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3 years ago

Describe the shape of the distribution below.

Mathematics

2 answers:

alexandr402 [8]3 years ago

6 0

Answer:

orange_08............................

Ahat [919]3 years ago

3 0

The distribution is approximately symmetrical

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What is the approximate area of the triangle below ?

aleksandrvk [35]

Answer:

72.8

Step-by-step explanation:

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4 years ago

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What is the value of x? A. 142 B. 152 C. 76 D. 71

Nadusha1986 [10]

Answer:

x has a value of 71, making D the correct answer.

Step-by-step explanation:

Because this is an isosceles triangle, we know that the bottom two angles are equal. We also know that the sum of the angles inside a triangle will always come to 180°. With that in mind, we can simply subtract 38° from 180° and divide by two to get the correct answer.

$x = (180 - 38) / 2\\= 142 / 2\\= 71$

So the correct answer is D, 71 degrees.

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3 years ago

Determine the area of the circular floor space inside the igloo, given the walls were 0.3m thick

Vanyuwa [196]

More details are required.

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3 years ago

Someone please help. Will mark brainliest!

marusya05 [52]

Step 2, replace the perpendicular bisector with angle bisector.

Note that in Step 3, it says "bisectors" with an s. This means that there are more than one angle bisector. Step 1 only states 1 bisector, so above answer should be correct.

hope this helps

5 0

3 years ago

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Verify that the function u(x, y, z) = log x^2 + y^2 is a solution of the two dimensional Laplace equation u_xx + u_yy = 0 everyw

Daniel [21]

Answer:

The function $u(x,y,z)=log ( x^{2} +y^{2})$ is indeed a solution of the two dimensional Laplace equation $u_{xx} +u_{yy} =0$ .

The wave equation $u_{tt} =u_{xx}$ is satisfied by the function $u(x,t)=cos(4x)cos(4t)$ but not by the function $u(x,t)=f(x-t)+f(x+1)$ .

Step-by-step explanation:

To verify that the function $u(x,y,z)=log ( x^{2} +y^{2})$ is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.

$u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})$

$u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})$

then we introduce it in the equation $u_{xx} +u_{yy} =0$

we get that $\frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0$

To see if the functions 1) $u(x,t)=cos(4x)cos(4t)$ and 2) $u(x,t)=f(x-t)+f(x+1)$ solve the wave equation we have to calculate the second partial derivative with respect to x and the with respect to t for each function. Then we see if they are equal.

1) $u_{xx}=-16 cos (4x) cos (4t)$

$u_{tt}=-16cos(4x)cos(4t)$

we see for the above expressions that $u_{tt} =u_{xx}$

2) with this function we will have to use the chain rule

If we call $s=x-t$ and $w=x+1$ then we have that

$u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)$

So $\frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}$

because we have $\frac{\partial s}{\partial x} =1$ and $\frac{\partial w}{\partial x} =1$

then $\frac{\partial u}{\partial x} =f'(s)+f'(w)$

⇒ $\frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)$

Regarding the derivatives with respect to time

$\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)$

then $\frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)$

we see that $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }$

$u(x,t)=f(x-t)+f(x+1)$ doesn´t satisfy the wave equation.

4 0

3 years ago