Answer:
Each student ticket costs $8.33
Each adult ticket costs $15.34
Step-by-step explanation:
At Niagra High, Mr. Borton bought 4 student tickets and 2 adult tickets for the high school musical which cost $64. then Mrs. Gelvoria bought 3 student tickets and 3 adult tickets for the show and it cost her $72. How much are each type of tickets?
s = cost of each student ticket
a = cost of adult ticket
Our system of equations:
4s + 2a = 64
3s + 3a = 71
-3(4s + 2a = 64) ==> -12s - 6a = -192
2(3s + 3a = 71) ==> 6s + 6a = 142
-12s - 6a = -192
6s + 6a = 142
-6s = -50
/-6 /-6
s = $8.33 (the cost of each student ticket)
Now, let's find the cost of each adult ticket:
4s + 2a = 64
4(8.33) + 2a = 64
33.32 + 2a = 64
-33.32 -33.32
2a = 30.68
/2 /2
a = 15.34 (the cost of each adult ticket)
(x, y) ==> (8.33, 15.34)
Check your answer:
4s + 2a = 64
4(8.33) + 2(15.34) = 64
33.32 + 30.68 = 64
64 = 64
This statement is true
Hope this helps!
Answer:2/15 hope this helps
Step-by-step explanation:
Answer:
242
Step-by-step explanation:
Simplify the following:
11 ((9^2 - 5^2)/2^2 + 8)
Hint: | Evaluate 2^2.
2^2 = 4:
11 ((9^2 - 5^2)/4 + 8)
Hint: | Evaluate 5^2.
5^2 = 25:
11 ((9^2 - 25)/4 + 8)
Hint: | Evaluate 9^2.
9^2 = 81:
11 ((81 - 25)/4 + 8)
Hint: | Subtract 25 from 81.
| 7 | 11
| 8 | 1
- | 2 | 5
| 5 | 6:
11 (56/4 + 8)
Hint: | Reduce 56/4 to lowest terms. Start by finding the GCD of 56 and 4.
The gcd of 56 and 4 is 4, so 56/4 = (4×14)/(4×1) = 4/4×14 = 14:
11 (14 + 8)
Hint: | Evaluate 14 + 8 using long addition.
| 1 |
| 1 | 4
+ | | 8
| 2 | 2:
11×22
Hint: | Multiply 11 and 22 together.
| 2 | 2
× | 1 | 1
| 2 | 2
2 | 2 | 0
2 | 4 | 2:
Answer: 242
Let's put it this way.
It's a ratio of 6:72 and we need to change it in a way to find number of rooms cleaned in 9 days.
Divide 6 & 72 both by 2
Ratio is now 3:36
Multiply 3 & 36 both by 3
Ratio is now 9:108
The number of rooms in which Robin can clean in 9 days is 108 rooms.
