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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Could 10.6\text{ cm}, 5.6\text{ cm},10.6 cm,5.6 cm,10, point, 6, space, c, m, comma, 5, point, 6, space, c, m, comma and 4.0\tex
vladimir1956 [14]
Since we know that triangle inequality theorem states that sum of any two sides of a triangle must be greater than third side.
Now let us see if it is true for our given side lengths.
Now let us try with another pair.
We can see that sum of 5.6 and 4 is less than 10.6. Therefore, 10.6 cm, 5.6 cm and 4.0 cm can not be side lengths of a triangle.
Y=1/x solve for x
x=1/y now reverse lables...
y=1/x
Yes that is the same equation you started with. What that means is the the inverse function in this case produces the same exact graph :)
I believe that the answer to the equation is 10
