To paint a house, two painters are hired. the first painter can paint the entire house in ten hours. the second painter will need six hours to paint the entire house. if the two painters start painting at 8: 00 am, at what time will they finish painting the house?

Simple,

take both the hours, and the beginning time they start at,

8:00 AM

For the first painter:

8:00 AM+10 hours...

Beginning at 8..

9,10,11,12,1,2,3,4,5,6

Thus, the FIRST painter will end at 6 P.M.

Now, the second painter:

8:00 AM+6 hours

Beginning at 8

9,10,11,12,1,2

Thus, the second painter will end at 2 P.M.

First painter can do house in 10 hours
in 1 hour, does 1/10 house

second is 6 hours
in 1 hour, does 1/6 house

if both start at same time, then how long, then add to 8:00

1st+2nd rate per hour=1/10+1/6=6/60+10/60=16/60 per hour=4/15
4/15 times x hours=1/1house
multiply both sides by 15/4 to clear fraction (15/4 times 4/15=60/60=1)
x hours=15/4=3 and 3/4 hours=3 housr and 45 mins

8:00+3 hours and 45mins=11:45

end at 11:45 AM
I have a feeling this question was asked about amont before

300b(x) is graphed by the functio

First subtract 10 from both side
y^2+6y-16=0
(y+8)(y-2)=0
-8 and 2