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NemiM [27]
4 years ago
6

What is the slope between (10,4) and (7,4)?

Mathematics
1 answer:
Inga [223]4 years ago
3 0
Slope is rise/run to find it we do (y1-y2)/(x1-x2) y1 means the 1st y and y2 means the second y in (x,y) form y1=4 y2=4 x1=10 x2=7 (4-4)/(10-7)=0/3=0 slope is 0
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Please help with this question ​
Elanso [62]

Answer:

bra no

Step-by-step explanation:

bra

4 0
3 years ago
Nhich equation results from taking the square root of both sides of (x+9)2=25
ale4655 [162]

Answer:

square root of 25=5 and square root of (x+9)=(x+9) so the answer is.... x+9=+-5

Step-by-step explanation:

3 0
3 years ago
Answer then get brainliest :)​
vfiekz [6]

Answer:

\frac{2}{5}  =  \frac{6}{15}  =  \frac{12}{45}

6 0
2 years ago
How many solutions does the system have?
nikdorinn [45]

Answer:

none

Step-by-step explanation:

there is no solutions to this system

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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