2) your class is raising money for a class trip. you make $10 on each pizza and $4 on each box of cookies that you sell.
how many items of each type must you sell to raise more than $100? write and graph an inequality to model the
situation. define the variables and state the constraints. give three possible combinations that you could sell.

The variables are 'p' and 'c'.

The inequality is:

The graph is plotted below.

Three possible solutions are: (0, 25), (10, 0) and (5, 20)

Step-by-step explanation:

Let the number of pizzas sold be 'p' and number of cookies sold be 'c'.

Given:

Price per pizza = $10

Price per cookie = $4

Minimum amount to be earned = $100

Price for 'p' pizzas sold =

Price for 'c' cookies sold =

As per question:

Also, number of pizzas and cookies can't be negative. So,

Plotting the above inequalities on a graph using DESMOS.

The region that is common to all the above inequalities is the solution region and is shown in the graph below.

The solution region also includes all the points on the line.

So, the three possible combinations of solutions can be any 3 points in the solution region. One such combination is:

(0, 25), (10, 0) and (5, 20)

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source