The pep squad sold c, cheeseburgers and h, hothogs at the friday night football game. A total of 220 were sold. There were 3 tim
es more hotdogs sold than cheeseburgers. Write a system of equations for this situation.
1 answer:
Answer:
Step-by-step explanation:
The total sold is the sum of the individual numbers sold, hence c+h.
We assume "3 times more" means "3 times as many", so the number of hotdogs sold (h) is 3 times the number of cheeseburgers sold (c), hence 3c.
c + h = 220
h = 3c
_____
55 cheeseburgers and 165 hotdogs were sold.
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Assume that the notation (a,r,s,t) means multiply q and r, then add the product to s and (2,6,4,8)+(4,7,2,6)? Please show work a
Mama L [17]
<h3>
Answer: 7.0 (choice E)</h3>
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Work Shown:
![\perp (q,r,s,t) = \frac{qr + s}{t}\\\\\perp (2,6,4,8) = \frac{2*6 + 4}{8}\\\\\perp (2,6,4,8) = \frac{12 + 4}{8}\\\\\perp (2,6,4,8) = \frac{16}{8}\\\\\perp (2,6,4,8) = 2\\\\](https://tex.z-dn.net/?f=%5Cperp%20%28q%2Cr%2Cs%2Ct%29%20%3D%20%5Cfrac%7Bqr%20%2B%20s%7D%7Bt%7D%5C%5C%5C%5C%5Cperp%20%282%2C6%2C4%2C8%29%20%3D%20%5Cfrac%7B2%2A6%20%2B%204%7D%7B8%7D%5C%5C%5C%5C%5Cperp%20%282%2C6%2C4%2C8%29%20%3D%20%5Cfrac%7B12%20%2B%204%7D%7B8%7D%5C%5C%5C%5C%5Cperp%20%282%2C6%2C4%2C8%29%20%3D%20%5Cfrac%7B16%7D%7B8%7D%5C%5C%5C%5C%5Cperp%20%282%2C6%2C4%2C8%29%20%3D%202%5C%5C%5C%5C)
Let m = 2 so we can use this result later.
![\perp (q,r,s,t) = \frac{qr + s}{t}\\\\\perp (4,7,2,6) = \frac{4*7 + 2}{6}\\\\\perp (4,7,2,6) = \frac{28 + 2}{6}\\\\\perp (4,7,2,6) = \frac{30}{6}\\\\\perp (4,7,2,6) = 5\\\\](https://tex.z-dn.net/?f=%5Cperp%20%28q%2Cr%2Cs%2Ct%29%20%3D%20%5Cfrac%7Bqr%20%2B%20s%7D%7Bt%7D%5C%5C%5C%5C%5Cperp%20%284%2C7%2C2%2C6%29%20%3D%20%5Cfrac%7B4%2A7%20%2B%202%7D%7B6%7D%5C%5C%5C%5C%5Cperp%20%284%2C7%2C2%2C6%29%20%3D%20%5Cfrac%7B28%20%2B%202%7D%7B6%7D%5C%5C%5C%5C%5Cperp%20%284%2C7%2C2%2C6%29%20%3D%20%5Cfrac%7B30%7D%7B6%7D%5C%5C%5C%5C%5Cperp%20%284%2C7%2C2%2C6%29%20%3D%205%5C%5C%5C%5C)
Let n = 5 so we can use this result later.
Add the two results.
m+n = 2 + 5 = 7.0 is the final answer.