Adiesel train traveled to the repair yards andback. it took two hours longer to go there thanit did to come back. the average speed on thetrip there was 70 km/h. the average speed onthe way back was 80 km/h. how many hoursdid the trip there take? a) 15 hoursc) 25 hoursb) 16 hoursd) 10 hours

B.  16 hrs

Step-by-step explanation:

Distance = rate × time

The best way to do this is to make a table with the info.  We are concerned with the trip There and the Return trip.  Set it up accordingly:

                    d     =     r     ×     t

There    

Return

The train made a trip from A to B and then back to A again, so the distances are both the same.  We don't know what the distance is, but it doesn't matter.  Just go with it for now.  It'll be important later.

               d     =     r     ×     t

There      d    

Return     d

We are also told the rates.  There is 70 km/hr and return is 80 km/hr

               d     =     r     ×     t

There      d     =    70

Return     d     =    80

All that's left is the time column now.  We don't know how long it took to get there or back, but if it took 2 hours longer to get There than on the Return, the Return trip took t and the There trip took t + 2:

                      d     =     r     ×     t

There             d     =    70   ×    t+2

Return            d     =    80   ×     t

The distances, remember, are the same for both trips, so that means that by the transitive property of equality, their equations can be set equal to each other:

70(t + 2) = 80t

70t + 140 = 80t

140 = 10t

14 = t

That t represents the Return trip's time.  Add 2 hours to it since the There trip's time is t+2.  So 14 + 2 = 16.  

B.  16 hours

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

jfkc

step-by-step explanation:

jhcxzbfzc