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sdas [7]
3 years ago
5

Please help me with this

Mathematics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

9x squared + 12x

Step-by-step explanation:

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Simplify 3/5(10g-5k)-(-3g+2k)
dybincka [34]

Answer:

9g -5k

Step-by-step explanation:

6 0
3 years ago
What is the square root of 149 rounded to the nearest tenth?
Ad libitum [116K]
Rounded to the nearest tenth, the square root of 149 is 12.2.
3 0
3 years ago
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A Speedboat increases in speed uniformly from 20m/s to 30m/s for 12.2 s. What is the acceleration of the speedboat?
Aloiza [94]

Answer:

0.82 m/s^2

Step-by-step explanation:

Given data

initial velocity=20m/s

Final velocity= 30m/s

Time = 12.2s

Applying the formula

a= v-u/t

a= 30-20/12.2

a= 10/12.2

a= 0.819

a=0.82 m/s^2

6 0
2 years ago
Given that the measurement is in centimeters, find the area of the circle to the nearest tenth.(Use 3.14 for n) Radius is 3
olasank [31]
List the a b c d platform if you have it
5 0
3 years ago
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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.

3 0
3 years ago
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