Question

Brooklyn bought a snowcone from the local shop. It is shaped like a cone topped with a half-sphere.
The cone has a height of 6 in. and a radius of 2 in.
What is the approximate volume of the whole shape? Round your answer to the nearest tenth.
Use 3.14 to approximate pi. (Show your work.)

Answer 1

Answer:

The approximate volume of the whole shape is 41.9 in

Step-by-step explanation:

Provided:

height = 6

radius = 2

The total volume:

Variables:

$Vt = total\;volume$

$Vcone = volume\:of\:a\:cone=\dfrac{ 1 }{ 3 } \pi { r }^{ 2 } h$

$Vhemisphere = volume\:of\:a\:hemisphere= \dfrac{ 2 }{ 3 } \pi { r }^{ 3 }$

$Vt = Vcone + Vhemisphere$

$Vt = \dfrac{ 1 }{ 3 } \pi { r }^{ 2 } h+ \dfrac{ 2 }{ 3 } \pi { r }^{ 3 }$

$Vt = \dfrac{ 1 }{ 3 } \left( 3.14 \right) { \left( 2 \right) }^{ 2 } \left( 6 \right) + \dfrac{ 2 }{ 3 } \left( 3.14 \right) { \left( 2 \right) }^{ 3 }$

$\mathrm{Remove\:all\:parenthesis}$

$Vt = \dfrac{ 1 }{ 3 }\times \left 3.14 \right\times { \left 2 \right }^{ 2 }\times \left 6 \right + \dfrac{ 2 }{ 3 } \times\left 3.14 \right\times { \left 2 \right }^{ 3 }$

$\mathrm{Do\:the\:exponents\:first}$

$Vt = \dfrac{ 1 }{ 3 }\times \left 3.14 \right\times 4\times \left 6 \right + \dfrac{ 2 }{ 3 } \times\left 3.14 \right\times 8$

$\mathrm{Multiply\:\dfrac{ 1 }{ 3 }\:and\:3.14}$

$Vt = 1.04666667 \right\times 4\times \left 6 \right + \dfrac{ 2 }{ 3 } \times\left 3.14 \right\times 8$

$\mathrm{Multiply\:\dfrac{ 2 }{ 3 }\:and\:3.14}$

$Vt = 1.04666667 \right\times 4\times \left 6 \right + 2.09333333\times 8$

$\mathrm{Multiply\:1.04666667\:by\:4\:and\:then\:by\:6}$

$Vt = 25.12 + 2.09333333\times 8$

$\mathrm{Multiply\:2.09333333\:by\:8}$

$Vt = 25.12 + 16.746667$

$\mathrm{Add\:25.12\:and\:16.746667}$

$Vt = 41.866667$

$\mathrm{41.866667\:rounded\:to\:the\:nearest\:tenth\:is\:41.9}$

$Vt = 41.9$