It’s BD because it need to be the opposite ray and this one is the opposite ray so yes this is it
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• Function: f(x) = 3x + 12.
A. Finding the inverse of f.
The composition of f with its inverse results in the identity function:
(f o g)(x) = x
f[ g(x) ] = x
3 · g(x) + 12 = x
3 · g(x) = x – 12
x – 12
g(x) = ⸺⸺
3
x g(x) = ⸺ – 4 <——— this is the inverse of f.
3________
B. Verifying that the composition of f and g gives us the identity function:
•

![\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Df%5Cbig%5Bg%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D3%5Ccdot%20%5Cleft%28%5Cdfrac%7Bx%7D%7B3%7D-4%5Cright%29%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%5Cdfrac%7Bx%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-3%5Ccdot%204%2B12%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx-12%2B12%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
and also
•

![\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3Dg%5Cbig%5Bf%28x%29%5Cbig%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7Bf%28x%29%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%3D%5Cdfrac%7B3x%2B12%7D%7B3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7B%5Cdiagup%5Chspace%7B-7%7D3%5Ccdot%20%28x%2B4%29%7D%7B%5Cdiagup%5Chspace%7B-7%7D3%7D-4%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%2B4-4%7D%5C%5C%5C%5C%0A%5Cmathsf%7B%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
________
C. Since f and g are inverse, then
f(g(– 2))
= (f o g)(– 2)
=
– 2 <span>✔
</span>
• Call h the compositon of f and g. So,
h(x) = (f o g)(x)
h(x) = x
As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).
I hope this helps. =)
To approximate the distance of points with three dimension, make use of the equation,
d = sqrt ((x2 - x1)^2 + (y2 - y1)^2 +(z2 - z1)^2)
Substituting all the data from the points given,
<span>d = sqrt ((2 - -2)^2 + (-7 - 3)^2 +(4 - -5)^2) = sqrt 197
</span>
Thus, the distance from the points is approximately 14.04 and that is letter D.
https://jhmsmath.weebly.com/uploads/2/7/7/5/27756807/8th_grade_countdown.pdf go to this link page 52-54