Answer:
t ’= 
, v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’= 
t ’=
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’= 
t ’= 547.19 s
Distance= speed * time
So multiply speed and time to get the value of distance
20*9.8 =196 meters
The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y
Find x:
x = y/tan(θ)
So x = 3/tan(π/6)
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt
Since altitude remains the same, dy/dt = 0. Now...
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min
Answer:
Need a picture
-Malignant tissue
-Over the lower abdomen
-during pregnancy
-Over an infection
Explanation:
IDK if this helps bc there is no picture, but here ya go
