Question 509175
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26.  Add -2 to both sides.  Then factor.  3 times 5 is 15, 2 times -1 is -2, 3 times -1 plus 5 times 2 is -3 plus 10 equals 7.


14.  Use distance = rate times time which can also be expressed as time equals distance divided by rate.  If you let *[tex \Large r] represent the speed in still water, then the upstream rate is *[tex \Large r\ -\ 5] because the current works against you going upstream and the downstream rate is *[tex \Large r\ +\ 5].  Then let *[tex \Large t] represent the time it took to go upstream.  Since the total trip took 3 hours, the time to go downstream must be *[tex \Large 3\ -\ t].


First describe the upstream trip using time equals distance divided by rate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{10}{r\ -\ 5}]


Then describe the downstream trip the same way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ -\ t\ =\ \frac{10}{r\ +\ 5}] 


A little Algebra music, Sammy (while we isolate *[tex \Large t] in that second equation):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 3\ -\ \frac{10}{r\ +\ 5}] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{3r\ +\ 15\ -\ 10}{r\ +\ 5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{3r\ +\ 5}{r\ +\ 5}] 


Now we have two expressions that equate *[tex \Large t] to something, so set the two right-hand sides equal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10}{r\ -\ 5}\ =\ \frac{3r\ +\ 5}{r\ +\ 5}]


A little more Algebra music while we cross-multiply the proportion and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10(r\ +\ 5)\ =\ (r\ -\ 5)(3r\ +\ 5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10r\ +\ 50\ =\ 3r^2\ -\ 10r\ -\ 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r^2\ -\ 20r\ -\ 75\ =\ 0]


This ugly thing will defy all of your efforts to factor it, so it is time for the -- Look!  Up in the sky!  It's a bird!  It's a plane!  It's - it's SuperQuadratic Formula! -- the Quadratic Formula.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \frac{-(-20)\ \pm\ \sqrt{-20^2\ -\ 4(3)(-75)}}{2(3)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{20\ \pm\ \sqrt{400\ +\ 900}}{6}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{10\ \pm\ 5\sqrt{13}}{3}]


A little calculator work will show you that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{10\ -\ 5\sqrt{13}}{3}\ < 0]


and since it is unlikely that Abby got anywhere rowing backwards, we can safely discard this root leaving us with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{10\ +\ 5\sqrt{13}}{3}]


As the exact answer.  Use your calculator to obtain a numerical approximation to the precision you require.
 

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