Answer:
Colleen alone to clear the yard of leaves in 40 minutes
Step-by-step explanation:
P.S - The exact question is -
Given - Seas and Colleen are raking leaves in their yard. Working together they can clear the yard of leaves in 24 minutes. What kind of long it would take Sean 20 minutes longer to clear the yard dinner with Colleen working alone
To find - When c is the number of minutes it would take Colleen to finish the job when working alone, the situation is modeled by this rational equation: How long would it take Colleen alone to clear the yard of leaves?
Proof -
Given that the equation is -
![\frac{1}{c} + \frac{1}{c+20} = \frac{1}{24} \\\frac{c + 20 + c}{c(c+20)} = \frac{1}{24}\\\frac{2c + 20}{c^{2} +20c} = \frac{1}{24}\\24(2c + 20) = c^{2} +20c\\48c + 480 = c^{2} +20c\\28c + 480 = c^{2}\\c^{2} - 28c - 480 = 0\\c^{2} - 40c + 12c - 480 = 0\\c(c - 40) + 12(c - 40) = 0\\(c+12)(c-40) = 0\\c = -12, 40](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bc%7D%20%2B%20%5Cfrac%7B1%7D%7Bc%2B20%7D%20%20%3D%20%5Cfrac%7B1%7D%7B24%7D%20%5C%5C%5Cfrac%7Bc%20%2B%2020%20%2B%20c%7D%7Bc%28c%2B20%29%7D%20%3D%20%5Cfrac%7B1%7D%7B24%7D%5C%5C%5Cfrac%7B2c%20%2B%2020%7D%7Bc%5E%7B2%7D%20%2B20c%7D%20%3D%20%5Cfrac%7B1%7D%7B24%7D%5C%5C24%282c%20%2B%2020%29%20%3D%20c%5E%7B2%7D%20%2B20c%5C%5C48c%20%2B%20480%20%3D%20c%5E%7B2%7D%20%2B20c%5C%5C28c%20%2B%20480%20%3D%20c%5E%7B2%7D%5C%5Cc%5E%7B2%7D%20-%2028c%20-%20480%20%3D%200%5C%5Cc%5E%7B2%7D%20-%2040c%20%2B%2012c%20-%20480%20%3D%200%5C%5Cc%28c%20-%2040%29%20%2B%2012%28c%20-%2040%29%20%3D%200%5C%5C%28c%2B12%29%28c-40%29%20%3D%200%5C%5Cc%20%3D%20-12%2C%2040)
As time can not be negative
So,
we get
c = 40 minutes
∴ we get
Colleen alone to clear the yard of leaves in 40 minutes
Answer:
20,602 ft
Step-by-step explanation:
The vertical distance between these two points is the sum of 282 ft and 20,320 ft. That comes to 20,602 ft.
Answer:Sixth grade
Step-by-step explanation:
Given
There are
sixth grade students out of which
Participated in math olympiad
also no of fifth grade students![=910](https://tex.z-dn.net/?f=%3D910)
Fifth graders who participated in math olympiad ![=745](https://tex.z-dn.net/?f=%3D745)
Since ![745>740](https://tex.z-dn.net/?f=745%3E740)
There are more fifth graders in Olympiad [Numerically]
Percentage wise
Percenteage of sixth graders who participated in Math olympiad![=\dfrac{700}{840}\times 100=83.33\ \%](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B700%7D%7B840%7D%5Ctimes%20100%3D83.33%5C%20%5C%25)
Percenteage of Fifth graders who participated in Math olympiad![=\dfrac{745}{910}\times 100=81.86\ \%](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B745%7D%7B910%7D%5Ctimes%20100%3D81.86%5C%20%5C%25)
So, a greater Population of Sixth graders participated in math olympiad compared to Fifth graders as ![83.33\ \%>81.86\ \%](https://tex.z-dn.net/?f=83.33%5C%20%5C%25%3E81.86%5C%20%5C%25)
(0,-4.69) i hope this is correct