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2 years ago

Please answer :puppy eyes emoji:

Mathematics

1 answer:

Anettt [7]2 years ago

8 0

Well ven diagrams help you visually compare and contrast two figures. They helps you note the similarities and the differences between two items. Hope this helps!

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Evaluate the definite integral. 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$

$= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C$

substitute u = 1-2x⁵ into the result

$= \dfrac{1}{40(1-2x^5)^4} +C$

Hence $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ $= \dfrac{1}{40(1-2x^5)^4} +C$

5 0

3 years ago

Need some help with this

sergij07 [2.7K]

Answer:

gradient 1/2, y-intercept 9

Step-by-step exp

4 0

3 years ago

Read 2 more answers

Which quadrilateral is a trapezoid? B. с А G

Romashka-Z-Leto [24]

Answer:

You didn't add a photo

Step-by-step explanation:

......

6 0

3 years ago

If the measurment of the angles of a triangle are 3b,2b,and 4b then what is the smallest angle

d1i1m1o1n [39]

Given:

The measurement of the angles of a triangle are 3b,2b,and 4b.

To find:

The smallest angle.

Solution:

According to the angle sum property, the sum of all angles of a triangle is 180 degrees.

$3b+2b+4b=180^{\circ}$ [Angle sum property]

$9b=180^{\circ}$

Divide both sides by 9.

$b=20^{\circ}$

Now,

$3b=3(20^{\circ})=60^{\circ}$

$2b=2(20^{\circ})=40^{\circ}$

$3b=4(20^{\circ})=80^{\circ}$

Therefore, the smallest angle is 40 degrees.

6 0

3 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