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3 years ago

How do I get the value of x

Mathematics

1 answer:

brilliants [131]3 years ago

3 0

Actually you should look it up there are so answers that can help you solve them

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At 2:00 PM a car's speedometer reads 30 mi/h. At 2:10 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:10 the acce

Daniel [21]

Answer: If we define 2:00pm as our 0 in time; then:

at t= 0. the velocity is 30 mi/h.

then at t = 10m (or 1/6 hours) the velocity is 50mi/h

Then, if we think in the "mean acceleration" as the slope between the two velocities, we can find the slope as:

a= (y2 - y1)/(x2 - x1) = (50 mi/h - 30 mi/h)/(1/6h - 0h) = 20*6mi/(h*h) = 120mi/ $h^{2}$

Now, this is the slope of the mean acceleration between t= 0h and t = 1/6h, then we can use the mean value theorem; who says that if F is a differentiable function on the interval (a,b), then exist at least one point c between a and b where F'(c) = (F(b) - F(a))/(b - a)

So if v is differentiable, then there is a time T between 0h and 1/6h where v(T) = 120mi/ $h^{2}$

5 0

3 years ago

The sum of an infinite geometric series is 1,280, while the first term of the series is 160. What is the common ratio of the ser

Arlecino [84]

The sum is 160/(1-r)=1280 where r is the common ratio,
1/(1-r)=1280/160=8
Inverting we get 1-r=1/8, r=7/8.

7 0

3 years ago

Simplify each expression by distributing and combining like terms.

solniwko [45]

Hello!

Answer:

1) 7k+35

2) 9n−36

3) 4x+22

Step-by-step explanation:

1) 7(k+5)

1) 7k+7×5

1) 7k+35

2) 9(n-4)

2) 9n+9×−4

2) 9n−36

3) 4(x+5)+2

3) 4x+20+2

3) 4x+(20+2)

3) 4x+22

Hope this helps!

5 0

2 years ago

Need help with problems 16 and 17, please? I'll do brainiac

natali 33 [55]

Answer: 16) y=-0.5x+3 and 17) y=2x-3

Step-by-step explanation:

16: The slope-intercept form of a line is y=mx+b, where m is the slope, and b is the y-value at the y-intercept.

Since y=3 at x=0, the y-intercept is (0,3), and b=3.

Slope= $\frac{y_{2} -y_{1} }{x_{2}-x_{1} }$ where ( $x_{1} ,y_{1}$ ) and ( $x_{2} ,y_{2}$ ) are two points on the line. Choose (0,3) and (2,2):

Slope= $\frac{2-3}{2-0}$ =-0.5

Plug m=-0.5 and b=3 into y=mx+b:

y=-0.5x+3

17: The slope-intercept form of a line is y=mx+b, where m is the slope, and b is the y-value at the y-intercept.

Because y=-3 at x=0, the y-intercept is (0,-3), and b=-3.

Slope= $\frac{y_{2} -y_{1} }{x_{2}-x_{1} }$ where ( $x_{1} ,y_{1}$ ) and ( $x_{2} ,y_{2}$ ) are two points on the line. Choose (0,-3) and (1,-1):

Slope= $\frac{-3-(-1)}{0-1}$ = $\frac{-2}{-1}$ =2

Plug m=2 and b=-3 into y=mx+b:

y=2x-3

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

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

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