Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bubble gum nested at the bottom of the cone. The radius of the snow cone is 2 inches, and the height of the cone is 3 inches. If the diameter of the bubble gum is 0.5 inches, which of the following can be used to calculate the volume of the cone that can be filled with flavored ice? 1 over 3(3.14)(22)(3) − 4 over 3(3.14)(0.253)
1 over 3(3.14)(32)(2) − 4 over 3(3.14)(0.253)
1 over 3(3.14)(22)(3) − 4 over 3(3.14)(0.53)
1 over 3(3.14)(32)(2) − 4 over 3(3.14)(0.53)

A. (3.14)( 3) - (3.14)

Step-by-step explanation:

volume of the cone that can be filled with flavored ice = volume of cone - volume of the bubble gum

volume of cone = h

where r is the radius of the cone and h is its height.

volume of cone = x 3.14 x x 3

                          = 12.56 cube inches

volume of a sphere =

where r is the radius of the sphere.

radius =

          =

radius = 0.25 inches

So that;

volume of the bubble gum = x 3.14 x

                                 = 0.06542 cube inches

Thus,

volume of the cone that can be filled with flavored ice = 12.56 - 0.06542

                                                            = 12.49458

The volume of the cone that can be filled with flavored ice is 12.5 cube inches.

could you give more information?

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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