Answer:
She can create a function for the total savings in terms of the time x by adding the two functions together. The new function should look like this:
T(x)=S(x)+a(x)
So now we can substitute the given functions:
T(x)=450+6(x-2)
But this function can be simplified for it to be easier for her to calculate her total amount of savings. We can do that by distributing the 6 into the parenthesis. This is, multiply the 6 by both the x and the -2, so we get:
T(x)=450+6x-12
and now we can combine like terms. We can subtract 12 from 450 so we get:
T(x)=6x+438
And that will be our final function.
Answer:
probably would say PS would be your best
Sine 35 degrees = x / 20
x = sine 35 degrees * 20
x = .57358 * 20
x = 11.4716
Answer:
The answer is below
Step-by-step explanation:
The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.
Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.

The loan must be at least for 5.72 years for an annual interest rate (r) of 10%
Answer:
w(C(L))= 0.25 |L - 105|
Step-by-step explanation:
you plug in C(L) into L in w(L)