You and your friends are starting a t-shirt business. The company you buy t-shirts from charges you $3 for each shirt plus a one-time screen print set up fee of $50. After surveying other students, you are going to sell the shirts for 7 each A. Will your company be successful if you sell 23 shirts
B. Will your company be successful if you sell 10 shirts
C. How many shirts will you need to break even

Part A. The company will be successful selling 23 shirts.

Part B. The company will not be successful selling 10 shirts.

Part C. 13 shirts

Step-by-step explanation:

initial cost: $50

cost per shirt: $3

sale price per shirt: $7

number of shirts: x

The cost, y, of x shirts is

y = 3x + 50

Part A.

cost of 23 shirts

y = 3(23) + 50

y = 119

revenue from selling 23 shirts:

23 * $7 = $161

It costs $119 to make 23 shirts. They are sold for $161.

Part A. The company will be successful selling 23 shirts.

Part B.

cost of 10 shirts

y = 3(10) + 50

y = 80

revenue from selling 10 shirts:

10 * $7 = $70

It costs $80 to make 10 shirts. They are sold for $70.

Part B. The company will not be successful selling 10 shirts.

Part C.

To break even, the cost and the revenue must be equal.

Cost of x shirts: 3x + 50

Revenue from selling x shirts: 7x

7x = 3x + 50

4x = 50

x = 12.5

The break even point is 12.5 shirts. Since the number of shirts sold must be a whole number, the break even number is 13.

Part C. 13 shirts.

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! : )

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