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3 years ago

Solve the equation step by step in form of rational and irrational numbersx^2=144

Mathematics

2 answers:

Blababa [14]3 years ago

8 0

X^2=144
12^2=144
x=12 or -12

Olenka [21]3 years ago

5 0

X^2=144 => x=√144 => x=12 or -12

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Please help me with this thank you !!

andreev551 [17]

I think the answer is b i've never learned this sorry :(

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3 years ago

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

irakobra [83]

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

Learn more here:

brainly.com/question/18591940

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2 years ago

This graph shows a proportional relationship.

Artemon [7]

It a keep on the work

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2 years ago

The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for hou

o-na [289]

This question s incomplete, the complete question is;

The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for 15 hours. The Thompson family's sprinkler was used for 30 hours.

There was a combined total output of 1050 of water. What was the water output rate for each sprinkler if the sum of the two rates was 55L per hour

Answer:

The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr

Step-by-step explanation:

Given the data in the question;

let water p rate for Watson family and the Thompson family sprinklers be represented by x and y respectively

x + y = 55 ----------------equ1

x = 55 - y ------------------qu2

also

15x + 30y = 1050

x + 2y = 70 --------------equ3

input equ2 into equ3

(55 - y) + 2y = 70

- y + 2y = 70 - 55

y = 15

input value of y into equ1

x + 15 = 55

x = 55 - 15

x = 40

Therefore, The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr

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3 years ago

In the milkshake example, the price ($) of the milkshake is the

vlabodo [156]

Independent variable (im pretty sure sorry if im wrong)

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3 years ago

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