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3 years ago

Please and thank you!! 25 points!!

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

6 0

Answer:

1. 5

2. 6

3. 1

4. 4

5. 2

6. 3

Step-by-step explanation:

→Number 1 is paired with explanation 5, because it explains how a certain pair can be associated with one another, and provides the example of corresponding angles.

→Number 2 is paired with explanation 6, because the explanation states how angles are paired with one another based on the correspondence between them.

→Number 3 is paired with explanation 1, because the explanation states how sides are paired with one another based on the correspondence between them.

→Number 4 is paired with explanation 4, because the explanation is talking about how figures can be congruent to one another.

→Number 5 is paired with explanation 2, because the explanation states the definition of an included angle.

→Number 6 is paired with explanation 3, because the explanation states the definition of an included side.

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The braking distance, in feet of a car a Travling at v miles per hour is given.

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The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

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