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nekit [7.7K]
2 years ago
6

How do i find the square root of a number that is not a perfect square?

Mathematics
1 answer:
nlexa [21]2 years ago
7 0
"Calculate the square root of 10 (√ 10) to 4 decimal places.<span>Find the perfect square number closer to 10. 32 = 9 and 42 = 16, so take 3.Divide 10 by 3. 10÷3 = 3.33 (you can round off the answer)Average 3.33 and 3. ( 3.33 + 3)÷2 = 3.1667."
Source-</span>http://burningmath.blogspot.com/2013/12/finding-square-roots-of-numbers-that.html
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Mia wants to buy 3 notebooks for $1.29 each. What is the expression for the total cost?
ololo11 [35]

Step-by-step explanation:

Mia want to buy 3 notebook for $1.29 each ,

then total cost =3×1.29

hence, 3×29

1\6+1/12+2/6=2+1+4/12

=7/12

8 0
3 years ago
What is true about the function graphed below?
Dafna1 [17]

Answer:

  D. The axis of symmetry is the y-axis.

Step-by-step explanation:

The vertex is at coordinates (0, 3). Since this parabola opens vertically, its axis of symmetry is the x=coordinate of the vertex: x = 0. That is the equation of the y-axis.

The axis of symmetry is the y-axis.

8 0
2 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

<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>

Point <em>D</em>'(0<em>,0.99)</em>

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.

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

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: <em>Find the volume change</em>

<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>

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
2 years ago
WILL GIVE BRAINLIEST!!
dexar [7]

Peter reflecting trapezoid ABCD across the y-axis would not change the degree measurement of angle A

The degree measurement of angle A is 115 degrees

<h3>How to determine the degree measurement of angle A?</h3>

From the question, we have:

A = 115 degrees

B = 65 degrees

The transformation is a reflection across the y-axis

Reflection is a rigid transformation; and it does not change the angle measure or side lengths.

After the transformation; we have:

A = 115 degrees

B = 65 degrees

Hence, the degree measurement of angle A is 115 degrees

Read more about transformation at:

brainly.com/question/4289712

8 0
2 years ago
PLEASE HELP ASAP I WILL MARK BRAINLIEST!!!
Setler79 [48]

the answer on number 14. is ×=3 your welcome now please give me brainlest

3 0
3 years ago
Read 2 more answers
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