Mathematics: Match each operation involving f(x) and g(x) to its answer

Match each operation involving f(x) and g(x) to - id1732495, destinystanley3794

Ответ:

gagem1278

30.12.2021

Mathematics

(g-f) (-1)= sqrt(15)

(f/g)(-1)= 0

(g+f)(2)=sqrt(3)-3

(g*f)(2)=-3*sqrt(3)

Step-by-step explanation:

We have to eval the expressions given in the point indicated.

Lets start by the first equation

(g-f)(-1)= g(-1) - f(-1)= $\sqrt{11-4*(-1)} - 1 +(-1)^{2}$ = $\sqrt{15}$

Now, lest continue with the others

(f/g)(-1)= f(-1)/g(-1)= (1-1)/sqrt(15)=0

(g+f)(2)=g(2)+f(2)=sqrt(3)-3

(g*f)(2)=g(2)*f(2)=sqrt(3)*(-3)=-3sqrt(3)

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Ответ:

lovemusic4

30.12.2021

Mathematics

(g-f) (-1)= $\sqrt{15}$

(f/g)(-1)= 0

(g+f)(2) = $\sqrt{3}-3$

(g*f)(2) = $-3 \sqrt{3}$

Step-by-step explanation:

Given :

$f(x)=1-x^2\\g(x)=\sqrt{11-4x}$

To find : (g-f)(-1)

Solution :

$g(x)-f(x)=\sqrt{11-4x}-1+x^2\\(g-f)(x)=\sqrt{11-4x}-1+x^2\\(g-f)(-1)=\sqrt{11+4}-1+1=\sqrt{15}$

To find : $\left ( \frac{f}{g} \right )(-1)$

Solution :

$\frac{f(x)}{g(x)}=\frac{1-x^2}{\sqrt{11-4x}}\\\frac{f(-1)}{g(-1)}=\frac{1-1}{\sqrt{11+4}}=0$

To find : (g+f)(2)

Solution:

$g(x)+f(x)=\sqrt{11-4x}+1-x^2\\(g+f)(x)=\sqrt{11-4x}+1-x^2\\(g+f)(2)=\sqrt{11-8}+1-4=\sqrt{3}-3$

To find: $(g\times f)(2)$

Solution:

$g(x)f(x)=(gf)(x)=\sqrt{11-4x}(1-x^2)\\(gf)(2)=\sqrt{11-8}(1-4)=-3\sqrt{3}$

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Ответ:

milkshakegrande101

18.08.2022

Mathematics

-6,0 so D. hope this helps (:

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