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Nuetrik [128]
3 years ago
14

Order from least to greatest: 15,21,15,15,18,19,14,20,95,18,21,14,15,20,16,14,22,21,15,19

Mathematics
1 answer:
EleoNora [17]3 years ago
8 0
1,2,14,14,14,15,15,15,15,15,16,18,18,19,19,20,20,21,21,22,95
Hope that helps!!!
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11 The net force on a vehicle that is accelerating at a rate of 1.5 m/s² is 2,800 newtons. What is
ololo11 [35]

Answer:

Mass = 1867kg

Step-by-step explanation:

Given

Force = 2800N

Acceleration = 1.5m/s^2

Required

Determine the mass of the vehicle

This question will be answered using Newton's second law

Force = Mass * Acceleration

Substitute values for Force and Acceleration

2800 = Mass * 1.5m/s^2

Make Mass the subject

Mass = \frac{2800N}{1.5m/s^2}

Mass = 1866.6667kg

Mass = 1867kg ---- approximated

<em>Hence, the mass of the vehicle is 1867kg</em>

4 0
2 years ago
14 a and b please and thank u
Murrr4er [49]

I don't know the answer, but I suggest downloading the app: PhotoMath. It scans a math problem and shows you how to complete the problem. I hope this helped~!

7 0
3 years ago
After Ricardo received his allowance for the week, he went to the mall with some friends. He spent half of his allowance on a ne
ad-work [718]
Lets work backwards, he had $5 after it all, and spent $1.25 on a snack, so we add that to the remainder, which is $6.25. then he spent half of that on whatever stuff he likes, so add $6.25 and $6.25, which is $12.50
6 0
2 years ago
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While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
tekilochka [14]

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

<h3>How to estimate the speed of the moving walkway relative to the airport terminal?</h3>

Let x be the speed of the walkway.

(2.8 + x) = speed of child moving in direction of the walkway

(2.8 - x) =  speed of child moving against the direction of the walkway

Travel time = distance/speed

Travel time of child moving in direction of walkway = 23/(2.8+x)

Total elapsed time given = 29s

23/(2.8 + x)+ 23 / (2.8-x) = 29

LCD = (2.8 + x)(2.8 - x)

23(2.8 - x) + 23(2.8 + x) = 29(2.8 + x)(2.8 -x)

simplifying the equation, we get

23*2.8-23x+23*2.8+23x=29(2.8^2-x^2)

23(2.8+2.8)/29=2.8^2-x^2

x^2=(2.8)^2-(23*5.6)/29)=3.4

x=\sqrt{3.4}=1.84m/s

Speed of walkway = 1.84 m/s

The speed of the moving walkway relative to the airport terminal exists at 1.84 m/s.

To learn more about Speed refer to:

brainly.com/question/4931057

#SPJ4

4 0
1 year ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
pav-90 [236]

Answer:

-3f - 10

Step-by-step explanation:

-1 times 10 = -10

-1f times 3f = -3f

Sorry if this isn't right.

4 0
3 years ago
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