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3 years ago

How to caculate area of a circle?

Mathematics

2 answers:

grigory [225]3 years ago

7 0

A=pi r elevate 2 A=πr2

Aleks [24]3 years ago

6 0

The area of a circle is pi times the radius squared (A = π r²). Example of calculating the area of a circle is in the image below.

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Please help.. the third and fourth choice are the same as the first two but positive

LenaWriter [7]

Answer:

Your answer is going to be one of the 1/3 I'm pretty sure it is the positive 1/3 because the line is not decreasing.

Step-by-step explanation:

4 0

3 years ago

You drive 140 miles in 2.5 hours and 300 miles in 5.5 hours. What is the rate of change ?

Sveta_85 [38]

The answer is one and five elevenths, or: 1 5/11

this is because 140/2.5 is 56 and 300/5.5 is 54 6/11.

56 - 54 6/11= 1 5/11

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

Let f(x)=x^3-x-1

alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us the function is decreasing at that point.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

6 0

3 years ago

A clothing store offers a free T-shirt when a customer spends $75 or more. Lyndon has already spent $36. 95. Which statement bes

Allisa [31]

Lyndon should spend $38.05 or more to get a free T-shirt.

We have to find the statement best represents all of the amounts he can spend to get a free T-shirt

Suppose that the amount Lyndon get free T shirt is x.

Given that a clothing store offers a free T-shirt when a customer spends $75 or more.

Lyndon has already spent $36.95.

$x+36.95\geq 75$