well, if that function f(x) were to be continuos on all subfunctions, that means that whatever value 7x + k has when x = 2, meets or matches the value that kx² - 6 has when x = 2 as well, so then 7x + k = kx² - 6 when f(2)
![f(x)= \begin{cases} 7x+k,&x\leqslant 2\\ kx^2-6&x > 2 \end{cases}\qquad \qquad f(2)= \begin{cases} 7(2)+k,&x\leqslant 2\\ k(2)^2-6&x > 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 7(2)+k~~ = ~~k(2)^2-6\implies 14+k~~ = ~~4k-6 \\\\\\ 14~~ = ~~3k-6\implies 20~~ = ~~3k\implies \cfrac{20}{3}=k](https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cbegin%7Bcases%7D%207x%2Bk%2C%26x%5Cleqslant%202%5C%5C%20kx%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20f%282%29%3D%20%5Cbegin%7Bcases%7D%207%282%29%2Bk%2C%26x%5Cleqslant%202%5C%5C%20k%282%29%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207%282%29%2Bk~~%20%3D%20~~k%282%29%5E2-6%5Cimplies%2014%2Bk~~%20%3D%20~~4k-6%20%5C%5C%5C%5C%5C%5C%2014~~%20%3D%20~~3k-6%5Cimplies%2020~~%20%3D%20~~3k%5Cimplies%20%5Ccfrac%7B20%7D%7B3%7D%3Dk)
Answer:4
Step-by-step explanation:
The reference angle is
pi/3.
At 180 it would be 9pi/3 so going past that is 10pi/3.
Answer:
x = 1178 games
Step-by-step explanation:
Let the number of games = x
Let the total cost = Tc
Let the total revenue = Tr
Given the following data;
Investment = $10,000
Cost of each game = $1.50
Selling cost = $9.99
Total cost, Tc = (Cost of each game * Number of games) + Investment
Tc = 1.50x + 10000
Total revenue, Tr = Selling cost * Number of games
Tr = 9.99x
Breakeven point is when total cost is equal to total revenue;
Tc = Tr
x = 1177.86 ≈ 1178 games.
<em>Therefore, the number of games that must be sold before the business breaks even is 1178 games. </em>