Question

y varies jointly as x and the inverse of the square of z. When y = 8 and x = 12, z = 6. Find the constant of variation and determine the value for z when y = 16 and x = 6.

Answer 1

wat grade are you in so i know what math textbook to look in

Answer 2

im a senior but this is an online math class making up for a math class I failed It is called Intregated Math or Math Topics

Answer 3

im sorry this problem im not sure i thought i had but it turned out 642352

Answer 4

@johnweldon1993 can you help with this one too??? ^_^

Answer 5

Of course :)

so varying jointly means multiply....and inversely...means divide...so

\[\large y = \frac{kx}{z^2}\]
Does that make sense? we have a 'k' in there because that is the constant of variation and is actually what we will be solving for first

Answer 6

Thanks :)
But i dont really understand

Answer 7

Not done yet :P

Answer 8

okay sweetheart lol

Answer 9

So the original equation for varying directly is

\[\large y = kx\]
and for varying inversely is
\[\large y = \frac{k}{x}\]

So we have
"y varies jointly as x" means

\[\large y = kx\]

and also we have "and the inverse of the square of z"

which means we have to combine that with the other one

\[\large y = \frac{kx}{z^2}\]

Answer 10

So now...

"When y = 8 and x = 12, z = 6. Find the constant of variation"

We plug all those numbers into that equation...

\[\large y = \frac{kx}{z^2}\]
will becomes
\[\large 8 = \frac{12k}{(6)^2}\]

Now we want to solve for 'k' can you ?

Answer 11

i think its 3?

Answer 12

So we have

\[\large 8 = \frac{12k}{36}\]multiply both sides by 36 and then divide by 12 to solve for 'k'

\[\large 8 \times 36 = 12k\]
\[\large 288 = 12k\]
\[\large k = \frac{288}{12}\]
\[\large k = 24\]

okay?

Answer 13

Dang I did that all wrong :/

Answer 14

so now we have our original equation again...but this time...instead of 'k' we write 24

\[\large y = \frac{24x}{z^2}\]

Answer 15

So for the second part of the problem
"determine the value for z when y = 16 and x = 6"

We have

\[\large y = \frac{24x}{z^2}\] and that will become\[\large 16 = \frac{24(6)}{z^2}\]

Can you solve that for 'z' ? :)

Answer 16

I think that was it

Answer 17

Nope ...not yet :P we want z = something...

\[\large 16 = \frac{24(6)}{z^2}\]

\[\large 16 = \frac{144}{z^2}\]
\[\large z^2 = \frac{144}{16}\]
\[\large z^2 = 9\]
\[\large z = \sqrt{9}\]
\[\large z = 3\]

make sense?

Answer 18

Wait I just realized before you said you thought it was 3!!! If that was the answer you got good job smarty pants ;)

Answer 19

lol Thanks you confused the heck out of me!

Answer 20

Aww sorry hun :( didnt mean to :( but that is the process you would use to solve it :)

Answer 21

Thanks so much sweetheart ^-^

Answer 22

Anytime beautiful ^_^