Answer: 25%
Step-by-step explanation:
Boy =50%
Girl =50%
Girl + Girl =25%
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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In mathematical form, it would be: 2x ≥ 40
x ≥ 20
Hope this helps!
Could you attach photo of a problem. It appears 2+..... =4 is missing
The equation is actually
. Free fall is always -16t^2 as the position function. We are looking for how long it takes the object to hit the ground. In other words, the height of an object is 0 when it is laying on the ground, so how long (t) did it take to get there? We will then set that position equal to 0 and solve for t.
. If we subtract 1437 from both sides and divide by -16, we have
. Taking the square root of both sides gives us, rounded to the nearest tenth, t = 9.5 or t=-9.5. The 2 things in math that will never EVER be negative are time and distance/length, so -9.5 is out. That means that it took just about 9.5 seconds for the object to fall to the ground from a height of 1437 feet when pulled on by the force of gravity.
