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= 9x+64
Play B, y= -x^2+26x+12

Play A: y=15x+76

Play B: y=-x^2 +36x-4

set them equal:

15x+76=-x^2 +36x-4

80=-x^2+21x

x^2-21x+80=0

-> solve for x

for example with the quadratic equation formula or solving the square

factorized:

(x-16)(x-5)=0

so x1=5 and x2=16

insert them into y=15x+76

y1=151, y2=316

so (5,151) and (16,316) are days with equal attendance for both plays which is option C

Step-by-step explanation:

math. mathematics.

question: NO VIRUS LINKS!!! 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= 9x+64

Play B, y= -x^2+26x+12

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8