Question 1013788
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The main rocket burns all of the fuel in 50 seconds, so it must burn *[tex \Large \frac{1}{50}] of the fuel in one second.  Likewise, the secondary rocket burns *[tex \Large \frac{1}{80}] of the fuel in one second.  Together they burn *[tex \Large \frac{1}{50}\ +\ \frac{1}{80}] of the fuel in one second.  Therefore, together they burn all of the fuel in *[tex \Large \frac{1}{\frac{1}{50}\ +\ \frac{1}{80}}] seconds.  Just do the arithmetic.  Correct arithmetic gets you an answer greater than 29 seconds but less than 32 seconds.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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