Problems with Transposing Equations

[heartywaffkes]

Hello. I hope this is in the right section. I understand the basics behind transposing equations (inverse operations to cancel out/balance the equation), and can do simple transposition problems, but when the equations become more complex, I'm not sure what steps to take. Here are two problems I've had difficulties with, including the steps I've taken:

1) A = πr[sqrt(r^2 + h^2)] Give in terms of h.

a) divide by πr -- A/πr = sqrt(r^2 + h^2)
b) square -- A^2/π^2r^2 = r^2 + h^2
c) subtract r^2 -- A^2/π^2r^2 - r^2 = h^2
d) square root -- h = sqrt(A^2/π^2r^2 - r^2)

Answer given: h = sqrt(A^2 - π^2r^4)/πr
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2) V = πh/6(3R^2 + h^2) Give in terms of R.

a) multiply by 6 -- 6V = πh(3R^2 + h^2)
b) divide by πh -- 6V/πh = 3R^2 + h^2
c) subtract h^2 -- 6V - h^2/πh = 3R^2
d) divide by 3 -- 6V - h^2/3πh = R^2
e) square root -- R = sqrt(6V - h^2/3πh)

Answer given: R = sqrt(6V - πh^3/3πh)​

I have the answers for these questions (listed above), and, in both cases, the answers I have are wrong. But, I don't know what I did wrong, or what I need to do to get the right answer. In the second question, I had the answer in front of me while I worked the problem to see if I could finally figure it out. Any help is appreciated. Thank you.

 

[tkhunny]

There is nothing WRONG with the first one. Get a little confidence from somewhere. The Answer Given is just in a little different form.

Try adding the terms inside the square root and then consider the denominator.

 

[mmm4444bot]

1) A = πr[sqrt(r^2 + h^2)] Give in terms of h.

Does the given instruction actually read "Give in terms of h"? :shock:

(Symbol A is already stated in terms of h. If they want you to solve the equation for h, then that is what they should say.)


It seems to me that their answer is incomplete. I get:

\(\displaystyle h = \frac{\sqrt{A^2 - Pi^2 r^4}}{\pi r} \; \text{or} \; h = -\frac{\sqrt{A^2 - Pi^2 r^4}}{\pi r}\)

 

[mmm4444bot]

2) v = πh/6(3r^2 + h^2) give in terms of r.

a) multiply by 6 -- 6v = πh(3r^2 + h^2)
b) divide by πh -- 6v/πh = 3r^2 + h^2
c) subtract h^2 -- 6v - h^2/πh = 3r^2 You may not subtract h^2 from the numerator; subtract it from the fraction, instead.

​

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