so we know that 24% of the students buy their lunch at the cafeteria, and 190 students brownbag.
well, 100% - 24% = 76%, so the remainder of the students, the one that is not part of the 24% is 76%, and we know that's 190 of them.
since 190 is 76%, how much is the 24%?
![\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 190&76\\ x&24 \end{array}\implies \cfrac{190}{x}=\cfrac{76}{24}\implies 4560=76x \\\\\\ \cfrac{4560}{76}=x\implies 60=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{total amount of students}}{190+60\implies 250}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20amount%26%5C%25%5C%5C%20%5Ccline%7B1-2%7D%20190%2676%5C%5C%20x%2624%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B190%7D%7Bx%7D%3D%5Ccfrac%7B76%7D%7B24%7D%5Cimplies%204560%3D76x%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4560%7D%7B76%7D%3Dx%5Cimplies%2060%3Dx%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btotal%20amount%20of%20students%7D%7D%7B190%2B60%5Cimplies%20250%7D)
Answer:
x + 5y = 0
Step-by-step explanation:
y = mx + b
y = -1/5 x + b
1 = -1/5 (-5) + b
1 = 1 + b
b = 0
y = -1/5 x
5y = -x
x + 5y = 0
1/9 + 1/2 = 1/x
Let x = length of time it will take both hoses to work together
Multiply both sides by 18x.
2x + 9x = 18
10x = 18
x = 18/10
x = 9/5
x = 1.8 hours
In other words, if both hoses work together, the pool can be filled in 1 hour and 48 minutes.
I believe the answer is 28.27, or 28.3
Degree of ploynominal is the largest exponent you find, which is here the 4x to the 6th. there are only two terms here.