Answer:
both kinds of tickets are $5 each
Step-by-step explanation:
Let s and c represent the dollar costs of a senior ticket and child ticket, respectively. The problem statement describes two relationships:
12s + 5c = 85 . . . . . revenue from the first day of sales
6s + 9c = 75 . . . . . . revenue from the second day of sales
Double the second equation and subtract the first to eliminate the s variable.
2(6s +9c) -(12s +5c) = 2(75) -(85)
13c = 65 . . . . . simplify
65/13 = c = 5 . . . . . divide by the coefficient of c
Substitute this value into either equation. Let's use the second one.
6s + 9·5 = 75
6s = 30 . . . . . . . subtract 45
30/6 = s = 5 . . . divide by the coefficient of s
The price of a senior ticket is $5; the price of a child ticket is $5.
Answer:
A. 5 g - 8 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
45/15= 3
So 1 minute=3 cents
75*3=225=2.25 converted
75 minutes=$2.25
Answer:
15 days
Step-by-step explanation:
Well, as you would be able to tell, you just have to find the LCM of 3 and 5. These do not have one before 3 x 5, so it is 15.
If r(x)= 11x, and c(x)=6x +20, then just put that into the equation.
So instead of p(x) = r(x) - c(x)
I would be p(x) = 11x - 6x + 20
Now solve.
p(x) = 5x + 20 is your final answer, so A.
