Answer:
For this case we know that the diameter and height of a cone are both equal. That means:
[tex] h = D=x[/tex]
With [tex] D =2r[/tex] with r the radius. The volume of a cone is given by:
[tex] V =\frac{1}{3} \pi r^2 h [/tex]
And for this case we know that h =x and [tex] x =2r[/tex] so then [tex] r= \frac{x}{2}[/tex]
And replacing we got:
[tex]V= \frac{1}{3} (\frac{x}{2})^2 (x)= \frac{1}{12} x^3[/tex]
And the best option would be:
1/12pix^3
Step-by-step explanation:
For this case we know that the diameter and height of a cone are both equal. That means:
[tex] h = D=x[/tex]
With [tex] D =2r[/tex] with r the radius. The volume of a cone is given by:
[tex] V =\frac{1}{3} \pi r^2 h [/tex]
And for this case we know that h =x and [tex] x =2r[/tex] so then [tex] r= \frac{x}{2}[/tex]
And replacing we got:
[tex]V= \frac{1}{3} (\frac{x}{2})^2 (x)= \frac{1}{12} x^3[/tex]
And the best option would be:
1/12pix^3
