Answer:
Step-by-step explanation:
To find : Acceleration in first 15 min . Distance between two cities Average speed of journey
Solution:
Each horizontal block is 1/8 hr = 7.5 min
Each vertical block is 10 km/hr
Time Velocity km/hr
0 Min ( 0 hr) 0
15 Min (1/4 hr) 50
45 Min (3/4 hr) 50
60 MIn ( 1 hr) 100
90 Min ( 3/2 hr) 100
120 Min ( 2hr) 0
Acceleration in first 15 min (1/4 hr) = (50 - 0)/(1/4 - 0) = 50/(1/4)
= 200 km/h²
Distance between two cities
= (1/2)(0 + 50)(1/4 - 0) + 50 * (3/4 - 1/4) + (1/2)(50 + 100)(1 - 3/4) + 100 * (3/2 - 1) + (1/2)(100 + 0)(2 - 3/2)
= 25/4 + 25 + 75/4 + 50 + 25
= 125
Distance between two cities = 125 km
Average Speed of journey = 125/2 = 62.5 km/hr
Acceleration in first 15 min = 200 km/h²
Distance between two cities = 125 km
Average Speed of journey = 62.5 km/hr
Hope this helps..
Let J = rate of jet in still airLet W = rate of wind Distance formula: d = rt / r = d/t Flying against the wind the jet flies at a rate of J - W = 1860 miles/3 hours = 620 miles per hourFlying with the wind the jet flies at a rate of J+W = 9180 miles/9 hours = 1020 miles per hour The average of these 2 rate is the speed of the jet in still air J = (620+1020)/2 = 820 miles per hour J - W = 620
820 - W = 620
W = 820 - 620 = 200 miles per hour The jet in still air flies at a rate of 820 mph(miles per hour)The wind speed is 200 mph(miles per hour)
Answer:
Sorry but your question is incomplete!
it's impossible to answer this question if we don't know the <u>total number of outcomes</u>
<em>I'm referring to writing the question instead of OCR-ing it</em>
Answer:
B
Step-by-step explanation:
THis is a single equation with a single variable [y].
We can separate the variable in one side and the numbers to another and solve for y. Shown below:
3y + 4 = -2
3y = -2 - 4
3y = -6
y = -6/3
y = -2
THus, this equation has only 1 solution. B is correct
Its 43 pushups cause the number of pushups are going up by 3