Answer:
c bro okkkkkkkkkkkkkkmmmmmmmkkkkkkkkkbey
We are going to use the basic form of an exponential function:
![y=ab^x](https://tex.z-dn.net/?f=y%3Dab%5Ex)
to solve this problem.
Since we have tow points
![(x_{1},y_{1})](https://tex.z-dn.net/?f=%28x_%7B1%7D%2Cy_%7B1%7D%29)
and
![(x_{2},y_{2})](https://tex.z-dn.net/?f=%28x_%7B2%7D%2Cy_%7B2%7D%29)
, we are going to have a system of equations:
![y_{1}=ab^{x_{1}](https://tex.z-dn.net/?f=y_%7B1%7D%3Dab%5E%7Bx_%7B1%7D)
equation (1)
![y_{2}=ab^{x_{2}}](https://tex.z-dn.net/?f=y_%7B2%7D%3Dab%5E%7Bx_%7B2%7D%7D)
equation (2)
We know for our problem that
![x_{1}=2](https://tex.z-dn.net/?f=x_%7B1%7D%3D2)
,
![y_{1}=8](https://tex.z-dn.net/?f=y_%7B1%7D%3D8)
,
![x_{2}=5](https://tex.z-dn.net/?f=x_%7B2%7D%3D5)
, and
![y_{2}=512](https://tex.z-dn.net/?f=y_%7B2%7D%3D512)
, so lets replace those values in our equations:
![8=ab^2](https://tex.z-dn.net/?f=8%3Dab%5E2)
equation (1)
![512=ab^5](https://tex.z-dn.net/?f=512%3Dab%5E5)
equation (2)
Now, we just need to find
![a](https://tex.z-dn.net/?f=a)
and
![b](https://tex.z-dn.net/?f=b)
:
Dividing equation (2) by equation (1):
![\frac{512=ab^5}{8=ab^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B512%3Dab%5E5%7D%7B8%3Dab%5E2%7D%20)
![64=b^3](https://tex.z-dn.net/?f=64%3Db%5E3)
![b= \sqrt[3]{64}](https://tex.z-dn.net/?f=b%3D%20%5Csqrt%5B3%5D%7B64%7D%20)
![b=4](https://tex.z-dn.net/?f=b%3D4)
equation (3)
Replacing equation (3) in equation (1):
![8=a(4)^2](https://tex.z-dn.net/?f=8%3Da%284%29%5E2)
![8=16a](https://tex.z-dn.net/?f=8%3D16a)
![a= \frac{8}{16}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B8%7D%7B16%7D%20)
![a= \frac{1}{2}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B1%7D%7B2%7D%20)
Finally, we can put our function together:
![y=ab^x](https://tex.z-dn.net/?f=y%3Dab%5Ex)
![y= \frac{1}{2} (4^x)](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B2%7D%20%284%5Ex%29)
We can conclude that the exponential function whose graph passes through the points (2,8) and (5,512) is
![y= \frac{1}{2} (4^x)](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B2%7D%20%284%5Ex%29)
.
Find the missing digit of the following UPC code<br>0, 2, 8, 5, 0 ,0, 1, 1, 0, 7, 0 ■
olga nikolaevna [1]
It’s the square and the number woidl be 0
1) is D.......$37.15
2) and I got 20.30 for the other one
Answer:
B.
Step-by-step explanation:
Assuming the cost increased linearly is equivalent to assuming the cost increased at a constant rate. Because the cost of the clothing increased from $620 to $1,000 in 10 years, the constant rate is equal to
= ($1,000 - $620)/10
= $38 per year
Therefore, the cost of the family’s clothing in 1991 (6 years after 1985) is:
$620 + (6 years)($38 per year)
= $620 + $228
= $848