the equilibrium of forces (more precisely: equilibrium of torques), the difference between the tight side force and the slack side force corresponds to the circumferential force to be transmitted. This means that the tight side force increases by just half the circumferential force and the slack side force decreases by half the circumferential force:
\begin{align}
&\boxed{F_t = F_p + \tfrac{F_c}{2}} \\[5px]
\label{F_L}
&\boxed{F_s = F_p – \tfrac{F_c}{2}} \\[5px]
&F_t – F_s = \left(F_p+\tfrac{F_c}{2} \right) – \left(F_p – \tfrac{F_c}{2} \right) = F_p + \tfrac{F_c}{2} – F_p + \tfrac{F_c}{2} = F_c \\[5px]
\end{align}
In accordance with the equation (\ref{leertrum}), the slack side force must under no circumstances drop to zero under load, as otherwise the belt tension would be lost. There would be no contact pressure between belt and pulley and there would be no friction that would be capable of transmitting circumferential forces.
The preload Fp in equation (\ref{F_L}) must therefore be chosen in such a way that the slack side force Fs does not fall below a critical value during operation, i.e. when the circumferential force Fc is transmitted. This critical case is reached when the slack side force has decreased to such an extent that the circumferential force Fc can just be transmitted with the maximum possible circumferential force according to the equation (\ref{leertrum}) [Fc,max=Fc]. The slack side force must therefore not fall below the following value:
\begin{align}
&F_{c} = F_{c,max} = F_{s,min} \cdot \left(e^{\mu \cdot \varphi} -1 \right) \\[5px]
& F_{s,min} = F_{c} \cdot \frac{1}{e^{\mu \cdot \varphi} -1} \\[5px]
\end{align}
With this slack side force Fs,min, which must be present at least to be able to transmit the given circumferential force Fc, the minimum preload Fp,min can then also be determined according to equation (\ref{F_L}):
\begin{align}
\label{F_V}
& F_{s,min} = F_{p,min} – \tfrac{F_c}{2} \\[5px]
& F_{p,min} = F_{L,min} + \tfrac{F_U}{2} \\[5px]
& F_{p,min} = F_{c} \cdot \frac{1}{e^{\mu \cdot \varphi} -1} + \tfrac{F_c}{2} \\[5px]
& F_{p,min} = F_{c} \cdot \left( \frac{1}{e^{\mu \cdot \varphi} -1} + \frac{1}{2} \right) \\[5px]
& \underline{F_{p,min} = F_{c} \cdot \frac{e^{\mu \cdot \varphi}+1}{2 \left(e^{\mu \cdot \varphi} -1 \right) }} ~~~\text{without taking centrifugal forces into account} \\[5px]
\end{align}
Preload with taking centrifugal forces into account
In the previous section, centrifugal forces have not yet been taken into account, but at high belt speeds v they lead to a reduction in contact pressure and thus to a reduction in frictional force. In such a case, the required circumferential force can no longer be transmitted. To compensate for this effect, the belt must be additionally preloaded by the amount of the centrifugal effect Fcf=m’⋅v² (with m’ as specific weight – “mass per unit belt length”):
\begin{align}
& F_{p,total,min} = F_{p,min} + F_cf \\[5px]
& \boxed{F_{p,total,min} = F_{c} \cdot \frac{e^{\mu \cdot \varphi}+1}{2 \left(e^{\mu \cdot \varphi} -1 \right)} + m’ \cdot v^2} ~~~\text{taking centrifugal forces into account} \\[5px]
\end{align}
Since the centrifugal force during operation is compensated by the additional preload Fcf, only the preload Fp remains as the relevant effect for contact pressure, which is therefore also referred to as dynamic preload. In contrast, the total (static) preload Fp,total also takes into account the centrifugal force to be compensated, which only acts as an additional contact force at rest. In a separate article, the concept of the belt centrifugal force is explained in more detail and the derivation of the corresponding formula is shown.
The sections above showed, that the higher the circumferential force to be transmitted and the higher the belt speed, the greater the total preload must be. However, excessive pre-tensioning forces should be avoided, as this not only leads to high bearing loads but also increases belt wear. In addition, only small circumferential forces can be transmitted with high pretensioning forces, as otherwise there is a risk that the belt will tear under the high tension.
In practice, the preload force of the belt can be measured and adjusted accordingly by vibration tests. For this the belt span is picked like the string of a guitar. The frequency at which the belt span now oscillates is called natural frequency f. In the same way that the oscillation frequency (pitch) of a guitar is determined by the tension of the string, the natural frequency of the oscillating string is also directly related to the pre-load tension.
The higher the tension of the belt, i.e. the higher the preload, the higher its natural frequency. The frequency of the oscillating belt span is finally measured with the aid of an optical measuring device. In addition to the preload Fp,total, the span length l and the specific weight m’ (“mass per meter belt length”) influence the natural frequency, but these variables are generally known in advance. The following formula can then be used to determine the preload from the measured natural frequency:
\begin{align}
&F_{p,total} = 4 \cdot f^2 \cdot m’ \cdot l^2 \\[5px]
\end{align}
Or at a desired preload, the natural frequency to be set applies:
\begin{align}
& \boxed{f = \sqrt{\frac{F_{p,total}}{4 \cdot m’ \cdot l^2}} } \\[5px]
\end{align}
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# Difference Quotient
## Difference Quotient Lesson
### What is the Difference Quotient?
The difference quotient is a method for finding the average rate of change of a function over an interval. It calculates an approximated form of a derivative.
The difference quotient is given as:
$$\frac{f(x+h)-f(x)}{h}$$
Where f(x) is the function and h is the step size. This calculates the average rate of change of the function f(x) over the interval [x, x + h]. We apply the difference quotient to our function, which creates a new function of the variables x and h.
INTRODUCING
### How to use the Difference Quotient
Let's step through an example of using the difference quotient.
Find the difference quotient of the function f(x) = 3x2 + 4. Then, determine the average rate of change on the interval x = [2, 4] and x = [5, 11].
1. First we plug the function into the difference quotient.
f(x + h)h - f(x)h = [3(x + h)2 + 4]h - [3x2 + 4]h
= [3(x + h)2 - 3x2]h
2. [3(x + h)2 - 3x2]h is our new function that we can use for calculating the average rate of change for 3x2 + 4.
3. Let's calculate for the interval x = [2, 4] first.
[3(x + h)2 - 3x2]h[3(2 + 2)2 - 3(22)]2 = 18.
4. Now, let's do the same for the interval x = [5, 11].
[3(x + h)2 - 3x2]h[3(5 + 6)2 - 3(52)]6 = 48.
### When to use the Difference Quotient
If we are given a function and must find the slope at a point, we can make an approximation by using the difference quotient. To approximate the slope, we pick our x limits on either side of the point. Imagine the point is right in the middle of the interval. The closer the interval x limits are to the point, the more accurately the difference quotient will approximate the slope at that point. In other words, a narrower interval = a more accurate approximation.
As shown earlier in the example, we also use the difference quotient to find the average rate of change over a range of x values for a function. Look out for questions that give a function and ask to find the average slope or average rate of change over an interval or range of x values. The difference quotient is especially useful when there are multiple points to perform this with because it saves time compared to using the slope formula.
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Volume of tank in pounds
E
Thread Starter
Elliott, Shane
We have a tank that is shaped like a cylinder in the lower half, and then straightens out vertically the last 12 inches. The measurements are 9.5 feet long, by 6 foot in diameter wide, and
29.5 inches tall. I need to find the volume in pounds for this item. If you could help us out it would be greatly appreciated.
e-mail us back at [email protected]--thank you.
T
Tony Firth
Shane,
Need better description of tank geometry - can only visualize tank as being rectangular with a curved bottom - in which case need to know if the curve of the bottom has a radius of 3 feet, or is it a longitudinal slice of a cylinder of width 6 feet. (In which case 6 ft is not the diameter but is the length of a chord of the circle). In any case, break the tank up into standard geometric shapes, calculate the volume of each and add them all together. e.g. The upper rectangular prism part of the tank is 9.5 x 6.0 x 1.0 = 57.0 cu ft. Multiply this volume by the weight per cubic ft, (which you do not state) and voila - you have your answer!
Hope this helps,
Tony Firth, EE,
Quester Technology Inc. Fremont CA.
D
Dave Ferguson
Pounds of what ?
Fill it with maple syrup and it has one weight and fill it with gasoline and it has another.
The second question is at what temperature would you like this calculation.
The usual example used to illustrate this is fill a 55 gallon barrel with oil in the morning and for round numbers say it weighs 100 pounds (example) and it is 60 degrees F. The barrel is full........
Now at 2pm in the afternoon it is 90 Degrees out and 5% of the "volume" is on the ground and the barrell only weighs 90 pounds but the volume would have filled a barrell plus 5%.
These numbers are not real but you get the point. This is what makes mass flowmeters so popular, weight and volume are products of temperature........
Hope I did not oversimplify this, it is a concept that I struggle with on occasion and if someone has a better explanation I am a lifetime student of my craft.
Hope this helps......
Dave Ferguson
Blandin Papser Company
UPM-Kymmene
DAVCO Automation
A
Alan Bell
Elliot,
Hi, Alan Bell Here.
I did some years ago come accross a table and calculations for volume of a dished end horizontal cylinder. I will try and find these for you. But you do need ti know the radius of curvature of the dished parts.
Bottom line was we found it so cumbersome to use that next time we filled the tank with a safe fluid, ran it off to containers on scales and kept note of the level change with each filled container. Turned out to be a lot more accurate.
Unless you are up against a real nasty compound - I'd use the fluid an scales routine. Old fashioned, but functional.
I'll get back to you
Alan
Alan N. Bell TD BSc
Process Control Engineer
+44(0)1977 712403
[email protected]
Hickson & Welch Ltd
Castleford
West Yorkshire
WF10 2JT
+44(0)1977 556565
L
Lou Heavner
Shane,
Would it be easier just to fill it with water? Meter/totalize the water as you add it and build a strapping table. Do a density
conversion to get from water weight to whatever you are trying to measure. Or are you trying to measure weight of the tank itself? Oh and just a small nit... pounds is a unit of mass, volume is measured as gallons or liters or cubic feet or cubic meters or etc....
Regards,
Lou Heavner
Emerson Performance Solutions
Austin, TX, USA
E
Edward Shmulevich The
Hello,
The simple way to measure volume of tanks of any size/shape and open/closed tanks is to sue ultrasonic portable level gage UUP1 and AUP4.
They're intended for a measurement of liquid level through one-sheeted wall of 50mm in closed/open containers.
Brief Characteristics:
Range of Levels -.2-10m
Limit of supposed intrinsic error +-3mm (.2m<H<5m)
+-(3mm+2(H-5)) (5m<H)
Method of strengthening of a converter-- magnetic
Dimensions- 200x150x100mm
Weight 2kgs
One of the advantages of this device is that it can measure thickness of liquid layer located under the layer of another liquid (gas condensate-
water, edge water-- oil , and so on). It is the only portable gage system available.
Some other gages are manufactured by Milltronics (Echomax) or BM-70 by firm Krochne. Their accuracy of level measurement of .25% and .3% with respect to thee measured level.
If you'd like more information, let me know and I'll fax it to you.
Regards,
Edward.
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# math
posted by on .
word problem:
Q1.there were 60 birds at three treesin some moment 6 birds flew away from the first tree,8 birds flew away from the second tree, and 4 birds flew away from the third tre. then there were same number of birds at each of the three trees. How many birds were there at the second tree at the begning?
(a)24 (b)22 (c) 21 (d)20
thanks,
Rosalia
• math - ,
T1: x
T2: y
T3: 60-x-y
after fly-away:
T1: x-6
T2: y-8
T3: 60-x-y - 4
so now x-6 = y-4 or y = x+2
and x-6 = 60-x-y - 4
2x + y = 62
sub in y=x+2
2x + (x+2) = 62
x = 20
then y=22
so T1 = 20
T2 = 22
T3 = 60-20-22 = 18
(nice little morning work-out)
• small typo - math - ,
line
<< so now x-6 = y-4 or y = x+2 >>
should of course say
so now x-6 = y-8 or y = x+2
### Related Questions
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# Formula to guess Month of Birthday from Singapore NRIC
## Latest Update: We have created a JavaScript App to Guess Birthday Month from NRIC
Here is a Math Formula trick to have fun with your friends, to guess their Month of Birthday given their NRIC, within two tries.
(only works for Singapore citizens born after 1970)
## The formula is: take the 3rd and 4th digit of the NRIC, put them together, divide by 10, and multiply by 3.
For an example, if a person’s NRIC is S8804xxxx, we take 04, divide by 10 to get 0.4
Then, 0.4 multiplied by 3 gives 1.2
Then, guess that the person is either born in January (round down 1.2 to 1) or February (round up 1.2 to 2). There is a high chance that you are right! Usually, round up for the first six months (Jan to Jun), and round down for the last six months (Jul to Dec).
This formula was developed and tested by me. There are some exceptions to the rule, but generally it works fine especially for people born from 1980 to 2000.
Hope you have fun with maths, and impress your friends!
## Author: mathtuition88
Math and Education Blog
## 8 thoughts on “Formula to guess Month of Birthday from Singapore NRIC”
1. Tony says:
Do you know how to derive day of birthday from singapore nric?
Liked by 1 person
1. Hi, thanks for your comment!
Unfortunately deriving the day of birthday from Singapore NRIC is not possible to my knowledge.
For deriving month of birthday, you can follow the formula listed on this site, it gives a high percentage of accuracy.
Like
2. anon says:
I tried 4 different ic number and all was wrong
Liked by 1 person
1. Ok, thanks for your feedback. The formula works best for Singapore citizens born in 1980s or 1990s. There is no 100% accurate formula to guess month of birthday.
Like
3. You need to explain how the “theory” works, or how you derived it. eg. How did you determine when to round up/round down? Is there an official link describing how the 3rd and 4th digit is derived to get an explicit month? Are you just implementing “magic” (coincidental) numbers here without any actual basis?
Anyway, i have another method using some real sg-birth stats and averages. (again, speculation only). See link below to compare:
http://jsfiddle.net/zresgru9/
Liked by 1 person
1. Nice method you have there. No doubt yours is more accurate as it uses more information. My method is a “party trick” formula that can be calculated mentally without computer, derived based on empirical evidence in a job tabulating thousands of NRIC and birth dates.
Like
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# Algebraic operations
Best of all, Algebraic operations is free to use, so there's no sense not to give it a try!
• Download full explanation
• Figure out mathematic equation
• Download full solution
## Algebraic operations
In mathematics, a basic algebraic operation is any one of the common operations of arithmetic, which include addition, subtraction, multiplication, division, raising to a whole number power, and taking roots (fractional power). These operations may be performed on numbers, in which case they are often called See more
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Algebraic Operations Addition. When two or more terms in an algebraic equation are separated by a plus sign +, the algebraic operation is Subtraction. When
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Step 1: The multiplication is done by multiplying each term of the first expression with each term of the second Step 2: Add the powers and express as exponent with the
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Operations on Algebraic Expressions Fig. 2 Math operators used for algebraic expressions (created by Geogebra) Arithmetic operators combine the letters and numbers for
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Math is the study of numbers, shapes, and patterns. It is used to solve problems and to understand the world around us.
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## Random connected area
Published on Saturday, 8th February 2020, 04:00 pm; Solved by 129
### Problem 701
Consider a rectangle made up of $W \times H$ square cells each with area 1.
Each cell is independently coloured black with probability 0.5 otherwise white. Black cells sharing an edge are assumed to be connected.
Consider the maximum area of connected cells.
Define $E(W,H)$ to be the expected value of this maximum area. For example, $E(2,2)=1.875$, as illustrated below.
You are also given $E(4, 4) = 5.76487732$, rounded to 8 decimal places.
Find $E(7, 7)$, rounded to 8 decimal places.
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# Riddle #821
## A Liar OR a Truth Teller
You're walking down a path and come to two doors. One of the doors leads to a life of prosperity and happiness, and the other door leads to a life of misery and sorrow. You don't know which door is which. In front of the door is ONE man. You know that this man either always lies, or always tells the truth, but you don't know which. The man knows which door is which. You are allowed to ask the man ONE yes-or-no question to figure out which door to go through. To make things more difficult, the man is very self-centered, so you are only allowed to ask him a question about what he thinks or knows; your question cannot involve what any other person or object (real or hypothetical) might say. What question should you ask to ensure you go through the good door?
You should ask: "If I asked you if the good door is on the left, would you say yes?" Notice that this is subtly different than asking "Is the good door on the left?", in that you are asking him IF he would say yes to that question, not what his answer to the question would be. Thus you are asking a question about a question, and if it ends up being the liar you are talking to, this will cause him to lie about a lie and thus tell the truth. The four possible cases are: The man is a truth-teller and the good door is on the left. He will say "yes". The man is a truth-teller and the good door is on the right. He will say "no". The man is a liar and the good door is on the left. He will say "yes" because if you asked him "Is the good door on the left?", he would lie and say "no", and so when you ask him if he would say "yes", he will lie and say "yes". The man is a liar and the good door is on the right. Similar to the previous example, he'll say "no". So regardless of whether the man is a truth-teller or a liar, this question will get a "yes" if the door on the left is the good door, and a "no" if it's not.
75.59 %
## Similar riddles
See also best riddles or new riddles.
## Getting wet
What gets wet while it's drying?
A towel.
78.45 %
## How many apples?
If there're 3 apples and you take away 2, how many do you have?
If you take 2 apples, than you have of course 2.
78.40 %
## You bury me when I am alive
You bury me when I am alive, and dig me up when I die. What am I?
A plant.
78.04 %
## Pirate Pete
Pete told them to form a circle around him. All the squad was facing in at Pete, ready to shoot, when they realized that everyone who missed would likely end up shooting another squad member. So no one dared to fire, knowing the risk. Thus at sundown he was released.
77.54 %
## The watchmen
One morning an airline president is leaving on a business trip and finds he left some paperwork at his office. He runs into his office to get it and the night watchman stops him and says, "Sir, don't get on the plane. I had a dream last night that the plane would crash and everyone would die!" The man takes his word and cancels his trip. Sure enough, the plane crashes and everyone dies. The next morning the man gives the watchman a \$1,000 reward for saving his life and then fires him. Why did he fire the watchman that saved his life?
He was fired from sleeping on his job.
77.49 %
## An ancient invention
What is the ancient invention that allows people to see through walls?
A window.
77.40 %
## Volume of the pizza
If you had a pizza with crust thickness 'a' and radius 'z', what's the volume of the pizza?
pi * z * z * a
77.23 %
## Mt. Everest
Before Mt. Everest was discovered, what was the highest mountain in the world?
Mt. Everest; it just wasn’t discovered yet.
77.17 %
## Take off my skin
Take off my skin - I won't cry, but you will! What am I?
An onion.
77.06 %
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Melbourne University Mathematics and Statistics Society
Sample Questions
In addition to these sample questions, you may also wish to look over the questions asked in previous Schools Maths Olympics or University Maths Olympics.
Questions
Question 1. (5 marks)
Each of the angles of a triangle when expressed in degrees is a perfect square. Find the three angles.
Question 2. (5 marks)
Call a four-digit number balanced if the sum of the two left-most digits equals the sum of the two right-most digits. For example, 2754 is balanced. How many balanced numbers are there between 1000 and 2000?
Question 3. (5 marks)
There are exactly four positive integers n such that (n+7) is a factor of (n+1)2. Find the largest such n.
Question 4. (10 marks)
Ten (not necessarily distinct) integers have the property that if all but one of them are added the possible results are: 82, 83, 84, 85, 87, 89, 90, 91, 92. What is the smallest of the integers?
Question 5. (10 marks)
A circle is inscribed in a triangle whose sides are 40, 40 and 48 cm respectively. A smaller circle is tangent to the two equal sides of the triangle and to the first circle. Find the radius, in cm, of the smaller circle.
Question 6. (10 marks)
Find the sum of all values of x which satisfy the following system of equations:
• y = x2 - 5x + 5
• z = x2 - 12x + 35
• yz = 1
where x, y, z are real numbers.
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# 16.6. Math Precision¶
## 16.6.1. Minimal and Maximal Values¶
Maximal and minimal float values:
import sys
sys.float_info.min # 2.2250738585072014e-308
sys.float_info.max # 1.7976931348623157e+308
## 16.6.2. Infinity¶
Infinity representation:
1e308 # 1e+308
-1e308 # -1e+308
1e309 # inf
-1e309 # -inf
float('inf') # inf
float('-inf') # -inf
float('Infinity') # inf
float('-Infinity') # -inf
## 16.6.3. Not-a-Number¶
float('nan')
# nan
float('-nan')
# nan
## 16.6.4. NaN vs Inf¶
float('inf') + float('inf') # inf
float('inf') + float('-inf') # nan
float('-inf') + float('inf') # nan
float('-inf') + float('-inf') # -inf
float('inf') - float('inf') # nan
float('inf') - float('-inf') # inf
float('-inf') - float('inf') # -inf
float('-inf') - float('-inf') # nan
float('inf') * float('inf') # inf
float('inf') * float('-inf') # -inf
float('-inf') * float('inf') # -inf
float('-inf') * float('-inf') # inf
float('inf') / float('inf') # nan
float('inf') / float('-inf') # nan
float('-inf') / float('inf') # nan
float('-inf') / float('-inf') # nan
## 16.6.5. Floating Numbers Precision¶
>>> 0.1
0.1
>>>
>>> 0.2
0.2
>>>
>>> 0.3
0.3
>>>
>>> 0.1 + 0.2 == 0.3
False
>>> round(0.1+0.2, 16) == 0.3
True
>>>
>>> round(0.1+0.2, 17) == 0.3
False
>>> 0.1 + 0.2
0.30000000000000004
## 16.6.6. IEEE 754 standard¶
>>> a = 1.234
>>> b = 1234 * 10e-4
>>>
>>> a == b
True
>>> 1234 * 10e-4
1.234
>>> 1.234 == 1234 * 10e-4
True
## 16.6.7. Floats in Doctest¶
def add(a, b):
"""
3.0
0.30000000000000004
>>> add(0.1, 0.2) # doctest: +ELLIPSIS
0.3000...
"""
return a + b
## 16.6.8. Decimal Type¶
from decimal import Decimal
a = Decimal('0.1')
b = Decimal('0.2')
a + b
# Decimal('0.3')
from decimal import Decimal
a = Decimal('0.3')
float(a)
# 0.3
## 16.6.9. Solutions¶
• Round values to 4 decimal places (generally acceptable)
• Store values as int, do operation and then divide. For example instead of 1.99 USD, store price as 199 US cents
• Use Decimal type
• Decimal type is much slower
Problem:
>>> candy = 0.10 # price in dollars
>>> cookie = 0.20 # price in dollars
>>>
>>> result = candy + cookie
>>> print(result)
0.30000000000000004
Round values to 4 decimal places (generally acceptable):
>>> candy = 0.10 # price in dollars
>>> cookie = 0.20 # price in dollars
>>>
>>> result = round(candy + cookie, 4)
>>> print(result)
0.3
Store values as int, do operation and then divide:
>>> candy = 10 # price in cents
>>> cookie = 20 # price in cents
>>>
>>> result = (candy + cookie) / 100 # divide by 100 (number of cents in dollar)
>>> print(result)
0.3
Use Decimal type:
>>> from decimal import Decimal
>>>
>>>
>>> candy = Decimal('0.10') # price in dollars
>>> cookie = Decimal('0.20') # price in dollars
>>>
>>> result = candy + cookie
>>> print(result)
0.30
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06 Jun 2023
# 14 C to F – Introduction, And More.
14 C to F
Introduction
Learn the temperature change of 14 C to F (Degrees Celsius to Fahrenheit) now! Converting degrees from Metric to Majestic is easy with our simple-to-use change calculator, or keep reading to learn how to change these units yourself!
14 Degrees Celsius (C) Equals 57.2 Degrees Fahrenheit (F)
or 14 C = 57.2 F
How to Convert 14 C to F (Celsius to Fahrenheit)
## C to F calculation:
• Conversion influence:
• 1 C = (1 C * 1.8) + 32 = 33.8 F
• 14 C to F Conversion Equation
• 14 C = (14 C * 1.8) +32 = 57.2 F
## Convert 14 C to Other Temperature Units
Want to convert 14 C to other temperature units? Well, here are some additional helpful temperature conversions:
• Unit 14 Celsius (C) =
• Kelvin (K) 287.15 K
• Rankine (R) 516.87 R
• Rømer (Rø) 14.85 Rø
• Newton (N) 4.62 N
## What is Celsius (C)?
Celsius (C), also referred to as centigrade, is the unit for temperature in the Metric System of Measurement. This temperature scale remains based on the freezing point of water at 0 degrees C and the boiling point at 100 degrees C.
The abbreviation symbol for Celsius is “C”. For example, 14 degrees Celsius can remain written as 14 C
## What is Fahrenheit (F)?
Fahrenheit (F) is the Imperial System of Measurement temperature unit. This system remains based on the temperature scale from physicist Daniel Gabriel Fahrenheit. In the Fahrenheit system, the freezing point of water is 32 degrees F, and the boiling point of water is 212 degrees F.
The abbreviation symbol for Fahrenheit is “F”. So, for example, 14 degrees Fahrenheit can remain written as 14 F.
Celsius is or tells to the Celsius temperature scale (previously known as the centigrade scale). The degree Celsius (symbol: °C) can mention an exact temperature on the Celsius scale and serve as a unit increment to indicate a temperature interval (a difference between two temperatures or an uncertainty). “Celsius” remains named after the Swedish astronomer Anders Celsius (1701 – 1744), who developed a similar temperature scale two years before his death. See all conversions for Celsius here.
Fahrenheit is a temperature scale called after the Polish-German physicist Daniel Gabriel Fahrenheit (1686–1736), who future it in 1724. In this scale, the freezing opinion of water is 32 degrees Fahrenheit (written “32 °F”), and the sweltering point is 212 degrees, introduction the boiling and freezing points of water are precisely 180 degrees apart. On the Celsius scale, water’s freezing and boiling opinions are specifically 100 degrees separately.
the component of this scale, a degree Fahrenheit, is 5⁄9 of a degree Celsius. Negative 40 degrees Fahrenheit (-40 °F) equals negative 40 degrees Celsius (-40 °C). See all conversions for Fahrenheit here.
Temperatures worldwide are slow either in Fahrenheit(the US temperature scale) or Celsius(the metric scale), two of the most popular and extensively used scales.
## Temperature Conversion – Degrees Celsius into Grades Fahrenheit
Celsius to Fahrenheit conversion formulation is about converting the temperature denoting Celsius to Fahrenheit. As mentioned previous, the temperature of boiling (hot) water in Celsius is 0 degrees and in Fahrenheit is 21 degrees; the formula to change C to F is
F = C x (9/5) + 32
The math here is pretty simple and can remain easily understood by an example. Let’s say we requirement to 14 Celsius to Fahrenheit!
Also Read: Is Model Cycles Polyphonic? Can you Load Samples on Model Cycles?
## How To Convert 14 C to F?
Change 14 degrees Celsius to Fahrenheit; all one requirement is to put in the standards in the converter equation- F = 14 x (9/5) +32. F = 57.2 degrees Thus, after putting on the formula to convert 14 Celcius to Fahrenheit, the answer is – 14°C = 57.2°F
Fourteen degrees Celsius generations 57.2 degrees Fahrenheit! Commonly asked questions about converting 14 Degrees Celsius into Degrees Fahrenheit How far is 14 degrees in Celsius to Fahrenheit? 14C to F = 57.2 °F.
### Conclusion
What is the formula to calculate Celsius to Fahrenheit? The C to F formula is (C × 9/5) + 32 = F When we enter 14 for C in the formula, we get (14 × 9/5) + 32 = 57.2 F To solve (14 × 9/5) + 32 = F, we first multiply nine by 14, divide the product by 5, and improve 32 to the quotient to get the response.
Also Read: Samsung SSD 2tb – Introduction, And Mores.
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A127165 a(n) = the maximum prime S possible, if S = product of b(k)'s + product of c(k)'s, where the distinct positive integers <= n are partitioned into the two sets {b(k)} and {c(k)}. a(n) = 0 if no prime S exists for that n. 3
2, 2, 3, 7, 11, 43, 149, 1013, 8069, 0, 0, 39916801, 43545611, 566092811, 7925299211, 118879488011, 1609445376013, 32335220736011, 44771844096143, 582033973248209, 221172909834240011, 3930072474746880013 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(0)=a(1)=2 because the product over the empty set is defined here as 1. For S to be a prime, the positive integers <= n, except 1 and the primes > n/2, must all be together in either {b(k)} or {c(k)}. If p is a prime where n/2 < p <= n, then it is possible that p is in either product of the S sum, as can 1. Terms calculated by W. Edwin Clark. LINKS Ray Chandler, Table of n, a(n) for n=0..100 EXAMPLE For n = 6 we have the only prime S (and so the maximum prime S) with S = 1*2*3*4*6 + 5 = 149. MATHEMATICA f[n_] := Block[{d = Divisors[Times @@ Select[Range[n], PrimeQ[ # ] && 2# > n &]]}, Select[Union[d + n!/d], PrimeQ]]; If[ # == {}, 0, Last[ # ]] & /@ Array[f, 30, 0] (* Ray Chandler, Feb 14 2007 *) CROSSREFS Cf. A127166, A128199. Sequence in context: A087384 A179283 A234850 * A100683 A153940 A049905 Adjacent sequences: A127162 A127163 A127164 * A127166 A127167 A127168 KEYWORD nonn AUTHOR Leroy Quet, Jan 06 2007 EXTENSIONS a(21)-a(35) from Ray Chandler, Feb 14 2007 STATUS approved
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Last modified April 16 16:44 EDT 2021. Contains 343050 sequences. (Running on oeis4.)
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