1 x 2 x 2 x 2 x 3 x 3 x 7
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:
Therefore, the pairs of twin-prime numbers are (101,103) , (107,109) , (137,139) , (149,151) , (179,181) , (191,193) , (197,199) . So, the correct answer is “ (101,103) , (107,109) , (137,139) , (149,151) , (179,181) , (191,193) , (197,199) ."
Answer:
12
Step-by-step explanation:
This is assuming that x2 means x to the power of 2 and that the x on the left is to represent multiplying. First, plug in all of the number to get 2*3+3^2-3.Using PEMDAS, you should solve the exponent to get 9. Then, multiply 2 by 3 to get 6. Add 6 to 9 to get 15. Subtract 2 to 15 to get 12.
TL;DR
2*3+3^2-3
6+9-3
15-3
12
37.5% is what I think the correct answer is sorry if its wrong
