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Elenna [48]
4 years ago
6

WILL GIVE BRAINLIEST!! PLEASE HELP ASAP !!! 20 POINTSSSSS

Mathematics
1 answer:
grin007 [14]4 years ago
5 0

Answer:

She can buy 5 rolls of crepe paper with 2ft of paper left.

Step-by-step explanation:

If you multiply 12ft by 4 sides you will get 48ft so then 5 rolls of paper that is 10ft equals 50ft so you will have more than enough paper

You might be interested in
In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + ta
Maurinko [17]

Answer:

\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c

Step-by-step explanation:

The question is not properly formatted. However, the integral of \int {7 \sec(\theta) } \, d\theta is as follows:

<h3></h3>

\int {7 \sec(\theta) } \, d\theta

Remove constant 7 out of the integrand

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta

Multiply by 1

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta

Express 1 as: \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}

\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta

Expand

\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta

Let

u = \sec(\theta) + \tan(\theta)

Differentiate

\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)

Make d\theta the subject

d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}

So, we have:

\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}

Cancel out \sec(\theta)\tan(\theta) + sec^2(\theta)

\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}

Integrate

\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c

Recall that: u = \sec(\theta) + \tan(\theta)

\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c

8 0
3 years ago
Timmy takes out a loan for $750 for 15 months, but only receives $725 into his bank account. What is the simple interest rate ad
12345 [234]

Answer:

<u>The annual interest rate advertised by the bank is 2.67%  or 0.22% per month.</u>

Step-by-step explanation:

Let's evaluate the information given:

Amount of the loan = US$ 750

Duration of the loan = 15 months

Net amount received by Timmy into his bank account = US$ 725

Interests for the loan = Amount of the loan - Net amount received into bank account

Interests for the loan = 750 - 725

Interests for the loan = US$ 25

Now for calculating the interest rate, we do this operation:

Interest rate = Interests for the loan / Amount of the loan

Interest rate = 25/750 = 1/30 = 0.033 * 100 = 3.3%

But let's remember that this is for the total duration of the loan, 15 months. For calculating the annual interest rate, we do the following operation:

Annual interest rate = Loan interest rate/15 * 12 (We use 12 because the year has 12 months)

Annual interest rate = (1/30) /15 * 12

Annual interest rate = (1/30) * 1/15 * 12 = 1/450 * 12 = 12/450 = 4/150 = 2/75 = 0.0267 * 100 = 2.67% (Rounding to two decimals)

<u>The annual interest rate advertised by the bank is 2.67%  or 0.22% per month.</u>

5 0
4 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
How many 1 x 2 shelfs cleats 8" long can be cut from a 1 x 2 board 16' long? how much is left?
inn [45]
8 i think so it might 
8 0
3 years ago
What is the sum of the exterior angles of a 21 sided polygon
svet-max [94.6K]

Answer:

360°

Step-by-step explanation:

The sum of the exterior angles for any polygon is 360°

6 0
3 years ago
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