Who wants a scoop?

let $\large V_s$ be the volume of the sphere and $\large V_c$ be the volume of the cone

since their heights are equal, the height of the cone must be $\large 2r$

$\large V_s=\dfrac{4}{3}\pi r^3$

$\large V_c=\dfrac{1}{3}\pi r^2(h)=\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3$

let $n$ be the number of ice cream cones, so

$\large n= \dfrac{V_s}{V_c}=\dfrac{\frac{4}{3}}{\frac{2}{3}}=\dfrac{4}{3} \times \dfrac{3}{2}=\dfrac{4}{2}=$ $\color{#D61F06}\large \boxed{2}$

Because the ice cream is a sphere with radius $r$ , it has volume $\dfrac{4}{3}\pi r^3$ .

On the other hand, the volume of the cone is $\dfrac{1}{3}\pi r^2h = \dfrac{1}{3}\pi r^2 (2r) = \dfrac{2}{3}\pi r^3$ , exactly half that of the volume of the ice cream. Therefore, two ice cream cones can be filled with melted ice cream.

Let $x$ be the number of required ice cream cones. Then we have

$x=\dfrac{\text{volume of sphere}}{\text{volume of cone}}=\dfrac{\frac{4}{3}\pi (r^3)}{\frac{1}{3}\pi (r^2)(2r)}=\dfrac{\frac{4}{3}}{\frac{2}{3}}=\boxed{2}$

$V_{sphere}=\frac{4}{3}πr^3$

$V_{cone}=\frac{1}{3}πr^2(2r)=\frac{2}{3}πr^3$

$\frac{V_{sphere}}{V_{cone}}=(4/3)(2/3)=2$

r=radius which equals or half height .

pi r^3 4/3 is the volume of the sphere and the volume of the cone is pi r^2 h/3 .

we can divide each equation by pi and have the same ratio so we divide them and get r^3 4/3 for the sphere and r^2 h/3 for the cone .

we can then divide each equation by r^2 and have the same ratio so we divide them and get r 4/3 for the sphere and h 1/3 or 2 r 13 but we can change that to r*2/3 for the cone .

so we have 4 thirds of the radius for the sphere and 2 thirds for the cone so the volume of the sphere is double the volume of the cone

Volume of the scoop is 4/3 * phi * r^3

Volume of cone is 1/3 * phi * r^2 * t, t = 2 * r

So 4/3 * phi * r^3 = n * 1/3 * phi * r^2 * (2 * r)

                            = n * 2/3 * phi * r^3       ===>>> n = 2

vol of cone = 1/3πr²h;h=d=2r; Therefore V=1/3πr²x2r = 2/3πr³=: Vol of sphere =4/3πr³= 2x2/3πr³ So 2 cones.