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[MSE] Help on a special kind of functional equation. MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 06/10/2021, 12:21 PM Hi, I know this is not directly related to tetration. It is related to iteration theory and solving functional equations. https://math.stackexchange.com/questions...alpha-beta The question is the following $f(g,\beta\alpha,x+y)=f(g^\alpha,\beta,x)f(g,\alpha,y) $ Mind that here x,y can be thought to be numbers. But the crucial part is that $\alpha$, $\beta$ and $g$ are intended to be matrices, formal powerseries or functions (non commutative objects in general) under composition. I ask also here because with JmsNxn deep familiarity with iterated composition (in the question i rendered the Omega notation as a product to not scare away normies) I could get some knowledge. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 748 Threads: 104 Joined: Dec 2010 06/11/2021, 02:10 AM (This post was last modified: 06/11/2021, 03:50 AM by JmsNxn.) Is this some kind of semi-direct product? Jesus; that would be cool... Let me think on this... I'll see if I can think of anything. Is $g^\alpha$ really supposed to be an element of $\text{Aut}$; rather than necessarily a conjugation? ........ Update: So I'm not certain I can answer this; but, did you understand when I introduced the congruent integral (if you read the paper)? This is a different form of the compositional integral that is in a "modded out space". I can reduce (I think); your question into the abelian case; but I need to know how much of the congruent integral I should explain. Also; what the solutions $f$ are referred to as is homormorphisms of the semi-direct products between two groups. In this case you are taking an inner automorphism $g^{\alpha}\in \text{InnAut}(G) \subset \text{Aut}(G)$. Then you are constructing what is typically written, $ G \propto_{g^\alpha} N\\$ And you are looking for $ f : G \propto_{g^\alpha} N \to G$ There's a word for this; I can't remember it exactly. Also, this forum doesn't have the best latex implementation; so $\propto$ should actually be the symbol $\rtimes$; which is more angular. It should look more like this ..... I'll explain this better tomorrow. Long night; but it's at least SOMETHING like this. Nonetheless; your answer lies in semi-direct products. JmsNxn Long Time Fellow Posts: 748 Threads: 104 Joined: Dec 2010 06/13/2021, 08:38 AM (This post was last modified: 06/13/2021, 09:42 AM by JmsNxn.) I finally can answer your question. I was a little confused about the details before; but it's definitely a semi-direct product. Let's first of all, ignore the function $f$. Let's write a product, $ (xy,\beta \alpha) = (\varphi(\alpha)(x),\beta)(y, \alpha)\\ x,y \in N\,\,\alpha,\beta \in G\\$ And call this group, $ N \propto_\varphi G\\$ Where $\varphi(\alpha) : G \to \text{Aut}(N)$. Where $\text{Aut}(N)$ is the group of isomorphisms of $N$. Now; there are many choices of $\varphi(\alpha)$ (something to do with Euler's phi function will be involved in the estimate of how many). Now when you write $g^{\alpha} = \alpha g \alpha^{-1} \in \text{Aut}(G)$, you are choosing an isomorphism of $\text{Aut}(G) \to \text{Aut}(N)$; let's call this $\psi : \text{Aut}(G) \to \text{Aut}(N)$. We can do this because we're only going to care about $f$ in the final result. And this is just considering $f(g,\alpha,x)$ at an implicit level in the preimage and considering it equivalent. Now, here is where I wasn't making any sense before. You wrote your equation backwards from the usual semi-direct product. The right way I should've written; which I apologize for saying. Is that it's, $ N \propto_{\psi(g)} G = N \propto_{\varphi} G\\$ You are now looking for projections, $ f(N \propto_{\psi(g)} G) \to G\\$ I fucking knew it was semi-direct products! Took me a while to think about it though... Regards, James. I hope this helps. Long story short; you have a lot of group theory at your disposal, Mphlee. May I recommend dummit & foote. MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 06/14/2021, 01:55 PM Yes! That's very promising. I'll have to work hard on this. Thank you so much! I believe that it is the solution: I just need to rephrase it, play with it a bit and prove it with my hands. This problem popped out at the intersection between the abstract Jabotinky foundation and superfunction-spaces. Can't wait to find time to work on the details. Thank you again.... nobody there on MSE even had the kindness to drop a single keyword. For group theorist this should be the abc... for sure my question is not well written, my English su*x, but I don't feel is a bad question. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 748 Threads: 104 Joined: Dec 2010 06/14/2021, 09:52 PM I think your question was written very well. It was probably more suited to Mathoverflow though; stack exchange tends to be mostly for undergraduate homework problems. But even then, the community can be a tad abrasive; and they always poke holes in technicalities while ignoring the larger picture. I deleted my stack exchange accounts a long time ago; bored by how little they actually help. Sometimes, you have to be left to your own wits . I suspect though, no one saw that it was a semi-direct product... just a weird one. Only reason I saw it was because I was thinking about bullet notation. And I was initially trying to give an example of your function. Regards, James « Next Oldest | Next Newest »
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## What is a Byte?
Keywords: Memory
Tags: General Tutorial
Views: 1734
In computer memory terms, the definition of a byte is a collection of eight bits. Unlike a bit that can hold the value of zero or one, a byte of memory can hold a value from 0-255.
Think of the below number as eight individual bits making up a byte, all turned off. The value is 0.
00000000
If we turn on the right-most bit in the byte, the 0th bit (computer scientists love to count starting from the number 0), the value is now one.
00000001
What if we want to add 1 to the value? Remember that a bit can only hold values of 0 or 1. So how is the number 2 represented? We turn on the bit to the left, the 1st bit, like so:
00000010
If the first bit is a 1 and the 0th bit is 0, the total value of all bits is 2. Now what if we want to add 1 to create a byte value of 3? Turn that 1st bit on again.
00000011
What happens if we want to create a value of 4? Remember how the 1st bit represented the number 2? The 2nd bit represents the number 4. Turn it on and turn the other two bits off.
00000100
And so forth. Here is a table:
Bit 7 - Value of 128
Bit 6 - Value of 64
Bit 5 - Value of 32
Bit 4 - Value of 16
Bit 3 - Value of 8
Bit 2 - Value of 4
Bit 1 - Value of 2
Bit 0 - Value of 1
An aside - If you know what exponentiation means, think of each bit taking on the value of two to the power of the bit number. 2 to the power of 0 is 1, 2 to the power of 1 is 2, etc.
So to get a byte value of 255, let's turn all of the bits on.
11111111
Several bytes can be strung together to form kilobytes, megabytes, gigabytes, and terabytes.
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# Solving the cubic
Document Sample
``` Solving the cubic
• Three people were involved in the solution of the cubic:
del Ferro (1465-1526), Niccolò Tartaglia (1499-1557)
and Girolamo Cardano (1501-1557). Del Ferro and
Tartaglia. There were thought to be two cases x3+cx=d
called the cosa and x3=cx+d with positive c and d.
• Del Ferro solved the cosa and kept his result secret.
• Tartaglia then first claimed to be able to solve equations
of the kind x3+bx2=d, but then also solved all cubic
equations.
• Cardano was given the formula by Tartaglia and found
and published proofs for the formulas for all cubic
equations. He was also aware of del Ferro’s solution.
Girolamo Cardano (1501-1557)
Ars Magna (The great art)
1. Let the cube be equal to the first 1. x3=cx+d
power and constant 2. Let DC, DF be cubes with sides AB
2. and let DC and DF be two cubes and BC.
3. the product of the sides of which, AB 3. s.t. (a) AB BC=1/3c.
and BC, is equal to one-third the 4. and (b) DC+DF=AB3+BC3=d.
coefficient of x, 5. Then AC=x!
4. and let the sum of these cubes be Proof:
equal to the constant.
5. I say that AC is the value of x. 6. AB BC=1/3c 3AB BC=c
6. Now since AB x BC equals one-third 7. and under the assumption AC=x,
the coefficient of x, 3(AB x BC) will AC(3ABxBC)=cx.
equal the coefficient of x, 8. Claim:
7. and the product of AC and 3(AB x BC) AC(3ABxBC)=3ABxBC2+3BCxAB2.
is the whole first power, AC having Notice that (AC=AB+BC) so that the
been assumed to be x. equation is actually clear
algebraically.
8. But AC x 3(AB x BC) makes six 9. By 7. AC(3ABxBC)=cx and by 8.
bodies, three of which are AB x BC2 AC(3ABxBC)=3ABxBC2+3BCxAB2 so
and the other three, BC x AB2. 3ABxBC2+3BCxAB2=cx
9. Therefore these six bodies are equal
to the whole first power, F E
D
A C
B
Girolamo Cardano (1501-1557)
Ars Magna (The great art)
10. and these [six bodies] plus the cubes 10. 3ABxBC2+3BCxAB2+CD+DF=AE or
DC and DF constitute the cube AE, (u+v)(3uv)=3u2v+3uv2+u3+v3=(u+v)3.
according to the first proposition of 11. DC+DF=d, this was assumption 3.
Chapter VI. 12. So from 9,10,12 AE=cx+d and if
11. The cubes DC and DF are also equal AC=x then x3=cx+d.
to the given number. 13. Remains to prove the claim 8.
12. Therefore the cube AE is equal to the 14. Enough to show
given first power and number, which AB(BC x AC) = AB x BC2+ BC x AB2
was to be proved. [by eliminating the common factor
13. It remains to be shown that 3AC(AB x 3].
BC) is equal to the six bodies. 15. Since BE=ACxBC:
14. This is clear enough if I prove that AC(ABxBC)= ABxBE
AB(BC x AC) equals the two bodies 16. Since BE=CD+DE:
ABxBC2 and BC x AB2, ABxBE=AB(CD+DE)
15. for the product of AC and (AB x BC) is 17. Since GE=AB & DG=CB:
equal to the product of AB and the AB x DE= ABxGExDG=CB x
surface BE — since all sides are equal AB 2
to all sides —
16. but this [i.e., AB x BE] is equal to the 18. So AC(ABxBC) = AB(CD+DE) =
product of AB and (CD + DE); ABxBC2+ BC x AB2 [since CD=BC2]
F E
17. the product AB x DE is equal to the
product CB x AB2, since all sides are
equal to all sides; D G
18. and therefore AC(AB x BC) is equal to
AB x BC2 plus BC x AB2, as was
proposed. A C
B
The Great Art
Short version and the rule: As Cardano says:
1. Set x=u+v The rule, therefore, is:
2. x3=(u+v)3=u3+3u2v+3uv2+v3. When the cube of one-third
3. If (1) u3+v3=d and (2) 3uv=c, so the coefficient of x is not
3u2v+3uv2=(u+v)3uv=cx. greater than the square of
4. Then x3=cx+d. one-half the constant of the
equation, subtract the former
5. From (2) v=c/(3u). from the latter and add the
6. Substituted in (1) yields square root of the remainder
u3+(c/(3u))3=d u6-du3+(c/3)3=0 to one-half the constant of
7. Set U=u3, then U2-dU+(c/3)3=0 and the equation and, again,
subtract it from the same
U d ( d ) 2 ( c )3
2 2 3
half, and you will have, as
was said, a binomium and its
8. Let V=v3, then V=d-U, so apotome, the sum of the
cube roots of which
V d ( d ) 2 ( c )3
2 2 3 constitutes the value of x
9. Finally
x 3 U 3 V 3 d
2 ( d ) 2 ( c )3 3
2 3
d
2 ( d ) 2 ( c )3
2 3
Binomial Theorem for Exponent 3
in Terms of Solids
• (u+v)3=u3+3u2v+3uv2+v3=u3+3uv(u+v)+v3
u3
v u v3 uv(u+v)
Examples
• x3=6x+40 • Problems with x3=15x+4,
– c=6, (c/3)3=23=8 the formula yields
– d=40, (d/2)2=202=400 – c=15, (c/3)3=53=125
– 400-8=392 and – d=4, (d/2)2=22=4
– 125-4=121 and se
x 3 20 392 3 20 392
x 3 2 121 3 2 121
• x3=6x+6 which contains complex
numbers. But actually there
– c=6, (c/3)3=23=8
are three real solutions:
– d=6, (d/2)2=32=9
4,-2+ √3 and -2-√3.
– 9-8=1 and thus
x 3 3 1 3 3 1 3 4 3 2 • Also what about the other
possible solutions?
Rafaeleo Bombelli (1526-1572) started to calculate cube roots of complex numbers
```
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# Amina buys a book for Rs. 275 and sells it at a loss of 15 %. By how much does she sell it for (a) Rs. 230 (b) Rs. 233.75 (c) Rs. 240 (d) Rs. 245
Grade:12th pass
## 1 Answers
Pawan Prajapati
askIITians Faculty 60796 Points
2 months ago
Our expert is working on this Class VIII Maths answer. We will update the answer very soon.
## ASK QUESTION
Get your questions answered by the expert for free
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# What do voltage limits on MOSFETs mean?
I want to include a MOSFET in a schematic I'm designing. I want to manufacture the finished board at JLCPCB, but it seems they they have a very limited set of options for MOSFETs or transistors in general.
This is a part of my schematic:
I want to turn on 4 LEDs with an Arduino. I tried doing this without a MOSFET but the output-wattage of about 2 W is way too low for driving all my LEDs (250 mW each). Here is the datasheet of the MOSFET I chose for now.
I'm not sure if this MOSFET suits my needs. The input to the gate (C01) comes from the Arduino, so 5 V. The VCC would be the same power adapter as the Arduino, so also 5 V, but with way more current available. Each LED would consume 20 mA of current according to the datasheet.
Would the AO3414 be suitable for this? The datasheet states it can handle 20 V, but I'm not sure if that means it can handle 5 V as well? The voltage from gate to source is ±8 V. Is 5 V still possible? If not, could you maybe suggest another MOSFET? If the voltage from gate to source is below 5 V, what would happen if I'd use 5 V anyway?
Edit: I changed the orientation of the MOSFET, would this setup work:
Edit: Since PStechPaul pointed out that the voltage drop is too much, I gave each LED a resistor. Would it be OK this way:
I calculated the resistance with this calculator.
• 20V is a maximum, as is 3A. Because 5 < 20, this is a safe application. Also notice Vgs(on) at Rds(on) is given for 4.5, 2.5 and 1.8V. This means it will also turn on strongly (at least up to the current listed there) even as low as 1.8V. 5 > 1.8 so it will be well turned on. Any Vgs(on) from 1.8 to 8V will do the trick, it would seem. Remember to add a pull-up resistor so the gate voltage is defined during reset/boot (before pin states have been assigned). Jul 24, 2022 at 7:14
• Not what you asked about but why not bog-standard low side MOSFET? It will greatly simplify your driving requirements. Jul 24, 2022 at 7:30
• @TimWilliams please note that the OP is proposing to put an N-channel mosfet on the high side. So I don't think the schematic as shown is correct. It will not result in the mosfet being turned on. Jul 24, 2022 at 7:43
• I see that your white LEDs have a forward voltage of 2.6 to 3.6 volts. So you can't use two in series with a 5V supply. Jul 24, 2022 at 8:12
• Could you explain what you mean by "output wattage"? The power rating of a transistor rarely matters when just using it as a switch, if that's what you mean. That refers to how much power can be safely dissipated in the transistor, not how much power it can switch. Jul 24, 2022 at 13:56
simulate this circuit – Schematic created using CircuitLab
That MOSFET will work fine but you should put it on the low side between the LEDs and GND. Not only was it upside down (source/drain swapped) but it was also on the high side of the load when it should be on the low side of the load. After you build it, you can fine-tune the LED current by adjusting R1-R4. If the current is too high, use a larger resistor. If the current is too low use a smaller resistor.
Hope that helps.
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## Operation Research with IML
Started 02-08-2017 by
Modified 02-09-2017 by
Views 2,536
Rick Wicklin's blog has encouraged me to solve via IML problems that appear in the book "Schaum's Outline of Operations Research".
I would like to share the problem expressed in the book and the IML approach I chose to solve it.
The attached images describe the problem and its solution following the logic and the recipe of the book. Furthermore they serve as a verifier for the solution I get running the IML code.
And here comes the code:
``````proc iml;
/* information about the variables */
LowerB = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; /* lower bound constraints on x */
UpperB = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}; /* upper bound constraints on x */
/* define the objective function */
c = {65, 73, 63, 57,
67, 70, 65, 58,
68, 72, 69, 55,
67, 75, 70, 59,
71, 69, 75, 57,
69, 71, 66, 59} ; /* vector for objective function c*x */
/* control vector for optimization */
ctrl = {1, /* maximize objective */
1}; /* print some details */
/* specify linear constraints */
A = {1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, /* matrix of constraint coefficients */
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,
0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0,
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0,
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0,
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1,
1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0,
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0,
0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0,
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1};
b = { 1, /* RHS of constraint eqns (column vector) */
1,
1,
1,
1,
1,
1,
1,
1,
1};
LEG = {L, L, L, L, L, L, G, G, G, G}; /* specify symbols for constraints:
'L' for less than or equal
'E' for equal
'G' for greater than or equal */
/* solve the LP problem */
CALL LPSOLVE(rc, objVal, result, dual, reducost, /* output variables */
c, A, b, /* objective and linear constraints */
ctrl, /* control vector */
LEG, /*range*/ , LowerB, UpperB);
print rc objVal, result[rowname={x11 x12 X13 X14 x21 x22 X23 X24 x31 x32 X33 X34 x41 x42 X43 X44 x51 x52 X53 X54 x61 x62 X63 X64}];``````
And it works giving me one of the optimal solutions.
I have managed to solve some other problems of the book with IML.
But I need some help figuring out how to blend the Travelling Salesman Problem with the networking optimisation described on page
http://support.sas.com/documentation/cdl/en/imlug/66845/HTML/default/viewer.htm#imlug_genstatexpls_s...
Use case is the best route, minimizing the travelled miles, that a bakery business should define taking into consideration the supply and shortage of bread in their network of bakery shops. The factory has all the output that needs to be distributed to their network depending on the demand and stock at each shop.
Help would be very welcome. On reply I'll share the code I've tried so far.
Bye, Arne
Great. Thanks for posting. I think @Ksharp would be interested in this.
If I might offer a small suggestion, you have several constant vectors that you typed explicitly. You can use the J function or the REPEAT function to simplify the construction of a constant vector. For example:
``````LowerB = j(24, 1, 0); /* lower bound constraints on x */
UpperB = j(24, 1, 1); /* upper bound constraints on x */
...
b = j(nrow(A), 1, 1); /* RHS of constraint eqns (column vector) */``````
I was curious as to whether I could generate the RONAMES= vector automatically, without typing the numbers. It turns out you can use the ROWCATC function to concatenate character matrices, and you can generate the sequence of numbers by using the EXPANDGRID function, like this:
``````rownames = "X" + strip(rowcatc(char( expandgrid(1:6, 1:4) )));
print rownames;``````
```If your OR problems were simple, you can use two functions in IML to solve it, like linear, integral programming.
Otherwise, you need the help of GA code. GA almost suited for any OR problem.
I suggest you to learn SAS/OR which can guarantee give you a solution for all the solvable OR problem.
```
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https://egvideos.com/video/georgia/grade-1/math/mgse1.nbt.6/subtract-multiples-of-10
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# Georgia - Grade 1 - Math - Numbers and Operations in Base Ten - Subtract Multiples of 10 - MGSE1.NBT.6
### Description
MGSE1.NBT.6 Subtract multiples of 10 in the range 10-90 from multiples of 10 in the range of 10-90 (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. (e.g.,70 – 30, 30 – 10, 60 – 60) MGSE1.NBT.7 Identify dimes, and understand ten pennies can be thought of as a dime. (Use dimes as manipulatives in multiple mathematical contexts.)
• State - Georgia
• Standard ID - MGSE1.NBT.6
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More Georgia Topics
MGSE1.OA.1 Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
MGSE1.G.2 Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.6 This is important for the future development of spatial relations which later connects to developing understanding of area, volume, and fractions.
MGSE1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
MGSE1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
MGSE1.OA.2 Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
MGSE1.NBT.1 Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.
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https://us.metamath.org/mpeuni/ovmptss.html
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ovmptss Structured version Visualization version GIF version
Theorem ovmptss 7778
Description: If all the values of the mapping are subsets of a class 𝑋, then so is any evaluation of the mapping. (Contributed by Mario Carneiro, 24-Dec-2016.)
Hypothesis
Ref Expression
ovmptss.1 𝐹 = (𝑥𝐴, 𝑦𝐵𝐶)
Assertion
Ref Expression
ovmptss (∀𝑥𝐴𝑦𝐵 𝐶𝑋 → (𝐸𝐹𝐺) ⊆ 𝑋)
Distinct variable groups: 𝑥,𝑦,𝐴 𝑦,𝐵 𝑥,𝑋,𝑦
Allowed substitution hints: 𝐵(𝑥) 𝐶(𝑥,𝑦) 𝐸(𝑥,𝑦) 𝐹(𝑥,𝑦) 𝐺(𝑥,𝑦)
Proof of Theorem ovmptss
Dummy variables 𝑣 𝑢 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ovmptss.1 . . . 4 𝐹 = (𝑥𝐴, 𝑦𝐵𝐶)
2 mpomptsx 7751 . . . 4 (𝑥𝐴, 𝑦𝐵𝐶) = (𝑧 𝑥𝐴 ({𝑥} × 𝐵) ↦ (1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶)
31, 2eqtri 2821 . . 3 𝐹 = (𝑧 𝑥𝐴 ({𝑥} × 𝐵) ↦ (1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶)
43fvmptss 6762 . 2 (∀𝑧 𝑥𝐴 ({𝑥} × 𝐵)(1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋 → (𝐹‘⟨𝐸, 𝐺⟩) ⊆ 𝑋)
5 vex 3444 . . . . . . . 8 𝑢 ∈ V
6 vex 3444 . . . . . . . 8 𝑣 ∈ V
75, 6op1std 7688 . . . . . . 7 (𝑧 = ⟨𝑢, 𝑣⟩ → (1st𝑧) = 𝑢)
87csbeq1d 3832 . . . . . 6 (𝑧 = ⟨𝑢, 𝑣⟩ → (1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶 = 𝑢 / 𝑥(2nd𝑧) / 𝑦𝐶)
95, 6op2ndd 7689 . . . . . . . 8 (𝑧 = ⟨𝑢, 𝑣⟩ → (2nd𝑧) = 𝑣)
109csbeq1d 3832 . . . . . . 7 (𝑧 = ⟨𝑢, 𝑣⟩ → (2nd𝑧) / 𝑦𝐶 = 𝑣 / 𝑦𝐶)
1110csbeq2dv 3835 . . . . . 6 (𝑧 = ⟨𝑢, 𝑣⟩ → 𝑢 / 𝑥(2nd𝑧) / 𝑦𝐶 = 𝑢 / 𝑥𝑣 / 𝑦𝐶)
128, 11eqtrd 2833 . . . . 5 (𝑧 = ⟨𝑢, 𝑣⟩ → (1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶 = 𝑢 / 𝑥𝑣 / 𝑦𝐶)
1312sseq1d 3946 . . . 4 (𝑧 = ⟨𝑢, 𝑣⟩ → ((1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋))
1413raliunxp 5675 . . 3 (∀𝑧 𝑢𝐴 ({𝑢} × 𝑢 / 𝑥𝐵)(1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋 ↔ ∀𝑢𝐴𝑣 𝑢 / 𝑥𝐵𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋)
15 nfcv 2955 . . . . 5 𝑢({𝑥} × 𝐵)
16 nfcv 2955 . . . . . 6 𝑥{𝑢}
17 nfcsb1v 3852 . . . . . 6 𝑥𝑢 / 𝑥𝐵
1816, 17nfxp 5553 . . . . 5 𝑥({𝑢} × 𝑢 / 𝑥𝐵)
19 sneq 4535 . . . . . 6 (𝑥 = 𝑢 → {𝑥} = {𝑢})
20 csbeq1a 3842 . . . . . 6 (𝑥 = 𝑢𝐵 = 𝑢 / 𝑥𝐵)
2119, 20xpeq12d 5551 . . . . 5 (𝑥 = 𝑢 → ({𝑥} × 𝐵) = ({𝑢} × 𝑢 / 𝑥𝐵))
2215, 18, 21cbviun 4924 . . . 4 𝑥𝐴 ({𝑥} × 𝐵) = 𝑢𝐴 ({𝑢} × 𝑢 / 𝑥𝐵)
2322raleqi 3362 . . 3 (∀𝑧 𝑥𝐴 ({𝑥} × 𝐵)(1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋 ↔ ∀𝑧 𝑢𝐴 ({𝑢} × 𝑢 / 𝑥𝐵)(1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋)
24 nfv 1915 . . . 4 𝑢𝑦𝐵 𝐶𝑋
25 nfcsb1v 3852 . . . . . 6 𝑥𝑢 / 𝑥𝑣 / 𝑦𝐶
26 nfcv 2955 . . . . . 6 𝑥𝑋
2725, 26nfss 3907 . . . . 5 𝑥𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋
2817, 27nfralw 3189 . . . 4 𝑥𝑣 𝑢 / 𝑥𝐵𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋
29 nfv 1915 . . . . . 6 𝑣 𝐶𝑋
30 nfcsb1v 3852 . . . . . . 7 𝑦𝑣 / 𝑦𝐶
31 nfcv 2955 . . . . . . 7 𝑦𝑋
3230, 31nfss 3907 . . . . . 6 𝑦𝑣 / 𝑦𝐶𝑋
33 csbeq1a 3842 . . . . . . 7 (𝑦 = 𝑣𝐶 = 𝑣 / 𝑦𝐶)
3433sseq1d 3946 . . . . . 6 (𝑦 = 𝑣 → (𝐶𝑋𝑣 / 𝑦𝐶𝑋))
3529, 32, 34cbvralw 3387 . . . . 5 (∀𝑦𝐵 𝐶𝑋 ↔ ∀𝑣𝐵 𝑣 / 𝑦𝐶𝑋)
36 csbeq1a 3842 . . . . . . 7 (𝑥 = 𝑢𝑣 / 𝑦𝐶 = 𝑢 / 𝑥𝑣 / 𝑦𝐶)
3736sseq1d 3946 . . . . . 6 (𝑥 = 𝑢 → (𝑣 / 𝑦𝐶𝑋𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋))
3820, 37raleqbidv 3354 . . . . 5 (𝑥 = 𝑢 → (∀𝑣𝐵 𝑣 / 𝑦𝐶𝑋 ↔ ∀𝑣 𝑢 / 𝑥𝐵𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋))
3935, 38syl5bb 286 . . . 4 (𝑥 = 𝑢 → (∀𝑦𝐵 𝐶𝑋 ↔ ∀𝑣 𝑢 / 𝑥𝐵𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋))
4024, 28, 39cbvralw 3387 . . 3 (∀𝑥𝐴𝑦𝐵 𝐶𝑋 ↔ ∀𝑢𝐴𝑣 𝑢 / 𝑥𝐵𝑢 / 𝑥𝑣 / 𝑦𝐶𝑋)
4114, 23, 403bitr4ri 307 . 2 (∀𝑥𝐴𝑦𝐵 𝐶𝑋 ↔ ∀𝑧 𝑥𝐴 ({𝑥} × 𝐵)(1st𝑧) / 𝑥(2nd𝑧) / 𝑦𝐶𝑋)
42 df-ov 7143 . . 3 (𝐸𝐹𝐺) = (𝐹‘⟨𝐸, 𝐺⟩)
4342sseq1i 3943 . 2 ((𝐸𝐹𝐺) ⊆ 𝑋 ↔ (𝐹‘⟨𝐸, 𝐺⟩) ⊆ 𝑋)
444, 41, 433imtr4i 295 1 (∀𝑥𝐴𝑦𝐵 𝐶𝑋 → (𝐸𝐹𝐺) ⊆ 𝑋)
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∀wral 3106 ⦋csb 3828 ⊆ wss 3881 {csn 4525 ⟨cop 4531 ∪ ciun 4882 ↦ cmpt 5111 × cxp 5518 ‘cfv 6327 (class class class)co 7140 ∈ cmpo 7142 1st c1st 7676 2nd c2nd 7677 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-sep 5168 ax-nul 5175 ax-pr 5296 ax-un 7448 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-eu 2629 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-ral 3111 df-rex 3112 df-rab 3115 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-un 3886 df-in 3888 df-ss 3898 df-nul 4244 df-if 4426 df-sn 4526 df-pr 4528 df-op 4532 df-uni 4802 df-iun 4884 df-br 5032 df-opab 5094 df-mpt 5112 df-id 5426 df-xp 5526 df-rel 5527 df-cnv 5528 df-co 5529 df-dm 5530 df-rn 5531 df-res 5532 df-ima 5533 df-iota 6286 df-fun 6329 df-fv 6335 df-ov 7143 df-oprab 7144 df-mpo 7145 df-1st 7678 df-2nd 7679 This theorem is referenced by: relmpoopab 7779 relxpchom 17430 reldv 24487
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https://schoollearningcommons.info/question/write-9-9-9-in-eponential-form-with-base-3-23110659-88/
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## Write 9×9×9 in exponential form with base 3 ?
Question
Write 9×9×9 in exponential form with base 3 ?
in progress 0
5 days 2021-09-14T11:00:27+00:00 2 Answers 0 views 0
9×9×9
3²×3²×3²
3²+²+²
3⁶
hope it helps ☺️…….
2. 9×9×9= (3)^6.
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https://a4accounting.com.au/treating-text-as-zero-in-excel/
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# Treating text as zero in Excel
Let’s say you are getting inputs you can’t control and in some cases you get text and others you get numbers. You want the numbers, but you need to treat text as zero. Here’s the easy way to do that.
There is a little used function in Excel called N. It converts text into a zero. Numbers remain the same. See the image below.
If you are dealing with either numbers or text then the humble, and extremely short, N function does the trick. The formula in cell C2 is
`=N(B2)`
### Converting text number and text dates
If you need to handle text numbers and text dates as, numbers and dates respectively, the N function doesn’t provide the solution – see image below.
The number in cell C4 is the underlying number for the date 1/1/2020. If you apply a date format to cell C4 it will look correct.
Note: Left aligned entries (when no alignment formatting has been applied) like B5 and B6 are treated as text by Excel.
As you can see rows 5 and 6 are treated as text and return zeros by the N function.
To have rows 5 and 6 return their respective number/date you need to use the IFERROR function. See image below.
The formula in cell D2 is
`=IFERROR(B2*1,0)`
Whilst not as short, this function will convert anything that looks like a number or a date into a number or a date. Multiplying a text number/date by 1 ”coerces” (forces) it into number.
If you multiply text by 1 it returns an error. Hence we used a zero on the end of the IFERROR function. Any text that can’t be converted into a number or a date will return a zero.
### Other Text Dates
Not all text dates will be converted when you multiply by 1 – see row 7 in the image below.
The IFERROR technique handles most situations, but not all.
Please note: I reserve the right to delete comments that are offensive or off-topic.
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https://www.ask.com/youtube?q=Figure+Square+Feet&qo=relatedSearchNarrow&o=0&l=dir
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To measure a floor in square feet, measure the length and width of the room, multiply those together, and subtract any islands or cut-ins that appear in the room.
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Hindi/Urdu How to Calculate Land Area in Square feet Square Yard, Square Meter ==How to Calculate Field or Land Area in Feet square (feet2), Yard Square ...
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Using basic math, you can convert inches to square feet. The principle is similar to converting feet to square feet, except you need to divide your measurement ...
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Watch more How to Buy a Home videos: http://www.howcast.com/videos/437149-How-to-Calculate-Square-Footage Whether you are calculating the square ...
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This is a video about How Many Square Feet In An Acre. 1 acre. Did I translate "How Many Square Feet In An Acre" correctly in your language? كم قدم مربع في ...
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Friday
December 2, 2016
# Posts by Rachal
Total # Posts: 97
Math 10
Cashew nuts are sold at 15 per kg. Walnuts are sold at 12 per kg. What quantities of each nut would a store owner put into a 100-kg barrel so that it could be sold for 13.20 per kg?
June 1, 2016
Algebra
Thanks!
June 8, 2011
Algebra
The function g is defined below. g(x)=((x)^(2)-9x+14)/((x)^2-11x+24) Find all values of x that are NOT in the domain of g. If there is more than one value, separate them with commas.
June 8, 2011
Algebra
Boyle's Law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4L, the pressure is 165 kPa(kilopascals). What is the volume when the pressure is 60 kPa?
June 7, 2011
Physics
Boyle's Law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4L , the pressure is 165kPa (kilopascals). What is the volume when the pressure is 60kPa?
May 23, 2011
Algebra
Miguel is choosing between two exercise routines. In Routine #1, he does only running, burning 15 calories per minute. In Routine #2, he burns 12 calories walking. He then runs at a rate that burns 12 calories per minute. For what amounts of time spent running will Routine #1 ...
May 18, 2011
Algebra
Miguel is choosing between two exercise routines. In Routine #1, he does only running, burning 15 calories per minute. In Routine #2, he burns 12 calories walking. He then runs at a rate that burns 12 calories per minute. For what amounts of time spent running will Routine #1 ...
May 18, 2011
Algebra
Austin is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Company A charges \$104.00 and allows unlimited mileage. Company B has an initial fee of \$65.00 and charges an additional \$.60 for every mile driven. For...
May 18, 2011
Algebra restate problem
The function h is is defined below h(x)=(x+6)/(x^2+9x+8) Find all values of x that are NOT in the domain of h. If there is more than one value, separate them with commas.
May 18, 2011
Algebra
The function h is is defined below. h(x)=48x^2+ 12x - 16 Find all values of x that are NOT in the domain of h. If there is more than one value, separate them with commas. I'm lost.
May 18, 2011
Algebra
The polynomial function is defined by f(x)=4x^4+2x^3-8x^-5x+2. Use a calculator to find all the points (x,f(x)) where there is a local maximum. Round to the nearest hundredth.
May 18, 2011
Algebra
Find a polynomial f(x) of degree 3 with real coefficients and the following zeros. -1,3+i f(x)=
May 17, 2011
Trig
A ramp 38' long rises to a platform. The bottom of the platform is 20' from the foot of the ramp. Find x , the angle of elevation of the ramp. Round your answer to the nearest hundredth of a degree. After this question I'm done with TRIG!!! Thanks
May 15, 2011
Trig
Find all solutions of the equation in the interval [0,2pi). sqrt(3)cottheta-1= 0 Write your answer in radians in terms of pi. If there is more than one solution, separate them with commas.
May 15, 2011
Trig
Find all solutions of the equation in the interval [0,2pi) . cos(theta)=0.7125 If there is more than one solution, separate them with commas. Do not round any intermediate computations, and round your answer(s) to the nearest hundredth. Can someone help me with this? Thanks
May 15, 2011
Trig
The polynomial function f(x) is defined by f(x)=-4x^4+9x^3+2x^2-7x-2. Use a graphing calculator to find all the points where there is a local minimum. Round to the nearest hundredth.
May 15, 2011
Trig
thanks!
May 14, 2011
Trig
I need help, I don't remember what to do. Find all solutions of the equation in the interval [0,2pi]. (2cosè+sqrt3)(cscè+1)=0 Write your answer in radians in terms of pi. If there is more than one solution, separate them with commas.
May 14, 2011
Find the unit vector in the direction of <-1,-2>. Do not approximate any numbers in your answer. I get (-1/sqrt10) and (-3/sqrt10)
May 14, 2011
I now think its 36 degrees
May 14, 2011
Find the reference angle for -144 degrees. I'm coming up with 504 degrees, but I think I'm missing a step.
May 14, 2011
Trig
Thanks!!!!
May 13, 2011
Trig
I need help solving the two problems below. Thanks For each equation, determine whether its graph is symmetric with respect to the -axis, the -axis, and the origin. Check all symmetries that apply. 1. y=-sqrt(4-x^(2)) 2. 34x^(2)+12y^(2)=18
May 13, 2011
Trig
thats supposed to be a neg in front of the 4 in problem 1.
May 13, 2011
Trig
I need help solving the two problems below. Thanks For each equation, determine whether its graph is symmetric with respect to the -axis, the -axis, and the origin. Check all symmetries that apply. 1. y=-ã(4-x^(2)) 2. 34x^(2)+12y^(2)=18
May 13, 2011
Trig
Thank you I see where I got stuck
May 12, 2011
Trig
How can I find the exact value of Sin 480 degrees? Thanks
May 12, 2011
Algebra
Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 25 liters per minute. There are 700 liters in the pond to start. Let W represent the amount of water in the pond (in liters), and let T represent the number of minutes that ...
May 12, 2011
Algebra- I get 1 out of two parts
Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 25 liters per minute. There are 700 liters in the pond to start. Let W represent the amount of water in the pond (in liters), and let T represent the number of minutes that ...
May 11, 2011
Use a cofunction to write an expression equal to csc(4ð/11). Is the answer csc(4ð/11)= 1/sin(4ð/11)
May 11, 2011
Trig
Use a cofunction to write an expression equal to csc(4ð/11). I need help solving this problem. Thank You
May 11, 2011
math- check me
I got: 2873 = 2600 e^(2.5r) ln(2873 / 2600) = 2.5r [ln(2873 / 2600)] / 2.5 = r r=0.0399381 it says to round the percentage to the nearest hundreth. any help
February 21, 2011
Math
Suppose that the number of bacteria in a certain population increases according to an exponential growth model. A sample of 2600 bacteria selected from this population reached the size of 2873 bacteria in two and a half hours. Find the continuous growth rate per hour. Write ...
February 21, 2011
Math
I got 2/21 thanks for the input
February 21, 2011
Math
A department store is holding a drawing to give free shopping sprees to two lucky customers. There are 15 customers who have entered the drawing: 4 live in the town of Gaston, 6 live in Pike, and 5 live in Wells. In the drawing, the first customer will be selected at random, ...
February 21, 2011
Math
Solve for x. 14^(-3x)=11^(-x-3) Write the exact answer using base-10 logarithms How do I start this problem? Thank You
February 16, 2011
Math
Solve for x. log3(x+7)=2-log3(x-1) Write the exact answer using base-10 logarithms.
February 15, 2011
Math
I figured it out. Equation: time + time = 60 hrs. --- x/30 + (1425-x)/15 = 60 --- x + 2*1425- 2x = 30*60 --- -x = -1050 x = 1050mL --- x/30 = 35 hrs (time of 1st storm) ----- (1425-x)/15 = 375/15 = 25 hr (time of 2nd storm)
February 15, 2011
Math
Can anybody help me start this? Thanks
February 15, 2011
math- check me
f(x)=3 -4|-11+x| -4|-11+3| -4|8| Do I divide or multiply?
February 15, 2011
math- check me
Solve the inequality for x. 16 <= 4x+4 I get x>=3 Is that right?
February 15, 2011
Math
Thank You
February 15, 2011
Math
Evaluate the following: |6-10|-|5| I get: |-4|-|5| 4-5 Is the answer -1
February 15, 2011
intermediate algebra
Write the equation of the line: The points are at (-8,-6) and (0,0). The line passing through these two points has a y-intercept of 0 since it passes through the origin. The slope is -6-0/-8-0=3/4 How would I finish this? Thanks
February 15, 2011
intermediate algebra-can somebody check my answer
Evaluate the expression m^2+93+7 when m=-4 -4^2+9*(-4)+7 does it = -13
February 15, 2011
intermediate algebra
Write the equation of the graph. Points are (-8,-6) and (0,0). I think you start at rise over run. But I'm confused thanks
February 15, 2011
Statistics- I'm stuck and its due tonight
In a particular hospital, 6 newborn babies were delivered yesterday. Here are their weights (in ounces): 112, 96, 109, 126, 99, 106 Assuming that these weights constitute an entire population, find the standard deviation of the population. Round your answer to at least two ...
February 13, 2011
Math
Frank has a 65% chance of receiving an A grade in geology, a 60% chance of receiving an A grade in mathematics, and a 77% chance of receiving an A grade in geology or mathematics (or both). Find the probability that he receives A grades in both geology and mathematics. I get: ...
February 12, 2011
Statistics - can someone check me
Frank has a 65% chance of receiving an A grade in geology, a 60% chance of receiving an A grade in mathematics, and a 77% chance of receiving an A grade in geology or mathematics (or both). Find the probability that he receives A grades in both geology and mathematics. P(geo ...
February 12, 2011
Math
You can answer any 8 questions from the 12 questions on an exam. In how many different ways can you choose the 8 questions, assuming that the order in which you choose the questions is irrelevant?
February 10, 2011
Math
I get this but i can't remember what else I do The "AND" are is shared. This is the key. Abandoning the lengthy names: A + S = 0.34 ==> S = 0.34 - A A + B = 0.30 ==> B = 0.30 - A S + B - A = 0.35
February 10, 2011
Math
Suppose that 34% of the people who inquire about investments at a certain brokerage firm end up investing in stocks, 30% end up investing in bonds, and 35% end up investing in stocks or bonds (or both). What is the probability that a person who inquires about investments at ...
February 10, 2011
Statistics-desperate
Its impossible I have alrady tried and it inserts funky symbols
February 9, 2011
Statistics-desperate
I don't know how to post some of the question is there a way I can upload it?
February 9, 2011
beginning algebra
The length of a rectangle is 1ft more than twice its width, and the area of the rectangle is 28ft^2 . Find the dimensions of the rectangle. l= w=
February 9, 2011
Rewrite the expression without using a negative exponent. 4z^-4 Simplify your answer as much as possible. is this right? (1/4z^4)
February 9, 2011
E=IR E=15*8 E=120V -------------- I=E/R I=120/4 I=30 amperes
February 9, 2011
Physics
In an electrical circuit, the current passing through a conductor varies inversely with the resistance. Suppose that when the current is 8A (amperes), the resistance is 15 ohms. What is the current when the resistance is 4 ohms?
February 9, 2011
Math
I get add -2,82,87,-1,-51,-2,53,87,-20,32,35=300 300/49=6.12244 Round two decimals = .061244 or 61.22? I don't know which way I should go.
February 8, 2011
Math
The following is a list of 11 measurements: -2,82,87,-1,-51,-2,53,87,-20,32,35 Suppose that these 11 measurements are respectively labeled. x1,x2¡¦¡¦¡¦¡¦¡¦x11.(Thus,-2 is labeled x1,82 is labeled x2 and so on.) ...
February 8, 2011
beginning algebra
Well this is a question within the subject I posted. Sorry for not labeling it right
February 8, 2011
beginning algebra
In an electrical circuit, the current passing through a conductor varies inversely with the resistance. Suppose that when the current is 8A (amperes), the resistance is 15 ohms. What is the current when the resistance is 4 ohms?
February 8, 2011
beginning algebra
Factor: -8w+uw+vu-8v
February 8, 2011
[(x^2-1)/(x-3)] / [(x^2-x-2)/(3x-9)] I get (3x-3/x-2)
February 8, 2011
can anyone clarify
if the answer should be m<=80 I don't have an equal answer option. Thanks
February 5, 2011
Charlie is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Company A charges \$127 and allows unlimited mileage. Company B has an initial fee of \$55.00 and charges an additional \$0.90 for every mile driven. For ...
February 5, 2011
a=127 b=55+.9m 127=55+.9m 72=.9m m=80
February 5, 2011
Math
Charlie is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Company A charges \$127 and allows unlimited mileage. Company B has an initial fee of \$55.00 and charges an additional \$0.90 for every mile driven. For ...
February 5, 2011
Algebra
A 3 inch thick slice is cut off the top of a cube, resulting in a rectangular box that has volume 128in^3 .Round your answer to two decimal places
February 4, 2011
Algebra
Below is the graph of a polynomial function f with real coefficients. Use the graph to answer the following questions about f. All local extrema of f are shown in the graph. I really need help with this one but I can't post the graph.
February 4, 2011
Algebra
I don't even know where to start, teacher gave us this with no direction. Suppose the polynomial R(x)=a9x^9+a8x^8+...+a,x+a0 has real coefficients with a9 cannot equal 0. Suppose also that R(x)has the following zeros: 5, 6, 5+5i. Using this information, answer the ...
February 4, 2011
Algebra
Thanks
February 3, 2011
Algebra
The number of bacteria in a certain population increases according to an exponential growth model, with a growth rate of 11% per hour. An initial sample is obtained from this population, and after four hours, the sample has grown to 4130 bacteria. Find the number of bacteria ...
February 3, 2011
Got it thanks
February 3, 2011
Choose the end behavior of the graph of each polynomial function. A f(x)= -5x^3-4x^2+8x+5 B f(x)= -4x^6+6x^4-6x^3-2x^2 C f(x)= 2x(x-1)^2(x+3) A= falls to the left and rises to the right B= Falls to the left and right C=Rises to the right and left.
February 3, 2011
Algebra
Find all complex solutions of 4x^2+5x+2=0.
February 3, 2011
Algebra-I'm still stuck
I want to thank all of you for taking the time to help me understand the problem
February 3, 2011
Algebra-I'm still stuck
The one-to-one function f is defined by f(x)=(4x-1)/(x+7). Find f^-1, the inverse of f. Then, give the domain and range of f^-1 using interval notation. f^-1(x)= Domain (f^-1)= Range (f^-1)= Any help is greatly appreciated. Algebra - helper, Wednesday, February 2, 2011 at 7:...
February 3, 2011
The one-to-one function f is defined by f(x)=(4x-1)/(x+7). Find f^-1, the inverse of f. Then, give the domain and range of f^-1 using interval notation. f^-1(x)= Domain (f^-1)= Range (f^-1)= Any help is greatly appreciated. Algebra - helper, Wednesday, February 2, 2011 at 7:...
February 2, 2011
Algebra
I don't know if this is right but this is what I came up with. f^-1=(-7x+1)/(x-4) domain f(^-1)=(-inf,-7)U(-7,inf) range f(^-1)=(-inf,4)U(4,inf) Let me know if it looks right. Thanks
February 2, 2011
Algebra
Ok I got another one I can figure out The one-to-one function f is defined by f(x)=(4x-1)/(x+7). Find f^-1, the inverse of f. Then, give the domain and range of f^-1 using interval notation. f^-1(x)= Domain (f^-1)= Range (f^-1)= Any help is greatly appreciated.
February 2, 2011
Solve the following inequality. (x-4)/(-x+2)>0 Write your answer using interval notation. x<2 or x>4 Thanks
February 2, 2011
Algebra
Graph the equations y=2|x-5|+2
January 26, 2011
Algebra
I don't understand what else do I need to do?
January 25, 2011
Algebra
Find the equation of the quadratic function f whose graph is shown below. Ponts on the graph are (-3,-1)(-2,-4) f(x)=
January 25, 2011
Math
The polynomial function f is defined by f(x)x^4-3x^3-2x^2+4x+5. Use a graphing calculator to find all the points (x,f(x)) where there is a local maximum. Round to the nearest hundredth
January 25, 2011
Algebra
The polynomial function f is defined by f(x)x^4-3x^3-2x^2+4x+5. Use a graphing calculator to find all the points (x,f(x)) where there is a local maximum. Round to the nearest hundredth.
January 24, 2011
Math-word problem
I got Let t = number of hours hall is rented then 180t + 4200 <= 5100 . Solving for t: 180t <= 900 t <= 5 hours
January 24, 2011
Math-word problem
To rent a certain banquet hall, there is a reservation fee of \$4200.00 and an additional fee of \$180.00 per hour. The Russell family wants to spend at most \$5100.00 on renting the hall. What are the possible amounts of time for which they could rent the hall? Use t for the ...
January 24, 2011
Algebra
January 23, 2011
Algebra
To rent a certain banquet hall, there is a reservation fee of \$4200.00 and an additional fee of \$180.00 per hour. The Russell family wants to spend at most \$5100.00 on renting the hall. What are the possible amounts of time for which they could rent the hall? Use t for the ...
January 23, 2011
Algebra
thanks
January 23, 2011
Algebra
The function h is defined by h(x) . 6+x/4+2x Find h(2x)
January 23, 2011
Algebra
The function h is defined by h(x) . 6+x/4+2x Find h(2x)
January 23, 2011
Algebra
Multiply. x-(6+2i)][x-(6-2i)] Note that these expressions contain complex numbers. Simplify your answer as much as possible.
January 23, 2011
Math ratio word problem
The ratio of horseshoes to shamrocks was 11 to 5, and two times the number of shamrocks was 12 less than the number of horseshoes. How many were shamrocks? thank you :)
January 3, 2009
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# 4K
Sollicitatievragen voor Management Consultant gedeeld door sollicitanten
## Meest gestelde sollicitatievragen
Sorteren: Relevantie|Populair|Datum
Er werd een Project Manager/Implementation Consultant gevraagd...19 maart 2011
### You have a bouquet of flowers. All but two are roses, all but two are daisies, and all but two are tulips. How many flowers do you have?
44 antwoorden
3 flowers - 1 rose, 1 daisy and 1 tulip
The solution is quite simple, if you start with the “All but 2” first: Roses = All but 2 = Two flowers are not a rose; one tulip, one daisy Daisies = All but 2 = Two flowers are not a daisy; one rose, one tulip Tulips = All but 2 = Two flowers are not a tulip; one rose, one daisy Answer: One rose, one daisy, one tulip. Minder
I would say, "Do you consider three flowers to be a bouquet?"
Meer reacties weergeven
### How many minutes before 5pm is it if 30 mins ago it was four times as many minutes after 3pm?
11 antwoorden
The answer is 18 minutes. It made sense to me to sketch a timeline showing the 3 components of time given in the problem that add up to the 120 minute total span. (X = minutes before 5pm, 30 min gap, and 4X is time between 3pm and the start of the 30 min gap.) Visually and chronologically it would look something like: 3pm --> 4X --> 30 min --> X --> 5pm. So then algebraically, the equation is 4X + 30 min + X = 120 min. Therefore 5X = 90 or X = 18. Minder
18 mins before 5 = 4:42. 30 before 4:42 puts the time at 4:12. There are 72 minutes between 3 and 4:42 divided by 4 is 18. So the answer is 18 mins before 5pm. Minder
That last explanation seems like you need to know the answer before you even start trying to solve. My solution is as follows: 30 minutes before 5 is 4:30 leaving 90 minutes between 3 and then. The remaining time needs to be split into an interval so that x4 exists. The most logical interval would be in 5ths because the 4 proceeding intervals would be 4x greater then the following. 90/5=18 for each interval. 18 being four times less then 72 minutes proceeding it. This literally look me about a minute and a half to reason through, which I'm assuming the interviewer would not want to sit through. Guess I would fail. Minder
Meer reacties weergeven
### An apple costs 40 cents, a banana costs 60 cents, and grapefruit costs 80 cents. Under the same circumstances, how much does a pear cost?
8 antwoorden
40 cents... it's 20 cents per vowel, not 10.
40 cents. 20 cents for each cents.
I would ask the interviewer to rank his fruit in order of what his favorite was. If a pear was his favorite I would charge 20% price increase on the grapefruit, which would put the pear at 96 cents. If he is willing to pay for grapefruit at 80 cents but he would rather have a pear, he would most likely be happy to purchase the pear at a margin increase. Pricing depends on the who, why, and where of sell/buying the product. "Under the same circumstances” can be taken creatively or mathematically depending on how you look at the circumstances. There is no right or wrong answer, it a question to see how your mind works when asked to solve a problem. I base the circumstances on the environment, the people and problem on hand. So I would want to understand more about why he wanted to know how much a pear cost, whether he was hungry, if it was for him, ect… and then appropriately price the pear based on the demand of the individual the environment and the situation. Minder
Meer reacties weergeven
### All the questions were around project management.
5 antwoorden
The position was for project analyst so yes, the questions were around project management. We went to bat all around and 2 weeks is not a long time for an offer when there are other candidates to interview. The offer wasnt declined. When we sent the offer, we were ghosted and never received a decline. ARRAY is an amazing company and I feel lucky to be here. Management is the best I have ever worked under. They make you feel like family. Minder
Shared my experience with other potential candidates who are serious about their PERSONAL INFORMATION. Minder
People make mistake. Sounds like you are hyper sensitive. My offer was a great experience. With ARRAY Minder
Meer reacties weergeven
### How did you hear about us?
3 antwoorden
Through a job board.
Again not selling yourself. Make your answers sound more enthusiastic about the job. How did you hear about us? I've seen your adds on TV, billboards, In magazines at the Dr 's office. Freestyle Cruising right? I started to wonder more about your company. Then I saw your post on (job board name here) and I got excited. This is my chance I thought. I could get my foot in the door and make a career change. I've always wondered what it would be like working for a company as well known as yours. I think that the travel and tourism industry is a perfect place to go for a career change. Minder
With that answer you might have just killed two birds with one stone. Her next question would be, why do you want to work for us? If she needs to ask that question see my comment on the next question. Remember.. you are the product.. you need to sell yourself.. It's a cruise line... they are a form of entertainment.. so be entertaining. Minder
### There are three buckets, one with apples, one with oranges, and one with a mixture of both. They are all labeled wrong. You can pick one piece of fruit from one bucket, what would you pick to determine what is in all three buckets?
4 antwoorden
I posted the question, sorry, I should have answered it. You pick a piece from the basket labeled "mixed." This is because you know it is labeled wrong (every basket is). So, if you pull out an orange you know that the basket holds only oranges. Now you have one basket figured out, and you know the remaining two are also mislabeled, so you switch their labels and you're done. Minder
The question should state that the label always lie. If the label says "Apple Only" it could be "Mixture" or "Orange only" So the one labeled "Mixture" is either "Apple Only" or "Orange Only". So you choose Mixture and what ever fruit you get is the label that is correct. Then switch the other two. Minder
All of the answers are close but not entirely correct. Any box is labeled incorrectly. Choose the mixture box. If you pick out an orange, it is necessarily an orange only box. The other two must only be the mixture or apples only. The mislabeld apples only box must be the mixture because it is mislabeled and there are only 2 other choices left. The last box is the mixture. This is the order in which you must think, although the particular fruit you pick up first could be either apple or orange. Minder
Meer reacties weergeven
3 antwoorden
Told them what I thought they wanted to hear. Exactly, what they were doing to me. Minder
I'm work on the board, as a waiter
I have (BST) basic safety training, PASSPORT, and seaman book. 6 year i'm work on the board as a waiter and second chef Minder
### A man just finished painting his house and goes to the hardware store to get something more. He finds what he is looking for and the clerk says "One for \$1." The man replies, "I need 600, so here's \$3." What did he buy?
3 antwoorden
600 is the number of his house. He bought 3 numbers.
Apparently they bought 3 of whatever it was and stole 597. This was just his fancy one-liner before he held up the store. Haha, I feel like this answer fits the description really well but it probably isn't the right one. Minder
The man needed to put his address on his house. 600 was his address. So he bought 3 numbers. A six and two zeros. Each number cost his one dollar. Minder
### "What else can you tell me that would make me interested in hiring you?"
3 antwoorden
I gave examples of previous experiences and how I, step by step, addressed a specific business problem from start to finish and I emphasized that these business skills would translate over to the interviewer's company. Minder
*fit not "fie".
Ask yourself a question: Have you ever behaved like you believe a BAH employee would? I have. I work with BAH employees, and from the outside looking in I can name a few behavioral traits that make me a good fie: Integrity, Professionalism and Competence. Read the company's core value statements and make sure to take ownership of them. Minder
### Situational questions and assessment of maturity/seniority. Questions on how I dealt with project members who did not pull their weight.
2 antwoorden
What other factors are present in their personal/ home life?
explain the steps to help resolve the problem. is it due to wrong fit/skill set? disinterest? is this their normal behavior? Minder
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# Normal
If one line is normal to another, then it is at right angles. In geometry, the normal (or a normal) refers to a line drawn perpendicular to and intersecting another line, plane or surface.
###### Light and geometry
• Light travels in a straight line through a vacuum or a transparent medium such as air, glass, or still water.
• When light encounters an obstacle or passes from one transparent medium to another, it can result in a variety of optical phenomena including absorption, dispersion, diffraction, polarization, reflection, refraction, scattering or transmission.
• Geometry can be used to calculate the outcome of light encountering different optical phenomena.
• When a normal is drawn on a ray-tracing diagram, it provides a reference against which changes in direction of light can be measured.
###### Drawing a line normal to another
• When light strikes a flat surface or plane, or the boundary between two surfaces, the normal is drawn perpendicular to, so at a right angle (900) to the surface.
• When light strikes a curved surface or plane, the normal is drawn at a tangent to the surface.
• When light strikes a sphere, the normal is drawn from the centre of the sphere and through the strike-point on the circumference.
###### References
• https://en.wikipedia.org/wiki/Geometrical_optics
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# Decrease time to upload a figure
조회 수: 2(최근 30일)
Jorge Pascual 2019년 12월 12일
댓글: Philippe Lebel 2019년 12월 12일
Hi everyone,
I am traying to develop an instrumented insole. I have 5 sensor, where I am measuring the change of voltage that it is generated when the pressure in the sensor change. As the sensor is obviolsy only in one pixel, I want to simulate that the pressure is decreasing around 3-4cm, that following my calcules are about 100pixels.
Pressure_points is a function where I enter the foot image, the radius that I have mentioned (95 pixels in this case), Paux are the coordinates of the five sensors and S are their voltage value, then I obtain an image which name es Pedobarography.
So first of all I start reading the voltage values, and if they are cero, I represent the simple foot, waiting in the second 'while', in order to not repeat every time the same figure. When a chagne in the voltage values is deected in any sensor, the process is repeated, but here is the problem. I have cheked that the function Pressure_points is not the problem because without drawing the heatmap it runs quickly, but when I try to upload the foot heatmap with the sensor changes, it runs very slow because it has to draw everything again.
To sum up, is there any way to draw the changes in the same heatmap quickly, to have these change in real time.
Jorge.
while bt.Status == 'open'
M=0;
[s1,s2,s3,s4,s5]=Voutput(bt);
if s1==0 && s2==0 && s3==0 && s4==0 && s5==0
S=zeros(1,5);
%Locate the pressure sensor, and the pressure created in each of them
R=1;
[Pedobarography]=Pressure_Points(K1,R,Paux,S);
%Pressure is drawn in mV
[X, h] = contourf(Pedobarography,'LevelList',[0.1:1]);
set(gca, 'YDir','reverse')
colorbar
caxis([0 100])
drawnow
end
while s1==0 && s2==0 && s3==0 && s4==0 && s5==0
[s1,s2,s3,s4,s5]=Voutput(bt);
M=M+1
end
while s1>0 || s2>0 || s3>0 || s4>0 || s5>0
[s1,s2,s3,s4,s5]=Voutput(bt);
S=[s1;s2;s3;s4;s5];
% Locate the pressure sensor, and the pressure created in each of them
R=95;
[Pedobarography]=Pressure_Points(K1,R,Paux,S);
% Pressure is drawn in mV
[X, h] = contourf(Pedobarography,'LevelList',[0.1:20:max(S)]);
set(gca, 'YDir','reverse')
colorbar
caxis([0 max(S)])
drawnow
end
end
##### 댓글 수: 4표시숨기기 이전 댓글 수: 3
Philippe Lebel 2019년 12월 12일
Starts with "pedo" and ends in "ography". read quickly skipped the middle of the word and made me read twice haha
댓글을 달려면 로그인하십시오.
### 답변(1개)
Philippe Lebel 2019년 12월 12일
편집: Philippe Lebel 2019년 12월 12일
set(gca, 'YDir','reverse')
from the while loop, I'm pretty sure you only need to do it once.
Maybe set a flag in the while loop to call it on the first iteration.
Samething for
colorbar
and
caxis
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Philippe Lebel 2019년 12월 12일
Publish sample data that works with your code. Ill look into it tomorrow.
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Translated by
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# Quick answer: Will The ISS Ever Fall Out Of Orbit?
## What would happen if the ISS fell out of orbit?
If total control of the ISS were lost, and it’s orbit somehow decayed to the point where it would fall into the atmosphere, the first thing that would happen is the crew would gather up their belongings and whatever materials were deemed important enough to salvage/rescue and would board one or both of the two attached ….
## At what height Earth gravity is zero?
Near the surface of the Earth (sea level), gravity decreases with height such that linear extrapolation would give zero gravity at a height of one half of the Earth’s radius – (9.8 m·s−2 per 3,200 km.)
## Is space debris dangerous?
Although debris smaller than 1 mm in size does not generally pose a hazard to spacecraft, it can still damage optics and solar arrays. So while a spacecraft may survive being hit by tiny debris, such hits can still result in catastrophe and mission failure.
## What force keeps the ISS from drifting away in space?
That’s the conceptual essence. The ISS doesn’t fall to Earth because it is moving forward at exactly the right speed that when combined with the rate it is falling, due to gravity, produces a curved path that matches the curvature of the Earth. Newton went a bit further, though. He figured out the math.
## Would a dead body decay in space?
If you do die in space, your body will not decompose in the normal way, since there is no oxygen. If you were near a source of heat, your body would mummify; if you were not, it would freeze. If your body was sealed in a space suit, it would decompose, but only for as long as the oxygen lasted.
## Why do satellites not collide with each other?
Why Don’t Satellites Crash Into Each Other? … Collisions are rare because when a satellite is launched, it is placed into an orbit designed to avoid other satellites. But orbits can change over time. And the chances of a crash increase as more and more satellites are launched into space.
## Can you see SpaceX satellites?
SpaceX, the space exploration company founded by Elon Musk, plans to put thousands of satellites into space to beam internet access down to remote parts of the Earth. There are already almost 300 Starlink satellites in orbit right now, and some are visible to the naked eye, depending on where they are.
## How long will ISS stay in orbit?
International Space StationStation statisticsOrbital period92.68 minutesOrbits per day15.54Orbit epoch14 May 2019 13:09:29 UTCDays in orbit21 years, 6 months, 27 days (16 June 2020)25 more rows
## Does the ISS get hit by debris?
In 1989, the ISS panels were predicted to degrade approximately 0.23% in four years due to the “sandblasting” effect of impacts with small orbital debris. An avoidance maneuver is typically performed for the ISS if “there is a greater than one-in-10,000 chance of a debris strike”.
## How many dead bodies are in space?
As of 2020, there have been 14 astronaut and 4 cosmonaut fatalities during spaceflight. Astronauts have also died while training for space missions, such as the Apollo 1 launch pad fire which killed an entire crew of three. There have also been some non-astronaut fatalities during spaceflight-related activities.
## Who is the owner of ISS?
The ISS programme is a joint project between five participating space agencies: NASA (United States), Roscosmos (Russia), JAXA (Japan), ESA (Europe), and CSA (Canada). The ownership and use of the space station is established by intergovernmental treaties and agreements.
## Is dying in space painful?
Asphyxiation There’s no pressure in space, so air expands and would painfully tear through the tissue of your lungs as this happened.
## Why does the ISS not fall?
The ISS doesn’t fall to Earth because it is moving forward at exactly the right speed that when combined with the rate it is falling, due to gravity, produces a curved path that matches the curvature of the Earth.
## Where is ISS now?
The International Space Station orbits 248 miles (400 kilometers) above Earth, and can be seen from the ground using a new interactive map called Spot the Station. The International Space Station completes multiple orbits around Earth every day, and now you can track the space lab as it passes overhead.
## What will happen to ISS after 2024?
After 2024, the ISS will be in the hands of NASA’s commercial and international partners, but the agency isn’t retreating from the station entirely. … NASA has published detailed guidelines for private companies wishing to apply for use of the ISS, which you can read here.
## Can a man jump from space to earth?
The Red Bull Stratos “space jump” planned by Austrian skydiver Felix Baumgartner, 43, won’t actually be from space. The Oct. 8 stunt takes aim at an altitude of almost 23 miles, or 120,000 feet (36 kilometers) — well short of the altitude where space begins, 62 miles, or 327,000 feet (100 km), above Earth.
## How does NASA avoid space junk?
The space station has orbital debris shields in place to protect from debris less than 1.5 centimeters in size. Larger debris pieces are tracked by ground control, and if needed, the space station thrusters can be used to safely move station away from the debris.
## What will replace the ISS?
Lunar Orbital Platform-Gateway (LOP-G) Formerly named the Deep Space Gateway, the LOP-G is the proposed direct replacement for the outgoing ISS. The LOP-G will orbit the moon, which will it take it far further from the Earth’s surface than the ISS’s mere 400km.
## Has anyone been lost in space?
As of 2020, there have been 15 astronaut and 4 cosmonaut fatalities during spaceflight. Astronauts have also died while training for space missions, such as the Apollo 1 launch pad fire which killed an entire crew of three.
## What would kill you first in space?
Left unchecked, the inflated bubbles will cause significant tissue damage, but other things will kill you first. In space there’s nothing to insulate you, so eventually you’ll freeze to death. But fortunately, that loss of 100 watts of heat isn’t all that much compared to the sheer mass of your body.
## Why do astronauts need to exercise two hours a day while they’re living in the space station?
On Earth, each time we move, gravity provides resistance to the muscles and bones of our body. It’s like we’re exercising without even realizing it! … That means that without exercise, the astronauts’ bones would be more fragile and their muscles weaker after time spent in space.
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https://math.stackexchange.com/questions/2012215/finding-the-limit-lim-n-to-infty-sqrtn4n9n-using-lhopitals-rule
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# Finding the limit $\lim_{n\to\infty}\sqrt[n]{4^n+9^n}$ using L'Hopital's rule [duplicate]
The problem was to find
$$\lim_{n\to\infty}\sqrt[n]{4^n+9^n}$$
So after a couple tries what I did was to take the natural logarithm of the limit so
$$\lim_{n\to\infty}\sqrt[n]{4^n+9^n}=L$$ $$\downarrow$$ $$\lim_{n\to\infty}\ln(\sqrt[n]{4^n+9^n})=\ln L$$ $$\downarrow$$ $$\lim_{n\to\infty}\frac{\ln({4^n+9^n})}{n}=\ln L$$ $$\downarrow L'Hopital$$ $$\lim_{n\to\infty}\frac{4^n\ln 4+9^n\ln 9}{4^n+9^n}=\ln L$$
And there I'm stuck. I checked in Wolfram and $\lim_{n\to\infty}$ of both the initial function and the one after L'Hopital's rule is $9$. ($\ln L=\ln 9\rightarrow L=9$).
I'd like to know how to find the limit from the last step I made, and if there's a more elegant way of solving the problem (which I'm sure there is), maybe without using L'Hôpital's rule.
Thanks.
## marked as duplicate by Guy Fsone, Parcly Taxel, muaddib, Chris Godsil, Arnaud D.Jan 28 '18 at 15:21
• If you don't have to use L'hopital, then have a look at this. – StubbornAtom Nov 13 '16 at 17:41
• In the last step, just split the thing you have on the left hand side. – 3x89g2 Nov 13 '16 at 17:42
• Easier to use $2\cdot 9^n> 4^n+9^n> 9^n$ and apply the squeeze theorem. – Thomas Andrews Nov 13 '16 at 17:42
• – Workaholic Nov 13 '16 at 18:22
If you put a factor of $9$ outside the root sign, we get $$\sqrt[n]{4^n+9^n} = 9 \cdot \sqrt[n]{(4/9)^n+1}$$
Here $(4/9)^n$ goes to $0$, and taking an $n$th root yields something even closer to $1$ than $(4/9)^n+1$.
Divide top and bottom by $9^n$ to get:
$$\lim_{n\to\infty} \frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}$$
$$\lim_{n\to\infty}\frac{4^n\ln 4+9^n\ln 9}{4^n+9^n}=\lim_{n\to\infty}\frac{(4/9)^n\ln 4+\ln 9}{(4/9)^n+1}=\ln 9$$
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## Is there something out there?
### Introduction: you said “space”?
high-dimensional sensory input vector$\qquad \overset{\text{Brain}}{\rightsquigarrow} \qquad \underbrace{\textit{space, attributes, ...}}_{\text{easier to visualize}}$
All the brain can do:
1. issue motor commands
2. observe the resulting environmental changes
then collect sensory inputs
## I. - Exteroception & Compensation
### I.A Exteroception vs. Proprioception
Sensory input Definition
Proprioceptive independent of the environment
Exteroceptive dependent of the environment
### I.B - Compensated movements
Compensated movements:
Variations of the motor command and the environment that compensate one another.
Relative distance between them is the same at steps 1 & 3
### Organism 1
Compensable movements: exactly what stems from the notion of the physical space in the sensory inputs
So the true goal: computing the dimension of the rigid group of compensated movements.
### II. Mathematical formulation
\begin{align*} \mathcal{E} &≝ \lbrace E ∈ \text{environmental states}\rbrace\\ \mathcal{M} &≝ \lbrace M ∈ \text{motor commands}\rbrace\\ \mathcal{S} &≝ \lbrace S ∈ \text{sensory inputs}\rbrace \end{align*}
are manifolds of dimension $e, m$ and $s$ respectively such that:
$$\mathcal{S} = ψ(\mathcal{M} × \mathcal{E})$$
NB: We are only considering exteroceptive inputs, i.e. points $S^e ∈ \mathcal{S}$ s.t.:
$$∃ \mathcal{M}' ⊆ \mathcal{M}; \; ψ^{-1}(S^e) = \mathcal{M}' × \mathcal{E}$$
Pushforward of $(M_0, E_0)$ by $ψ$
⟹ Tangent space at $S_0 ≝ ψ(M_0, E_0)$:
$$\lbrace dS \rbrace = \lbrace dS \rbrace_{dE=0} + \lbrace dS \rbrace_{dM=0}$$
Moreover:
• $\lbrace dS \rbrace_{dE=0}$ is the tangent space of $ψ(E_0, \mathcal{M})$ at $S_0$
• $\lbrace dS \rbrace_{dM=0}$ is the tangent space of $ψ(\mathcal{E}, M_0)$ at $S_0$
$$\mathcal{C}(M_0, E_0) ≝ ψ(\mathcal{E}, M_0) ∩ ψ(\mathcal{E}, M_0)$$
Along $\mathcal{C}(M_0, E_0)$: exteroceptive changes obtained by adding
• either $dE$
• or $dM$.
Compensated (infinitesimal) movements:
when infinitesimal changes along $\lbrace dS \rbrace_{dE=0}$ and $\lbrace dS \rbrace_{dM=0}$ compensate one another
Dimension of the space of compensated movements:
$$d ≝ \dim \underbrace{\lbrace dS_{dM=0} \mid ∃ dS_{dE=0}; dS_{dM=0} + dS_{dE=0} = 0 \rbrace}_{= \; \lbrace dS \rbrace_{dE=0} ∩ \lbrace dS \rbrace_{dM=0}} = \dim \mathcal{C}(M_0, E_0)$$
So by Grassmann formula:
\begin{align*} d \quad &≝ \quad \dim \lbrace dS \rbrace_{dE=0} ∩ \lbrace dS \rbrace_{dM=0}\\ \quad &= \quad \dim \lbrace dS \rbrace_{dE=0} + \dim \lbrace dS \rbrace_{dM=0} \\ \quad & \qquad - \dim \Big( \underbrace{\lbrace dS \rbrace_{dE=0} +\lbrace dS \rbrace_{dM=0}}_{= \lbrace dS \rbrace} \Big)\\ \\ \quad &= \quad \dim \lbrace dS \rbrace_{dE=0} + \dim \lbrace dS \rbrace_{dM=0} - \dim (\lbrace dS \rbrace) \end{align*}
## Algorithm
Get rid of proprioceptive inputs # (these don't change when no motor command is issued and the environment changes)
for "source" in [motor commands, environment, both]:
Estimate dim(space of sensory inputs resulting from "source" variations)
dim(compensated movements) = dim(inputs resulting from motor commands variations)
+ dim(inputs resulting from environment variations)
- dim(inputs resulting from both variations)
### Principal Component Analysis
Goal: Find orthogonal axes onto which the variance of the data points under projection is maximal, i.e. find the best possible “angles” from which the data points are the most spread out.
### Implementation
Neurorobotics_Project
│ index.md
│
└───sensorimotor_dependencies
│ │ __init__.py
│ │ utils.py
│ │ organisms.py
│
└───docs
│ ...
where
- utils.py ⟹ utility functions, among which dimension reduction algorithms
- organisms.py ⟹ Organism1(), Organism2(), Organism3()
### Object-Oriented Programming
class Organism1:
def __init__(self, seed=1, retina_size=1., M_size=M_size, E_size=E_size,
nb_joints=nb_joints, nb_eyes=nb_eyes, nb_lights=nb_lights,
extero=extero, proprio=proprio,
nb_generating_motor_commands=nb_generating_motor_commands,
nb_generating_env_positions=nb_generating_env_positions,
neighborhood_size=neighborhood_size, sigma=σ):
self.random = np.random.RandomState(seed)
# [...]
self.random_state = self.random.get_state()
def get_sensory_inputs(self, M, E, QPaL=None):
# [...]
def get_proprioception(self):
# [...]
def get_variations(self):
self.env_variations = ...
self.mot_variations = ...
self.env_mot_variations = ...
def get_dimensions(self, dim_red='PCA'):
self.get_proprioception()
self.get_variations()
# Now the number of degrees of freedom!
self.dim_env = dim_reduction_dict[dim_red](self.env_variations)
self.dim_extero = ...
self.dim_env_extero = ...
self.dim_rigid_group = ...
return self.dim_rigid_group, self.dim_extero, self.dim_env, self.dim_env_extero
### Other organisms
>>> O = organisms.Organism1(); O.get_dimensions()
(4, 10, 5, 11)
>>> print(str(O))
Characteristics Value
Dimension of motor commands 40
Dimension of environmental control vector 40
Dimension of proprioceptive inputs 16
Dimension of exteroceptive inputs 40
Number of eyes 2
Number of joints 4
Diaphragms None
Number of lights 3
Light luminance Fixed
Dimension for body § 10
Dimension for environment (e) 5
Dimension for both (b) 11
Dimension of group of compensated movements 4
## Some results
Varying retina_size (called var) values for different random seeds.
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https://www.studymode.com/essays/Financial-Statement-Analysis-Of-Gul-Ahmed-984581.html
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# Financial Statement Analysis of Gul Ahmed
Topics: Financial ratio, Financial ratios, Balance sheet Pages: 21 (5895 words) Published: April 27, 2012
COURSE
Financial Statement Analysis
Report on
Financial Statement Analysis of Gull Ahmed Company
SUBMITTED TO
Sir Raza Haider Sawal
SUBMITTED BY
Imran khan
Raza Khan
Taimur Shareef
BBA Hons Semester Vii Section A
INSTITUTE OF MANAGEMENT SCIENCES | PESHAWAR
Financial statement analysis
Introduction: Financial statement analysis is the art of transforming data from financial statement into information that is useful for informed decision making. Financial analysis involves the use of various financial statements. These statements do several things. First the balance sheet summarizes the asset, liabilities, and owners equity of a business at a moment in time, usually the end of a quarter or year. Income statement summarizes the revenues and expenses of the firm over a particular period of time, again usually year or at the end of a quarter. Simply the balance sheet represents the company position at a given point in time and income statement shows the summary of the firm profitability over time. The tool we are using in this project is Ratio analysis, Index analysis, and common size analysis. Ratio Analysis: Ratio analysis involves methods of calculating and interpreting financial ratios to analyze and monitor the firm performance. Types of Ratio: Financial ratio can be divided for convenience into four categories: liquidity, activity, debt, and profitability. Activity liquidity and debt ratios primarily measure the risk and profitability ratios measure the return. We are using these ratios to analyze the financial reports and the current position of the GULL AHMED Company. Liquidity Ratios: The liquidity of a firm is measured by its ability to satisfy its short term obligations as they come due or refers to the solvency of the firm overall financial position. In liquidity ratios we are finding the liquidity of the proposed company through these two ratios 1) Current ratio the most commonly cited financial ratio measure the firm ability to meet its short term obligation, its expressed as follows: Current ratio = Current assets
Current liabilities
The current ratio of the Gull Ahmed Company of year 2009 is
7359272000
7749618000
= 0.90
Time series or Trend analysis: The remaining year’s current ratios are obtained through this method. The table of five year ratio is given below: Year| 2009| 2008| 2007| 2006| 2005|
Current Ratio| 0.95| 0.9| 0.95| 1| 1.05|
Generally the higher the current ratio the more liquid the firm is considered to be but in many cases it depends upon the nature of the firm. Now it is clearly shows from the above table that the company current ratio is not good. It’s continuously declining but after year 2008 there is some little increase in current ratio. The main reason of this decline or poor performance is increasing in current liabilities. However there is also increase in current assets but the proportion of liabilities is greater than the current assets. In 2007 current liabilities is the 54.75% of the total liabilities and equity, and in 2008 it was 57.45 and in 2009 it was 57.05 %.The reason of this increase in current liabilities is short term borrowing, current maturity of long term loans and trade and other payable which is continuously increasing. The overall decrease in current asset ratios during these five years is -9.5%.
Industry analysis: the industry comparison for Gull Armed company current ratio is given below with table. years| Gull Ahmad| Industry average|
2005| 1.05| 1.10|
2006| 1| 1.15|
2007| 0.95| 1.25|
2008| 0.9| 0.95|
2009| 0.95| 0.87|
In industry analysis which is the combination of cross sectional and time series makes us to assess the trend in the behavior of the ratio in relation to the trend for the industry. As we know that higher the current ratio the better is considered. So from...
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# Number theory explained
Number theory should not be confused with Numerology.
Number theory (or arithmetic or higher arithmetic in older usage) is a branch of pure mathematics devoted primarily to the study of the integers and arithmetic functions. German mathematician Carl Friedrich Gauss (1777–1855) said, "Mathematics is the queen of the sciences—and number theory is the queen of mathematics." Number theorists study prime numbers as well as the properties of mathematical objects constructed from integers (for example, rational numbers), or defined as generalizations of the integers (for example, algebraic integers).
Integers can be considered either in themselves or as solutions to equations (Diophantine geometry). Questions in number theory are often best understood through the study of analytical objects (for example, the Riemann zeta function) that encode properties of the integers, primes or other number-theoretic objects in some fashion (analytic number theory). One may also study real numbers in relation to rational numbers, for example, as approximated by the latter (Diophantine approximation).
The older term for number theory is arithmetic. By the early twentieth century, it had been superseded by "number theory".[1] (The word "arithmetic" is used by the general public to mean "elementary calculations"; it has also acquired other meanings in mathematical logic, as in Peano arithmetic, and computer science, as in floating-point arithmetic.) The use of the term arithmetic for number theory regained some ground in the second half of the 20th century, arguably in part due to French influence.[2] In particular, arithmetical is commonly preferred as an adjective to number-theoretic.
## History
### Origins
#### Dawn of arithmetic
The earliest historical find of an arithmetical nature is a fragment of a table: the broken clay tablet Plimpton 322 (Larsa, Mesopotamia, ca. 1800 BC) contains a list of "Pythagorean triples", that is, integers
(a,b,c)
such that
a2+b2=c2
.The triples are too many and too large to have been obtained by brute force. The heading over the first column reads: "The takiltum of the diagonal which has been subtracted such that the width..."[3]
The table's layout suggests[4] that it was constructed by means of what amounts, in modern language, to the identity
\left( 1 2
\left(x-
1 x
\right)\right)2+1=\left(
1 2
\left(x+
1 x
\right)\right)2,
which is implicit in routine Old Babylonian exercises. If some other method was used,[5] the triples were first constructed and then reordered by
c/a
, presumably for actual use as a "table", for example, with a view to applications.
It is not known what these applications may have been, or whether there could have been any; Babylonian astronomy, for example, truly came into its own only later. It has been suggested instead that the table was a source of numerical examples for school problems.[6]
While Babylonian number theory—or what survives of Babylonian mathematics that can be called thus—consists of this single, striking fragment, Babylonian algebra (in the secondary-school sense of "algebra") was exceptionally well developed. Late Neoplatonic sources[7] state that Pythagoras learned mathematics from the Babylonians. Much earlier sources[8] state that Thales and Pythagoras traveled and studied in Egypt.
Euclid IX 21–34 is very probably Pythagorean;[9] it is very simple material ("odd times even is even", "if an odd number measures [= divides] an even number, then it also measures [= divides] half of it"), but it is all that is needed to prove that
is irrational. Pythagorean mystics gave great importance to the odd and the even.The discovery that
\sqrt{2}
is irrational is credited to the early Pythagoreans (pre-Theodorus).[10] By revealing (in modern terms) that numbers could be irrational, this discovery seems to have provoked the first foundational crisis in mathematical history; its proof or its divulgation are sometimes credited to Hippasus, who was expelled or split from the Pythagorean sect. This forced a distinction between numbers (integers and the rationals—the subjects of arithmetic), on the one hand, and lengths and proportions (which we would identify with real numbers, whether rational or not), on the other hand.
The Pythagorean tradition spoke also of so-called polygonal or figurate numbers. While square numbers, cubic numbers, etc., are seen now as more natural than triangular numbers, pentagonal numbers, etc., the study of the sums of triangular and pentagonal numbers would prove fruitful in the early modern period (17th to early 19th centuries).
We know of no clearly arithmetical material in ancient Egyptian or Vedic sources, though there is some algebra in each. The Chinese remainder theorem appears as an exercise[11] in Sunzi Suanjing (3rd, 4th or 5th century CE).[12] (There is one important step glossed over in Sunzi's solution:[13] it is the problem that was later solved by Āryabhaṭa's Kuṭṭaka – see below.)
There is also some numerical mysticism in Chinese mathematics,[14] but, unlike that of the Pythagoreans, it seems to have led nowhere. Like the Pythagoreans' perfect numbers, magic squares have passed from superstition into recreation.
#### Classical Greece and the early Hellenistic period
Aside from a few fragments, the mathematics of Classical Greece is known to us either through the reports of contemporary non-mathematicians or through mathematical works from the early Hellenistic period. In the case of number theory, this means, by and large, Plato and Euclid, respectively.
While Asian mathematics influenced Greek and Hellenistic learning, it seems to be the case that Greek mathematics is also an indigenous tradition.
Eusebius, PE X, chapter 4 mentions of Pythagoras:
"In fact the said Pythagoras, while busily studying the wisdom of each nation, visited Babylon, and Egypt, and all Persia, being instructed by the Magi and the priests: and in addition to these he is related to have studied under the Brahmans (these are Indian philosophers); and from some he gathered astrology, from others geometry, and arithmetic and music from others, and different things from different nations, and only from the wise men of Greece did he get nothing, wedded as they were to a poverty and dearth of wisdom: so on the contrary he himself became the author of instruction to the Greeks in the learning which he had procured from abroad."[15]
Aristotle claimed that the philosophy of Plato closely followed the teachings of the Pythagoreans,[16] and Cicero repeats this claim: Platonem ferunt didicisse Pythagorea omnia ("They say Plato learned all things Pythagorean").[17]
Plato had a keen interest in mathematics, and distinguished clearly between arithmetic and calculation. (By arithmetic he meant, in part, theorising on number, rather than what arithmetic or number theory have come to mean.) It is through one of Plato's dialogues—namely, Theaetetus—that we know that Theodorus had proven that
\sqrt{3},\sqrt{5},...,\sqrt{17}
are irrational. Theaetetus was, like Plato, a disciple of Theodorus's; he worked on distinguishing different kinds of incommensurables, and was thus arguably a pioneer in the study of number systems. (Book X of Euclid's Elements is described by Pappus as being largely based on Theaetetus's work.)
Euclid devoted part of his Elements to prime numbers and divisibility, topics that belong unambiguously to number theory and are basic to it (Books VII to IX of Euclid's Elements). In particular, he gave an algorithm for computing the greatest common divisor of two numbers (the Euclidean algorithm; Elements, Prop. VII.2) and the first known proof of the infinitude of primes (Elements, Prop. IX.20).
In 1773, Lessing published an epigram he had found in a manuscript during his work as a librarian; it claimed to be a letter sent by Archimedes to Eratosthenes. The epigram proposed what has become known asArchimedes's cattle problem
its solution (absent from the manuscript) requires solving an indeterminate quadratic equation (which reduces to what would later be misnamed Pell's equation). As far as we know, such equations were first successfully treated by the Indian school. It is not known whether Archimedes himself had a method of solution.
#### Diophantus
Very little is known about Diophantus of Alexandria; he probably lived in the third century AD, that is, about five hundred years after Euclid. Six out of the thirteen books of Diophantus's Arithmetica survive in the original Greek and four more survive in an Arabic translation. The Arithmetica is a collection of worked-out problems where the task is invariably to find rational solutions to a system of polynomial equations, usually of the form
f(x,y)=z2
or
f(x,y,z)=w2
. Thus, nowadays, we speak of Diophantine equations when we speak of polynomial equations to which rational or integer solutions must be found.
One may say that Diophantus was studying rational points, that is, points whose coordinates are rational—on curves and algebraic varieties; however, unlike the Greeks of the Classical period, who did what we would now call basic algebra in geometrical terms, Diophantus did what we would now call basic algebraic geometry in purely algebraic terms. In modern language, what Diophantus did was to find rational parametrizations of varieties; that is, given an equation of the form (say)
f(x1,x2,x3)=0
, his aim was to find (in essence) three rational functions
g1,g2,g3
such that, for all values of
r
and
s
, setting
xi=gi(r,s)
for
i=1,2,3
gives a solution to
f(x1,x2,x3)=0.
Diophantus also studied the equations of some non-rational curves, for which no rational parametrisation is possible. He managed to find some rational points on these curves (elliptic curves, as it happens, in what seems to be their first known occurrence) by means of what amounts to a tangent construction: translated into coordinate geometry(which did not exist in Diophantus's time), his method would be visualised as drawing a tangent to a curve at a known rational point, and then finding the other point of intersection of the tangent with the curve; that other point is a new rational point. (Diophantus also resorted to what could be called a special case of a secant construction.)
While Diophantus was concerned largely with rational solutions, he assumed some results on integer numbers, in particular that every integer is the sum of four squares (though he never stated as much explicitly).
#### Āryabhaṭa, Brahmagupta, Bhāskara
While Greek astronomy probably influenced Indian learning, to the point of introducing trigonometry, it seems to be the case that Indian mathematics is otherwise an indigenous tradition;[18] in particular, there is no evidence that Euclid's Elements reached India before the 18th century.
Āryabhaṭa (476–550 AD) showed that pairs of simultaneous congruences
n\equiva1\bmodm1
,
n\equiva2\bmodm2
could be solved by a method he called kuṭṭaka, or pulveriser;[19] this is a procedure close to (a generalisation of) the Euclidean algorithm, which was probably discovered independently in India. Āryabhaṭa seems to have had in mind applications to astronomical calculations.
Brahmagupta (628 AD) started the systematic study of indefinite quadratic equations—in particular, the misnamed Pell equation, in which Archimedes may have first been interested, and which did not start to be solved in the West until the time of Fermat and Euler. Later Sanskrit authors would follow, using Brahmagupta's technical terminology. A general procedure (the chakravala, or "cyclic method") for solving Pell's equation was finally found by Jayadeva (cited in the eleventh century; his work is otherwise lost); the earliest surviving exposition appears in Bhāskara II's Bīja-gaṇita (twelfth century).
Indian mathematics remained largely unknown in Europe until the late eighteenth century; Brahmagupta and Bhāskara's work was translated into English in 1817 by Henry Colebrooke.
#### Arithmetic in the Islamic golden age
In the early ninth century, the caliph Al-Ma'mun ordered translations of many Greek mathematical works and at least one Sanskrit work (the Sindhind, which may[20] or may not[21] be Brahmagupta's Brāhmasphuṭasiddhānta).Diophantus's main work, the Arithmetica, was translated into Arabic by Qusta ibn Luqa (820–912).Part of the treatise al-Fakhri (by al-Karajī, 953 – ca. 1029) builds on it to some extent. According to Rashed Roshdi, Al-Karajī's contemporary Ibn al-Haytham knew what would later be called Wilson's theorem.
#### Western Europe in the Middle Ages
Other than a treatise on squares in arithmetic progression by Fibonacci—who traveled and studied in north Africa and Constantinople—no number theory to speak of was done in western Europe during the Middle Ages. Matters started to change in Europe in the late Renaissance, thanks to a renewed study of the works of Greek antiquity. A catalyst was the textual emendation and translation into Latin of Diophantus' Arithmetica.[22]
### Early modern number theory
#### Fermat
Pierre de Fermat (1607–1665) never published his writings; in particular, his work on number theory is contained almost entirely in letters to mathematicians and in private marginal notes. In his notes and letters, he scarcely wrote any proofs—he had no models in the area.[23]
• One of Fermat's first interests was perfect numbers (which appear in Euclid, Elements IX) and amicable numbers;[24] these topics led him to work on integer divisors, which were from the beginning among the subjects of the correspondence (1636 onwards) that put him in touch with the mathematical community of the day.[25]
• In 1638, Fermat claimed, without proof, that all whole numbers can be expressed as the sum of four squares or fewer.[26]
• Fermat's little theorem (1640):[27] if a is not divisible by a prime p, then
ap-1\equiv1\bmodp.
[28]
a2+b2
is not divisible by any prime congruent to −1 modulo 4;[29] and every prime congruent to 1 modulo 4 can be written in the form
a2+b2
. These two statements also date from 1640; in 1659, Fermat stated to Huygens that he had proven the latter statement by the method of infinite descent.
• In 1657, Fermat posed the problem of solving
x2-Ny2=1
as a challenge to English mathematicians. The problem was solved in a few months by Wallis and Brouncker. Fermat considered their solution valid, but pointed out they had provided an algorithm without a proof (as had Jayadeva and Bhaskara, though Fermat was not aware of this). He stated that a proof could be found by infinite descent.
• Fermat stated and proved (by infinite descent) in the appendix to Observations on Diophantus (Obs. XLV) that
x4+y4=z4
has no non-trivial solutions in the integers. Fermat also mentioned to his correspondents that
x3+y3=z3
has no non-trivial solutions, and that this could also be proven by infinite descent. The first known proof is due to Euler (1753; indeed by infinite descent).
xn+yn=zn
for all
n\geq3
; this claim appears in his annotations in the margins of his copy of Diophantus.
#### Euler
The interest of Leonhard Euler (1707–1783) in number theory was first spurred in 1729, when a friend of his, the amateur[30] Goldbach, pointed him towards some of Fermat's work on the subject. This has been called the "rebirth" of modern number theory, after Fermat's relative lack of success in getting his contemporaries' attention for the subject.[31] Euler's work on number theory includes the following:[32]
• Proofs for Fermat's statements. This includes Fermat's little theorem (generalised by Euler to non-prime moduli); the fact that
p=x2+y2
if and only if
p\equiv1\bmod4
; initial work towards a proof that every integer is the sum of four squares (the first complete proof is by Joseph-Louis Lagrange (1770), soon improved by Euler himself); the lack of non-zero integer solutions to
x4+y4=z2
(implying the case n=4 of Fermat's last theorem, the case n=3 of which Euler also proved by a related method).
• Pell's equation, first misnamed by Euler.[33] He wrote on the link between continued fractions and Pell's equation.
• First steps towards analytic number theory. In his work of sums of four squares, partitions, pentagonal numbers, and the distribution of prime numbers, Euler pioneered the use of what can be seen as analysis (in particular, infinite series) in number theory. Since he lived before the development of complex analysis, most of his work is restricted to the formal manipulation of power series. He did, however, do some very notable (though not fully rigorous) early work on what would later be called the Riemann zeta function.[34]
• Quadratic forms. Following Fermat's lead, Euler did further research on the question of which primes can be expressed in the form
x2+Ny2
, some of it prefiguring quadratic reciprocity.
• Diophantine equations. Euler worked on some Diophantine equations of genus 0 and 1. In particular, he studied Diophantus's work; he tried to systematise it, but the time was not yet ripe for such an endeavour—algebraic geometry was still in its infancy. He did notice there was a connection between Diophantine problems and elliptic integrals, whose study he had himself initiated.
#### Lagrange, Legendre, and Gauss
Joseph-Louis Lagrange (1736–1813) was the first to give full proofs of some of Fermat's and Euler's work and observations—for instance, the four-square theorem and the basic theory of the misnamed "Pell's equation" (for which an algorithmic solution was found by Fermat and his contemporaries, and also by Jayadeva and Bhaskara II before them.) He also studied quadratic forms in full generality (as opposed to
mX2+nY2
)—defining their equivalence relation, showing how to put them in reduced form, etc.
Adrien-Marie Legendre (1752–1833) was the first to state the law of quadratic reciprocity. He also conjectured what amounts to the prime number theorem and Dirichlet's theorem on arithmetic progressions. He gave a full treatment of the equation
ax2+by2+cz2=0
and worked on quadratic forms along the lines later developed fully by Gauss. In his old age, he was the first to prove Fermat's Last Theorem for
n=5
(completing work by Peter Gustav Lejeune Dirichlet, and crediting both him and Sophie Germain).
In his Disquisitiones Arithmeticae (1798), Carl Friedrich Gauss (1777–1855) proved the law of quadratic reciprocity and developed the theory of quadratic forms (in particular, defining their composition). He also introduced some basic notation (congruences) and devoted a section to computational matters, including primality tests. The last section of the Disquisitiones established a link between roots of unity and number theory:
The theory of the division of the circle...which is treated in sec. 7 does not belong by itself to arithmetic, but its principles can only be drawn from higher arithmetic.[35]
In this way, Gauss arguably made a first foray towards both Évariste Galois's work and algebraic number theory.
### Maturity and division into subfields
Starting early in the nineteenth century, the following developments gradually took place:
• The rise to self-consciousness of number theory (or higher arithmetic) as a field of study.[36]
• The development of much of modern mathematics necessary for basic modern number theory: complex analysis, group theory, Galois theory—accompanied by greater rigor in analysis and abstraction in algebra.
• The rough subdivision of number theory into its modern subfields—in particular, analytic and algebraic number theory.
Algebraic number theory may be said to start with the study of reciprocity and cyclotomy, but truly came into its own with the development of abstract algebra and early ideal theory and valuation theory; see below. A conventional starting point for analytic number theory is Dirichlet's theorem on arithmetic progressions (1837), whose proof introduced L-functions and involved some asymptotic analysis and a limiting process on a real variable.[37] The first use of analytic ideas in number theory actually goes back to Euler (1730s), who used formal power series and non-rigorous (or implicit) limiting arguments. The use of complex analysis in number theory comes later: the work of Bernhard Riemann (1859) on the zeta function is the canonical starting point; Jacobi's four-square theorem (1839), which predates it, belongs to an initially different strand that has by now taken a leading role in analytic number theory (modular forms).[38]
The history of each subfield is briefly addressed in its own section below; see the main article of each subfield for fuller treatments. Many of the most interesting questions in each area remain open and are being actively worked on.
## Main subdivisions
### Elementary number theory
The term elementary generally denotes a method that does not use complex analysis. For example, the prime number theorem was first proven using complex analysis in 1896, but an elementary proof was found only in 1949 by Erdős and Selberg. The term is somewhat ambiguous: for example, proofs based on complex Tauberian theorems (for example, Wiener–Ikehara) are often seen as quite enlightening but not elementary, in spite of using Fourier analysis, rather than complex analysis as such. Here as elsewhere, an elementary proof may be longer and more difficult for most readers than a non-elementary one.Number theory has the reputation of being a field many of whose results can be stated to the layperson. At the same time, the proofs of these results are not particularly accessible, in part because the range of tools they use is, if anything, unusually broad within mathematics.[39]
### Analytic number theory
See main article: Analytic number theory.
Analytic number theory may be defined
• in terms of its tools, as the study of the integers by means of tools from real and complex analysis; or
• in terms of its concerns, as the study within number theory of estimates on size and density, as opposed to identities.[40]
Some subjects generally considered to be part of analytic number theory, for example, sieve theory,[41] are better covered by the second rather than the first definition: some of sieve theory, for instance, uses little analysis,[42] yet it does belong to analytic number theory.
The following are examples of problems in analytic number theory: the prime number theorem, the Goldbach conjecture (or the twin prime conjecture, or the Hardy–Littlewood conjectures), the Waring problem and the Riemann hypothesis. Some of the most important tools of analytic number theory are the circle method, sieve methods and L-functions (or, rather, the study of their properties). The theory of modular forms (and, more generally, automorphic forms) also occupies an increasingly central place in the toolbox of analytic number theory.[43]
One may ask analytic questions about algebraic numbers, and use analytic means to answer such questions; it is thus that algebraic and analytic number theory intersect. For example, one may define prime ideals (generalizations of prime numbers in the field of algebraic numbers) and ask how many prime ideals there are up to a certain size. This question can be answered by means of an examination of Dedekind zeta functions, which are generalizations of the Riemann zeta function, a key analytic object at the roots of the subject.[44] This is an example of a general procedure in analytic number theory: deriving information about the distribution of a sequence (here, prime ideals or prime numbers) from the analytic behavior of an appropriately constructed complex-valued function.[45]
### Algebraic number theory
See main article: Algebraic number theory.
An algebraic number is any complex number that is a solution to some polynomial equation
f(x)=0
with rational coefficients; for example, every solution
x
of
x5+(11/2)x3-7x2+9=0
(say) is an algebraic number. Fields of algebraic numbers are also called algebraic number fields, or shortly number fields. Algebraic number theory studies algebraic number fields. Thus, analytic and algebraic number theory can and do overlap: the former is defined by its methods, the latter by its objects of study.
It could be argued that the simplest kind of number fields (viz., quadratic fields) were already studied by Gauss, as the discussion of quadratic forms in Disquisitiones arithmeticae can be restated in terms of ideals andnorms in quadratic fields. (A quadratic field consists of allnumbers of the form
a+b\sqrt{d}
, where
a
and
b
are rational numbers and
d
is a fixed rational number whose square root is not rational.)For that matter, the 11th-century chakravala method amounts—in modern terms—to an algorithm for finding the units of a real quadratic number field. However, neither Bhāskara nor Gauss knew of number fields as such.
The grounds of the subject as we know it were set in the late nineteenth century, when ideal numbers, the theory of ideals and valuation theory were developed; these are three complementary ways of dealing with the lack of unique factorisation in algebraic number fields. (For example, in the field generated by the rationalsand
\sqrt{-5}
, the number
6
can be factorised both as
6=23
and
6=(1+\sqrt{-5})(1-\sqrt{-5})
all of
2
,
3
,
1+\sqrt{-5}
and
1-\sqrt{-5}
are irreducible, and thus, in a naïve sense, analogous to primes among the integers.) The initial impetus for the development of ideal numbers (by Kummer) seems to have come from the study of higher reciprocity laws, that is, generalisations of quadratic reciprocity.
Number fields are often studied as extensions of smaller number fields: a field L is said to be an extension of a field K if L contains K.(For example, the complex numbers C are an extension of the reals R, and the reals R are an extension of the rationals Q.)Classifying the possible extensions of a given number field is a difficult and partially open problem. Abelian extensions—that is, extensions L of K such that the Galois group[46] Gal(L/K) of L over K is an abelian group—are relatively well understood.Their classification was the object of the programme of class field theory, which was initiated in the late 19th century (partly by Kronecker and Eisenstein) and carried out largely in 1900–1950.
An example of an active area of research in algebraic number theory is Iwasawa theory. The Langlands program, one of the main current large-scale research plans in mathematics, is sometimes described as an attempt to generalise class field theory to non-abelian extensions of number fields.
### Diophantine geometry
See main article: Diophantine geometry.
The central problem of Diophantine geometry is to determine when a Diophantine equation has solutions, and if it does, how many. The approach taken is to think of the solutions of an equation as a geometric object.
For example, an equation in two variables defines a curve in the plane. More generally, an equation, or system of equations, in two or more variables defines a curve, a surface or some other such object in n-dimensional space. In Diophantine geometry, one asks whether there are any rational points (points all of whose coordinates are rationals) orintegral points (points all of whose coordinates are integers) on the curve or surface. If there are any such points, the next step is to ask how many there are and how they are distributed. A basic question in this direction is if there are finitelyor infinitely many rational points on a given curve (or surface).
x2+y2=1,
we would like to study its rational solutions, that is, its solutions
(x,y)
such that x and y are both rational. This is the same as asking for all integer solutionsto
a2+b2=c2
; any solution to the latter equation gives us a solution
x=a/c
,
y=b/c
to the former. It is also thesame as asking for all points with rational coordinates on the curve described by
x2+y2=1
. (This curve happens to be a circle of radius 1 around the origin.)
The rephrasing of questions on equations in terms of points on curves turns out to be felicitous. The finiteness or not of the number of rational or integer points on an algebraic curve—that is, rational or integer solutions to an equation
f(x,y)=0
, where
f
is a polynomial in two variables—turns out to depend crucially on the genus of the curve. The genus can be defined as follows:[47] allow the variables in
f(x,y)=0
to be complex numbers; then
f(x,y)=0
defines a 2-dimensional surface in (projective) 4-dimensional space (since two complex variables can be decomposed into four real variables, that is, four dimensions). If we count the number of (doughnut) holes in the surface; we call this number the genus of
f(x,y)=0
. Other geometrical notions turn out to be just as crucial.
There is also the closely linked area of Diophantine approximations: given a number
x
, then finding how well can it be approximated by rationals. (We are looking for approximations that are good relative to the amount of space that it takes to write the rational: call
a/q
(with
\gcd(a,q)=1
) a good approximation to
x
if
|x-a/q|< 1 qc
, where
c
is large.) This question is of special interest if
x
is an algebraic number. If
x
cannot be well approximated, then some equations do not have integer or rational solutions. Moreover, several concepts (especially that of height) turn out to be critical both in Diophantine geometry and in the study of Diophantine approximations. This question is also of special interest in transcendental number theory: if a number can be better approximated than any algebraic number, then it is a transcendental number. It is by this argument that and e have been shown to be transcendental.
Diophantine geometry should not be confused with the geometry of numbers, which is a collection of graphical methods for answering certain questions in algebraic number theory. Arithmetic geometry, however, is a contemporary term for much the same domain as that covered by the term Diophantine geometry. The term arithmetic geometry is arguably used most often when one wishes to emphasise the connections to modern algebraic geometry (as in, for instance, Faltings's theorem) rather than to techniques in Diophantine approximations.
## Other subfields
The areas below date from no earlier than the mid-twentieth century, even if they are based on older material. For example, as is explained below, the matter of algorithms in number theory is very old, in some sense older than the concept of proof; at the same time, the modern study of computability dates only from the 1930s and 1940s, and computational complexity theory from the 1970s.
### Probabilistic number theory
See main article: Probabilistic number theory.
Much of probabilistic number theory can be seen as an important special case of the study of variables that are almost, but not quite, mutually independent. For example, the event that a random integer between one and a million be divisible by two and the event that it be divisible by three are almost independent, but not quite.
It is sometimes said that probabilistic combinatorics uses the fact that whatever happens with probability greater than
0
must happen sometimes; one may say with equal justice that many applications of probabilistic number theory hinge on the fact that whatever is unusual must be rare. If certain algebraic objects (say, rational or integer solutions to certain equations) can be shown to be in the tail of certain sensibly defined distributions, it follows that there must be few of them; this is a very concrete non-probabilistic statement following from a probabilistic one.
At times, a non-rigorous, probabilistic approach leads to a number of heuristic algorithms and open problems, notably Cramér's conjecture.
### Arithmetic combinatorics
See main article: Arithmetic combinatorics and Additive number theory.
A
, does it contain many elements in arithmetic progression:
a
,
a+b,a+2b,a+3b,\ldots,a+10b
, say? Should it be possible to write large integers as sums of elements of
A
?
These questions are characteristic of arithmetic combinatorics. This is a presently coalescing field; it subsumes additive number theory (which concerns itself with certain very specific sets
A
of arithmetic significance, such as the primes or the squares) and, arguably, some of the geometry of numbers, together with some rapidly developing new material. Its focus on issues of growth and distribution accounts in part for its developing links with ergodic theory, finite group theory, model theory, and other fields. The term additive combinatorics is also used; however, the sets
A
being studied need not be sets of integers, but rather subsets of non-commutative groups, for which the multiplication symbol, not the addition symbol, is traditionally used; they can also be subsets of rings, in which case the growth of
A+A
and
A
·
A
may be compared.
### Computational number theory
See main article: Computational number theory. While the word algorithm goes back only to certain readers of al-Khwārizmī, careful descriptions of methods of solution are older than proofs: such methods (that is, algorithms) are as old as any recognisable mathematics—ancient Egyptian, Babylonian, Vedic, Chinese—whereas proofs appeared only with the Greeks of the classical period.An early case is that of what we now call the Euclidean algorithm. In its basic form (namely, as an algorithm for computing the greatest common divisor) it appears as Proposition 2 of Book VII in Elements, together with a proof of correctness. However, in the form that is often used in number theory (namely, as an algorithm for finding integer solutions to an equation
ax+by=c
, or, what is the same, for finding the quantities whose existence is assured by the Chinese remainder theorem) it first appears in the works of Āryabhaṭa (5th–6th century CE) as an algorithm called kuṭṭaka ("pulveriser"), without a proof of correctness.
There are two main questions: "Can we compute this?" and "Can we compute it rapidly?" Anyone can test whether a number is prime or, if it is not, split it into prime factors; doing so rapidly is another matter. We now know fast algorithms for testing primality, but, in spite of much work (both theoretical and practical), no truly fast algorithm for factoring.
The difficulty of a computation can be useful: modern protocols for encrypting messages (for example, RSA) depend on functions that are known to all, but whose inverses are known only to a chosen few, and would take one too long a time to figure out on one's own. For example, these functions can be such that their inverses can be computed only if certain large integers are factorized. While many difficult computational problems outside number theory are known, most working encryption protocols nowadays are based on the difficulty of a few number-theoretical problems.
Some things may not be computable at all; in fact, this can be proven in some instances. For instance, in 1970, it was proven, as a solution to Hilbert's tenth problem, that there is no Turing machine which can solve all Diophantine equations.[48] In particular, this means that, given a computably enumerable set of axioms, there are Diophantine equations for which there is no proof, starting from the axioms, of whether the set of equations has or does not have integer solutions. (We would necessarily be speaking of Diophantine equations for which there are no integer solutions, since, given a Diophantine equation with at least one solution, the solution itself provides a proof of the fact that a solution exists. We cannot prove that a particular Diophantine equation is of this kind, since this would imply that it has no solutions.)
## Applications
The number-theorist Leonard Dickson (1874–1954) said "Thank God that number theory is unsullied by any application". Such a view is no longer applicable to number theory.[49] In 1974, Donald Knuth said "virtually every theorem in elementary number theory arises in a natural, motivated way in connection with the problem of making computers do high-speed numerical calculations".[50] Elementary number theory is taught in discrete mathematics courses for computer scientists; on the other hand, number theory also has applications to the continuous in numerical analysis.[51]
Number theory has now several modern applications spanning diverse areas such as:
• Cryptography: Public-key encryption schemes such as RSA are based on the difficulty of factoring large composite numbers into their prime factors.[52]
• Computer science: The fast Fourier transform (FFT) algorithm, which is used to efficiently compute the discrete Fourier transform, has important applications in signal processing and data analysis.[53]
• Physics: The Riemann hypothesis has connections to the distribution of prime numbers and has been studied for its potential implications in physics.[45]
• Error correction codes: The theory of finite fields and algebraic geometry have been used to construct efficient error-correcting codes.[54]
• Communications: The design of cellular telephone networks requires knowledge of the theory of modular forms, which is a part of analytic number theory.
• Study of musical scales: the concept of "equal temperament", which is the basis for most modern Western music, involves dividing the octave into 12 equal parts.[55] This has been studied using number theory and in particular the properties of the 12th root of 2.
## Prizes
The American Mathematical Society awards the Cole Prize in Number Theory. Moreover, number theory is one of the three mathematical subdisciplines rewarded by the Fermat Prize.
## Sources
• Book: Apostol , Tom M. . Tom M. Apostol. 1976. Introduction to analytic number theory. Undergraduate Texts in Mathematics. Springer. 978-0-387-90163-3. 2016-02-28.
• Apostol. Tom M.. n.d.. An Introduction to the Theory of Numbers. (Review of Hardy & Wright.) Mathematical Reviews (MathSciNet). 0568909. American Mathematical Society. 2016-02-28. 2016-03-01. https://web.archive.org/web/20160301103120/http://www.ams.org/mathscinet/. live. (Subscription needed)
• Becker. Oskar. 1936. Oskar Becker. de. Die Lehre von Geraden und Ungeraden im neunten Buch der euklidischen Elemente. Quellen und Studien zur Geschichte der Mathematik, Astronomie und Physik. Abteilung B:Studien. 3. 533–553.
• Book: Boyer. Carl Benjamin. Merzbach. Uta C.. Uta Merzbach. 1991. Carl Benjamin Boyer. A History of Mathematics. 2nd. 1968. New York. Wiley. 978-0-471-54397-8. 1968 edition at archive.org
• Book: Clark. Walter Eugene (trans.) . Aryabhata. Aryabhata-->. 1930. The Āryabhaṭīya of Āryabhaṭa: An ancient Indian work on Mathematics and Astronomy. University of Chicago Press. 2016-02-28.
• Book: Colebrooke , Henry Thomas . 1817. Henry Thomas Colebrooke. Algebra, with Arithmetic and Mensuration, from the Sanscrit of Brahmegupta and Bháscara.. London. J. Murray. 2016-02-28.
• Book: Davenport. Harold. Harold Davenport. 2000. Montgomery. Hugh L.. Hugh Montgomery (mathematician). Multiplicative Number Theory. revised 3rd. Graduate Texts in Mathematics. 74. Springer. 978-0-387-95097-6.
• Edwards. Harold M.. Harold Edwards (mathematician). November 1983. Euler and Quadratic Reciprocity. Mathematics Magazine. 56. 5. 285–291. 2690368. 10.2307/2690368.
• Book: Edwards , Harold M. . 2000. 1977. Fermat's Last Theorem: a Genetic Introduction to Algebraic Number Theory. reprint of 1977. Graduate Texts in Mathematics. 50. Springer Verlag. 978-0-387-95002-0.
• Book: Fermat , Pierre de . 1679. Pierre de Fermat. fr, la. Varia Opera Mathematica. Toulouse. Joannis Pech. 2016-02-28.
• Friberg. Jöran. August 1981. Methods and Traditions of Babylonian Mathematics: Plimpton 322, Pythagorean Triples and the Babylonian Triangle Parameter Equations. Historia Mathematica. 8. 3. 277–318. 10.1016/0315-0860(81)90069-0. free.
• Book: von Fritz , Kurt . Christianidis. J.. 2004. The Discovery of Incommensurability by Hippasus of Metapontum. Classics in the History of Greek Mathematics. Berlin. Kluwer (Springer). 978-1-4020-0081-2.
• Book: Gauss. . Carl Friedrich. Waterhouse. William C. (trans.). 1966. Carl Friedrich Gauss. Disquisitiones Arithmeticae. 1801. Springer. 978-0-387-96254-2.
• Web site: Goldfeld. Dorian M.. Dorian M. Goldfeld. 2003. Elementary Proof of the Prime Number Theorem: a Historical Perspective. 2016-02-28. 2016-03-03. https://web.archive.org/web/20160303234413/http://www.math.columbia.edu/~goldfeld/ErdosSelbergDispute.pdf. live.
• Book: Goldstein. Catherine. Catherine Goldstein. Schappacher. Norbert. 2007. Goldstein. C.. Schappacher. N.. Schwermer. Joachim. A book in search of a discipline. The Shaping of Arithmetic after C.F. Gauss's "Disquisitiones Arithmeticae". Berlin & Heidelberg. Springer. 978-3-540-20441-1. https://books.google.com/books?id=IUFTcOsMTysC. 3–66. 2016-02-28.
• Book: Granville , Andrew . Andrew Granville. 2008. Gowers. Timothy. Timothy Gowers. Barrow-Green. June. Leader. Imre. Imre Leader. Analytic number theory. The Princeton Companion to Mathematics. Princeton University Press. 978-0-691-11880-2. https://books.google.com/books?id=ZOfUsvemJDMC&pg=PA332. 2016-02-28. The Princeton Companion to Mathematics.
• Book: . Guthrie. K.S. (trans.). Kenneth Sylvan Guthrie. 1920. Porphyry. Porphyry (philosopher). Life of Pythagoras. Alpine, New Jersey. Platonist Press. 2012-04-10. 2020-02-29. https://web.archive.org/web/20200229061904/http://www.tertullian.org/fathers/porphyry_life_of_pythagoras_02_text.htm. live.
• Book: Guthrie , Kenneth Sylvan . 1987. Kenneth Sylvan Guthrie. The Pythagorean Sourcebook and Library. Grand Rapids, Michigan. Phanes Press. 978-0-933999-51-0.
• Book: Hardy. Godfrey Harold. G. H. Hardy. Wright. E.M.. An Introduction to the Theory of Numbers. 1938. Oxford University Press. 6th. 978-0-19-921986-5. 2445243. 2008. An Introduction to the Theory of Numbers.
• Book: Heath , Thomas L. . 1921. Thomas Little Heath. A History of Greek Mathematics, Volume 1: From Thales to Euclid. Oxford. Clarendon Press. 2016-02-28.
• Book: Hopkins , J.F.P. . Young. M.J.L.. Latham. J.D.. Serjeant. R.B.. 1990. Geographical and Navigational Literature. Religion, Learning and Science in the 'Abbasid Period. The Cambridge history of Arabic literature. Cambridge University Press. 978-0-521-32763-3.
• Encyclopedia: Huffman. Carl A.. Zalta. Edward N.. 8 August 2011. Pythagoras. Stanford Encyclopaedia of Philosophy. Fall 2011. 7 February 2012. 2 December 2013. https://web.archive.org/web/20131202071830/http://plato.stanford.edu/archives/fall2011/entries/pythagoras/. live.
• Book: Iwaniec. Henryk. Henryk Iwaniec. Kowalski. Emmanuel. 2004. Analytic Number Theory. American Mathematical Society Colloquium Publications. 53. Providence, RI. American Mathematical Society. 978-0-8218-3633-0.
• Book: . Jowett. Benjamin (trans.). Benjamin Jowett. 1871. Plato. Plato. Theaetetus. 2012-04-10. 2011-07-09. https://web.archive.org/web/20110709194524/http://classics.mit.edu/Plato/theatu.html. live.
• Book: Lam. Lay Yong. Ang. Tian Se. 2004. Lam Lay Yong. Fleeting Footsteps: Tracing the Conception of Arithmetic and Algebra in Ancient China. revised. Singapore. World Scientific. 978-981-238-696-0. 2016-02-28.
• Book: Long , Calvin T. . 1972. Elementary Introduction to Number Theory. 2nd. D.C. Heath and Company. Lexington, VA. 77171950 .
• Book: Mahoney , M.S. . 1994. The Mathematical Career of Pierre de Fermat, 1601–1665. Reprint, 2nd. Princeton University Press. 978-0-691-03666-3. 2016-02-28.
• Book: . Morrow. Glenn Raymond (trans., ed.). 1992. Proclus. Proclus. A Commentary on Book 1 of Euclid's Elements. Princeton University Press. 978-0-691-02090-7.
• Mumford. David. David Mumford. Mathematics in India: reviewed by David Mumford. Notices of the American Mathematical Society. March 2010. 57. 3. 387. 1088-9477. 2021-04-28. 2021-05-06. https://web.archive.org/web/20210506220302/https://www.ams.org/notices/201003/rtx100300385p.pdf. live.
• Book: Neugebauer , Otto E. . 1969. Otto E. Neugebauer. The Exact Sciences in Antiquity. Acta Historica Scientiarum Naturalium et Medicinalium. 9. 1–191. corrected reprint of the 1957. New York. Dover Publications. 14884919. 978-0-486-22332-2. 2016-03-02. 2023-03-01. https://web.archive.org/web/20230301144241/https://books.google.com/books?id=JVhTtVA2zr8C. live.
• Book: . Neugebauer. Otto E.. Sachs. Abraham Joseph. Götze. Albrecht. 1945. Otto E. Neugebauer. Abraham Sachs. Mathematical Cuneiform Texts. American Oriental Series. 29. American Oriental Society etc..
• Web site: O'Grady. Patricia. Patricia O'Grady. September 2004. Thales of Miletus. The Internet Encyclopaedia of Philosophy. 7 February 2012. 6 January 2016. https://web.archive.org/web/20160106182825/http://www.iep.utm.edu/thales/. live.
• . Pingree. David. Ya'qub. ibn Tariq. David Pingree. Yaʿqūb ibn Ṭāriq. 1968. The Fragments of the Works of Ya'qub ibn Tariq. Journal of Near Eastern Studies. 26.
• . Pingree. D.. David Pingree. al-Fazari. 1970. al-Fazari. The Fragments of the Works of al-Fazari. Journal of Near Eastern Studies. 28.
• Book: Plofker, Kim. Kim Plofker. 2008. Mathematics in India. Mathematics in India (book). Princeton University Press. 978-0-691-12067-6.
• Book: Qian. Baocong. 1963. zh. Suanjing shi shu (Ten Mathematical Classics). Beijing. Zhonghua shuju. 2016-02-28. 2013-11-02. https://web.archive.org/web/20131102154812/http://www.scribd.com/doc/53797787/Jigu-Suanjing%E3%80%80%E7%B7%9D%E5%8F%A4%E7%AE%97%E7%B6%93-Qian-Baocong-%E9%8C%A2%E5%AF%B6%E7%90%AE. live.
• Rashed. Roshdi. 1980. Ibn al-Haytham et le théorème de Wilson. Archive for History of Exact Sciences. 22. 4. 305–321. 10.1007/BF00717654. 120885025.
• Robson. Eleanor. Eleanor Robson. 2001. Neither Sherlock Holmes nor Babylon: a Reassessment of Plimpton 322. 28. Historia Mathematica. 3. 167–206. 10.1006/hmat.2001.2317. https://web.archive.org/web/20141021070742/http://www.hps.cam.ac.uk/people/robson/neither-sherlock.pdf. 2014-10-21.
• Book: Sachau. Eduard. Eduard Sachau. Bīrūni. ̄Muḥammad ibn Aḥmad. Abū Rayḥān al-Bīrūnī. 1888. Alberuni's India: An Account of the Religion, Philosophy, Literature, Geography, Chronology, Astronomy and Astrology of India, Vol. 1. London. Kegan, Paul, Trench, Trübner & Co.. 2016-02-28. 2016-03-03. https://web.archive.org/web/20160303235035/http://onlinebooks.library.upenn.edu/webbin/book/lookupname?key=Sachau,%20Eduard,%201845-1930. live.
• Book: Serre , Jean-Pierre . 1996. 1973. Jean-Pierre Serre. A Course in Arithmetic. Graduate Texts in Mathematics. 7. Springer. 978-0-387-90040-7.
• Book: Smith , D.E. . 1958. History of Mathematics, Vol I. New York. Dover Publications.
• Book: . Tannery. Paul. Paul Tannery. Charles Henry . Charles Henry (librarian). 1891. Fermat. Pierre de. Pierre de Fermat. fr, la. Oeuvres de Fermat. (4 Vols.). Paris. Imprimerie Gauthier-Villars et Fils. Volume 1 Volume 2 Volume 3 Volume 4 (1912)
• Book: . Taylor. Thomas (trans.). Thomas Taylor (neoplatonist). 1818. Iamblichus. Iamblichus. Life of Pythagoras or, Pythagoric Life. London. J.M. Watkins. bot: unknown. https://web.archive.org/web/20110721184914/http://www.aurumsolis.info/index.php?option=com_phocadownload&view=category&download=1%3Aiamblichus-the-pythagorean-life&id=19%3Awritings-from-the-founders&Itemid=143&lang=en. 2011-07-21. For other editions, see Iamblichus#List of editions and translations
• Book: Truesdell , C.A. . Clifford Truesdell. 1984. Hewlett. John (trans.). Leonard Euler, Supreme Geometer. Leonard Euler, Elements of Algebra. reprint of 1840 5th. New York. Springer-Verlag. 978-0-387-96014-2. https://books.google.com/books?id=mkOhy6v7kIsC. This Google books preview of Elements of algebra lacks Truesdell's intro, which is reprinted (slightly abridged) in the following book:
• Book: Truesdell , C.A. . Clifford Truesdell. 2007. Dunham. William. Leonard Euler, Supreme Geometer. The Genius of Euler: reflections on his life and work. Volume 2 of MAA tercentenary Euler celebration. New York. Mathematical Association of America. 978-0-88385-558-4. https://books.google.com/books?id=M4-zUnrSxNoC. 2016-02-28.
• Book: Varadarajan , V.S. . 2006. Euler Through Time: A New Look at Old Themes. American Mathematical Society. 978-0-8218-3580-7. 2016-02-28.
• Vardi. Ilan. Archimedes' Cattle Problem. April 1998. American Mathematical Monthly. 105. 4. 305–319. 10.2307/2589706. 2589706. 10.1.1.383.545. 2012-04-08. 2012-07-15. https://web.archive.org/web/20120715031904/https://www.cs.drexel.edu/~crorres/Archimedes/Cattle/cattle_vardi.pdf. live.
• Book: . van der Waerden. Bartel L.. Dresden. Arnold (trans). 1961. Bartel Leendert van der Waerden. Science Awakening. 1 or 2. New York. Oxford University Press.
Two of the most popular introductions to the subject are:
• Book: G.H. Hardy . G. H. Hardy. E.M. Wright . An introduction to the theory of numbers . 2008 . 1938. . rev. by D.R. Heath-Brown and J.H. Silverman, 6th. 978-0-19-921986-5 . 2016-03-02.
• Book: Vinogradov , I.M. . Ivan Matveyevich Vinogradov. Elements of Number Theory . Mineola, NY. Dover Publications . 2003. 1954 . reprint of the 1954.
Hardy and Wright's book is a comprehensive classic, though its clarity sometimes suffers due to the authors' insistence on elementary methods (Apostol n.d.).Vinogradov's main attraction consists in its set of problems, which quickly lead to Vinogradov's own research interests; the text itself is very basic and close to minimal. Other popular first introductions are:
Popular choices for a second textbook include:
• Book: Borevich . A. I. . Shafarevich . Igor R. . Borevich . Igor Shafarevich . Number theory. 20 . 1966 . . Boston, MA . Pure and Applied Mathematics . 978-0-12-117850-5 . 0195803.
• Book: Serre, Jean-Pierre . 1996 . 1973 . Jean-Pierre Serre . A course in arithmetic . . 7 . Springer . 978-0-387-90040-7 .
## Notes and References
1. Already in 1921, T. L. Heath had to explain: "By arithmetic, Plato meant, not arithmetic in our sense, but the science which considers numbers in themselves, in other words, what we mean by the Theory of Numbers."
2. Take, for example, . In 1952, Davenport still had to specify that he meant The Higher Arithmetic. Hardy and Wright wrote in the introduction to An Introduction to the Theory of Numbers (1938): "We proposed at one time to change [the title] to An introduction to arithmetic, a more novel and in some ways a more appropriate title; but it was pointed out that this might lead to misunderstandings about the content of the book."
3. . The term takiltum is problematic. Robson prefers the rendering "The holding-square of the diagonal from which 1 is torn out, so that the short side comes up...".
4. . Other sources give the modern formula
(p2-q2,2pq,p2+q2)
. Van der Waerden gives both the modern formula and what amounts to the form preferred by Robson.
5. Neugebauer discusses the table in detail and mentions in passing Euclid's method in modern notation .
6. . This is controversial. See Plimpton 322. Robson's article is written polemically with a view to "perhaps [...] knocking [Plimpton 322] off its pedestal" ; at the same time, it settles to the conclusion that
[...] the question "how was the tablet calculated?" does not have to have the same answer as the question "what problems does the tablet set?" The first can be answered most satisfactorily by reciprocal pairs, as first suggested half a century ago, and the second by some sort of right-triangle problems .
Robson takes issue with the notion that the scribe who produced Plimpton 322 (who had to "work for a living", and would not have belonged to a "leisured middle class") could have been motivated by his own "idle curiosity" in the absence of a "market for new mathematics".
7. [Iamblichus]
8. Herodotus (II. 81) and Isocrates (Busiris 28), cited in: . On Thales, see Eudemus ap. Proclus, 65.7, (for example,) cited in: . Proclus was using a work by Eudemus of Rhodes (now lost), the Catalogue of Geometers. See also introduction, on Proclus's reliability.
9. , cited in: .
10. Plato, Theaetetus, p. 147 B, (for example,), citedin : "Theodorus was writing out for us something about roots, such as the roots of three or five, showing that they are incommensurable by the unit;..." See also Spiral of Theodorus.
11. Sunzi Suanjing, Chapter 3, Problem 26. This can be found in, which contains a full translation of the Suan Ching (based on). See also the discussion in .
12. The date of the text has been narrowed down to 220–420 CE (Yan Dunjie) or 280–473 CE (Wang Ling) through internal evidence (= taxation systems assumed in the text). See .
13. Sunzi Suanjing, Ch. 3, Problem 26,in :
[26] Now there are an unknown number of things. If we count by threes, there is a remainder 2; if we count by fives, there is a remainder 3; if we count by sevens, there is a remainder 2. Find the number of things. Answer: 23.
Method: If we count by threes and there is a remainder 2, put down 140. If we count by fives and there is a remainder 3, put down 63. If we count by sevens and there is a remainder 2, put down 30. Add them to obtain 233 and subtract 210 to get the answer. If we count by threes and there is a remainder 1, put down 70. If we count by fives and there is a remainder 1, put down 21. If we count by sevens and there is a remainder 1, put down 15. When [a number] exceeds 106, the result is obtained by subtracting 105.
14. See, for example, Sunzi Suanjing, Ch. 3, Problem 36, in :
[36] Now there is a pregnant woman whose age is 29. If the gestation period is 9 months, determine the sex of the unborn child. Answer: Male.
Method: Put down 49, add the gestation period and subtract the age. From the remainder take away 1 representing the heaven, 2 the earth, 3 the man, 4 the four seasons, 5 the five phases, 6 the six pitch-pipes, 7 the seven stars [of the Dipper], 8 the eight winds, and 9 the nine divisions [of China under Yu the Great]. If the remainder is odd, [the sex] is male and if the remainder is even, [the sex] is female.
This is the last problem in Sunzi's otherwise matter-of-fact treatise.
15. Web site: Eusebius of Caesarea: Praeparatio Evangelica (Preparation for the Gospel). Tr. E.H. Gifford (1903) – Book 10 . 2017-02-20 . 2016-12-11 . https://web.archive.org/web/20161211194042/http://www.tertullian.org/fathers/eusebius_pe_10_book10.htm . live .
16. Metaphysics, 1.6.1 (987a)
17. Tusc. Disput. 1.17.39.
18. Any early contact between Babylonian and Indian mathematics remains conjectural .
19. Āryabhaṭa, Āryabhatīya, Chapter 2, verses 32–33, cited in: . See also . A slightly more explicit description of the kuṭṭaka was later given in Brahmagupta, Brāhmasphuṭasiddhānta, XVIII, 3–5 (in, cited in).
20. , cited in . See also the preface in cited in
21. , and, cited in .
22. [Claude Gaspard Bachet de Méziriac|Bachet]
23. . This was more so in number theory than in other areas (remark in). Bachet's own proofs were "ludicrously clumsy" .
24. Perfect and especially amicable numbers are of little or no interest nowadays. The same was not true in medieval times—whether in the West or the Arab-speaking world—due in part to the importance given to them by the Neopythagorean (and hence mystical) Nicomachus (ca. 100 CE), who wrote a primitive but influential "Introduction to Arithmetic". See .
25. . The initial subjects of Fermat's correspondence included divisors ("aliquot parts") and many subjects outside number theory; see the list in the letter from Fermat to Roberval, 22.IX.1636,, cited in .
26. Book: Numbers and Measurements. Faulkner. Nicholas. Hosch. William L.. 2017. Encyclopaedia Britannica. 978-1538300428. en. 2019-08-06. 2023-03-01. https://web.archive.org/web/20230301144254/https://books.google.com/books?id=5tFFDwAAQBAJ. live.
27. , Letter XLVI from Fermat to Frenicle, 1640,cited in
28. Here, as usual, given two integers a and b and a non-zero integer m, we write
a\equivb\bmodm
(read "a is congruent to b modulo m") to mean that m divides a − b, or, what is the same, a and b leave the same residue when divided by m. This notation is actually much later than Fermat's; it first appears in section 1 of Gauss's Disquisitiones Arithmeticae. Fermat's little theorem is a consequence of the fact that the order of an element of a group divides the order of the group. The modern proof would have been within Fermat's means (and was indeed given later by Euler), even though the modern concept of a group came long after Fermat or Euler. (It helps to know that inverses exist modulo p, that is, given a not divisible by a prime p, there is an integer x such that
xa\equiv1\bmodp
); this fact (which, in modern language, makes the residues mod p into a group, and which was already known to Āryabhaṭa; see above) was familiar to Fermat thanks to its rediscovery by Bachet . Weil goes on to say that Fermat would have recognised that Bachet's argument is essentially Euclid's algorithm.
29. , cited in . All of the following citations from Fermat's Varia Opera are taken from . The standard Tannery & Henry work includes a revision of Fermat's posthumous Varia Opera Mathematica originally prepared by his son .
30. Up to the second half of the seventeenth century, academic positions were very rare, and most mathematicians and scientists earned their living in some other way . (There were already some recognisable features of professional practice, viz., seeking correspondents, visiting foreign colleagues, building private libraries . Matters started to shift in the late 17th century ; scientific academies were founded in England (the Royal Society, 1662) and France (the Académie des sciences, 1666) and Russia (1724). Euler was offered a position at this last one in 1726; he accepted, arriving in St. Petersburg in 1727 (and).In this context, the term amateur usually applied to Goldbach is well-defined and makes some sense: he has been described as a man of letters who earned a living as a spy ; cited in). Notice, however, that Goldbach published some works on mathematics and sometimes held academic positions.
31. and
32. and
33. . Euler was generous in giving credit to others, not always correctly.
35. From the preface of Disquisitiones Arithmeticae; the translation is taken from
36. See the discussion in section 5 of . Early signs of self-consciousness are present already in letters by Fermat: thus his remarks on what number theory is, and how "Diophantus's work [...] does not really belong to [it]" (quoted in).
37. See the proof in
38. See the comment on the importance of modularity in
39. See, for example, the initial comment in .
40. "The main difference is that in algebraic number theory [...] one typically considers questions with answers that are given by exact formulas, whereas in analytic number theory [...] one looks for good approximations."
41. Sieve theory figures as one of the main subareas of analytic number theory in many standard treatments; see, for instance, or
42. This is the case for small sieves (in particular, some combinatorial sieves such as the Brun sieve) rather than for large sieves; the study of the latter now includes ideas from harmonic and functional analysis.
43. See the remarks in the introduction to : "However much stronger...".
44. "[Riemann] defined what we now call the Riemann zeta function [...] Riemann's deep work gave birth to our subject [...]"
45. See, for example,, p. 1.
46. The Galois group of an extension L/K consists of the operations (isomorphisms) that send elements of L to other elements of L while leaving all elements of K fixed.Thus, for instance, Gal(C/R) consists of two elements: the identity element(taking every element x + iy of C to itself) and complex conjugation(the map taking each element x + iy to x − iy).The Galois group of an extension tells us many of its crucial properties. The study of Galois groups started with Évariste Galois; in modern language, the main outcome of his work is that an equation f(x) = 0 can be solved by radicals(that is, x can be expressed in terms of the four basic operations togetherwith square roots, cubic roots, etc.) if and only if the extension of the rationals by the roots of the equation f(x) = 0 has a Galois group that is solvablein the sense of group theory. ("Solvable", in the sense of group theory, is a simple property that can be checked easily for finite groups.)
47. If we want to study the curve
y2=x3+7
. We allow x and y to be complex numbers:
(a+bi)2=(c+di)3+7
. This is, in effect, a set of two equations on four variables, since both the realand the imaginary part on each side must match. As a result, we get a surface (two-dimensional) in four-dimensional space. After we choose a convenient hyperplane on which to project the surface (meaning that, say, we choose to ignore the coordinate a), we canplot the resulting projection, which is a surface in ordinary three-dimensional space. Itthen becomes clear that the result is a torus, loosely speaking, the surface of a doughnut (somewhatstretched). A doughnut has one hole; hence the genus is 1.
48. Book: Felix E. Browder . Felix Browder . Mathematical Developments Arising from Hilbert Problems . . XXVIII.2 . 1976 . . 978-0-8218-1428-4 . 323–378 . Martin . Davis . Martin Davis (mathematician) . Yuri . Matiyasevich . Yuri Matiyasevich . Julia . Robinson . Julia Robinson . Hilbert's Tenth Problem: Diophantine Equations: Positive Aspects of a Negative Solution . 0346.02026 . Reprinted in The Collected Works of Julia Robinson, Solomon Feferman, editor, pp. 269–378, American Mathematical Society 1996.
49. The Unreasonable Effectiveness of Number Theory, Stefan Andrus Burr, George E. Andrews, American Mathematical Soc., 1992,
50. Computer science and its relation to mathematics" DE Knuth – The American Mathematical Monthly, 1974
51. "Applications of number theory to numerical analysis", Lo-keng Hua, Luogeng Hua, Yuan Wang, Springer-Verlag, 1981,
52. Book: An Introduction to Number Theory with Cryptography. 2nd . 2018 . Chapman and Hall/CRC . 978-1-351-66411-0 . 10.1201/9781351664110 . 2023-02-22 . 2023-03-01 . https://web.archive.org/web/20230301144259/https://www.taylorfrancis.com/books/mono/10.1201/9781351664110/introduction-number-theory-cryptography-james-kraft-lawrence-washington . live .
53. Book: Krishna, Hari . Digital Signal Processing Algorithms : Number Theory, Convolution, Fast Fourier Transforms, and Applications . 2017 . 978-1-351-45497-1 . London . 1004350753 . 2023-02-22 . 2023-03-01 . https://web.archive.org/web/20230301144250/https://www.worldcat.org/title/1004350753 . live .
54. Book: Baylis, John . Error-Correcting Codes: A Mathematical Introduction . 2018 . Routledge . 978-0-203-75667-6 . en . 10.1201/9780203756676 . 2023-02-22 . 2023-03-01 . https://web.archive.org/web/20230301144305/https://www.taylorfrancis.com/books/mono/10.1201/9780203756676/error-correcting-codes-baylis . live .
55. Cartwright . Julyan H. E. . Gonzalez . Diego L. . Piro . Oreste . Stanzial . Domenico . 2002-03-01 . Aesthetics, Dynamics, and Musical Scales: A Golden Connection . Journal of New Music Research . 31 . 1 . 51–58 . 10.1076/jnmr.31.1.51.8099 . 10261/18003 . 12232457 . 0929-8215. free .
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ACCT 2300_chapter 2 in class - Exercise 5-1 Preparing a Contribution Format Income Statement Wheeler Corporation's most recent income statement follows
# ACCT 2300_chapter 2 in class - Exercise 5-1 Preparing a...
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Exercise 5-1 Preparing a Contribution Format Income Statement Wheeler Corporation's most recent income statement follows: Total Per Unit Sales (8,000 units) 211,200 26.40 Variable expenses 140,000 17.50 Contribution margin 71,200 8.90 Fixed expenses 54,800 Net operating income 16,400 Required: Prepare a new contribution format income statement under each of the following conditions (consider each case independently): 1. The sales volume increases by 30 units. The new income statement would be: Total Per unit Sales (8030 units) ### 26.40 Variable expenses ### 17.50 Contribution margin 71,467.00 8.90 Fixed expenses 54,800.00 Net Operating Income 16,667.00 As an alternative, you could have found the net income using the following method: Original net operating income \$8,000 Change in contribution margin (50 units * \$8.00 per unit) 400 New net operating income \$8,400 2. The sales volume declines by 50 units. Sales (7950 units) \$206,700 \$26.00 Variable expenses 143,100 18.00 Contribution margin \$63,600 \$8.00 Fixed expenses 56,000 Net Operating Income \$7,600 As an alternative, you could have found the net income using the following method: Original net operating income \$8,000 Change in contribution margin (-50 units * \$8.00 per unit) -400 New net operating income \$7,600 3. The sales volume is 7,0000. Sales (7000 units) \$182,000 \$26.00 Variable expenses 126,000 18.00 Contribution margin \$56,000 \$8.00 Fixed expenses 56,000 Net Operating Income \$0 This is the company's break even point because it it the point where the contribution margin covers the fixed expenses and there is no net income/net loss (it is zero).
Exercise 5-2 Prepare a Cost-Volume-Profit (CVP) Graph Katara Enterprises distributes a single product whose selling price is \$36 and whose variable cost is \$24 per unit. The company's monthly fixed expense is \$12,000. Required: 1. Prepare a cost-volume-profit graph for the company up to a sales level of 2,000 units. Note: The CVP graph is plotted using the three steps in the text. Step 1: Draw a line parallel to the volume axis to represent the total fixed expense (See purple line). For this company, the total fixed expense is \$12,000. Step 2: Choose some volume of sales and plot the point representing total expenses (fixed and variable) at the activity level you have selected. (Part (a) of this problem wants us to find the sales level at 2,000 units) so: Fixed expense \$12,000 Variable expense (\$2,000 units * \$24/ unit) 48,000 Total expense \$60,000 2. Estimate the company's break-even point in unit sales using your cost-volume profit graph. The break-even point where the total sales revenue and the total expense lines intersect. This occurs at sales of 1,000 units. This can be verified by solving for the break-even point in unit sales, Q, using the equation method as follows: Sales = Variable expenses + Fixed expenses + Profits \$36Q = \$24Q + \$12,000 + \$0 \$12Q = \$12,000 Q = \$12,000/ \$12 per unit Q = 1,000 units 0 \$20,000 500 1,000 0 1,500 2,000 \$40,000 \$60,000 \$80,000 Fixed costs \$12,000 Total costs: Variable (\$24 * 2000) + \$12,000 = \$60,000 Total sales revenue: 2,000 units * \$36 = \$72,000. Break-even point: 1,000 units
Exercise 5-3 Computing and Using the CM Ratio Last month when Harrison Creations, Inc., sold 40,000 units, total sales were \$300,000, total variable expenses were \$240,000, and total fized expenses were \$45,000.
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Algebra - Polynomials
If a manufacturer charges 'p' dollars each for shirts, then he expects to sell 2000-100p shirts per week. What polynomial represents the total revenue expected for a week? How many shirts will be sold if the manufacturer charges \$20 each for the shirts. Find total revenue when the shirts are sold for \$20 each.
Solution Preview
(a) Total revenue = price per shirt * number of shirts sold
= p * (2000 - 100p)
= 2000p - 100 ...
Solution Summary
Neat and step-wise solutions are provided.
\$2.19
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# Number 12499488
### Properties of number 12499488
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 17 * 23 * 37
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
beba20
Base 32:
bteh0
sin(12499488)
0.94846523234271
cos(12499488)
-0.31688121281813
tan(12499488)
-2.9931254803896
ln(12499488)
16.341198241434
lg(12499488)
7.0968922239418
sqrt(12499488)
3535.4614974569
Square(12499488)
### Number Look Up
Look Up
12499488 which is pronounced (twelve million four hundred ninety-nine thousand four hundred eighty-eight) is a very amazing figure. The cross sum of 12499488 is 45. If you factorisate the figure 12499488 you will get these result 2 * 2 * 2 * 2 * 2 * 3 * 3 * 3 * 17 * 23 * 37. The figure 12499488 has 192 divisors ( 1, 2, 3, 4, 6, 8, 9, 12, 16, 17, 18, 23, 24, 27, 32, 34, 36, 37, 46, 48, 51, 54, 68, 69, 72, 74, 92, 96, 102, 108, 111, 136, 138, 144, 148, 153, 184, 204, 207, 216, 222, 272, 276, 288, 296, 306, 333, 368, 391, 408, 414, 432, 444, 459, 544, 552, 592, 612, 621, 629, 666, 736, 782, 816, 828, 851, 864, 888, 918, 999, 1104, 1173, 1184, 1224, 1242, 1258, 1332, 1564, 1632, 1656, 1702, 1776, 1836, 1887, 1998, 2208, 2346, 2448, 2484, 2516, 2553, 2664, 3128, 3312, 3404, 3519, 3552, 3672, 3774, 3996, 4692, 4896, 4968, 5032, 5106, 5328, 5661, 6256, 6624, 6808, 7038, 7344, 7548, 7659, 7992, 9384, 9936, 10064, 10212, 10557, 10656, 11322, 12512, 13616, 14076, 14467, 14688, 15096, 15318, 15984, 16983, 18768, 19872, 20128, 20424, 21114, 22644, 22977, 27232, 28152, 28934, 30192, 30636, 31968, 33966, 37536, 40848, 42228, 43401, 45288, 45954, 56304, 57868, 60384, 61272, 67932, 81696, 84456, 86802, 90576, 91908, 112608, 115736, 122544, 130203, 135864, 168912, 173604, 181152, 183816, 231472, 245088, 260406, 271728, 337824, 347208, 367632, 390609, 462944, 520812, 543456, 694416, 735264, 781218, 1041624, 1388832, 1562436, 2083248, 3124872, 4166496, 6249744, 12499488 ) whith a sum of 41368320. The figure 12499488 is not a prime number. 12499488 is not a fibonacci number. The figure 12499488 is not a Bell Number. The number 12499488 is not a Catalan Number. The convertion of 12499488 to base 2 (Binary) is 101111101011101000100000. The convertion of 12499488 to base 3 (Ternary) is 212112001002000. The convertion of 12499488 to base 4 (Quaternary) is 233223220200. The convertion of 12499488 to base 5 (Quintal) is 11144440423. The convertion of 12499488 to base 8 (Octal) is 57535040. The convertion of 12499488 to base 16 (Hexadecimal) is beba20. The convertion of 12499488 to base 32 is bteh0. The sine of 12499488 is 0.94846523234271. The cosine of 12499488 is -0.31688121281813. The tangent of the figure 12499488 is -2.9931254803896. The root of 12499488 is 3535.4614974569.
If you square 12499488 you will get the following result 156237200262144. The natural logarithm of 12499488 is 16.341198241434 and the decimal logarithm is 7.0968922239418. that 12499488 is impressive figure!
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11 Nov 2010, 00:30
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No only smoking cigarettes but also cigar smoking has been banned now from many places.
(A) No only smoking cigarettes but also cigar smoking has been banned now
(B) Cigarette smoking and cigar smoking are both banned now
(C) Not only has smoking cigarettes been banned but so has cigar smoking
(D) Both smoking cigarettes and cigar smoking is now banned
(E) Smoking cigarettes as well as cigars is now banned
[Reveal] Spoiler: OA
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Re: not only but also [#permalink]
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11 Nov 2010, 00:57
niraj92 wrote:
No only smoking cigarettes but also cigar smoking has been banned now from many places.
(A) No only smoking cigarettes but also cigar smoking has been banned now
(B) Cigarette smoking and cigar smoking are both banned now
(C) Not only has smoking cigarettes been banned but so has cigar smoking
(D) Both smoking cigarettes and cigar smoking is now banned
(E) Smoking cigarettes as well as cigars is now banned
A Incorrect => Not only x but also y - x & y are not parallel in the given sentence.
B Correct => Cigrette smoking and cigar smoking are parallel.
C Incorrect => Obviously
D Incorrect => Smoking Cigrette & Cigar smoking connected with AND, are not parallel.
E incorrect => Smoking cigarettes as well as cigars notparallel. Should have been smoking cigarettes as well as smoking cigars
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Re: not only but also [#permalink]
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12 Nov 2010, 13:51
Great work to everybody on this one. The fundamental rule has two-parts:
1. If you say "not only" you are required to also say "but also." You can never have the construction "not only...also" or "not...but also" or anything else of the sort. "Not only" and "but also" are bff, and you can't have one without the other.
2. What follows "not only" must be structurally and logically similar to what follows "but also." For example:
"I bought not only bread, but also milk." --> Right! Bread and milk are both foods and both nouns.
"I bought not only bread, but also considered buying milk." --> Wrong! "Bread" is a noun, but "considered buying milk" is a more complex clause.
Isn't parallelism fun?
Brett
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Re: not only but also [#permalink]
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12 Nov 2010, 22:01
BKimball wrote:
Great work to everybody on this one. The fundamental rule has two-parts:
1. If you say "not only" you are required to also say "but also." You can never have the construction "not only...also" or "not...but also" or anything else of the sort. "Not only" and "but also" are bff, and you can't have one without the other.
2. What follows "not only" must be structurally and logically similar to what follows "but also." For example:
"I bought not only bread, but also milk." --> Right! Bread and milk are both foods and both nouns.
"I bought not only bread, but also considered buying milk." --> Wrong! "Bread" is a noun, but "considered buying milk" is a more complex clause.
Isn't parallelism fun?
Brett
Thanks for a great explanation. It's always comforting when one sees answers confirmed by experts. There are instances when two different people have different OAs.
I'm not too sure but I think I remember that Tommy Wallach (one of Manhattan Instructors) once said that there are constructions that don't always follow the '...not only...but also...' rule.
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Re: not only but also [#permalink]
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15 Nov 2010, 07:26
Werewolf,
Grammatically, it is possible (though HIGHLY suspect) to have a "not only...but" structure. However, the GMAT loves to test your understanding of why "not only...but also" is necessary for purposes of parallelism.
As with "numbers" (i.e. "My favorite numbers are six and twelve."), and "being" (i.e. "My uncle Charlie is a human being.") there are times when you can use "not only...but." Still, they VERY rarely include GMAT questions, so it's pretty safe to forget about them for our purposes.
Brett
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Re: not only but also [#permalink]
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16 Nov 2010, 02:49
Thank you Brett. That's a relief!
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Re: not only but also [#permalink]
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16 Nov 2010, 03:54
niraj92 wrote:
No only smoking cigarettes but also cigar smoking has been banned now from many places.
(A) No only smoking cigarettes but also cigar smoking has been banned now
(B) Cigarette smoking and cigar smoking are both banned now
(C) Not only has smoking cigarettes been banned but so has cigar smoking
(D) Both smoking cigarettes and cigar smoking is now banned
(E) Smoking cigarettes as well as cigars is now banned
In E, I took "Smoking" as a gerund thus is is in SV with "is" and thus I chose E as it is concise. In B "Cigarette smoking and cigar smoking" is wordy. Also the placement of NOW seems correct in E.
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Re: not only but also [#permalink]
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16 Nov 2010, 08:13
puneetpratik wrote:
niraj92 wrote:
No only smoking cigarettes but also cigar smoking has been banned now from many places.
(A) No only smoking cigarettes but also cigar smoking has been banned now
(B) Cigarette smoking and cigar smoking are both banned now
(C) Not only has smoking cigarettes been banned but so has cigar smoking
(D) Both smoking cigarettes and cigar smoking is now banned
(E) Smoking cigarettes as well as cigars is now banned
In E, I took "Smoking" as a gerund thus is is in SV with "is" and thus I chose E as it is concise. In B "Cigarette smoking and cigar smoking" is wordy. Also the placement of NOW seems correct in E.
I was also under the same understanding that E was more concise....
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Re: not only but also [#permalink]
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16 Nov 2010, 15:54
You are right that (E) is more concise. However, we need to remember that concision is ONLY a tie-breaker on SC.
Far more important than concision is GRAMMAR. If an answer is grammatically incorrect, it must be crossed off.
Once you've dealt with all the grammatical issues, you can then look at the MEANING of the sentence. If the meaning is changed in an answer choice, then that answer is incorrect.
Only after you've dealt with GRAMMAR and MEANING should you consider CONCISION. Concision is very much a tiebreaker on the GMAT, and in my experience the shortest answer tends to be incorrect because it leaves out something important. Remember that the GMAT writers know people are looking for concise answers, so concise answers are a great place for a trap.
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Re: not only but also [#permalink]
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17 Nov 2010, 00:18
Thanks for the advice Brett, will keep this one in mind.
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Re: not only but also [#permalink]
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18 Nov 2010, 00:08
In this problem answer choice A is not correct only because it is in passive voice is that correct. Otherwise there is no gramatical mistake?
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Re: not only but also [#permalink]
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18 Nov 2010, 15:25
Good question! The answer has two parts:
1. General: No SC answer choice will ever be wrong simply on the basis of passive vs. active verbs. Although we tend to prefer the active voice (many of your future Business school professors will point this out regularly), there is no rule that says "passive voice is wrong."
2. Specific to this question: "No only smoking cigarettes but also cigar smoking has been banned now..." Although you are correct that this is passive voice ("has been banned"), the reason this answer choice is wrong really relates to parallelism. First, you have the "not only...but also" issue. When you have "but also" you really should have "not only" also. Here, you have "no only." This might be a typo on the part of the original poster. Second, and perhaps even more grammatically sound: The two things that follow each of these signal phrases should be parallel. "Smoking cigarettes" and "cigar smoking" are not parallel; the first phrase starts with the verb "smoking" and the second starts with the noun "cigar."
Does that all make sense?
Brett
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Re: not only but also [#permalink] 18 Nov 2010, 15:25
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# algebra
posted by .
Can someone show me step by step examples of simplifying
3√5/64
thank you
• algebra -
If you mean √5 / 64 there is nothing left to do.
If you meant √(5/64), you can use the fact that √64=8 to reduce the fraction to 3√5 / 8
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# Piano Chord Diagrams
Piano chord diagrams can be represented in many different ways. But, the main concept stays the same. They are used to show what notes go along with a certain chord and sometimes have a picture and the fingering.
Piano Chord Diagrams - Pictures
Probably the easiest type of chord diagram to use would be an image. These diagrams show an actual picture of the keyboard and have a some way of representing which notes go along with that chord.
These are helpful because they show the actual image of the piano and gives and easy to read way to find the notes.
Here is a list of the piano chord diagrams for all the notes and keys for most chords.
Number System
Another way of displaying chords is through the number system. This can be helpful for the more advanced piano player who already knows piano scales.
To use this method you need to know the key signature of the chord you are finding. For example lets use the key of C. C Major is made up of the notes C, D, E, F, G, A, B. Now that we have the notes of the scale we need to give each note in the scale a number.
We would say that C is 1, D would be 2, and E is 3. You would go up the scale giving each note name a letter.
Here is the numbering for the key of C.
Now that each note has a letter we can go ahead and look at the diagram.
These diagrams are made up of numbers that correlate with the note in that key and the note that is played in that chord.
The b in the diagram represents a flat note or a lowered half step. A # represents a sharp, or a raised half step.
For example, in a major chord in the key of C you would play the 5th, 3rd, and 1st note in the scale. In the key of C that would be G, E, and C.
This simple diagram is an easy way to find any chord in any key on the piano as long as you know the scale.
Let’s do one more example!
Let’s use the key of G. The notes in the key of G are G, A, B, C, D, E, F#. Give each note name a number and it will look like this.
If we wanted to find the notes in a minor chord for this key you have to find the 1st note, b flat 3rd, and 5th.
This tells us that the notes for a G minor chord would be G, Bb, and D.
Use these piano chord diagrams to find the notes for any key on the piano. You can print these off for reference until they become second nature. It will take time but the longer you use these the quicker you will be able to quickly remember the notes of any chord on the piano.
Need more help learning chords? Learn almost every chord you'll ever need with our Piano Chord Encyclopedia.
Go from Piano Chord Diagrams back to Piano Chords
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# A certain aquarium holds three types of fish: angelfish, swordtails an
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A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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24 Mar 2015, 03:23
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A certain aquarium holds three types of fish: angelfish, swordtails and guppies. What is the ratio of the number of guppies to the number of angelfish?
(1) There are 200 fish in the aquarium
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
Kudos for a correct solution.
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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24 Mar 2015, 03:32
Bunuel wrote:
A certain aquarium holds three types of fish: angelfish, swordtails and guppies. What is the ratio of the number of guppies to the number of angelfish?
(1) There are 200 fish in the aquarium
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
Kudos for a correct solution.
Qu: G:A=?
A+S+G=Total
(1) There are 200 fish in the aquarium
We know Total=200 , no information on distribution among A, S and G. NOT SUFFICIENT
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
45%T =S
S=2A
so we know G+A combine to 55% and A is $$45/2 %$$ so G will be $$100-45-22.5$$
we cannot find individual number of fish of each type but we can surely find G:A.
Option B : Sufficient .
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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24 Mar 2015, 19:05
1
Bunuel wrote:
A certain aquarium holds three types of fish: angelfish, swordtails and guppies. What is the ratio of the number of guppies to the number of angelfish?
(1) There are 200 fish in the aquarium
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
Kudos for a correct solution.
Statement 1:
We are not given a breakdown of the 3 types of fish
Insufficient
Statement 2:
(9/20)(a+g+s)=s
s=2a
Combining the equations together, we get g/a = 9/13
Sufficient
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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24 Mar 2015, 22:55
1
Let A,S & G are # of fishes in the tank; # of G/# of A =?
1. Total = 200, no idea about of G & A, insufficient
2. Let total is x, so s=0.45x, a= 0.45x/2 and G= x-3*0.45x/2; so #of G/# of A can be determined by given values. Sufficient
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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30 Mar 2015, 03:22
Bunuel wrote:
A certain aquarium holds three types of fish: angelfish, swordtails and guppies. What is the ratio of the number of guppies to the number of angelfish?
(1) There are 200 fish in the aquarium
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:
Attachment:
guppiesangelfishswordtails_text.PNG [ 19.41 KiB | Viewed 4445 times ]
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink]
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24 Nov 2019, 07:08
Top Contributor
Bunuel wrote:
A certain aquarium holds three types of fish: angelfish, swordtails and guppies. What is the ratio of the number of guppies to the number of angelfish?
(1) There are 200 fish in the aquarium
(2) 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
Given: A certain aquarium holds three types of fish: angelfish, swordtails, and guppies.
Target question: What is the ratio of the number of guppies to the number of angelfish?
Statement 1: There are 200 fish in the aquarium.
Since we are not giving any information about the number of each type of fish, statement 1 is NOT SUFFICIENT
Statement 2: 45 percent of the fish are swordtails, and there are twice as many swordtails as there are angelfish.
There are twice as many swordtails as there are angelfish. 45% ÷ 2 = 22.5%
So, 45% are swordtails and 22.5% are angelfish.
Since all three percentages must add to 100%, we can conclude that the remaining 32.5% are guppies.
We can write: swordtails : angelfish : guppies = 45 : 22.5 : 32.5
The answer to the target question is the ratio of guppies to angelfish = 32.5 : 22.5
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Cheers,
Brent
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Re: A certain aquarium holds three types of fish: angelfish, swordtails an [#permalink] 24 Nov 2019, 07:08
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# what is freshnel and fraunhoffer diffraction ? where it is useful ?
BOLLU SRINIVAS
420 Points
8 years ago
Freshnal difraction: both the point source and screen is relatively close to the obstacle then forms difraction pattern. this situation is described as near field diffraction or freshnal diffraction.
Fraunhoffer diffraction: IF the source, obstacle and screen are far enough away that all lines from source to the obstacle can conside parllal,this pnenomenon is called fraunhoffer diffraction.
We will use fraunhofffer diffrction is usually simpler to analyze.
THANK YOU
T.kumar
281 Points
8 years ago
Freshnal difraction: both the point source and screen is relatively close to the obstacle then forms difraction pattern. this situation is described as near field diffraction or freshnal diffraction.
Fraunhoffer diffraction: IF the source, obstacle and screen are far enough away that all lines from source to the obstacle can conside parllal,this pnenomenon is called fraunhoffer diffraction.
We will use fraunhofffer diffrction is usually simpler to analyze.
THANK YOU
317 Points
8 years ago
the diffraction in which the source obstacle and the screen are of enough distance. in this diffraction the waves are considered parallel. this is known as fraunhoffer or field diffraction
N JYOTHEESWAR
342 Points
8 years ago
THE diffraction in which the source obstacle and the screen are of enough distance. in this diffraction the waves are considered parallel. this is known as fraunhoffer or field diffraction
salvathshaik
164 Points
8 years ago
fraun hoffer diffraction is known as a abstacle and source and the screen are the close ly placed..and it is also known as the near field differaction...
frehnal diffraction: it is known as the obsatacle and source and screen are to be far away to the each other and it is also known as the far field differacton....
Prabhakar ch
577 Points
8 years ago
The diffraction in which the source obstacle and the screen are of enough distance. in this diffraction the waves are considered parallel. this is known as fraunhoffer or field diffraction.
All the best for your bright future
RAMCHANDRARAO
159 Points
8 years ago
in freshnel difraction source and obstacle and the screen are close to each other this is also called near fieald difraction in fraunhoffer diffraction source and obstacle and screen are far apart to each other this also called farfield diffraction
Gowri sankar
292 Points
8 years ago
In freshnel diffraction source and obstacle and the screen are close to each other this is called near field diffraction in fraunhoffer diffraction source and obstacle and screen are far apart to each other this also called farfield diffraction.
raj
383 Points
8 years ago
The diffraction in which the source obstacle and the screen are of enough distance. in this diffraction the waves are considered parallel. this is known as fraunhoffer or field diffraction.
In freshnel diffraction source and obstacle and the screen are close to each other this is called near field diffraction in fraunhoffer diffraction source and obstacle and screen are far apart to each other this also called farfield diffraction.
E.Nandhini ketha
115 Points
6 years ago
In optics, the Fraunhofer diffraction equation is used to model the diffraction of waves when the diffraction pattern is viewed at a long distance
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# Surface Area and Friction
• stu12345
However, there are cases where μ does depend on surface area, most notably in the case of materials which deform easily.
## Homework Statement
How does the surface area of an object affect the force of static friction? I'm trying to figure out whether two different objects with equal mass and different surface areas requires the same amount of applied force or different amounts.
## Homework Equations
I know of this one.
Ff = coefficient of friction(Fn)
## The Attempt at a Solution
I'm not sure how to answer this.
Does Fn depend on area?
Well each object in contact with a flat surface will have equal masses just different surface areas. I'm using a same flat surface for each object with equal masses but different surface areas.
stu12345 said:
Well each object in contact with a flat surface will have equal masses just different surface areas. I'm using a same flat surface for each object with equal masses but different surface areas.
Right, well in most cases, the normal force would be the weight right? So would the frictional force formula have area in it?
Ya it would have area.
stu12345 said:
Ya it would have area.
No, it would not, if Fn=mg and Ff=μFn then area does not appear.
Once your materials are the same, then μ is the same and if the masses are the same, then the frictional force produced by them are the same, regardless of area.
Ah ok. Thanks so much.
rock.freak667 said:
No, it would not, if Fn=mg and Ff=μFn then area does not appear.
Once your materials are the same, then μ is the same and if the masses are the same, then the frictional force produced by them are the same, regardless of area.
As long as you can treat the coefficient of friction, μ, as simply being a constant which depends only upon the two materials which are present, then surface area will not make any difference. That's the simple model used in physics courses.
However, coefficient of friction, μ, actually does depend on the temperature of the materials. We seldom include that in our model. The greater the surface area, the less the increase in temperature (The thermal energy is dissipated over a wider area.) so that μ will tend to change less than in the case less surface area. Of course, this is more a factor with sliding friction, which tends to produce heat.
It's also the case that for materials which deform rather easily, like rubber used in tires, the pressure at the point of contact affects μ in a rather complicated way.
This is just "scratching the surface". - pun intended
Generally, when solving a physics problem with friction involved, frictional force does not depend on surface area. In any case, Ff = μ Fn works very well.
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http://oeis.org/A249605
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A249605 Dissectible numbers in the sense of Gunjikar and Kaprekar. 3
9, 18, 27, 36, 45, 54, 63, 72, 81, 108, 117, 126, 135, 144, 153, 162, 207, 216, 225, 234, 243, 306, 315, 324, 405 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The numbers really have exactly three digits, allowing leading zeros. The full list is: 009, 018, 027, 036, 045, 054, 063, 072, 081, 108, 117, 126, 135, 144, 153, 162, 207, 216, 225, 234, 243, 306, 315, 324, 405. A dissectible number is a three-digit number abc with the property that when multiplied by any Wonderful Demlo number (A002477) the product has the form axx..xbyy..yc, with the same number of x digits as y digits. LINKS K. R. Gunjikar and D. R. Kaprekar, Theory of Demlo numbers, J. Univ. Bombay, Vol. VIII, Part 3, Nov. 1939, pp. 3-9. [Annotated scanned copy] EXAMPLE 9*12321 = 009*12321 = 0110889 (a=b=0, c=9, x=1, y=8). 162*121 = 19602 (here x=9, y=0). 162*12321 = 1996002 (again x=9, y=0). CROSSREFS Cf. A002477. Sequence in context: A248050 A044052 A131418 * A130692 A043453 A028439 Adjacent sequences: A249602 A249603 A249604 * A249606 A249607 A249608 KEYWORD nonn,fini,full,base AUTHOR N. J. A. Sloane, Nov 04 2014 STATUS approved
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https://ch.mathworks.com/matlabcentral/cody/problems/672-longest-run-of-consecutive-numbers/solutions/1757620
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Cody
# Problem 672. Longest run of consecutive numbers
Solution 1757620
Submitted on 22 Mar 2019 by Daniel Moran
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [1 2 2 2 1 3 2 1 4 5 1]; y_correct = 2; assert(isequal(longrun(a),y_correct))
n = 11
2 Pass
a = [1 1 1 2 2 2 3 3 3]; y_correct = [1 2 3]; assert(isequal(longrun(a),y_correct))
n = 9
3 Pass
a = [-2 -2 -2 -2 -1 0 3]; y_correct = -2; assert(isequal(longrun(a),y_correct))
n = 7
4 Pass
a=[0 1 1 1 0 2 2 0 1 1 1 0]; y_correct = [1 1]; assert(isequal(longrun(a),y_correct))
n = 12
5 Pass
a=[3 3 3 2 2 1 6]'; y_correct=3; assert(isequal(longrun(a),y_correct))
n = 7
6 Pass
a=[3 3 3 2 2 2 1 6]'; y_correct=[3 2]'; assert(isequal(longrun(a),y_correct))
n = 8
7 Pass
a=[1 2 3 4 5]'; y_correct=a; assert(isequal(longrun(a),y_correct))
n = 5
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https://socratic.org/questions/what-is-the-slope-intercept-form-of-the-equation-of-the-line-that-passes-through
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# What is the slope-intercept form of the equation of the line that passes through the points (2, -1) and (-3, 4)?
Apr 5, 2016
$\textcolor{b l u e}{y = - x + 1}$
#### Explanation:
$\text{standard form } \to y = m x + c$
Where $m$ is the gradient and $c$ is the ${y}_{\text{intercept}}$
$m = \left(\text{change in y-axis")/("change in x-axis}\right)$
Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(2 , - 1\right)$
Let point 2 be${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(- 3 , 4\right)$
Then $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{4 - \left(- 1\right)}{- 3 - 2}$
$\textcolor{b l u e}{\implies m = \frac{5}{- 5} = - 1}$
This means that as you move from left to right; for one along you go down 1 (negative incline).
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So the equation becomes
$\textcolor{b r o w n}{y = - x + c}$
At P_1"; "color(brown)(y=-x+c)color(green)( " "->" "-1=-2+c)
$\implies c = 2 - 1 = 1$
So the equation becomes
$\textcolor{b l u e}{y = - x + 1}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~
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https://forum.nengo.ai/t/is-it-possible-to-use-snn-as-q-function-approximation/338
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# Is it possible to use SNN as Q_function approximation?
Hello everyone,
I am now doing a project, our team try to solve a robot control problem, we will use reinforcement learning to solve it.
And for the algorithm we will use Q_learning, and for the Q_function approximation we will try to use spiking neural network.
As we all know, usually we will use artificial neural network as Q_function approximation, so the idea is directly replace the ann by our spiking neural network.
But it does not work, and we test our algorithm on normal artificial neural network, it works well.
So my question is,
1, can we use snn as Q-function approximation.
2, If not, what is the standard way to solve reinforcement learning using spiking neural networks.
best greetings
You can check out this thread for links to various papers on RL in Nengo, as well as code examples.
1 Like
If youâre doing robot control, you may be more interested in @travis.dewolfâs adaptive control work?
Hello drasmuss,
Thanks for your answer, actually we have already make the robot work if we use normal artificial neural network as Q_function approximation. So it will be easy for us, if we can replace the ANN by Spiking neural network.
And now we try to use nengo_dl, because this simulator is quit similar to the ANN framework.
We have few questions, hope you can help us:
1, I do not understand the âminibatch_sizeâ in nengo_dl simulator. For example, the mnist dataset has training images 60000, when I start the training, I will feed whole data into nengo model. I should set the minibatch_size to 60000, or I can give it a arbitrary number, for example 60?
And there is another argument n_epochs, I guess we can give a arbitrary number to minibatch_size, for example 60, and if we train the n_epochs = 1000, the whole dataset 60000 will be trained, am I right?
2, After training, I print the sim.loss, it is 0.1. But when I do prediction, the accuracy is very low, so I think I didât find the rights way to do prediction. Please tell me the right way to do it. My code is there:
import nengo
import nengo_dl
import numpy as np
import tensorflow as tf
class Spiking_Qnetwork:
``````def __init__(self, input_shape, output_shape, nb_hidden, weights_path):
'''
Spiking neural network as the Q value function approximation
:param input_shape: the input dimension without batch_size, example: state is 2 dimension,
action is 1 dimenstion, the input shape is 3.
:param output_shape: the output dimension without batch_size, the dimenstion of Q values
:param nb_hidden: the number of neurons in ensemble
:param weights_path: the path to save all parameters
'''
self.input_shape = input_shape
self.output_shape = output_shape
self.nb_hidden = nb_hidden
self.weights_path = weights_path
self.model = self.build()
def encoder_decoder_initialization(self, shape):
'''
:return: initialised encoder or decoder
'''
rng = np.random.RandomState(1)
coders = rng.normal(size=shape)
return coders
def build(self):
encoders = self.encoder_decoder_initialization(shape=(self.nb_hidden, self.input_shape))
decoders = self.encoder_decoder_initialization(shape=((self.nb_hidden, self.output_shape)))
model = nengo.Network(seed=3)
with model:
self.input_neuron = nengo.Ensemble(n_neurons=self.nb_hidden,
dimensions=self.input_shape,
neuron_type=nengo.LIFRate(),
intercepts=nengo.dists.Uniform(-1.0, 1.0),
max_rates=nengo.dists.Choice([100]),
encoders=encoders,
)
output = nengo.Node(size_in=self.output_shape)
self.output_p = nengo.Probe(output)
nengo.Connection(self.input_neuron.neurons,
output,
synapse=None,
transform=decoders.T
)
return model
def training(self, input_data, label, batch_size, nb_epochs):
sim = nengo_dl.Simulator(self.model,
minibatch_size=batch_size,
step_blocks=1,
device="/gpu:0"
)
sim.train({self.input_neuron: input_data},
{self.output_p: label},
tf.train.MomentumOptimizer(5e-2, 0.9),
n_epochs=nb_epochs
)
sim.save_params(self.weights_path)
sim.close()
def predict(self, input_data, batch_size=1):
sim = nengo_dl.Simulator(self.model,
minibatch_size=batch_size,
step_blocks=1,
device="/gpu:0")
sim.step(input_feeds={self.input_neuron: input_data})
output = sim.data[self.output_p]
sim.close()
return output
``````
if name == âmainâ:
``````from keras.datasets import mnist
from keras.utils import np_utils
from sklearn.metrics import accuracy_score
(X_train, y_train), (X_test, y_test) = mnist.load_data()
# data pre-processing
X_train = X_train.reshape(X_train.shape[0], -1) / 255. # normalize
X_test = X_test.reshape(X_test.shape[0], -1) / 255. # normalize
y_train = np_utils.to_categorical(y_train, nb_classes=10)
y_test = np_utils.to_categorical(y_test, nb_classes=10)
X_train_ = np.expand_dims(X_train, axis=1)
X_test_ = np.expand_dims(X_test, axis=1)
y_train_ = np.expand_dims(y_train, axis=1)
y_test_ = np.expand_dims(y_test, axis=1)
model = Spiking_Qnetwork(input_shape=28*28,
output_shape=10,
nb_hidden=1000,
weights_path="/home/huangbo/SpikingDeepRLControl/huangbo_ws/"
"networks/saved_weights/snn_weights")
# training
model.training(input_data=X_train_, label=y_train_, batch_size=10000, nb_epochs=5)
output = model.predict(batch_size=X_test.shape[0], input_data=X_test_)
prediction = np.squeeze(output, axis=1)
# evaluate the model
from sklearn.metrics import accuracy_score
acc = accuracy_score(np.argmax(y_test, axis=1), np.argmax(prediction, axis=1))
print "the test acc is:", acc``````
Just to be clear, the papers in that thread I linked to all use spiking neural networks to approximate the Q function.
There is a good explanation of the batch/minibatch/epoch terminology of machine learning here.
You need an input node in your model in order to pass in `input_data` (you canât pass it directly into an Ensemble). So youâd need something like
``````with model:
input_node = nengo.Node(...)
nengo.Connection(input_node, self.input_neuron.neurons, ...)
...
sim.train({input_node: input_data}, ...)
``````
Right now `nengo_dl` is just ignoring invalid input feeds â it should give an error to help identify problems like this, Iâll make that change. Thanks for bringing that to light!
Also, keep in mind that if you want to train something with gradient descent (e.g., `tf.train.MomentumOptimizer`), then your network needs to be differentiable (which `nengo.LIFRate` is not). However, you can try training with `nengo_dl.SoftLIFRate` (a differentiable approximation of LIF neurons), and then switching to spiking neurons when you do your prediction.
2 Likes
Hello drasmuss,
First I change my neuron type to SoftLIFRate, and I add a input node to get the input.
There are still some thing confused me,
1, the training time.
One of the strength of nengo_dl is using tensorflow as backend, so that make the training very fast. But I found the training time actually is very long. For example the mnist case, when I use normal nengo simulator and NEF to train a nengo network, I directly input whole mnist dataset into the network, the training takes 17s. Now I use nengo_dl, the trianing time is depend on the number_epochs we set. If we set batch_size as 60, and for 60000 mnist dataset we will train it 1000 epochs. It takes 8 hours.
2, save the parameters.
I think, this function is very similar as we do in tensorflow, we canât change the architecture of network, and this function will save all the trainable parameters. Is that right?
3, prediction
After the training, we want to test the performance of the network. We want to input a single image. So we change the batch_size of simulator to 1, and run one step of simulation to get the output, we not sure if this is the right way to do prediction.
sim.step(input_feeds={self.input_neuron: input_data})
output = sim.data[self.output_p]
And in normal nengo, we do this to get the output of network:
_, acts = nengo.utils.ensemble.tuning_curves(input_neuron, sim, inputs=input)
The reason we ask is, in nengo_dl, after training, the loss is very small, but the prediction performance is very bad, it likes nothing is trained.
The new code is like this:
import nengo
import nengo_dl
import numpy as np
import tensorflow as tf
class Spiking_Qnetwork:
``````def __init__(self, input_shape, output_shape, nb_hidden, weights_path):
'''
Spiking neural network as the Q value function approximation
:param input_shape: the input dimension without batch_size, example: state is 2 dimension,
action is 1 dimenstion, the input shape is 3.
:param output_shape: the output dimension without batch_size, the dimenstion of Q values
:param nb_hidden: the number of neurons in ensemble
:param weights_path: the path to save all parameters
'''
self.input_shape = input_shape
self.output_shape = output_shape
self.nb_hidden = nb_hidden
self.weights_path = weights_path
self.model = self.build()
def encoder_decoder_initialization(self, shape):
'''
:return: initialised encoder or decoder
'''
rng = np.random.RandomState(1)
coders = rng.normal(size=shape)
return coders
def build(self):
encoders = self.encoder_decoder_initialization(shape=(self.nb_hidden, self.input_shape))
decoders = self.encoder_decoder_initialization(shape=((self.nb_hidden, self.output_shape)))
model = nengo.Network(seed=3)
with model:
nengo_dl.configure_trainable(model, default=False)
model.config[nengo.Ensemble].neuron_type = nengo_dl.neurons.SoftLIFRate()
model.config[nengo.Ensemble].gain = nengo.dists.Choice([1])
model.config[nengo.Ensemble].bias = nengo.dists.Uniform(-1, 1)
model.config[nengo.Connection].synapse = None
self.input_node = nengo.Node(size_in=self.input_shape)
layer = nengo.Ensemble(n_neurons=self.nb_hidden,
dimensions=self.input_shape,
encoders=encoders,
)
nengo.Connection(self.input_node, layer)
output = nengo.Node(size_in=self.output_shape)
self.output_p = nengo.Probe(output)
conn = nengo.Connection(layer.neurons,
output,
transform=decoders.T
)
model.config[conn].trainable = True
return model
def training(self, input_data, label, total_nb_dataset, batch_size, nb_epochs=None):
if nb_epochs==None:
nb_epochs = total_nb_dataset//batch_size
sim = nengo_dl.Simulator(self.model,
minibatch_size=batch_size,
step_blocks=1,
device="/gpu:0",
seed=2,
)
sim.train({self.input_node: input_data},
{self.output_p: label},
tf.train.MomentumOptimizer(5e-2, 0.9),
n_epochs=nb_epochs
)
sim.save_params(self.weights_path)
sim.close()
def predict(self, input_data, batch_size=1):
sim = nengo_dl.Simulator(self.model,
minibatch_size=batch_size,
step_blocks=1,
device="/gpu:0",
seed=1)
sim.step(input_feeds={self.input_node: input_data})
output = sim.data[self.output_p]
sim.close()
return output
``````
if name == âmainâ:
``````from keras.datasets import mnist
from keras.utils import np_utils
from sklearn.metrics import accuracy_score
(X_train, y_train), (X_test, y_test) = mnist.load_data()
# data pre-processing
X_train = X_train.reshape(X_train.shape[0], -1) / 255. # normalize
X_test = X_test.reshape(X_test.shape[0], -1) / 255. # normalize
y_train = np_utils.to_categorical(y_train, nb_classes=10)
y_test = np_utils.to_categorical(y_test, nb_classes=10)
X_train_ = np.expand_dims(X_train, axis=1)
X_test_ = np.expand_dims(X_test, axis=1)
y_train_ = np.expand_dims(y_train, axis=1)
y_test_ = np.expand_dims(y_test, axis=1)
model = Spiking_Qnetwork(input_shape=28*28,
output_shape=10,
nb_hidden=1000,
weights_path="/home/huangbo/SpikingDeepRLControl/huangbo_ws/"
"networks/saved_weights/snn_weights")
# training
model.training(input_data=X_train_, label=y_train_, total_nb_dataset=60000, batch_size=600, nb_epochs=50)
output = model.predict(batch_size=X_test.shape[0], input_data=X_test_)
prediction = np.squeeze(output, axis=1)
# evaluate the model
from sklearn.metrics import accuracy_score
acc = accuracy_score(np.argmax(y_test, axis=1), np.argmax(prediction, axis=1))
print "the test acc is:", acc``````
The normal NEF training time will definitely be much faster than gradient descent techniques; that is the main advantage of the NEF optimization, that it is very quick to compute. However, 1000 epochs is almost certainly more than you would need, something on the order of 10 should be enough for MNIST (you are doing the NEF optimization with 1 epoch).
That is correct.
You need to feed the input into the Node, not the Ensemble (the same idea as when you are doing the training). It looks like you are doing that in the code you posted though, in which case that should be correct. One thing to keep in mind is that the loss you are training on is mean squared error, not classification accuracy, so the values you see during training will be different than the test error youâre computing at the end.
I would actually be kind of surprised if that training was working (a single hidden layer with random encoders is much simpler than most mnist networks successfully trained via gradient descent). So my first guess would be that the training isnât actually succeeding, which is why your prediction performance is poor. However, if you are seeing MSE values that do correspond to good performance after training, then there could be something wrong with the weight saving/loading. You could try just doing your prediction right after the `sim.train` (rather than saving the weights and then loading them in `predict`), just to see if that is what is causing the problem.
Hello drasmuss,
Thanks for the help!
First of all, our reinforcement learning algorithm now work with normal nengo simulator. But I am really interested in nengo_dl, so I still want to fix the problem.
I made the changes you mentioned, I use a Node to input data for both prediction and training.
And I also test not save the parameter but directly do the prediction. But unfortunately it still does not work.
Here are the results:
# ----------------------------------------------#
Building networkBuilding completed in 0:00:00
Optimizing graphOptimization completed in 0:00:00
Construction completed in 0:00:00
Using TensorFlow backend.
[##############################] ETA: 0:00:00 (Training)
Training completed in 0:03:23
Simulation startedSimulation completed in 0:00:00
the test acc is: 0.1
And here are the code:
# ----------------------------------------------#
import nengo
import nengo_dl
import numpy as np
import tensorflow as tf
with nengo.Network(seed=0) as model:
``````model.config[nengo.Ensemble].neuron_type = nengo_dl.neurons.SoftLIFRate()
model.config[nengo.Ensemble].gain = nengo.dists.Choice([1])
model.config[nengo.Ensemble].bias = nengo.dists.Uniform(-1, 1)
model.config[nengo.Connection].synapse = None
# initialize encoder and decoder
rng = np.random.RandomState(1)
encoders = rng.normal(size=(1000, 784))
decoders = rng.normal(size=(1000, 10))
# network
input_node = nengo.Node(size_in=784)
layer = nengo.Ensemble(n_neurons=1000,
dimensions=784,
encoders=encoders,
)
nengo.Connection(input_node, layer)
output = nengo.Node(size_in=10)
output_p = nengo.Probe(output)
conn = nengo.Connection(layer.neurons,
output,
transform=decoders.T
)
``````
with nengo_dl.Simulator(model, minibatch_size=60, step_blocks=1, device="/gpu:0", seed=2) as sim:
``````from keras.datasets import mnist
from keras.utils import np_utils
(X_train, y_train), (X_test, y_test) = mnist.load_data()
# data pre-processing
X_train = X_train.reshape(X_train.shape[0], -1) / 255. # normalize
X_test = X_test.reshape(X_test.shape[0], -1) / 255. # normalize
y_train = np_utils.to_categorical(y_train, nb_classes=10)
y_test = np_utils.to_categorical(y_test, nb_classes=10)
X_train_ = np.expand_dims(X_train, axis=1)
X_test_ = np.expand_dims(X_test, axis=1)
y_train_ = np.expand_dims(y_train, axis=1)
y_test_ = np.expand_dims(y_test, axis=1)
sim.train({input_node: X_train_},
{output_p: y_train_},
tf.train.MomentumOptimizer(5e-2, 0.9),
n_epochs=10
)
sim.step(input_feeds={input_node: X_test_[0:60,:,:]})
output = sim.data[output_p]
prediction = np.squeeze(output, axis=1)
# evaluate the model
from sklearn.metrics import accuracy_score
acc = accuracy_score(np.argmax(y_test[0:60,:], axis=1), np.argmax(prediction, axis=1))
print "the test acc is:", acc
``````
Further question:
1, Do we always create a new simulator when the batch_size changes?
When we create the nengo_dl simulator, the batch_size has to be set up. For training we may use bigger batch_size, but for prediction, we just input a single image into the network, so we need set the batch_size to 1.
2, Is the way I do prediction right?
#-------------------------------
sim.step(input_feeds={input_node: X_test_[1,:,:]})
output = sim.data[output_p]
#-------------------------------
I test the save and load parameters by using the loss: print sim.loss({input_node: X_test_}, {output_p: y_test_}, âmseâ).
After training, I create a new simulator, and first load the parameter, the loss also be 0.1.
If I donât load the parameters, the loss will be very big.
#------------------------------------------------------#
0.10000000149
Building completed in 0:00:00
Optimization completed in 0:00:00
Construction completed in 0:00:00
#------------------------------------------------------#
So I think the training actually works, so I guess the problem is the way I do prediction, because this is the only differences between what I have done before with normal nengo simulator.
An MSE of 0.1 probably means that the training was not very successful. E.g., if your network just outputs [0 0 0 0 0 0 0 0 0 0] for every training example, then the MSE will be 0.1. So the problem is in the training, not the prediction. Also note that using MSE as the loss function when performing classification is usually not recommended, for exactly this reason (an apparently low error can actually correspond to poor classification). Youâd probably get better results using cross entropy. You can read more about this here.
However, as I mentioned, training a one-layer network with random encoders to perform MNIST classification is still probably not going to work great. If that is what you want to do, youâre better off using the NEF optimization, because it is optimal for a single layer optimization like that. If you want to see the advantages of deep learning methods, youâll probably need to use a more complex, multi-layer system.
Yes, you have to rebuild the simulator to change the batch size (since it is built into the graph, for performance reasons). However, if you donât want to rebuild the simulator you can just pass in `n` images, and ignore the ones youâre not interested in (e.g., use `sim.data[output_p][0]` to just look at the results for the first image).
Hello drasmuss,
I made the changes you mentioned, I add 2 more layers and change the loss_function to cross entropy. But it still does not works. And I also test different optimizer, the result is still bad.
The output are all 0, and it seems the training does not happens, and I have already set everything to trainable.
#---------------------------------------------------
cross_entropy as loss is: 2.30258536339
the test acc is: 0.098
#---------------------------------------------------
import nengo
import nengo_dl
import numpy as np
import tensorflow as tf
def cross_entropy(prediction, label):
return tf.reduce_mean(tf.nn.softmax_cross_entropy_with_logits(labels=label, logits=prediction))
with nengo.Network(seed=0) as model:
``````nengo_dl.configure_trainable(model, default=True)
model.config[nengo.Ensemble].neuron_type = nengo_dl.neurons.SoftLIFRate()
model.config[nengo.Ensemble].gain = nengo.dists.Choice([1])
model.config[nengo.Ensemble].bias = nengo.dists.Uniform(-1, 1)
model.config[nengo.Ensemble].trainable = True
model.config[nengo.Connection].trainable = True
model.config[nengo.Connection].synapse = None
# initialize encoder and decoder
rng = np.random.RandomState(1)
encoders = rng.normal(size=(1000, 784))
decoders = rng.normal(size=(1000, 10))
# network
input_node = nengo.Node(size_in=784)
layer_1 = nengo.Ensemble(n_neurons=1000,
dimensions=784,
encoders=encoders
)
layer_2 = nengo.Ensemble(n_neurons=1000,
dimensions=784
)
layer_3 = nengo.Ensemble(n_neurons=1000,
dimensions=784,
)
output = nengo.Node(size_in=10)
conn_1 = nengo.Connection(input_node, layer_1)
conn_2 = nengo.Connection(layer_1, layer_2)
conn_3 = nengo.Connection(layer_2, layer_3)
conn_4 = nengo.Connection(layer_3.neurons, output, transform=decoders.T)
output_p = nengo.Probe(output)
``````
with nengo_dl.Simulator(model, minibatch_size=60, step_blocks=1, device="/gpu:0", seed=2) as sim:
``````from keras.datasets import mnist
from keras.utils import np_utils
(X_train, y_train), (X_test, y_test) = mnist.load_data()
# data pre-processing
X_train = X_train.reshape(X_train.shape[0], -1) / 255. # normalize
X_test = X_test.reshape(X_test.shape[0], -1) / 255. # normalize
y_train = np_utils.to_categorical(y_train, nb_classes=10)
y_test = np_utils.to_categorical(y_test, nb_classes=10)
X_train_ = np.expand_dims(X_train, axis=1)
X_test_ = np.expand_dims(X_test, axis=1)
y_train_ = np.expand_dims(y_train, axis=1)
y_test_ = np.expand_dims(y_test, axis=1)
sim.train({input_node: X_train_},
{output_p: y_train_},
#tf.train.MomentumOptimizer(5e-2, 0.9),
n_epochs=1,
objective =cross_entropy
)
print "cross_entropy as loss is:", sim.loss({input_node: X_test_}, {output_p: y_test_}, cross_entropy)
sim.save_params("/home/huangbo/SpikingDeepRLControl/huangbo_ws/networks/saved_weights/snn_weights")
``````
with nengo_dl.Simulator(model, minibatch_size=10000, step_blocks=1, device="/gpu:0", seed=1) as sim:
``````sim.step(input_feeds={input_node: X_test_})
output = sim.data[output_p]
prediction = np.squeeze(output, axis=1)
# evaluate the model
from sklearn.metrics import accuracy_score
acc = accuracy_score(np.argmax(y_test, axis=1), np.argmax(prediction, axis=1))
print "the test acc is:", acc``````
You arenât doing anything wrong, youâre just running into the fact that training deep networks is complicated. There are a lot of hyperparameters to play around with (number of layers, size of each layer, convolutional/fully connected layers, pooling, neuron parameters, learning rates, weight initialization, etc.). You need to have all those parts working well together, or the training wonât succeed (and itâll settle on some low-energy solution like all-zero output, as youâre seeing).
My advice would be to start with an architecture that you know can succeed (there are a lot of MNIST tutorials out there that should give you a place to get started). Then make sure that when you recreate that architecture within your framework, it still works. Then start introducing new elements, like SoftLIFRate neurons. If things stop working, make small adjustments to the architecture to try to find some settings that do work.
2 Likes
Also, Iâve been working on some new features this week that should make it easier to build different kinds of networks in `nengo_dl`, which includes an MNIST example. It should be finished early next week.
2 Likes
Hello drasmuss,
When you finish the new features, please let us know.
And I will try your suggestions.
best greetings.
You can see the example here. Note that that example includes some features that werenât in the latest official release, so you will need to do a developer installation.
1 Like
Hello drasmuss,
Thank for your example of mnist. It really helps, it not only solves the mnist problem, but also make me more clear about how to use nengo_dl.
And now I know why my early training failed. I think the problem is the way I define the input Node.
If I just define the input node like:
input_node = nengo.Node(size_in = self.input_shape)
It seems that data are not feed into the network, I need use nengo.processes.PresentInput.
After I change it to:
input = nengo.Node(nengo.processes.PresentInput(np.zeros(shape=(1, self.input_shape)), 0.1))
The problem is solved.
Ah yes, thatâs an example of the same problem as we identified previously, where if you try to feed input into an invalid Node it just silently fails instead of giving an error message. Iâll fix that tomorrow!
Hello dear drasmuss,
I donât know do not use nengo_dl in the right way or any reason, I found the memory leak when I use nengo_dl.
For our reinforcement learning project, we will collect the state and action Paars, but our code die after collect 800 data, and we checked the memory, it is fulled.
and you can also see from this mnist example. After we load the mnist_data, the memory should not increased that much during the training. Here is the code, can you please test it:
``````import nengo
import nengo_dl
import numpy as np
import tensorflow as tf
from copy import deepcopy
class Deep_qNetwork_snn:
'''
Q_function approximation using Spiking neural network
'''
def __init__(self, input_shape, output_shape, save_path):
'''
:param input_shape: the input shape of network, a number of integer
:param output_shape: the output shape of network, a number of integer
:param save_path: the path to save network parameters, in the prediction, network will load the weights in
this path.
example: '/home/huangbo/Desktop/weights/mnist_parameters'
'''
self.input_shape = input_shape
self.output_shape = output_shape
self.save_path = save_path
self.softlif_neurons = nengo_dl.SoftLIFRate(tau_rc=0.02, tau_ref=0.002, sigma=0.002)
self.ens_params = dict(max_rates=nengo.dists.Choice([100]), intercepts=nengo.dists.Choice([0]))
self.amplitude = 0.01
def build_network(self):
# input_node
input = nengo.Node(nengo.processes.PresentInput(np.zeros(shape=(1, self.input_shape)), 0.1))
# layer_1
x = nengo_dl.tensor_layer(input, tf.layers.dense, units=100)
x = nengo_dl.tensor_layer(x, self.softlif_neurons, **self.ens_params)
# layer_2
x = nengo_dl.tensor_layer(x, tf.layers.dense, transform=self.amplitude, units=100)
x = nengo_dl.tensor_layer(x, self.softlif_neurons, **self.ens_params)
# output
x = nengo_dl.tensor_layer(x, tf.layers.dense, units=self.output_shape)
return input, x
def choose_optimizer(self, opt, learning_rate=1):
elif opt == "rms":
optimizer = tf.train.RMSPropOptimizer(learning_rate=learning_rate)
elif opt == "sgd":
return optimizer
def objective(self, x, y):
return tf.nn.softmax_cross_entropy_with_logits(logits=x, labels=y)
def training(self, minibatch_size, train_whole_dataset, train_whole_labels, num_epochs):
'''
Training the network, objective will be the loss function, default is 'mse', but you can alse define your
own loss function, weights will be saved after the training.
:param minibatch_size: the batch size for training.
:param train_whole_dataset: whole training dataset, the nengo_dl will take minibatch from this dataset
:param train_whole_labels: whole training labels
:param num_epochs: how many epoch to train the whole dataset
:param pre_train_weights: if we want to fine-tuning the network, load weights before training
:return: None
'''
with nengo.Network(seed=0) as model:
nengo_dl.configure_trainable(model, default=True)
input, output = self.build_network()
out_p = nengo.Probe(output)
train_inputs = {input: train_whole_dataset}
train_targets = {out_p: train_whole_labels}
with nengo_dl.Simulator(model, minibatch_size=minibatch_size) as sim:
if self.save_path is not None:
try :
except:
pass
# construct the simulator
sim.train(train_inputs, train_targets, optimizer, n_epochs=num_epochs, objective='mse')
# save the parameters to file
sim.save_params(self.save_path)
'''
prediction of the network
:param prediction_input: a input data shape = (minibatch_size, 1, input_shape)
:param minibatch_size: minibatch size, default = 1
:return: prediction with shape = (minibatch_size, output_shape)
'''
with nengo.Network(seed=0) as model:
nengo_dl.configure_trainable(model, default=False)
input, output = self.build_network()
out_p = nengo.Probe(output)
with nengo_dl.Simulator(model, minibatch_size=minibatch_size) as sim:
try:
except:
pass
input_data = {input: prediction_input}
sim.step(input_feeds = input_data)
output = np.squeeze(sim.data[out_p], axis=1)
return deepcopy(output)
if __name__ == '__main__':
import matplotlib.pyplot as plt
from tensorflow.examples.tutorials.mnist import input_data
from sklearn.metrics import accuracy_score
X_test = mnist.test.images
y_test = mnist.test.labels
deep_qNetwork = Deep_qNetwork_snn(input_shape=784,
output_shape=10,
save_path='/home/huangbo/Desktop/weights/mnist_parameters'
)
for i in range(10):
deep_qNetwork.training(minibatch_size=32,
train_whole_dataset = mnist.train.images[:, None, :],
train_whole_labels = mnist.train.labels[:, None, :],
num_epochs = 1
)
test_input = X_test[:, None, :]
In the meantime, you can avoid the problem by not closing and reopening the session every time you call `predict/training`. For example, just do `self.my_predict_sim = nengo_dl.Simulator(...)` once, and then you can use `self.my_predict_sim.load_params(...)` or `self.my_predict_sim.step(...)` within your `prediction` function.
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## Monday, April 18, 2016
### Connect 4 With My Kid
Inspired by Joe Schwartz's Connect 4 using a multiplication grid, I tried it out on my unsuspecting 6-year-old.
A little background. K's understanding of multiplication is in the frame of repeated addition. She can count by 2, 5 and 10.
The beginning of the game was quick. K was able to calculate multiples of ten quickly, even commenting "Oh, this is easy". When she got to her first entry in the 9 row, she interpreted the multiplication as 9 groups of whatever number is at the top of the slot. So she saw 9x4 as 4+4+... 9 times. She used this strategy for most of row of 9s, but when she did 9x7, she suddenly switched strategies and added 9 to 54. I asked her about the strategy and she said each spot is 9 more than the one before it. As the game progressed, she used a similar strategy vertically, recognizing that each column had its own progression.
Obviously, I was very pleased, as she was both enjoying the game and working on some important foundation arithmetic. Her strategies revealed a lot about how she interpreted the game. She noticed patterns and created her own meanings. I was also really intrigued by her choosing to consider the multiplication m*n as m groups of size n, but then proceed in rows by adding a group of m.
Overall, the game is a win for math, but I learned that my kid needs to work on her Connect 4 strategy...
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Drake
random_piecewise_polynomial.h
Go to the documentation of this file.
1 #pragma once
2
3 #include <vector>
4
5 #include <Eigen/Core>
6
8
9
10 namespace drake {
11 namespace trajectories {
12 namespace test {
13
14 /**
15 * Obtains a random PiecewisePolynomial with the given @p segment_times. Each
16 * segment will have a matrix of random Polynomials of the specified size.
17 */
18 template<typename CoefficientType = double>
20 MakeRandomPiecewisePolynomial(Eigen::Index rows, Eigen::Index cols,
21 Eigen::Index num_coefficients_per_polynomial,
22 const std::vector<double> &segment_times) {
23 Eigen::Index num_segments =
24 static_cast<Eigen::Index>(segment_times.size() - 1);
25 typedef Polynomial<CoefficientType> PolynomialType;
26 typedef Eigen::Matrix<PolynomialType, Eigen::Dynamic, Eigen::Dynamic>
27 PolynomialMatrix;
28 std::vector<PolynomialMatrix> polynomials;
29 for (Eigen::Index segment_index = 0; segment_index < num_segments;
30 ++segment_index) {
31 polynomials.push_back(
32 drake::test::RandomPolynomialMatrix<CoefficientType>(
33 num_coefficients_per_polynomial, rows, cols));
34 }
35 return PiecewisePolynomial<CoefficientType>(polynomials,
36 segment_times);
37 }
38
39 } // namespace test
40 } // namespace trajectories
41 } // namespace drake
Definition: automotive_demo.cc:105
A scalar multi-variate piecewise polynomial.
Definition: piecewise_polynomial.h:47
PiecewisePolynomial< CoefficientType > MakeRandomPiecewisePolynomial(Eigen::Index rows, Eigen::Index cols, Eigen::Index num_coefficients_per_polynomial, const std::vector< double > &segment_times)
Obtains a random PiecewisePolynomial with the given segment_times.
Definition: random_piecewise_polynomial.h:20
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Help implementing finite difference scheme for heat equation
I am trying to solve the following problem via a finite difference approximation:
$u_t = k \, u_{xx}$, on $0 < x < L$ and $t > 0$;
$u(0,t) = u(L,t) = 0$;
$u(x,0) = f(x)$.
I take $u(x,0) = f(x) = x^2$ for my problem.
Programming is not my forte at all, so I am having some issues with the implementation. Here is my code:
## This program is to implement a Finite Difference method approximation
## to solve the Heat Equation, u_t = k * u_xx,
## in 1D w/out sources & on a finite interval 0 < x < L. The PDE
## is subject to B.C: u(0,t) = u(L,t) = 0,
## and the I.C: u(x,0) = f(x).
import numpy as np
import matplotlib.pyplot as plt
# definition of solution to u_t = k * u_xx
def u(x,t):
return u
# definition of initial condition function
def f(x):
return x^2
# parameters
L = 1
T = 10
N = 10
M = 10
s = 0.25
# uniform mesh
x_init = 0
x_end = L
dx = float(x_end - x_init) / N
# time discretization
t_init = 0
t_end = T
dt = float(t_end - t_init) / M
t = np.zeros(M+1)
t = np.arange(t_init, t_end, dt)
# Boundary Conditions
for m in xrange(0, M):
t[m] = m * dt
u(0, t[m]) = 0
u(N, t[m]) = 0
# Initial Conditions
for j in xrange(0, N):
x[j] = j * dx
u(x[j], 0) = f(x[j])
# finite difference scheme
for j in xrange(1, N-1):
u(x[j],t[m+1]) = u(x[j],t[m]) + s * ( u(x[j+1],t[m]) -
2 * u(x[j],t[m]) + u(x[j-1],t[m]) )
So in particular, I am wondering (1) if my function definitions are okay? (2) is my uniform mesh and time discretization okay? (3) am I selecting reasonable values for the parameters? (I know that the approximation is more accurate for smaller $\Delta x$ and $\Delta t$, with the disadvantage of increased computing time..)
(4) how are my loops (for the initial/boundary conditions)? I get the feeling I need a nested loop for the finite difference scheme.. One for loop for the values of j and another for loop going through each value m in time. Is this correct? ... Also, I keep getting the error: u(0, t[m]) = 0 "can't assign to function call." What is wrong there?
Thanks in advance for the help! As I mentioned, my programming skills are not very strong so I feel like a fish out of water doing this.
• def u(x, t): return u, while it happens to be valid python, is not at all meaningful here. Perhaps it might be easier to start with a beginner programming textbook of some sort? – Kirill Jul 5 '16 at 2:05
• Javier can you please post your final code – user22301 Nov 10 '16 at 8:56
So your first big issue is one you bring up in point (4), where you say you get the error 'u(0, t[m]) = 0 "can't assign to function call"'. You're trying to store data by assigning data to a function. That won't work. The way you're implementing this code, you actually should make u(x,t) be represented by a 2-D array that can be accessed via u[m][j], where m is associated with the $m^{th}$ time value, and j is associated with the $j^{th}$ spatial value.
As you also note, you should have a nested for loop. Do an outside loop for time, and then an inner loop for looping through the spatial discretization.
Now assuming your value s makes sense for your problem, the other potential issue I see is your step size, dt. I am pretty sure this time stepsize won't be stable based on the value of dx since it seems you're just using an Explicit Euler time stepping scheme. Try picking some dt that is proportional to dx*dx, like dt=0.1*dx*dx, and see how it fairs for you.
• Hey, just wanted to say thanks so much for your help. You were very helpful in getting my program to work :) – Javier Jul 12 '16 at 0:31
• @Javier no problem man! – spektr Jul 12 '16 at 0:37
Don't worry about programming skills, everyone's a beginner at some point. It's a good start actually. I'm just going to give some hints assuming that this is an exercise. Actually, choward's answer is complete, this is just a rephrasing of his suggestions in, I hope, a more beginner friendly language. So, if you do accept an answer it should be his ;)
1. Your definition of the solution does not make sense in Python. I would suggest storing it in a NxM numpy array. That way you can easily define the initial and boundary conditions with the slice notation (i.e. U[:,0] = x**2 where x is an 1-dim array, see numpy.linspace).
2. It seems that you are using the central difference scheme for the second order derivative in space. The definition is usually $$u_{xx}(x_i) = \frac{u_{i+1} - 2u_i + u_{i-1}}{h^2}$$ where $h$ is the step size in space (your dx).
3. You are using a explicit time stepping scheme called explicit Euler or forward Euler which is conditionally stable. This means that your discretisation has to fulfil a certain condition in order for your solution to remain bounded or, in other words, that your time step has to be small enough. Another around this would be to use an implicit method such as backward Euler which is unconditionally stable. Note that, while this is fine for the heat equation which is intrinsically stable, it's maybe not the best choice for neutral or unstable problems since your numerical solution may suffer from over-stabilisation (a.k.a. numerical damping).
4. You have only implemented the inner loop. You would need an outer loop to advance in time as well.
• Hi, thank you so much. Your comment has been very helpful. I am still having some issues though. First, may I define the solution as follows: $u = np.zeros((N,M))$? I think it defines an $NxM$ matrix with zeros for all entries, but I am not sure. Then my loop should just fill in the values as it iterates.... Also you said I can set the initial condition with $u[:,0] = x**2$. So can I similarly define the B.C. with $u[0,:] = 0$ and $u[N,:] = 0$? When I run it this way I get the error that for $u[N,:] = 0$ the "index 10 is out of bounds for axis 0 with size 10". How can I correct this? – Javier Jul 6 '16 at 19:46
• Hey, no worries man! Yeah, using numpy.zeros is one way to do it. What you wrote actually creates an array of size $N \times M$ so the indices go from 0 to $N-1$ and $M-1$ respectively. Therefore, indices $N$ or $M$ are out of bounds :) – Chris Jul 11 '16 at 12:42
• I actually got it to work. Thanks so much for your help :) – Javier Jul 12 '16 at 0:30
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# Derivatives of Log
1. Nov 30, 2008
### 10min
1. The problem statement, all variables and given/known data
Question is
f(x) = log10(x/x-1), find f `(x)
2. Relevant equations
so I used the formula
d/dx loga(u) = 1/uLNa . (d/dx U)
3. The attempt at a solution
(1/(x/x-1)*ln10)*1/(x-1)^2 I used quotent rule for d/dx U part.
editing
1/x(x-1)ln10
or
1/x2-x ln10
thank you
Last edited: Nov 30, 2008
2. Nov 30, 2008
### lurflurf
it may be easier to use the log property
log(ab)=log(a)+log(b)
log(x/(x-1))=log(x)-log(x-1)
[log10(x/(x-1))]'=[log(x/(x-1))]'/log(10)
=[log(x)-log(x-1)]'/log(10)
=([log(x)]'-[log(x-1)]')/log(10)
3. Nov 30, 2008
### 10min
so many answer is not right
4. Nov 30, 2008
### lurflurf
Just a sign error
Quotient rule is
(u/v)'=(u'v-uv')/v^2
so
(x/(x-1))'=(x'(x-1)-x(x-1)')/(x-1)^2
=(1(x-1)-x(1))/(x-1)^2
=-1/(x-1)^2
not
1/(x-1)
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Use the saturation framework, others
May 2, 2017
Isomorphic problems are those that have the same underlying structure. Recognizing that that new problem you’re facing is similar to one that has been solved before, or a simpler one, can go a long way in improving your creativity.
Meet the omnipresent saturation framework
Perhaps the best-known example of an issue map/logic tree is the profitability problem. If you have been to business school or are preparing for a case interview, you’ve seen it. It starts with a variation of “how can we increase our profitability?,” and breaks down into “by increasing revenues” and “by decreasing costs.” It then breaks down these components into further details. As you drill deeper and organize elements in a MECE way, you may end up with something like this:
[IMAGE MISSING: saturation-2-582x1024.jpg]
A profitability issue map relies on a saturation framework as it separates supply and demand.
If you step back from the specificities of this problem, you may find that the framework supporting this issue map is rather common; I’ve come to call it the “saturation framework.”
In a saturation situation, there’s a mismatch between supply and demand. Too small a parking lot? That’s a saturation problem. Figuring out your Ironman race day nutrition? Saturation problem. Congested driveways in Stockholm? Saturation. Not enough people to do the job in your team? Same. United Airlines dragging passengers off their plane? A disastrous solution to a saturation problem. Worrying about having enough money to retire? Yep. Controlling your weight? Check. Planning to drive your Tesla to that place that’s 10% outside its advertised range? You get the idea.
The saturation framework might be one of the most useful out there, but it certainly isn’t the only one. So there’s value in seeing past the surface features of your problem to check whether its structure is analogous to one that you’ve already solved or that is easier to solve.
When I teach this stuff, I advise my students to consider using an existing framework as a starting point (for examples for standard frameworks, see Chevallier, 2016, pp. 72–75). However, I urge them to not assume that the framework is perfect as is. Indeed, many of those aren’t MECE, for instance, including such household standards as the 4P / marketing mix. So those existing frameworks might provide a good basis on which to build, but chances are that you’ll need to adapt them, sometimes to correct an inherent weakness—say, poor MECEness—and sometimes to make them more insightful for your specific problem.
Leverage analogous problems
If you can step back and realize that the structure of your problem is analogous to another one, you may gain insight that can dramatically improving your understanding of the problem (see literature review in Lin).To use analogies for solving unfamiliar problems, identify a familiar source and map out the correspondence between its components and those of your target problem. Let’s say that you need to diagnose why your company cannot deliver your products on time but that you know nothing about logistics. An analogous problem might be “why am I late to work?” If you can solve that simple, more-familiar problem—not that am I implying anything about your punctuality or lack thereof, dear reader, this is just a hypothetical example; you, of course, are never late—then you can map the solution back to your original (target) problem (Chevallier, 2016, pp. 70–72).
[IMAGE MISSING: saturation-3-1024x415.jpg]
Using an analogous, simpler problem may be an effective way of thinking about one that you’re not familiar with
As with all things in problem solving, leveraging analogous problems doesn’t come for free. Here, I can think of three ways in which the cost manifests itself:
1. You need to invest effort into thinking about a source problem and, if it’s not already solved, solve it.
2. Analogies can be restricting (see Thibodeau, 2011; Gilovich, 1981; Chevallier, 2016, pp. 22–23), so, ideally, you’ll consider at least two.
3. Analogous thinking is an inductive process and, therefore, it is uncertain (Holyoak, p. 235, 239).
But even considering those drawbacks, I find that actively trying to step back from my problem and look past its surface features to see its underlying structure is usually a useful exercise that usually results in more creativity.
References:
Chevallier, A. (2016). Strategic Thinking in Complex Problem Solving. Oxford, UK, Oxford University Press.Gilovich, T. (1981). “Seeing the past in the present: The effect of associations to familiar events on judgments and decisions.” Journal of Personality and Social Psychology
40(5): 797.Holyoak, K. J. (2012). Analogy and relational reasoning. The Oxford Handbook of Thinking and Reasoning. K. J. Holyoak and R. G. Morrison. New York, Oxford University Press
234-259.Lin, S.-Y. and C. Singh (2011). “Using isomorphic problems to learn introductory physics.” Physical Review Special Topics-Physics Education Research
7(2): 020104.Thibodeau, P. H. and L. Boroditsky (2011). “Metaphors we think with: The role of metaphor in reasoning.” PLoS One
6(2): e16782.
Van Waterschoot, W., & Van den Bulte, C. (1992). The 4P classification of the marketing mix revisited. The Journal of Marketing, 83–93.
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# Parametric equations and a value for the parameter t are given x = (80cos 45^o)t, y = 6
Parametric equations and a value for the parameter t are given $$\displaystyle{x}={\left({80}{\cos{{45}}}^{{o}}\right)}{t},{y}={6}+{\left({80}{\sin{{45}}}^{{o}}\right)}{t}-{16}{t}^{{2}}$$. t = 2.
Find the coordinates of the point on the plane curve described by the parametric equations corresponding to the given value of t.
• Questions are typically answered in as fast as 30 minutes
### Plainmath recommends
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.
Theodore Schwartz
Solution:
1.For t=2, $$\displaystyle{x}={\left({80} \cos{{45}}^{o}\right)}{t}={\left({80} \cos{{45}}^{o}\right)}{\left({2}\right)}={80}\cdot\frac{\sqrt{{2}}}{{2}}\cdot{2}={80}\sqrt{{2}}$$
For t=2, $$\displaystyle{y}={6}+{\left({80} \sin{{45}}^{o}\right)}{t}-{16}$$
$$\displaystyle{t}^{2}={6}+{\left({80} \sin{{45}}\right)}{\left({2}\right)}={16}{\left({2}\right)}^{2}={6}{\left({80}\cdot\frac{\sqrt{{2}}}{{2}}\cdot{2}\right)}-{16}{\left({4}\right)}={6}+{80}\sqrt{{2}}-{64}={80}\sqrt{{2}}-{58}$$
2. So the coordinate of the point is=(x,y)=
$$\displaystyle{\left({80}\sqrt{{2}},{8}-\sqrt{{2}}-{58}\right)}$$
Answer:$$\displaystyle{\left({80}\sqrt{{2}},{8}-\sqrt{{2}}-{58}\right)}$$
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# Probabilistic Momentum with Intraday data
March 30, 2014
By
(This article was first published on Systematic Investor » R, and kindly contributed to R-bloggers)
I want to follow up the Intraday data post with testing the Probabilistic Momentum strategy on Intraday data. I will use Intraday data for SPY and GLD from the Bonnot Gang to test the strategy.
```##############################################################################
# Load Systematic Investor Toolbox (SIT)
# http://systematicinvestor.wordpress.com/systematic-investor-toolbox/
###############################################################################
setInternet2(TRUE)
con = gzcon(url('http://www.systematicportfolio.com/sit.gz', 'rb'))
source(con)
close(con)
#*****************************************************************
#******************************************************************
# data from http://thebonnotgang.com/tbg/historical-data/
# please save SPY and GLD 1 min data at the given path
spath = 'c:/Desktop/'
data1 <- new.env()
data1\$FI = data\$GLD
data1\$EQ = data\$SPY
data = data1
bt.prep(data, align='keep.all', fill.gaps = T)
lookback.len = 120
confidence.level = 60/100
prices = data\$prices
ret = prices / mlag(prices) - 1
models = list()
#*****************************************************************
# Simple Momentum
#******************************************************************
momentum = prices / mlag(prices, lookback.len)
data\$weight[] = NA
data\$weight\$EQ[] = momentum\$EQ > momentum\$FI
data\$weight\$FI[] = momentum\$EQ <= momentum\$FI
models\$Simple = bt.run.share(data, clean.signal=T)
#*****************************************************************
# Probabilistic Momentum + Confidence Level
# http://cssanalytics.wordpress.com/2014/01/28/are-simple-momentum-strategies-too-dumb-introducing-probabilistic-momentum/
#******************************************************************
ir = sqrt(lookback.len) * runMean(ret\$EQ - ret\$FI, lookback.len) / runSD(ret\$EQ - ret\$FI, lookback.len)
momentum.p = pt(ir, lookback.len - 1)
data\$weight[] = NA
data\$weight\$EQ[] = iif(cross.up(momentum.p, confidence.level), 1, iif(cross.dn(momentum.p, (1 - confidence.level)), 0,NA))
data\$weight\$FI[] = iif(cross.dn(momentum.p, (1 - confidence.level)), 1, iif(cross.up(momentum.p, confidence.level), 0,NA))
models\$Probabilistic = bt.run.share(data, clean.signal=T)
data\$weight[] = NA
data\$weight\$EQ[] = iif(cross.up(momentum.p, confidence.level), 1, iif(cross.up(momentum.p, (1 - confidence.level)), 0,NA))
data\$weight\$FI[] = iif(cross.dn(momentum.p, (1 - confidence.level)), 1, iif(cross.up(momentum.p, confidence.level), 0,NA))
models\$Probabilistic.Leverage = bt.run.share(data, clean.signal=T)
#*****************************************************************
# Create Report
#******************************************************************
strategy.performance.snapshoot(models, T)
```
Next, let’s examine the hourly perfromance of the strategy.
``` #*****************************************************************
# Hourly Performance
#******************************************************************
strategy.name = 'Probabilistic.Leverage'
ret = models[[strategy.name]]\$ret
ret.number = 100*as.double(ret)
dates = index(ret)
factor = format(dates, '%H')
layout(1:2)
par(mar=c(4,4,1,1))
boxplot(tapply(ret.number, factor, function(x) x),outline=T, main=paste(strategy.name, 'Distribution of Returns'), las=1)
barplot(tapply(ret.number, factor, function(x) sum(x)), main=paste(strategy.name, 'P&L by Hour'), las=1)
```
There are lots of abnormal returns in the 9:30-10:00am box due to big overnight returns. I.e. a return from today’s open to prior’s day close. If we exclude this observation every day, the distribution each hour is more consistent.
``` #*****************************************************************
# Hourly Performance: Remove first return of the day (i.e. overnight)
#******************************************************************
ret.number[day.stat\$day.start] = 0
layout(1:2)
par(mar=c(4,4,1,1))
boxplot(tapply(ret.number, factor, function(x) x),outline=T, main=paste(strategy.name, 'Distribution of Returns'), las=1)
barplot(tapply(ret.number, factor, function(x) sum(x)), main=paste(strategy.name, 'P&L by Hour'), las=1)
```
The strategy performs best in the morning and dwindles down in the afternoon and overnight.
These hourly seasonality plots are just a different way to analyze performance of the strategy based on Intraday data.
To view the complete source code for this example, please have a look at the bt.strategy.intraday.thebonnotgang.test() function in bt.test.r at github.
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### 物理代写|核物理代写nuclear physics代考|ERICE2022
statistics-lab™ 为您的留学生涯保驾护航 在代写核物理nuclear physics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写核物理nuclear physics代写方面经验极为丰富，各种代写核物理nuclear physics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 物理代写|核物理代写nuclear physics代考|Inconsistency of the “Plum Pudding” Model
Let us consider what would be expected if the “Plum Pudding” model were indeed correct.
We know from Gauss’ law that at a distance $r$ from the centre of the atom, the electric field is determined by the charge enclosed in a sphere of radius $r$ surrounding the centre of the atom.
The volume of a sphere of radius $r$ is proportional to $r^{3}$. Therefore for $r$ smaller than the radius, $R$, of the atom, the electric charge enclosed with a sphere of radius $r$ is a fraction $r^{3} / R^{3}$ of the total electric charge (assuming a uniform distribution of electric charge throughout the “dough”), so that the magnitude of the electric field at a distance $r$ from the centre of the atom is given by
$$\left(\frac{r^{3}}{R^{3}}\right) \frac{Z e}{4 \pi \varepsilon_{0} r^{2}},(r \leq R) .$$
This is a maximum for $r=R$. This means that the scattering angle cannot be larger than the scattering angle corresponding to impact parameter $b=R$. For values of impact parameter $b<R$, the scattering angle decreases as $b$ decreases.
We have seen above that for $\alpha$-particles with typical kinetic energy of $5 \mathrm{MeV}$, this corresponds to a maximum scattering angle of around $3 \times 10^{-4}$ radians $\left(\approx 0.017^{\circ}\right)$. Such an angle would have been far too small to be observed in any of the GeigerMarsden experiments and they certainly would not have observed any scattering exceeding $90^{\circ}$.
## 物理代写|核物理代写nuclear physics代考|Confirmation of Rutherford Scattering Cross Section
In 1913, Geiger and Marsden [18] performed a far more accurate experiment to check the details of Rutherford’s formula (1.12). They checked the dependence of the rate on the scattering angle and found consistency with the prediction
$$N(\theta) \propto \frac{1}{\sin ^{4}(\theta / 2)}$$
Their results, shown in Fig. 1.5, agree remarkably well.
By using foils of different thickness, they showed that the number of particles scattered through a given angle was proportional to the thickness of the foil, and by using foils made from different metals (tin, silver, copper and aluminium) they were able to show that this number was proportional to the square of the atomic number, $Z$, of the material of the foil.
They were able to slow down the incident $\alpha$-particles, by placing thin sheets of mica immediately in front of the radioactive source. From this they were able to verify that the number of scattered particles was inversely proportional to the fourth power of their velocity, as indicated in (1.12).
## 物理代写|核物理代写nuclear physics代考|Inconsistency of the “Plum Pudding” Model
$$\left(\frac{r^{3}}{R^{3}}\right) \frac{Z e}{4 \pi \varepsilon_{0} r^{2}},(r \leq R)$$
## 物理代写|核物理代写nuclear physics代考|Confirmation of Rutherford Scattering Cross Section
1913 年，Geiger 和 Marsden [18] 进行了更准确的实验来检查卢瑟福公式 (1.12) 的细节。他们检查了速率对散 射角的依赖性，并发现与预测的一致性
$$N(\theta) \propto \frac{1}{\sin ^{4}(\theta / 2)}$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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# Math--Statistics
posted by on .
Two questions that I'm unsure of are below:
1). The variability of a statistic is described by:
a) the spread of its sampling distribution.
b) the amount of bias present
c) the vagueness in the wording of the question used to collect the sample data
d) the stability of the population it describes
(I chose letter "d"--but I'm not sure if this is correct.)
2) The law of large numbers states that as the number of observations drawn at random from a population with finite mean μ increases, the mean x̄ of the observed values:
a) gets larger and larger
b) gets smaller and smaller
c) tends to get closer and closer to the population mean μ
d) fluctuates steadily between one standard deviation above and one standard deviation below the mean
(I chose choice "c"--but again, am unsure of my answer)
• Math--Statistics - ,
1. A, variability is expressed in terms of range, standard deviation or variance.
2. C is correct.
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SOLVED
# Help with what formula to use
Copper Contributor
# Help with what formula to use
Hello everyone
I need help from this community.
I have table with data like that.
Column A - there are dates from 1.1. 2023 till today and every day new day is added
Comun B, C, D - there are diferent numbers like value, quantity etc...
My ideal set is that I have 2 scroll down list where I will pick up starting date and ending date. And formula will sum up all number from column B, C or D between those two picked dates.
Is it possible to do ??
2 Replies
best response confirmed by Spanek1740 (Copper Contributor)
Solution
# Re: Help with what formula to use
it is possible to achieve your desired functionality using Excel formulas. You can use the SUMIFS function to sum values in a range based on multiple criteria, in this case, the date range selected from the drop-down lists.
Here's how you can set it up:
1. Create two drop-down lists for selecting the start date and end date. You can use Data Validation to create these lists. Let's say these drop-down lists are located in cells F1 and G1, respectively.
2. In cell H1, you can use the following formula to sum values in column B between the selected start and end dates:
SUMIFS(B:B, A:A, ">="&F1, A:A, "<="&G1)
This formula sums values in column B (change B:B to C:C for column C, and D:D for column D) where the date in column A is greater than or equal to the start date (F1) and less than or equal to the end date (G1).
3. Drag or copy the formula in cell H1 to cells I1 and J1 to calculate the sums for columns C and D, respectively.
Now, when you select the start date and end date from the drop-down lists in cells F1 and G1, the sums for the selected date range will be calculated in cells H1, I1, and J1 for columns B, C, and D, respectively.
Remember to adjust the cell references and ranges as needed based on your actual data layout. Also, make sure your dates in column A are formatted as dates, not text. The text was created with the help of AI.
My answers are voluntary and without guarantee!
Was the answer useful? Mark as best response and like it!
This will help all forum participants.
# Re: Help with what formula to use
Checkout the solution provided using Filter function in attached worksheet.
1 best response
Accepted Solutions
best response confirmed by Spanek1740 (Copper Contributor)
Solution
# Re: Help with what formula to use
it is possible to achieve your desired functionality using Excel formulas. You can use the SUMIFS function to sum values in a range based on multiple criteria, in this case, the date range selected from the drop-down lists.
Here's how you can set it up:
1. Create two drop-down lists for selecting the start date and end date. You can use Data Validation to create these lists. Let's say these drop-down lists are located in cells F1 and G1, respectively.
2. In cell H1, you can use the following formula to sum values in column B between the selected start and end dates:
SUMIFS(B:B, A:A, ">="&F1, A:A, "<="&G1)
This formula sums values in column B (change B:B to C:C for column C, and D:D for column D) where the date in column A is greater than or equal to the start date (F1) and less than or equal to the end date (G1).
3. Drag or copy the formula in cell H1 to cells I1 and J1 to calculate the sums for columns C and D, respectively.
Now, when you select the start date and end date from the drop-down lists in cells F1 and G1, the sums for the selected date range will be calculated in cells H1, I1, and J1 for columns B, C, and D, respectively.
Remember to adjust the cell references and ranges as needed based on your actual data layout. Also, make sure your dates in column A are formatted as dates, not text. The text was created with the help of AI.
My answers are voluntary and without guarantee!
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# Properties
Label 2888.1.f.d Level $2888$ Weight $1$ Character orbit 2888.f Self dual yes Analytic conductor $1.441$ Analytic rank $0$ Dimension $3$ Projective image $D_{9}$ CM discriminant -8 Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [2888,1,Mod(723,2888)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(2888, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([1, 1, 0]))
N = Newforms(chi, 1, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("2888.723");
S:= CuspForms(chi, 1);
N := Newforms(S);
Level: $$N$$ $$=$$ $$2888 = 2^{3} \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 2888.f (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$1.44129975648$$ Analytic rank: $$0$$ Dimension: $$3$$ Coefficient field: $$\Q(\zeta_{18})^+$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{3} - 3x - 1$$ x^3 - 3*x - 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 152) Projective image: $$D_{9}$$ Projective field: Galois closure of 9.1.69564674215936.1 Artin image: $D_9$ Artin field: Galois closure of 9.1.69564674215936.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + q^{2} - \beta_1 q^{3} + q^{4} - \beta_1 q^{6} + q^{8} + (\beta_{2} + 1) q^{9}+O(q^{10})$$ q + q^2 - b1 * q^3 + q^4 - b1 * q^6 + q^8 + (b2 + 1) * q^9 $$q + q^{2} - \beta_1 q^{3} + q^{4} - \beta_1 q^{6} + q^{8} + (\beta_{2} + 1) q^{9} + ( - \beta_{2} + \beta_1) q^{11} - \beta_1 q^{12} + q^{16} - q^{17} + (\beta_{2} + 1) q^{18} + ( - \beta_{2} + \beta_1) q^{22} - \beta_1 q^{24} + q^{25} + ( - \beta_1 - 1) q^{27} + q^{32} + ( - \beta_{2} + \beta_1 - 1) q^{33} - q^{34} + (\beta_{2} + 1) q^{36} + \beta_{2} q^{41} - q^{43} + ( - \beta_{2} + \beta_1) q^{44} - \beta_1 q^{48} + q^{49} + q^{50} + \beta_1 q^{51} + ( - \beta_1 - 1) q^{54} + ( - \beta_{2} + \beta_1) q^{59} + q^{64} + ( - \beta_{2} + \beta_1 - 1) q^{66} + \beta_{2} q^{67} - q^{68} + (\beta_{2} + 1) q^{72} + ( - \beta_{2} + \beta_1) q^{73} - \beta_1 q^{75} + (\beta_1 + 1) q^{81} + \beta_{2} q^{82} + \beta_{2} q^{83} - q^{86} + ( - \beta_{2} + \beta_1) q^{88} - q^{89} - \beta_1 q^{96} + \beta_{2} q^{97} + q^{98} + (\beta_1 - 1) q^{99}+O(q^{100})$$ q + q^2 - b1 * q^3 + q^4 - b1 * q^6 + q^8 + (b2 + 1) * q^9 + (-b2 + b1) * q^11 - b1 * q^12 + q^16 - q^17 + (b2 + 1) * q^18 + (-b2 + b1) * q^22 - b1 * q^24 + q^25 + (-b1 - 1) * q^27 + q^32 + (-b2 + b1 - 1) * q^33 - q^34 + (b2 + 1) * q^36 + b2 * q^41 - q^43 + (-b2 + b1) * q^44 - b1 * q^48 + q^49 + q^50 + b1 * q^51 + (-b1 - 1) * q^54 + (-b2 + b1) * q^59 + q^64 + (-b2 + b1 - 1) * q^66 + b2 * q^67 - q^68 + (b2 + 1) * q^72 + (-b2 + b1) * q^73 - b1 * q^75 + (b1 + 1) * q^81 + b2 * q^82 + b2 * q^83 - q^86 + (-b2 + b1) * q^88 - q^89 - b1 * q^96 + b2 * q^97 + q^98 + (b1 - 1) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$3 q + 3 q^{2} + 3 q^{4} + 3 q^{8} + 3 q^{9}+O(q^{10})$$ 3 * q + 3 * q^2 + 3 * q^4 + 3 * q^8 + 3 * q^9 $$3 q + 3 q^{2} + 3 q^{4} + 3 q^{8} + 3 q^{9} + 3 q^{16} - 3 q^{17} + 3 q^{18} + 3 q^{25} - 3 q^{27} + 3 q^{32} - 3 q^{33} - 3 q^{34} + 3 q^{36} - 3 q^{43} + 3 q^{49} + 3 q^{50} - 3 q^{54} + 3 q^{64} - 3 q^{66} - 3 q^{68} + 3 q^{72} + 3 q^{81} - 3 q^{86} - 3 q^{89} + 3 q^{98} - 3 q^{99}+O(q^{100})$$ 3 * q + 3 * q^2 + 3 * q^4 + 3 * q^8 + 3 * q^9 + 3 * q^16 - 3 * q^17 + 3 * q^18 + 3 * q^25 - 3 * q^27 + 3 * q^32 - 3 * q^33 - 3 * q^34 + 3 * q^36 - 3 * q^43 + 3 * q^49 + 3 * q^50 - 3 * q^54 + 3 * q^64 - 3 * q^66 - 3 * q^68 + 3 * q^72 + 3 * q^81 - 3 * q^86 - 3 * q^89 + 3 * q^98 - 3 * q^99
Basis of coefficient ring in terms of $$\nu = \zeta_{18} + \zeta_{18}^{-1}$$:
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$\nu^{2} - 2$$ v^2 - 2
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$\beta_{2} + 2$$ b2 + 2
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2888\mathbb{Z}\right)^\times$$.
$$n$$ $$1445$$ $$2167$$ $$2529$$ $$\chi(n)$$ $$-1$$ $$-1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
723.1
1.87939 −0.347296 −1.53209
1.00000 −1.87939 1.00000 0 −1.87939 0 1.00000 2.53209 0
723.2 1.00000 0.347296 1.00000 0 0.347296 0 1.00000 −0.879385 0
723.3 1.00000 1.53209 1.00000 0 1.53209 0 1.00000 1.34730 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
8.d odd 2 1 CM by $$\Q(\sqrt{-2})$$
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 2888.1.f.d 3
8.d odd 2 1 CM 2888.1.f.d 3
19.b odd 2 1 2888.1.f.c 3
19.c even 3 2 2888.1.k.b 6
19.d odd 6 2 2888.1.k.c 6
19.e even 9 2 152.1.u.a 6
19.e even 9 2 2888.1.u.b 6
19.e even 9 2 2888.1.u.g 6
19.f odd 18 2 2888.1.u.a 6
19.f odd 18 2 2888.1.u.e 6
19.f odd 18 2 2888.1.u.f 6
57.l odd 18 2 1368.1.eh.a 6
76.l odd 18 2 608.1.bg.a 6
95.p even 18 2 3800.1.cv.c 6
95.q odd 36 4 3800.1.cq.b 12
152.b even 2 1 2888.1.f.c 3
152.k odd 6 2 2888.1.k.b 6
152.o even 6 2 2888.1.k.c 6
152.t even 18 2 608.1.bg.a 6
152.u odd 18 2 152.1.u.a 6
152.u odd 18 2 2888.1.u.b 6
152.u odd 18 2 2888.1.u.g 6
152.v even 18 2 2888.1.u.a 6
152.v even 18 2 2888.1.u.e 6
152.v even 18 2 2888.1.u.f 6
456.bu even 18 2 1368.1.eh.a 6
760.bz odd 18 2 3800.1.cv.c 6
760.cp even 36 4 3800.1.cq.b 12
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
152.1.u.a 6 19.e even 9 2
152.1.u.a 6 152.u odd 18 2
608.1.bg.a 6 76.l odd 18 2
608.1.bg.a 6 152.t even 18 2
1368.1.eh.a 6 57.l odd 18 2
1368.1.eh.a 6 456.bu even 18 2
2888.1.f.c 3 19.b odd 2 1
2888.1.f.c 3 152.b even 2 1
2888.1.f.d 3 1.a even 1 1 trivial
2888.1.f.d 3 8.d odd 2 1 CM
2888.1.k.b 6 19.c even 3 2
2888.1.k.b 6 152.k odd 6 2
2888.1.k.c 6 19.d odd 6 2
2888.1.k.c 6 152.o even 6 2
2888.1.u.a 6 19.f odd 18 2
2888.1.u.a 6 152.v even 18 2
2888.1.u.b 6 19.e even 9 2
2888.1.u.b 6 152.u odd 18 2
2888.1.u.e 6 19.f odd 18 2
2888.1.u.e 6 152.v even 18 2
2888.1.u.f 6 19.f odd 18 2
2888.1.u.f 6 152.v even 18 2
2888.1.u.g 6 19.e even 9 2
2888.1.u.g 6 152.u odd 18 2
3800.1.cq.b 12 95.q odd 36 4
3800.1.cq.b 12 760.cp even 36 4
3800.1.cv.c 6 95.p even 18 2
3800.1.cv.c 6 760.bz odd 18 2
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{3} - 3T_{3} + 1$$ acting on $$S_{1}^{\mathrm{new}}(2888, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T - 1)^{3}$$
$3$ $$T^{3} - 3T + 1$$
$5$ $$T^{3}$$
$7$ $$T^{3}$$
$11$ $$T^{3} - 3T + 1$$
$13$ $$T^{3}$$
$17$ $$(T + 1)^{3}$$
$19$ $$T^{3}$$
$23$ $$T^{3}$$
$29$ $$T^{3}$$
$31$ $$T^{3}$$
$37$ $$T^{3}$$
$41$ $$T^{3} - 3T + 1$$
$43$ $$(T + 1)^{3}$$
$47$ $$T^{3}$$
$53$ $$T^{3}$$
$59$ $$T^{3} - 3T + 1$$
$61$ $$T^{3}$$
$67$ $$T^{3} - 3T + 1$$
$71$ $$T^{3}$$
$73$ $$T^{3} - 3T + 1$$
$79$ $$T^{3}$$
$83$ $$T^{3} - 3T + 1$$
$89$ $$(T + 1)^{3}$$
$97$ $$T^{3} - 3T + 1$$
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# A line segment has endpoints at (6 ,5 ) and (8 ,7 ). If the line segment is rotated about the origin by pi , translated horizontally by 2 , and reflected about the x-axis, what will the line segment's new endpoints be?
Jul 18, 2018
color(violet)("After all 3 transformations " color(green)((7, 4)to(-4, -5)" and "(-7, -8)to(-7, 8)
#### Explanation:
$\text{since there are 3 transformations to be performed}$
$\text{label the endpoints}$
$A \left(6 , 5\right) \text{ and } B \left(8 , 7\right)$
$\textcolor{m a r \infty n}{\text{transformation of rotation about the origin of }} \pi$
$\text{ a point } \left(x , y\right) \to \left(- x , - y\right)$
$\Rightarrow A \left(6 , 5\right) \to A ' \left(- 6 , - 5\right)$
$\Rightarrow B \left(8 , 7\right) \to B ' \left(- 7 , - 8\right)$
$\textcolor{m a r \infty n}{\text{next transformation under a horizontal translation }} \left(\begin{matrix}2 \\ 0\end{matrix}\right)$
• " a point "(x,y)to(x + 2, y)
$\Rightarrow A ' \left(- 6 , - 5\right) \to A ' ' \left(- 4 , - 5\right)$
$\Rightarrow B ' \left(- 7 , - 8\right) \to B ' ' \left(- 5 , - 8\right)$
$\textcolor{m a r \infty n}{\text{last transformation under a reflection in the x-axis}}$
• " a point "(x,y)to(x,-y)
$\Rightarrow A ' ' \left(- 4 , - 5\right) \to A ' ' ' \left(- 4 , 5\right)$
$\Rightarrow B ' ' \left(- 7 , - 8\right) \to B ' ' ' \left(- 7 , 8\right)$
color(violet)("After all 3 transformations " color(green)((7, 4)to(-4, -5)" and "(-7, -8)to(-7, 8)
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# 2004 AMC 10B Problems/Problem 24
In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$?
$\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$
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# Subtracting and transforming shapes
## Video: Subtracting and transforming shapes
I've saved my progress as Shapes so far, found inside the 27_pen_tool folder. Thus far, we have added a series of shapes in order to fill out the light bulb. We've got two rectangles, one vertical, one horizontal. We've got one ginormous circle, and then two smaller circles on either side of the vertical rectangle. In this exercise, we are going to begin subtracting from our shapes in order to create this contoured edge along the left side and the right side of the light bulb. Now, this is going to involve not only subtracting one shape from another, but it's also going to involve a little bit of transforming shapes, that is, scaling and rotating them.
## Subtracting and transforming shapes
I've saved my progress as Shapes so far, found inside the 27_pen_tool folder. Thus far, we have added a series of shapes in order to fill out the light bulb. We've got two rectangles, one vertical, one horizontal. We've got one ginormous circle, and then two smaller circles on either side of the vertical rectangle. In this exercise, we are going to begin subtracting from our shapes in order to create this contoured edge along the left side and the right side of the light bulb. Now, this is going to involve not only subtracting one shape from another, but it's also going to involve a little bit of transforming shapes, that is, scaling and rotating them.
When we get into the transformation territory, Photoshop starts having screen redraw problems. So, I'm just warning you about that upfront. If somehow you're not seeing what you should see on screen, just bear in mind that it's fairly normal behavior for the program right now. Anyway, I'm going to go ahead and scroll over to the left-hand side of the image a little bit. I'm going to grab my Ellipse tool, because we're going to be cutting a circle out of this area right there for starters. I'm going to switch to the Subtract Mode, so that we'd go about subtracting right off the bat from our existing shape.
So I'll switch to Subtract Mode. I should still see the Paths option selected. If it isn't selected for you, it may be because your vector mask is not active. So, make sure that it is here inside the Layers panel. Then, I'm going to drag with the tool like so, and I'm going to press the Shift key as I come in closer on that side of the shape, and I just forced an auto-scroll, which is unfortunate, because now I can't really see what I'm doing. But that's okay, once I get it more or less aligned like so, I'll go ahead and release, and I had the Shift key down the entire time, by the way, so I released the mouse button first, and then released the Shift key.
Sure enough, that circle is not big enough. I'm going to go ahead and select the Black Arrow tool, the Path Selection tool if you like. I'll click on the circle to select it. Then I'll go up to the Edit menu and I'll choose Free Transform Path. So notice as soon as you're working with a path outline, whether it's a shape outline what have you, it's going to say Free Transform Path to indicate that you're transforming a vector outline as opposed to pixels. It still has the same keyboard shortcut, Ctrl+T, Cmd+T on the Mac. I'm going to move the transformation origin over here to the right-hand point, so it snaps into alignment.
Then, while I drag this upper-left point, and I'm going to give myself a little more room to work, I'm going to press the Shift and Alt keys. The Shift key because I want to constrain the proportions to a circle, the Alt key or the Option key on a Mac, because I want to scale with respect to that transformation origin. All right, once I get about there, it looks like I have matched the edge pretty well. Now, this is a great example of how Photoshop does not keep up with your path outline transformations. It just doesn't. It's completely baffling, but it doesn't show things properly.
We are doing things properly, don't worry about that. It's just that the screen redraw is not keeping up. Now, you would expect that when you press the Enter key or the Return key in order to apply the transformation, then Photoshop would get with the program, but it doesn't. Now, notice over here in this little thumbnail, if you're seeing your thumbnails, and it very much helps to see them, which means that if you have smaller thumbnails going inside the Layers panel, I very much urge you to right-click in this lower region, where there are no layers, and choose Large Thumbnails. That way you can keep track of what's going on, at least a little bit, with your vector outlines.
Notice that I am cutting in just fine, so the screen redraw is out of whack. When I zoom out to a certain level, notice it shapes up. It says, oh yes! I'm sorry. That's where your edge really is. Unfortunately, when you zoom back in, it gets it wrong again. So, that's a little bit of a problem. We'll take care of it in a moment, but for now, we're just going to abide. I need to move this path outline over a little bit. I don't know what yours looks like, but mine is too far over to the left. So, I'm going to press the arrow key a couple of times to nudge it to the right like so, and then I'll zoom back out.
It looks like we've got this guy pretty much in the right position. We can always adjust that later once we get things roughed in a little bit, if we need to take that edge farther inward. Okay, what about this edge right there? Well, that remaining area from about here to about here is more or less a straight line, actually. So, I'm going to try to render it out using a rotated rectangle, which means it's going to require another transformation, which means we can anticipate further screen redraw issues.
But as I say, we'll take care of that momentarily. All right, so I'll switch over to the Rectangle tool, and then I'll just draw me a rectangle. It doesn't matter where it is, because it's nowhere near right. It should be set to Subtract though, which it is, so everything's good there. We're further subtracting from the shapes. I'll go back to my Black Arrow tool. I'll click on that rectangle in order to select it, and then I'll move it into position right there. It's that intersection that I'm looking for, I think. I think this is going to work out for us. Actually, it could be a little lower, couldn't it? I guess I'm going from the wrong area, I guess about right there is where the light bulb starts cutting away from that circle.
So that's what we want them at. We essentially want to create a straight edge that's tangent. In the case you don't know what tangent means, imagine that you have a straight line resting on a ball. That would be a tangent line, because it's just sitting on the ball, and the line is straight. So it's barely touching the circle of the ball. Anyway, it's right there. It'll cleave off of that circle and then meet onto that circle there. We're going to do that using, of course, the transformation. Now, you'll just have to take my word for it. This rectangle is still the same, because we can't see the corner points right now.
I'm going to press Ctrl+T or Cmd+T on the Mac to re-invoke that Free Transform command. I'm going to move my transformation origin right there, and then I'll move my cursor outside the rectangle and drag in order to rotate it until it comes into position at that location. So, this is a good tangent line, it looks to me. It just barely touches the top circle, and it just barely touches the bottom circle. Now, it goes too far. It cuts into the light bulb down right here, but I'm unconcerned with that because we can move the rectangle to a different location.
So, I'm going to press the Enter key or the Return key on the Mac, and watch in disappointment and horror, as the screen does not redraw properly. All right, anyway, I'm now going to move my shape upward, because I still have the Black Arrow tool selected. I'll move it to about there, so that we still have a nice tangent edge, and lo and behold we still have a little bit of an issue up here. Now that's not Photoshop's problem. That's not a redraw problem. That's my problem. I need to solve it. I'm going to do so by transforming a shape, and I'll show you which shape it is that we're going to transform, as well as show you how to go about flipping and duplicating these shapes over here on the left-hand side in order to cut away this region on the right- hand side, in the next exercise.
Show transcript
#### This video is part of
Photoshop CS5 One-on-One: Mastery
192 video lessons · 43836 viewers
Author
Expand all | Collapse all
1. ### Introduction
40m 45s
1. Welcome
2m 45s
2. Making Photoshop your default image editor
7m 43s
3. Installing the dekeKeys keyboard shortcuts
8m 10s
4. Remapping OS shortcuts
7m 37s
5. Installing the Best Workflow color settings
4m 31s
6. The color settings explained
6m 54s
3m 5s
2. ### 25. What Photoshop Can Do, Pt. 3: The Face Paint
1h 11m
1. Your creative range continues to expand
1m 46s
2. The Avatar project so far
2m 38s
3. Painting on a photograph
7m 50s
6m 14s
5. Simulating chalky white paint
7m 23s
6. Masking and placing an image
7m 20s
7. Upsampling and Lens Blur
5m 9s
8. Blending blurry elements
3m 48s
9. Making a Smart Object
6m 46s
10. Placing an image as a Smart Object
3m 22s
11. Blending away a background
5m 56s
12. Applying Smart Filters
4m 34s
13. Creating a glow with Lens Flare
3m 45s
14. Blending and masking a glow
5m 3s
1h 26m
1. Using the image to select itself
1m 53s
6m 32s
3. Making an alpha channel
6m 54s
4. Using the Calculations command
6m 48s
5. Add, Subtract, Offset, and Scale
5m 54s
6. Prepping an image with the Dodge tool
6m 55s
7. Fixing mistakes before they get too big
6m 32s
8. Painting in the Overlay mode
5m 51s
9. Exaggerating and selecting flesh tones
7m 39s
10. Smudge, Median, and the Blur tool
6m 59s
6m 7s
5m 45s
13. Masking and compositing the foreground
5m 27s
14. Finessing the final composition
7m 39s
4. ### 27. Everything About the Pen Tool
2h 24m
1. Connecting the dots
1m 40s
2. The Pen tool and the Paths panel
6m 32s
3. Drawing a straight-sided outline
6m 25s
4. Editing a path outline
6m 36s
5. Adding and editing smooth points
5m 35s
6. Creating vector masks with the shape tools
4m 59s
7. Building a complex outline from shapes
4m 26s
8. Subtracting and transforming shapes
6m 45s
9. Cloning, flipping, and combining shapes
8m 58s
10. Roughing in non-symmetrical paths
7m 41s
11. Finessing a complex outline
9m 15s
8m 26s
13. Isolating an image element
6m 8s
14. Smooth points and control handles
9m 3s
15. Stretching curved segments
7m 49s
16. Using the Rubber Band option
9m 33s
17. Drawing smooth points with the Pen tool
6m 59s
3m 45s
19. Drawing cusp points
7m 14s
20. Setting points in the pasteboard
9m 57s
21. Using the Convert Point tool
6m 42s
5. ### 28. Blend Modes Revealed
2h 57m
1. Everything you need to know about blending
1m 45s
2. Photoshop CS5's blend modes
7m 21s
3. Cycling between blend modes
6m 15s
4. Darken and Lighten and their derivatives
6m 3s
5. The blend mode shortcuts
8m 6s
6. The Multiply and Burn modes
4m 28s
7. The Screen and Dodge modes
6m 0s
8. How opposite blend modes work
8m 24s
9. Why Multiply darkens and Divide lightens
5m 23s
10. Cleaning up a client's bad art
5m 3s
11. Dropping out a white background
5m 56s
12. Blending inside blend modes
8m 3s
13. Overlay, Soft Light, and Hard Light
6m 26s
14. Vivid, Linear, and Pin Light (and Hard Mix)
6m 35s
15. Difference, Exclusion, Subtract, and Divide
7m 34s
16. Great uses for the Difference mode
6m 18s
17. Promising uses for the Divide mode
9m 6s
18. Hue, Saturation, Color, and Luminosity
7m 0s
19. Blending an inverted layer
3m 32s
20. The "Fill Opacity Eight"
7m 25s
21. Making bad blend modes good
5m 16s
22. Making a knockout layer
6m 53s
23. Blending in the CMYK mode
8m 3s
24. Overprinting black text
8m 29s
25. Using the Luminance slider
5m 24s
6m 21s
27. Adjusting the behavior of luminance effects
10m 8s
6. ### 29. The Power of Smart Objects
2h 2m
1. Smart Objects = protective containers
1m 35s
2. Placing an Illustrator graphic
6m 30s
3. Vector copy and paste options
6m 56s
4. Applying Puppet Warp to vectors
8m 9s
5. "Gluing" vector art for Puppet Warp
5m 50s
6. Warping art onto the surface of an image
8m 7s
7. Blending a Smart Object
4m 30s
8. Blurring and blending a Smart Object
6m 8s
9. Making changes in Illustrator
5m 57s
10. Creating "true clones"
7m 18s
11. Double-flipping text
4m 44s
12. Applying effects to multiple layers
3m 24s
13. Updating true clones in one operation
7m 36s
14. Editing JPEGs as Camera Raw objects
5m 49s
15. Creating a double-exposure effect
7m 15s
7m 47s
17. Applying and repeating Camera Raw edits
6m 9s
18. Copying vs. cloning a Smart Object
5m 18s
19. Flipping a Smart Object and its mask
3m 42s
20. Adjusting multiple Camera Raw clones
3m 53s
21. Text that inverts everything behind it
5m 34s
7. ### 30. Nondestructive Smart Filters
1h 59m
1. This time, "smart" means dynamic
1m 37s
2. Introducing Smart Filters
6m 28s
5m 17s
4. Smart High Pass in the Lab mode
7m 57s
5. Sharpening a high-frequency image
7m 46s
6. Retroactively reducing noise
7m 31s
7. Which filters are Smart Filters?
6m 20s
8. Shadows/Highlights as a Smart Filter
4m 37s
9. Nesting one Smart Object inside another
7m 11s
10. Drawing a mask from a nested Smart Object
8m 7s
9m 16s
12. Tempering saturation values in Lab
7m 0s
13. Filtering live, editable text
9m 2s
14. Enhancing filters with layer effects
4m 33s
15. Applying a filter multiple times
5m 0s
16. Creating a synthetic star field
7m 7s
17. Making a stucco or drywall pattern
6m 28s
18. Land, sea, and clouds
8m 30s
8. ### 31. The Bristle and Mixer Brushes
2h 50m
2m 3s
2. Canvas texture and brush libraries
6m 40s
3. Painting with a predefined custom brush
9m 21s
4. Dissecting a custom brush
11m 9s
5. Designing and using a custom brush
4m 54s
5m 27s
7. The ten styles of bristle brushes
9m 47s
8. Size, Spacing, and Angle
7m 2s
9. Using the Bristle Brush preview
7m 53s
10. Bristles, Length, Thickness, and Stiffness
6m 53s
11. Stylus tilt and mouse behavior
5m 25s
12. Stroking a path outline with a brush
4m 0s
13. Troubleshooting a stylus
5m 49s
14. Introducing the Mixer Brush
7m 22s
15. The Load, Mix, and Wet values
5m 1s
6m 26s
17. Shading a piece of graphic art
6m 34s
7m 53s
19. Mixing a photographic portrait
6m 11s
20. Tracing the fine details in an image
5m 52s
21. Crosshatching and brush size
5m 53s
22. Covering up and augmenting details
7m 36s
23. Painting in hair and fabric
5m 54s
24. Painting and scaling very fine hairs
8m 7s
25. Adding texture with the Emboss filter
8m 31s
26. Exploiting a "happy accident"
2m 46s
9. ### 32. Auto-Align, Auto-Blend, and Photomerge
1h 40m
1. Artificial intelligence that works
1m 22s
2. The Auto-Align Layers command
7m 25s
3. The Auto-Blend Layers command
3m 54s
4m 50s
5. The Geometric Distortion setting
6m 44s
6. The Seamless Tones and Colors checkbox
4m 8s
7. Creating the best possible layer mask
9m 18s
8. Auto-blending depths of field
5m 54s
6m 29s
10. Shooting and downsampling panorama images
5m 54s
11. Introducing the Photomerge command
6m 40s
12. Evaluating the Layout settings
6m 47s
5m 36s
14. Tracing and extracting seams
7m 18s
5m 55s
16. Simplifying and correcting a panorama
5m 58s
17. Smart Filters and nondestructive cropping
6m 43s
10. ### 33. High Dynamic Range (HDR Pro)
1h 18m
1. The most mysterious of mysterious topics
2m 29s
2. Introducing HDR Toning
6m 43s
3. Reigning in clipped highlights
5m 54s
9m 5s
5. Nondestructive editing with HDR Toning
8m 22s
6. Using the HDR Toning Curve
7m 2s
6m 0s
8. Merging multiple exposures
7m 14s
9. A first look at HDR Pro
6m 24s
10. Removing ghosts, correcting backlighting
7m 11s
11. Generating and editing an HDR comp
7m 0s
12. HDR rendered to completion
5m 19s
11. ### 34. Actions and Batch Processing
1h 27m
1. Processing hundreds of files in no time
1m 43s
2. Creating an action set
6m 37s
3. Making an action
7m 7s
4. Stop, Delete, and Record
7m 12s
6m 40s
6. Playing and testing an action
6m 31s
7. Playing and editing a specific operation
6m 39s
8. Permitting the user to change settings
4m 58s
9. Explaining an action with a custom stop
5m 0s
10. Batch-processing multiple images
7m 22s
11. Adding a Save As operation
6m 34s
12. Creating an action to save web graphics
7m 59s
13. Batching two actions into one
7m 15s
5m 30s
1m 19s
1. See ya
1m 19s
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# A newspaper recently reported the story of a group of
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A newspaper recently reported the story of a group of hunters who found a heavily damaged mobile home suspended in the treetops in a forest. Although mobile homes in many towns in the surrounding area had recently been swept away by tornados, officials suspect that this particular mobile home came from nearby New City, which had been almost completely destroyed by a tornado three months earlier, on March 10th.
Which of the following, if true, most strongly supports the officials' hypothesis about the origin of the mobile home?
-Many residents of New City and the surrounding areas live in trailer homes.
-A calendar found inside the mobile home was turned to March 10th.
-Weather stations widely reported the tornado of March 10th.
-Sales of this particular brand of mobile home have been very strong in New City and other surrounding towns for the past six months.
-Decals supporting the nearby State University football team were found on the mobile home.
[Reveal] Spoiler:
interested in the answer choice D. any explanation?
[Reveal] Spoiler: OA
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 11:21
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LalaB wrote:
A newspaper recently reported the story of a group of hunters who found a heavily damaged mobile home suspended in the treetops in a forest. Although mobile homes in many towns in the surrounding area had recently been swept away by tornados, officials suspect that this particular mobile home came from nearby New City, which had been almost completely destroyed by a tornado three months earlier, on March 10th.
Which of the following, if true, most strongly supports the officials' hypothesis about the origin of the mobile home?
-Many residents of New City and the surrounding areas live in trailer homes.
-A calendar found inside the mobile home was turned to March 10th.
-Weather stations widely reported the tornado of March 10th.
-Sales of this particular brand of mobile home have been very strong in New City and other surrounding towns for the past six months.
-Decals supporting the nearby State University football team were found on the mobile home.
[Reveal] Spoiler:
interested in the answer choice D. any explanation?
there is two problems with alternative D: 1) "other surrounding towns" means that the house can possibly been carried from any town on the vicinities. 2) "past six months" indicate that it is a brand new house, and the facts presented on the questions don't support any assumption that it is new home.
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 11:32
lamanxa, thank you for ur reply.
I agree with you about "other surrounding towns", but disagree about periods.
since the q.stem stated that a tornado destroyed the city three months earlier, I thought it is ok to think that these homes have been there for the past six months. dont u think so?
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 12:28
LalaB wrote:
lamanxa, thank you for ur reply.
I agree with you about "other surrounding towns", but disagree about periods.
since the q.stem stated that a tornado destroyed the city three months earlier, I thought it is ok to think that these homes have been there for the past six months. dont u think so?
LalaB, this information only would be relevant if it was stated that the house were brought AFTER the tornado. Since the tornado destroyed the city 3 months earlier, it is irrelevant if the houses are 6 months old or 50 years old. All of them were possibly carried away by the tornado.
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 14:21
This is an interesting question
Initially voted for D - but yes agree with why it should not be the best option "New City and other surrounding towns "
and then by POE B supports the hypothesis - we have exact date "March 10" and on March 10 - the tornado destroyed New City
"New City, which had been almost completely destroyed by a tornado three months earlier, on March 10th"
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 20:20
Went with B, because New City was screwed on March 10, and the calendar had March 10 on it.
There was a mention that trailers were in the surrounding area which includes New city, but also other cities too, which muddies up choice D. We know New City was bye bye on March 10, but we know nothing on the other cities. We know the calendar showed March 10 inside, so that's how 2 and 2 went together here.
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Re: A newspaper recently reported the story [#permalink]
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27 Mar 2012, 20:47
What if the calendar page turned during the tornado? Its a storm & the mobile home is hanging in a tree. I would suspect that a calendar can move be tossed around. This seems to be a very loose argument
Posted from my mobile device
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Re: A newspaper recently reported the story [#permalink]
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28 Mar 2012, 00:58
boomtangboy wrote:
What if the calendar page turned during the tornado? Its a storm & the mobile home is hanging in a tree. I would suspect that a calendar can move be tossed around. This seems to be a very loose argument
Posted from my mobile device
good point! I thought the same , when crossed out this answer choice
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Re: A newspaper recently reported the story [#permalink]
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02 Apr 2012, 04:39
fell for D... Thought B was weak... cos ppl stopped using the trailer home and since tornado is common in the area... maybe the home was crashed by a tornado after March.. But When comparing B and D,... Now I cld see B is way better option than D...
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Re: A newspaper recently reported the story of a group of [#permalink]
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28 Nov 2012, 05:42
1
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+1 B. D does not address the problem
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Re: A newspaper recently reported the story [#permalink]
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28 Nov 2012, 21:31
LalaB wrote:
boomtangboy wrote:
What if the calendar page turned during the tornado? Its a storm & the mobile home is hanging in a tree. I would suspect that a calendar can move be tossed around. This seems to be a very loose argument
Posted from my mobile device
good point! I thought the same , when crossed out this answer choice
B) A calendar found inside the mobile home was turned to March 10th.
D) Sales of this particular brand of mobile home have been very strong in New City and other surrounding towns for the past six months.
I guess ur point is valid...
But between B and D... If u add weights for both choices B will win...
So actual contenders are B and D.
First of all D says this home has proof that it would have been sold in past 6 months... This itself fails... This weakens argument...
There is no possibility here.. Its viable...
On other hand, B says about date in calendar.. There is possibility here..
IMO tornado will turn calendar to March 10th no it might be some person...
Anyways it strengthen our argument ....
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Re: A newspaper recently reported the story of a group of [#permalink]
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28 Nov 2012, 23:56
Really interesting problem
I also chose D first, by eliminating others that indicate anything other than place. Agree with all that Correct Answer is B.
I just listed down the facts.
1. Damaged mobile home found suspended in the treetops.
2. Mobile home came from nearby New City.
3. New City had been destroyed by a tornado on March 10.
B & D are possible contenders.
Option (D) says sales have been strong in New City and other surrounding towns. So, It's not strongly supports that its surely from New City.
Option (B) There is weak link between premises and Option B.
[New city destroyed on March 10] + [10 March cal date in the Home]=> [Place of Home is from New City].
Its sufficient to prove the place as the New City.
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Re: A newspaper recently reported the story of a group of [#permalink]
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12 Jan 2013, 09:32
I'm still not convinced by B. It is entirely possible for the page to be turned to March 10 because of the tornado
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Re: A newspaper recently reported the story of a group of [#permalink]
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13 Jan 2013, 03:03
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mahendru1992 wrote:
I'm still not convinced by B. It is entirely possible for the page to be turned to March 10 because of the tornado
B is not pretty, but D is just plain nasty. D does not weaken or strengthen, the answer choice implies that the trailer could have come from one of multiple locations.
B is the lesser of all the evils, but at least it strengthens by implying that the trailer may have come from the location where a tornado occurred 3 months ago. Pretty Loosy Goosy IMO
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Re: A newspaper recently reported the story of a group of [#permalink]
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13 Jan 2013, 07:21
jgomey wrote:
mahendru1992 wrote:
I'm still not convinced by B. It is entirely possible for the page to be turned to March 10 because of the tornado
B is not pretty, but D is just plain nasty. D does not weaken or strengthen, the answer choice implies that the trailer could have come from one of multiple locations.
B is the lesser of all the evils, but at least it strengthens by implying that the trailer may have come from the location where a tornado occurred 3 months ago. Pretty Loosy Goosy IMO
hahaha, ok i guess your explanation does make sense. Thanks
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Re: A newspaper recently reported the story of a group of [#permalink]
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24 Jan 2014, 23:12
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How can one say by choosing option B that the tailor van only came from nearby New City?
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Re: A newspaper recently reported the story of a group of [#permalink]
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25 Jan 2014, 08:37
WE NEED A CLEAR EVIDENCE TO RELATE THE heavily damaged mobile home TO ONE DESTROYED IN NEW CITY..........SINCE HOMES WERE DESTROYED ON MARCH 10 IN NEW CITY ONLY AND NO OTHER CITY........THIS HOME WITH calendar turned to March 10th- IS A CLOSE EVIDENCE OF BEING FROM NEW CITY....................HENCE " B" CORRECT......
"D" IS NOT AS STRONG AN EVIDENCE, AS THE DAMAGED HOME DESCRIBED COULD BE FROM NEW CITY OR ANY OTHER SURROUNDING TOWNS........
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Re: A newspaper recently reported the story of a group of [#permalink]
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15 Mar 2014, 00:35
semwal wrote:
WE NEED A CLEAR EVIDENCE TO RELATE THE heavily damaged mobile home TO ONE DESTROYED IN NEW CITY..........SINCE HOMES WERE DESTROYED ON MARCH 10 IN NEW CITY ONLY AND NO OTHER CITY........THIS HOME WITH calendar turned to March 10th- IS A CLOSE EVIDENCE OF BEING FROM NEW CITY....................HENCE " B" CORRECT......
"D" IS NOT AS STRONG AN EVIDENCE, AS THE DAMAGED HOME DESCRIBED COULD BE FROM NEW CITY OR ANY OTHER SURROUNDING TOWNS........
The argument says that " Although mobile homes in many towns in the surrounding area had recently been swept away by tornados". Hence there is no evidence that it only came from "New City".
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Re: A newspaper recently reported the story of a group of [#permalink]
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15 Mar 2014, 15:43
1
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email2vm wrote:
semwal wrote:
WE NEED A CLEAR EVIDENCE TO RELATE THE heavily damaged mobile home TO ONE DESTROYED IN NEW CITY..........SINCE HOMES WERE DESTROYED ON MARCH 10 IN NEW CITY ONLY AND NO OTHER CITY........THIS HOME WITH calendar turned to March 10th- IS A CLOSE EVIDENCE OF BEING FROM NEW CITY....................HENCE " B" CORRECT......
"D" IS NOT AS STRONG AN EVIDENCE, AS THE DAMAGED HOME DESCRIBED COULD BE FROM NEW CITY OR ANY OTHER SURROUNDING TOWNS........
The argument says that " Although mobile homes in many towns in the surrounding area had recently been swept away by tornados". Hence there is no evidence that it only came from "New City".
I think you want a perfect link between the mobile home from New City and the tornado on March 10. To support the statement, the evidence that you want would be ideal, but since we don't have the perfect answer, option B leaves us with less doubt than option D. After all, how can you be so sure the mobile home came from New City after sales of those mobile homes have been equal among the other cities? At least option B links the date of the tornado in New City and the last date on the calendar in mobile home.
I hope this helps.
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# The Number 922827 : Square Root, Cube Root, Factors, Prime Checker
You can find Square Root, Cube Root, Factors, Prime Check, Binary, Octal, Hexadecimal and more of Number 922827. 922827 is written as Nine Hundred And Twenty Two Thousand, Eight Hundred And Twenty Seven. You can find Binary, Octal, Hexadecimal Representation and sin, cos, tan values and Multiplication, Division tables.
• Number 922827 is an odd Number.
• Number 922827 is a Prime Number
• Sum of all digits of 922827 is 30.
• Previous number of 922827 is 922826
• Next number of 922827 is 922828
## Square, Square Root, Cube, Cube Root of 922827
• Square Root of 922827 is 960.63884993269
• Cube Root of 922827 is 97.358400634772
• Square of 922827 is 851609671929
• Cube of 922827 is 785888398717223283
## Numeral System of Number 922827
• Binary Representation of 922827 is 11100001010011001011
• Octal Representation of 922827 is 3412313
• Hexadecimal Representation of 922827 is e14cb
## Sin, Cos, Tan of Number 922827
• Sin of 922827 is 0.54463903501547
• Cos of 922827 is -0.83867056794514
• Tan of 922827 is -0.64940759319826
## Multiplication Table for 922827
• 922827 multiplied by 1 equals to 922,827
• 922827 multiplied by 2 equals to 1,845,654
• 922827 multiplied by 3 equals to 2,768,481
• 922827 multiplied by 4 equals to 3,691,308
• 922827 multiplied by 5 equals to 4,614,135
• 922827 multiplied by 6 equals to 5,536,962
• 922827 multiplied by 7 equals to 6,459,789
• 922827 multiplied by 8 equals to 7,382,616
• 922827 multiplied by 9 equals to 8,305,443
• 922827 multiplied by 10 equals to 9,228,270
• 922827 multiplied by 11 equals to 10,151,097
• 922827 multiplied by 12 equals to 11,073,924
## Division Table for 922827
• 922827 divided by 1 equals to 922827
• 922827 divided by 2 equals to 461413.5
• 922827 divided by 3 equals to 307609
• 922827 divided by 4 equals to 230706.75
• 922827 divided by 5 equals to 184565.4
• 922827 divided by 6 equals to 153804.5
• 922827 divided by 7 equals to 131832.42857143
• 922827 divided by 8 equals to 115353.375
• 922827 divided by 9 equals to 102536.33333333
• 922827 divided by 10 equals to 92282.7
• 922827 divided by 11 equals to 83893.363636364
• 922827 divided by 12 equals to 76902.25
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# How much mortgage can i afford with my salary
## How much mortgage can I afford if I make 300 000 a year?
Multiply Your Annual Income By 2.5 or 3
Simply take your gross income and multiply it by 2.5 or 3, to get the maximum value of the home you can afford. For somebody making \$100,000 a year, the maximum purchase price on a new home should be somewhere between \$250,000 and \$300,000.
## What house can I afford based on my salary?
To calculate ‘how much house can I afford,’ a good rule of thumb is using the 28%/36% rule, which states that you shouldn’t spend more than 28% of your gross monthly income on home-related costs and 36% on total debts, including your mortgage, credit cards and other loans like auto and student loans.
## How much house can I afford with my salary Philippines?
Instead, look at your budget first to find out how much house you can afford. Most lenders suggest home expenses should be a maximum of 28% of your gross monthly income. So if you make P50,000 gross a month, your budget for monthly mortgage payments for your new house shouldn’t go over P14,000.
## How much mortgage can I afford if I make 110 000 per year?
Some experts suggest that you can afford a mortgage payment as high as 28% of your gross income. If true, a couple who earn a combined annual salary of \$100,000 can afford a monthly payment of about \$2,300/month. That could translate to a \$450,000 loan, assuming a 4.5% 30-year fixed rate.22 мая 2012 г.
## How much do I need to make to afford a 250k house?
How much do you need to make to be able to afford a house that costs \$250,000? To afford a house that costs \$250,000 with a down payment of \$50,000, you’d need to earn \$43,430 per year before tax. The monthly mortgage payment would be \$1,013.
You might be interested: How to reaffirm a mortgage after chapter 7
## What mortgage can I afford on 60k?
The usual rule of thumb is that you can afford a mortgage two to 2.5 times your annual income. That’s a \$120,000 to \$150,000 mortgage at \$60,000.
## How much should you make to buy a 500000 house?
A generally accepted rule of thumb is that your mortgage shouldn’t be more than three times your annual income. So if you make \$165,000 in household income, a \$500,000 house is the very most you should get.
## How much of a down payment do you need for a house?
Lenders require 5% to 15% down for other types of conventional loans. When you get a conventional mortgage with a down payment of less than 20%, you have to get private mortgage insurance, or PMI. The monthly cost of PMI varies, depending on your credit score, the size of the down payment and the loan amount.
## How do you know how much to spend on a house?
To determine how much house you can afford, most financial advisers agree that people should spend no more than 28 percent of their gross monthly income on housing expenses and no more than 36 percent on total debt — that includes housing as well as things like student loans, car expenses, and credit card payments.
## Can I build a house for 100k in Philippines?
If the amount is in pesos, 100k would hardly be enough for a small house. For one, to legally build a house, that’s not a residential nipa hut or bahay kubo, you have to have plans drawn and get a building permit. If you have US\$100k, yes, you can build a small house. …24 мая 2017 г.
You might be interested: How to find a broker as a real estate agent
## What salary do you need to live comfortably in the Philippines?
The Philippines has a generally low cost of living. International Living reports that you could comfortably live on \$800 to \$1200 a month, covering housing, utilities, food, healthcare and taxes.
## How can I save money for a house in the Philippines?
How to Save up for your house down payment.
1. Track your monthly expenses. …
2. Create your realistic budget and stick to it. …
3. Take out a loan to finance the down payment for your home. …
4. Apply for a housing loan with low monthly charges equivalent to your installment fee capacity as mentioned in the step above.
## How much do you have to make to afford a \$650000 house?
Income to Afford a \$650,000 House. How much do you need to make to be able to afford a house that costs \$650,000? To afford a house that costs \$650,000 with a down payment of \$130,000, you’d need to earn \$112,918 per year before tax. The monthly mortgage payment would be \$2,635.
## How much do I have to make to afford a 2 million dollar house?
Required income to afford a 2 or 3 million dollar housePurchase Price\$1 million\$2 millionDown Payment\$200,000\$400,000Total Cash on Hand\$240,000\$480,000Required Income\$175,230\$340,275
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Momepi
Registered: 22.07.2004
From: Ireland
Posted: Saturday 30th of Dec 08:47 Thanks for the details . I have bought the Algebra Professor from https://softmath.com/comparison-algebra-homework.html and I happened to go through function domain yesterday. It is pretty cool and easily readable . I was impressed by the detailed explanations offered on multiplying matrices. Rather than being exam oriented, the Algebra Professor aims at equipping you with the basic principles of College Algebra. The payment guarantee and the unbelievable rebates that they are currently offering makes the purchase particularly attractive.
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https://web2.0calc.com/questions/need-help-with-square
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# need help with square
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Points A,B,C and D are midpoints of the sides of the larger square. If the smaller square has area 60, what is the area of the bigger square?
Jul 23, 2020
#1
0
Points A,B,C and D are midpoints of the sides of the larger square. If the smaller square has area 60, what is the area of the bigger square?
First, for reference, let's name one of the corners of the large square.
Any of the corners will do, so let's just choose the top left one. Call it V
Since the area of the small square is 60, then each side is length sqrt(60).
DVA is an isoceles right triangle, with hypotenuse length sqrt(60).
Therefore, VA is length sqrt(60) divided by sqrt(2), and the entire side is twice that.
2 • sqrt(60)
The side of the large square is ——————
sqrt(2)
4 • 60
Square that side, and you have the area of the large square ———— thus Alargesquare = 120
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Jul 23, 2020
#2
0
Hey, it's me again. I was looking at the picture and an idea fell out of the sky.
I don't know how to draw on this site, so I'll try to explain it in words.
You have the large square which is the outside one.
You have the small square which is the inside one.
Draw a line from point A to point C and
draw a line from point D to point B.
Now you have four tiny squares, each of which is cut into two equal triangles by a diagonal line.
The small square contains four of those triangles and
the large square contains eight of those triangles.
Thus, the large square contains twice the area of the small square.
The area of the small square is 60, so twice that is the area of the large square, i.e., 120
Neat, huh? Even if I do say so myself. TTFN.
.
Guest Jul 24, 2020
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https://www.unitconverters.net/energy/therm-ec-to-millijoule.htm
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Home / Energy Conversion / Convert Therm (EC) to Millijoule
# Convert Therm (EC) to Millijoule
Please provide values below to convert therm (EC) to millijoule [mJ], or vice versa.
From: therm (EC) To: millijoule
### Therm (EC) to Millijoule Conversion Table
Therm (EC)Millijoule [mJ]
0.01 therm (EC)1055056000 mJ
0.1 therm (EC)10550560000 mJ
1 therm (EC)105505600000 mJ
2 therm (EC)211011200000 mJ
3 therm (EC)316516800000 mJ
5 therm (EC)527528000000 mJ
10 therm (EC)1055056000000 mJ
20 therm (EC)2110112000000 mJ
50 therm (EC)5275280000000 mJ
100 therm (EC)10550560000000 mJ
1000 therm (EC)1.055056E+14 mJ
### How to Convert Therm (EC) to Millijoule
1 therm (EC) = 105505600000 mJ
1 mJ = 9.4781698791344E-12 therm (EC)
Example: convert 15 therm (EC) to mJ:
15 therm (EC) = 15 × 105505600000 mJ = 1582584000000 mJ
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https://www.coursehero.com/file/6868903/2516-eg-1/
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2.5.16-eg-1
# 2.5.16-eg-1 - Example Where is the function f(x =...
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Unformatted text preview: Example Where is the function f (x) = continuous on the interval − x2 + 1 sin 2x , x∈ − 5π 5π , , 44 5π 5π ? , 44 Solution: The function f is continuous where its denominator is nonzero. We know that sin y = 0 at y = 0, ±kπ, k = 1, 2, . . . , so sin 2x = 0 at 2x = 0, ±kπ, k = 1, 2, . . . , that is, for x values satisfying π x = 0, ± k · , k = 1, 2, . . . . 2 5π 5π , , it follows that sin 2x = 0 at x = −π, −π/2, 0, π/2, π . So the 44 function f is continuous for So in the interval − x∈ − 5π 5π π π , , x = −π, − , 0, − , π, 44 2 2 or rewriting the above as a union of intervals, x∈ − 5π π π π 5π π , −π ∪ −π, − ∪ − , 0 ∪ 0, ∪ , π ∪ π, . 4 2 2 2 2 4 Note: In WeBWorK notation, the answer is: [-5pi/4,-pi) U (-pi,-pi/2) U (-pi/2,0) U (0,pi/2) U (pi/2,pi) U (pi, 5pi/4] ...
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## This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
Ask a homework question - tutors are online
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https://javatutoring.com/java-hours-into-seconds/
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# Java: Convert Hours To Seconds & Minutes | Vice Versa
Java program to convert hours to minutes & seconds vice versa. Its a simple code to convert hours to seconds, this java program has been written in 4 ways for easy understanding. If you have any queries just do leave a comment here.
In this tutorial, we learn how to use the Java program to effectively to calculate and convert the number of hours into seconds and it will involve us using the scanner class and a few tweaks here and there. So let’s quickly delve into the problem solving with Java.
## Java to convert Hours into Seconds
Converting hours to seconds is quite simple. All we require is an input denoting the hours.
We are well aware that, one minute is 60 seconds and one hour is 60 minutes which will make, one hour to 3600 seconds (60*60). This same logic is used to convert hours to seconds.
For this, we need to read our input hours at runtime. This can be done by using Scanner class which is well known for reading any primitive datatype at runtime from the console screen.
So, we’ll first create an object of this Scanner class and then invoke the required method to read our input. Since our input is hours, which is generally an integer only, we will be making use of the nextInt() method of the class by writing the following statements:
Now that we have our input in hand, since 1 hour == 60*60 seconds or 3600 seconds
x hours == x*60*60 seconds by cross multiplication.
Therefore our output seconds will be calculated as:
sec=hrs*60*60;
Output:
Using Method
In this method, we’ll be making use of another method which will consist of the main logic for converting hours to seconds i.e.,
sec=h*60*60;
We are making use of another separate method so that, we can make the main logic of the code reusable.
So, we first read the input hours using Scanner class and then create an object (sc) of this method.
Since it is not within main method, object is required to invoke it.
Then with this created object (res) of the class (hoursToSceonds) we invoke the seconds() method by passing hours as parameter which performs calculation and stores resultant is a variable (sec).
This variable (sec) is then called in the main method using the same object to be printed on console screen using println() method.
Output:
Using Static Method
Similar to the above method, even here we write the code is a separate method. The only difference is that, we make use of static method here instead.
The advantage of static method is that, we do not have the need to create an object for static method even though,
the static method is outside main method because, it belongs to the class instead of the instance of the class.
So now, the static method (hrsToSec) contains the logic while the main method takes care of input output operations and call this static method (hrsToSec) by passing input hours as argument.
### Java Code Hours To Minutes – Using Command Line Arguments
In all the above methods, we have seen the use of Scanner class to read input at runtime.
Similarly, we can also make use of command line arguments to read input at runtime.
Here, while we write the run command itself we send our input as argument with a space in between.
This argument is generally of type string. So, we first convert it to desired type (integer) by parsing. This is then stored in a variable followed by which, the same logic as above is used.
Since there is only one argument (arg[0]), we parse this as above which is our required input hours.
x
## Merge Sort Java – Program 2 Ways | Sortings
Merge Sort Java – Java program to implement merge sort using array & Buffered reader. ...
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1 Farady (C12)/Square Meter = 9648530900 Abcoulomb/Square Kilometer
One Farady (C12)/Square Meter is Equal to How Many Abcoulomb/Square Kilometer?
The answer is one Farady (C12)/Square Meter is equal to 9648530900 Abcoulomb/Square Kilometer and that means we can also write it as 1 Farady (C12)/Square Meter = 9648530900 Abcoulomb/Square Kilometer. Feel free to use our online unit conversion calculator to convert the unit from Farady (C12)/Square Meter to Abcoulomb/Square Kilometer. Just simply enter value 1 in Farady (C12)/Square Meter and see the result in Abcoulomb/Square Kilometer.
Manually converting Farady (C12)/Square Meter to Abcoulomb/Square Kilometer can be time-consuming,especially when you don’t have enough knowledge about Surface Charge Density units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Farady (C12)/Square Meter to Abcoulomb/Square Kilometer converter tool to get the job done as soon as possible.
We have so many online tools available to convert Farady (C12)/Square Meter to Abcoulomb/Square Kilometer, but not every online tool gives an accurate result and that is why we have created this online Farady (C12)/Square Meter to Abcoulomb/Square Kilometer converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
How to Convert Farady (C12)/Square Meter to Abcoulomb/Square Kilometer (F/m2 to abC/km2)
By using our Farady (C12)/Square Meter to Abcoulomb/Square Kilometer conversion tool, you know that one Farady (C12)/Square Meter is equivalent to 9648530900 Abcoulomb/Square Kilometer. Hence, to convert Farady (C12)/Square Meter to Abcoulomb/Square Kilometer, we just need to multiply the number by 9648530900. We are going to use very simple Farady (C12)/Square Meter to Abcoulomb/Square Kilometer conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Farady (C12)/Square Meter} = 1 \times 9648530900 = \text{9648530900 Abcoulomb/Square Kilometer}$$
What Unit of Measure is Farady (C12)/Square Meter?
Farady (C12) per square meter is a unit of measurement for surface charge density. It is defined as one farady (C12) of electric charge per one square meter of surface of an object.
What is the Symbol of Farady (C12)/Square Meter?
The symbol of Farady (C12)/Square Meter is F/m2. This means you can also write one Farady (C12)/Square Meter as 1 F/m2.
What Unit of Measure is Abcoulomb/Square Kilometer?
Abcoulomb per square kilometer is a unit of measurement for surface charge density. It is defined as one abcoulomb of electric charge per one square kilometer of surface of an object.
What is the Symbol of Abcoulomb/Square Kilometer?
The symbol of Abcoulomb/Square Kilometer is abC/km2. This means you can also write one Abcoulomb/Square Kilometer as 1 abC/km2.
How to Use Farady (C12)/Square Meter to Abcoulomb/Square Kilometer Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Farady (C12)/Square Meter and in the first input field, enter a value.
• From the second dropdown, select Abcoulomb/Square Kilometer.
• Instantly, the tool will convert the value from Farady (C12)/Square Meter to Abcoulomb/Square Kilometer and display the result in the second input field.
Example of Farady (C12)/Square Meter to Abcoulomb/Square Kilometer Converter Tool
1
Abcoulomb/Square Kilometer
9648530900
Farady (C12)/Square Meter to Abcoulomb/Square Kilometer Conversion Table
Farady (C12)/Square Meter [F/m2]Abcoulomb/Square Kilometer [abC/km2]Description
1 Farady (C12)/Square Meter9648530900 Abcoulomb/Square Kilometer1 Farady (C12)/Square Meter = 9648530900 Abcoulomb/Square Kilometer
2 Farady (C12)/Square Meter19297061800 Abcoulomb/Square Kilometer2 Farady (C12)/Square Meter = 19297061800 Abcoulomb/Square Kilometer
3 Farady (C12)/Square Meter28945592700 Abcoulomb/Square Kilometer3 Farady (C12)/Square Meter = 28945592700 Abcoulomb/Square Kilometer
4 Farady (C12)/Square Meter38594123600 Abcoulomb/Square Kilometer4 Farady (C12)/Square Meter = 38594123600 Abcoulomb/Square Kilometer
5 Farady (C12)/Square Meter48242654500 Abcoulomb/Square Kilometer5 Farady (C12)/Square Meter = 48242654500 Abcoulomb/Square Kilometer
6 Farady (C12)/Square Meter57891185400 Abcoulomb/Square Kilometer6 Farady (C12)/Square Meter = 57891185400 Abcoulomb/Square Kilometer
7 Farady (C12)/Square Meter67539716300 Abcoulomb/Square Kilometer7 Farady (C12)/Square Meter = 67539716300 Abcoulomb/Square Kilometer
8 Farady (C12)/Square Meter77188247200 Abcoulomb/Square Kilometer8 Farady (C12)/Square Meter = 77188247200 Abcoulomb/Square Kilometer
9 Farady (C12)/Square Meter86836778100 Abcoulomb/Square Kilometer9 Farady (C12)/Square Meter = 86836778100 Abcoulomb/Square Kilometer
10 Farady (C12)/Square Meter96485309000 Abcoulomb/Square Kilometer10 Farady (C12)/Square Meter = 96485309000 Abcoulomb/Square Kilometer
100 Farady (C12)/Square Meter964853090000 Abcoulomb/Square Kilometer100 Farady (C12)/Square Meter = 964853090000 Abcoulomb/Square Kilometer
1000 Farady (C12)/Square Meter9648530900000 Abcoulomb/Square Kilometer1000 Farady (C12)/Square Meter = 9648530900000 Abcoulomb/Square Kilometer
Farady (C12)/Square Meter to Other Units Conversion Table
ConversionDescription
1 Farady (C12)/Square Meter = 96485.31 Coulomb/Square Meter1 Farady (C12)/Square Meter in Coulomb/Square Meter is equal to 96485.31
1 Farady (C12)/Square Meter = 964.85 Coulomb/Square Decimeter1 Farady (C12)/Square Meter in Coulomb/Square Decimeter is equal to 964.85
1 Farady (C12)/Square Meter = 9.65 Coulomb/Square Centimeter1 Farady (C12)/Square Meter in Coulomb/Square Centimeter is equal to 9.65
1 Farady (C12)/Square Meter = 0.096485309 Coulomb/Square Millimeter1 Farady (C12)/Square Meter in Coulomb/Square Millimeter is equal to 0.096485309
1 Farady (C12)/Square Meter = 9.6485309e-8 Coulomb/Square Micrometer1 Farady (C12)/Square Meter in Coulomb/Square Micrometer is equal to 9.6485309e-8
1 Farady (C12)/Square Meter = 9.6485309e-14 Coulomb/Square Nanometer1 Farady (C12)/Square Meter in Coulomb/Square Nanometer is equal to 9.6485309e-14
1 Farady (C12)/Square Meter = 96485309000 Coulomb/Square Kilometer1 Farady (C12)/Square Meter in Coulomb/Square Kilometer is equal to 96485309000
1 Farady (C12)/Square Meter = 80674.01 Coulomb/Square Yard1 Farady (C12)/Square Meter in Coulomb/Square Yard is equal to 80674.01
1 Farady (C12)/Square Meter = 8963.78 Coulomb/Square Foot1 Farady (C12)/Square Meter in Coulomb/Square Foot is equal to 8963.78
1 Farady (C12)/Square Meter = 62.25 Coulomb/Square Inch1 Farady (C12)/Square Meter in Coulomb/Square Inch is equal to 62.25
1 Farady (C12)/Square Meter = 249895803132.1 Coulomb/Square Mile1 Farady (C12)/Square Meter in Coulomb/Square Mile is equal to 249895803132.1
1 Farady (C12)/Square Meter = 96.49 Kilocoulomb/Square Meter1 Farady (C12)/Square Meter in Kilocoulomb/Square Meter is equal to 96.49
1 Farady (C12)/Square Meter = 0.96485309 Kilocoulomb/Square Decimeter1 Farady (C12)/Square Meter in Kilocoulomb/Square Decimeter is equal to 0.96485309
1 Farady (C12)/Square Meter = 0.0096485309 Kilocoulomb/Square Centimeter1 Farady (C12)/Square Meter in Kilocoulomb/Square Centimeter is equal to 0.0096485309
1 Farady (C12)/Square Meter = 0.000096485309 Kilocoulomb/Square Millimeter1 Farady (C12)/Square Meter in Kilocoulomb/Square Millimeter is equal to 0.000096485309
1 Farady (C12)/Square Meter = 9.6485309e-11 Kilocoulomb/Square Micrometer1 Farady (C12)/Square Meter in Kilocoulomb/Square Micrometer is equal to 9.6485309e-11
1 Farady (C12)/Square Meter = 9.6485309e-17 Kilocoulomb/Square Nanometer1 Farady (C12)/Square Meter in Kilocoulomb/Square Nanometer is equal to 9.6485309e-17
1 Farady (C12)/Square Meter = 96485309 Kilocoulomb/Square Kilometer1 Farady (C12)/Square Meter in Kilocoulomb/Square Kilometer is equal to 96485309
1 Farady (C12)/Square Meter = 80.67 Kilocoulomb/Square Yard1 Farady (C12)/Square Meter in Kilocoulomb/Square Yard is equal to 80.67
1 Farady (C12)/Square Meter = 8.96 Kilocoulomb/Square Foot1 Farady (C12)/Square Meter in Kilocoulomb/Square Foot is equal to 8.96
1 Farady (C12)/Square Meter = 0.06224846195444 Kilocoulomb/Square Inch1 Farady (C12)/Square Meter in Kilocoulomb/Square Inch is equal to 0.06224846195444
1 Farady (C12)/Square Meter = 249895803.13 Kilocoulomb/Square Mile1 Farady (C12)/Square Meter in Kilocoulomb/Square Mile is equal to 249895803.13
1 Farady (C12)/Square Meter = 96485309 Millicoulomb/Square Meter1 Farady (C12)/Square Meter in Millicoulomb/Square Meter is equal to 96485309
1 Farady (C12)/Square Meter = 964853.09 Millicoulomb/Square Decimeter1 Farady (C12)/Square Meter in Millicoulomb/Square Decimeter is equal to 964853.09
1 Farady (C12)/Square Meter = 9648.53 Millicoulomb/Square Centimeter1 Farady (C12)/Square Meter in Millicoulomb/Square Centimeter is equal to 9648.53
1 Farady (C12)/Square Meter = 96.49 Millicoulomb/Square Millimeter1 Farady (C12)/Square Meter in Millicoulomb/Square Millimeter is equal to 96.49
1 Farady (C12)/Square Meter = 0.000096485309 Millicoulomb/Square Micrometer1 Farady (C12)/Square Meter in Millicoulomb/Square Micrometer is equal to 0.000096485309
1 Farady (C12)/Square Meter = 9.6485309e-11 Millicoulomb/Square Nanometer1 Farady (C12)/Square Meter in Millicoulomb/Square Nanometer is equal to 9.6485309e-11
1 Farady (C12)/Square Meter = 96485309000000 Millicoulomb/Square Kilometer1 Farady (C12)/Square Meter in Millicoulomb/Square Kilometer is equal to 96485309000000
1 Farady (C12)/Square Meter = 80674006.69 Millicoulomb/Square Yard1 Farady (C12)/Square Meter in Millicoulomb/Square Yard is equal to 80674006.69
1 Farady (C12)/Square Meter = 8963778.52 Millicoulomb/Square Foot1 Farady (C12)/Square Meter in Millicoulomb/Square Foot is equal to 8963778.52
1 Farady (C12)/Square Meter = 62248.46 Millicoulomb/Square Inch1 Farady (C12)/Square Meter in Millicoulomb/Square Inch is equal to 62248.46
1 Farady (C12)/Square Meter = 249895803132100 Millicoulomb/Square Mile1 Farady (C12)/Square Meter in Millicoulomb/Square Mile is equal to 249895803132100
1 Farady (C12)/Square Meter = 96485309000 Microcoulomb/Square Meter1 Farady (C12)/Square Meter in Microcoulomb/Square Meter is equal to 96485309000
1 Farady (C12)/Square Meter = 964853090 Microcoulomb/Square Decimeter1 Farady (C12)/Square Meter in Microcoulomb/Square Decimeter is equal to 964853090
1 Farady (C12)/Square Meter = 9648530.9 Microcoulomb/Square Centimeter1 Farady (C12)/Square Meter in Microcoulomb/Square Centimeter is equal to 9648530.9
1 Farady (C12)/Square Meter = 96485.31 Microcoulomb/Square Millimeter1 Farady (C12)/Square Meter in Microcoulomb/Square Millimeter is equal to 96485.31
1 Farady (C12)/Square Meter = 0.096485309 Microcoulomb/Square Micrometer1 Farady (C12)/Square Meter in Microcoulomb/Square Micrometer is equal to 0.096485309
1 Farady (C12)/Square Meter = 9.6485309e-8 Microcoulomb/Square Nanometer1 Farady (C12)/Square Meter in Microcoulomb/Square Nanometer is equal to 9.6485309e-8
1 Farady (C12)/Square Meter = 96485309000000000 Microcoulomb/Square Kilometer1 Farady (C12)/Square Meter in Microcoulomb/Square Kilometer is equal to 96485309000000000
1 Farady (C12)/Square Meter = 80674006692.95 Microcoulomb/Square Yard1 Farady (C12)/Square Meter in Microcoulomb/Square Yard is equal to 80674006692.95
1 Farady (C12)/Square Meter = 8963778521.44 Microcoulomb/Square Foot1 Farady (C12)/Square Meter in Microcoulomb/Square Foot is equal to 8963778521.44
1 Farady (C12)/Square Meter = 62248461.95 Microcoulomb/Square Inch1 Farady (C12)/Square Meter in Microcoulomb/Square Inch is equal to 62248461.95
1 Farady (C12)/Square Meter = 249895803132100000 Microcoulomb/Square Mile1 Farady (C12)/Square Meter in Microcoulomb/Square Mile is equal to 249895803132100000
1 Farady (C12)/Square Meter = 96485309000000 Nanocoulomb/Square Meter1 Farady (C12)/Square Meter in Nanocoulomb/Square Meter is equal to 96485309000000
1 Farady (C12)/Square Meter = 964853090000 Nanocoulomb/Square Decimeter1 Farady (C12)/Square Meter in Nanocoulomb/Square Decimeter is equal to 964853090000
1 Farady (C12)/Square Meter = 9648530900 Nanocoulomb/Square Centimeter1 Farady (C12)/Square Meter in Nanocoulomb/Square Centimeter is equal to 9648530900
1 Farady (C12)/Square Meter = 96485309 Nanocoulomb/Square Millimeter1 Farady (C12)/Square Meter in Nanocoulomb/Square Millimeter is equal to 96485309
1 Farady (C12)/Square Meter = 96.49 Nanocoulomb/Square Micrometer1 Farady (C12)/Square Meter in Nanocoulomb/Square Micrometer is equal to 96.49
1 Farady (C12)/Square Meter = 0.000096485309 Nanocoulomb/Square Nanometer1 Farady (C12)/Square Meter in Nanocoulomb/Square Nanometer is equal to 0.000096485309
1 Farady (C12)/Square Meter = 96485309000000000000 Nanocoulomb/Square Kilometer1 Farady (C12)/Square Meter in Nanocoulomb/Square Kilometer is equal to 96485309000000000000
1 Farady (C12)/Square Meter = 80674006692954 Nanocoulomb/Square Yard1 Farady (C12)/Square Meter in Nanocoulomb/Square Yard is equal to 80674006692954
1 Farady (C12)/Square Meter = 8963778521439.4 Nanocoulomb/Square Foot1 Farady (C12)/Square Meter in Nanocoulomb/Square Foot is equal to 8963778521439.4
1 Farady (C12)/Square Meter = 62248461954.44 Nanocoulomb/Square Inch1 Farady (C12)/Square Meter in Nanocoulomb/Square Inch is equal to 62248461954.44
1 Farady (C12)/Square Meter = 249895803132100000000 Nanocoulomb/Square Mile1 Farady (C12)/Square Meter in Nanocoulomb/Square Mile is equal to 249895803132100000000
1 Farady (C12)/Square Meter = 96485309000000000 Picocoulomb/Square Meter1 Farady (C12)/Square Meter in Picocoulomb/Square Meter is equal to 96485309000000000
1 Farady (C12)/Square Meter = 964853090000000 Picocoulomb/Square Decimeter1 Farady (C12)/Square Meter in Picocoulomb/Square Decimeter is equal to 964853090000000
1 Farady (C12)/Square Meter = 9648530900000 Picocoulomb/Square Centimeter1 Farady (C12)/Square Meter in Picocoulomb/Square Centimeter is equal to 9648530900000
1 Farady (C12)/Square Meter = 96485309000 Picocoulomb/Square Millimeter1 Farady (C12)/Square Meter in Picocoulomb/Square Millimeter is equal to 96485309000
1 Farady (C12)/Square Meter = 96485.31 Picocoulomb/Square Micrometer1 Farady (C12)/Square Meter in Picocoulomb/Square Micrometer is equal to 96485.31
1 Farady (C12)/Square Meter = 0.096485309 Picocoulomb/Square Nanometer1 Farady (C12)/Square Meter in Picocoulomb/Square Nanometer is equal to 0.096485309
1 Farady (C12)/Square Meter = 9.6485309e+22 Picocoulomb/Square Kilometer1 Farady (C12)/Square Meter in Picocoulomb/Square Kilometer is equal to 9.6485309e+22
1 Farady (C12)/Square Meter = 80674006692954000 Picocoulomb/Square Yard1 Farady (C12)/Square Meter in Picocoulomb/Square Yard is equal to 80674006692954000
1 Farady (C12)/Square Meter = 8963778521439400 Picocoulomb/Square Foot1 Farady (C12)/Square Meter in Picocoulomb/Square Foot is equal to 8963778521439400
1 Farady (C12)/Square Meter = 62248461954440 Picocoulomb/Square Inch1 Farady (C12)/Square Meter in Picocoulomb/Square Inch is equal to 62248461954440
1 Farady (C12)/Square Meter = 2.498958031321e+23 Picocoulomb/Square Mile1 Farady (C12)/Square Meter in Picocoulomb/Square Mile is equal to 2.498958031321e+23
1 Farady (C12)/Square Meter = 6.0221393167565e+23 Elementary Charge/Square Meter1 Farady (C12)/Square Meter in Elementary Charge/Square Meter is equal to 6.0221393167565e+23
1 Farady (C12)/Square Meter = 6.0221393167565e+21 Elementary Charge/Square Decimeter1 Farady (C12)/Square Meter in Elementary Charge/Square Decimeter is equal to 6.0221393167565e+21
1 Farady (C12)/Square Meter = 60221393167565000000 Elementary Charge/Square Centimeter1 Farady (C12)/Square Meter in Elementary Charge/Square Centimeter is equal to 60221393167565000000
1 Farady (C12)/Square Meter = 602213931675650000 Elementary Charge/Square Millimeter1 Farady (C12)/Square Meter in Elementary Charge/Square Millimeter is equal to 602213931675650000
1 Farady (C12)/Square Meter = 602213931675.65 Elementary Charge/Square Micrometer1 Farady (C12)/Square Meter in Elementary Charge/Square Micrometer is equal to 602213931675.65
1 Farady (C12)/Square Meter = 602213.93 Elementary Charge/Square Nanometer1 Farady (C12)/Square Meter in Elementary Charge/Square Nanometer is equal to 602213.93
1 Farady (C12)/Square Meter = 6.0221393167565e+29 Elementary Charge/Square Kilometer1 Farady (C12)/Square Meter in Elementary Charge/Square Kilometer is equal to 6.0221393167565e+29
1 Farady (C12)/Square Meter = 5.0352754484718e+23 Elementary Charge/Square Yard1 Farady (C12)/Square Meter in Elementary Charge/Square Yard is equal to 5.0352754484718e+23
1 Farady (C12)/Square Meter = 5.594750498302e+22 Elementary Charge/Square Foot1 Farady (C12)/Square Meter in Elementary Charge/Square Foot is equal to 5.594750498302e+22
1 Farady (C12)/Square Meter = 388524340159860000000 Elementary Charge/Square Inch1 Farady (C12)/Square Meter in Elementary Charge/Square Inch is equal to 388524340159860000000
1 Farady (C12)/Square Meter = 1.5597269229186e+30 Elementary Charge/Square Mile1 Farady (C12)/Square Meter in Elementary Charge/Square Mile is equal to 1.5597269229186e+30
1 Farady (C12)/Square Meter = 9648.53 EMU of Charge/Square Meter1 Farady (C12)/Square Meter in EMU of Charge/Square Meter is equal to 9648.53
1 Farady (C12)/Square Meter = 96.49 EMU of Charge/Square Decimeter1 Farady (C12)/Square Meter in EMU of Charge/Square Decimeter is equal to 96.49
1 Farady (C12)/Square Meter = 0.96485309 EMU of Charge/Square Centimeter1 Farady (C12)/Square Meter in EMU of Charge/Square Centimeter is equal to 0.96485309
1 Farady (C12)/Square Meter = 0.0096485309 EMU of Charge/Square Millimeter1 Farady (C12)/Square Meter in EMU of Charge/Square Millimeter is equal to 0.0096485309
1 Farady (C12)/Square Meter = 9.6485309e-9 EMU of Charge/Square Micrometer1 Farady (C12)/Square Meter in EMU of Charge/Square Micrometer is equal to 9.6485309e-9
1 Farady (C12)/Square Meter = 9.6485309e-15 EMU of Charge/Square Nanometer1 Farady (C12)/Square Meter in EMU of Charge/Square Nanometer is equal to 9.6485309e-15
1 Farady (C12)/Square Meter = 9648530900 EMU of Charge/Square Kilometer1 Farady (C12)/Square Meter in EMU of Charge/Square Kilometer is equal to 9648530900
1 Farady (C12)/Square Meter = 8067.4 EMU of Charge/Square Yard1 Farady (C12)/Square Meter in EMU of Charge/Square Yard is equal to 8067.4
1 Farady (C12)/Square Meter = 896.38 EMU of Charge/Square Foot1 Farady (C12)/Square Meter in EMU of Charge/Square Foot is equal to 896.38
1 Farady (C12)/Square Meter = 6.22 EMU of Charge/Square Inch1 Farady (C12)/Square Meter in EMU of Charge/Square Inch is equal to 6.22
1 Farady (C12)/Square Meter = 24989580313.21 EMU of Charge/Square Mile1 Farady (C12)/Square Meter in EMU of Charge/Square Mile is equal to 24989580313.21
1 Farady (C12)/Square Meter = 288878170658680 ESU of Charge/Square Meter1 Farady (C12)/Square Meter in ESU of Charge/Square Meter is equal to 288878170658680
1 Farady (C12)/Square Meter = 2888781706586.8 ESU of Charge/Square Decimeter1 Farady (C12)/Square Meter in ESU of Charge/Square Decimeter is equal to 2888781706586.8
1 Farady (C12)/Square Meter = 28887817065.87 ESU of Charge/Square Centimeter1 Farady (C12)/Square Meter in ESU of Charge/Square Centimeter is equal to 28887817065.87
1 Farady (C12)/Square Meter = 288878170.66 ESU of Charge/Square Millimeter1 Farady (C12)/Square Meter in ESU of Charge/Square Millimeter is equal to 288878170.66
1 Farady (C12)/Square Meter = 288.88 ESU of Charge/Square Micrometer1 Farady (C12)/Square Meter in ESU of Charge/Square Micrometer is equal to 288.88
1 Farady (C12)/Square Meter = 0.00028887817065868 ESU of Charge/Square Nanometer1 Farady (C12)/Square Meter in ESU of Charge/Square Nanometer is equal to 0.00028887817065868
1 Farady (C12)/Square Meter = 288878170658680000000 ESU of Charge/Square Kilometer1 Farady (C12)/Square Meter in ESU of Charge/Square Kilometer is equal to 288878170658680000000
1 Farady (C12)/Square Meter = 241538942194470 ESU of Charge/Square Yard1 Farady (C12)/Square Meter in ESU of Charge/Square Yard is equal to 241538942194470
1 Farady (C12)/Square Meter = 26837660243830 ESU of Charge/Square Foot1 Farady (C12)/Square Meter in ESU of Charge/Square Foot is equal to 26837660243830
1 Farady (C12)/Square Meter = 186372640582.16 ESU of Charge/Square Inch1 Farady (C12)/Square Meter in ESU of Charge/Square Inch is equal to 186372640582.16
1 Farady (C12)/Square Meter = 748191027341600000000 ESU of Charge/Square Mile1 Farady (C12)/Square Meter in ESU of Charge/Square Mile is equal to 748191027341600000000
1 Farady (C12)/Square Meter = 288878170658680 Franklin/Square Meter1 Farady (C12)/Square Meter in Franklin/Square Meter is equal to 288878170658680
1 Farady (C12)/Square Meter = 2888781706586.8 Franklin/Square Decimeter1 Farady (C12)/Square Meter in Franklin/Square Decimeter is equal to 2888781706586.8
1 Farady (C12)/Square Meter = 28887817065.87 Franklin/Square Centimeter1 Farady (C12)/Square Meter in Franklin/Square Centimeter is equal to 28887817065.87
1 Farady (C12)/Square Meter = 288878170.66 Franklin/Square Millimeter1 Farady (C12)/Square Meter in Franklin/Square Millimeter is equal to 288878170.66
1 Farady (C12)/Square Meter = 288.88 Franklin/Square Micrometer1 Farady (C12)/Square Meter in Franklin/Square Micrometer is equal to 288.88
1 Farady (C12)/Square Meter = 0.00028887817065868 Franklin/Square Nanometer1 Farady (C12)/Square Meter in Franklin/Square Nanometer is equal to 0.00028887817065868
1 Farady (C12)/Square Meter = 288878170658680000000 Franklin/Square Kilometer1 Farady (C12)/Square Meter in Franklin/Square Kilometer is equal to 288878170658680000000
1 Farady (C12)/Square Meter = 241538942194470 Franklin/Square Yard1 Farady (C12)/Square Meter in Franklin/Square Yard is equal to 241538942194470
1 Farady (C12)/Square Meter = 26837660243830 Franklin/Square Foot1 Farady (C12)/Square Meter in Franklin/Square Foot is equal to 26837660243830
1 Farady (C12)/Square Meter = 186372640582.16 Franklin/Square Inch1 Farady (C12)/Square Meter in Franklin/Square Inch is equal to 186372640582.16
1 Farady (C12)/Square Meter = 748191027341600000000 Franklin/Square Mile1 Farady (C12)/Square Meter in Franklin/Square Mile is equal to 748191027341600000000
1 Farady (C12)/Square Meter = 0.096485309 Megacoulomb/Square Meter1 Farady (C12)/Square Meter in Megacoulomb/Square Meter is equal to 0.096485309
1 Farady (C12)/Square Meter = 0.00096485309 Megacoulomb/Square Decimeter1 Farady (C12)/Square Meter in Megacoulomb/Square Decimeter is equal to 0.00096485309
1 Farady (C12)/Square Meter = 0.0000096485309 Megacoulomb/Square Centimeter1 Farady (C12)/Square Meter in Megacoulomb/Square Centimeter is equal to 0.0000096485309
1 Farady (C12)/Square Meter = 9.6485309e-8 Megacoulomb/Square Millimeter1 Farady (C12)/Square Meter in Megacoulomb/Square Millimeter is equal to 9.6485309e-8
1 Farady (C12)/Square Meter = 9.6485309e-14 Megacoulomb/Square Micrometer1 Farady (C12)/Square Meter in Megacoulomb/Square Micrometer is equal to 9.6485309e-14
1 Farady (C12)/Square Meter = 9.6485309e-20 Megacoulomb/Square Nanometer1 Farady (C12)/Square Meter in Megacoulomb/Square Nanometer is equal to 9.6485309e-20
1 Farady (C12)/Square Meter = 96485.31 Megacoulomb/Square Kilometer1 Farady (C12)/Square Meter in Megacoulomb/Square Kilometer is equal to 96485.31
1 Farady (C12)/Square Meter = 0.080674006692954 Megacoulomb/Square Yard1 Farady (C12)/Square Meter in Megacoulomb/Square Yard is equal to 0.080674006692954
1 Farady (C12)/Square Meter = 0.0089637785214394 Megacoulomb/Square Foot1 Farady (C12)/Square Meter in Megacoulomb/Square Foot is equal to 0.0089637785214394
1 Farady (C12)/Square Meter = 0.00006224846195444 Megacoulomb/Square Inch1 Farady (C12)/Square Meter in Megacoulomb/Square Inch is equal to 0.00006224846195444
1 Farady (C12)/Square Meter = 249895.8 Megacoulomb/Square Mile1 Farady (C12)/Square Meter in Megacoulomb/Square Mile is equal to 249895.8
1 Farady (C12)/Square Meter = 288878170658680 Statcoulomb/Square Meter1 Farady (C12)/Square Meter in Statcoulomb/Square Meter is equal to 288878170658680
1 Farady (C12)/Square Meter = 2888781706586.8 Statcoulomb/Square Decimeter1 Farady (C12)/Square Meter in Statcoulomb/Square Decimeter is equal to 2888781706586.8
1 Farady (C12)/Square Meter = 28887817065.87 Statcoulomb/Square Centimeter1 Farady (C12)/Square Meter in Statcoulomb/Square Centimeter is equal to 28887817065.87
1 Farady (C12)/Square Meter = 288878170.66 Statcoulomb/Square Millimeter1 Farady (C12)/Square Meter in Statcoulomb/Square Millimeter is equal to 288878170.66
1 Farady (C12)/Square Meter = 288.88 Statcoulomb/Square Micrometer1 Farady (C12)/Square Meter in Statcoulomb/Square Micrometer is equal to 288.88
1 Farady (C12)/Square Meter = 0.00028887817065868 Statcoulomb/Square Nanometer1 Farady (C12)/Square Meter in Statcoulomb/Square Nanometer is equal to 0.00028887817065868
1 Farady (C12)/Square Meter = 288878170658680000000 Statcoulomb/Square Kilometer1 Farady (C12)/Square Meter in Statcoulomb/Square Kilometer is equal to 288878170658680000000
1 Farady (C12)/Square Meter = 241538942194470 Statcoulomb/Square Yard1 Farady (C12)/Square Meter in Statcoulomb/Square Yard is equal to 241538942194470
1 Farady (C12)/Square Meter = 26837660243830 Statcoulomb/Square Foot1 Farady (C12)/Square Meter in Statcoulomb/Square Foot is equal to 26837660243830
1 Farady (C12)/Square Meter = 186372640582.16 Statcoulomb/Square Inch1 Farady (C12)/Square Meter in Statcoulomb/Square Inch is equal to 186372640582.16
1 Farady (C12)/Square Meter = 748191027341600000000 Statcoulomb/Square Mile1 Farady (C12)/Square Meter in Statcoulomb/Square Mile is equal to 748191027341600000000
1 Farady (C12)/Square Meter = 9648.53 Abcoulomb/Square Meter1 Farady (C12)/Square Meter in Abcoulomb/Square Meter is equal to 9648.53
1 Farady (C12)/Square Meter = 96.49 Abcoulomb/Square Decimeter1 Farady (C12)/Square Meter in Abcoulomb/Square Decimeter is equal to 96.49
1 Farady (C12)/Square Meter = 0.96485309 Abcoulomb/Square Centimeter1 Farady (C12)/Square Meter in Abcoulomb/Square Centimeter is equal to 0.96485309
1 Farady (C12)/Square Meter = 0.0096485309 Abcoulomb/Square Millimeter1 Farady (C12)/Square Meter in Abcoulomb/Square Millimeter is equal to 0.0096485309
1 Farady (C12)/Square Meter = 9.6485309e-9 Abcoulomb/Square Micrometer1 Farady (C12)/Square Meter in Abcoulomb/Square Micrometer is equal to 9.6485309e-9
1 Farady (C12)/Square Meter = 9.6485309e-15 Abcoulomb/Square Nanometer1 Farady (C12)/Square Meter in Abcoulomb/Square Nanometer is equal to 9.6485309e-15
1 Farady (C12)/Square Meter = 9648530900 Abcoulomb/Square Kilometer1 Farady (C12)/Square Meter in Abcoulomb/Square Kilometer is equal to 9648530900
1 Farady (C12)/Square Meter = 8067.4 Abcoulomb/Square Yard1 Farady (C12)/Square Meter in Abcoulomb/Square Yard is equal to 8067.4
1 Farady (C12)/Square Meter = 896.38 Abcoulomb/Square Foot1 Farady (C12)/Square Meter in Abcoulomb/Square Foot is equal to 896.38
1 Farady (C12)/Square Meter = 6.22 Abcoulomb/Square Inch1 Farady (C12)/Square Meter in Abcoulomb/Square Inch is equal to 6.22
1 Farady (C12)/Square Meter = 24989580313.21 Abcoulomb/Square Mile1 Farady (C12)/Square Meter in Abcoulomb/Square Mile is equal to 24989580313.21
1 Farady (C12)/Square Meter = 26.8 Ampere Hour/Square Meter1 Farady (C12)/Square Meter in Ampere Hour/Square Meter is equal to 26.8
1 Farady (C12)/Square Meter = 0.26801474722222 Ampere Hour/Square Decimeter1 Farady (C12)/Square Meter in Ampere Hour/Square Decimeter is equal to 0.26801474722222
1 Farady (C12)/Square Meter = 0.0026801474722222 Ampere Hour/Square Centimeter1 Farady (C12)/Square Meter in Ampere Hour/Square Centimeter is equal to 0.0026801474722222
1 Farady (C12)/Square Meter = 0.000026801474722222 Ampere Hour/Square Millimeter1 Farady (C12)/Square Meter in Ampere Hour/Square Millimeter is equal to 0.000026801474722222
1 Farady (C12)/Square Meter = 2.6801474722222e-11 Ampere Hour/Square Micrometer1 Farady (C12)/Square Meter in Ampere Hour/Square Micrometer is equal to 2.6801474722222e-11
1 Farady (C12)/Square Meter = 2.6801474722222e-17 Ampere Hour/Square Nanometer1 Farady (C12)/Square Meter in Ampere Hour/Square Nanometer is equal to 2.6801474722222e-17
1 Farady (C12)/Square Meter = 26801474.72 Ampere Hour/Square Kilometer1 Farady (C12)/Square Meter in Ampere Hour/Square Kilometer is equal to 26801474.72
1 Farady (C12)/Square Meter = 22.41 Ampere Hour/Square Yard1 Farady (C12)/Square Meter in Ampere Hour/Square Yard is equal to 22.41
1 Farady (C12)/Square Meter = 2.49 Ampere Hour/Square Foot1 Farady (C12)/Square Meter in Ampere Hour/Square Foot is equal to 2.49
1 Farady (C12)/Square Meter = 0.017291239431789 Ampere Hour/Square Inch1 Farady (C12)/Square Meter in Ampere Hour/Square Inch is equal to 0.017291239431789
1 Farady (C12)/Square Meter = 69415500.87 Ampere Hour/Square Mile1 Farady (C12)/Square Meter in Ampere Hour/Square Mile is equal to 69415500.87
1 Farady (C12)/Square Meter = 96485.31 Ampere Second/Square Meter1 Farady (C12)/Square Meter in Ampere Second/Square Meter is equal to 96485.31
1 Farady (C12)/Square Meter = 964.85 Ampere Second/Square Decimeter1 Farady (C12)/Square Meter in Ampere Second/Square Decimeter is equal to 964.85
1 Farady (C12)/Square Meter = 9.65 Ampere Second/Square Centimeter1 Farady (C12)/Square Meter in Ampere Second/Square Centimeter is equal to 9.65
1 Farady (C12)/Square Meter = 0.096485309 Ampere Second/Square Millimeter1 Farady (C12)/Square Meter in Ampere Second/Square Millimeter is equal to 0.096485309
1 Farady (C12)/Square Meter = 9.6485309e-8 Ampere Second/Square Micrometer1 Farady (C12)/Square Meter in Ampere Second/Square Micrometer is equal to 9.6485309e-8
1 Farady (C12)/Square Meter = 9.6485309e-14 Ampere Second/Square Nanometer1 Farady (C12)/Square Meter in Ampere Second/Square Nanometer is equal to 9.6485309e-14
1 Farady (C12)/Square Meter = 96485309000 Ampere Second/Square Kilometer1 Farady (C12)/Square Meter in Ampere Second/Square Kilometer is equal to 96485309000
1 Farady (C12)/Square Meter = 80674.01 Ampere Second/Square Yard1 Farady (C12)/Square Meter in Ampere Second/Square Yard is equal to 80674.01
1 Farady (C12)/Square Meter = 8963.78 Ampere Second/Square Foot1 Farady (C12)/Square Meter in Ampere Second/Square Foot is equal to 8963.78
1 Farady (C12)/Square Meter = 62.25 Ampere Second/Square Inch1 Farady (C12)/Square Meter in Ampere Second/Square Inch is equal to 62.25
1 Farady (C12)/Square Meter = 249895803132.1 Ampere Second/Square Mile1 Farady (C12)/Square Meter in Ampere Second/Square Mile is equal to 249895803132.1
1 Farady (C12)/Square Meter = 1608.09 Ampere Minute/Square Meter1 Farady (C12)/Square Meter in Ampere Minute/Square Meter is equal to 1608.09
1 Farady (C12)/Square Meter = 16.08 Ampere Minute/Square Decimeter1 Farady (C12)/Square Meter in Ampere Minute/Square Decimeter is equal to 16.08
1 Farady (C12)/Square Meter = 0.16080884833333 Ampere Minute/Square Centimeter1 Farady (C12)/Square Meter in Ampere Minute/Square Centimeter is equal to 0.16080884833333
1 Farady (C12)/Square Meter = 0.0016080884833333 Ampere Minute/Square Millimeter1 Farady (C12)/Square Meter in Ampere Minute/Square Millimeter is equal to 0.0016080884833333
1 Farady (C12)/Square Meter = 1.6080884833333e-9 Ampere Minute/Square Micrometer1 Farady (C12)/Square Meter in Ampere Minute/Square Micrometer is equal to 1.6080884833333e-9
1 Farady (C12)/Square Meter = 1.6080884833333e-15 Ampere Minute/Square Nanometer1 Farady (C12)/Square Meter in Ampere Minute/Square Nanometer is equal to 1.6080884833333e-15
1 Farady (C12)/Square Meter = 1608088483.33 Ampere Minute/Square Kilometer1 Farady (C12)/Square Meter in Ampere Minute/Square Kilometer is equal to 1608088483.33
1 Farady (C12)/Square Meter = 1344.57 Ampere Minute/Square Yard1 Farady (C12)/Square Meter in Ampere Minute/Square Yard is equal to 1344.57
1 Farady (C12)/Square Meter = 149.4 Ampere Minute/Square Foot1 Farady (C12)/Square Meter in Ampere Minute/Square Foot is equal to 149.4
1 Farady (C12)/Square Meter = 1.04 Ampere Minute/Square Inch1 Farady (C12)/Square Meter in Ampere Minute/Square Inch is equal to 1.04
1 Farady (C12)/Square Meter = 4164930052.2 Ampere Minute/Square Mile1 Farady (C12)/Square Meter in Ampere Minute/Square Mile is equal to 4164930052.2
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# Query regarding interference
In Young's Double Slit experiment, why is it that two waves vibrating perpendicular to each other doesn't show interference?
I know that for interference to happen, the waves must be coherent ( i.e., they maintain a constant phase w.r.t each other) and if the waves meet at same point in the screen it results in constructive interference and destructive if there is a phase change of lambda/2 and so on.
Any help regarding the query would be nice.
• Are you talking about polarization? If so, that's simply an observed phenomenon. – Carl Witthoft Mar 3 '14 at 13:15
• I'm talking about interference from two slits. – Albert Einstein Mar 3 '14 at 13:40
There's no way that you can add a vector pointing along the $x$ axis, and have it cancel a vector pointing in the $y$ direction. So the amplitude of the light is zero nowhere, and there are no intensity fringes. Instead, the polarization of the wave changes at the screen. As you study the light along the direction that you expect to see fringes, you would see the light changing polarization rather than amplitude: from linear (inclined 45${}^\circ$) where you expect maxima, to circular where you would expect minima. I suppose you could observe intensity fringes if you put a polarizer in from of the screen. I've never seen that done. So there is interference of a sort, it's just of a different kind.
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# Object COMP Xform Page
## Parameters - Xform Page
The Xform parameter page controls the object component's transform in world space.
Transform Order `xord` - The menu attached to this parameter allows you to specify the order in which the changes to your Component will take place. Changing the Transform order will change where things go much the same way as going a block and turning east gets you to a different place than turning east and then going a block. In matrix math terms, if we use the 'multiply vector on the right' (column vector) convention, a transform order of Scale, Rotate, Translate would be written as `translate * rotate * scale * vectorOrPosition`.
Rotate Order `rord` - The rotational matrix presented when you click on this option allows you to set the transform order for the Component's rotations. As with transform order (above), changing the order in which the Component's rotations take place will alter the Component's final position.
Translate / Rotation / Scale `t[xyz] r[xyz] s[xyz]` - The three fields allow you to specify the amount of movement along any of the three axes; the amount, in degrees, of rotation around any of the three axes; and a non-uniform scaling along the three axes. As an alternative to entering the values directly into these fields, you can modify the values by manipulating the Component in the Viewport with the Select & Transform state.
Pivot `p[xyz]` - The Pivot point edit fields allow you to define the point about which a Component scales and rotates. Altering the pivot point of a Component produces different results depending on the transformation performed on the Component.
For example, during a scaling operation, if the pivot point of an Component is located at `-1, -1, 0` and you wanted to scale the Component by `0.5` (reduce its size by 50%), the Component would scale toward the pivot point and appear to slide down and to the left.
In the example above, rotations performed on an Component with different pivot points produce very different results.
Uniform Scale `scale` - This field allows you to change the size of an Component uniformly along the three axes.
Note: Scaling a camera's channels is not generally recommended. However, should you decide to do so, the rendered output will match the Viewport as closely as possible when scales are involved.
Constrain To `constrain` - Allows the location of the object to be constrained to any other object whose path is specified in this parameter.
Look At `lookat` - Allows you to orient your Component by naming the Component you would like it to Look At, or point to. Once you have designated this Component to look at, it will continue to face that Component, even if you move it. This is useful if, for instance, you want a camera to follow another Component's movements. The Look At parameter points the Component in question at the other Component's origin.
Tip: To designate a center of interest for the camera that doesn't appear in your scene, create a Null Component and disable its display flag. Then Parent the Camera to the newly created Null Component, and tell the camera to look at this Component using the Look At parameter. You can direct the attention of the camera by moving the Null Component with the Select state. If you want to see both the camera and the Null Component, enable the Null Component's display flag, and use the Select state in an additional Viewport by clicking one of the icons in the top-right corner of the TouchDesigner window.
Look At Up Vector `lookup` - When specifying a Look At, it is possible to specify an up vector for the lookat. Without using an up vector, it is possible to get poor animation when the lookat Component passes through the Y axis of the target Component.
• Don't Use Up Vector - Use this option if the look at Component does not pass through the Y axis of the target Component.
• Use Up Vector - This precisely defines the rotates on the Component doing the looking. The Up Vector specified should not be parallel to the look at direction. See Up Vector below.
• Use Quaternions - Quaternions are a mathematical representation of a 3D rotation. This method finds the most efficient means of moving from one point to another on a sphere.
Path SOP `pathsop` - Names the SOP that functions as the path you want this Component to move along. For instance, you can name an SOP that provides a spline path for the camera to follow.
Production Tip: For Smooth Motion Along a Path - Having a Component follow an animation path is simple. However, when using a NURBS curve as your path, you might notice that the Component speeds up and slows down unexpectedly as it travels along the path. This is usually because the CVs are spaced unevenly. In such a case, use the Resample SOP to redistribute the CVs so that they are evenly spaced along the curve. A caution however - using a Resample SOP can be slow if you have an animating path curve.
An alternative method is to append a Basis SOP to the path curve and change it to a `Uniform Curve`. This way, your Component will move uniformly down the curve, and there is no need for the Resample SOP and the unnecessary points it generates.
Roll `roll` - Using the angle control you can specify a Component's rotation as it animates along the path.
Position `pos` - This parameter lets you specify the Position of the Component along the path. The values you can enter for this parameter range from `0` to `1`, where `0` equals the starting point and `1` equals the end point of the path. The value slider allows for values as high as `10` for multiple "passes" along the path.
Orient Along Path `pathorient` - If this option is selected, the Component will be oriented along the path. The positive Z axis of the Component will be pointing down the path.
Up Vector `up` - When orienting a Component, the Up Vector is used to determine where the positive Y axis points.
Auto-Bank Factor `bank` - The Auto-Bank Factor rolls the Component based on the curvature of the path at its current position. To turn off auto-banking, set the bank scale to `0`.
An Operator Family that contains its own Network inside. There are twelve 3D Object Component and eight 2D Panel Component types. See also Network Path.
The location of an operator within the TouchDesigner environment, for example, `/geo1/torus1`, a node called `torus1` in a component called `geo1`. The path `/` is called Root. To refer instead to a filesystem folder, directory, disk file or `http:` address, see Folder.
An Operator Family that reads, creates and modifies 3D polygons, curves, NURBS surfaces, spheres, meatballs and other 3D surface data.
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# CCF to MCF Conversion
••• ChiccoDodiFC/iStock/GettyImages
Print
CCF and MCF are standardized units of measurement of natural gas. The initial C in the term "CCF" is the Roman numeral for 100; "CCF" means 100 cubic feet. The initial M in the term "MCF" is the Roman numeral for 1,000: "MCF" means 1,000 cubic feet. Here's a quick primer of Roman numerals: I = 1; V = 5; X = 10; L = 50; C = 100; D = 500; and M = 1,000. If you know this, and keep in mind that the "CF" in "CCF" and "MCF" stands for "cubic feet," you're already halfway to making the conversion from CCF to MCF.
Write down the CCF number that you want to convert. For an example, let's use 1,000 CCF.
Divide this number by 10 and this will yield the MCF. For example, 1000/10 = 100.
To convert the MCF figure to CCF, multiply the MCF number by 10.
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The push for rigorous evaluations of the impact of interventions has led to an increase in the use of randomized trials (RTs). In practice, it is often the case that interventions are delivered at the cluster level, such as a whole school reform model or a new curriculum. In these cases, the cluster (i.e., the school), is the logical unit of random assignment and I hereafter refer to these as cluster randomized trials (CRTs).
Designing a CRT is necessarily more complex than a RT for several reasons. First, there are two sample sizes, i.e., the number of students per school and the total number of schools. Second, the greater the variability in the outcome across schools, the more schools you will need to detect an effect of a given magnitude. The percentage of variance in the outcome that is between schools is commonly referred to as the intra-class correlation (ICC). For example, suppose I am testing an intervention and the outcome of interest is math achievement, there are 500 students per school, and a school level covariate explain 50 percent of the variation in the outcome. If the ICC is 0.20 and I want to detect an effect size difference of 0.2 standard deviations between the treatment and comparison conditions, 82 total schools, or 41 treatment and 41 comparison schools, would be needed to achieve statistical power equal to 0.80, the commonly accepted threshold. Instead, if the ICC is 0.05, the total number of schools would only be 24, a reduction of 54. Hence an accurate estimate of the ICC is critical in planning a CRT as it has a strong impact on the number of schools needed for a study.
The challenge is that the required sample size needs to be determined prior to the start of the study, hence I need to estimate the ICC since the actual data has not yet been collected. Recently there has been an increase in empirical studies which seek to estimate ICCs for different contexts. The findings suggest that the ICC varies depending on outcome type, unit of the clusters (i.e., schools, classrooms, etc.), grade and other features.
Resources:
Resources have started popping up to help evaluators planning CRTs find accurate estimates of the ICC. Two widely used in education include:
1. The Online Variance Almanac: http://stateva.ci.northwestern.edu/
2. The Optimal Design Plus Software: http://wtgrantfoundation.org/resource/optimal-design-with-empirical-information-od*
*Note that Optimal Design Plus is a free program that calculates power for CRTs. Embedded within the program is a data repository with ICC estimates.
In the event that empirical estimates are not available for your particular outcome type a search of the relevant literature may uncover estimates or a pilot study may be used to generate reasonable values. Regardless of the source, accurate estimates of the ICC are critical in determining the number of clusters needed in a CRT.
Jessaca Spybrook
Associate Professor, Education, Leadership, Research, and Technology, WMU
Dr. Jessaca Spybrook is an associate professor of educational leadership, research, and technology at Western Michigan University, specializing in evaluation, measurement and research. She earned her Ph.D. in education from the University of Michigan, where she also received an M.A. in applied statistics and B.A. in elementary education. She teaches, among other courses, Data Analytics I and II and hierarchical linear modeling. Her research focuses on improving the quality of the designs and power analyses of group randomized trials in education. She coauthored the software and documentation for Optimal Design Plus, a program that assists researchers in planning adequately powered experiments. She has published in journals such as Educational Evaluation and Policy Analysis, Journal for Research on Educational Effectiveness, and Evaluation Review. Spybrook's work has been funded by the National Science Foundation, the Institute of Education Sciences, and the William T. Grant Foundation. She received a National Academy of Education/Spencer Postdoctoral Fellowship in 2010. Prior to attending graduate school she was a seventh grade math teacher.
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Class 7
MATHS
Comparing Quantities
# Amina buys a book for Rs 275 and sells it at a loss of 15%. How much does she sellit for?
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 10-7-2020
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Text Solution
Solution :
Amina buys book for = rs 275 <br> sells it at loss of = 15 % <br> selling price = CP - loss <br> SP = CP - loss <br> = 275 - (15 % of 275) <br> = 275 - (15 xx 275)/100 <br> = 275 -165/4 <br> = 275 - 41.25 = 233.75 rs <br> answer
Image Solution
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## Latest Questions
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### Geometry chapter 2 resource book lesson 2 1 practice a
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An obtuse angle is an angle that measures more than 908 and less than 1808. The car runs when there is gas in the tank. Chapter 4continued quiz 1 ( geometry chapter 2 resource book lesson 2 1 practice a p. Obtuse isosceles triangle 2. Acute scalene triangle 3. Right scalene triangle 4. Mama1 65 lesson 4.
Practice a continued for use with the lesson “ reason using properties from algebra”. Geometry chapter resource book 2- 63 lesson 2. Answer key practice c geometry chapter 2 resource book lesson 2 1 practice a 1. Slope of slope and slope slope so abcd is by definition 10. And so abcd is a since both. Geometry chapter 4 resource book. 2 date m 1200 practice a continued for use with pages 225— find the value of y. Practice a geometry chapter 2 resource book lesson 2 1 practice a for use with the lesson “ use inductive reasoning” sketch the next figure in the pattern. Geometry chapter resource book a13 2. Can you find your fundamental truth using slader as a completely free mcdougal littell geometry chapter 2 resource book lesson 2 1 practice a geometry practice workbook solutions manual? Now is the time to redefine your true self using slader’ s free mcdougal littell geometry practice workbook answers.
Practice geometry chapter 2 resource book lesson 2 1 practice a b for use with the lesson “ use postulates and diagrams” draw a sketch to illustrate each postulate. Geometry a20 chapter resource book 2. Geometry 11- 22 chapter resource book lesson 11. 2 practice b for use with pages 729– 736 find the area of the trapezoid. 6 find the area of the rhombus or geometry chapter 2 resource book lesson 2 1 practice a kite.
16 7 use the given information to fi nd the value of x. Area 5 196 ft2 12.
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Essential elements violin book 1 with cd
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### Stata Data Analysis Examples Multiple Regression Power Analysis
#### Introduction
Power analysis is the name given to the process for determining the sample size for a research study. The technical definition of power is that it is the probability of detecting a "true" effect when it exists. Many students think that there is a simple formula for determining sample size for every research situation. However, the reality is that there are many research situations that are so complex that they almost defy rational power analysis. In most cases, power analysis involves a number of simplifying assumptions, in order to make the problem tractable, and running the analyses numerous times with different variations to cover all of the contingencies.
In this unit we will try to illustrate how to do a power analysis for multiple regression model that has two control variables, one continuous research variable and one categorical research variable (three levels).
#### Description of the Experiment
A school district is designing a multiple regression study looking at the effect of gender, family income, mother's education and language spoken in the home on the English language proficiency scores of Latino high school students. The variables gender and family income are control variables and not of primary research interest. Mother's education is a continuous research variable that measures the number of years that the mother attended school. The range of this variable is expected to be from 4 to 20. The variable language spoken in the home is a categorical research variable with three levels: 1) Spanish only, 2) both Spanish and English, and 3) English only. Since there are three levels, it will take two dummy variables to code language spoken in the home.
The full regression model will look something like this,
engprof = b0 + b1(gender) + b2(income) + b3(momeduc) + b4(homelang1) + b5(homelang2)
Thus, the primary research hypotheses are the test of b3 and the joint test of b4 and b5. These tests are equivalent the testing the change in R2 when momeduc (or homelang1 & homelang2) are added last to the regression equation.
#### The Power Analysis
We will make use of the Stata program powerreg (findit powerreg) (see How can I use the findit command to search for programs and get additional help? for more information about using findit) to do the power analysis. To begin with, we believe, from previous research, that the R2 for the full-model (r2f) with five predictor variables (2 control, 1 continuous research, and 2 dummy variables for the categorical variable) will be will be about 0.48.
Let's start with the continuous predictor (momeduc). We think that it will add about 0.03 to the R2 when it is added last to the model. This means that the R2 for the model without the variable (the reduced model, r2r) would be about 0.45. The total number of variables (nvar) is 5 and the number being tested (ntest) is one. We will run powerreg three times with power equal to .7, .8 and .9.
powerreg, r2f(.48) r2r(.45) nvar(5) ntest(1) power(.7)
Linear regression power analysis
alpha=.05 nvar=5 ntest=1
R2-full=.48 R2-reduced=.45 R2-change=0.0300
nominal actual
power power n
0.7000 0.6959 108
powerreg, r2f(.48) r2r(.45) nvar(5) ntest(1) power(.8)
Linear regression power analysis
alpha=.05 nvar=5 ntest=1
R2-full=.48 R2-reduced=.45 R2-change=0.0300
nominal actual
power power n
0.8000 0.7998 138
powerreg, r2f(.48) r2r(.45) nvar(5) ntest(1) power(.9)
Linear regression power analysis
alpha=.05 nvar=5 ntest=1
R2-full=.48 R2-reduced=.45 R2-change=0.0300
nominal actual
power power n
0.9000 0.8966 182
This gives us a range of sample sizes ranging from 108 to 182 depending on power.
Let's see how this compares with the categorical predictor (homelang1 & homelang2) which uses two dummy variables in the model. We believe that the change in R2 attributed to the two dummy variables will be about 0.025. This would give an r2r of 0.455. The nvar stays at 5 while the ntest is now 2.
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) power(.7)
Linear regression power analysis
alpha=.05 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.7000 0.7021 164
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) power(.8)
Linear regression power analysis
alpha=.05 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.8000 0.7990 203
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) power(.9)
Linear regression power analysis
alpha=.05 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.9000 0.8997 266
This series of power analyses yielded sample sizes ranging from 164 to 266. These sample sizes are larger than those for the continuous research variable.
If it is the case that both of these research variables are important, we might want to take into that we are testing two separate hypotheses (one for the continuous and one for the categorical) by adjusting the alpha level. The simplest but most draconian method would be to use a Bonferroni adjustment by dividing the nominal alpha level, 0.05, by the number of hypotheses, 2, yielding an alpha of 0.025. We will rerun the categorical variable power analysis using the new adjusted alpha level.
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.025) power(.7)
Linear regression power analysis
alpha=.025 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.7000 0.7000 199
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.025) power(.8)
Linear regression power analysis
alpha=.025 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.8000 0.8042 245
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.025) power(.9)
Linear regression power analysis
alpha=.025 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.9000 0.8971 308
The Bonferroni adjustment assumes that the tests of the two hypotheses are independent which is, in fact, not the case. The squared correlation between the two sets of predictors is about .2 which is equivalent to a correlation of approximately .45. Using an internet applet to compute a Bonferroni adjusted alpha taking into account the correlation gives us an adjusted alpha value of 0.034 to use in the power analysis.
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.034) power(.7)
Linear regression power analysis
alpha=.034 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.7000 0.6966 182
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.034) power(.8)
Linear regression power analysis
alpha=.034 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.8000 0.8030 227
powerreg, r2f(.48) r2r(.455) nvar(5) ntest(2) alpha(.034) power(.9)
Linear regression power analysis
alpha=.034 nvar=5 ntest=2
R2-full=.48 R2-reduced=.455 R2-change=0.0250
nominal actual
power power n
0.9000 0.9031 294
Based on the series of power analyses the school district has decided to collect data on a sample of about 225 students. This sample size should yield a power of around 0.8 in testing hypotheses concerning both the continuous research (momeduc) variable and the categorical research variable language spoken in the home (homelang1 and homelang2). The nominal alpha level is 0.05 but has been adjusted to .034 to take into account the number of hypotheses tested and the correlation between the predictors.
• Data Analysis Examples
• References
• Cohen, J. 1988. Statistical Power Analysis for the Behavioral Sciences, Second Edition. Mahwah, NJ: Lawrence Erlbaum Associates.
The content of this web site should not be construed as an endorsement of any particular web site, book, or software product by the University of California.
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Question
# The distance of a galaxy from the earth is $$5.6 \times 10^{25}m$$. Assuming the speed of light to be $$3 \times 10^8 ms^{-1}$$ find the time taken by light to travel this distance.$$[Hint: Time taken$$= $$\frac{distance travelled}{speed}$$]
Solution
## Given:distance = $$5.6 \times 10^{25}m$$speed = $$3\times 10^{8} ms^{-1}$$Time = ?Time taken = $$\frac{distance travelled}{speed}$$$$= \frac{5.6 \times 10^{25}}{3 \times 10^{8}}$$$$= 1.867 \times 10^{17}s$$Physics
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Cody
# Problems
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4
302
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2
281
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280
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256
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1
249
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3
222
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214
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6
190
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174
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162
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3
161
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1
137
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1
122
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114
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0
112
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109
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106
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101
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87
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Tags easy, basic, matlab
3
86
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2
85
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2
85
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Tags string
0
81
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80
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79
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72
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Tags string
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#### Problem 43660. Use R2016b Text Manipulations to Fix These Addresses (Part 1)
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0
63
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0
63
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Created by: Sourav Mondal
1
63
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0
60
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1
57
#### Problem 563. How to add?
Created by: AMITAVA BISWAS
4
54
#### Problem 43661. Use R2016b Text Manipulations to Fix These Addresses (Part 2)
Created by: Dave Bergstein
0
54
Created by: goc3
0
54
#### Problem 43662. Use R2016b Text Manipulations to Fix These Addresses (Part 3)
Created by: Dave Bergstein
0
51
#### Problem 489. Bang Bang in Bangalore
Created by: AMITAVA BISWAS
Tags string, edit, find
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47
#### Problem 723. QWERTY Shift Encoder
Created by: Richard Zapor
Tags lexi, code, string
0
46
Created by: goc3
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45
#### Problem 43009. Back to Basics - New Data Type in R2016b - convert a char to a string
Created by: Gareth Thomas
0
44
Created by: Siva
0
44
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Created by: Jamil Kasan
9
43
Created by: goc3
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42
#### Problem 44444. Problem 44444 !!! free beer everyone
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42
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#### Problem 1682. Make a list string
Created by: Chris
Tags string, list
0
39
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https://www.teacherspayteachers.com/Product/3NF3Fractions-Part-1-3rd-Grade-Common-Core-Math-Worksheets-3rd-9-Weeks-399755
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# (3.NF.3)Fractions Part 1 3rd Grade Common Core Math Worksheets 3rd 9 Weeks
Common Core Standards
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You are purchasing Common Core math practice sheets aligned to assessment tasks. There is a practice sheet for every day of the week and a test for Friday.
The attachment consists of 10 worksheets designed to reinforce math skills. There are two sheets for each day that can be copied front and back to save paper.
The front sheet focuses on the weekly target skill and the back side of the sheet is a review of previously taught skills. This particular set of sheets focuses on:
3.FT.3 a Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
3.FT.3b Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3). Explain why the fractions are equivalent, e.g., by using a visual fraction model.
3.FT.3 c Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram.
3.FT.3 d Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
The back side of the paper is a REVIEW of the following skills:
3.NF.2 Fractions and number lines
3.OA.4 multiplication and division
The last two sheets in the set are designed to be an assessment and are aligned to the other practice sheets in the set.
Other Related Products:
3rd Common Core Math Worksheets - 1st 9 weeks
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3rd Common Core Math Worksheets - 3rd 9 weeks
3rd Common Core Math Worksheets - 4th 9 weeks
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https://www.physicsforums.com/threads/bungee-jumping-question.747724/
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Homework Help: Bungee Jumping Question
1. Apr 8, 2014
patrickmoloney
1. The problem statement, all variables and given/known data
A bungee jumper of mass m drops off a bridge and falls vertically downwards.
The bungee cord is elastic with natural length L and stiffness k. Deduce that
at the lowest point of the fall, the cord is stretched by an amount $$x = \frac {mg}{k} \Big( 1 + \sqrt {1+ \frac {2kL}{mg}} \Big)$$
2. Relevant equations
$mgh = \frac{1}{2} k x^2$
3. The attempt at a solution
Energy considerations dictate that the gravitational potential energy of the jumper in the initial state is equal to the elastic potential of the cord in the final state
$mg ( L + x ) = \frac{1}{2} k x^2$
$mgL + mgx = \frac{1}{2} k x^2$
$2mgL + 2mgx = k x^2$
$x^2 - \Big( \frac{2mg}{k} \Big )x - \Big( \frac{2mgL}{k} \Big) = 0$
$x = \frac{- \Big(- \frac{2mg}{k} \Big) + \sqrt{ \Big( - \frac{2mg}{k} \Big)^2 - 4 \Big( - \frac{2mgL}{k} \Big )}}{2}$
$x = \frac{1}{2} \Big( \frac{2mg}{k} + \sqrt{ \frac{4m^2g^2}{k^2} + \frac{8mgL}{k}} \Big)$
$x = \frac{mg}{k} + \frac{1}{2} \Big ( \sqrt{ \frac{4m^2g^2}{k^2} \Big ( 1 + \frac{2kL}{mg} \Big )} \Big )$
$x = \frac{mg}{k} + \frac{1}{2} \Big( \sqrt{\frac{4m^2g^2}{k^2}} \sqrt{1+\frac{2kL}{mg}} \Big )$
$x = \frac{mg}{k} + \frac{1}{2} \Big( \frac{2mg}{k} \Big) \sqrt{1 + \frac{2kL}{mg}}$
$x = \frac{mg}{k} + \frac{mg}{k} \sqrt{1 + \frac{2kL}{mg}}$
$x = \frac{mg}{k} \Big( 1 + \sqrt{1+ \frac{2kL}{mg}} \Big)$
Is this the correct method? There is no solution online to this problem.
2. Apr 8, 2014
CWatters
Seems reasonable to me.
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https://www.physicsforums.com/threads/how-do-you-calculate-the-ph-of-a-sodium-ethanoate-solution.75972/
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# How Do You Calculate the pH of a Sodium Ethanoate Solution?
• josephcollins
In summary, the person is asking for help with calculating the pH of a solution of sodium ethanoate, given the pKa of ethanoic acid/ethanoate ions. They explain their method of using the Ka to find the concentration of H+ ions, but there may be more unknown concentrations in the system. They also request clarification on whether they should be using ethanoic acid or ethanoate.
josephcollins
Hi ppl. I have a question. I'm given the pka of ethanoic acid/ethanoate ions at 4.75 and i have to work out the ph of a 5*10^-2 moldm3 solution of sodium ethanoate. I worked it out by getting the Ka from 10^-4.75 and then saying that the concentration of Ch3COO- in the dissociation of ethanoic acid is also 5*10^-2 moldm3. Then i used the equation for the Ka to get the corresponding concentration of H+ ions in this dissociation and hence the ph, is this method of reasoning correct? Thanks, Joe
The question itself seems to indicate an initial concentration of ethanoate, however, your discussion implies ethanoic acid? Please be more clear...
## 1. What is the formula for calculating the pH of CH3COONa?
The formula for calculating the pH of CH3COONa is pH = -log[H+], where [H+] is the concentration of hydrogen ions in the solution.
## 2. How do I determine the concentration of hydrogen ions in a solution of CH3COONa?
The concentration of hydrogen ions can be determined using the dissociation constant (Ka) of CH3COONa, which is 1.8 x 10^-5 at 25°C. The formula is [H+] = √(Ka*[CH3COONa]).
## 3. Can I use a pH meter to measure the pH of CH3COONa?
Yes, a pH meter can be used to measure the pH of CH3COONa. However, it is important to calibrate the pH meter using standard solutions of known pH values before use.
## 4. How does the pH of CH3COONa change when diluted with water?
When CH3COONa is diluted with water, the concentration of hydrogen ions decreases, leading to an increase in pH. This is because the dissociated ions of CH3COONa can interact with water molecules, reducing the number of free hydrogen ions in the solution.
## 5. What is the range of pH values for a solution of CH3COONa?
The pH range for a solution of CH3COONa is from about 8-9, making it slightly basic. This is due to the presence of the acetate ion, which is a weak base and can act as a buffer to maintain a constant pH in the solution.
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Cody
# Problem 1239. Decoding : Find the value
Solution 201882
Submitted on 7 Feb 2013 by Freddy
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
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%% Test 1 u = 3; v= 0.05; y_correct = 1020; assert(isequal(your_fcn_name(u,v),y_correct));
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%% Test 2 u=30; v = 0.01; y_correct = [5100 4800]; assert(isequal(your_fcn_name(u,v),y_correct));
3 Pass
%% Test 3 u=3; v = 0.03; y_correct = 1700; assert(isequal(your_fcn_name(u,v),y_correct));
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https://www.thefreelibrary.com/Bifurcation+study+of+thin+plate+with+an+all-over+breathing+crack-a0509160865
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# Bifurcation study of thin plate with an all-over breathing crack.
1. Introduction
The plate is widely used in engineering practices such as the aircraft structure and body, concrete floor slabs, and marine industry, to name a few. The existence of a crack in a plate aflects its stiffness, mass, and damping properties and then changes its vibration characteristics. For a vibration plate, a crack not only effects the free vibration of plate, including natural frequency and mode shape, but also changes the forced vibration, especially the response of nonlinear dynamics. The decrease of stiffness increases the vibration amplitude of the cracked plate with large deflection. Furthermore, the crack alternately opens and closes during vibration cycle, which forms the cracked structure with nonsmooth characteristics. Then, complex nonlinear dynamics behaviors will appear, which can seriously affect the safety of structures.
An earlier extensive literature review on the vibration of cracked structures could be found in the paper of Dimarogonas [1] in 1996. Rice and Levy [2] developed a line-spring model for approximate solution of a plate containing a part-through crack in 1972. Numerical results for natural frequencies of both isotropic and orthotropic plates with internal cracks were presented using subdomain method by Lee et al. [3]. Khadem and Rezaee [4] developed an analytical approach for crack detection in rectangular thin plates with an all-over part-through crack and subjected to uniform external loads using vibration analysis. Khadem and Rezaee [5] employed the modified comparison functions to obtain the natural frequencies of a simply supported rectangular cracked plate using the Rayleigh-Ritz method. Based on Mindlin plate theory (MPT), Hosseini-Hashemi et al. [6] proposed a set of exact closed-form characteristic equations to analyze free vibration problem of moderately thick rectangular plates with an arbitrary number of all-over part-through cracks.
For the nonlinear dynamic investigation of crack structures, many researchers considered cracks as open-crack models. In 2005, Fu et al. [7] derived the nonlinear equations of motion for the moderate thickness rectangular plates with transverse surface penetrating crack on the two-parameter foundation and analyzed the nonlinear vibration behaviors of the plate with different crack parameters. In 2009, Israr and Atepor [8] investigated nonlinear vibration of an isotropic cracked plate subjected to transverse harmonic excitation, both analytically and experimentally. In the same year, Yang et al. [9] established nonlinear governing equations of motion for the simply supported functionally graded rectangular plate with a through-width surface crack to investigate the influences of material property gradient, crack depth, crack location, and plate thickness ratio on the vibration frequencies and transient response of the surface-racked FGM plate. Saito et al. [10] investigated veering phenomena in the nonlinear vibration frequencies of a cantilevered cracked plate and proposed an efficient method for estimating these frequencies. In 2012, Ismail and Cartmell [11] presented the forced vibration analysis of a plate with an inclined surface crack based on the classical plate theory. They formulated the inclined surface crack using a simplified line-spring model and utilized the multiple scale perturbation method to derive the amplitude-frequency equation of the cracked plate. In 2013, Bose and Mohanty [12] studied the vibration of a rectangular thin isotropic plate with a part-through surface crack of arbitrary orientation and position by using the Kirchhoff plate theory, and the problem of [11] was reconsidered and rigorously modified. AsadiGorgi et al. [13] investigated the flutter, limit cycle oscillation, and different nonlinear oscillations of rectangular panels with an all-over part-through crack in 2015. They performed these analyses for panels according to the first-order shear deformation or Mindlin plate theory and accurately studied the effects of the depths and locations of cracks on various aeroelastic characteristics of moderately thick rectangular panels.
In addition to the open crack, the surface crack breathing is a more practical and common situation during the vibration of the cracked structures [14]. The crack will stay open during one part of the vibration cycle and closed during the rest of the cycle which is usually called breathing.
In 2000, Pugno et al. [15] presented a technique capable of evaluating the dynamic response of a beam with several breathing cracks, which was based on the assumption of periodic response and the fact that cracks open and close continuously. The vibrational response to harmonic force of a cantilever beam with cracks of different sizes and locations was analyzed using this "harmonic balance" approach. In 2010, Caddemi et al. [16] investigated nonlinear dynamic response of beam with some switching cracks. They regarded the overall behavior of a beam with switching cracks as a sequence of linear phases, each of them characterised by different number and positions of the cracks in open state, and investigated the behavior under different boundary conditions for both harmonic loading and free vibrations. In 2016, Dotti et al. [17] studied the nonlinear dynamic response to simple harmonic excitation of a thin-walled beam with a breathing crack by employing a refined one-dimensional model.
The mentioned researches show that breathing crack model could reveal the real nonlinear dynamic behaviors of surface crack structures. Most of the researches on the crack of breathing are focused on the beam, and the nonlinear dynamics study of plate with surface crack employed the open-crack model.
The aim of this paper is to study the bifurcations of thin plate with an all-over breathing crack. The piecewise model is used to describe the opening and closing of breathing crack during the vibration: the crack is taken as open crack when crack opens; otherwise, the cracked plate is regarded as intact plate when crack closes. Based on the von Karman large deflection theory, Hamilton's principle is used to establish the nonlinear governing equations of motion for the cracked plate. The mode shape functions are derived using the geometric boundary conditions and the boundary conditions along the crack line of thin plate. The effect of location and depth of crack on mode shapes and free vibration frequencies of cracked plate are analyzed, which is assessed with FEM results. The stress functions of cracked and intact plate with large deflection are obtained by the equations of compatibility. The partial differential equation is discretized via the Galerkin method. The Runge-Kutta method is utilized to investigate the bifurcations and chaotic motions of cracked plate. Nonlinear dynamic behaviors of cracked plate are studied to analyze the effect of external excitation amplitudes and frequencies on plate with different crack parameters which consist of crack depth and crack location.
2. Modal Analysis
In this section, the mode shape functions of cracked thin plate are obtained by using the simply supported boundary conditions and internal boundary conditions along the crack.
The simply supported rectangular plate with an all-over breathing crack can be illustrated by Figure 1. The thickness of the plate is H. The dimensions in x and y direction are a and b, respectively. Assume that ([u.sub.0], [v.sub.0], [w.sub.0]) represent the displacements of a point in a mid-plane. The displacements of arbitrary point in the cracked plate are obtained as
u(x, y, z, t) = [u.sub.0] (x, y, t) - z [[partial derivative][w.sub.0]/[partial derivative]x], (1a)
v(x, y, z, t) = [v.sub.0] (x, y, t) - z [[w.sub.0]/[partial derivative]y], (1b)
w(x, y, z, t) = [w.sub.0] (x, y, t). (1c)
Using the Hamilton theory ([[integral].sup.T.sub.0] ([delta]U + [delta]V - [delta]K)dt = 0) and the von Karman large deflection theory, the nonlinear dynamic equations of the thin plate with large deflection are gained [18]:
[[partial derivative][N.sub.x]/[partial derivative]x] + [[partial derivative][N.sub.xy]/[partial derivative]y] = [I.sub.0][[[partial derivative].sup.2][u.sub.0]/[partial derivative][t.sup.2], (2a)
[[partial derivative][N.sub.xy]/[partial derivative]x] + [[partial derivative][N.sub.y]/[partial derivative]y] = [I.sub.0][[[partial derivative].sup.2][v.sub.0]/[partial derivative][t.sup.2], (2b)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (2c)
where [I.sub.0] = [rho]H. The quantities ([N.sub.x], [N.sub.y], [N.sub.xy]) are the in-plane force resultants, and ([M.sub.x], [M.sub.y], [M.sub.xy]) are the moment resultants. P is external uniform load in the z direction. The expression N([w.sub.0]) is as follows:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (2d)
For the convenience of solving dynamic equations, a stress function [phi](x, y, t) is introduced to eliminate the in-plane displacements ([u.sub.0], [v.sub.0]), and its relations with internal forces are obtained as
[N.sub.x] = H[[[partial derivative].sup.2][phi]/[partial derivative][y.sup.2], (3a)
[N.sub.y] = H[[[partial derivative].sup.2][phi]/[partial derivative][x.sup.2], (3b)
[N.sub.xy] = -[[[partial derivative].sup.2][phi]/[partial derivative]x[partial derivative]y. (3c)
The compatibility equation in the form of stress function and transverse displacement is received:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (4)
Substituting (3a), (3b), and (3c) into the nonlinear dynamic equations (2a), (2b), (2c), and (2d), (2a) and (2b) will automatically meet balance conditions after ignoring the inertia terms. The dynamic equation of the stress function and deflection for (2c) is obtained as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (5)
The simply supported boundary conditions will be
x = 0 or a: [u.sub.0] = 0, w = 0, [M.sub.x] = 0, [N.sub.xy] = 0 (6a)
y = 0 or b: [v.sub.0] = 0, w = 0, [M.sub.y] = 0, [N.sub.xy] = 0. (6b)
In order to enable the boundary conditions to be expressed in terms of the stress function, [u.sub.0] = 0 and [v.sub.0] = 0 are substituted by [[DELTA].sub.u] = 0 and [[DELTA].sub.v] = 0, where [19]
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (7a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (7b)
As to the thin plate with an all-over crack, the authors divide the thin plate into two parts, plate I and plate II, at the location y = [y.sub.0], as shown in Figure 1.
Breathing crack is continuously open-closed during vibration. When the crack is open, at the crack location, the continuity and discontinuity conditions can be written along the crack line as follows:
[w.sup.I] = [w.sup.II], (8a)
[u.sup.I.sub.0] = [u.sup.II.sub.0], (8b)
[v.sup.I.sub.0] = [v.sup.II.sub.0], (8c)
[M.sup.I.sub.y] = [M.sup.II.sub.y], (8d)
[V.sup.I.sub.y] = [V.sup.II.sub.y], (8e)
[N.sup.I.sub.y] = [N.sup.II.sub.y], (8f)
[N.sup.I.sub.Xy] = [N.sup.II.sub.Xy], (8g)
[[partial derivative][w.sup.I]/[partial derivative]y] - [[partial derivative][w.sup.II]/[partial derivative]y] = [theta], (8h)
where (8a)~(8g) represent displacement (w, [u.sub.0], [v.sub.0]) and internal forces ([M.sub.y], [V.sub.y], [N.sub.y],[N.sub.xy]) are equal on both sides of the crack, respectively. Equation (8h) represents the discontinuity condition of rotation angle along the crack. A line-spring model is used to describe the elastic behavior of the crack. d is the rotation due to the presence of the crack and can be expressed as
[theta] = [12(1 - [v.sup.2])/E] x [[sigma].sub.b] x [[alpha].sub.bb], (9)
where [[sigma].sub.b] is the nominal stress at a point far from the crack and [[alpha].sub.bb] is the dimensionless bending compliance factor. [[sigma].sub.b] and [[alpha].sub.bb] are given by
[[sigma].sub.b] = 6[[M.sub.y]/[H.sup.2]], (10a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (10b)
where [g.sub.b] are dimensionless compliance coefficients, depending on the crack depth to thickness ratio [xi] = [h.sub.0]/H, and defined by [2]
[g.sub.b] = [square root of ([xi])](1.99 - 2.47[xi] + 12.97[[xi].sup.2] - 23.11[[xi].sup.3] + 24.80[[xi].sup.4]) .
The mode shape functions for w of cracked thin plate are obtained by using the simply supported boundary conditions and the boundary conditions along the crack line. The dimension in the y direction is longer than that in the x direction, which means it is more possible that there will be higher modes in the y direction than in x direction. Based on Levy method, the solutions for the differential equations of the plate are assumed as
w(x, y, t) = [N[summation over (n=1)][Y.sub.n](y) x sin ([pi]x/a) x [e.sup.i[omega]t]. (12)
[Y.sub.n](y) can be written as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (13)
where [[beta].sub.n] = (b/a)[square root of ([[lambda].sup.2.sub.n] + [([pi]).sup.2])], [[gamma].sub.n] = (b/a) [square root of ([[lambda].sup.2.sub.n] + [([pi]).sup.2])], [[lambda].sup.4.sub.n] = [a.sup.4][[omega].sup.2.sub.n][rho]H/D, and [A.sub.n], [B.sub.n], [C.sub.n], and [D.sub.n] are undetermined coefficients.
According to boundary conditions (6a) and (6b), we can obtain [A.sub.n] = [D.sub.n] = 0. Then, the modal functions of transversal displacement w in the y direction for plate I and plate II are obtained:
[Y.sup.I.sub.n](y) = [B.sup.I.sub.n] sinh ([[beta].sup.n]y/b) + [C.sup.I.sub.n] sin([[gamma].sub.n]y/b), (0 [less than or equal to] y [less than or equal to] [y.sub.0]), (14a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (14b)
Substituting (14a) and (14b) into inner boundary condition equations (8a), (8d), (8e), and (8h), Homogeneous algebraic equations of the undetermined coefficients ([B.sup.I.sub.n], [B.sup.II.sub.n], and [C.sup.II.sub.n]) are derived, which results in an eigenvalue problem with the eigenvalues related to the natural frequencies of plate. Then, substituting the result of each order frequency to Homogeneous algebraic equations, the undetermined coefficients can be obtained by solving these equations and given in terms of expression relevant to [C.sup.I.sub.n], respectively.
[B.sup.I.sub.n] = [e.sup.1.sub.n][C.sup.I.sub.n], (15a)
[C.sup.II.sub.n] = [e.sup.2.sub.n][C.sup.I.sub.n], (15b)
[B.sup.II.sub.n] = [e.sup.3.sub.n][B.sup.I.sub.n] = [e.sup.3.sub.n][e.sup.1.sub.n][C.sup.I.sub.n], (15c)
where parameters [e.sup.1.sub.n], [e.sup.2.sub.n], and [e.sup.3.sub.n] can be expressed as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (16a)
[e.sup.2.sub.n] = sin([[gamma].sub.n][y.sub.0]/b)/sin[[[gamma].sub.n]([y.sub.0] - b)/b], (16b)
[e.sup.3.sub.n] = sinn([[beta].sub.n][y.sub.0]/b)/sinh[[[beta].sub.n]([y.sub.0] - b)/b]. (16c)
Substituting (15a), (15b), and (15c) into (14a) and (14b), we obtained the modal function of plate I and plate II in the y direction.
[Y.sup.I.sub.n](y) = [a.sup.I.sub.n] sinh([[beta].sub.n]y/b) + sin ([[gamma].sub.n]y/b), (17a)
[Y.sup.II.sub.n](y) = [a.sup.II.sub.n] sinh [[[beta].sub.n](y - b)/b] + sin [[gamma](y - b)/b], (17b)
where [a.sup.1.sub.n] = [e.sup.1.sub.n] and [a.sup.II.sub.n] = [e.sup.3.sub.n][e.sup.1.sub.n]/[e.sup.2.sub.n].
The first three-order modes and natural frequencies of the cracked plate can be obtained by the method mentioned above, which is verified by finite element method. We chose parameters of material and size as a = 1 m, b = 2 m, [mu] = 0.33 and H = 0.008 m E = 2.1 x [10.sup.5] MPa, and p = 7860 kg/[m.sup.3].
Tables 1 and 2 show the first three-order frequencies and modes for plates with cracks at different location (p = [y.sub.0]/b), respectively. They show that the crack almost has no influence on natural frequencies and modes when the crack coincides with a line of node, while the crack has more effect at the maximum defection position. For example, the crack at [y.sub.0]/b = 0.5 does not affect the second frequency and mode, because this location is the node line of second mode; however, it has more influence on the first- and the third-order frequency and mode. When the crack moves to the position ([y.sub.0]/b = 0.25), the effects on the second mode are more evident than that on the first and third mode.
Theoretical results and simulation data are in good agreement, and the accuracy of the theoretical method is verified in this section. Comparing theoretical results to simulation data, the first three-order modal figures obtained by theoretical calculation are consistent with FEM simulations. Thus, the modal functions corresponding to the first three-order modal figures are used to construct the solutions of dynamic equations.
3. Stress Functions
The stress functions of cracked thin plate with large deflection are obtained by the equations of compatibility. Using the first three modes of mode shape function (17a) and (17b), the solutions of dynamic equations are assumed as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (18a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (18b)
where [w.sub.1](t), [w.sub.2](t), and [w.sub.3](t) are time-related variables. Introducing (18a) and (18b) into the compatibility equations (4), and using the boundary conditions (6a) and (6b), the stress function expression of thin plate with an all-over surface crack is gained
[[phi].sup.r](x, y, t) = [12.summation over (i=1)][[phi].sup.r.sub.i], (r = I, II), (19)
where the parameter [[phi].sup.r.sub.i] is given in Appendix.
During vibration, surface crack will be continuously open-closed. When the crack is open, the modes of crack thin plate are adopted and the stress function of crack thin plate considering large deflection is obtained as (19). When the crack is closed, the modes of intact plate are adopted. Using the mode functions of intact plate, the authors get the stress function expression of intact plate considering large deflection. As to intact plate, the first three modes are taken and the solutions of dynamic equations are
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (20)
Substituting expression (2[degrees]) into the compatibility equations (4), the stress function of intact plate with large deflection is obtained
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (21)
where the coefficients can be found in Appendix.
4. Discrete Equation
Having obtained the mode shape functions and stress function expression of crack thin plate and intact plate, the authors use Galerkin method to discrete the dynamic equation (5).
As to crack thin plate, the authors adopt the mode shape function and stress function of crack thin plate. Substituting expressions (18a) and (18b) and (19) into (5), the discrete dynamic equations of open-crack thin plate are obtained as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (22a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (22b)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (22c)
where a with subscript are coefficients determined by material, geometry, and crack parameters of the plate.
As to intact plate, the modal function and stress function of intact plate are adopted. Putting expressions (20) and (21) into (5), the discrete dynamic equations of intact plate are gained:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (23a)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (23b)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (23c)
where b with subscript are coefficients in relation to material and geometry of the plate.
The opening and closing of breathing crack are described as piecewise model during the vibration. The authors assume that the crack is on the top surface of the plate and the transversal displacement w is positive when w is below origin of coordinates. During vibration, the dynamic equations are discrete dynamic equations which are (22a)-(22c), when w [less than or equal to] 0 (crack is open); the dynamic equations are (23a)-(23c), when w >0 (crack is closed).
5. Effect of Crack Parameters on Nonlinear Dynamic Behaviors
The Runge-Kutta algorithm is utilized to numerically analyze the nonlinear dynamical behaviors of the cracked plate under different excitation amplitude and frequency. The nonlinear dynamical behaviors of the plate with different crack location and depth are studied by analyzing the bifurcation diagrams. To compare with the effect of breathing crack on the plate, the nonlinear dynamic responses of the plate with an open crack and the intact plate are analyzed too. In numerical simulation, we chose parameters of material and size as [rho] = 7860 kg/[m.sup.3], E = 2.1 x [10.sup.5] MPa, a = 1m, b = 2 m, [mu] = 0.33, and H = 0.008 m.
For intact plate, as shown in Figure 2(a), the excitation frequencies change from 30 rad/s to 90 rad/s, when the external excitation amplitude is P = 3000 N/[m.sup.2]. The excitation amplitude changes from 1000 N/[m.sup.2] to 4000 N/[m.sup.2], while the excitation frequency is w = 80 rad/s, such as Figure 2(b). Figures 2(a) and 2(b) represent the bifurcation diagram of intact plate via the different excitation frequencies and amplitudes on transversal displacement w, respectively. The coordinate and the variation of external excitation for the following bifurcation figures are the same as those in Figure 2.
5.1. Effect of Crack Depth on Nonlinear Dynamic Behavior. The nonlinear dynamic behaviors of cracked plate are studied to analyze the effect of external excitation amplitudes and frequencies on plate with different crack depth. The rectangular plates have a crack at % = 0.5, where % = [y.sub.0]/b. We chose the crack depths as [h.sub.0]/H = 0.2,0.4, and 0.6, respectively.
Figures 3, 4, and 5 are the bifurcation diagrams for plates with breathing cracks at different crack depth. Figures 6, 7, and 8 demonstrate bifurcation diagrams for open-crack plates at various crack depth.
The above bifurcation diagrams show that breathing crack plate has complex nonlinear phenomenon: quasi-periodic motion, bifurcation motion, and chaotic motion. With the change of crack depth, the bifurcation diagrams of breathing crack plates represent different nonlinear dynamic behaviors. However, the motion of open-crack plates is relatively simple, which merely demonstrates the single-cycle movement and the movement of doubling the cycle.
As for bifurcation diagram 4, the corresponding waveform and phase diagrams are gained when the external excitation amplitudes are 2500, 3100, 3300, 3400, 3700, and 3800 N/[m.sup.2], respectively, as shown in Figure 9. Figures 9(a)-9(d) show the periodic motion of cycle 1 ,2, 5, and 7 times the cycle of movement, while Figures 9(e) and 9(f) represent the quasi-periodic and chaotic motions.
For one cycle motion, as shown in Figure 9(a), the authors find that the transversal displacement amplitudes in positive and negative directions are, respectively, 4.3 mm and 5 mm. It represents that when w > 0, the crack is closed and the stiffness of plate is relatively large, so the amplitude is 4.3 mm; when w [less than or equal to] 0, the crack is open and the stiffness decreases, so the amplitude increases to 5 mm.
5.2. Effect of Crack Location on Nonlinear Dynamic Behaviors. In this section, the effect of crack location on nonlinear dynamic behavior is studied to analyze the bifurcation diagrams of plates subject to external excitation. The simply supported rectangular plates have a crack at depth ([h.sub.0]/H = 0.65) and various locations ([[eta].sub.0] = 0.15, 0.30, and 0.45).
Figures 10, 11, and 12 are the bifurcation diagrams of plates with breathing cracks at different crack location. Figures 13, 14, and 15 demonstrate bifurcation diagrams for open-crack plates at various crack locations.
6. Conclusions
The piecewise model is used to describe the opening and closing of crack during the vibration. The dynamic equations are solved by using the stress function and the mode functions derived by this paper. Bifurcation diagrams, wave-shape diagrams, and phase diagrams of simply supported rectangular thin plate with an all-over breathing crack are investigated in this paper. The following conclusions have been obtained:
(1) It is complex to derive the mode functions of in-plane displacements ([u.sub.0], [v.sub.0]) for large deflection cracked plate. This paper introduces the stress functions obtained by compatibility equation of cracked plate to eliminate the in-plane displacement ([u.sub.0], [v.sub.0]), which provide a new solution for the problem of large deflection vibration in cracked plates.
(2) The previous study of the nonlinear dynamics of surface crack plate usually uses open-crack model. However, the authors find that the nonlinear dynamic behaviors of open-crack model are similar to those of the intact plate, which does not reveal the effect of crack on nonlinear dynamic behaviors of plate. Breathing crack model reflects the real process of opening-closing of surface plate and presents more complex phenomena. Through three sets of bifurcation diagrams, we find that breathing crack plate has complex nonlinear phenomena such as quasi-periodic motion, bifurcation motion, and chaotic motion, and the nonlinear dynamic behaviors of open-crack model are similar to intact plate model, which merely presents the single-cycle motion and the double periodic motion.
(3) The deeper the crack is, the smaller the critical excitation frequency will be, which means that the increase of the crack depth will make the motion of crack plate more complicated. There are two reasons for this phenomenon. Firstly, the increase of the crack depth reduces the stiffness of plate, resulting in the increase of vibration amplitude for the cracked plate which leads to complex nonlinear dynamic behavior. Secondly, the crack alternately opens and closes during vibrational cycle and the increase of the crack depth forms the cracked structure with strong nonsmooth characteristics.
(4) When the crack is near the center of the plate, complex nonlinear dynamic behavior more easily occurs in the vibration of cracked plate subject to certain excitation amplitude and frequency. This is because the maximum transversal displacement is near the center of the plate, and the farther away the crack is from the locations of zero displacement (w = 0) which conclude nodal line and boundary, the greater the influence of the stiffness is.
Appendix
The coefficients given in (16a), (16b), and (16c) can be written as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (A.1)
where
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (A.2)
The coefficients given in (18a) and (18b) can be expressed as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (A.3)
http://dx.doi.org/10.1155/2016/1509384
Competing Interests
The authors declare that there is no conflict of interests regarding the publication of this article.
Acknowledgments
The authors gratefully acknowledge the National Natural Science Foundation of China (NNSFC) through Grant nos. 11172011 and 11472019.
References
[1] A. D. Dimarogonas, "Vibration of cracked structures: a state of the art review," Engineering Fracture Mechanics, vol. 55, no. 5, pp. 831-857, 1996.
[2] J. R. Rice and N. Levy, "The part-through surface crack in an elastic plate," Journal of Applied Mechanics, vol. 3, pp. 183-194, 1972.
[3] H. P. Lee, S. P. Lim, and S. T. Chow, "Prediction of natural frequencies of cracked isotropic and orthotropic composite plate," in Proceedings of the International Conference on Computational Engineering Mechanics, pp. 632-648, Beijing, China, 1987
[4] S. E. Khadem and M. Rezaee, "An analytical approach for obtaining the location and depth of an all-over part-through crack on externally in-plane loaded using vibration analysis," Journal of Sound and Vibration, vol. 230, pp. 291-308, 1999.
[5] S. E. Khadem and M. Rezaee, "Introduction of modified comparison functions for vibration analysis of a rectangular cracked plate," Journal of Sound and Vibration, vol. 236, no. 2, pp. 245-258, 2000.
[6] Sh. Hosseini-Hashemi, Gh. Heydar Roohi, and D. T. Hossein-Rokni, "Exact free vibration study of rectangular Mindlin plates with all-over part-through open cracks," Computers and Structures, vol. 88, no. 17-18, pp. 1015-1032, 2010.
[7] Y. Fu, Y. Xiao, and X. Zha, "Nonlinear vibration for moderate thickness rectangular cracked plates including coupled effect of elastic foundation," Applied Mathematics and Mechanics, vol. 8, no. 26, pp. 963-972, 2005.
[8] A. Israr and L. Atepor, "Investigation of the nonlinear dynamics of a partially cracked plate," in Proceedings of the 7th International Conference on Modern Practice in Stress and Vibration Analysis, vol. 181, 2009.
[9] J. Yang, Y. X. Hao, W. Zhang, and S. Kitipornchai, "Nonlinear dynamic response of a functionally graded plate with a through-width surface crack," Nonlinear Dynamics, vol. 59, no. 1-2, pp. 207-219, 2010.
[10] A. Saito, M. P. Castanier, and C. Pierre, "Estimation and veering analysis of nonlinear resonant frequencies of cracked plates," Journal of Sound and Vibration, vol. 326, no. 3-5, pp. 725-739, 2009.
[11] R. Ismail and M. P. Cartmell, "An investigation into the vibration analysis of a plate with a surface crack of variable angular orientation," Journal of Sound and Vibration, vol. 331, no. 12, pp. 2929-2948, 2012.
[12] T. Bose and A. R. Mohanty, "Vibration analysis of a rectangular thin isotropic plate with a part-through surface crack of arbitrary orientation and position," Journal of Sound and Vibration, vol. 332, no. 26, pp. 7123-7141, 2013.
[13] H. AsadiGorgi, M. Dardel, and M. H. Pashaei, "Effects of allover part-through cracks on the aeroelastic characteristics of rectangular panels," Applied Mathematical Modelling, vol. 39, no. 23-24, pp. 7513-7536, 2015.
[14] M.-H. H. Shen and Y. C. Chu, "Vibrations of beams with a fatigue crack," Computers and Structures, vol. 45, no. 1, pp. 79-93, 1992.
[15] N. Pugno, C. Surace, and R. Ruotolo, "Evaluation of the nonlinear dynamic response to harmonic excitation of a beam with several breathing cracks," Journal of Sound and Vibration, vol. 235, no. 5, pp. 749-762, 2000.
[16] S. Caddemi, I. Calio, and M. Marletta, "The non-linear dynamic response of the Euler-Bernoulli beam with an arbitrary number of switching cracks," International Journal of Non-Linear Mechanics, vol. 45, no. 7, pp. 714-726, 2010.
[17] F. E. Dotti, V. H. Cortinez, and F. Reguera, "Non-linear dynamic response to simple harmonic excitation of a thin-walled beam with a breathing crack," Applied Mathematical Modelling, vol. 40, no. 1, pp. 451-467, 2016.
[18] J. N. Reddy, Mechanics of Laminated Composite Plates and Shells: Theory and Analysis, CRC Press, Boca Raton, Fla, USA, 2nd edition, 2004.
[19] G. Z. Harris, "The buckling and post-buckling behaviour of composite plates under biaxial loading," International Journal of Mechanical Sciences, vol. 17, no. 3, pp. 187-202, 1975.
Lihua Chen, Jian Xue, Zhijie Zhang, and Wei Zhang
Beijing University of Technology, No. 100, Pingleyuan, Chaoyang District, Beijing, China
Correspondence should be addressed to Lihua Chen; chenlihua@bjut.edu.cn
Received 17 June 2016; Revised 17 September 2016; Accepted 16 October 2016
Caption: Figure 1: A simply supported rectangular thin plate with and all-over breathing crack.
Caption: Figure 2: The bifurcation diagram of intact plate for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 3: The bifurcation diagram of plate with a breathing crack ([h.sub.0]/H = 0.2) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 4: The bifurcation diagram of plate with a breathing crack ([h.sub.0]/H = 0.4) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 5: The bifurcation diagram of plate with a breathing crack ([h.sub.0]/H = 0.6) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 6: The bifurcation diagram of plate with an open crack ([h.sub.0]/H = 0.2) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 7: The bifurcation diagram of plate with an open crack ([h.sub.0]/H = 0.4) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 8: The bifurcation diagram of plate with an open crack ([h.sub.0]/H = 0.6) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 9: Waveforms and bifurcation of the plate with breathing crack; relative crack depth [h.sub.0]/H = 0.6; crack location [[eta].sub.0] = 0.5.
Caption: Figure 10: The bifurcation diagram of plate with a breathing crack ([[eta].sub.0] = 0.15) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 11: The bifurcation diagram of plate with a breathing crack ([[eta].sub.0] = 0.30) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 12: The bifurcation diagram of plate with a breathing crack ([[eta].sub.0] = 0.45) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 13: The bifurcation diagram of plate with an open crack ([[eta].sub.0] = 0.15) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 14: The bifurcation diagram of plate with an a open crack ([[eta].sub.0] = 0.30) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
Caption: Figure 15: The bifurcation diagram of plate with an open crack ([[eta].sub.0] = 0.45) for transverse displacement w, (a) via the external excitation of changing frequency and (b) via the external excitation of changing amplitude.
```Table 1: First three-order frequencies of cracked and intact plates.
First-order natural frequency (Hz)
Theoretical results FEM
Intact plate 23.9874 25.101
[eta] = 0.25 23.9279 24.987
[eta] = 0.5 23.8698 24.941
Second-order natural frequency (Hz)
Theoretical results FEM
Intact plate 38.3798 41.853
[eta] = 0.25 38.1120 41.359
[eta] = 0.5 38.3798 41.749
Third-order natural frequency (Hz)
Theoretical results FEM
Intact plate 62.3672 68.485
[eta] = 0.25 62.1117 68.071
[eta] = 0.5 61.8510 67.468
```
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# Thread: Explicit functions with discontinous intervals?
1. ## Explicit functions with discontinous intervals?
I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:
$g(x) = \frac{(x-2)(x+3)}{(x-2)}$
will obviously appear as the graph of $y = x+3$ with a hole at $x=2$. But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values
$2 \leq x \leq 6$
or more generally
$a \leq x \leq b$
is there a manner in which it is possible to construct a function that would produce this property?
Recap:
[1]Explicit, non-piece-wise function $f(x)$. No using set theory directly "cut and delete". No given domain restrictions
[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$, but rather is discontinuos for
$a \leq x \leq b$
How would one go about creating such a function? Thanks in advance for any advice related to this topic
2. Originally Posted by mfetch22
I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:
$g(x) = \frac{(x-2)(x+3)}{(x-2)}$
will obviously appear as the graph of $y = x+3$ with a hole at $x=2$. But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values
$2 \leq x \leq 6$
or more generally
$a \leq x \leq b$
is there a manner in which it is possible to construct a function that would produce this property?
Recap:
[1]Explicit, non-piece-wise function $f(x)$. No using set theory directly "cut and delete". No given domain restrictions
[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$, but rather is discontinuos for
$a \leq x \leq b$
How would one go about creating such a function? Thanks in advance for any advice related to this topic
Dear mfetch22,
Let me give you an idea. If we define a function such that $f:\Re\rightarrow{\Re}~and~f(x)=0~\forall~x\in[a,b]$ then, the function, $h(x)=\frac{f(x).g(x)}{f(x)}$ is a function which is not defined for $x\in[a,b]$ (here g(x) is another arbitary function)
I hope that the function h(x) satisfies all the conditions you have mentioned.
3. I don't think sudharaka's function works, because it's not defined anywhere, and therefore isn't really a function. We need a well-defined function that is discontinuous everywhere in an interval. So, mfetch, is the characteristic function out of bounds for your problem? You could easily define $f(x)=\chi_{\mathbb{Q}}(x)$. There's a function that is discontinuous everywhere. Trouble is, the characteristic function is itself defined piecewise. So I don't know if this "once removed" type of function is allowed or not.
Your problem sounds like it's part of a bigger problem. Any chance you'd be willing to state this bigger problem?
4. Originally Posted by Ackbeet
I don't think sudharaka's function works, because it's not defined anywhere, and therefore isn't really a function. We need a well-defined function that is discontinuous everywhere in an interval. So, mfetch, is the characteristic function out of bounds for your problem? You could easily define $f(x)=\chi_{\mathbb{Q}}(x)$. There's a function that is discontinuous everywhere. Trouble is, the characteristic function is itself defined piecewise. So I don't know if this "once removed" type of function is allowed or not.
Your problem sounds like it's part of a bigger problem. Any chance you'd be willing to state this bigger problem?
Its not neccesarily a bigger problem, its just a conecpt that came to my mind while watching this science channell program about finding new planets by observing wobbling stars, or by ovserving tiny shawdows or gaps in which the sun is partially blocked out. Then I, for some reason that I am unaware, I pictured an ellipse tracing out the path of a planet. I pictured viewing this planet orbit a lonely star, from a view point for which the planet dissipears for some period of its orbit behind this star and then reappears as it comes back into view. This gap in the visibility of the planet made me consider dropping that section of the ellipse when I viewed it from above. Then, I wondered if there was a way to make this "cut" without simply just stating its there. Maybe this means that I should also allow implicit relationships. So add that to the list, any implicit or explicit relationship that may arise in "continous discontinuities"..... I like this as name for this problem, it might not be mathematically proper but I've always loved oxy-morons.
Sudharaka, thank you for your post:
SUDHARAKA:
I hope that the function h(x) satisfies all the conditions you have mentioned.
I dont mean to be spliting hairs here, but the manner in which you defined $f(x)$ falls under the use of set theory to me. I dont mean to make any general statements or negitive comments about set theory, I'm simply trying to find a way to solve this problem with out it. Let me give an explaination of why..
Say I want to find an equalitiy with a graphical property named $P$. Lets, in addition call, this equality $S$. Lets put some restriction on this equality:
[1] $S$ must be explicitally or implicitally expressed as an equality of the standard two rectangluar axis variables $x$ and $y$, in $\mathbb{R}^2$
[2] $S$ must exhibit property $P$ as desribed or defined mathematically, but it must exhibit this propery as a natural consequence of the manner in which the equation was expressed. I would consider most all domain restrictions given as a supplement to the original equation to be beyond the "natural consequences" of the form of its expression.
[3] $S$ may not be expressed in peice-wise form, or in the langue of set theory.
The reason I make this restriction is because without some of them, this question becomes extremely simple. Like the set theory restriction, say I wanted to define discontinuity from a to b. Without the set theory restriction, this can easiliy be done, like Sudharaka showed. One simply defines $P$, a discontinuity in this case, for all values $f(x) \in [a, b]$.
Does that clear things up? The only thing I could think of that would satisfy this definition (actually I think it doesnt fit into my bounds since its an infinite expression) is the following progression for a discontinuity for $[a,b]$, I think you'll catch the pattern:
$f(x) = \frac{1}{(a-x)(b-x)(a+1-x)(b-1-x)(a+2-x)(b-2-x).......(a+d_n-x)(b-d_n-x)}$
$\;\;untill\;\; d_n=\frac{b-a}{2}\;\; and\;\; d_n = 0,1,2,3,4....\;\;\;for\;\;n=1,2,3,4.....$
next step:
$f(x) = \frac{1}{(a-x)(b-x)(a+0.5-x)(b-0.5-x)(a+1-x)(b-1-x)(a+1.5-x)(b-1.5-x)......
.(a+d_n-x)(b-d_n-x)}$
$\;\;untill\;\; d_n = \frac{b-a}{2}\;\;and\;\;d_n = 0, 0.5, 1.0, 1.5,.....\;\;\;for\;\;n=1,2,3,4$
Can you see the pattern? This is the only manner in which I can concieve of a "continous discontinuity" from a to b. I dont think this fits into my restrictions, so I need help finding a much simpler form. As for the above, if the process is continued, would this be correct?
$\lim_{(d_{n+1}-d_n)\to 0}f(x) = (continous\;\;discontinuity\;\;on\;\;[a, b])???$
5. Originally Posted by mfetch22
I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:
$g(x) = \frac{(x-2)(x+3)}{(x-2)}$
will obviously appear as the graph of $y = x+3$ with a hole at $x=2$. But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values
$2 \leq x \leq 6$
or more generally
$a \leq x \leq b$
is there a manner in which it is possible to construct a function that would produce this property?
Recap:
[1]Explicit, non-piece-wise function $f(x)$. No using set theory directly "cut and delete". No given domain restrictions
[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$, but rather is discontinuos for
$a \leq x \leq b$
How would one go about creating such a function? Thanks in advance for any advice related to this topic
Not sure that I understand completely what you are looking for ...
Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?
6. Originally Posted by earboth
Not sure that I understand completely what you are looking for ...
Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?
Of course! I didnt even consider the square root! Yes, this has a "continuous discontinuity", as I like to call it. Do you know if there is a way to use the property of this function you gave, to define some implict expression of an elipse that has a "continuous discontinuity" over some interval on the curve? Just asking, in the off chance that you know the answer.
7. Originally Posted by mfetch22
Of course! I didnt even consider the square root! Yes, this has a "continuous discontinuity", as I like to call it. Do you know if there is a way to use the property of this function you gave, to define some implict expression of an elipse that has a "continuous discontinuity" over some interval on the curve? Just asking, in the off chance that you know the answer.
$\left\{\begin{array}{l}x=5 \cdot \cos\left(\sqrt{t^2-9}\right) \\ y= 3 \cdot \sin\left(\sqrt{t^2-9}\right)\end{array} , t\in [-2\pi, 2\pi]\right.$
will produce the graph in the attachment.
8. Originally Posted by earboth
Not sure that I understand completely what you are looking for ...
Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?
wow...that was easy rofl... here's another one although it has no values $f(x)=\sqrt{-(x^2+1)}$
9. Originally Posted by earboth
$\left\{\begin{array}{l}x=5 \cdot \cos\left(\sqrt{t^2-9}\right) \\ y= 3 \cdot \sin\left(\sqrt{t^2-9}\right)\end{array} , t\in [-2\pi, 2\pi]\right.$
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# Thermo physics question
1. Jun 1, 2008
### mcewendavid
I am having trouble with this question...It is from "applied thermodynamics for engineering technologists" by Eastop (Q5.2 if anyone has it).....
The Question:
Two reversible heat engines operate in series between a source at 527°C and a sink at 17°C. If the engines have equal efficiencies and the first rejects 400kJ to the second, calculate:
(i) the temp at which heat is supplied to the 2nd engine ALL DONE :-) =208.7°C
(ii) The heat taken from the source
(iii) The work done by each engine
I have done part (i) and I know it's right but I have no idea what equations to use for the next 2 parts because heat and work are dependant on each other so I dont understand how I can find one without the other....
Dave :-)
2. Jun 1, 2008
$$\eta_{rev}=1-\frac{T_L}{T_H}[/itex] [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}[/itex] (1st law) [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]\frac{Q_{1_{in}}}{400}-.319=1[/itex] Using the above formulas, I ended up with the work for each engine at 127.6kJ. Total heat from the source 527.6kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7? 3. Jun 1, 2008 ### Jupiter6 [tex]\eta_{rev}=1-\frac{T_L}{T_H}[/itex] [tex]\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}[/itex] (1st law) [tex]\eta_{rev}=1-\frac{T_L}{T_H}=.319=\eta_{actual}=\frac{W}{Q_{in}}[/itex] [tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.319=\frac{400}{Q_{in}}[/itex] I MESSED my first response up. Using the above formulas, I ended up with the work for each engine at 187.4kJ. Total heat from the source 587.4kJ. However, I took the mid resovoir temp to be 272 deg C, the average of the high and low. Can you tell me how you got 208.7? 4. Jun 1, 2008 ### mcewendavid The answer in the book is 208.7 Deg C....but I got it using:- [tex]\eta = 1-(T2/T1)$$
1 - (T2/T3) = 1 - (T3/T1)
1 - (800/T3) = 1 - (T3/290) - Figures converted to Kelvin
232000 = T3$$^{}2$$
T3 = 481.66K = 208.66 Deg C
5. Jun 1, 2008
### Jupiter6
Ohh....derrr. My bad (I guess I've been paying more attention to the latex than the math). One more time.
[tex]\eta_{rev}=1-\frac{T_L}{T_H}=1-\frac{481.7K}{800K}=.398=\eta_{actual}=\frac{W}{Q_{in}}[/itex]
[tex]Q_{in}=W+Q_{out}----->Q_{in}-W=400kJ --->1-.398=\frac{400kJ}{Q_{in}}[/itex]
[tex]Q_{Total Input}=664.5kJ[/itex]
[tex]W_{1}=664.5kJ-400kJ=264.5kJ[/itex]
[tex]W_{2}=400kJ*.398=159.2kJ[/itex]
Last edited: Jun 1, 2008
6. Jun 1, 2008
### mcewendavid
omg you're a champion!
I totally didnt get the -W = 1 - 0.398 by using the T3 from part one
Now it just all fits together...it's great
Thankyou so much :-)
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# Homework Help: Quantum optics - dirac notation
1. Dec 12, 2011
### sam_021
1. The problem statement, all variables and given/known data
http://quantum.leeds.ac.uk/~almut/section2.pdf
Please note i am refering to the above notes
I basically dont get how the maths works to get
(eq(25))(eq(22))(eq(24)) = eq(26)
am i missing something interms of the commutator relations ?
2. Relevant equations
3. The attempt at a solution
2. Dec 12, 2011
### e.bar.goum
Ah, those equations are familiar. Did you try just multiplying through, remembering to keep everything in the right order, and recalling that <1|2> = <2|1> = 0 and <1|1> = <2|2> = 1, and that some of those exponentials will cancel?
3. Dec 13, 2011
^^ yes
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## Design a stack that supports getMin() in O(1) time and O(1) extra space
Question: Design a Data Structure SpecialStack that supports all the stack operations like push(), pop(), isEmpty(), isFull() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialStack must be O(1). To implement SpecialStack, you should only use standard Stack data structure and no other data structure like arrays, list, .. etc.
Example:
Consider the following SpecialStack
16 --> TOP
15
29
19
18
When getMin() is called it should return 15,
which is the minimum element in the current stack.
If we do pop two times on stack, the stack becomes
29 --> TOP
19
18
When getMin() is called, it should return 18
which is the minimum in the current stack.
• 解答
# Class to make a Node
class Node:
# Constructor which assign argument to nade's value
def __init__(self, value):
self.value = value
self.next = None
# This method returns the string representation of the object.
def __str__(self):
return "Node({})".format(self.value)
# __repr__ is same as __str__
__repr__ = __str__
class Stack:
# Stack Constructor initialise top of stack and counter.
def __init__(self):
self.top = None
self.count = 0
self.minimum = None
# This method returns the string representation of the object (stack).
def __str__(self):
temp = self.top
out = []
while temp:
out.append(str(temp.value))
temp = temp.next
out = '\n'.join(out)
return ('Top {} \n\nStack :\n{}'.format(self.top,out))
# __repr__ is same as __str__
__repr__=__str__
# This method is used to get minimum element of stack
def getMin(self):
if self.top is None:
return "Stack is empty"
else:
print("Minimum Element in the stack is: {}" .format(self.minimum))
# Method to check if Stack is Empty or not
def isEmpty(self):
# If top equals to None then stack is empty
if self.top == None:
return True
else:
# If top not equal to None then stack is empty
return False
# This method returns length of stack
def __len__(self):
self.count = 0
tempNode = self.top
while tempNode:
tempNode = tempNode.next
self.count+=1
return self.count
# This method returns top of stack
def peek(self):
if self.top is None:
print ("Stack is empty")
else:
if self.top.value < self.minimum:
print("Top Most Element is: {}" .format(self.minimum))
else:
print("Top Most Element is: {}" .format(self.top.value))
# This method is used to add node to stack
def push(self,value):
if self.top is None:
self.top = Node(value)
self.minimum = value
elif value < self.minimum:
temp = (2 * value) - self.minimum
new_node = Node(temp)
new_node.next = self.top
self.top = new_node
self.minimum = value
else:
new_node = Node(value)
new_node.next = self.top
self.top = new_node
print("Number Inserted: {}" .format(value))
# This method is used to pop top of stack
def pop(self):
if self.top is None:
print( "Stack is empty")
else:
removedNode = self.top.value
self.top = self.top.next
if removedNode < self.minimum:
print ("Top Most Element Removed :{} " .format(self.minimum))
self.minimum = ( ( 2 * self.minimum ) - removedNode )
else:
print ("Top Most Element Removed : {}" .format(removedNode))
# Driver program to test above class
stack = Stack()
stack.push(3)
stack.push(5)
stack.getMin()
stack.push(2)
stack.push(1)
stack.getMin()
stack.pop()
stack.getMin()
stack.pop()
stack.peek()
# This code is contributed by Blinkii
• 分析
https://www.geeksforgeeks.org/design-a-stack-that-supports-getmin-in-o1-time-and-o1-extra-space/
posted on 2019-10-22 10:01 星辰之衍 阅读(49) 评论(0编辑 收藏
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Cody
# Problem 761. Create a matrix X, where each column is a shifted copy of the vector v
Solution 1171926
Submitted on 26 Apr 2017 by Arief Anbiya
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = (1:5)'; y_correct = [1 5 4 3 2;2 1 5 4 3;3 2 1 5 4;4 3 2 1 5;5 4 3 2 1]; assert(isequal(shifted(x),y_correct))
2 Pass
x = (1:3)'; y_correct = [1 3 2;2 1 3;3 2 1]; assert(isequal(shifted(x),y_correct))
3 Pass
x = (1:4)'; y_correct = [1 4 3 2;2 1 4 3;3 2 1 4;4 3 2 1]; assert(isequal(shifted(x),y_correct))
4 Pass
x = (1:7)'; y_correct = [1 7 6 5 4 3 2;2 1 7 6 5 4 3;3 2 1 7 6 5 4;... 4 3 2 1 7 6 5;5 4 3 2 1 7 6; 6 5 4 3 2 1 7; 7 6 5 4 3 2 1]; assert(isequal(shifted(x),y_correct))
5 Pass
x = (1:6)'; y_correct = [1 6 5 4 3 2;2 1 6 5 4 3;3 2 1 6 5 4;... 4 3 2 1 6 5;5 4 3 2 1 6; 6 5 4 3 2 1]; assert(isequal(shifted(x),y_correct))
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# Is quadrilateral ABCD a rhombus? 1. AC is perpendicular to
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Is quadrilateral ABCD a rhombus? 1. AC is perpendicular to [#permalink]
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01 Nov 2008, 07:26
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Is quadrilateral ABCD a rhombus?
1. AC is perpendicular to BD
2. AB + CD = BC + AD
_________________
"You have to find it. No one else can find it for you." - Bjorn Borg
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Re: DS : A Rhombus [#permalink]
### Show Tags
01 Nov 2008, 07:47
Properties of rhombus: All opposite sides are parallel, opposite angles are equal, and diagnols bisect each other at 90 degress.
Stmt1: Suff
Stmt2: Could be a rectangle
Answer: A
PS: Some of you thinking, why can't it be a square ? well, a square can be a rhombus right ?
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Joined: 30 Jun 2008
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Re: DS : A Rhombus [#permalink]
### Show Tags
01 Nov 2008, 07:50
LiveStronger wrote:
Properties of rhombus: All opposite sides are parallel, opposite angles are equal, and diagnols bisect each other at 90 degress.
Stmt1: Suff
Stmt2: Could be a rectangle
Answer: A
PS: Some of you thinking, why can't it be a square ? well, a square can be a rhombus right ?
try again
_________________
"You have to find it. No one else can find it for you." - Bjorn Borg
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Joined: 21 Apr 2008
Posts: 269
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Re: DS : A Rhombus [#permalink]
### Show Tags
01 Nov 2008, 07:58
oh shoot
it is AB+CD = AD+BC
awww.. : definitely not a rectangle
It can't be C for sure, but I am not sure how I can use stmt2 to prove it as D ???
Oh! you have put me in a confused state Amit
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Re: DS : A Rhombus [#permalink]
### Show Tags
01 Nov 2008, 08:59
i think its E..
Rhoumbus has all sides equal and the angles might be different..
from either statement i cant tell if AB=BC=CD=AD...
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Re: DS : A Rhombus [#permalink]
### Show Tags
01 Nov 2008, 09:56
E it is
Statement 1 says that the 2 diagonals are perpendicular .... The quadrilateral could be a rhombus or a kite (one diagonal longer than other, but perpendicular to each other)
_________________
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Re: DS : A Rhombus [#permalink] 01 Nov 2008, 09:56
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# Is quadrilateral ABCD a rhombus? 1. AC is perpendicular to
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# Solving Systems of Linear Equations
In this example we solve a system of linear equations $$Ax = b$$ were the coefficient matrix is symmetric positiv definite and banded. We first compute the Cholesky factorization $$S = L L^T$$ with lapack::pbtrf which is the FLENS port of LAPACK's dpbtrf. We then solve $$Lu=b$$ and then $$L^T x = u$$ using lapack::pbtrs which is the port of dpbtrs.
Note that we might rename lapack::pbtrf to lapack::potrf and lapack::pbtrs to lapack::potrs.
## Example Code
#include <iostream>
#include <flens/flens.cxx>
using namespace std;
using namespace flens;
typedef double T;
int
main()
{
typedef SbMatrix<BandStorage<T> > SymmetricBandMatrix;
typedef DenseVector<Array<T> > Vector;
typedef SymmetricBandMatrix::IndexType IndexType;
const IndexType n = 4;
SymmetricBandMatrix A(n, Upper, 1);
Vector b(n);
A.general().diag( 1) = -1;
A.general().diag( 0) = 2;
b = 1,
2,
3,
4;
cout << "A = " << A << endl;
cout << "b = " << b << endl;
lapack::pbtrf(A);
cout << "L^T = " << A.general().upper() << endl;
lapack::pbtrs(A, b);
cout << "x = " << b << endl;
}
Define some convenient typedefs for the matrix/vector types of our system of linear equations.
typedef SbMatrix<BandStorage<T> > SymmetricBandMatrix;
typedef DenseVector<Array<T> > Vector;
We also need an extra vector type for the pivots. The type of the pivots is taken for the system matrix.
typedef SymmetricBandMatrix::IndexType IndexType;
Set up the baby problem ...
const IndexType n = 4;
We allocate a symmetric band matrix with one off-diagonal.
SymmetricBandMatrix A(n, Upper, 1);
We initialize the positiv definite $$n \times n$$ tridiagonal matrix.
A.general().diag( 1) = -1;
A.general().diag( 0) = 2;
We factorize $$A$$ with lapack::pbtrf
lapack::pbtrf(A);
In the upper part of $$A$$ now the triangular factor $$L^T$$ is stored.
cout << "L^T = " << A.general().upper() << endl;
Solve it with lapack::pbtrs
lapack::pbtrs(A, b);
## Compile
Note that we need to link against an external LAPACK implementation:
$shell> cd flens/examples$shell> g++ -std=c++11 -Wall -I../.. -DUSE_CXXLAPACK -framework vecLib -o lapack-pbtrs lapack-pbtrs.cc
## Run
$shell> cd flens/examples$shell> ./lapack-pbtrs
A =
2.000000 -1.000000
-1.000000 2.000000 -1.000000
-1.000000 2.000000 -1.000000
-1.000000 2.000000
b =
1.000000 2.000000 3.000000 4.000000
L^T =
1.414214 -0.707107
1.224745 -0.816497
1.154701 -0.866025
1.118034
x =
4.000000 7.000000 8.000000 6.000000
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# Batched coupon collector with quota
Assume that you draw coupons uniformly at random from a collection of $n$ coupons and you want to collect $m_i$ coupons of type $i$. This is referred to as the coupon collector with quota (http://www.combinatorics.org/ojs/index.php/eljc/article/download/v15i1n31/pdf)
Assume now that you draw your coupon by batches of $k\le n$ distinct coupons and let $T_{\vec{m},k})$ be the number of coupons that one has to buy in order to collect $m_i$ coupons of type $i$ (for each $i$).
Intuitively (and numerically), the expectation $\mathbb{E}[T_{\vec{m},k}]$ is decreasing in $k$ but I am not able to find a reference (or a proof for general $k$).
My questions are:
• does anyone know a reference?
• is it still true if the batches are of random size or if the coupons have non uniform probabilities?
Note that the $T_{\vec{m},k}$ are not stochastically ordered. For example when $\vec{m}=(2,0)$, we have $T_{\vec{m},1}$ and $T_{\vec{m},2}$ have the same mean but none of them is stochastically greater than the other one.
## Edit: answer for $k=2$ (11 april)
For $k=1$, the function $E(m,k)=\mathbb{E}[T_{k,m}]$ is uniquely defined by: $$E(x,1) = \left\{ \begin{array}{ll} 0 & if x =0\\ 1+\frac{1}{N}\sum_{i=1}^n E((x-e_{i})^+,1) & \end{array}\right.$$ where $e_i$ is the vector with a "1" on its $i$th coordinate and 0 otherwise; $(x)^+=max(x,0)$ (coordinate-wise).
More generally, for $k\ge 1$, we have: $$E(x,k) = \left\{ \begin{array}{cc} 0 & if x =0\\ 1+\frac{(N-k)!}{N!}\sum_{i_1\dots i_k distincts} E((x-e_{i_1}-\dots-e_{i_k})^+,k) & \end{array}\right.$$
Using the above equation, I am able to show by induction on $x$ that $E(x,2)\le E(x,1)$ by showing that $2E(x-e_i-e_j) \le E(x-2e_i) + E(x-2e_j)$.
• You say that you don't have a reasonably small proof. Do you have a long proof? – Douglas Zare Apr 7 '16 at 18:29
• I have a proof that $k=2$ requires less coupon than $k=1$. This proof is based on using a recurrence equation (on $m$) for $\mathbb{E}[T_{k,\vec{m}}]$. – N. Gast Apr 7 '16 at 18:55
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1. ## BADLY NEED HELP MASTERS. Thanks!
Here is my problem, I can't figure out this simple tasking job. here it is
Problem #1
Ask the user for a number(1-5), say X. Check for input validity. In cases of an invalid input, continuously ask the user to enter a number until he finally enters a valid one. Print the numbers from1 to X for each row starting from row 1 to row X. (It is required to use looping constructs)
sample run:
Enter a number (1-5):0
0 is an invalid entry. Try another.
Enter a number (1-5): 3
row 1: 1 2 3
row 2: 1 2 3
row 3: 1 2 3
here's my problematic code for that problem:
Code:
```#include<stdio.h>
int main (void)
{ int num;
do {
printf("Enter a number (1-5):");
scanf("%d", &num);
if((num<1)||(num>5))
printf("%d is an invalid entry. Please enter another one.\n", num);
} while ((num<1)||(num>5));
for(int i=0; i<1; i++){
printf("\nrow%d", i+1);
for(int j=0; j<0; j--){
printf("%d", j+1);
}
}
return 0;
}```
PROBLEM 2:
write a program that continually asks the user to enter an integer until he/she enters the number 0. Display the maximum of these numbers and the number of times it appeared.
Sample run:
Enter a number: 2
Enter a number: 3
Enter a number: 5
Enter a number: 1
Enter a number: 8
Enter a number: 8
The largest number is 8 and it was entered 2 time(s).
here is my sample problematic program
Code:
```#include<stdio.h>
int main(void)
{
int a, i, m, j, n;
printf("enter a number: ");
scanf("%d", &number1);
max=count+1;
count=1;
void largest(int a[][MAXCOLS],int m,int n)
{
int i,j,largest;
largest = a[0][0];
for (i=0;i<m;++i)
{
for (j=0;j<n;++j)
{
if (a[i][j]>largest)
largest=a[i][j];
}
}
printf("
The largest element of the matrix is %d",largest);
printf("\n\nThe largest number is %d and it appeared %d time(s).", max, count);
return 0;
}```
by noobprogrammer
2. Code:
``` for(int i=0; i<1; i++){
printf("\nrow%d", i+1);
for(int j=0; j<0; j--){
printf("%d", j+1);
}
}```
look at your limits for the for loops...shouldn't they have something to do with num?
Don't know what you're doing for #2...it should be much shorter than that. All you need to do is keep asking until the input is 0, otherwise the only other thing that should be going on is:
if the last input is bigger than the largest, set the largest to the last input and reset the number of times the largest has been counted (or increment if you want to keep track, might want to use an array). otherwise, if the input matches the largest, then increment the number of times it's been entered.
3. sufoode, that's using C++, not C, and you shouldn't do his homework for him/her, that's not letting him/her learn.
Besides, void main doesn't follow the standard, they should be int mains, <stupid mistake>and your solution to #2 doesn't correctly count how many times a number has been entered. What if the user entered 5 5 5 1 5? Your program would say that 5 has been entered once.</stupid mistake>
4. Hell, I went and confused myself. You're right.
Still, don't do homework for people, and this particular forum section is for C, not C++.
5. wow, thanks for replies, maybe the C++ can't help me we are doing it in C our professor forbids using c++ anyway thanks i'll ask you some more questions as i go through your answers and understand it. Thanks by the way. for your replies ^_^
6. I forgot in problem number two after entering 0 the screen will print the largest number and its recurrence in the whole program. Thanks for your help again masters!
7. Code:
``` for(int i=0; i<1; i++){
printf("\nrow%d", i+1);
for(int j=0; j<0; j--){
printf("%d", j+1);
}
}```
Wah! what should I do with the loops? Can someone with a good heart explain to me this please because honestly our professor's mouth is racing with the spped of light. Totally, I'm a beginner in programming.
8. Just put some thought into it!!!! Don't be such a lazy twit.
Code:
```for(int i=0; i<1; i++){ // How many times will this be executed if i starts at 0 and must be < 1?
printf("\nrow%d", i+1);
for(int j=0; j<0; j--){ // How many times will this be executed if j starts at 0 and must be < 0???
printf("%d", j+1);
}
}```
9. Code:
``` #include<stdio.h>
int main (void)
{ int num;
do {
printf("Enter a number (1-5):");
scanf("%d", &num);
if((num<1)||(num>5))
printf("%d is an invalid entry. Please enter another one.\n", num);
} while ((num<1)||(num>5));
for(int i=1; i<=num; i++){
printf("\nrow %d:", i);
}
return 0;
}
===========
#include<stdio.h>
int main (void)
{ int num;
do {
printf("Enter a number (1-5):");
scanf("%d", &num);
if((num<1)||(num>5))
printf("%d is an invalid entry. Please enter another one.\n", num);
} while ((num<1)||(num>5));
for(int i=0; i<num; i++){
printf("\nrow %d: ", i+1);
for(int j=0; j<0; j--){
printf("%d", j+1);
}
}
return 0;
}```
Masters, I've tried this code but all that appears is for example:
row 1:
row 2:
row 3:
what I need is
row 1: 123
row 2: 123
row 3: 123
Can anyone pinpoint the mistakes in my program? thank you in advance.
10. thanks for the answers, I'm really sorry because I really didn't understand this thing completely.
11. Code:
```for(int i=0; i<1; i++){ // How many times will this be executed if i starts at 0 and must be < 1?
printf("\nrow%d", i+1);
for(int j=0; j<0; j--){ // How many times will this be executed if j starts at 0 and must be < 0???
printf("%d", j+1);
}
}```
Am I right the first one will be executed at exactly one time only? while the 2nd one will not be executed. That's wrong in'st it?
12. Masters! Thanks for your words of wisdoms and sermons, i've solved the 1st problem, the 2nd one is the only problem. Yahoo!
13. Code:
``` #include<stdio.h>
int main (void)
{ int num;
do {
printf("Enter a number (1-5):");
scanf("%d", &num);
if((num<1)||(num>5))
printf("%d is an invalid entry. Please enter another one.\n", num);
} while ((num<1)||(num>5));
for(int i=1; i<=num; i++){
printf("\nrow %d:", i);
}
return 0;
}
===========
#include<stdio.h>
int main (void)
{ int num;
do {
printf("Enter a number (1-5):");
scanf("%d", &num);
if((num<1)||(num>5))
printf("%d is an invalid entry. Please enter another one.\n", num);
} while ((num<1)||(num>5));
for(int i=0; i<num; i++){
printf("\nrow %d: ", i+1);
for(int j=0; j<0; j--){
printf("%d", j+1);
}
}
return 0;
}```
Masters, I've tried this code but all that appears is for example:
row 1:
row 2:
row 3:
what I need is
row 1: 123
row 2: 123
row 3: 123
Can anyone pinpoint the mistakes in my program? thank you in advance.
14. Look again at this for-loop:
Code:
`for(int j=0; j<0; j--){`
How many times will the loop be executed?
Think about how you could rewrite this to make it work.
| 2,031
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| 2.984375
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CC-MAIN-2019-51
|
latest
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en
| 0.721611
|
https://discusstest.codechef.com/t/promel-name-is-spam_classification_using_neural_net-can-you-find-what-wrong-in-this-program/19871
| 1,685,889,671,000,000,000
|
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| 242,984,942
| 5,445
|
# promel name is (Spam_Classification_Using_Neural_Net) can you find what wrong in this program
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main() {
int t;
cin>>t;
while(t–){
ll n,min,max;
cin>>n>>min>>max;
vector<pair<ll,ll>>V;
ll w,b;
for(int i=0;i<n;i++){
cin>>w>>b;
V.push_back(make_pair(w,b));
}
ll total,count=0;
bool even=false;
bool odd=false;
for(int i=1;i<3;i++){
total=i;
if(i%2==0){
even=true;
odd=false;
}else{
even=false;
odd=true;
}
for(ll j=0;j<n;j++){
if(even){
if(V[j].second%2==0){
odd=false;
}else{
odd=true;
even=false;
}
}
else if(odd){
if(((V[j].first%2==0)&&(V[j].second%2==0)) || ((V[j].first%2!=0)&&(V[j].second%2!=0)) ){
odd=false;
even=true;
}else{
even=false;
}
}
// total = (total*V[j].first)+V[j].second;
}
if(even)
count++;
}
ll spammer,nonspamer;
if(count==1){
ll total=(max-min+1);
if(min%2!=0)
spammer=ceil(float(total)/2);
else{
spammer=floor(float(total)/2);
}
nonspamer= total - spammer;
cout<<nonspamer<<" “<<spammer<<endl;
}else if(count==0){
cout<<0<<” “<<(max-min+1)<<endl;
}else if(count==2){
cout<<(max-min+1)<<” "<<0<<endl;
}
``````}
``````
}
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|
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| 2.515625
| 3
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CC-MAIN-2023-23
|
latest
|
en
| 0.207541
|
https://inperc.com/wiki/index_title_Algorithm_for_Grayscale_Images.html
| 1,652,831,354,000,000,000
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|
This page is a part of the Computer Vision Wiki. The wiki is devoted to computer vision, especially low level computer vision, digital image analysis, and applications. The exposition is geared towards software developers, especially beginners. The wiki contains discussions, mathematics, algorithms, code snippets, source code, and compiled software. Everybody is welcome to contribute with this in mind - all links are no-follow.
Algorithm for Grayscale Images
In this article the algorithm for analysis for gray scale images is explained in detail and illustrated with C++ source code.
The data structure of the algorithm
The input image of the algorithm is a gray scale 2D image, i.e., an array of of integers from 0 to 255 with dimensions dim1 and dim2.
```int image[ dim1*dim2 ];
```
The algorithm first creates binary frames for each value of the threshold and then partitions the image by providing boundaries of the objects. Geometrically, these boundaries are circular sequences of directed edges of pixels in the image. In other words, these are 0- and 1-cycles. The first task is to count them.
```long betti[2] = {0,0}; // Betti numbers = number of current cycles of each dimension
```
For each value of the parameter, the gray level, the image is binary. Two Boolean arrays of the same size are used: the previous frame (PrevFrame) and the current frame (CurrFrame).
```bool *CurrFrame = new bool [ dim1*dim2 ];
bool *PrevFrame = new bool [ dim1*dim2 ];
```
The binary frame is created by thresholding the image.
```for(int k=0;k < dim1;k++) // start the (k,l) pixel
for(int l=0;l < dim2;l++)
if( image[l+dim2*k] < frame )
CurrFrame[l+dim2*k] = 0;
else
CurrFrame[l+dim2*k] = 1;
```
As pixels are added, this array is incrementally changed - pixels are added - and analyzed by the algorithm. Recall that at every step the cycles that have no descendants are called current cycles. Similarly, the edges that represent current cycles are called current edges. Cycles in the image are represented by nodes in the graph. The structure of the graph is the same as before - an acyclic binary directed graph. Except this time this graph spans through the frames. For each frame, the regions and holes are represented as nodes of this graph. Each node representing a region or hole is connected to the nodes representing regions and holes that existed previously (its parents) or appear later (its children) by directed edges, or arrows, of the graph. The arrows (one or two from each node) indicate merging and splitting of the regions as we add vertices, edges, and pixels, Adding Pixels. There are no circular sequences of arrows. Each node of the graph contains a pair of links (pointers) to next nodes. If the cycle is merged with another, one link used, if it splits, two. This information is used even after the cycle ceases to be current, i.e., after a merge or a split.
```int frame = 0; // current frame number, 0-255
```
In the end of processing a frame, CurrFrame, it is copied to PrevFrame.
```for(int k=0; k < dim1; k++) // start the (k,l) pixel
for(int l=0; l < dim2; l++)
PrevFrame[l+dim2*k] = CurrFrame[l+dim2*k];
```
The run through pixels of the image
At every step of the run through the image, the value of the pixel in the image is considered. If the color is white, skip to the next step. If the color is black, execute AddCell. Every time a cell is added, the topology of the image changes, the graph, the list of current cycles, and the Betti numbers are updated.
```for( k=0; k < dim1; k++ )// start the (k,l) pixel
for( l=0; l < dim2; l++ )
{
if( CurrFrame[ l+dim2*k ] == 0 )// if the new pixel is black..
}
```
Cycles in the image are represented by nodes in a directed graph as before. In this graph, the cycles are related to each other to record the changes in the topology of the image in terms of merging and splitting of components and holes as pixels are added. In particular, the 0- and 1-Betti numbers are updated at every step. In addition to the structure defined in Chapter 2, we need to record the frame when the cycle was created.
```struct Cycle {
...
int Birth; // the frame of birth
};
```
Each node of the graph contains only a pair of links (pointers) to next nodes. If the cycle is merged with another, one link used, if it splits, two. The node may contain additional information: the dimension of the node, the frame of birth (gray level), a pointer to the edge that created the cycle, the area, the perimeter, etc.
CreateCycle is almost the same as before. The only new command is
``` Cycle0->Birth = frame;
```
At every step of the run through the image, a pixel in PrevFrame is compared to the corresponding one in CurrFrame. If the colors are the same, skip to the next step. If the color changes from white to black, execute AddCell. Observe that in the case of gray scale images, the color can never change from black to white (RemoveCell will used in the future).
```// processing the frame
for( int k=0; k < dim1; k++ ) // start the (k,l) pixel
for( int l=0; l < dim2; l++ )
{
// if the new pixel is black and the color of its position is white
if( CurrFrame[l+dim2*k] == 0 && PrevFrame[l+dim2*k] == 1 )
}
```
``` The AddCell command is exactly the same.
```
Exercise. Create a simple interface for the program without simplification and run it. The output should be a sequence of the Betti numbers of the frames. Find or create a simple gray scale image and verify the correctness of the Betti numbers.
Exercise. Design a character recognition system. Answer
Continue to Homology in 2D.
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| 3.765625
| 4
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CC-MAIN-2022-21
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en
| 0.885303
|
http://www.claymath.org/library/historical/euclid/files/elem.8.26.html
| 1,448,741,801,000,000,000
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| 351,884,241
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|
## Book VIII, Proposition 26
Similar plane numbers have to one another the ratio which a square number has to a square number.
Οἱ ὅμοιοι ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. Ἔστωσαν ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β: λέγω, ὅτι ὁ Α πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. Ἐπεὶ γὰρ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν, τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐμπιπτέτω καὶ ἔστω ὁ Γ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Γ, Β οἱ Δ, Ε, Ζ: οἱ ἄρα ἄκροι αὐτῶν οἱ Δ, Ζ τετράγωνοί εἰσιν. καὶ ἐπεί ἐστιν ὡς ὁ Δ πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Β, καί εἰσιν οἱ Δ, Ζ τετράγωνοι, ὁ Α ἄρα πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ὅπερ ἔδει δεῖξαι. Similar plane numbers have to one another the ratio which a square number has to a square number. Let A, B be similar plane numbers; I say that A has to B the ratio which a square number has to a square number. For, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [VIII. 18] Let it so fall, and let it be C; and let D, E, F, the least numbers of those which have the same ratio with A, C, B, be taken; [VII. 33 or VIII. 2] therefore the extremes of them D, F are square. [VIII. 2, Por.]
| 612
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| 4.0625
| 4
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CC-MAIN-2015-48
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longest
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en
| 0.653698
|
https://www.aqua-calc.com/calculate/food-calories/substance/mixed-blank-apple-blank-slices-coma-and-blank-upc-column--blank-074641052575
| 1,627,539,913,000,000,000
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|
Calories of MIXED APPLE SLICES, UPC: 074641052575
mixed apple slices, upc: 074641052575: calories and nutrients
See how many calories in 100 g (3.5 oz) of MIXED APPLE SLICES, UPC: 074641052575
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 50 209.2 Total 50 209.2
Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
See how many nutrients in 100 g (3.5 oz) of MIXED APPLE SLICES, UPC: 074641052575
Nutrient (find foods
rich in nutrients)
UnitValue /
100 g
Proximates
Energykcal50
Carbohydrate, by differenceg13.57
Fiber, total dietaryg2.1
Sugars, total including NLEAg10.71
Minerals
Calcium, Camg43
Iron, Femg0.26
Potassium, Kmg107
Vitamins
Vitamin C, total ascorbic acidmg132.9
Vitamin A, IUIU71
• About MIXED APPLE SLICES, UPC: 074641052575
• MIXED APPLE SLICES, UPC: 074641052575 contain(s) 50 calories per 100 grams (≈3.53 ounces) [ price ]
• Ingredients: APPLES WITH CALCIUM ASCORBATE TO PROMOTE WHITENESS AND MAINTAIN TEXTURE.
• Manufacturer: Country Fresh Inc.
• Food category: Pre-Packaged Fruit and Vegetables
• A food with a name containing, like or similar to MIXED APPLE SLICES, UPC: 074641052575:
• MIXED APPLE SLICES, UPC: 074641052629 contain(s) 50 calories per 100 grams (≈3.53 ounces) [ price ]
• The calories and nutrients calculator answers questions like these: How many nutrients (amino acids, lipids, minerals, proximates and vitamins) in a selected food per given weight? How many total calories (calories from carbohydrates, fats and proteins) in a selected food per given weight? The total number of calories and amount of nutrients are calculated based on the selected food and its given weight, and using the USDA Food Composition Databases. Visit our food calculations forum for more details.
Foods, Nutrients and Calories
PROVOLONE NATURAL CHEESE SLICES WITH SMOKE FLAVOR, PROVOLONE, UPC: 046100002216 contain(s) 344 calories per 100 grams (≈3.53 ounces) [ price ]
23525 foods that contain Magnesium, Mg. List of these foods starting with the highest contents of Magnesium, Mg and the lowest contents of Magnesium, Mg, and Recommended Dietary Allowances (RDAs) for Magnesium
Gravels, Substances and Oils
CaribSea, Freshwater, Flora Max, Volcano Red weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Lead carbonate [PbCO3] weighs 6 582 kg/m³ (410.90084 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-437A, liquid (R437A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
Weights and Measurements
A zeptoweber is a SI-multiple (see prefix zepto) of the magnetic flux unit weber and equal to equal to 1.0 × 10-21 weber
The length measurement was introduced to measure distance between any two objects.
st/m³ to kg/ft³ conversion table, st/m³ to kg/ft³ unit converter or convert between all units of density measurement.
Calculators
Calculate gas pressure at different temperatures
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https://www.coursehero.com/file/9036380/6-Solvethesystembelow-6x2y18-6x7y3-6x-2y-18-16x/
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Systems of Equations
# 6 solvethesystembelow 6x2y18 6x7y3 6x 2y 18 16x
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Unformatted text preview: -4y = 4 -4 -4 y = -1 -4x + 6y = -18 -4x + 6(-1) = -18 -4x - 6 = -18 +6 +6 -4x = -12 -4 -4 x=3 2 CHAPTER 7 TEST REVIEW ANSWER KEY PART II: Answer all questions in this part. Clearly indicate the necessary steps, including appropriate formula substitutions, calculations, etc. 6.) Solve the system below. 6x 2y = 18 6x 7y = 3 6x - 2y = 18 -1(6x - 7y = 3) (4, 3) 8.) 6x - 2y = 18 -6x + 7y = -3 5y = 15 5 5 y=3 Solve the system of equations below. 7.) 4u + 9v = 9 6x - 2y = 18 u 3v = 6 6x - 2(3) = 18 6x - 6 = 18 +6 +6 6x = 24 6 6 x=4 2(4x + 3y = 7) -4u + 9v = 9 3u - 9v = -18 -1u = -9 -1 -1 u=9 u=9 v=5 3x + 2y = 4 4x + 3y = 7 -3(3x + 2y = 4) What is the solution of the system of equations below? -9x - 6y = -12 8x + 6y = 14 -1x = 2 -1 -1 x = -2...
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## This document was uploaded on 03/19/2014 for the course MATH Integrated at Brewster High School.
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Mathematics
djmatthies12:
Zoya has to earn at least \$300 to meet her fundraising goal. She has only 100 bracelets that she plans to sell at \$5 each. Which inequality shows the number of bracelets, x, Zoya can sell to meet her goal?
jhonyy9:
do you have any idea about this ? what mean inequality ? using these given details what inequality you can write one bracelet cost \$5 so x=5 do you can continue it ?
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# An insulating rod of length l carries a charge q distributed uniformly on it.The rod is pivoted at an end and is rotated at a frequency f about a fixed perpendicular axis.The magnetic moment of the system is?
Pratham Ashish
17 Points
14 years ago
let the charge density on rod be λ = q/l
consider a small element of rod dx at a distance x from the fixed end
charge on it dq = λ dx
when the rod rotates the current due to this element , di = f dq
= f λ dx
magnetic moment of this current element dM = di * area
= f λ dx * ∏ x^2
total magnetic moment M = ∫ dM =
= ∫0 l f λ dx * ∏ x^2
= fλ∏ ∫0( x^2 ) dx
= fλ∏/3 * x^3 ]0 l
= fλ∏ l^3 /3
λ l = q , so
M = f∏q l^2 /3
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## Lifespan Estimate of Damped Semilinear Wave Equation in Exterior Domain with Neumann Boundary Condition
Zhao Jinglei,1, Lan Jiacheng,1, Yang Shanshan,2
Abstract
This paper concerns about the upper bound of lifespan estimate to damped semilinear wave equations in exterior domain with vanishing Neumann boundary condition. We find that the initial boundary value problem with Neumann boundary condition admits the same upper bound of lifespan as that of the Cauchy problem in $\mathbb{R}^n (n\ge 1)$. This fact is different from the zero Dirichlet boundary value problem in 2-D exterior domain for lifespan estimate, compared to the corresponding result in [6], and is also different from the zero Dirichlet boundary value problem on half line for critical power, compared to the result in [16].
Keywords： Lifespan ; Damped semilinear wave equations ; Neumann boundary condition ; Exterior problem
Zhao Jinglei, Lan Jiacheng, Yang Shanshan. Lifespan Estimate of Damped Semilinear Wave Equation in Exterior Domain with Neumann Boundary Condition. Acta Mathematica Scientia[J], 2021, 41(4): 1033-1041 doi:
## 1 引言
$$$\left\{ \begin{array}{ll} u_{tt} - \Delta u +u_t = |u_t|^p, \quad (t, x)\in[0, T)\times B_1^c, \\ u(0, x) = \varepsilon f(x), \quad u_t(0, x) = \varepsilon g(x), \quad x\in B_1^c, \\ \partial_ru(t, x)|_{\partial B_1^c} = 0, \\ \end{array} \right.$$$
## 4 定理2.1的证明
$\begin{eqnarray} && \varepsilon\left(\int_{B_1^c}(f(x)+g(x)){\rm d}x\right)+ \int_0^{T_{\varepsilon}}\int_{B_1^c}|u|^p\Phi(t, x){\rm d}x{\rm d}t{}\\ & = &\int_0^{T_{\varepsilon}}\int_{B_1^c}u\Phi_{tt}{\rm d}x{\rm d}t-\int_0^{T_{\varepsilon}}\int_{B_1^c}u\Phi_{t}{\rm d}x{\rm d}t-\int_0^{T_{\varepsilon}}\int_{B_1^c}u\Delta\Phi {\rm d}x{\rm d}t{}\\ &\triangleq&I+II+III. \end{eqnarray}$
$\begin{eqnarray} Y_2(R)& = &\int_0^{T_{\varepsilon}}\int_{B_1^c}|u|^p\bigg(\int_0^R\eta^{2p'}(t/\sigma)\theta^{*2p'}(|x|^2/\sigma)\sigma^{-1}{\rm d}\sigma\bigg){\rm d}x{\rm d}t{}\\ & = &\int_0^{T_{\varepsilon}}\int_{B_1^c}|u|^p\bigg(\int_{\frac{|x|^2}{R}}^\infty\eta^{2p'}(s)\theta^{*2p'}\big(\frac{|x|^2s}{t}\big)s^{-1}{\rm d}s\bigg){\rm d}x{\rm d}t{}\\ &\le& \log2\int_0^{T_{\varepsilon}}\int_{B_1^c}|u|^p\eta_R^{2p'}(t)\theta_R^{2p'}(x){\rm d}x{\rm d}t. \end{eqnarray}$
$$$\left(C_2\varepsilon+Y_1(R)+Y_2(R)\right)^p\le C_3R(Y_1'(R)+Y_2'(R)),$$$
$$$\left(C_2\varepsilon+\overline{Y}(R)\right)^p\le C_3R\overline{Y}'(R),$$$
## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子
Crispo F , Maremonti P .
An interpolation inequality in exterior domains
Rend Semi Mate Univ Padova, 2004, 112, 11- 39
Fujita H .
On the blowing up of solutions of the Cauchy problem for utu+u1+α
Journal of the Faculty of Science University Tokyo Sect I, 1966, 13, 109- 124
Geng J B , Yang Z Z , Lai N A .
Blow-up and lifespan estimates for initial boundary value problems for semilinear Schrödinger equations on half-line
Acta Math Sci, 2016, 36A (6): 1186- 1195
Huang S J , Meng X W .
Improved ordinary differential inequality and its application to semilinear wave equations
Acta Math Sci, 2020, 40A (5): 1319- 1332
Ikeda M , Ogawa T .
Lifespan of solutions to the damped wave equation with a critical nonlinearity
Journal of Differential Equations, 2016, 261 (3): 1880- 1903
Ikeda M , Sobajima M .
Remark on upper bound for lifespan of solutions to semilinear evolution equations in a two-dimensional exterior domain
Journal of Mathematical Analysis and Applications, 2019, 470, 318- 326
Ikehata R .
Global existence of solutions for semilinear damped wave equation in 2-D exterior domain
Journal of Differential Equations, 2004, 200, 53- 68
Ikehata R .
Two dimensional exterior mixed problem for semilinear damped wave equations
Journal of Mathematical Analysis and Applications, 2005, 301, 366- 377
Lai N A , Takamura H , Wakasa K .
Blow-up for semilinear wave equations with the scale invariant damping and super-Fujita exponent
Journal of Differential Equations, 2017, 263, 5377- 5394
Lai N A , Yin S L .
Finite time blow-up for a kind of initial-boundary value problem of semilinear damped wave equation
Mathematical Methods in the Applied Sciences, 2017, 40 (4): 1223- 1230
Lai N A , Zhou Y .
The sharp lifespan estimate for semilinear damped wave equation with Fujita critical power in higher dimensions
Journal De Mathematiques Pures Et Appliquees, 2018, 123, 229- 243
Li T , Zhou Y .
Breakdown of solutions to $□u+u_t=|u|.{1+α}$
Discrete Contin Dyn Syst, 1995, 1 (4): 503- 520
Li Y C .
Classical solutions for fully nonlinear wave equations with dissipation
Chinese Annals of Mathematics Series A, 1996, 17 (4): 451- 466
Lin Y H , Jiang H B , Yin S L .
Global existence for damped semilinear wave equations outside obstacles in high dimensions(in Chinese)
Scientia Sinica Mathematica, 2018, 48 (4): 507- 518
Nishihara K .
Lp-Lq estimates of solutions to the damped wave equation in 3-dimensional space and their application
Mathematische Zeitschrift, 2003, 244 (3): 631- 649
Nishihara K , Zhao H J .
Existence and nonexistence of time-global solutions to damped wave equation on half-line
Nonlinear Analysis: Theory Methods Applications, 2005, 61 (6): 931- 960
Ogawa T , Takeda H .
Non-existence of weak solutions to nonlinear damped wave equations in exterior domains
Nonlinear Analysis: Theory Methods Applications, 2009, 70 (10): 3696- 3701
Todorova G , Yordanov B .
Critical exponent for a nonlinear wave equation with damping
Journal of Differential Equations, 2001, 174 (2): 464- 489
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Limits: 277h46m40s, 512 MB
Today is your best friend SJ’s birthday. You want to buy a birthday present for her. You want to buy such a present that she likes the most. You are very superstitious. You think that, SJ will love your gift, if the price of the present you buy is an interesting number (pretty weird isn’t it 😛).
This is a companion discussion topic for the original entry at https://toph.co/p/birthday-gift-for-sj
what means by 4=22 and so on here?
@Being_Gorom
4 = 22 Its means 4 = 2 to the power of 2, 8 = 23 Its means 8 = 2 to power of 3 , 9 = 32 Its means 9 = 3 to power of 2.
1 Like
thanks man appreciate that
you are most welcome
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# Thread: Binomial & L'Hospital math...
1. ## Binomial & L'Hospital math...
Determine lim[h→0] [(x+h)4 -(x4)]/ (h)
1) By using binomial theorem
2) By L'Hospitals rule
$\displaystyle (x+h)^4=x^4+4x^3h+6x^2h^2+4xh^3+h^4$
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# How to Prepare a Statement of Cash Flows
Co-authored by Jill Newman, CPA
Updated: March 29, 2019
A statement of cash flows is one of the four major financial statements prepared by corporations at the end of each accounting period (the others being a balance sheet, income statement, and statement of retained earnings). The goal of the cash flow statement is to provide an accurate picture of the cash inflows, outflows, and net changes of cash during the accounting period. The statement is prepared by calculating net changes to cash from operating, investing, and financing activities. The total increase or decrease in cash for the current year is added to the ending cash from the prior year to calculate the ending cash and cash equivalents for the current year. Keep in mind that the ending cash amount on the statement of cash flows should be equal to the ending cash amount on the balance sheet. If the amounts are not equal, then there has been an error.
### Part 1 of 4: Calculating Beginning Cash and Cash Equivalents
1. 1
Determine the ending cash balance from the prior year. If the company prepared a statement of cash flows for the prior year, you can find this information there. If not, you will have to find information from the prior year's ending balance sheet and calculate the ending cash balance. Include cash and cash equivalents that can be converted into cash within one year. Cash equivalents include money market funds, certificates of deposit and savings accounts.[1]
2. 2
Add up the value of all of the cash and cash equivalents. On the balance sheet, find the value of the cash and cash equivalents. Suppose, for example, at the end of the prior year, the company had \$800,000 in cash. In addition, it had money market funds worth \$2,500,000 and CDs worth \$1,500,000. Finally, there were savings accounts worth \$1,200,000.
• Add all of these amounts together to determine the ending cash balance for the prior year.
• \$800,000 (cash) + \$2,500,000 (money market funds) + \$1,500,000 (CDs) + \$1,200,000 (savings) = \$6,000,000 (prior year ending balance).
3. 3
Establish the beginning cash balance for the current year. The ending balance from the prior year becomes the beginning balance for the current year. Using the above example, the ending balance from the prior year was \$6,000,000. Use this as the beginning balance for the current year.
• The beginning balance of cash and cash equivalents for the current year is \$6,000,000.
### Part 2 of 4: Calculating Cash Generated from Operations
1. 1
Start with net income. Net income is total revenues less operating expenses, depreciation, amortization and taxes. It is the company’s profit for the year. It includes all of the money that is left over after expenses have been paid. It is found on the company’s income statement.[2]
• The company in the above example reported a net income of \$8,000,000.
2. 2
Adjust for depreciation and amortization. Depreciation and amortization are non-cash expenses that record the decrease in value of assets over time.[3] They are calculated based on the original value of the asset and its useful life. But, since these expenses do not require an expenditure or receipt of cash, the amounts must be added back to the cash balances.[4]
• The company in the above example reported \$4,000,000 in depreciation and amortization expenses. As a consequence, \$4,000,000 would be added back to the cash balance.
3. 3
Make adjustments for accounts payable and accounts receivable. Accounts payable is money the company owes to pay its creditors. Accounts receivable is money owed to the company for goods and services. For the income statement, accruals for accounts payable and accounts receivable are entered for the time period in which they occurred, whether or not cash has actually been paid or received. However, these accruals are non-cash transactions, so they must be adjusted for the statement of cash flows.[5]
• Make sure to check the balance sheet for accrued liability accounts, such as Accrued Taxes or Accrued Payroll. These are expenses that will occur in the future, but that are not cash expenses right now. However, you will still need to adjust for these on the statement of cash flows. On the other hand, if you have any Prepaid Assets on the balance sheet, then these are expenses that have already been paid but that have not been incurred. You do not need to adjust these.
• The Accounts receivable balance at the end of the prior year is the beginning balance of the current year. For example, imagine that the beginning balance was \$6 million. At the end of the period, the accounts receivable balance is \$8 million, an increase of \$2 million during the year. Accounts receivable is income that has been earned, but not transferred into cash.
• Accordingly, an increase in AR during the period means that the company has used cash during the year to finance its sales and requires deducting the increase from the cash balance. A decrease in AR means that customers have paid down the amounts previously owed and requires adding the decrease back in the cash balance.
• For the company in the above example, net change to accounts receivable was \$2,000,000. The money is still owed by customers, but it has not been paid. So, this must be subtracted.
• Net change to accounts payable was \$1,000,000. This is money the company owes but has not yet paid. So this must be added.
4. 4
Calculate net cash generated from operations. Start with net income. Add back in depreciation and amortization expense. Reverse accruals for accounts payable and accounts receivable.
• \$8 million (net income) + \$4 million (depreciation & Amortization expense) - \$2 million (increase in Accounts Receivable) + \$1 million (increase in Accounts Payable) = \$11 million (net cash generated from operations).
• The net cash flow provided by operating activities is \$11,000,000.
### Part 3 of 4: Calculating Cash Flows from Investing and Financing Activities
1. 1
Review investments in capital. Capital investments are all of the funds the company used to buy equipment that can produce goods or services.[6] When a company purchases equipment, it exchanges one asset (cash) for another asset (capital equipment). As a consequence, the purchase of the equipment is a use of cash. Similarly, if a company sold capital equipment, it would also be an exchange of one asset for another (receiving cash or an account receivable for the equipment). If a company purchases capital equipment with cash during the time period for which it is preparing the statement of cash flows, this outflow of cash must be included.[7]
2. 2
Determine the impact of financing activities. Financing activities include issuing and redemption of long and short term debt, issuing and retirement of stock and payment of stock dividends. These activities can have positive and negative effects on cash flow. Issuing debt and stock increase the company’s cash. Redeeming debt and paying stock dividends decreases cash.
3. 3
Make adjustments for investments and financing. Deduct cash paid for purchasing capital equipment. Subtract cash paid to redeem debt or pay dividends. Add in cash raised by issuing stock or new debt. Imagine that the example company performed the following transactions:
• They purchased new computer equipment and assembly line machinery for a total of \$4,000,000. This must be subtracted.
• They increased short term debt by \$500,000 and issued \$250,000 in stock. These must be added.
• Finally, they redeemed \$3,000,000 in long-term debt and paid \$2,000,000 in dividends. These must be subtracted.
• -\$4 million (equipment purchases for cash) + \$0.5 million (sale of debt for cash) + \$0.25 million (sale of stock for cash) - \$3 million (redemption of long-term debt) - \$2 million (payment of dividends) = \$8.25 million (reduction in cash during period due to investment and financing activities).
• The net adjustment to cash for investing and financing activities is -\$8,250,000.
### Part 4 of 4: Calculating Ending Cash and Cash Equivalents
1. 1
Determine the net increase or decrease to cash. This means figuring out if there was a net increase or decrease to cash for the current year. Start with total cash flows from operating activities. Add adjustments to cash flows for investment and financing activities. The end result is the total net increase or decrease to cash for the year.
• In the above example, net cash flow from operating activities was \$11,000,000.
• The net change to cash from investing and financing activities was -\$8,250,000.
• The net increase or decrease to cash is ${\displaystyle \11,000,000-\8,250,000=\2,750,000}$.
2. 2
Calculate ending cash and cash equivalents. Start with the ending cash balance from the prior year. Add the net increase or decrease to cash from the current year. The end result is the total ending cash and cash equivalents for this year.
• For the company in the above example, the ending cash balance from the prior year was \$6,000,000.
• The net increase or decrease to cash for the current year was \$2,750,000.
• The ending cash and cash equivalents for the current year is ${\displaystyle \6,000,000+\2,750,000=\8,750,000}$.
3. 3
Use the cash flow statement to evaluate the company’s financial health. The cash flow statement removes accounting methods such as accruals, depreciation and amortization. Therefore, it provides a more accurate statement of how cash is flowing in and out of the company. This allows investors to get a clear picture of the company’s earning power and operating success.[8]
• A net increase in cash usually means that the company is running its operations efficiently and responsibly managing its investing and financing activities.
• A net decrease in cash might indicate problems with the company’s operating, investing or financing activities. It would signal that the company needs to decrease expenses somewhere in order to improve its financial health.
• Keep in mind that cash flow analysis is only a small part of analyzing a company's financial health. A net decrease in cash might also be concurrent with a major investment in the company's future growth. Similarly, a net increase in cash might reflect that management is getting lazy about reinvesting in the company.
## Community Q&A
Search
• Question
How do I adjust for WHT and VAT in cash flow statements?
Jill Newman, CPA
Certified Public Accountant
Jill Newman is a Certified Public Accountant in Ohio, with over 20 years of accounting experience. She received her CPA from the Accountancy Board of Ohio in 1994.
Certified Public Accountant
WHT is withholding tax, so if this is a liability on the balance sheet, it would be employees' withholding taxes that have not yet been paid to local/state/federal agencies. The taxes are taken out of the employees gross pay, which was a cash expense, so nothing needs to be added back. If it is the employer's portion of payroll tax, it would be a payable account and this would be added back because it was recorded as an expense but not yet paid. VAT is value added tax and this would be treated like any other expense if it was accrued or booked as a payable.
• Question
What can you do when ending cash is less than opening cash in the cash flow statement?
Jill Newman, CPA
Certified Public Accountant
Jill Newman is a Certified Public Accountant in Ohio, with over 20 years of accounting experience. She received her CPA from the Accountancy Board of Ohio in 1994.
Certified Public Accountant
If ending cash is less than opening cash, it means that the cash balance decreased through the year, which is a possibility. What matters is if the ending cash balance on the cash flow statement ties to the actual ending cash balance. This is how you will know if an error has been made.
• Question
Why are depreciation and amortization non-cash expenses? Why are they added back to cash balances?
Depreciation and amortization are non-cash expenses, because no physical cash went out and so it does not affect the cash account of a business. They are added back, because no cash went out despite the fact that it affected the net income.
• Question
Is taxation included in the statement of cash flows?
Yes, if the company paid tax on its profit because it is an outflow of cash from business to tax authorities.
• How can I account for cash and cash equivalents of two years comparatively?
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Certified Public Accountant
This article was co-authored by Jill Newman, CPA. Jill Newman is a Certified Public Accountant in Ohio, with over 20 years of accounting experience. She received her CPA from the Accountancy Board of Ohio in 1994.
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Updated: March 29, 2019
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Categories: Accounting
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To prepare a statement of cash flows, find out how much money the company had last year by checking the prior year’s ending balance sheet. Then, add the company’s net income, which is its revenue minus its expenses, taxes, and the depreciation of its assets. Make sure you include the amount the company owes other, and what others owe the company. Record those debts now, even though they haven’t been paid yet. Add those figures up to get your ending cash. Keep reading for tips from our Accounting reviewer on including investing expenses and income.
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# Using a Microscope
In this worksheet, students will review their understanding of how to use a microscope, what can go wrong and how to measure specimens.
Key stage: KS 3
Curriculum topic: Biology: Structure and Function of Living Organisms
Curriculum subtopic: Cells and Organisation
Difficulty level:
### QUESTION 1 of 10
Microscopes have helped us, as humans, to find out so much more about our world and about ourselves than people living just a couple of hundred years ago could possibly have imagined. They truly have opened up a whole new world to us!
So, how does a microscope work? How does it help you to see tiny things? Join Ben as he learns how to use his microscope, and see whether you can help him to improve his techniques and maybe learn something new yourself along the way!
Here is a typical light microscope:
So why is it called a light microscope?
because it's not very heavy
because the specimens you look at are very shiny
because it uses light to shine on the specimens
Here's the microscope again:
Which part of the microscope magnifies the object you're looking at?
magnifying glass
lens
focusing knob
Ben has been given a tiny letter on a microscope slide that he wants to look at under the microscope. He will need to do things in the correct order if he wants to see anything. Here are four things that Ben needs to do - put them in the correct order by matching each operation to the correct number.
## Column B
1
focus the specimen by using the focusing knob
2
look from the side and move the lens down until it...
3
place the microscope slide on to the stage
4
turn on the light or angle the mirror so you can s...
On Ben's microscope his eyepiece lens has a magnification of x10.
His objective lens has a magnification of x20.
What is the TOTAL MAGNIFICATION he is using?
x20
x30
x200
Ben gets the tiny letter on his microscope slide into focus and it looks like this:
If he changes his x20 objective lens for a x40 lens, what will happen to the image of the letter?
it will get bigger
he won't be able to see anything
it will get smaller
Ben's friend Tom was having difficulty seeing anything at all down his microscope.
Here is a list of things that Tom might have got wrong. Tick the FIVE that would mean he couldn't see anything clearly down his microscope.
he forgot to put the slide on the microscope
he hasn't switched the microscope's light on
he hasn't got the specimen properly under the lens
his microscope isn't angled properly
the lenses are dirty
he hasn't focused correctly on the specimen
his magnification is too high
A microscope is brilliant at making tiny things visible.
Ben puts a little microscopic ruler under his microscope and this is what he sees:
The distance between each mark is 0.5mm. About how big is Ben's 'field of view' - that is, how far is it from one side to the other?
3.5mm
7mm
35mm
Ben was looking at a tiny beetle under his microscope and he used his ruler (marked in 0.5mm divisions) to measure how long it was. Look at the picture and work out the answer.
2mm
3mm
4mm
All living organisms are made of at least one little box-like structure - in most cases, millions of them! What are these little 'boxes' called?
unit
cube
cell
Ben wanted to use his microscope to look at some of those little 'boxes' from the inside of his cheek.
What's the best way for him to collect some of them to look at?
carefully scrape some off with a knife
wipe a cotton bud around the inside of his cheek
use his finger to get some from the inside of his mouth
• Question 1
Here is a typical light microscope:
So why is it called a light microscope?
because it uses light to shine on the specimens
EDDIE SAYS
The light microscope uses light to make the specimen (you are hoping to see) visible. The light can either come from an electric light attached to the microscope or it can be reflected light from a mirror. Either way, it's light that shines on the specimen - whereas with an eye-wateringly expensive electron microscope, it's a beam of atomic particles that makes the specimen visible.
• Question 2
Here's the microscope again:
Which part of the microscope magnifies the object you're looking at?
lens
EDDIE SAYS
The two lenses in the microscope contain small 'magnifying glasses' - that is convex lenses made of glass - which bend the light and so make objects appear larger than they really are.
• Question 3
Ben has been given a tiny letter on a microscope slide that he wants to look at under the microscope. He will need to do things in the correct order if he wants to see anything. Here are four things that Ben needs to do - put them in the correct order by matching each operation to the correct number.
## Column B
1
turn on the light or angle the mi...
2
place the microscope slide on to ...
3
look from the side and move the l...
4
focus the specimen by using the f...
EDDIE SAYS
Phew - that was a toughie! Having turned on the microscope light, or adjusted the little mirror, you will be able to see down the microscope, so next you pop the slide on to the stage (trying to get the specimen under the lens). Then you look from the side as you move the lens DOWN to just above the slide (because it's easy to break the slide with the lens) and then you look down the microscope and focus up, away from the slide, until the specimen comes into focus (if you're lucky!).
• Question 4
On Ben's microscope his eyepiece lens has a magnification of x10.
His objective lens has a magnification of x20.
What is the TOTAL MAGNIFICATION he is using?
x200
EDDIE SAYS
This isn't too hard: you multiply (times) the two lenses together. So that's 10 x 20 = x200.
• Question 5
Ben gets the tiny letter on his microscope slide into focus and it looks like this:
If he changes his x20 objective lens for a x40 lens, what will happen to the image of the letter?
it will get bigger
EDDIE SAYS
When Ben INCREASES his magnification, the letter will seem to INCREASE in size (it's been magnified!). Mind you, he probably won't be able to see much unless he changes the focus to get it nice and sharp!
• Question 6
Ben's friend Tom was having difficulty seeing anything at all down his microscope.
Here is a list of things that Tom might have got wrong. Tick the FIVE that would mean he couldn't see anything clearly down his microscope.
he hasn't switched the microscope's light on
he hasn't got the specimen properly under the lens
the lenses are dirty
he hasn't focused correctly on the specimen
his magnification is too high
EDDIE SAYS
Microscopes are always tricky to start off with - there's so much you need to get right to see anything! If Tom forgot to put the slide on or if he didn't angle the microscope, that wouldn't stop him seeing anything. However, without light he will see nothing, the specimen has to be under the lens and if he's way out of focus or his lens is too powerful it's going to be a struggle to see much.
• Question 7
A microscope is brilliant at making tiny things visible.
Ben puts a little microscopic ruler under his microscope and this is what he sees:
The distance between each mark is 0.5mm. About how big is Ben's 'field of view' - that is, how far is it from one side to the other?
3.5mm
EDDIE SAYS
OK, so how many gaps are there between the marks? 7, right? Each one's 0.5mm, so that's 7 x 0.5 = 3.5mm (or thereabouts!).
• Question 8
Ben was looking at a tiny beetle under his microscope and he used his ruler (marked in 0.5mm divisions) to measure how long it was. Look at the picture and work out the answer.
2mm
EDDIE SAYS
OK, so the beetle covers FOUR of the gaps, right? They are 0.5mm each, so that's 4 x 0.5 = 2mm. Simple!
• Question 9
All living organisms are made of at least one little box-like structure - in most cases, millions of them! What are these little 'boxes' called?
cell
EDDIE SAYS
They are cells - at least that is the name given to the tiny 'boxes' that every living thing is made of. Robert Hooke (who lived in the 1600s) was the first person to see them and thought that they looked like little prison cells.
• Question 10
Ben wanted to use his microscope to look at some of those little 'boxes' from the inside of his cheek.
What's the best way for him to collect some of them to look at?
wipe a cotton bud around the inside of his cheek
EDDIE SAYS
That knife\'s going to be painful! And his finger ... where\'s that been since he last washed it? Ugh!! No, a cotton bud is best - he can carefully wipe that around the inside of his cheek, smear it on to his microscope slide, and then pop the bud into some disinfectant (just in case!).
---- OR ----
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Do Now
1 / 14
# Do Now - PowerPoint PPT Presentation
6. 18. __. __. t. 45. 6. 42. __. __. 8. n. 21. x. __. __. 11. 44. Do Now. Find the missing value in each ratio. 1. = 2 . = 3 . =. t = 15. n = 56. x = 84. 10/28/2013 3-3 C Ratio and Rate Problems. 2. 4. __. __. =. 1. 2.
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Presentation Transcript
6
18
__
__
t
45
6
42
__
__
8
n
21
x
__
__
11
44
Do Now
Find the missing value in each ratio.
1. =
2. =
3.=
t = 15
n = 56
x = 84
### 10/28/2013 3-3 C Ratio and Rate Problems
2
4
__
__
=
1
2
To solve ratio and rate problems, we can use
• Equivalent ratios/ Proportions
• Multiplication and Division
in a club
in a club
?
2
_____
___
=
total
total
300
5
Example 1
Pine Hill Middle School has 300 students. In Zoey’s class, two out of five students belong to a club. Use this ratio to predict how many students at Pine Hill Middle School belong to a club.
We can predict that 120 students belong to a club.
Example 2
In a survey, the ratio of students who prefer soccer to baseball is 2 to 3. The number of people who prefer soccer is 36. How many people in the survey prefer baseball?
36
soccer
soccer
2
2
_____
=
___
___
baseball
baseball
?
3
3
×18
36
_____
=
54
×18
Or use equivalent fractions (proportions)
Since 2 18 = 36, multiply 3 by 18.
54 people in the survey prefer baseball.
Example 3
There are 57 ounces of dog biscuits in 5 boxes. At this rate, how many ounces of dog biscuits are in 8 boxes?
÷5
11.4 ounces
______________
× 8 boxes = 91.2 ounces
1 box
57 ounces
57 ounces
11.4 ounces
? ounces
____________
____________
___________
___________
=
=
÷5
5 boxes
5 boxes
1 box
1 box
In a survey, seven out of ten people preferred apple pie over cherry pie. There are 250 people shopping at the grocery store. Use the survey to predict how many people in the store would prefer apple pie.
1.)
2.)
175 people
A recipe for lasagna calls for 15 ounces of ricotta cheese for 12 servings. How many ounces of cheese are needed for 20 servings?
25 ounces
LUNCH PERIODS A survey found that 5 out of every 8 students would prefer an earlier lunch period. If 120 students chose an earlier lunch on the survey, how many students took the survey?
Homework
Pg. 183 #1-22 Evens
1 Typing Test
Test on Wednesday
Quiz Corrections due next Monday
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Question
# A chemist needs to determine the concentration of a sulfuric acid solution by titration with a...
A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1294 M0.1294 M standard sodium hydroxide solution. He takes a 25.00 mL25.00 mL sample of the original acid solution and dilutes it to 250.0 mL.250.0 mL. Then, he takes a 10.00 mL10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 17.43 mL17.43 mL of the standard solution. What is the concentration of the original sulfuric acid solution?
H2SO4 + 2NaOH = Na2SO4 + 2H20
USING M1V1=M2V2,
M1 = X (TO BE CALCULATED)
V1 = 25ML
V2 = 250 ML
M2 = M1V1/V2
M2 = X*25/250
M2 = 0.1X
CONCENTRATION OF H2SO4 AFTER DILUTION IS 0.1X
GIVEN CONC. OF NaOH = 0.1294 M
NOW TITRATING THE SULPHURIC ACID WITH 0.1294M NaOH, END POINT REACHED AFTER USING 17.43 ML OF NaOH
M1 = 0.1X
M2 = 0.1294 M
V1 = 10 ML
V2 = 17.43 ML
AGAIN USING, M1V1 = M2V2
0.1X x 10 = 0.1294 x 17.43
X = 0.1294 x 17.43
= 2.255 M
THE CONCENTRATION OF THE ORIGINAL SULPHURIC ACID IS 2.255 M
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# Applied Calculus (4th Edition) View more editions Solutions for Chapter 8.3
• 3203 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 2
The estimates for of daily catch for the fishing data is tabulated as below:
t (tons of fish) (fraction of fishing days) 2 0 3 0.10 4 0.24 5 0.42 6 0.64 7 0.85 8 1
The objective is to approximate median daily catch for this data.
• Step 2 of 2
As the median is the middle value of the range values, the median is about half the fishes are to be sold.
So, plot the points for the above data and join them with a smooth curve as shown in the below figure:
To find median of the data, draw a horizontal line at which is the middle value of the range values. Then draw a vertical line from the point on the curve. The line touches the axis at about 5.35.
Therefore, the median daily catch for this fishing data is
Corresponding Textbook
Applied Calculus | 4th Edition
9780470170526ISBN-13: 0470170522ISBN:
Alternate ISBN: 9780470170533, 9780470615348, 9780470927731
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Proposition: Complex Numbers are a Field Extension of Real Numbers
Let $$(\mathbb R,\oplus,\odot)$$ be the field of real numbers and let $$(\mathbb C, + ,\cdot)$$ be the field of complex numbers. We define a function. $f:=\cases{\mathbb R\to\mathbb C,\\ x\to (x,0).}$
Then $$f$$ is an bijective field homomorphism, i.e. for all $$a,b\in\mathbb R$$, we have
$\begin{array}{rcl} f(a\oplus b)&=&f(a) + f(b),\\ f(a\odot b)&=&f(a) \cdot f(b). \end{array}$
The following statements are equivalent:
• "The field of real numbers is embedded in the field of complex numbers."
• "The field of complex numbers is a field extension of the field of real numbers".
Proofs: 1
Chapters: 1
Definitions: 2 3
Proofs: 4 5 6
Thank you to the contributors under CC BY-SA 4.0!
Github:
References
Bibliography
1. Timmann, Steffen: "Repetitorium der Funktionentheorie", Binomi-Verlag, 2003
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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• Index
• » Help Me !
• » !!!!help Me Please - Word Problem Chemistry Related!!!!
## #1 2013-02-18 19:33:21
Stella198392
Guest
### !!!!help Me Please - Word Problem Chemistry Related!!!!
A chemist has 500ml of a salt solution that contains 2% salt in units of g/cm squared. How much water must be added to reduce the concentration to 0.5%
SORRY
Guest
Its G/ CM CUBED
## #3 2013-02-21 13:15:59
John E. Franklin
Star Member
Offline
### Re: !!!!help Me Please - Word Problem Chemistry Related!!!!
(49/50)(500ml)(3) should work. The reason is that if 49 parts water out of 50 make 98% water, then that's 2% salt by weight.
so therefore you want to quadruple the water to go from 2 to 1 to 1/2 of a percent salt. So add 3 more times the 49 parts.
Am i right?
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
## #4 2013-02-21 18:17:02
anonimnystefy
Real Member
Online
### Re: !!!!help Me Please - Word Problem Chemistry Related!!!!
Actually, I think 1500ml must be added.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #5 2013-02-22 16:31:54
bob bundy
Moderator
Offline
### Re: !!!!help Me Please - Word Problem Chemistry Related!!!!
Hi
Mrs b agrees with 1500 & with g per cm cubed.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
• Index
• » Help Me !
• » !!!!help Me Please - Word Problem Chemistry Related!!!!
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Require Export algebra.cmra. Require Import algebra.functor. (* COFE *) Section cofe. Context {A : cofeT}. Inductive option_dist : Dist (option A) := | option_0_dist (x y : option A) : x ={0}= y | Some_dist n x y : x ={n}= y → Some x ={n}= Some y | None_dist n : None ={n}= None. Existing Instance option_dist. Program Definition option_chain (c : chain (option A)) (x : A) (H : c 1 = Some x) : chain A := {| chain_car n := from_option x (c n) |}. Next Obligation. intros c x ? n i ?; simpl; destruct (decide (i = 0)) as [->|]. { by replace n with 0 by lia. } feed inversion (chain_cauchy c 1 i); auto with lia congruence. feed inversion (chain_cauchy c n i); simpl; auto with lia congruence. Qed. Instance option_compl : Compl (option A) := λ c, match Some_dec (c 1) with | inleft (exist x H) => Some (compl (option_chain c x H)) | inright _ => None end. Definition option_cofe_mixin : CofeMixin (option A). Proof. split. * intros mx my; split; [by destruct 1; constructor; apply equiv_dist|]. intros Hxy; feed inversion (Hxy 1); subst; constructor; apply equiv_dist. by intros n; feed inversion (Hxy n). * intros n; split. + by intros [x|]; constructor. + by destruct 1; constructor. + destruct 1; inversion_clear 1; constructor; etransitivity; eauto. * by inversion_clear 1; constructor; apply dist_S. * constructor. * intros c n; unfold compl, option_compl. destruct (decide (n = 0)) as [->|]; [constructor|]. destruct (Some_dec (c 1)) as [[x Hx]|]. { assert (is_Some (c n)) as [y Hy]. { feed inversion (chain_cauchy c 1 n); try congruence; eauto with lia. } rewrite Hy; constructor. by rewrite (conv_compl (option_chain c x Hx) n); simpl; rewrite Hy. } feed inversion (chain_cauchy c 1 n); auto with lia congruence; constructor. Qed. Canonical Structure optionC := CofeT option_cofe_mixin. Global Instance Some_ne : Proper (dist n ==> dist n) (@Some A). Proof. by constructor. Qed. Global Instance is_Some_ne n : Proper (dist (S n) ==> iff) (@is_Some A). Proof. inversion_clear 1; split; eauto. Qed. Global Instance Some_dist_inj : Injective (dist n) (dist n) (@Some A). Proof. by inversion_clear 1. Qed. Global Instance None_timeless : Timeless (@None A). Proof. inversion_clear 1; constructor. Qed. Global Instance Some_timeless x : Timeless x → Timeless (Some x). Proof. by intros ?; inversion_clear 1; constructor; apply timeless. Qed. End cofe. Arguments optionC : clear implicits. (* CMRA *) Section cmra. Context {A : cmraT}. Instance option_validN : ValidN (option A) := λ n mx, match mx with Some x => ✓{n} x | None => True end. Global Instance option_empty : Empty (option A) := None. Instance option_unit : Unit (option A) := fmap unit. Instance option_op : Op (option A) := union_with (λ x y, Some (x ⋅ y)). Instance option_minus : Minus (option A) := difference_with (λ x y, Some (x ⩪ y)). Lemma option_includedN n (mx my : option A) : mx ≼{n} my ↔ n = 0 ∨ mx = None ∨ ∃ x y, mx = Some x ∧ my = Some y ∧ x ≼{n} y. Proof. split. * intros [mz Hmz]; destruct n as [|n]; [by left|right]. destruct mx as [x|]; [right|by left]. destruct my as [y|]; [exists x, y|destruct mz; inversion_clear Hmz]. destruct mz as [z|]; inversion_clear Hmz; split_ands; auto; cofe_subst; eauto using cmra_includedN_l. * intros [->|[->|(x&y&->&->&z&Hz)]]; try (by exists my; destruct my; constructor). by exists (Some z); constructor. Qed. Lemma None_includedN n (mx : option A) : None ≼{n} mx. Proof. rewrite option_includedN; auto. Qed. Lemma Some_Some_includedN n (x y : A) : x ≼{n} y → Some x ≼{n} Some y. Proof. rewrite option_includedN; eauto 10. Qed. Definition option_cmra_mixin : CMRAMixin (option A). Proof. split. * by intros n [x|]; destruct 1; constructor; repeat apply (_ : Proper (dist _ ==> _ ==> _) _). * by destruct 1; constructor; apply (_ : Proper (dist n ==> _) _). * destruct 1 as [[?|] [?|]| |]; unfold validN, option_validN; simpl; intros ?; auto using cmra_validN_0; eapply (_ : Proper (dist _ ==> impl) (✓{_})); eauto. * by destruct 1; inversion_clear 1; constructor; repeat apply (_ : Proper (dist _ ==> _ ==> _) _). * intros [x|]; unfold validN, option_validN; auto using cmra_validN_0. * intros n [x|]; unfold validN, option_validN; eauto using cmra_validN_S. * intros [x|] [y|] [z|]; constructor; rewrite ?associative; auto. * intros [x|] [y|]; constructor; rewrite 1?commutative; auto. * by intros [x|]; constructor; rewrite cmra_unit_l. * by intros [x|]; constructor; rewrite cmra_unit_idempotent. * intros n mx my; rewrite !option_includedN;intros [|[->|(x&y&->&->&?)]];auto. do 2 right; exists (unit x), (unit y); eauto using cmra_unit_preservingN. * intros n [x|] [y|]; rewrite /validN /option_validN /=; eauto using cmra_validN_op_l. * intros n mx my; rewrite option_includedN. intros [->|[->|(x&y&->&->&?)]]; [done|by destruct my|]. by constructor; apply cmra_op_minus. Qed. Definition option_cmra_extend_mixin : CMRAExtendMixin (option A). Proof. intros n mx my1 my2; destruct (decide (n = 0)) as [->|]. { by exists (mx, None); repeat constructor; destruct mx; constructor. } destruct mx as [x|], my1 as [y1|], my2 as [y2|]; intros Hx Hx'; try (by exfalso; inversion Hx'; auto). * destruct (cmra_extend_op n x y1 y2) as ([z1 z2]&?&?&?); auto. { by inversion_clear Hx'. } by exists (Some z1, Some z2); repeat constructor. * by exists (Some x,None); inversion Hx'; repeat constructor. * by exists (None,Some x); inversion Hx'; repeat constructor. * exists (None,None); repeat constructor. Qed. Canonical Structure optionRA := CMRAT option_cofe_mixin option_cmra_mixin option_cmra_extend_mixin. Global Instance option_cmra_identity : CMRAIdentity optionRA. Proof. split. done. by intros []. by inversion_clear 1. Qed. Lemma op_is_Some mx my : is_Some (mx ⋅ my) ↔ is_Some mx ∨ is_Some my. Proof. destruct mx, my; rewrite /op /option_op /= -!not_eq_None_Some; naive_solver. Qed. Lemma option_op_positive_dist_l n mx my : mx ⋅ my ={n}= None → mx ={n}= None. Proof. by destruct mx, my; inversion_clear 1. Qed. Lemma option_op_positive_dist_r n mx my : mx ⋅ my ={n}= None → my ={n}= None. Proof. by destruct mx, my; inversion_clear 1. Qed. Lemma option_updateP (P : A → Prop) (Q : option A → Prop) x : x ~~>: P → (∀ y, P y → Q (Some y)) → Some x ~~>: Q. Proof. intros Hx Hy [y|] n ?. { destruct (Hx y n) as (y'&?&?); auto. exists (Some y'); auto. } destruct (Hx (unit x) n) as (y'&?&?); rewrite ?cmra_unit_r; auto. by exists (Some y'); split; [auto|apply cmra_validN_op_l with (unit x)]. Qed. Lemma option_updateP' (P : A → Prop) x : x ~~>: P → Some x ~~>: λ y, default False y P. Proof. eauto using option_updateP. Qed. Lemma option_update x y : x ~~> y → Some x ~~> Some y. Proof. rewrite !cmra_update_updateP; eauto using option_updateP with congruence. Qed. Lemma option_update_None `{Empty A, !CMRAIdentity A} : ∅ ~~> Some ∅. Proof. intros [x|] n ?; rewrite /op /cmra_op /validN /cmra_validN /= ?left_id; auto using cmra_empty_valid. Qed. End cmra. Arguments optionRA : clear implicits. (** Functor *) Instance option_fmap_ne {A B : cofeT} (f : A → B) n: Proper (dist n ==> dist n) f → Proper (dist n==>dist n) (fmap (M:=option) f). Proof. by intros Hf; destruct 1; constructor; apply Hf. Qed. Instance option_fmap_cmra_monotone {A B : cmraT} (f: A → B) `{!CMRAMonotone f} : CMRAMonotone (fmap f : option A → option B). Proof. split. * intros n mx my; rewrite !option_includedN. intros [->|[->|(x&y&->&->&?)]]; simpl; eauto 10 using @includedN_preserving. * by intros n [x|] ?; rewrite /cmra_validN /=; try apply validN_preserving. Qed. Definition optionC_map {A B} (f : A -n> B) : optionC A -n> optionC B := CofeMor (fmap f : optionC A → optionC B). Instance optionC_map_ne A B n : Proper (dist n ==> dist n) (@optionC_map A B). Proof. by intros f f' Hf []; constructor; apply Hf. Qed. Program Definition optionF (Σ : iFunctor) : iFunctor := {| ifunctor_car := optionRA ∘ Σ; ifunctor_map A B := optionC_map ∘ ifunctor_map Σ |}. Next Obligation. by intros Σ A B n f g Hfg; apply optionC_map_ne, ifunctor_map_ne. Qed. Next Obligation. intros Σ A x. rewrite /= -{2}(option_fmap_id x). apply option_fmap_setoid_ext=>y; apply ifunctor_map_id. Qed. Next Obligation. intros Σ A B C f g x. rewrite /= -option_fmap_compose. apply option_fmap_setoid_ext=>y; apply ifunctor_map_compose. Qed.
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https://republicofsouthossetia.org/question/maria-put-trim-around-a-banner-that-is-the-shape-of-a-triangle-each-side-is-21-inches-long-maria-16245539-93/
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## Maria put trim around a banner that is the shape of a triangle. Each side is 21 inches long. Maria has 3 4 foot of trim left. What was
Question
Maria put trim around a banner that is the shape of a triangle. Each side is 21 inches long. Maria has 3 4 foot of trim left. What was the length of the trim when she started? Enter your answer in yards.
in progress 0
2 months 2021-10-12T01:45:59+00:00 1 Answer 0 views 0
1. We have been given that Maria put trim around a banner that is the shape of a triangle. Each side is 21 inches long.
The amount of trim will be equal to perimeter of triangle. The perimeter of given triangle will be 3 times each side length that is inches.
Now we need to convert 63 inches into feet.
12 inches = 1 feet
63 inches = feet = 5.25 feet.
We are also told that Maria has foot of trim left, so total length of trim would be trim used plus trim left that is feet.
Since we need to find length of trim in yards, so we will convert 6 feet into yards.
3 feet = 1 yard
6 feet = yards = 2 yards
Therefore, the length of the trim was 2 yards.
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https://www.convert-measurement-units.com/convert+Volt-ampere+reactive+to+Boiler+horsepower.php
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Convert var to Boiler horsepower (Volt-ampere reactive to Boiler horsepower)
## Volt-ampere reactive into Boiler horsepower
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Volt-ampere+reactive+to+Boiler+horsepower.php
## How many Boiler horsepower make 1 Volt-ampere reactive?
1 Volt-ampere reactive [var] = 0.000 101 910 828 025 48 Boiler horsepower - Measurement calculator that can be used to convert Volt-ampere reactive to Boiler horsepower, among others.
# Convert Volt-ampere reactive to Boiler horsepower (var to Boiler horsepower):
1. Choose the right category from the selection list, in this case 'Power'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Volt-ampere reactive [var]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Boiler horsepower'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '236 Volt-ampere reactive'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Volt-ampere reactive' or 'var'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Power'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '32 var to Boiler horsepower' or '38 var into Boiler horsepower' or '19 Volt-ampere reactive -> Boiler horsepower' or '74 var = Boiler horsepower' or '79 Volt-ampere reactive to Boiler horsepower' or '5 Volt-ampere reactive into Boiler horsepower'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(13 * 47) var'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '236 Volt-ampere reactive + 708 Boiler horsepower' or '73mm x 49cm x 35dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 9.999 999 909 ×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 9.999 999 909. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 9.999 999 909 E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 999 999 990 900 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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