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Setler [38]
3 years ago
6

The line plot shows the distances some students jumped during a Field Day competition.

Mathematics
1 answer:
trapecia [35]3 years ago
5 0
To answer this question here is what you would do. I can't give you the exact answer because there is not a line plot attached.

Count the number of X's above the number 5 1/2 feet. Put this number over the total number of X's you count (including the 5 1/2s).

This is your answer when you write these 2 pieces of information in fraction form.
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What did the 800lb monster say to the 400lb monater ??
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Some one please help!
nasty-shy [4]

Answer:

The answers are 23, 9,200 and 26, 10,400

Step-by-step explanation:


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3 years ago
Read 2 more answers
PLEASE HELP ASAAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER
Nutka1998 [239]

Let i = sqrt(-1) which is the conventional notation to set up an imaginary number

The idea is to break up the radicand, aka stuff under the square root, to simplify

sqrt(-8) = sqrt(-1*4*2)

sqrt(-8) = sqrt(-1)*sqrt(4)*sqrt(2)

sqrt(-8) = i*2*sqrt(2)

sqrt(-8) = 2i*sqrt(2)

<h3>Answer is choice A</h3>
5 0
3 years ago
A ½in diameter rod of 5in length is being considered as part of a mechanical linkagein which it can experience a tensile loading
Evgesh-ka [11]

Answer:

a. Maximum Load = Force = 27085.09 N

b. Maximum Energy = 3.440 Joules

Step-by-step explanation:

Given

Rod diameter = ½in = 0.5in

Length = 5in

Young's modulus = 15.5Msi

By applying the 0.2% offset rule,

The maximum load the rod can hold before it gets to breaking point is given as follows by taking the strain as 0.2%

Young Modulus = Stress/Strain ------- Make Stress the Subject of Formula

Stress = Strain * Young Modulus

Stress = 0.2% * 15.5

Stress = 0.002 * 15.5

Stress = 0.031Msi

Calculating the area of the rod

Area = πr² or πd²/4

Area, A = 22/7 * 0.5^4 / 4

A = 22/7 * 0.25 / 4

A = 5.5/28

A = 0.1964in²

The maximum load that the rod would take before it starts to permanently elongate is given by

Force = Stress * Atea

Force = 0.031Msi * 0.1964in²

Force = 31Ksi * 0.1964in²

Force = 6.089Ksi in²

Force = 6.089 * 1000lbf

Force = 6089 lbf

1 lbf = 4.4482N

So, Force = 6089 * 4.4482N

Force = 27085.09 N

b.

Using Strain to Energy Formula

U = V×σ²/2·E

Where V = Volume

V = Length * Area

V = 5 in * 0.1964in²

V = 0.982in³

σ = Stress = 0.031Msi

= 0.031 * 1000Ksi

= 31Ksi

= 31 * 1000psi

= 31000psi

E = Young Modulus = 15.5Msi

= 15.5 * 1000Ksi

= 15.5 * 1000 * 1000psi

= 15500000psi

So,

Energy = 0.982 * 31000²/ ( 2 * 15500000)

Energy = 943,702,000/31000000

Energy = 30.442in³psi

------- Converted to ftlbf

Energy = 2.537 ftlbf

-------- Converted to Joules

Energy = 3.440 Joules

7 0
4 years ago
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