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3 years ago

In this diagram, which equation could you prove to be true in order to conclude that the lines are parallel?

Mathematics

2 answers:

yan [13]3 years ago

7 0

Answer: Hey, so the answer is the second choice, b/a=c/d. I just took the quiz, it's correct.

Y_Kistochka [10]3 years ago

6 0

<span>b/a=c/d Two lines are parallel to each other if they have the same slope. So let's look at the options and see which one matches that statement. b/aâ‹…c/d=â’1 * This is checking to see if the two lines are perpendicular to each other. Almost the exact opposite of parallel. So this is the wrong answer. b/a=c/d * b/a is a legitimate slope. c/d is also a good slope. And they're being checked to see if they're equal to each other. So this is the correct answer. But let's see if the next two options will give us a chuckle or two. b/a=d/c * b/a is OK. d/c almost looks ok, check the figure. And d/c is NOT a slope. The slope is delta Y over delta X. And d/c is delta X over delta Y. The inverse of the slope. So comparing those two values is meaningless. So it's a wrong answer. b/aâ‹…d/c=â’1 * And they're trying the inverse of the slope trick again. WRONG. And they're checking if it's perpendicular. Also wrong. So definitely not a good choice.</span>

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balu736 [363]

Answer:

Option A. one rectangle and two triangles

Option E. one triangle and one trapezoid

Step-by-step explanation:

step 1

we know that

The area of the polygon can be decomposed into one rectangle and two triangles

see the attached figure N 1

therefore

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$A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2$

step 2

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The area of the polygon can be decomposed into one triangle and one trapezoid

see the attached figure N 2

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3 years ago

Help! Please:) Amd Thank you!!!

Mrrafil [7]

They get 2/3 because 2 divided by 3=2/3.

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3 years ago

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

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3 years ago

Will give the brainliest answer

jok3333 [9.3K]

Answer:

I think it's C, because the figures are similar by THAT statement

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