If $f (x) = \sqrt{25 - x^{2}}$ $f(x) = sqrt(25-x^2)$ , what is the value of $f (3)$ $f(3)$ ?

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Answer 1 · 2016-01-29T16:33:49

4 :)

since $f (3) = \sqrt{25 - x^{2}}$ $f(3) = sqrt(25 - x^2)$ ,

we subtitute the value of x for 3 since f(3)
$f (3) = \sqrt{25 - 3^{2}}$ $f(3) = sqrt(25 - 3^2)$

then evaluate the terms inside the radical sign (square root)
$f (3) = \sqrt{25 - 9}$ $f(3) = sqrt(25 - 9)$
$f (3) = \sqrt{16}$ $f(3) = sqrt(16)$

then finalize the answer, since $4 \cdot 4 = 16$ $4*4 = 16$ , the square root of $16$ $16$ is 4, hence the answer is:
$f (3) = 4$ $f(3) = 4$

If f(x) = sqrt(25-x^2)f(x)=√25−x2f(x) = sqrt(25-x^2), what is the value of f(3)f(3)f(3)?