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3 years ago

5. The graph of a system of two equations is a pair of parallel lines. Does the system have a solution? Explain

1 answer:

8 0

An inconsistent system has no solution. The lines are exactly the same, so every coordinate pair on the line is a solution to both equations.

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Factor: k^2 + 7x + 10 A:(k+10)(k+4) B:(k+2)(k+5) C:(k+2)(k-5) D:(k-2)(k-5)

Nata [24]

Answer:

the answer is B. (k+2)(k+5)

7 0

3 years ago

Read 2 more answers

Drag the tiles to the boxes to form correct pairs.

FrozenT [24]

Answer:

1 .4x2-9= 2x+3,2x-3

2 .16x2-1=4x-1,4x+1

3 .16x2-4=4(2x+1)(2x-1)

4 .4x2-1=(2x+1)(2x-1)

Step-by-step explanation:

16x² − 1 = (4x − 1)(4x + 1) ; 16x² − 4 = 4(2x + 1)(2x − 1); 4x² − 1 = (2x + 1)(2x − 1) ;

4x² − 9 = (2x + 3)(2x − 3)

16x² − 1 is the difference of squares. This is because 16x² is a perfect square, as is 1. To find the factors of the difference of squares, take the square root of each square; one factor will be the sum of these and the other will be the difference.

The square root of 16x² is 4x and the square root of 1 is 1; this gives us (4x-1)(4x+1).

16x² − 4 is also the difference of squares. The difference of 16x² is 4x and the square root of 4 is 2; this gives us (4x-2)(4x+2). However, we can also factor a 2 out of each of these binomials; this gives us

2(2x-1)(2)(2x+1) = 2(2)(2x-1)(2x+1) = 4(2x-1)(2x+1)

4x² − 1 is also the difference of squares. The square root of 4x² is 2x and the square root of 1 is 1; this gives us (2x-1)(2x+1).

4x² − 9 is also the difference of squares. The square root of 4x² is 2x and the square root of 9 is 3; this gives us (2x-3)(2x+3).

3 0

3 years ago

X = 0.19

Luda [366]

Step-by-step explanation:

.19 = 19/100

.864 = 86.4/100

.5348 = 53.48 /100

I multiplied the decimal by 100 to get what it was before which is the same as moving two decimal places to the right and divided by 100

if any confusion ask

4 0

2 years ago

Read 2 more answers

What is the equation of the line that passes through the point (-6, -6) and has a slope of - 1/3

pickupchik [31]

Answer:

The equation of the line is $y = -\frac{1}{3}x - 8$

Step-by-step explanation:

Equation of a line:

The equation of a line has the following format:

$y = mx + b$

In which m is the slope and b is the y-intercept.

Slope of - 1/3

This means that $m = -\frac{1}{3}$

$y = -\frac{1}{3}x + b$

Through the point (-6, -6)

This means that when $x = -6, y = -6$ . So

$y = -\frac{1}{3}x + b$

$-6 = -\frac{1}{3}(-6) + b$

$b + 2 = -6$

$b = -8$

The equation of the line is $y = -\frac{1}{3}x - 8$

5 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

5. The graph of a system of two equations is a pair of parallel lines. Does the system have a solution? Explain​

5. The graph of a system of two equations is a pair of parallel lines. Does the system have a solution? Explain