A power boat travels along a river in which the flow of water is at a constant in one direction. If the boat travels at 40 mph downstream and 30 mph upstream. What is the speed of the boat in still water and the speed of the river?

speed of boat is 35 mph

speed of river is 5mph.

Step-by-step explanation:

Let the speed of boat be u mph

let the speed of river be v mph

When boat travel downstream

in that case flow  of river will support the speed of boat hence

speed of river will be added to speed of boat

Thus,

speed of boat in downstream = speed of boat + speed of river = u + v

Given

the boat travels at 40 mph downstream

thus,

u + v = 40           1

When boat travel upstream

in that case flow  of river will oppose the speed of boat hence

speed of river will be subtracted from  speed of boat

Thus,

speed of boat in downstream = speed of boat -speed of river = u - v

Given

the boat travels at 30 mph upstream

thus,

u - v = 30  2

adding 1 + 2

we have

u + v = 40  

+ u - v = 30

2u = 70

u = 70/2 = 35

Thus,

speed of boat is 35 mph

using u = 35 in u + v = 40  

we have

35 + v = 40

v = 40 - 35 = 5

Thus, speed of river is 5mph.

step-by-step explanation:

; fx

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)