This is a distance = rate * time problem. The easiest way to solve these is to make a table with the information. Since d= rt, we will set up the table like that:
distance rate time
Mason d 5 t + 8
Nora d 9 t
Let me explain the values in the table. The problem says "...when the swimmers had swam exactly the same distance"; therefore, we put a d there to indicate that, although we have no idea the distance they swam, both distances were the same. The rates are easy; they are self-explanatory. The time could be a little tricky too though. If Mason began swimming 8 seconds earlier than Nora, Nora's time is t and Mason's time is t + 8, which is Nora's time with 8 seconds added to it. Because d = rt, we set up the equations like that: d = 5(t+8), and d = 9t. Because the 2 d's are the same, we set them equal to each other: 5(t+8) = 9t. Simplifying that gives you 5t + 40 = 9t and 40 = 4t and t = 10. Now put that t value of 10 into Mason's time to solve the question they are asking you: t + 8 with the substitution is 10 + 8 = 18. So Mason had been swimming for 18 seconds when they had both swam the same distance.
Lets say y is number of adult tickets and x is number of student tickets
x+y=348
2x+y=348
(2x since I know that the student tickets are twice as much as adult tickets)
I now know that that student tickets make up 2/3 of the total number of tickets and 1/3 make up the number of adults tickets. So...
1/3 x 348= 116 adult tickets
2/3 x 348=232 student tickets
Answer:
Step-by-step explanation:
what is the question?
Answer:
88 + 33v
Step-by-step explanation:
9+2(8+10v-7v)
11(8+3v) = 88+33v
