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2 years ago

What is the value of the rational expression 2x+1 when x = 5?

Mathematics

2 answers:

Nataly [62]2 years ago

6 0

Step-by-step explanation:

2x+1

x = 5

2(5)+1

=10+1=11

Stolb23 [73]2 years ago

6 0

Answer:

2(5)+1

10+1

Hope This Helps!!!

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Jon has 6 kites he and his friends will each fly one kite how many people in all will fly a kit

True [87]

Answer:

Step-by-step explanation:

One person will fly a kite for each of the 6 kites Jon has.

6 people in all will fly a kite.

6 0

3 years ago

Which equation represents a line that passes through (2,-1/2) and has a slope of 3 ?

Anna11 [10]

Answer:

C. $y + \frac{1}{2} = 3(x - 2)$

Step-by-step explanation:

Given

$Point: (2,\frac{-1}{2})$

$Slope: 3$

Required

Equation of line

Let m represents the slope of the line;

m is calculated as thus

$m = \frac{y - y_1}{x - x_1}$

$where (x_1,y_1) = (2,\frac{-1}{2})$

$So; x_1 = 2; y_1 = \frac{-1}{2}$

$m = 3$

By substituting the right values in the formula above;

$m = \frac{y - y_1}{x - x_1}$ becomes

$3 = \frac{y - \frac{-1}{2}}{x - 2}$

Multiply both sides by $(x - 2)$

$3 *(x - 2) = \frac{y - \frac{-1}{2}}{x - 2} *(x - 2)$

$3 *(x - 2) = (y - \frac{-1}{2})$

$3 *(x - 2) = (y + \frac{1}{2})$

$3(x - 2) = (y + \frac{1}{2})$

$3(x - 2) = y + \frac{1}{2}$

Reorder

$y + \frac{1}{2} = 3(x - 2)$

Hence, the equation that represents the line is $y + \frac{1}{2} = 3(x - 2)$

4 0

3 years ago

HELP ME PLEASE! tell me if 11,12 are right and help w/ 13

Marysya12 [62]

Answer: I know 12 is right

Step-by-step explanation:

7 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

Dividing rational expressions

Galina-37 [17]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
\cfrac{x+3}{x^2-2x}\cdot \cfrac{(x-2)^2}{x^2-9}\implies \cfrac{x+3}{x^2-2x}\cdot \cfrac{(x-2)(x-2)}{x^2-3^2}
\\\\\\
\cfrac{x+3}{x(x-2)}\cdot \cfrac{(x-2)(x-2)}{x^2-3^2}\implies \cfrac{\underline{x+3}}{x\underline{(x-2)}}\cdot \cfrac{\underline{(x-2)}(x-2)}{(x-3)\underline{(x+3)}}
\\\\\\
\cfrac{x-2}{x(x-3)}\iff\cfrac{x-2}{x^2-3x}$

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3 years ago