The cost for a biusness to make greeting cards can be divided into one-time cost (e. g., a printing machine) and repeated costs ( e. g., inc and paper ). suppose the total cost to make 300 cards is $900.00, and the total cost to make 650 cards is $1,600.00. what is the total cost to make 1,000 cards? round your answer to the nearest dollar.

It would cost $2300 to print 1000 cards.

I set up a system of equations for this problem.  Let x be the one-time cost (such as a printer) and y be the repeated costs (such as ink and paper).  The one-time cost will be the same no matter how many cards you print.  However the repeated costs will change based on how many cards are printed at a time.

The first equation looks like this:
x+300y=900
because the one time cost, added to the repeated costs for 300 cards, was 900.

The second equation looks like this:
x+650y=1600
because the one time cost, added to the repeated costs for 650 cards, was 1600.

Together the system is:
x+300y=900
x+650y=1600

Since the coefficients of x are the same, we can eliminate x by subtracting the bottom equation:
   x+300y=900
-(x+650y=1600)

Which gives us:
-350y=-700

Divide both sides by -350:
-350y/-350 = -700/-350
y = 2

Substitute that into the first equation:
x+300*2 = 900
x+600=900

Subtract 600 from both sides:
x+600-600=900-600
x=300

The one-time cost is $300 and the repeated costs are $2 per card.

This means that for 1000 cards we have
300+2(1000)=300+2000=2300.

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.