From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.
The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:
3.63 t = 1.055 t^2 ==> 1.055 t^2 - 3.63 t = 0
==> t = 0 s or t = 3.44 s
Answer:
Hi
Step-by-step explanation:
Why u do dis
Answer:
The equation of the relation ⇒ 
y = 18
Step-by-step explanation:
∵ y ∝ x/z
∴ y = kx/z
∵ y = 18 , x = 15 and z = 5
∴ 18 = 15k/5 ⇒ 18 = 3k
∴ k = 18/3 = 6
∴ y = 6x/z
∵ x = 21 and z = 7
∴ y = (6 × 21)/7 = 18
If you just want to know what is the least common multiple of 20 and 15, it is 60. Usually, this is written as
lcm(20,15) = 60
Answer:
(3x+1)^2
Step-by-step explanation: