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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1.) Not a sequence
2.) 2^n-1
3.) Not a sequence
4.) 81/3^n-1
5.) -(-3)^n-1
6.) Not a sequence
Answer 1a 18 1b 36 1c 648
1a : if you look at the shape you will see its made of squares. I counted the squares, its 18.
1b : one side is 6 in. since its a square all sides are equal. so multiply 6 by 6 to get 36.
1c : this one is easy. We take the number of squares across the surface (18) by the area of each square (36) it is 648.
Also sorry I didn't make this clearer earlier.
Answer:
it's B, 39 people.
Step-by-step explanation:
so 39 x 12 = 468
486+275 = $743
yayy :)
