If UT is the midsegment of QRS, then SQ = 2(2x + 2) = 4x + 4
Then,
(3x + 1)/(7x - 1) = (2x + 2)/(4x + 4) = 1/2
2(3x + 1) = 7x - 1
6x + 2 = 7x - 1
7x - 6x = 2 + 1
x = 3
Therefore, SQ = 4(3) + 4 = 12 + 4 = 16.
We have:
Initial velocity (u) = 32 m/s
Final velocity (v) = 0 m/s ⇒ The value is zero because the car comes to stationary position when it stops
Time = 14 seconds
We can use one of the constant acceleration equation:

where

is the acceleration



The acceleration is 2.3 m/s⁻² and the negative sign shows deceleration
Answer:
2
Step-by-step explanation:
2+2 is 4 so therefore the answer is 2.00.
Answer:
A b c d
Step-by-step explanation:
pertaining to or using a rule or procedure that can be applied repeatedly. Mathematics, Computers. pertaining to or using the mathematical process of recursion: a recursive function; a recursive procedure.