The revenue from selling movie tickets is given by the function r(t) = 10t, where "t" is the number of tickets sold. the theater holds a maximum of 200 people. the profit is a function of the revenue "r" and is given by the function p(r) = 0.20r. what is the domain of the profit function p(r)?

The domain of the profit function P(R) is

D = {R|R∈W|0≤R≤2000}

Step-by-step explanation:

Find the highest and lowest R values for R(t).

R(t) = 10t

Since the maximum number of people is 200, t ≤ 200.

You can't have a negative number of people, so t ≥ 0.

Sub t for 0

R(t) = 10t

R(0) = 10(0)

R(0) = 0

(0,0)

Sub t for 200

R(t) = 10t

R(200) = 10(200)

R(200) = 2000

In R(t), 0 ≤ R and R ≤ 2000.

In R(t), Range = {R|R∈W|0≤R≤2000}

Use 0 ≤ R ≤ 2000 to find the domain of P(R).

The domain in P(R) is the possible values of R because it's in the brackets.

We already know the possible values of R.

Domain = {R|R∈W|0≤R≤2000}

It sets the variable for domain, states the type of numbers R includes, and the specific numbers.

Whole numbers are numbers that are not partials starting from 0.

If you wanted to find the range of P(R), you would substitute R for 0 and 2000 in the equation to calculate the highest and lowest numbers of P.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: