Better recheck your numbers. C is correct. If the attendance was equal
15x + 76 = –x2 + 36x – 4
x^2-21x+80 = 0
(x-5)(x-16) = 0
Play A: y = 15x + 76
Play B: y = –x2 + 36x – 4
a. The attendance was the same on day 5. The attendance was 151 at both plays on that day.
b. The attendance was the same on day 16. The attendance was 316 at both plays on that day.
c. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.
d. The attendance was never the same at both plays.
I think the answer is D.
15x + 76 = –x2 + 36x – 4
x^2-21x+80 = 0
(x-5)(x-16) = 0
15x + 76 = -x^2 + 36x - 4
Rearranging the equation, we have:
x^2 + 21x - 80 = 0
Factoring the quadratic equation, we get:
(x + 16)(x - 5) = 0
This gives us two possible values for x: x = -16 and x = 5.
However, we are only interested in positive values for x since it represents the number of days since opening night. Therefore, the attendance was the same on day 5.
To find the attendance on day 5, we substitute x = 5 into either equation:
Play A: y = 15(5) + 76 = 151
Play B: y = -(5^2) + 36(5) - 4 = 151
Thus, the correct answer is:
c. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.
15x + 76 = -x^2 + 36x - 4
First, let's rearrange the equation:
x^2 - 21x + 80 = 0
Now, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -21, and c = 80. Plugging these values into the quadratic formula:
x = (-(-21) ± √((-21)^2 - 4(1)(80))) / (2(1))
x = (21 ± √(441 - 320)) / 2
x = (21 ± √121) / 2
x = (21 ± 11) / 2
So, we have two possible values for x: (21 + 11) / 2 = 16 and (21 - 11) / 2 = 5.
Now, let's substitute these values back into the equations to find the attendance. Starting with Play A:
y = 15x + 76
For x = 5: y = 15(5) + 76 = 151
For x = 16: y = 15(16) + 76 = 316
Now, let's substitute these values back into the equation for Play B:
y = -x^2 + 36x - 4
For x = 5: y = -(5^2) + 36(5) - 4 = 151
For x = 16: y = -(16^2) + 36(16) - 4 = 316
So, the attendance was the same at both plays on day 5 and day 16. The attendance on those days was 151 and 316, respectively.
Therefore, the correct answer is c. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316, respectively.