<span>Traveled Downstream a distance of 33 Mi and then came right back. If the speed of the current was 12 mph and the total trip took 3 hours and 40 minutes.
Let S = boat speed in still water then (s + 12) = downstream speed (s -12) = upstream speed
Given Time = 3 hours 40 minutes = 220 minutes = (220/60) h = (11/3) h Time = Distance/Speed
33/(s +12) + 33/(s-12) = 11/3 3{33(s-12) + 33(s +12)} = 11(s+12) (s -12) 99(s -12 + s + 12) = 11(</span> s^{2} + 12 s -12 s -144) 99(2 s) = 11(s^{2} -144) 198 s/11 = (s^{2} -144) 18 s = (s^{2} -144) (s^{2} - 18 s - 144) = 0 s^{2} - 24 s + 6 s -144 =0 s(s- 24) + 6(s -24) =0 (s -24) (s + 6) = 0 s -24 = 0, s + 6 =0 s = 24, s = -6 Answer) s = 24 mph is the average speed of the boat relative to the water.
If its circular, hyperbolic or ellipsoid
otherwise it will violate the rule of the verticle line test
Answer:
Charles practiced for the relay race for D. 9 hours last week.
Step-by-step explanation:
First, I would find the ratios the ratios for converting hurdle to javelin and javelin to relay.
Hurdle : Javelin ; 5 : 1.5
Javelin : Relay ; 2.5 : 5 or 1 : 2
Next, find the factors compared to the original numbers for hurdle to javelin.
15 / 5 = 3
To find the amount of javelin time they did, multiply 1.5 by the factor we got, 3.
1.5 * 3 = 4.5
Finally, double 4.5, since Charles does twice as much relay than he does javelin.
4.5 * 2 = 9 hours
The equation would be -9+-9 = -18
