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In one design being considered for the container shaped like a cylinder, the container will have a height of 12 inches. What will be the radius of the container, to the nearest tenth of an inch? volume being 258.75 cubic inches

 Mar 16, 2016

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 #1
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In one design being considered for the container shaped like a cylinder, the container will have a height of 12 inches. What will be the radius of the container, to the nearest tenth of an inch? volume being 258.75 cubic inches

 

r = radius

h = height

V = volume

 

\(\begin{array}{rcll} V_{\text{cylinder}} &=& \pi \cdot r^2 \cdot h\\\\ pi \cdot r^2 \cdot h &=& V_{\text{cylinder}} \qquad & | \qquad :h \\ pi \cdot r^2 &=& \frac{ V_{\text{cylinder}} } {h} \qquad & | \qquad :\pi \\ r^2 &=& \frac{ V_{\text{cylinder}} } {h \cdot \pi} \qquad & | \qquad \sqrt{} \\ \end{array}\)

 

\(\boxed{~ \begin{array}{rcll} r &=& \sqrt{ \frac{ V_{\text{cylinder}} } { \pi \cdot h } } \qquad & | \qquad h = 12\ \text{in} \qquad V_{\text{cylinder}} = 258.75\ \text{in}^3 \\\\ r &=& \sqrt{ \frac{ 258.75\ \text{in}^3 } { \pi \cdot 12\ \text{in} } } \\ r &=& \sqrt{ \frac{ 258.75\ \text{in}^2 } { \pi \cdot 12\ } } \\ r &=& \sqrt{ \frac{ 258.75 } { \pi \cdot 12\ } }\ \text{in} \\ r &=& \sqrt{ 6.86355692084 }\ \text{in} \\ r &=& 2.61983910209\ \text{in} \\ \mathbf{r} & \mathbf{=} & \mathbf{2.6 \ \text{in} } \qquad (\text{rounded})\\ \end{array} ~}\)

 

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 Mar 17, 2016
 #1
avatar+26387 
+25
Best Answer

In one design being considered for the container shaped like a cylinder, the container will have a height of 12 inches. What will be the radius of the container, to the nearest tenth of an inch? volume being 258.75 cubic inches

 

r = radius

h = height

V = volume

 

\(\begin{array}{rcll} V_{\text{cylinder}} &=& \pi \cdot r^2 \cdot h\\\\ pi \cdot r^2 \cdot h &=& V_{\text{cylinder}} \qquad & | \qquad :h \\ pi \cdot r^2 &=& \frac{ V_{\text{cylinder}} } {h} \qquad & | \qquad :\pi \\ r^2 &=& \frac{ V_{\text{cylinder}} } {h \cdot \pi} \qquad & | \qquad \sqrt{} \\ \end{array}\)

 

\(\boxed{~ \begin{array}{rcll} r &=& \sqrt{ \frac{ V_{\text{cylinder}} } { \pi \cdot h } } \qquad & | \qquad h = 12\ \text{in} \qquad V_{\text{cylinder}} = 258.75\ \text{in}^3 \\\\ r &=& \sqrt{ \frac{ 258.75\ \text{in}^3 } { \pi \cdot 12\ \text{in} } } \\ r &=& \sqrt{ \frac{ 258.75\ \text{in}^2 } { \pi \cdot 12\ } } \\ r &=& \sqrt{ \frac{ 258.75 } { \pi \cdot 12\ } }\ \text{in} \\ r &=& \sqrt{ 6.86355692084 }\ \text{in} \\ r &=& 2.61983910209\ \text{in} \\ \mathbf{r} & \mathbf{=} & \mathbf{2.6 \ \text{in} } \qquad (\text{rounded})\\ \end{array} ~}\)

 

laugh

heureka Mar 17, 2016

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