9514 1404 393
Answer:
Step-by-step explanation:
Let a and s represent the prices of adult and student tickets, respectively.
13a +12s = 211 . . . . . . ticket sales the first day
5a +3s = 65 . . . . . . . ticket sales the second day
Subtracting the first equation from 4 times the second gives ...
4(5a +3s) -(13a +12s) = 4(65) -(211)
7a = 49 . . . . . . . simplify
a = 7 . . . . . . . divide by 7
5(7) +3s = 65 . . . . substitute into the second equation
3s = 30 . . . . . . . subtract 35
s = 10 . . . . . . . divide by 3
The price of one adult ticket is $7; the price of one student ticket is $10.
To solve this question, first cross multiply
Let ? = x
4/9 = x/81
4(81) = 9(x)
Simplify.
81 x 4 = 324
324 = 9x
Isolate the x, divide 9 from both sides
324/9 = 9x/9
x = 324/9
x = 36
4/9 = 36/81
36 is your answer
hope this helps!
Answer: It's Already simplified!
Answer:
19t + 100
Step-by-step explanation:
19 per hour so multiplication
extra 100 so add that
Answer:
$3 max
Step-by-step explanation:
Charge= b, Customer= c, Revenue= r
r= bc, currently, r= 16*10= $160
We know that: b+1 ⇒ c-2 and the target is r ≥ 130
So, this will all be reflected as:
b=10+x ⇒ c= 16-2x
- (10+x)(16-2x) ≥ 130
- 160 -20x +16x - 2x² ≥ 130
- -2x² - 4x + 30 ≥ 0
- x² + 2x -15 ≤ 0
- (x+1)² ≤ 4²
- x+1 ≤ 4 (negative value not considered)
- x ≤ 3
As we see the max increase amount is $3, when the revenue will be:
(10+3)*(16-3*2)= 13*10= $130
