Answer:
s = 22.5 m
Step-by-step explanation:
the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.
v (t) = 25 - 5 t
at t = 0 , v = 20 m/s
Let the distance is s.
Let at t = t, the v = 20
So,
20 = 25 - 5 t
t = 1 s
So, s = 25 x 1 - 2.5 x 1 = 22.5 m
Answer: -3/2x + 3 ⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀
The answer is 6/45. Ask photomath
∠EGF = 65°
Since EF = EG the triangle is isosceles and the base angles are equal, that is
∠EGF = ∠GEF
∠EGF = = = 65°
B)113.04 cm^2