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 = 8x + 191 play b: y = -x^2 + 26x + 126

On 5th day and 13th day Attendance was equal for both plays and it was

231 for each .

Step-by-step explanation:

In order to find same attendance ,we need  plug both equation equal

         -x^2+26x+126 = 8x+191

        -x^2+26x-8x+126-191 = 0

        -x^2 +18x-65 =0

      multiplying both sides by -1 ,we get

      x^2 -18x+ 65 =0

        (x-5) (x-13) =0

         x =5 and x =13

On 5th day and 13th  day attendance was same.

And attendance can be obtained by plugging x =5 in either of equation

 y = 8(5)+191

   = 40 +191

  = 231

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

ian gone lie this got me confused

step-by-step explanation: