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NotSoSmart  Feb 22, 2018

Best Answer 

 #1
avatar+7266 
+2

volume of candle   =   \(\frac13\pi r^2h\)          Since  h = r , we can replace  h  with  r .

volume of candle   =   \(\frac13\pi r^2r\)

volume of candle   =   \(\frac13\pi r^3\)

 

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+h^2}\)        Since  h = r , we can replace  h  with  r .

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\sqrt{2r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\cdot r\sqrt{2}\)

surface area of candle   =   \(\pi r^2+\pi r^2\sqrt{2}\)

 

ratio of the volume of candle to its surface area  =   \(\frac{\text{volume of candle}}{\text{surface area of candle}}\)

 

\(\frac{\text{volume of candle}}{\text{surface area of candle}}\,=\,\frac{\frac13\pi r^3}{\pi r^2+\pi r^2\sqrt2}\\~\\ =\,\frac{\frac13r^3}{r^2+r^2\sqrt2} \\~\\ =\,\frac{\frac13r}{1+\sqrt2} \\~\\ =\,\frac{r}{3(1+\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1+\sqrt2)(1-\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1-2)} \\~\\ =\,\frac{r(1-\sqrt2)}{-3} \)

hectictar  Feb 22, 2018
 #1
avatar+7266 
+2
Best Answer

volume of candle   =   \(\frac13\pi r^2h\)          Since  h = r , we can replace  h  with  r .

volume of candle   =   \(\frac13\pi r^2r\)

volume of candle   =   \(\frac13\pi r^3\)

 

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+h^2}\)        Since  h = r , we can replace  h  with  r .

surface area of candle   =   \(\pi r^2+\pi r\sqrt{r^2+r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\sqrt{2r^2}\)

surface area of candle   =   \(\pi r^2+\pi r\cdot r\sqrt{2}\)

surface area of candle   =   \(\pi r^2+\pi r^2\sqrt{2}\)

 

ratio of the volume of candle to its surface area  =   \(\frac{\text{volume of candle}}{\text{surface area of candle}}\)

 

\(\frac{\text{volume of candle}}{\text{surface area of candle}}\,=\,\frac{\frac13\pi r^3}{\pi r^2+\pi r^2\sqrt2}\\~\\ =\,\frac{\frac13r^3}{r^2+r^2\sqrt2} \\~\\ =\,\frac{\frac13r}{1+\sqrt2} \\~\\ =\,\frac{r}{3(1+\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1+\sqrt2)(1-\sqrt2)} \\~\\ =\,\frac{r(1-\sqrt2)}{3(1-2)} \\~\\ =\,\frac{r(1-\sqrt2)}{-3} \)

hectictar  Feb 22, 2018
 #2
avatar+88899 
+1

Nice, hectictar   !!!!

 

 

cool cool cool

CPhill  Feb 22, 2018
 #3
avatar+7266 
+2

Gracias laughlaugh

hectictar  Feb 22, 2018
 #4
avatar+88899 
+1

Didn't know that you were fluent in Spanish!!!!....LOL!!!!

 

 

 

cool cool cool

CPhill  Feb 22, 2018

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