so, we'll end up with two parts, one being 4 times larger than the other, namely if the other is 1 slice, then the large part is 4 slices, meaning both parts will be on a 4:1 ratio.
well, simply enough, we can simply split 90 in (4+1) pieces, and then distribute accordingly to each part
![\bf \cfrac{\textit{large part}}{\textit{small part}}\qquad \qquad \stackrel{ratio}{4:1}\qquad \cfrac{4\cdot \frac{90}{4+1}}{1\cdot \frac{90}{4+1}}\implies \cfrac{4\cdot 18}{1\cdot 18}\implies \boxed{\cfrac{72}{18}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Blarge%20part%7D%7D%7B%5Ctextit%7Bsmall%20part%7D%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7Bratio%7D%7B4%3A1%7D%5Cqquad%20%5Ccfrac%7B4%5Ccdot%20%5Cfrac%7B90%7D%7B4%2B1%7D%7D%7B1%5Ccdot%20%5Cfrac%7B90%7D%7B4%2B1%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%2018%7D%7B1%5Ccdot%2018%7D%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B72%7D%7B18%7D%7D)
Answer:
Step-by-step explanation:
Let the fund raised in dollars be represented by R.
Let the number of lawns aerated during the day be represented by l.
The team has decided to charge $35/lawn for city-sized lots. This means that for l lawns, the fund raised would be
R = 35l
Since they were able to aerate 26 lawns in one day and the rental cost for two aerating machines is $215, then the amount if money that they were able to raise is
(35 × 26) - 215
= 910 - 215 = $695
The amount of money that the team will lose if they were unable to aerate due to weather conditions is the rental cost for two aerating machines which is $215
The only way they could avoid this loss is by taking note of the weather before renting the machine.
Answer:
River
Step-by-step explanation:
ok dear
River can run the water fast but can't walk
Answer:
slope=-3
y-intercept=6
x-intercept=2
Step-by-step explanation:
Check the forward differences of the sequence:
22 - 8 = 14
50 - 22 = 28 = 2*14
106 - 50 = 56 = 4*14
218 - 106 = 112 = 8*14
442 - 218 = 224 = 16*14
The differences are the products of increasing powers of 2 and 14:
![a_2-a_1=14\cdot2^0](https://tex.z-dn.net/?f=a_2-a_1%3D14%5Ccdot2%5E0)
![a_3-a_2=14\cdot2^1](https://tex.z-dn.net/?f=a_3-a_2%3D14%5Ccdot2%5E1)
![a_4-a_3=14\cdot2^2](https://tex.z-dn.net/?f=a_4-a_3%3D14%5Ccdot2%5E2)
and so on, with
![a_n-a_{n-1}=14\cdot2^{n-2}](https://tex.z-dn.net/?f=a_n-a_%7Bn-1%7D%3D14%5Ccdot2%5E%7Bn-2%7D)
![\implies a_n=a_{n-1}+7\cdot2^{n-1}](https://tex.z-dn.net/?f=%5Cimplies%20a_n%3Da_%7Bn-1%7D%2B7%5Ccdot2%5E%7Bn-1%7D)
Then the sequence has the recursive definition,
![\boxed{\begin{cases}a_1=8\\a_n=a_{n-1}+7\cdot2^{n-1}&\text{for }n>1\end{cases}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bcases%7Da_1%3D8%5C%5Ca_n%3Da_%7Bn-1%7D%2B7%5Ccdot2%5E%7Bn-1%7D%26%5Ctext%7Bfor%20%7Dn%3E1%5Cend%7Bcases%7D%7D)