Answer:
1. loans
2. credit
3. interest
4. total value
5.
6.
7. compound
8. principal
9.
10.
Step-by-step explanation:
I don't know what 5, 6, 9, or 10 is
first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.
now, at -1, the graph doesn't cross the x-axis, instead it <u>simply bounces off</u> of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.
at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.
![\bf \begin{cases} x=-1\implies &x+1=0\\ x=3\implies &x-3=0 \end{cases}~\hspace{5em}\stackrel{\textit{even multiplicity}}{(x+1)^2}\qquad \stackrel{\textit{odd multiplicity}}{(x-3)^1}=\stackrel{y}{0} \\\\\\ (x^2+2x+1)(x-3)=y\implies x^3+2x^2+x-3x^2-6x-3=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-5x-3=y~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D-1%5Cimplies%20%26x%2B1%3D0%5C%5C%20x%3D3%5Cimplies%20%26x-3%3D0%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7D%5Cstackrel%7B%5Ctextit%7Beven%20multiplicity%7D%7D%7B%28x%2B1%29%5E2%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bodd%20multiplicity%7D%7D%7B%28x-3%29%5E1%7D%3D%5Cstackrel%7By%7D%7B0%7D%20%5C%5C%5C%5C%5C%5C%20%28x%5E2%2B2x%2B1%29%28x-3%29%3Dy%5Cimplies%20x%5E3%2B2x%5E2%2Bx-3x%5E2-6x-3%3Dy%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20x%5E3-x%5E2-5x-3%3Dy~%5Chfill)
Answer:
-9=-1/3-13 is it in slope intercept form, look up ones to put it in different forms
Step-by-step explanation:
Answer: false
Step-by-step explanation:
If f and g are increasing on I, this implies that f' > 0 on I and g' > 0 on I. That is both f' and g' have a positive slope. However,
Using product rule;
(fg)' = fd(g) + gd(f)
(fg)' = f * g' + f' * g
and although it is given that g' and f' are both positive we don't have any information about the sign of the values of the functions themselves(f and g). Therefore, if at least one of the functions has negative values there is the possibility that the derivative of the product will be negative. For example;
f = x, g = 5x on I = (-5, -2)
f' = 1 and g' =5 both greater than 0
f and g are both lines with positive slopes therefore they are increasing, but f * g = 5x^2 is decreasing on I.
