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?
y=7x+84
y=-x^2+40x-6
The attendance was the same on day 30. The attendance was 294 at both plays that day.
The attendance was the same on day 3. The attendance was 105 at both plays that day.
The attendance was the same on days 3 and 30. The attendance at both plays on those days was 105 and 294 respectively.****
The attendance was never the same at both plays.
HOW DO I SOLVE THIS THOUGH
7x + 84 = - x^2 + 40 x - 6
x^2 - 33 x + 90 = 0
(x - 30)(x - 3) = 0
To solve this problem, we need to find the values of x (number of days) where the attendance at Play A (y=7x+84) is equal to the attendance at Play B (y=-x^2+40x-6).
To find the day(s) when the attendance was the same at both plays, we can set the two equations equal to each other and solve for x.
Setting y=7x+84 equal to y=-x^2+40x-6:
7x+84 = -x^2+40x-6
Rearranging the equation to bring all terms to one side:
x^2 - 33x + 90 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring in this case:
(x - 3)(x - 30) = 0
Now we have two possible values for x: x = 3 and x = 30.
So, the attendance was the same on days 3 and 30.
To find the attendance on those days, substitute the values of x into either equation.
For day 3:
y = 7(3) + 84
y = 21 + 84
y = 105
For day 30:
y = 7(30) + 84
y = 210 + 84
y = 294
Therefore, the attendance at both plays on days 3 and 30 was 105 and 294 respectively.