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Evaluate h(x)=12+x2
for x=−5

1 Answer

5 votes

First, you need to evaluate the function correctly.


\tt \: h(x) = 12 \: + {x}^(2)

  • when x = -5, we need to substitute -5 into the function

So ,


\tt \: h(-5) = 12 + (-5)^2

Now, solve the equation


\tt \: {h(-5)= 12 + (-5)²} \\ \tt \: {h(-5) = 12 + (25)} \\ \tt \: {h(-5) = \green{37}}

Therefore, when x is equal to -5, the value of h(x) is 37.

User Ffledgling
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