I need help please.. x varies inversely as v, and x=48 when v=8. Find x when v is 64. A. x=6 B. x=64 C. x=8 D. x=48

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I need help please.. x varies inversely as v, and x=48 when v=8. Find x when v is 64. A. x=6 B. x=64 C. x=8 D. x=48

Mathematics
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inversely think divides x=k/v where k is the constant of variation
use (x,v)=(48,8) to find the constant k
then once we obtain k we will go back to x=k/v where we know k now and then replace v with 64 and find x

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Other answers:

v = k/x 8 = k/48 8*48 = k
k = 384
Now, v = k/x Taking v as 64 and k as 384 v = k/x 64 = 384/x 64x = 384 x = 384/64
Okay. I see what you are saying.
64 times 6 is equal to 384
Therefore, x would 6.
Getting it?
Thank you. I understood well! It made more sense. Now, would this still imply to x varies inversely as y^2, & x=4 when y=10. Find x when y=2.
x = y^2/k
2 = 10^2/k
2 = 100/k 2k = 100 k = 100/2 k = 50
It would be 50
I think I made a mistake.
I am understanding better. But my answer choices are a. X=16 b. X=5 c. X=80 d. X=100
What are your options?
Those are my options.
Wait! Its x = k/y^2
x is 4 and y is 10
4 = k/10^2 4 = k/100 4*100 = k 400 = k
Now, we have to find x but y is given as 2.. Can you find x ?? k = 400 and y = 2
So, do I divide?
x = k/y^2
k is 400 and y is 2
So, my answer is d?
Yes absolutely right... Good Job..
Thank you!!!!
I think I did this one right. But I got a little confused. ~ You have 332 of fencing to enclose a rectangular region. What is the maximum area? I got a the first time and b the second time.
A. 6889 square feet B. 6885 square feet C. 110,224 square feet D. 27,556 square feet
2L + 2W = 332 and L x W = y so from eqn 1, L = 332/2 + W so (332/2 + W) x W = y 166W + W^2 = y when derivative of y = 0 = maximum area y' = 2W + 166 0 = 2W +166 -166 = 2W W = 83 therefore as 2L +2W =332 and w = 83, L must =83 also so L x W =A 83 x 83 = 6889 A
So, it is A and I was right the first time?
Yes...!
Thank you! I am understanding much much better.
:)

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