Cassie and Nicholas each rented a movie from two different stores. Cassie paid $4 for the video plus $1.50 for each day it was late. Nicholas spent $6 for his video plus $1.00 for each day it was late. If they both spent the same amount of money after the video was returned, how many days after the due date did they keep the video?
Since Cassie spent an extra $0.50 per day, how many days late would she need to be so that $0.50t = $2 (which is the difference in the base rental price)?
so t = 4?
In terms of equations, solve \[1.5t + 4 = t+6\]for t
Very good! :)
y = 4 + 1.5x y = 6 + x x gives number of days rented
thank you !!
yw
Adam and Gloria traveled in a sailboat for 4 hours with a 6 mph current in the water. Against the same current, it took them 6 hours to return home. How fast was the boat traveling in still water?
want to try this one?
ahh...the classic "boat in the current" problem :)
i know the difference between the time and mph is 2
Let r be the speed of the boat in still water. THen r+6 is their speed with the current and r-6 is their speed against the current. The distance they traveled is the same and so it's really not needed. 4(r+6) = 6(r-6)...solve for r
THis uses the d=rt formula (distance = rate * time) and we're setting the two distances (there and back) equal to each other
so it should look like (4r-24) = (6r-36) ?
4r+24=6r-36
ok now i add 6r = 4r?
or do i add 36 to 24 to get r alone?
Get all the r's on one side and all the numbers on the other. So move the 36 to the left (by adding to both sides) Then move the 4r to the right (by subtracting from both sides)
ok i got 60 = 2r
and 60 divided by 2 is 30 so 30r?
Close...60 = 2r is right...then you divide by 2 (which you did). That gives you 30 = r. So the rate of the boat in still water is 30 mph
ok got it
\[\sqrt{5}(8+3\sqrt{6})\]
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