Let
M---------------> money borrowed -------------> <span>$9,850
r--------------> </span>discounted rate--------> <span>9 ¼=9.25-------> 0.0925
t---------------> time--------> </span><span>9 months=9*30=270 days
D-------------> </span><span>amount of the discount
we know that
D=M*r*t/360=(9850)*(0.0925)*(270/360)=683.34
the answer is $683.34</span>
19 if u add all separately but it is 89 it u don't
Answer:
5a. -0.4 m/s²
5b. 290 m
6. 12.9 s
7. 100 s
8. 17.2 km/hr
Step-by-step explanation:
5. "While approaching a police officer parked in the median, you accelerate uniformly from 31 m/s to 27 m/s in a time of 10 s.
a. What is your acceleration?
b. How far do you travel in that time?"
Given:
v₀ = 31 m/s
v = 27 m/s
t = 10 s
Find: a and Δx
v = at + v₀
(27 m/s) = a (10 s) + (31 m/s)
a = -0.4 m/s²
Δx = ½ (v + v₀) t
Δx = ½ (27 m/s + 31 m/s) (10 s)
Δx = 290 m
6. "If a pronghorn antelope accelerates from rest in a straight line with a constant acceleration of 1.7 m/s², how long does it take for the antelope to reach a speed of 22 m/s?"
Given:
v₀ = 0 m/s
v = 22 m/s
a = 1.7 m/s²
Find: t
v = at + v₀
(22 m/s) = (1.7 m/s²) t + (0 m/s)
t = 12.9 s
7. "A 1200 kg airplane starts from rest and moves forward with a constant acceleration of 5 m/s² along a runway that is 250 m long. How long does it take the plane to travel the 250 m?"
Given:
v₀ = 0 m/s
a = 5 m/s²
Δx = 250 m
Find: t
Δx = v₀ t + ½ at²
(250 m) = (0 m/s) t + ½ (5 m/s²) t²
t = 100 s
8. "During a marathon, a runner runs the first 10 km in 0.58 hours, the next 10 km in 0.54 hours and the last 10 km in 0.62 hours. What is the average speed of the runner during that marathon?"
This isn't a constant acceleration problem, so there's no need for a chart.
Average speed = total distance / total time
v = (10 km + 10 km + 10 km) / (0.58 hr + 0.54 hr + 0.62 hr)
v = 30 km / 1.74 hr
v = 17.2 km/hr
130+45= 175
she took out $175
Answer:
f(x) = -√(x-4) + 2
Step-by-step explanation:
(a) Start with the basic function: f(x) = √x.
This would be the top half of a horizontal parabola opening to the right (always positive).
(b) The range is y < 2. This includes negative numbers, so the function must be f(x) = -√x.
(c) The function starts at x = -4. We must shift it four units to the left. We do that by adding four units to x. The function must be f(x) = -√(x + 4). The domain is x > -4.
(d) The function starts at y = 2. We must shift it up two units. We do that by adding two units to y. The function must be f(x) = -√(x+4) + 2. The range is y < 2.
The figure below is the graph of f(x) = -√(x+4) + 2.
