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forsale [732]
3 years ago
13

2(n+7)=14+2n and 1-x=7x-(8x-6)

Mathematics
1 answer:
myrzilka [38]3 years ago
3 0
2n+14=14+2n so this one as an infinite number of solutions. The second one 1-x=-x-6 so you add x on both sides which results in it having no solutions.
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On a coordinate graph,which of the following is true of Quadrant 3?
Nonamiya [84]
D. The x and y values are both negative

5 0
3 years ago
Simplify square root of 24 times square root of 2x times square root of 3x
sveticcg [70]
\sqrt{24} * \sqrt{2x} * \sqrt{3x} =  \sqrt{2*2*2*3}* \sqrt{2*x} * \sqrt{3*x} =  \sqrt{2*2*2*2*3*3*x*x} = 2*2*3*x = 12x
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When a sum of rupees hundred is divided in the ratio of 2 is to 3 find the share of each part​
yKpoI14uk [10]

2x + 3x = 100

5x=100

x=100/5

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4 0
3 years ago
Read 2 more answers
In which expression is the number 5 a factor?<br>A. z+5<br>B. z-5<br>C.5z<br>D.5/z​
LenaWriter [7]

The correct answer is: C.5z

Further explanation:

A factor is a number that is multiplied with a number

Let us look at the options one by one

<u>A. z+5</u>

5 is not a factor as it is being added to z.

<u>B. z-5</u>

5 is not a factor as it is being subtracted from z.

<u>C.5z</u>

5 is a factor as it is being multiplied with the z and can be separated easily

Hence,

The correct answer is: C.5z

Keywords: Factors, Multiple Choice

Learn more about factors at:

  • brainly.com/question/12884373
  • brainly.com/question/12896802

#LearnwithBrainly

4 0
3 years ago
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
3 years ago
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