Can someone check my answer? Also show you work so I know you didn't choose a random answer. I Pick C for the following problem:
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 = 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.
set both equations equal:
15x + 76 = –x2 + 36x – 4
x^2-21x+80=0
(x-5)(x-16)=0
x=5, and x=16
attendance at x=16 ...since you already know attendance was the same on day 5 and 16, you dont need to check attendance with both formulas
attendance16= 15*16+76=240+76=316
attendance5= 15*5+76=151
Thank you so much!
Just checking because I'm stupid. C is correct yes?
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.
Setting the equations equal to each other:
15x + 76 = -x^2 + 36x - 4
Rearranging the equation:
x^2 - 21x + 80 = 0
Factoring the quadratic equation:
(x - 5)(x - 16) = 0
Setting each factor equal to zero:
x - 5 = 0 or x - 16 = 0
Solving for x:
x = 5 or x = 16
Therefore, the attendance was the same on day 5 and day 16. To find the attendance at both plays on those days, substitute the values of x into either equation.
For day 5:
Attendance at Play A: y = 15(5) + 76 = 75 + 76 = 151
Attendance at Play B: y = -5^2 + 36(5) - 4 = -25 + 180 - 4 = 151
For day 16:
Attendance at Play A: y = 15(16) + 76 = 240 + 76 = 316
Attendance at Play B: y = -16^2 + 36(16) - 4 = -256 + 576 - 4 = 316
Therefore, the correct answer is C. The attendance was the same on days 5 and 16, and the attendance at both plays on those days was 151 and 316, respectively.
To find the day(s) when the attendance was the same at both plays, we need to set the attendance equations for Play A and Play B equal to each other and solve for x.
The equations are:
Play A: y = 15x + 76
Play B: y = –x^2 + 36x – 4
Setting them equal to each other:
15x + 76 = –x^2 + 36x – 4
Now, we need to solve this quadratic equation for x. To do this, we can set the equation equal to zero:
x^2 - 21x + 80 = 0
To factorize the equation, we need to find two numbers whose product is 80 and sum is -21. The numbers are -16 and -5:
(x - 16)(x - 5) = 0
Setting each factor equal to zero, we get:
x - 16 = 0 or x - 5 = 0
Solving for x, we find:
x = 16 or x = 5
So, the attendance was the same on day 5 and day 16.
To find the attendance on those days, we can substitute the values of x into either equation. Let's use Play A:
For day 5:
y = 15(5) + 76
y = 75 + 76
y = 151
For day 16:
y = 15(16) + 76
y = 240 + 76
y = 316
So, the 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.