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Y varies inversely as the square of x. If y = 12 when x = 2, then what is x when y = 3?

User Amar
by
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1 Answer

3 votes
Remark
When you start out, you should do this problem by solving for k

Formula
y = k/x^2

Givens
y = 12
x = 2

Solve for k
12 = k /2^2
12 = k / 4 Multiply by 4
12*4 = k
48 = k

Problem
k = 48
y = 3
x = ??
y = k/x^2
3 = 48 / x^2 Multiply both sides by x^2

3x^2 = 48 Divide by 3
x^2 = 48/3 Take the square root of both sides.
x^2 = 16
sqrt(x^2) = sqrt(16)
x = +/- 4 Both work

If only 1 answer is permitted, use plus 4
User Basma
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7.1k points