(x - 10) × 2
You have to multiply it by 2 so that you get the area
Answer:
The length of the flag is 730 ft and the width is 310 ft
Step-by-step explanation:
<u><em>The correct question is</em></u>
The perimeter of the flag is 2080 ft what is the flags width and length <u>if</u> the length is 420 ft greater than the width
Let
L ----> the length of the flag
W ---> the width of the flag
we know that
The perimeter of the flag (rectangle) is equal to

we have

so

----> equation A
---> equation B
Solve the system by substitution
substitute equation B in equation A
solve for W
<em>Find the value of L</em>
----> 
therefore
The length of the flag is 730 ft and the width is 310 ft
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)
The easiest way to do this is to plug in the numbers for the variebles and see if they equal the same in both sides. lets try the first one 5(1)+2(-3)=-1, multiply the nmbers to get 5-6=-1 now simplify to get the answer of -1=-1, they both equal the same so this means that the first option is the correct one
Hope this helps
The answer would be i<span>n step 4, he made an error in determining which value is closer to 82.5</span>