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: a²+b² = -99/2
Step-by-step explanation:
Since we are given two equations, this equations will be solved simultaneously to get a² and b²
a³ - 3ab² = 47 ... 1
b³ - 3a² b = 52... 2
From 1, a(a² - 3b²) = 47...3
From 2, b(b² - 3a²) = 52... 4
Adding 3 and 4, we have;
a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)
a²+b²-(3b²+3a²) = 99
(a²+b²) -3(b²+a²)= 99
Taking the difference we have
- 2(a²+b²) = 99
a²+b² = -99/2
Answer:
table 1
Step-by-step explanation:
table one, because you have different value of x (inputs) with no repetition for the same value
I can help with the formula just do the math on a calculator you multiply length x width x height
