Will give brainliest will give brainliest and 50 points out of a port departed a steam boat with a speed of 36.8km/h. in 1.75 hours out of the same port and going the sam direction departed a speed boat. in 1.8 hours after the the department of the speed boat, the distance between the speed boat and the steam boat was 86.9km(behind the steam boat)what was the speed boats speed in km/h?

Step-bysteam boat (b) = speed of 36.8km/h. In 1.75 hours out same port and going same direction departed speed boat (s). In 1.8 hours later distance between the speed boat (s) and steam boat (b) was 86.9km

Scenario A) the steam boat (b) stays ahead of the speed boat (s)

Start by first finding the total distance (d) that the steam boat (b) traveled. That would be 1.75+1.8 = 3.55 hrs

distance (d) = rate (r) × time (t)

d(s) = rt = (36.8km/h)(3.55h) = 130.64 km

Secondly, let's subtract from d the distance between b and speed boat (s):

d(s) = 130.64 - 86.9km = 43.74 km

Thirdly, since s traveled 43.74 km and left 1.8 hrs ago, let's set-up the rate of b [r(b)] equal to that distance [d(s)] divided by the time 1.8 hrs.

**Since d = rt, r = d/t

r(s) = d(s)/t = 43.74km/1.8h = 24.3km/h

Scenario B) the speed boat (s) moves ahead of the steam boat (b), this is more likely!!

b moves before s starts for 1.75 hrs, going 36.8km/h, that distance ahead + distance it travels after the next 1.8 hrs = 130.64 km

Now s starts and travels d(s) km at r(s) km/h

If s speeds past b in 1.8 hrs, and is AHEAD by the given distance of 86.9km, then total distance of s [d(s)] = the 130.64 b traveled PLUS the 86.9 it is now ahead

d(s) = 130.64+86.9 = 217.54 km

s still only traveled that in 1.8 hrs, so the speed/rate of s is:

r(s) = d(s)/t = 217.54km/1.8h = 120.86km/h

** I think that makes more sense, that a speed boat would move MUCH faster than a steam boat, if 120.9... than over 3 times faster!!

do i do it on a graph?

step-by-step explanation: