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Shalnov [3]
2 years ago
8

Find the area of the regular 18-gon with radius 11 mm

Mathematics
1 answer:
Anna35 [415]2 years ago
3 0

Answer:

36

Step-by-step explanation:

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Jay on started off his penny collection with 1 penny. He then adds 5 pennies to his collection each day. How could you change th
nata0808 [166]

Answer:

By doubling each day

Step-by-step explanation:

Jay is increasing  daily a fixed amount of pennies to his collection.  That's an arithmetic series, because to obtain the next term you have to do an addition.

To get a geometric series or progression you have to find the next term through multiplication.

So, if he wants to convert his series into a geometric series, he'd have to double his contribution to his collection each day (as an example).

4 0
3 years ago
A line has a slope of -5 and passes through (1,-1). What is the equation in line point-slope form?
valentinak56 [21]
The slope of the line is Y=-5
4 0
2 years ago
Read 2 more answers
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

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4 years ago
Helpp plzzz I will mark brainliestttt
tangare [24]
294cm^2
49 x 6 (sides)
3 0
3 years ago
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HELP WILL GIVE BRAINLIEST FOR ANSWER QUICKLY PLEASE HELP
Helen [10]

Answer:

We have sinθ = 12/13

The  method here is to figure out the value of θ

Using a calculator sin^(-1)(12/13) =67.38°

67.38° is in quadrant 1  so we must substract 67.38° from 180° wich is π

  • 180-67.38= 112.61° ⇒ θ= 112.61°

Now time to calculate cos2θ and cosθ using a calculator

  • cosθ = -5/13
  • cos2θ =  -0.7

The values we got make sense since θ is in quadrant 2 and 2θ in quadrant 3

5 0
3 years ago
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