and word prob#2
- Thread starter kristen
- Start date
When we see 'they run in the same amount of time', is a signal to solve the distance formula for t and set them equal.
d=rt, t=d/r
\(\displaystyle \frac{25}{r-3}=\frac{40}{r}\)
Solve for r, which is the rate of B. A is 3 less than that.
d=rt, t=d/r
\(\displaystyle \frac{25}{r-3}=\frac{40}{r}\)
Solve for r, which is the rate of B. A is 3 less than that.
- Joined
- Feb 4, 2004
- Messages
- 16,582
To learn how to set up and solve "uniform rate" exercises such as this, try here. :wink:
Once you have learned the basic terms and techniques, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your steps and reasoning so far, such as:
. . .A:
. . . . .distance: 25
. . . . .rate: b - 3
. . . . .time: 25/(b - 3)
. . .B:
. . . . .distance: 40
. . . . .rate: b
. . . . .time: 40/b
...and so forth.
Once you have learned the basic terms and techniques, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your steps and reasoning so far, such as:
. . .A:
. . . . .distance: 25
. . . . .rate: b - 3
. . . . .time: 25/(b - 3)
. . .B:
. . . . .distance: 40
. . . . .rate: b
. . . . .time: 40/b
...and so forth.