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3 years ago

Pleaseee answer correctly !!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

tester [92]3 years ago

8 0

Answer:

<h to <g

Step-by-step explanation:

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Hannah has $2 worth of dimes and quarters. She has a total of 17 dimes and quarters altogether. By following the steps below, de

sineoko [7]

Answer:

2 quarters and 15 dimes

Step-by-step explanation:

15 dimes = $1.50

2 quarters = $0.50

$1.50+$0.50=$2.00

4 0

3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

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

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