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Contact [7]
3 years ago
15

Ma poate ajuta cineva la acesta problema: pe o drepta g considerati punctele D, R,P si T coliniare in aceasta ordine astfel inca

t DR=PT=3 cm, iar RP=4cm

Mathematics
1 answer:
valina [46]3 years ago
8 0

Answer:

Givens

DR=PT=3cm and RP=4cm

Points D, R, P and T are collinear.

The image attached shows a representation of this problem.

By sum of segments, we have

DT=DR+RP+PT

Replacing all given values, we have

DT=3+4+3=10 \ cm

Therefore, the segment DT is 10 centimeters long.

Now, we know that point E is the center of segment RP, which means is a middle point, divides the segment equally. So, to prove that segment DT is symmetrical regarding point E, we need to prove that both sides are equal.

RE=EP=2cm, by definition of middle point.

By sum of segments, we have

DR+RE=3+2=5cm\\EP+PT=5cm

Which means DR+RE=\frac{DT}{2}\\ EP+PT=\frac{DT}{2}.

Therefore, segment DT is symmetrical regarding point E.

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What is the interquartile range of this data set?<br> 2,5,9,11,18, 30, 42, 48, 55, 73, 81
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Answer two questions about Equations A and B: A. 5x = 20
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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.

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3 years ago
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