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If
f(x)=(-3)/(-20+sqrtx), what is the value of
f(3), to the nearest tenth (if necessary)?
Answer:

If $f(x)=\frac{-3}{-20+\sqrt{x}}$ , what is the value of $f(3)$ , to the nearest tenth (if necessary)? $\newline$ Answer:

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Evaluate natural logarithms

Full solution

Q. If

f(x)=\frac{-3}{-20+\sqrt{x}}

, what is the value of

f(3)

, to the nearest tenth (if necessary)?

\newline

Answer:

Substitute $x$ with $3$ : To find the value of $f(3)$ , we need to substitute $x$ with $3$ in the function $f(x) = \frac{-3}{-20+\sqrt{x}}$ . $\newline$ So, $f(3) = \frac{-3}{-20+\sqrt{3}}$ .
Calculate square root of $3$ : First, calculate the square root of $3$ . $\sqrt{3}$ is approximately $1.732$ .
Substitute $\sqrt{3}$ : Now, substitute $\sqrt{3}$ into the function. $f(3) = \frac{-3}{-20+1.732}.$
Perform addition inside parentheses: Next, perform the addition inside the parentheses. $-20 + 1.732 = -18.268$ .
Divide $-3$ by $-18.268$ : Now, divide $-3$ by $-18.268$ . $f(3) = \frac{-3}{-18.268} = \frac{3}{18.268}.$
Simplify fraction and round: Finally, we simplify the fraction and, if necessary, round to the nearest tenth. $\newline$ $\frac{3}{18.268}$ is approximately $0.164$ to three decimal places. $\newline$ Rounded to the nearest tenth, this is $0.2$ .

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