The function that has an inverse that is also a function will have no repeated y-values. The appropriate choice is the third one.
{(-1, 3), (0, 4), (1, 14), (5, 6), (7, 2)}
Answer: -3y + 8z
Step-by-step explanation:
2x -2x = 0
-2y + - y= -3y
5z + 3z= 8z
-3y + 8z
i don't know sorry this is hard to do
Ooh, fun
geometric sequences can be represented as
![a_n=a(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da%28r%29%5E%7Bn-1%7D)
so the first 3 terms are
![a_1=a](https://tex.z-dn.net/?f=a_1%3Da)
![a_2=a(r)](https://tex.z-dn.net/?f=a_2%3Da%28r%29)
![a_2=a(r)^2](https://tex.z-dn.net/?f=a_2%3Da%28r%29%5E2)
the sum is -7/10
![\frac{-7}{10}=a+ar+ar^2](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3Da%2Bar%2Bar%5E2)
and their product is -1/125
![\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B125%7D%3D%28a%29%28ar%29%28ar%5E2%29%3Da%5E3r%5E3%3D%28ar%29%5E3)
from the 2nd equation we can take the cube root of both sides to get
![\frac{-1}{5}=ar](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D%3Dar)
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
![\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7Bar%7D%7Br%7D%2Bar%2B%28ar%29r)
subsituting -1/5 for ar
![\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7B%5Cfrac%7B-1%7D%7B5%7D%7D%7Br%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%28%5Cfrac%7B-1%7D%7B5%7D%29r)
which simplifies to
![\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7B-1%7D%7B5r%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%5Cfrac%7B-r%7D%7B5%7D)
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
![x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
so
for 2r²-5r+2=0
a=2
b=-5
c=2
![r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B-%28-5%29%20%5Cpm%20%5Csqrt%7B%28-5%29%5E2-4%282%29%282%29%7D%7D%7B2%282%29%7D)
![r=\frac{5 \pm \sqrt{25-16}}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B25-16%7D%7D%7B4%7D)
![r=\frac{5 \pm \sqrt{9}}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B9%7D%7D%7B4%7D)
![r=\frac{5 \pm 3}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%203%7D%7B4%7D)
so
![r=\frac{5+3}{4}=\frac{8}{4}=2](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%2B3%7D%7B4%7D%3D%5Cfrac%7B8%7D%7B4%7D%3D2)
or
![r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5-3%7D%7B4%7D%3D%5Cfrac%7B2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D)
use them to solve for the value of a
![\frac{-1}{5}=ar](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D%3Dar)
![\frac{-1}{5r}=a](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5r%7D%3Da)
try for r=2 and 1/2
![a=\frac{-1}{10}](https://tex.z-dn.net/?f=%20a%3D%5Cfrac%7B-1%7D%7B10%7D)
or
![a=\frac{-2}{5}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-2%7D%7B5%7D)
test each
for a=-1/10 and r=2
a+ar+ar²=
![\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B10%7D%2B%5Cfrac%7B-2%7D%7B10%7D%2B%5Cfrac%7B-4%7D%7B10%7D%3D%5Cfrac%7B-7%7D%7B10%7D)
it works
for a=-2/5 and r=1/2
a+ar+ar²=
![\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B5%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%5Cfrac%7B-1%7D%7B10%7D%3D%5Cfrac%7B-7%7D%7B10%7D)
it works
both have the same terms but one is simplified
the 3 numbers are
![\frac{-2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B5%7D)
,
![\frac{-1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D)
, and
Answer:
20
Step-by-step explanation: