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:
148 square inches
Step-by-step explanation:
Step 1: finding the area of the base (upper base plus lower base)
2( L x W).
2 (6in. x 5in.)
2( 30 square inches.)
60 square inches
Step 2: finding the area of first 2 sides
2 ( l x w)
2 ( 5in x 4in)
2 ( 20 square inches.)
40 square inches
step 3: finding the area of the last 2 sides
2(lxw)
2(4in.x 6in.)
2(24 square inches.)
48 square inches
step 4: combine all the answers.
60 square inches. + 40 square inches. + 48 square inches = 148 square inches
Answer:
Step-by-step explanation:
b=bench
852=b+(b-98) since the table is less than the bench by 98$
852-b=b-98
950-b=b
950=2b
475=b
