The f(x)=2^x and g(x)=f(x+k). If k=-5, what can be concluded about the graph of g(x).
2 answers:
Answer:
A and C
Step-by-step explanation:
so the final g(x) would be y=2^x+5
A because of the +5
C because there are no parenthesis and therefor cant be a horizontal shift
![\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B10em%7D%5Ctextit%7Bfunction%20transformations%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%20A%28%20Bx%2B%20C%29%5E2%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%5Csqrt%7B%20Bx%2B%20C%7D%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%28%5Cmathbb%7BR%7D%29%5E%7B%20Bx%2B%20C%7D%2B%20D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%5Ccfrac%7B1%7D%7BA%28Bx%2BC%29%7D%2BD%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D%20A%20sin%5Cleft%28%20B%20x%2B%20C%20%5Cright%29%2B%20D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20stretches%20or%20shrinks%20horizontally%20by%20%7D%20A%5Ccdot%20B%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20flips%20it%20upside-down%20if%20%7D%20A%5Ctextit%7B%20is%20negative%7D)


now, with that template in mind, let's take a look

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Both have the same denominator
Remove the denominator
5=n
Answer:
B
r=9.7
C=60.95
Step-by-step explanation:
Given diameter of 19.4
r = d/2
19.4/2
r=9.7
C= 2(pi)r
2(pi)(9.7)
60.95 = C
The answer is c I hope that helps
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