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Llana [10]
2 years ago
11

Figure ABCD was reflected across the x-axis to create

Mathematics
1 answer:
goldfiish [28.3K]2 years ago
5 0

Answer:

I think the answer is (8,2)

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Use distance formula d=√(X2 – X1 )2 + (X2 – X1 )2, calculate the distance r from the origin to the point r (-2, √5).
lesya692 [45]
d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \\d = \sqrt{(\sqrt{5} - (-2))^{2} + (\sqrt{5} - (-2))^{2}} \\d = \sqrt{(\sqrt{5} + 2)^{2} + (\sqrt{5} + 2)^{2}} \\d = \sqrt{(4 + 4\sqrt{5} + 5) + (4v + 4\sqrt{5} + 5)} \\d = \sqrt{0 + 0} \\d = \sqrt{0} \\d = 0
7 0
3 years ago
Evaluate the expression for x=6<br> 4×+6
alexira [117]

Answer:

30

Step-by-step explanation:

x=6

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7 0
3 years ago
The measure of one acute angle in this right triangle is 45°.
maks197457 [2]

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45°

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3 0
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Read 2 more answers
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
2 years ago
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