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solniwko [45]
3 years ago
12

How do you make a circle split into 3 equal parts

Mathematics
2 answers:
dlinn [17]3 years ago
5 0
U meet all 3 lines to the middle. Hope this helps
ValentinkaMS [17]3 years ago
5 0
Hey there!

So, when you do this, think like you were cutting  a pizza. Just all you do is first cut one down in the middle, and think of where the third slice would go, and then from there, you slice down the second slice, and then the is it. You only cut 2 times, not three.

Hope this helps you!
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Plz help ill give u brainlist
g100num [7]
Step One
Find the base area  of the large hexagon as though the smaller one was not removed.

Area = 3*Sqrt(3) * a^2 /2 where a is the length of one side of the hexagon
a = 5
Area = 3*sqrt(3) * 25/2 = 75 sqrt(3) / 2 of the large hexagon without the smaller one removed.

Step Two
Find the area of the smaller hexagon. In this case a = 4
Area2 = 3*sqrt(3)*16/2 = 3*sqrt(3)*8 = 24 sqrt(3)

Step Three
Find the area of the thick hexagonal area left by the removal of the small hexagon.

Area of the remaining piece = area of large hexagon - area of the small hexagon

Area of the remaining piece = 75 *sqrt(3)/2 - 24*sqrt(3)

Step Four
Find the volume of the results of the area from step 3
Volume = Area * h
h = 18
Volume = (75 * sqrt(3)/2 - 24*sqrt(3))* 18

I'm going to leave you with the job of changing all of this to a decimal answer. I get about 420 cm^3

3 0
3 years ago
Select all the fractions that are equivalent to 30%.
choli [55]
30% is equivalent to 3/10
So any fraction that can be simplified to 3/10.
Another way to to think about it if you multiply 3/10 by any number over it self because that is equivalent to 1.
Like 3/10 * 2/2 =6/20 is also equal to 30%
5 0
4 years ago
CHECK MY ANSWER ASAP
Butoxors [25]

Answer:

you goth or what dummy

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
When integrating polar coordinates, when should one use the polar differential element, <img src="https://tex.z-dn.net/?f=rdrd%2
vitfil [10]
To answer your first question: Whenever you convert from rectangular to polar coordinates, the differential element will *always* change according to

\mathrm dA=\mathrm dx\,\mathrm dy\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta

The key concept here is the "Jacobian determinant". More on that in a moment.

To answer your second question: You probably need to get a grasp of what the Jacobian is before you can tackle a surface integral.

It's a structure that basically captures information about all the possible partial derivatives of a multivariate function. So if \mathbf f(\mathbf x)=(f_1(x_1,\ldots,x_n),\ldots,f_m(x_1,\ldots,x_n)), then the Jacobian matrix \mathbf J of \mathbf f is defined as

\mathbf J=\begin{bmatrix}\mathbf f_{x_1}&\cdots&\mathbf f_{x_n}\end{bmatrix}=\begin{bmatrix}{f_1}_{x_1}&\cdots&{f_m}_{x_n}\\\vdots&\ddots&\vdots\\{f_m}_{x_1}&\cdots&{f_m}_{x_n}\end{bmatrix}

(it could be useful to remember the order of the entries as having each row make up the gradient of each component f_i)

Think about how you employ change of variables when integrating a univariate function:

\displaystyle\int2xe^{x^2}\,\mathrm dr=\int e^{x^2}\,\mathrm d(x^2)\stackrel{y=x^2}=\int e^y\,\mathrm dy=e^{r^2}+C

Not only do you change the variable itself, but you also have to account for the change in the differential element. We have to express the original variable, x, in terms of a new variable, y=y(x).

In two dimensions, we would like to express two variables, say x,y, each as functions of two new variables; in polar coordinates, we would typically use r,\theta so that x=x(r,\theta),y=y(r,\theta), and

\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}

The Jacobian matrix in this scenario is then

\mathbf J=\begin{bmatrix}x_r&y_\theta\\y_r&y_\theta\end{bmatrix}=\begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix}

which by itself doesn't help in integrating a multivariate function, since a matrix isn't scalar. We instead resort to the absolute value of its determinant. We know that the absolute value of the determinant of a square matrix is the n-dimensional volume of the parallelepiped spanned by the matrix's n column vectors.

For the Jacobian, the absolute value of its determinant contains information about how much a set \mathbf f(S)\subset\mathbb R^m - which is the "value" of a set S\subset\mathbb R^n subject to the function \mathbf f - "shrinks" or "expands" in n-dimensional volume.

Here we would have

\left|\det\mathbf J\right|=\left|\det\begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix}\right|=|r|

In polar coordinates, we use the convention that r\ge0 so that |r|=r. To summarize, we have to use the Jacobian to get an appropriate account of what happens to the differential element after changing multiple variables simultaneously (converting from one coordinate system to another). This is why

\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta

when integrating some two-dimensional region in the x,y-plane.

Surface integrals are a bit more complicated. The integration region is no longer flat, but we can approximate it by breaking it up into little rectangles that are flat, then use the limiting process and add them all up to get the area of the surface. Since each sub-region is two-dimensional, we need to be able to parameterize the entire region using a set of coordinates.

If we want to find the area of z=f(x,y) over a region \mathcal S - a region described by points (x,y,z) - by expressing it as the identical region \mathcal T defined by points (u,v). This is done with

\mathbf f(x,y,z)=\mathbf f(x(u,v),y(u,v),z(u,v))

with u,v taking on values as needed to cover all of \mathcal S. The Jacobian for this transformation would be

\mathbf J=\begin{bmatrix}x_u&x_v\\y_u&y_v\\z_u&z_v\end{bmatrix}

but since the matrix isn't square, we can't take a determinant. However, recalling that the magnitude of the cross product of two vectors gives the area of the parallelogram spanned by them, we can take the absolute value of the cross product of the columns of this matrix to find out the areas of each sub-region, then add them. You can think of this result as the equivalent of the Jacobian determinant but for surface integrals. Then the area of this surface would be

\displaystyle\iint_{\mathcal S}\mathrm dS=\iint_{\mathcal T}\|\mathbf f_u\times\mathbf f_v\|\,\mathrm du\,\mathrm dv

The takeaway here is that the procedures for computing the volume integral as opposed to the surface integral are similar but *not* identical. Hopefully you found this helpful.
5 0
3 years ago
Ali is reviewing family video. tape totaled 2 hours and 37 minutes. video of his children was 72 minutes. how much of video is n
notka56 [123]
85 min of the video was not of children.
8 0
3 years ago
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