inverse variations?

Question

determine c:

{(4, 3), (-2, c)} is a subset of {(x, y): y varies inversely as x^2}

how do i do this???

Answer 1

y = k / x ²
x² y = k
16 x 3 = k
k = 48

y = 48 / x ²
y = 48 / (- 2 ) ²
y = 48 / 4
y = 12
Thus c = 12

Answer 2

First determine the relationship.

y = a/x²

Solve for a by plugging in the first point.

3 = a/4² = a/16
a = 3*16 = 48

y = 48/x²

Now solve for c.

c = 48/(-2)² = 48/4 = 12

Answer 3

Inverse relationship means the product is a constant. Therefore,
y*x^2 = constant
3*4^2 = c*(-2)^2
Divide by 4,
c = 12

Answer 4

y varies inversely as x^2 so y = k/x^2

when y = 3, x = 4, so...
3 = k/4^2
k = 3(4^2) = 48

y = 48/x^2

when x=-2...

y = 48/-2^2 = 48/4 = 12

c = 12

Hope this helps! :) :)

Answer 5

c = 12
-------
When y varies inversely as x^2 our implicit equation is

y = k / (x^2)

From your first ordered pair 3 = k/(4^2).
Solve for the constant k.
Use k to determine c.

(Your equation will be c = k / [(-2)^2] )

Answer 6

y = k/(x^2)

3 = k/(4^2)
3 = k/16
k = 48

c = 48/((-2)^2)
c = 48/4
c = 12

ANS : c = 12