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3 years ago

What is the difference between 5x +3 and 5(x+3)?

Mathematics

1 answer:

ella [17]3 years ago

8 0

Step-by-step explanation:

for 5(x+3) you have to use the distributive property to find its standard form because it has parenthesis:

this would give you 5x+15 which is not the same as 5x+3.

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point I is on line segment HJ. GIVEN IJ = 3× + 3, HI = 3× - 1, and HJ = 3× + 8, determine the numerical length of HJ.

Vedmedyk [2.9K]

Given, IJ = 3x + 3, HI = 3x - 1, and HJ = 3x + 8.

Since I is a point on line segment HJ, we can write

$HJ=HI+IJ$ $\begin{gathered} 3x+8=(3x-1)+(3x+3) \\ 3x+8=6x+2 \\ 8-2=6x-3x \\ 6=3x \\ 2=x \end{gathered}$

Put x=2 in HJ=3x+8.

$\begin{gathered} HJ=3\times2+8 \\ HJ=6+8 \\ =14 \end{gathered}$

Therefore, the numerical length of HJ is 14.

4 0

1 year ago

Derivative f(x) = arcsin (2x-1)

sattari [20]

Answer:

x= 1/2

Step by Step explanation:

1- Substitute f (x) = 0

2- Determine the defined range

3- Swap the sides

4- Use the inverse trigonometric function

5- Evaluate the expression

6- Move the constant to the right

7- Divide both sides

8- Check the solution

5 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

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

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