Question

@rawr7548

Answer 1

Solve.
1. 3/y-1/4=1/y
2. 15/x+15/x+2=2

Answer 2

Okay ya

Answer 3

\(\large \dfrac{3}{y} - \dfrac{1}{4} = \dfrac{1}{y}\)

Answer 4

your first question is like this, rgiht ?

Answer 5

Yep

Answer 6

To solve means, you need to isolate the variable \(y\)

Answer 7

start by multiplying thru the entire equation by \(4y\)

Answer 8

^that gets rid of the denominator

Answer 9

\(\large \dfrac{3}{y} - \dfrac{1}{4} = \dfrac{1}{y}\)


Multiply \(4y\) both sides :

\(\large 4*3 - y = 4\)

\(\large 12 - y = 4\)

Answer 10

solve \(y\)

Answer 11

Can you just show work and put the answer also

Answer 12

it would be easy if we can work together :) me showing u everything is not that efficient way lol

Answer 13

\(\large 12 - y = 4\)

knw how to solve this one step equation ?

Answer 14

No

Answer 15

seems you just want the answer ?

Answer 16

I just want all the work and the answer

Answer 17

fair enough :) did u understand the work so far ?

Answer 18

Yes but please and actually lets skip that for now Can i have these first

Answer 19

Divide.
1. 12y^3+8y-3/3 2. (2x^2 +3x-20) divided (x+4) 3. (x^4-16) divided (x-2)

Answer 20

Simplify. 4. 1/x + 1 all over/ 1/x^2 - 1 5. x + 3/y all over/ x- 2/y^2 6. x/x-y all over/x^2/ x^2-y^2

Answer 21

Show work with the answer please? :)

Answer 22

lol I'm not going to do all these problems for u

Answer 23

if you want I'll help U do one problem. otherwise, bye bye

Answer 24

pick any one problem which you want me walk u through

Answer 25

@rawr7548

Answer 26

\(\large U = 0.5CV^2\)

Answer 27

\(\large \dfrac{U}{0.5V^2} = C\)

Answer 28

\(\large \dfrac{U}{\dfrac{1}{2}V^2} = C\)

Answer 29

Solve k = 0.5mv2 for v.

\(\ \sf k = 0.5mv^2 \)

I got \(\ \sf \sqrt{\dfrac{k}{1/2m}} = v \)

\(\ \sf \dfrac{k}{m} \times \dfrac{2}{1} = \sqrt{2k/m} = v \)

Answer 30

looks good !

Answer 31

:O

Answer 32

@AkashdeepDeb

Answer 33

max value of \(f : (a+b)c\)
subject to \(g : a^2+b^2+c^2=1\)

At max/min values the gradient vectors of both functions are proportional :
\(\nabla f = \lambda \nabla g\)
\(\implies \langle c, c, a+b \rangle = \lambda \langle 2a, 2b, 2c\rangle\)

so the equations to sovle are :
\(c = 2a\lambda\)
\(c = 2b\lambda\)
\(a+b = 2c\lambda\)
\(a^2+b^2+c^2=1\)

solving above four equations gives :
\(a = b = \dfrac{1}{2}\)
\(c = \dfrac{1}{\sqrt{2}}\)

Answer 34

evaluate the function \((a+b)c\) at the point \((\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}})\)

Answer 35

\((\frac{1}{2}+ \frac{1}{2})* \frac{1}{\sqrt{2}}\)

Answer 36

\((\frac{2}{2})* \frac{1}{\sqrt{2}}\)


\(\frac{1}{\sqrt{2}}\)

Answer 37

http://www.wolframalpha.com/input/?i=solve+c+%3D+2a%5Clambda%2C+c+%3D+2b%5Clambda%2C+a%2Bb+%3D+2c%5Clambda%2C+a%5E2%2Bb%5E2%2Bc%5E2%3D1

Answer 38

hehe. I am going crazy over all the latex I'm having to type now. T_T

Thanks! :)

Answer 39

np :) good luck !!

Answer 40

@ParthKohli is very much familiar with this solution.... lets see if he has any ideas on doing this using number theory...

Answer 41

@ParthKohli here is the actual problem :
http://assets.openstudy.com/updates/attachments/5375021ee4b0e27c43135f32-akashdeepdeb-1400177195036-screenshot20140515at9.55.44pm.png

Answer 42

Umm, OK. So in this problem, we'll introduce this thing which will help us to exploit the conditions.

First note that the condition is given as \(a^2 + b^2 + c^2 =1.\) Rewrite it as \(a^2 + b^2 + c^2 - 1 = 0\).

You can write \(a^2 + b^2 + c^2 - 1\) as \(\lambda \times (a^2 + b^2 + c^2 - 1)\) where \(\lambda \) is any real number. It doesn't matter which one because that expression is zero and you can multiply it by anything.

You can write \(f(a,b,c) = (a + b)c\) as \((a + b)c + \lambda (a^2 + b^2 + c^2 - 1)\).

For some stupid reason, the partial derivative of each of \(a,b,c\) must evaluate to zero.

Answer 43

I don't really remember the gradients technique. :P

Answer 44

by that I meant the partial derivative of this thing wrt each a, b, c would evaluate to zero.

Answer 45

ganeshie is lying though. I'm not familiar with this technique at all :P

we discussed it once, and I suck at it.

Answer 46

lol no, i had reviewed so much stuff and learned a great deal on that day :)

Akash wants some solution without using calculus i think...
if it exists.... im also trying but it feels hopeless :(

Answer 47

I am fine with the multi calc proof.
I understood all of what parth did up there ^

Then what?

Answer 48

then you'll get some simultaneous linear equations.

Answer 49

hmm, there are three equations and four variables? I think the fourth equation would be taking the partial derivative wrt \(\lambda\)

Answer 50

Partial derivative is different from taking a simple derivative yeah?

Answer 51

Yeah. You just assume that the others are constants.

Answer 52

What is the stupid reason btw ?

Answer 53

Ganesh knows the details, but it is just like how in a single-variable function, the maximum/minimum always has the slope of the tangent = 0 or derivative = 0.

Answer 54

So all the derivatives of f(a,b,c) must be zero, to maximize it?

Answer 55

Pretty much.