first off, is noteworthy that's the graph of an exponential function, thus the function will be along the lines of g(x) = abˣ , now, what's "a" and "b" values?
well, let's take a peek when x = 0 and x = 1.
![\bf g(x) = ab^x \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 0\\ y = 1 \end{cases}\implies 1=ab^0\implies 1=a(1)\implies \boxed{1=a} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 1\\ y = 4 \end{cases}\implies 4 = ab^1\implies 4=1b^1\implies \boxed{4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill g(x) = 4^x\qquad \qquad \qquad \begin{array}{|c|c|ll} \cline{1-2} x&y\\ \cline{1-2} -2&\frac{1}{4^2}\to \frac{1}{16}\\ -1&\frac{1}{4}\\ 0&1\\ 1&4\\ 2&16\\ \cline{1-2} \end{array}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20g%28x%29%20%3D%20ab%5Ex%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20x%20%3D%200%5C%5C%20y%20%3D%201%20%5Cend%7Bcases%7D%5Cimplies%201%3Dab%5E0%5Cimplies%201%3Da%281%29%5Cimplies%20%5Cboxed%7B1%3Da%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20x%20%3D%201%5C%5C%20y%20%3D%204%20%5Cend%7Bcases%7D%5Cimplies%204%20%3D%20ab%5E1%5Cimplies%204%3D1b%5E1%5Cimplies%20%5Cboxed%7B4%3Db%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20g%28x%29%20%3D%204%5Ex%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7B%7Cc%7Cc%7Cll%7D%20%5Ccline%7B1-2%7D%20x%26y%5C%5C%20%5Ccline%7B1-2%7D%20-2%26%5Cfrac%7B1%7D%7B4%5E2%7D%5Cto%20%5Cfrac%7B1%7D%7B16%7D%5C%5C%20-1%26%5Cfrac%7B1%7D%7B4%7D%5C%5C%200%261%5C%5C%201%264%5C%5C%202%2616%5C%5C%20%5Ccline%7B1-2%7D%20%5Cend%7Barray%7D~%5Chfill)
Answer: The teachers will get 2 and 1/4 of a sandwich
Step-by-step explanation:
Answer:
D’=(3,3); F’=(-2,-2); E’= (-5,0)
Step-by-step explanation:
To find the reflection over the y axis (the line x=0), you just need to change the x either from positive to negactive or the opposite.
Answer:
5/6
Step-by-step explanation:
On a standard 6-sided die, there is only one 2. That means that there are 5 other numbers that it could land on (1, 3, 4, 5, 6).
Using that information, the probability of it not landing on a 2 is 5 out of 6 or 5/6. This is because you must do part over whole. The "part" in this situation is the 5 "wanted" numbers, and the "whole" is 6 because there are six potential numbers that it could land on.
I hope this helps.
