H(x)=e^-x+3 Evaluate H(-2) H(3) H(5) Round nearest ten thousand

To evaluate the function H(x) = e^(-x) + 3, we substitute the given values into the function.

For H(-2):

$$H(-2) = e^{-(-2)} + 3$$
$$= e^2 + 3$$

For H(3):

$$H(3) = e^{-3} + 3$$

For H(5):

$$H(5) = e^{-5} + 3$$

To round to the nearest ten thousand, we need to evaluate the function to the fifth decimal place and round accordingly.

Let's calculate the values:

For H(-2):

$$H(-2) \approx e^2 + 3 \approx 7.38906 + 3 \approx 10.38906$$

For H(3):

$$H(3) \approx e^{-3} + 3 \approx 0.04979 + 3 \approx 3.04979$$

For H(5):

$$H(5) \approx e^{-5} + 3 \approx 0.00674 + 3 \approx 3.00674$$

Rounding to the nearest ten thousand:

For H(-2), rounded to the nearest ten thousandth, it is approximately 10.3891.

For H(3), rounded to the nearest ten thousandth, it is approximately 3.0498.

For H(5), rounded to the nearest ten thousandth, it is approximately 3.0067.

Answer:

H(-2) ≈ 10.3891
H(3) ≈ 3.0498
H(5) ≈ 3.0067