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3 years ago

HURRY!!!! NO SPAM!!!! Identify the area of the trapezoid.

Mathematics

2 answers:

saveliy_v [14]3 years ago

8 0

Answer:

$area = >$

a+b/2*h

Here,

a = 12xcm

b= 15xcm

h= 7cm

$= >$

(12+15)xcm/2*7

=27xcm*3.5

Virty [35]3 years ago

6 0

Answer:

94.5

Step-by-step explanation:

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Write the equations of three distinct lines that have the same y-intercept,-1.

guajiro [1.7K]

Three possible answers are:
y = 2x-1
y = 5x-1
y = -3x-1

All of them have the same y intercept of -1. The slopes are the only thing that differ.

Any linear equation can be written in the form y = mx+b where
m = slope
b = y intercept

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4 years ago

Simplify this expression 2(3x -2 + 2y) - (2 + 4x + y)

luda_lava [24]

Answer:

2 x + 3y − 6

Step-by-step explanation:

If you simplify you should get 2 x + 3y − 6

4 0

3 years ago

Read 2 more answers

Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ

tensa zangetsu [6.8K]

Answer: RQ= 8.99 ( approx)

Step-by-step explanation:

Let MR= x

Since, In triangle, PRQ, tan 75°= $\frac{18+x}{RQ}$

⇒ RQ= $\frac{18+x}{tan 75^{\circ}}$

Now, In triangle MRQ,

tan 60°= $\frac{18+x}{RQ}$

⇒ RQ= $\frac{x}{tan 60^{\circ}}$

On equating both values of RQ,

$\frac{18+x}{tan 75^{\circ}}=\frac{x}{tan 60^{\circ}}$

⇒ $\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}$

⇒ $\frac{18+x}{x}=2.15470053838$

⇒ $18=2.15470053838x-x$

⇒ $x=15.5884572681$ ≈15.60

Thus RQ=8.99999999999≈8.99

6 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

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

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

3 years ago