Let's solve this step by step.
1. Know the formula: The volume [tex]\( V \)[/tex] of a cone with radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex] is given by:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
2. Insert the known values: The volume [tex]\( V \)[/tex] is given as [tex]\( 20\pi \)[/tex] cubic meters and the height [tex]\( h \)[/tex] is [tex]\( 10 \)[/tex] meters. Plug these values into the formula:
[tex]\[ 20\pi = \frac{1}{3} \pi r^2 \cdot 10 \][/tex]
3. Simplify and solve for [tex]\( r^2 \)[/tex]:
[tex]\[ 20\pi = \frac{10}{3} \pi r^2 \][/tex]
[tex]\[ 20 = \frac{10}{3} r^2 \][/tex]
[tex]\[ r^2 = \frac{20}{\left(\frac{10}{3}\right)} \][/tex]
[tex]\[ r^2 = \frac{20 \cdot 3}{10} \][/tex]
[tex]\[ r^2 = \frac{60}{10} \][/tex]
[tex]\[ r^2 = 6 \][/tex]
4. Find the radius [tex]\( r \)[/tex] by taking the square root of [tex]\( r^2 \)[/tex]:
[tex]\[ r = \sqrt{6} \][/tex]
5. Calculate the diameter [tex]\( d \)[/tex], which is twice the radius:
[tex]\[ d = 2r = 2\sqrt{6} \][/tex]
6. Finding the numerical value for [tex]\( \sqrt{6} \)[/tex] and doubling it:
[tex]\[ \sqrt{6} \approx 2.449 \][/tex]
[tex]\[ d \approx 2 \times 2.449 \][/tex]
[tex]\[ d \approx 4.898 \][/tex]
7. Round the diameter to the nearest tenth:
[tex]\[ d \approx 4.9 \][/tex]
So the diameter of the cone is approximately [tex]\( 4.9 \)[/tex] meters when rounded to the nearest tenth.
