Answer:
ln (m^2n^9)
Step-by-step explanation:
Rule: ln a + ln b = ln ab
Rule: ln a^n = n * ln a
2 ln m + 9 ln n =
= ln m^2 + ln n^9
= ln (m^2n^9)
= 
Answer:the car was traveling at a speed of 80 ft/s when the brakes were first applied.
Step-by-step explanation:
The car braked with a constant deceleration of 16ft/s^2. This is a negative acceleration. Therefore,
a = - 16ft/s^2
While decelerating, the car produced skid marks measuring 200 feet before coming to a stop.
This means that it travelled a distance,
s = 200 feet
We want to determine how fast the car was traveling (in ft/s) when the brakes were first applied. This is the car's initial velocity, u.
Since the car came to a stop, its final velocity, v = 0
Applying Newton's equation of motion,
v^2 = u^2 + 2as
0 = u^2 - 2 × 16 × 200
u^2 = 6400
u = √6400
u = 80 ft/s
yeah there is u should be able to by clicking on the paper clip
Answer:
36π - 72 sq in.
Step-by-step explanation:
π×12² = 144π
90/360 × 144π = 36π
Triangle = ½ × 12 × 12 = 72
Segment = 36π - 72
