Question 653389
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You didn't share the value of the speed difference, so we'll just call that value *[tex \LARGE \delta].  We'll let *[tex \LARGE r] represent the faster speed and *[tex \LARGE t] represent the time taken by the faster aircraft.


Time is equal to distance divided by rate, so for the jet:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5940}{r}\ =\ t]


And for the slower plane:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5940}{r\ -\ \delta}\ =\ t\ +\ 6]


Add -6 to both sides and apply the LCD to combine the fractions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5940\ -\ 6\left(r\ -\ \delta\right)}{r\ -\ \delta}\ =\ t]


Now that you have two expressions both equal to *[tex \LARGE t], set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5940\ -\ 6\left(r\ -\ \delta\right)}{r\ -\ \delta}\ =\ \frac{5940}{r}]


Once you have your difference value, plug it in.  Then cross-multiply, collect like terms into the LHS, and solve the quadratic. Discard the negative root.


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