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Rama09 [41]
2 years ago
10

the braking distance of a vehicle is directly proportional to the square of its speed. when the speed of the vehicle is b m/s, i

t's breaking distance is d m. if the speed of the vehicle is increased by 200%. find the percentage increase in its breaking distance​
Mathematics
1 answer:
lorasvet [3.4K]2 years ago
6 0
“The breaking distance is directly proportional to the square of its speed”
Using your variables, this means: d = k•b^2
k here is some unknown (and for this question irrelevant) number.

Now if b (the speed) is increased by 200%, then your new speed is b+2b = 3b. The 2b is the 200% increase.

Throw 3b into the first equation we have the new d as:
d = k • (3b)^2

Simplifying this, you have d = k•9b or d=9kb

This means the new braking distance is 9 times the original braking distance.
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Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t
NISA [10]

Answer:

Class second have higher score and have greater spread.

Step-by-step explanation:

For first box plot

\text{Minimum value }= 53

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For second box plot

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First class has greater median.

second class has greater third quartile.

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Required [Missing from the question]

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