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4 years ago

I... I did it... Thanks-

Mathematics

2 answers:

umka21 [38]4 years ago

8 0

Answer:

YAYYYYYY!!!!!!

Step-by-step explanation:

Congrats?? Lol

sesenic [268]4 years ago

7 0

answer: good job

explanation:(ﾉ◕ヮ◕)ﾉ

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Can't seem to figure out this problem. Can anyone help?

Juli2301 [7.4K]

the third fifth and first

8 0

3 years ago

Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1

MA_775_DIABLO [31]

Answer:

$\int_R 15x+15y dA = \frac{8}{16875}$

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where $\frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}$ .

We will calculate the jacobian of the change variables. That is

$\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right|$ (we are calculating the determinant of this matrix). The matrix is

$\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75}$ (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

$15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v$

We will use the change of variables theorem. So,

$\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu$

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

$\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}$

5 0

4 years ago

The louvre museum is a metal and glass structure that serves as the main entrance to the louvre museum paris, france. The pyrami

Eddi Din [679]

Answer:

A≈3170.96m²

Step-by-step explanation:

5 0

3 years ago

Does (2, -1) make this a true statement? Why or why not?<br> 3x + 2y = 4

kvv77 [185]

Answer: Yes

Step-by-step explanation:

So a coordinate pair is always set up (x,y) so you plug the x term in the coordinate pair into the x in the equation and the y term in for the y. 3×2 + 2×-1. Multiply them together and you end up with 4.

3 0

3 years ago

Find x.<br><br> See attached picture.

pishuonlain [190]

Answer:

B. 22

Step-by-step explanation:

We know tan60=

$\sqrt{3}$

Let common side be x

Tan = opp/adj= sqrt(3)

$\frac{11 \sqrt{6} }{x} = \sqrt{3}$

Transposing we get :

$\frac{11 \sqrt{6} }{ \sqrt{3}} = x = 11 \sqrt{2}$

Now, we know that since the other triangle is a 45-45-90 triangle, we can state that the side adjacent to the given angle = 11 sqrt(2) as the two sides are equal due to the triangle being an isoceles triangle.

Solve for x by using Pythagorean theorem:

$\sqrt((11 \sqrt{2}^{2} + (11 \sqrt{2} )^{2} ) = x$

$= > \sqrt{2(11 \times 11 \times \sqrt{2} \times \sqrt{2}) } = x$

$= > \sqrt{484} = x = 22$

Hence answer is B. 22

3 0

2 years ago