need some math help?

Question

Suppose that w varies directly as z2 and inversely as xy. When w = 10, x = 15, y = 2, and z = 5. Find z when w = 2, x = 8, and y = 27.

Answer 1

Let w = kz^2 / xy.
Then:
10 = 25k / 30 = 5k/6
k = 60/5 = 12

For the 2nd case:
z^2 = xyw / k
= 8 * 27 * 2 / 12
= 2 * 9 * 2
= 36
z = +/- 6.

Answer 2

lets break this down.
w varies directly as z^2 and inversly as xy so we can write w=k*z^2/xy where k is a constantsthat we have to figure out.
so you are told when w=10, x=15, y=2 and z=5. this is enough information to find the constnat k

lets find k then, w=10, z=5, x=15, y=2
so 10=k*5*5/(15*2) by substituting values
or 10=25k/30=5k/6 so
60=5k or k=12

now we know that w=12*z^2/xy

Now you are told find z when w=2, x=8 and y=27. So substitute values
2=12*z^2/(8*27)
16*27=12z^2
16*27/12=4*27/3=4*9=z^2 so if we take the square root to get z, z=2*3=6 or -6

z=6, z=-6 are the answers

Answer 3

6

Answer 4

Assuming the z2 is a typo and should be z:
w=Kz/xy

10=5K/30
K=300/5=60

so w=60z/xy

2=60z/(8)(27)
2(8)(27)/60 =z
z=7.2

if that z2 in your question means z squared the solution is a little different:
w=Kz^2/xy
10=25K/30
300/25=K
K=12

w=12z^2/xy
2=12z^2/(8)(27)
2(8)(27)/12 = z^2
z= sqrt(36)

z=6

Answer 5

w = kz^2/xy
when w = 10, x = 15, y = 2 and z = 5,
10 = k(5^2)/(15)(2)
300 = 25k
k = 12
w = 12z^2/xy
when w = 2, x = 8 and y = 27,
2 = 12(z^2)/(8)(27)
z^2 = 36
z = 6 or z =-6

Answer 6

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Answer 7

w=kz^2/xy
10 = 25k/30 --> k = 12
So 2 = 12z^2/(8*27)
z^2= 8*27*2/12 = 1296
z = +/- 36