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# Help help ​

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Help help

Feb 22, 2018

#1
+7347
+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  =   $$\frac{\text{volume of candle}}{\text{surface area of candle}}$$

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

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Feb 22, 2018

#1
+7347
+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
#2
+94545
+1

Nice, hectictar   !!!!

Feb 22, 2018
#3
+7347
+2

Gracias

hectictar  Feb 22, 2018
#4
+94545
+1

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

Feb 22, 2018