The price that will maximize profit is $ 5.50.
Since an auditorium has seats for 1200 people, and for the past several days, the auditorium has been filled to capacity for each show, and tickets currently cost $ 5.00 and the owner wants to increase the ticket prices, and he estimates that for each $ 0.50 increase in price, 100 fewer people will attend, to determine which is the price that will maximize profit, the following calculation must be made:
- 1200 x 5 = 6000
- 1100 x 5.5 = 6050
- 1000 x 6 = 6000
Therefore, the price that will maximize profit is $ 5.50.
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Answer:
A. 10,000
B. 1,000,000
C. 10,000,000,000
D. 1,000,000,000,000
Step-by-step explanation:
10,000/ 10∧4 = 1
1,000,000/ 10∧6 = 1
10,000,000,000/ 10∧10 = 1
1,000,000,000,000/ 10∧12 = 1
Hope this helps! Good luck!
Answer:
13,000
Step-by-step explanation:
When rounding, first:
Find the place value that you are looking for, in this case, being the thousands place value. It is a 3:
1<u>3</u>,046
When you round, you will look at the place value directly next to the one you are trying to find (which is the thousands place value). In this case, it will be the hundreds place value, which has the value of 0.
When the value of the number is 5 or greater, you round up.
When the value of the number is 4 or less, you round down.
In this case, it is 0, which is less than 4, so you round down.
13,046 to the nearest thousand place value is 13,000.
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Answer:
D. Infinitely many solutions
Step-by-step explanation:
1.Since neither equation contains an isolated. However, we can isolate -y in the first equation by adding 
to both sides.
Like this: 
ending up with 
2.Now, we can change y to a positive y. By doing so, we divide -y by the entire equation.
Like this 
Ending with 
3.Now, we can plug the expression 
into the second equation as a substitute for y, and solve for x. Then, we can use x to calculate y.
Like this 
4. Since -2x+2x would cancel out and leave -30=-30. This is true because we know -30 equals -30 with no variable in sight.
At the start, the tank contains A(0) = 50 g of salt.
Salt flows in at a rate of
(1 g/L) * (5 L/min) = 5 g/min
and flows out at a rate of
(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 5 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 5
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)
(e^(t/40) A(t))' = 5e^(t/40)
e^(t/40) A(t) = 200e^(t/40) + C
A(t) = 200 + Ce^(-t/40)
Given that A(0) = 50, we find
50 = 200 + C ==> C = -150
so that the amount of salt in the tank at time t is
A(t) = 200 - 150 e^(-t/40)
