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Question #2c599

1 Answer

Apr 23, 2017

So we know that $\sqrt[3]{x} = c \times y^{2}$ $root 3x=cxxy^2$ where $c =$ $c=$ a constant.

Explanation:

Plug in what we know:

$\sqrt[3]{27} = c \times 3^{2} \to 3 = c \times 9 \to c = \frac{3}{9} = \frac{1}{3}$ $root3 27=cxx3^2->3=cxx9->c=3/9=1/3$

Now the new value of $y$ $y$ :

$\sqrt[3]{x} = \frac{1}{3} \times {1.5}^{2} = \frac{1}{3} \times 2.25 = 0.75 = \frac{3}{4}$ $root3x=1/3xx1.5^2=1/3xx2.25=0.75=3/4$

$\to x = {(\frac{3}{4})}^{3} = \frac{27}{64} \approx 0.42$ $->x=(3/4)^3=27/64~~0.42$

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