Find the volume of this cone.

To start calculating the volume of the cone, we obtain the following data:
To calculate the volume of a cone, we apply the following formula:
[tex]\boldsymbol{\sf{V=\dfrac{\pi r^{2}h }{3} }}[/tex], where
We solve, substituting our data in the formula:
taking the square root
Multiplying
Dividing
Therefore the volume of the cone is 144 cm³.
[tex]\huge\text{Hey there!}[/tex]
[tex]\huge\textbf{What is the formula for finding the}\\\\\huge\textbf{the volume of a cone?}[/tex]
[tex]\large\boxed{\mathsf{\dfrac{\pi \times r^2\times h}{3} = volume}}[/tex]
[tex]\bullet\large\textsf{ Whereas, \boxed{r} is your \underline{radius}, \boxed{h} is your \underline{height}, \& \boxed{\pi} is your pi.}[/tex]
[tex]\bullet\large\textsf{The pi }\boxed{(\pi)}\large\textsf{ is approximately equal to 3.14.}[/tex]
[tex]\huge\textbf{What are the labels in your equation?}[/tex]
[tex]\star\ \large\boxed{{Radius}}\rightarrow \textsf{\underline{4\ centimeters}}}}}[/tex]
[tex]\star\ \large\boxed{{Height}}\rightarrow \textsf{\underline{9\ centimeters}}}}}[/tex]
[tex]\star\ \large\boxed{{\pi}}\rightarrow \textsf{\underline{3}}}}}[/tex]
[tex]\huge\textbf{What does should the equation look}\\\\\huge\textbf{like?}[/tex]
[tex]\large\boxed{\mathsf{\dfrac{3 \times 4^2 \times9}{3}}}[/tex]
[tex]\huge\textbf{What are the steps to solve for the}\\\\\huge\textbf{question to get the result?}[/tex]
[tex]\large\boxed{\mathsf{\dfrac{3 \times 4^2 \times9}{3}}}[/tex]
[tex]\large\boxed{\mathsf{= \dfrac{3\times4\times4\times9}{3}}}[/tex]
[tex]\large\boxed{= \mathsf{\dfrac{3\times16\times9}{3}}}[/tex]
[tex]\large\boxed{\mathsf{= \dfrac{48\times9}{9}}}[/tex]
[tex]\large\boxed{\mathsf{= \dfrac{432}{3}}}[/tex]
[tex]\large\boxed{\mathsf{= \dfrac{432 \div 3}{3\div3}}}[/tex]
[tex]\large\boxed{= \mathsf{\dfrac{144}{1}}}[/tex]
[tex]\large\boxed{\mathsf{= 144\div1}}[/tex]
[tex]\large\boxed{= \textsf{144}}[/tex]
[tex]\huge\textbf{What is the result to this question?}[/tex]
[tex]\huge\boxed{\frak{144\ cm^3}}\huge\checkmark[/tex]
[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]