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3 years ago

Find the solutions to the following equation:

Mathematics

1 answer:

mojhsa [17]3 years ago

7 0

Answer:

x=1+/-4i

Step-by-step explanation:

(x+1)/3-(2x-1)/x=-1

find the common denominator,

in this case, it's going to be 3x.

(x^2+x)/3x-(6x-3)/3x=-1

(x^2+x-6x+3)/3x=-1

x^2-5x+3=-3x

x^2-5x-(-3x)+3=0

x^2-5x+3x+3=0

x^2-2x+3=0

apply the quadratic formula in order to find the roots for x.

Since ax^2+bx+c=0, in this case,

a=1, b=-2, c=3

--------------------------

x=(-b+/-sqrt(b^2-4ac))/2a

x=(2+/-sqrt(4-12))/2

x=(2+/-8i)/2

x=1+/-4i

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Calculate the flux of the vector field f⃗ (x,y,z)=cos(x2+y2)k⃗ through the disk x2+y2≤9 oriented upward in the plane z=4.

garik1379 [7]

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$\mathbf r_s\times\mathbf r_t=(\cos t\,\mathbf i+\sin t\,\mathbf j)\times(-s\sin t\,\mathbf i+s\cos t\,\mathbf j)=s\,\mathbf k$

The flux over the disk is given by the surface integral

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iint_T\mathbf f(x(s,t),y(s,t),z(s,t))\cdot(\mathbf r_s(s,t)\times\mathbf r_t(s,t))\,\mathrm ds\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=2\pi}\int_{s=0}^{s=3}(\cos(s^2)\,\mathbf k)\cdot(s\,\mathbf k)\,\mathrm ds\,\mathrm dt$
$=\displaystyle2\pi\int_{s=0}^{s=3}s\cos(s^2)\,\mathrm ds$
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(where we take $\sigma=s^2$ )

$=\pi\sin 9$

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3 years ago