Roger is paid $50 to sell videos of a play to an audience after the play is preformed. He also receives $2 for each video he sells. The equation e = 50 + 2v models the amount of money Roger earns selling v videos. The plays audience consists of 230 people.

What is the theoretical and practical domain of the equation?​

Roger can earn $510 at most.

Step-by-step explanation:

We are given the function



Which gives the earnings of Roger when he sells v videos. Since the play’s audience consists of 230 people and each one buys no more than one video, v can take values from 0 to 230, i.e.  



That is the practical domain of E(v)

If Roger is in bad luck and nobody is willing to purchase a video, v=0

If Roger is in a perfectly lucky night and every person from the audience wants to purchase a video, then v=230. It's the practical upper limit since each person can only purchase 1 video

In the above-mentioned case, where v=230, then



Roger can earn $510 at most.

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}

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!