Since the opening night, attendance at Play A has

increased steadily, while attendance at Play B first rose and then fell.
Equations modeling the daily attendance y at each play are shown below,
where x is the number of days since the opening night. On what day(S)
was the attendance the same at both plays? what was the attendance?

Play A: y = 16x + 150
Play B: y = -x^2 60x - 10

answer choices:

A)The attendance was the same on day 40. The attendance was 790 at both plays that day.
B)The attendance was the same on day 4. The attendance was 214 at both plays that day.
C)The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.****
D)The attendance was never the same at both plays.

My answer is C. Is that correct?

correct , see

http://www.wolframalpha.com/input/?i=plot+y+%3D+16x+%2B+150%2C+y+%3D+-x%5E2+%2B60x+-+10

Oh great! Thank you so much!

Yes, your answer is correct. To determine the day on which the attendance was the same at both plays, you need to find the values of x for which the attendance, y, is equal for Play A and Play B.

Setting the equations for Play A and Play B equal to each other, we have:

16x + 150 = -x^2 + 60x - 10

Rearranging the equation to set it equal to zero:

x^2 - 4x + 160 = 0

This is a quadratic equation. To solve, you can use factoring, the quadratic formula, or completing the square. In this case, factoring is the simplest method:

(x - 4)(x - 40) = 0

Setting each factor equal to zero:

x - 4 = 0 or x - 40 = 0

Solving for x, we find two possible values: x = 4 or x = 40.

Therefore, the attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790, respectively.

So, the correct answer is C) The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790, respectively.