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3 years ago

Given that f(x) = 2x + 5 and g(x) = x − 7, solve for f(g(x)) when x = −3. (1 point)

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Answer:

-15

Step-by-step explanation:

Given : $f(x) = 2x + 5$

$g(x) = x-7$

To Find: solve for f(g(x)) when x = −3

Solution:

$f(x) = 2x + 5$

$g(x) = x-7$

$f(g(x)) = 2(x-7)+ 5=2x-14+5=2x-9$ ---1

Now we are supposed to find f(g(x)) when x = −3

So, substitute x = -3 in 1

$f(g(-3)) =2(-3)-9$

$f(g(-3)) =-6-9$

$f(g(-3)) =-15$

Hence f(g(x)) when x = −3 is -15.

qaws [65]3 years ago

7 0

F(g(x))= 2(g(x))+5 = 2(x-7)+5 = 2x-14+5 = 2x-9
when x= -3
f(g(-3))= 2(-3)-9 = -6-9 = -15

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2 years ago

Fill in Sin, Cos, and tan ratio for angle x. <br> Sin X = 4/5 (28/35 simplified)

Fantom [35]

Answer:

Given: $\sin(x) = (4/5)$ .

Assuming that $0 < x < 90^{\circ}$ , $\cos(x) = (3/5)$ while $\tan(x) = (4/3)$ .

Step-by-step explanation:

By the Pythagorean identity $\sin^{2}(x) + \cos^{2}(x) = 1$ .

Assuming that $0 < x < 90^{\circ}$ , $0 < \cos(x) < 1$ .

Rearrange the Pythagorean identity to find an expression for $\cos(x)$ .

$\cos^{2}(x) = 1 - \sin^{2}(x)$ .

Given that $0 < \cos(x) < 1$ :

$\begin{aligned} &\cos(x) \\ &= \sqrt{1 - \sin^{2}(x)} \\ &= \sqrt{1 - \left(\frac{4}{5}\right)^{2}} \\ &= \sqrt{1 - \frac{16}{25}} \\ &= \frac{3}{5}\end{aligned}$ .