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OpenStudy (anonymous):
y varies directly as the square of x. When x = 1, y = 8. Find x when y = 128. Answer x = ±8 x = ±16 x = ±4 x = ±6
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OpenStudy (anonymous):
y=kx^2 k=8 x^2 = 128/8 = 16 \[x = \pm4\]
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