Stacy hits the jackpot one day at the gumball machine. She puts in a quarter and gets 444 gumballs rather than 111. The radius of each gumball is 666 mm.
What is the total volume of all 444 gumballs?
Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi.

3617.28 cubic millimeters.

Step-by-step explanation: I used the formula of a sphere since a gumball is a shape of a sphere and plugged the radius into the formula.

I got that the volume of a single gumball is 288π cubic millimeters.

Since there are 4 gumballs, i multiplied the volume by 4.

The total volume of the four gumballs add up to be 1152π .

Lastly, I multiplied 1152 by pi

and got the answer to be 3617.28 cubic millimeters

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

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.