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If
f(x)=(x^(2)-2)/(9), what is the value of
f(-10), to the nearest ten-thousandth (if necessary)?
Answer:

If $f(x)=\frac{x^{2}-2}{9}$ , what is the value of $f(-10)$ , to the nearest ten-thousandth (if necessary)? $\newline$ Answer:

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Find trigonometric functions using a calculator

Full solution

Q. If

f(x)=\frac{x^{2}-2}{9}

, what is the value of

f(-10)

, to the nearest ten-thousandth (if necessary)?

\newline

Answer:

Substitute $x$ with $-10$ : To find the value of $f(-10)$ , we need to substitute $x$ with $-10$ in the function $f(x)=\frac{x^{2}-2}{9}$ .
Calculate square of $-10$ : Substitute $x$ with $-10$ : $f(-10) = ((-10)^{2} - 2) / 9$ .
Substitute square into function: Calculate the square of $-10$ : $(-10)^{2} = 100$ .
Subtract $2$ from $100$ : Substitute the square of $-10$ into the function: $f(-10) = \frac{(100 - 2)}{9}$ .
Divide $98$ by $9$ : Subtract $2$ from $100$ : $100 - 2 = 98$ .
Round result to nearest ten-thousandth: Divide $98$ by $9$ : $98 / 9 \approx 10.8889$ .
Round result to nearest ten-thousandth: Divide $98$ by $9$ : $\frac{98}{9} \approx 10.88888888888889$ . Round the result to the nearest ten-thousandth: $10.88888888888889 \approx 10.8889$ .

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