Question

If h is directly proportional to x^2, but inversely proportional to y, and h= 9/8 when x=1/2 and y=1/3, what is the value of h when x=1/3 and y=1/2?

Answer 1

@Directrix @zzr0ck3r

Answer 2

hmm

Answer 3

I tried it, and got it wrong

Answer 4

h = kx^2/y
(9/8) = k* (1/4)*3
k = (9/8)*(4/3) = 3/2
so
h = (3/2)x^2/y
x=1/3 and y=1/2
h = (3/2)(1/9)*2 = 4/3
is that a choice?

Answer 5

@sakigirl

Answer 6

@sakigirl --> This is for you to work. See attachment.

http://tinyurl.com/kbcny58

Answer 7

I posted a problem for @sakigirl to do - a bona fide SAT practice problem when she returns with the correct answer to the previous problem.

Answer 8

c

Answer 9

@sakigirl While you were sleeping, you were asked a question:

h = kx^2/y
(9/8) = k* (1/4)*3
k = (9/8)*(4/3) = 3/2
so
h = (3/2)x^2/y
x=1/3 and y=1/2
h = (3/2)(1/9)*2 = 4/3
is that a choice?
That is from @zzr0ck3r

Answer 10

So, is 4/3 right or not?

Answer 11

@zzr0ck3r No , it isn't :(

Answer 12

hmm im not sure what the question is asking. I am not sure if they want the same proportional constant