Question

which expression is equivalent to $2\sqrt{25y^8} \cdot 4\sqrt{6y^3}$, if $y > 0$?\
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a. $17y^2\sqrt{6y^5}$\
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b. $8y^2\sqrt{150y}$\
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c. $40y^5\sqrt{6y}$\
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d. $14y^5\sqrt{6y^2}$

Explanation:

Step1: Simplify each radical term

First, simplify $\sqrt{25y^8}$: since $25=5^2$ and $y^8=(y^4)^2$, we get $\sqrt{25y^8}=5y^4$.
Then simplify $\sqrt{6y^3}$: rewrite $y^3$ as $y^2 \cdot y$, so $\sqrt{6y^3}=y\sqrt{6y}$.

Substitute back into the original expression:
$2 \cdot 5y^4 \cdot 4 \cdot y\sqrt{6y}$

Step2: Multiply the coefficients

Calculate the product of the numerical coefficients:
$2 \times 5 \times 4 = 40$

Step3: Multiply the variable terms

Use the exponent rule $y^a \cdot y^b = y^{a+b}$ for the $y$ terms:
$y^4 \cdot y = y^{4+1}=y^5$

Step4: Combine all terms

Combine the coefficient, variable term, and remaining radical:
$40y^5\sqrt{6y}$

Answer:

C. $40y^5\sqrt{6y}$