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melamori03 [73]
3 years ago
11

Which is true about the product of 3/8 and 7/2. 1.The product is less than 3/8. 2.The product is greater than 1/8 and less that

1/2. 3.The product is greater than 7/2. 4.The product is greater that 3/8 and less than 7/2.
Mathematics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

4.The product is greater that 3/8 and less than 7/2.

Step-by-step explanation:

The product of 3/8 and 7/2 is:

\frac{3}{8} * \frac{7}{2} \\\\= \frac{21}{16} = 1\frac{5}{16}

This number is greater than 3/8.

It is greater than 1/8 but not less than 1/2.

This number is less than 7/2.

This number is greater than 3/8 and it is less than 7/2.

Hence, the correct options is 4.

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jasmine took a cab home from her office. the cab charged a flat fee of $4, plus $2 per mile. Jasmine paid $32 for the trip. How
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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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