Answer:
4 and 5 hope this helps
Step-by-step explanation:
![\bf tan(x^o)=1.11\impliedby \textit{taking }tan^{-1}\textit{ to both sides} \\\\\\ tan^{-1}[tan(x^o)]=tan^{-1}(1.11)\implies \measuredangle x=tan^{-1}(1.11)](https://tex.z-dn.net/?f=%5Cbf%20tan%28x%5Eo%29%3D1.11%5Cimpliedby%20%5Ctextit%7Btaking%20%7Dtan%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Atan%5E%7B-1%7D%5Btan%28x%5Eo%29%5D%3Dtan%5E%7B-1%7D%281.11%29%5Cimplies%20%5Cmeasuredangle%20x%3Dtan%5E%7B-1%7D%281.11%29)
plug that in your calculator, make sure the calculator is in Degree mode
Answer:
sqrt166, 13, 14
Step-by-step explanation:
Answer:
<em>B. The graph of g is the graph of f shifted 2 units down</em>
Step-by-step explanation:
<u>Graph of Functions</u>
We have two functions:
f(x)=3^x
g(x)=3^x-2
Since g(x)=f(x)-2 it will be represented as an identical graph as that for f(x), but vertically displaced 2 units down. Let's check it by plugging some points
f(0)=3^0=1
g(0)=3^0-2=-1
f(1)=3^1=3
g(1)=3^1-2=1
f(3)=3^3=27
g(3)=3^3-2=25
We can notice the values of g(x) are always 2 units below f(x), thus the correct answer is
B. The graph of g is the graph of f shifted 2 units down