Please help...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 = –x2 + 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.
set the 2 equations equal to each other ...
16x + 150 = -x^2 + 60x - 10
x^2 - 44x + 160 = 0
(x-40)(x-4) = 0
x = 40 or x = 4
when x = 40, y = 16(40) + 150 = 790
when x = 4 , y = 16(4) + 150 = 214
Pick out the choice that matches my solution
Thank you
The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively. Wow, what a coincidence! It seems like both plays had a magical day where people couldn't resist attending. Maybe they were serving free popcorn or had acrobatic clowns performing during the intermission. Whatever the reason, those were the lucky days where both plays attracted the same number of attendees.
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 find the values of x (number of days) for which the attendance y is equal in both equations.
First, let's set the two equations equal to each other:
16x + 150 = -x^2 + 60x - 10
Now we have a quadratic equation. We can rearrange it to get it into the form of ax^2 + bx + c = 0:
x^2 + 44x - 160 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
x = (-44 ± √(44^2 - 4(1)(-160))) / (2(1))
Simplifying further:
x = (-44 ± √(1936 + 640)) / 2
x = (-44 ± √2576) / 2
x = (-44 ± 50.76) / 2
Now we have two possible values for x:
x = (-44 + 50.76) / 2 = 6.38
x = (-44 - 50.76) / 2 = -47.38
Since the number of days cannot be negative, we can discard the second solution. Therefore, the attendance was the same on day 6.38, which is not a feasible number of days. This means that the attendance was never the same at both plays.
Therefore, the correct answer is:
The attendance was never the same at both plays.