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 days was the attendance the same at both plays? What was the attendance?

Play A: y=7x+84

Play B: y=-x^2+40x-6

Answer Choices:

A.) The attendance was the same on day 30. The attendance was 294 at both plays that day.

B.) The attendance was the same on day 3. The attendance was 105 at both plays that day.

C.) The attendance was never the same at both plays.

D.) The attendance was the same on days 3 and 30. The attendanceat both plays on those days was 105 and 294 respectively.

i think the answer is D, any help please?

D looks good to me. I assume you selected it after solving

7x+84 =-x^2+40x-6

thank you!

To find the days when the attendance was the same at both plays, we need to find the values of x that make the equations equal. Let's set the two equations equal to each other:

7x + 84 = -x^2 + 40x - 6

Now we can simplify and solve for x:

x^2 + 33x - 90 = 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)

For our equation, a = 1, b = 33, and c = -90. Plugging these values into the quadratic formula:

x = (-33 ± √(33^2 - 4(1)(-90))) / (2(1))

Simplifying further:

x = (-33 ± √(1089 + 360)) / 2

x = (-33 ± √1449) / 2

Since attendance cannot be negative, we can ignore the negative root. So:

x = (-33 + √1449) / 2

Calculating this value approximately:

x ≈ 1.577

Since x represents the number of days since opening night, we can conclude that the attendance was the same on day 2 and day 3.

Now, let's find the attendance on these days by substituting the values of x into either of the original equations. Let's use the equation for Play A:

y = 7x + 84

On day 2:
y = 7(2) + 84
y = 14 + 84
y = 98

On day 3:
y = 7(3) + 84
y = 21 + 84
y = 105

So, the attendance at both plays on day 2 was 98, and on day 3 it was 105.

Therefore, the correct answer is option B:

The attendance was the same on day 3. The attendance was 105 at both plays that day.

To find out on which days the attendance was the same at both plays, we need to set the two equations equal to each other and solve for x:

7x + 84 = -x^2 + 40x - 6

Combining like terms, we get:

x^2 + 33x - 90 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 1, b = 33, and c = -90. Plugging in these values, we get:

x = (-33 ± √(33^2 - 4(1)(-90))) / (2(1))
x = (-33 ± √(1089 + 360)) / 2
x = (-33 ± √(1449)) / 2

Now we need to find the values of x that give us a real solution. Taking the positive square root:

x = (-33 + √(1449)) / 2 ≈ 3.383

This means that on the 3rd day after opening night, the attendance was the same at both plays.

Now let's substitute this value of x back into one of the equations to find the attendance on that day.

For Play A:

y = 7x + 84
y = 7(3) + 84
y = 21 + 84
y = 105

For Play B:

y = -x^2 + 40x - 6
y = -(3^2) + 40(3) - 6
y = -9 + 120 - 6
y = 105

So the attendance at both plays on the 3rd day after opening night was 105.

Now let's check the other potential solution:

x = (-33 - √(1449)) / 2 ≈ -36.383

This value of x is negative and does not represent a valid number of days since opening night. Thus, the attendance was never the same at both plays except on the 3rd day.

Therefore, the correct answer is D.) The attendance was the same on days 3 and 30. The attendance at both plays on those days was 105 and 294 respectively.