There's more than one way to combine them really
but an obvious one will be
![\bf \begin{array}{llll} h(x)&=&(f\circ g)(x)\\\\ &&\sqrt[3]{7x+1}\\\\ &&\sqrt[3]{g(x)}\leftarrow \begin{array}{llll} f(x)=\sqrt[3]{x }\\\\ g(x)=7x+1 \end{array} \end{array}\\\\ -----------------------------\\\\ (f\circ g)(x)\iff f[\quad g(x)\quad ]=\sqrt[3]{g(x)}\implies f[\quad g(x)\quad ]=\sqrt[3]{7x+1}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0Ah%28x%29%26%3D%26%28f%5Ccirc%20g%29%28x%29%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7B7x%2B1%7D%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cleftarrow%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7Bx%20%7D%5C%5C%5C%5C%0Ag%28x%29%3D7x%2B1%0A%5Cend%7Barray%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28f%5Ccirc%20g%29%28x%29%5Ciff%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cimplies%20%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7B7x%2B1%7D)
First we have to determine 30% of 12.
(30×12)÷100=3.6 so 3.6 is 30% of 12.
12 - 3.6= 8.4
Your answer is 8.4
Let us first find how much he is spending every year, to do this let's find his monthly expense and multiply it by 12.
Every month Cameron spends 1040 + 980 + 120 = 2140
To find out how much he spends yearly, multiply the monthly value by 12,
2140 x 12 = 25680
This value is more than his net income so he clearly has a surplus, but to check we can subtract 129 from every month to get:
(1040 + 980 + 120 - 129) = 2011
2011 x 12 = 24132, which shows that his budget is saving 129 surplus every month. Choice B is correct.
8 x 2 + 6x + 5=0
16 + 6x + 5=0
6x+21=0
6x+21-21=0-21
6x=-21
6x/6=-21/6
x=-7/2
Answer:
Given
f(x)=18,000(.88)^x
We need to find the value of x when f(x) <2000
2000 = 18000(0.88)^x
0.88^x 2000/18000
0.88^x 1/9
• x = log (1/9) / log (0.88)
• x= 17 (rounded down)
After 17 years or during year 18 the car value will drop below $2000
