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3 years ago

A 30cm ruler sometimes measures inches on the other side.

Mathematics

1 answer:

Alisiya [41]3 years ago

4 0

1inch = 2.5cm

∆ 30cm

5cm =2inch

so 30cm=12inch

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in the year 2001, a company made 7.7 million dollars in profit. For each consecutive year after that, their profit increased by

MatroZZZ [7]

Answer:

3.5 billions

Step-by-step explanation:

7 0

3 years ago

Write each number in standard form 62 hundreds and 30 tens

mestny [16]

We are given : 62 hundreds and 30 tens .

To find : the number in standard form.

Soltuion: We know hundreds represents times 100.

Also tens reprents times 10.

In order to find the values of 62 hundreds, we need to multiply 62 by 100.

In order to find the values of 32 tens, we need to multiply 32 by 10.

62 × 100 = 6200

32 × 10 = 320.

Now, we need to add those numbers above.

6200 + 320 = 6520.

In simple, we could also show as

62 hundreds and 30 tens = 62 × 100 = 30 × 10 = 6200+300 = 6520.

Therefore, standard form is 6520.

7 0

4 years ago

Read 2 more answers

What is the slope of a roof on a house that has a vertical height of 2.4 feet from the ceiling of the top floor to the top of th

Tems11 [23]

Answer:

Step-by-step explanation:

It is unclear from the phrasing what dimension 8.2 ft represents.

If 8.2 ft is the direct distance from the edge of the roof to the top of the pitch, then the <em>horizontal</em> distance from the edge to the top is √(8.2²-2.4²) ≅ 7.84 ft, and the slope is 2.4/7.84 ≅ 0.31

If 8.2 ft is the <em>horizontal </em>distance from the edge of the root to the top of the pitch, then the slope is 2.4/8.2 ≅ 0.29

7 0

3 years ago

write an exponential formula of the form f (t)= a×b^t to represent the following situation. assume t is in years. The beginning

nika2105 [10]

Answer:

$f(t)= 27000 * 0.965^t$

Step-by-step explanation:

Given

$a = 27000$ -- initial

$r = 3.5\%$

Required

The exponential equation

The exponential equation is

$y =ab^t$

Where

$b=1 -r$

We used minus, because the rate decreases..

So, we have:

$b=1 -3.5\%$

$b=1 -0.035$

$b=0.965$

So:

$y =ab^t$

$y = 27000 * 0.965^t$

Hence:

$f(t)= 27000 * 0.965^t$

5 0

3 years ago

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