I only need one system of inequalities with three points that satisfy both inequalities.

Mathematics

1 answer:

Ksju [112]3 years ago

4 0

Answer:

The given points plotted, only one of them is the solution:

(4, 2)

See the picture attached

Additional points to satisfy the system are:

(4, 3), (5, 1), (5, 2)

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1 point

-Dominant- [34]

39
Divide 78 by 2 to get 39 and add in to the number

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4 years ago

An electronics shop sells only TVs, phones and tablets. Number sold Select all of the statements below that are true. Number of

kondaur [170]

A graph is a way to represent a lot of data in such a visual format. The correct statements are A, B, and D.

<h3>What is a graph?</h3>

A graph is a way to represent a lot of data in such a visual format that it is easy for the user to understand the complete information in one go. Usually, the line of the graph is a function that follows the graph.

The statement that is correct about the given table is,

A.) The largest total number of electronic devices sold was on Wednesday.

B.) The smallest number of TVs sold was on Monday.

D.) On Tuesday, more tablets were sold than any other type of electronic device

hence, the correct statements are A, B, and D.

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2 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 C is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing C. Let x = 2 cos(t ) and y = 2 sin(t ), with 0 ≤ t ≤ 2π. 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 C nor in the region bounded by C, 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 D is the interior of C, 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: $-2\times \pi\times2^2 = -8\pi$ .