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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Answer:
1) A. -10/27
2) Either A or D whichever is supposed to be negative
3) -1 7/10
4) - 1 10/11
5) C. -1 31/33
Step-by-step explanation:
-3 1/3 divided by 9
-3 1/3 = -10/3
-10/3(1/9) = -10/27
-3 3/4 divided by 3
-3 3/4 = -15/4
-15/4(1/3) = -15/12 = -1 3/12 = -1 1/4
-2 1/8 divided by 1 1/4 WHAT IS A MULTI ANSWER?
- 17/8 divided by 5/4
- 17/8(4/5) = 68/40 = 1 7/10
-5 1/4 divided by 2 3/4 WHAT IS A MULTI ANSWER?
-21/4(4/11) = -21/11 = -1 10/11
-5 1/3 divided by 2 3/4
-16/3(4/11) = -64/33 = -1 31/33
Answer:
30%
Step-by-step explanation:
Answer:
10.75
Step-by-step explanation:
70%. I got this answer by starting with the fraction 875/1250, dividing the top of the fraction by the bottom (875÷1250) and then multiplying my answer by 100.
