f(n)=sqrt(10 n) g(n)=n^(2)-n Evaluate. (f@g)(10)=

Understand Composition of Functions: Understand the composition of functions. The composition of two functions, (f@g)(n), means that we first apply g to n, and then apply f to the result of g(n). So, (f@g)(10) means we need to find g(10) first and then apply f to that result.Calculate g(10): Calculate g(10). We have g(n) = n^2 - n. So, g(10) = 10^2 - 10 = 100 - 10 = 90.Apply f to g(10): Apply f to the result of g(10). Now we need to apply f to 90, since we found that g(10) = 90. The function f(n) is defined as f(n) = sqrt(10n). So, f(90) = sqrt(10 * 90).Calculate f(90): Calculate f(90). f(90) = sqrt(10 * 90) = sqrt(900) = 30.Conclude Final Answer: Conclude with the final answer. Therefore, the value of (f@g)(10) is 30.

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f(n)=sqrt(10 n)

g(n)=n^(2)-n
Evaluate.

(f@g)(10)=

$f(n)=\sqrt{10 n}$ $\newline$ $g(n)=n^{2}-n$ $\newline$ Evaluate. $\newline$ $(f \circ g)(10)=$

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Algebra 1

Domain and range of square root functions: equations

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f(n)=\sqrt{10 n}

\newline

g(n)=n^{2}-n

\newline

Evaluate.

\newline

(f \circ g)(10)=

Understand Composition of Functions: Understand the composition of functions. The composition of two functions, $f\circ g)(n)\, means that we first apply \$g$ to $n$ , and then apply $f$ to the result of $g(n)$ . So, $f\circ g)(10)\ means we need to find \$g(10)$ first and then apply $f$ to that result.
Calculate $g(10)$ : Calculate $g(10)$ . We have $g(n) = n^2 - n$ . So, $g(10) = 10^2 - 10 = 100 - 10 = 90$ .
Apply $f$ to $g(10)$ : Apply $f$ to the result of $g(10)$ . Now we need to apply $f$ to $90$ , since we found that $g(10) = 90$ . The function $f(n)$ is defined as $f(n) = \sqrt{10n}$ . So, $f(90) = \sqrt{10 \times 90}$ .
Calculate $f(90)$ : Calculate $f(90)$ . $f(90) = \sqrt{10 \times 90} = \sqrt{900} = 30$ .
Conclude Final Answer: Conclude with the final answer. $\newline$ Therefore, the value of $(f@g)(10)$ is $30$ .