Let p be the prize of a pen and m the prize of a mechanical pencils. If you buy six pens and one mechanical pencil, you spend 6p+m. We know that this equals 9, because you get 1$ change from a 10$ bill.
Similarly, if you buy four pens and two mechanical pencils, you spend 4p+2m, which is 8$, because now you get a $2 change. Put these equation together in a system:

Now, if you multiply the first equation by 2, the system becomes

Subtract the second equation from the first:

Plug this value into the first equation to get

Answer:
x=7/2 x=7/3
Step-by-step explanation:
Answer:
In the first week, Micah bought 11 gallons of gas.
In the second week, Micah bought 8 gallons of gas.
Step-by-step explanation:
Given:
First week
x gallons of gas at $2.39 per gallon
x gallons = $2.39x
Second week
3 fewer gallons of gas than the first week at $2.49 per gallon
x - 3 gallons = $2.49(x-3)
Total spent = $46.21
$2.39x + $2.49(x-3) = $46.21
2.39x + 2.49x - 7.47 = 46.21
4.88x - 7.47 = 46.21
Add 7.47 to both sides
4.88x - 7.47 + 7.47 = 46.21 + 7.47
4.88x = 53.68
Divide both sides by 4.88
x = 53.68/4.88
= 11
x = 11
First week = x = 11 gallons
Second week
= x - 3
= 11-3
= 8 gallons
Therefore,
In the first week, Micah bought 11 gallons of gas.
In the second week, Micah bought 8 gallons of gas.
Answer:
a) ![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000)
b) f(t=2015) = 264,034,317.7
Step-by-step explanation:
The rate of change in the number of hospital outpatient visits, in millions, is given by:

a) To find the function f(t) you integrate f(t):
![\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdf%28t%29%7D%7Bdt%7Ddt%3Df%28t%29%3D%5Cint%20%5B0.001155t%28t-1980%29%5E%7B0.5%7D%5Ddt)
To solve the integral you use:

Next, you replace in the integral:

Then, the function f(t) is:
![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2BC%27)
The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.
Hence C' = 264,034,000
The function is:
![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000)
b) For t = 2015 you have:
![f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7](https://tex.z-dn.net/?f=f%28t%3D2015%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7D%282015%29%282015-1980%29%5E%7B1%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%282015-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000%5C%5C%5C%5Cf%28t%3D2015%29%3D264%2C034%2C317.7)