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3 years ago

Please hurry I’m being timed!!

Mathematics

1 answer:

faust18 [17]3 years ago

4 0

gogle said:

The line about which a three-dimensional figure is rotated to obtain an object identical to the original is a(n) horizontal plane.

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What type of variable is the number of gallons of gasoline pumped by a filling station during a day? Select one: a. Qualitative

sammy [17]

Answer:

b. Continuous

Step-by-step explanation:

Continuous variable is a variable that can take on any value between its minimum value and its maximum value. A continuous variable is a type of quantitative variable used to describe data that is measurable while Discrete variables are countable in a finite amount of time. Now, justifying whether the gallons of gasoline pumped by a filling station during a day is continuous or discrete. The number of gallons of gasoline pumped by a filling during during a day is a continuous variable because it has a measurable volume which can take value from the minimum to maximum values of the total volume of gasoline the filling station have in the storage tank.

3 0

4 years ago

How would you write the equation with a slope of 2/3 and y-intercept of -3

Nuetrik [128]

Answer:

A) y=2/3x-3

Step-by-step explanation:

A is the answer

3 0

4 years ago

Larger of two numbers is nine less than four times the smaller number if the sum of two numbers is 76 find both numbers

makvit [3.9K]

x + y = 76

x = 4y - 9

4y - 9 + y = 76

5y = 65

y = 13

13 + x = 76

x = 63

8 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

On the first math test, Keiko answered 19 out of 25 questions correctly. On the

m_a_m_a [10]

Answer:

Keiko got a better answer on the test on her second test, and you could compare these two ratios as they are written.

Step-by-step explanation:

First: Convert each of the scores into percentages:

How do you convert ratios to percentages? Well, first we have to convert the ratio to a decimal or a fraction. Let's convert it to a fraction. So, 19 out of 25 will become $\frac{19}{25}$ , and 17 out of 20 will become $\frac{17}{20}$ . To convert a fraction to a percentage:

Divide the numerator by the denominator:

19 ÷ 25 = 0.76 AND 17 ÷ 20 = 0.85

Multiply your quotient by 100

0.76 x 100 = 76 AND 0.85 x 100 = 85

Add the percent sign

76% AND 85%

Now the two quantities are easier to compare. We can obviously see that 85% is bigger [and better] than 76%. So, the ratio that equals 85% - 17 out of 20 - is a better grade.

The second question was: Can you compare these two ratios as they are written? Explain why or why not.

Yes, you can compare these two ratios, just find the [positive] difference between the amount that she scored and the total number of scores that could be gained.

25 - 19 = 6

20 - 17 = 3

Since the equation with 20 and 17 in it has a smaller difference, it is most likely to have a greater value [because that means that Keiko got more right and less wrong, which means a greater mark which is also a greater quantity].

Thanks,

Topic: Quantity Relationships, Quantity Conversions

4 0

3 years ago