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evablogger [386]
3 years ago
7

landon wants to show that the product of rational numbers is always a rational number. complete his work and explanation by fill

ing in the boxes with values that support his conclusion​

Mathematics
1 answer:
lubasha [3.4K]3 years ago
7 0

Multiply <u>√2</u> by <u>√72</u>. The product is a <u>rational</u> number because <u>√144</u> can be simplified to an integer.

Step-by-step explanation:

As Landon has to prove that two product of two rational numbers, he has to choose two rational numbers from the list and then multiply and show that the product is also a rational number.

Let us define the rational numbers first

A number that can be written in the form of p/q where p,q are integers and q is not equal to zero, is called a rational number.

From the give =n list of rational numbers

Taking

√2 and √72

\sqrt{2} * \sqrt{72}\\=\sqrt{2*72}\\=\sqrt{144}\\=12\\=\frac{12}{1}

As we can see that the product of √2 and √72 is 12 which is also a rational number.

So,

Multiply <u>√2</u> by <u>√72</u>. The product is a <u>rational</u> number because <u>√144</u> can be simplified to an integer.

Keywords: Rational numbers, Product

Learn more about rational numbers at:

  • brainly.com/question/10879401
  • brainly.com/question/10940255

#LearnwithBrainly

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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

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

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+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>

Point A'(0<em>,0)</em>

Point B(x=1,y=0)

Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>

Point <em>B</em>'(1.03<em>,0)</em>

Point C(x=1,y=1)

Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>C</em>'(1.03<em>,0.99)</em>

Point D(x=0,y=1)

Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>

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The plot is attached with the solution.

<em>Part b: Calculate the six strain components.</em>

Solution

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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

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<em></em>\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\<em></em>

<em>Also the change in volume is 0.0197</em>

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

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