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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OpenStudy (freckles):
inversely think divides
x=k/v
where k is the constant of variation
OpenStudy (freckles):
use (x,v)=(48,8) to find the constant k
OpenStudy (freckles):
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
OpenStudy (aaronandyson):
v = k/x
8 = k/48
8*48 = k
OpenStudy (aaronandyson):
k = 384
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OpenStudy (aaronandyson):
Now,
v = k/x
Taking v as 64 and k as 384
v = k/x
64 = 384/x
64x = 384
x = 384/64
OpenStudy (anonymous):
Okay. I see what you are saying.
OpenStudy (aaronandyson):
64 times 6 is equal to 384
OpenStudy (aaronandyson):
Therefore, x would 6.
OpenStudy (aaronandyson):
Getting it?
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OpenStudy (anonymous):
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.
OpenStudy (aaronandyson):
x = y^2/k
OpenStudy (aaronandyson):
2 = 10^2/k
OpenStudy (aaronandyson):
2 = 100/k
2k = 100
k = 100/2
k = 50
OpenStudy (anonymous):
It would be 50
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OpenStudy (aaronandyson):
I think I made a mistake.
OpenStudy (anonymous):
I am understanding better. But my answer choices are a. X=16 b. X=5 c. X=80 d. X=100
OpenStudy (aaronandyson):
What are your options?
OpenStudy (anonymous):
Those are my options.
OpenStudy (aaronandyson):
Wait!
Its
x = k/y^2
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OpenStudy (aaronandyson):
x is 4 and y is 10
OpenStudy (aaronandyson):
4 = k/10^2
4 = k/100
4*100 = k
400 = k
OpenStudy (aaronandyson):
Now, we have to find x but y is given as 2..
Can you find x ??
k = 400 and y = 2
OpenStudy (anonymous):
So, do I divide?
OpenStudy (aaronandyson):
x = k/y^2
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OpenStudy (aaronandyson):
k is 400 and y is 2
OpenStudy (anonymous):
So, my answer is d?
OpenStudy (aaronandyson):
Yes absolutely right...
Good Job..
OpenStudy (anonymous):
Thank you!!!!
OpenStudy (anonymous):
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.
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OpenStudy (anonymous):
A. 6889 square feet
B. 6885 square feet
C. 110,224 square feet
D. 27,556 square feet
OpenStudy (aaronandyson):
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
OpenStudy (anonymous):
So, it is A and I was right the first time?
OpenStudy (aaronandyson):
Yes...!
OpenStudy (anonymous):
Thank you! I am understanding much much better.
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