Answer:
option C. Angle BTZ Is-congruent-to Angle BUZ
Step-by-step explanation:
Point Z is equidistant from the vertices of triangle T U V
So, ZT = ZU = ZV
When ZT = ZU ∴ ΔZTU is an isosceles triangle ⇒ ∠TUZ=∠UTZ
When ZT = ZV ∴ ΔZTV is an isosceles triangle ⇒ ∠ZTV=∠ZVT
When ZU = ZV ∴ ΔZUV is an isosceles triangle ⇒ ∠ZUV=∠ZVU
From the figure ∠BTZ is the same as ∠UTZ
And ∠BUZ is the same as ∠TUZ
So, the statement that must be true is option C
C.Angle BTZ Is-congruent-to Angle BUZ
The equation of a circle with center and radius is
In your case,
Plug these values into the generic formula and you'll get your equation.
Answer: The speed of car B is 6.44 m/s
Step-by-step explanation:
Let's define:
Sa = Speed of car A.
Sb = Speed of car B.
I assume that at t = 0s, the position of car A is 0m (then at t = 0s, the position of car B is 174m)
Then we can write the positions of each car as:
Pa(t) = Sa*t
Pb(t) = Sb*t + 174m.
We also know that Sa = 2*Sb.
And we know that at t= 27s, car A passes car B.
Then at t = 27s, the positions of both cars must be the same:
Pa(27s) = Pb(27s)
Sa*27s = Sb*27s + 174m.
Now we can replace Sa by 2*Sb (from the above equation)
2*Sb*27s = Sb*27s + 174m
Now we can solve this for Sb.
2*Sb*27s - Sb*27s = 174m
Sb*27s = 174m
Sb = 174m/27s = 6.44m/s.
The speed of car B is 6.44 m/s