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2 years ago

How do you solve it?

Mathematics

1 answer:

Basile [38]2 years ago

6 0

9514 1404 393

Answer:

a) NT = 8

b) RS = 8

c) RT = 16

d) ZR = 24

Step-by-step explanation:

All of the marked sides are congruent. (That's what the marks mean.) It might help you to put 8 next to each hash mark like it is on NS. So, you have ...

NT = RS = ST = TZ = NS = 8

Segments on the same line sum, so ...

RT = RS +ST = 2RS = 16

ZR = RS +ST +TZ = 3RS = 24

a) NT = 8

b) RS = 8

c) RT = 16

d) ZR = 24

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2 years ago

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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brainly.com/question/18591940

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