Let the number of adult tickets sold equal a.
Let the number of student tickets sold equal s.
The cost of the adult tickets sold is 6a.
The cost of the student tickets sold is 3s.
The total sales was $660, so we have our first equation:
6a + 3s = 660
25 more student tickets than adult tickets were sold.
That gives us our second equation
s = a + 25
We have a system of equations.
6a + 3s = 660
s = a + 25
Since the second equation is already solved for s, we can use the substitution method. Replace s of the first equation with a + 25 and solve for a.
6a + 3(a + 25) = 660
6a + 3a + 75 = 660
9a = 585
a = 65
65 adult tickets were sold.
Now we find the number of student tickets sold.
s = a + 25 = 65 + 25 = 90
Answer: The number of student tickets sold was 90.
Answer:
the ratio in lowest terms is: 7 knives to 8 spoons = 7/8
Step-by-step explanation:
First express the ratio 105 knives to 120 spoons as the quotient:

next write each of the numbers in prime-factor form:
105 = 3 x 5 x 7
120 = 2 x 2 x 2 x 3 x 5
Now notice that when we express the quotient sing the factor form of the numbers, the factor 3, and 5 can be cancelled out because they appear in numerator an in denominator at the same time. therefore our quotient becomes:

Therefore the ratio in lowest terms is: 7 knives to 8 spoons
<span>Since the 2 triangles are similar, their sides are proportional.
You also know AD is 20 yards, since it tells you that is the crosswalk length at A.
So the proportion is :
50 / 120 = 20 / BC
Or
if you flip it over,
120 / 50 = BC / 20
Then you can solve for BC.
120 / 50 = BC / 20
(20)(120/50) = BC
BC = 48</span>
Answer:
The answer is "Choice D".
Step-by-step explanation:
Please find the complete question in the attached file.
It might indeed multiply that each coordinated throughout this question and construct a delay. In this question, The c instant of differentiation to go and get a smaller image is less than 1, and construct a smaller polygon, that polygon would be extended on even metering systems. Its polygon is distributed out by using the source as the dilation core.