Question

y varies inversely as the square of x and the cube root of z, but inversely as the square root of w. variation equation if y=7 when x=3, z=27, and w=33

Answer 1

y varies directly as the square of x ----> \(y =k_1 x^2\)
y varies directly as the cube root of z---->\(y=k_2z^3\)
inversely as the square root of w -----> \(y=k_3/\sqrt w\)
combining all these 3 .
\(y=kx^2z^3/\sqrt w\)
gott his ?

Answer 2

You mean

\[y = k_2\sqrt[3]{z}\]

right?

Answer 3

oops, i missed the "root" part ...
y varies directly as the cube root of z---->\(y=k_2\sqrt[3]z\)
so,
\(y=kx^2\sqrt[3]z/\sqrt w\)

Answer 4

I'm with you so far. But don't I know have to solve for K, and plug that into my original equation?

Answer 5

yes, plug in y=7 when x=2, z=3, and w=25
in that equation!

Answer 6

And that's where it gets really funky for me.

Answer 7

ok, then take it one by one
x= 2, x^2 =.. ?

Answer 8

\[35 = 4K \sqrt[3]{3}\]

Answer 9

Is as far as I can seem to get.

Answer 10

ok, correct, now just isolate k
divide \(4\sqrt[3]3\) on both sides

Answer 11

if you need to rationalize the denominator, then use,
\(\dfrac{1}{\sqrt[3]3}=\dfrac{1}{3^{1/3}}=\dfrac{3^{-1/3}}{1}\)

Answer 12

Reviewing my notes for rationalizing denominators. BRB

Answer 13

i've used this rule
\(\large \dfrac{1}{x^n}=x^{-n}\)

Answer 14

um. I got \[\frac{ 140\sqrt[3]{3} }{ 48 }\]

Answer 15

Multiplied conjugates?

Answer 16

do you have choices ? then post them..

Answer 17

woops, 3 was supposed to be Z

Answer 18

Is that correct? No choices.

Answer 19

Ok thanks!

Answer 20

welcome ^_^

Answer 21

i missed the opportunity to welcome you here, sin you are new,so,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 22

thanks

Answer 23

go back to your recent question...