Find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation. If y
varies inversely as the square of x, and y=1/8 when x=1 find y when x=5
1 answer:
Y=k/x²; substitute x=1 and y=1/8: 1/8=k. So y=1/(8x²)
When x=5, y=1/(8×25)=1/200.
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