f(c)=3c-8 g(c)=sqrt(14-c) Evaluate. (f@g)(-11)=

Understand Function Composition: Understand the composition of functions. The composition of functions (f@g)(c) means that we first apply g to c, and then apply f to the result of g(c). So, (f@g)(c) = f(g(c)).Evaluate g(-11): Evaluate g(-11). We need to find the value of g(c) when c = -11. So, we calculate g(-11) = sqrt(14 - (-11)) = sqrt(14 + 11) = sqrt(25).Simplify g(-11): Simplify g(-11). The square root of 25 is 5, so g(-11) = 5.Evaluate f(g(-11)): Evaluate f(g(-11)). Now we need to apply f to the result of g(-11), which is 5. So, we calculate f(5) = 3*5 - 8.Simplify f(5): Simplify f(5). Multiplying 3 by 5 gives us 15, and subtracting 8 from 15 gives us 7. So, f(5) = 15 - 8 = 7.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

f(c)=3c-8

g(c)=sqrt(14-c)
Evaluate.

(f@g)(-11)=

$f(c)=3 c-8$ $\newline$ $g(c)=\sqrt{14-c}$ $\newline$ Evaluate. $\newline$ $(f \circ g)(-11)=$

View step-by-step help

Home

Math Problems

Algebra 1

Domain and range of square root functions: equations

Full solution

f(c)=3 c-8

\newline

g(c)=\sqrt{14-c}

\newline

Evaluate.

\newline

(f \circ g)(-11)=

Understand Function Composition: Understand the composition of functions. The composition of functions $(f \circ g)(c)$ means that we first apply $g$ to $c$ , and then apply $f$ to the result of $g(c)$ . So, $(f \circ g)(c) = f(g(c))$ .
Evaluate $g(-11)$ : Evaluate $g(-11)$ . We need to find the value of $g(c)$ when $c = -11$ . So, we calculate $g(-11) = \sqrt{14 - (-11)} = \sqrt{14 + 11} = \sqrt{25}$ .
Simplify $g(-11)$ : Simplify $g(-11)$ . The square root of $25$ is $5$ , so $g(-11) = 5$ .
Evaluate $f(g(-11))$ : Evaluate $f(g(-11))$ . Now we need to apply $f$ to the result of $g(-11)$ , which is $5$ . So, we calculate $f(5) = 3 \times 5 - 8$ .
Simplify $f(5)$ : Simplify $f(5)$ . $\newline$ Multiplying $3$ by $5$ gives us $15$ , and subtracting $8$ from $15$ gives us $7$ . So, $f(5) = 15 - 8 = 7$ .