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Question
- if \\(x = 5\\) and \\(y = -2\\), then \\(\frac{10y}{x} + 3y^2 =\\)
- 32
- 8
- -16
- \\(-\frac{8}{5}\\)
- 16
Substitute the given values into the expression
$$
\frac{10(-2)}{5} + 3(-2)^2
$$
Evaluate the first term
$$
\frac{-20}{5} = -4
$$
Evaluate the second term and sum the results
$$
3(4) = 12
$$
$$
-4 + 12 = 8
$$
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- (A) 32
- (B) 8 (Correct answer)
- (C) -16
- (D) \(-\frac{8}{5}\)
- (E) 16