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3 years ago

Sarah has sold $33 worth of tickets to the comedy show. She has 4 tickets left to sell for $11 per ticket?

Mathematics

1 answer:

pantera1 [17]3 years ago

6 0

Answer:

Part A) $f(x)=11x+33$

Part B) The domain is the interval [0,1,2,3,4] and the range is the interval [$33,$44,$66,$77]

Part C) This problem is a discrete function

Step-by-step explanation:

Part A) What is the function notation?

Let

x -----> the number of tickets left to sell

f(x) -----> amount of money for ticket sales

we know that

The amount of money for ticket sales is equal to the cost of each ticket by the number of tickets left to sell plus the amount of money from tickets already sold.

The function notation is

$f(x)=11x+33$

Part B) What is the domain and range?

A reasonable domain is the number of tickets left to sell

[0,1,2,3,4] -----> The domain is a set of input values consisting of only certain numbers in an interval

Find the range

For x=0 -----> $f(0)=11(0)+33=\$33$

For x=1 -----> $f(1)=11(1)+33=\$44$

For x=3 -----> $f(3)=11(3)+33=\$66$

For x=4 -----> $f(4)=11(4)+33=\$77$

The range is the interval -----> [$33,$44,$66,$77]

Part C) Is this problem continuous or discrete?

we know that

A discrete function allows the x-values to be only certain points in the interval, usually only integers or whole numbers, and the continuous function allows the x-values to be ANY points in the interval, including fractions, decimals, and irrational values.

In this problem

the x-values are only certain points in the interval

therefore

Is a discrete function

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2 years ago

Describe the steps you would take to sketch the graph p(x)=2x^(3)+7x^(2)-3x-18

mojhsa [17]

Answer:

When we have a given function, y = f(x), the general way of graphing the function is finding a lot of coordinate pairs (x, f(x)) and graphing them in a coordinate axis, and then connect them.

So here we have:

p(x) = 2x^(3)+7x^(2)-3x-18

Just let's find the value of p(x) for different values of x, and let's do it.

p(0) = 2*0^(3)+7*0^(2)-3*0-18 = -18

Then the point (0, -18) belongs to the graph

p(1) = 2*1^(3)+7*1^(2)-3*1-18 = 2 + 7 - 3 - 18 = -12

then the point (1, - 12) belongs to the graph.

p(-1) = 2*(-1)^(3)+7*(-1)^(2)-3*(-1)-18 = -2 + 7 + 3 - 18 = -10

then the point (-1, -10) belongs to the graph:

p(2) = 2*2^(3)+7*2^(2)-3*2-18 = 16 + 28 - 6 - 18 = 20

Then the point (2, 20) belongs to the graph

p(-2) = 2*(-2)^(3)+7*(-2)^(2)-3*(-2)-18 = -16 + 28 + 6 - 18 = 0

Then the point (-2, 0) belongs to the graph.

p(-3) = 2*(-3)^(3)+7*(-3)^(2)-3*(-3)-18 = -54 + 63 + 9 - 18 = 0

Then the point (-3, 0) belongs to the graph.

Now that we have 6 points we can graph them and try to connect them.

We also know that this is a cubic polynomial, so we can expect that it changes its direction two times (no more than that) then knowing that we can estimate a shape for the graph, and you will obtain something like the graph of the image below, where I just connected the known points.