Two cars are traveling north along a highway. The first drives at 40 mph, and the second, which leaves 3 hours later, travels at 60 mph. How long after the second car leaves will it take for the second car to catch the first?

Step-by-step explanation:

We will make a table and fill it in according to the information provided. What this question is asking us to find, in the end, is how long did it take the cars to travel the same distance. In other words, how long, t, til car 1's distance = car 2's distance. The table looks like this:

              d     =      r     *     t

car1

car2

We can fill in the rates right away:

             d       =      r     *     t

car1                       40

car2                       60

Now it tells us that car 2 leaves 3 hours after car 1, so logically that means that car 1 has been driving 3 hours longer than car 2:

            d     =      r      *      t

car1                     40         t + 3

car2                     60           t

Because distance = rate * time, the distances fill in like this:

                   d      =      r      *      t

car1    40(t + 3)    =      40        t+3

car2          60t     =       60         t

Going back to the interpretation of the original question, I am looking to solve for t when the distance of car 1 = the distance of car 2. Therefore,

40(t+3) = 60t and

40t + 120 = 60t and

120 = 20t so

t = 6 hours.

26

step-by-step explanation:

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%