Answer:
Im sorry but I am not really sure myself ! I hope you find or found your answer already !
Step-by-step explanation:
Answer:
Step-by-step explanation:
93+(75)
= 168
It looks like the integral is
![\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%2B4y%29%5C%2C%5Cmathrm%20dx%20%2B%20%282x-3y%29%5C%2C%5Cmathrm%20dy)
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%2B4y%29%5C%2C%5Cmathrm%20dx%20%2B%20%282x-3y%29%5C%2C%5Cmathrm%20dy%20%3D%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Cleft%28%283x%28t%29%2B4y%28t%29%29%5Cdfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dt%7D%20%2B%20%282x%28t%29-3y%28t%29%29%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E%7B2%5Cpi%7D%20%5Cbig%28%286%5Ccos%28t%29%2B8%5Csin%28t%29%29%28-2%5Csin%28t%29%29%20%2B%20%284%5Ccos%28t%29-6%5Csin%28t%29%29%282%5Ccos%28t%29%29%5Cbig%29%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E%7B2%5Cpi%7D%20%2812%5Ccos%5E2%28t%29-12%5Csin%5E2%28t%29-24%5Ccos%28t%29%5Csin%28t%29-4%29%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%204%20%5Cint_0%5E%7B2%5Cpi%7D%20%283%5Ccos%282t%29-3%5Csin%282t%29-1%29%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B-8%5Cpi%7D)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_C%20%283x%2B4y%29%5C%2C%5Cmathrm%20dx%20%2B%20%282x-3y%29%5C%2C%5Cmathrm%20dy%20%3D%20%5Ciint_D%5Cfrac%7B%5Cpartial%282x-3y%29%7D%7B%5Cpartial%20x%7D-%5Cfrac%7B%5Cpartial%283x%2B4y%29%7D%7B%5Cpartial%20y%7D%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%20%3D%20-2%5Ciint_D%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy)
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:
.
Answer:
No
Step-by-step explanation:
f(x) = x²
f(3) = 3²
= 3 × 3
= 9 Not 16
also
√16 = 4
Answer:
Can i have a picture of the model
Step-by-step explanation: