By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)
<h3>How many more distance does the average race car travels than the average consumer car?</h3>
In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:
s' = (v' - v) · t
Where:
- v' - Speed of the average race car, in miles per hour.
- v - Speed of the average consumer car, in miles per hour.
- t - Time, in hours.
Please notice that a hour equal 3600 seconds. If we know that v' = 210 mi / h, v = 120 mi / h and t = 30 / 3600 h, then the distance surplus of the average race car is:
s' = (210 - 120) · (30 / 3600)
s' = 3 / 4 mi
The distance surplus of the average race car is equal to 3 / 4 miles.
To learn more on uniform rectilinear motion: brainly.com/question/10153269
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Area of part 1:A1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2Area of part 2:A2=length*width=(8+2)*1=10 ft2Area of part 3:A3=length*width=8*2=16 ft2
Total shaded area : A1+A2+A3=41 ft2Total area of the reactangle : A=16*8=128 ft2Total area of the nonshaded region : 128-41=87 ft2
Answer:
you plug in the 3. so you get y= 3-4 which is -1
Answer:
numbers of tables on the y axis and the money earned on the x axis
Step-by-step explanation:
