Since 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 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
The attendance was the same on day 40. The attendance was 790 at both plays that day.
The attendance was the same on day 4. The attendance was 214 at both plays that day.
The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.
The attendance was never the same at both plays.
My answer B?
The two lines intersect at the point (4, 214). This means that the attendance was 214 for both plays on the 4th night right?
Your answer B is incorrect. The correct answer is:
The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790, respectively.
Your answer is incorrect. The correct answer is: The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.
To find the day(s) when the attendance was the same at both plays, we need to set the equations equal to each other and solve for x:
16x + 150 = -x^2 + 60x - 10
Combining like terms:
x^2 + 44x - 160 = 0
To solve this equation, we can use factoring or the quadratic formula. In this case, factoring is the easier option:
(x + 10)(x - 16) = 0
Setting each factor equal to zero and solving for x:
x + 10 = 0 or x - 16 = 0
x = -10 or x = 16
Since we're only interested in positive values for x (representing the number of days since opening night), we discard the negative solution.
Therefore, the attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.