(3n+2)(2n–3) find the product simplified

Mathematics

2 answers:

nlexa [21]3 years ago

4 0

Answer:

6n2−5n−6

Step-by-step explanation:

Sedbober [7]3 years ago

4 0

Answer:

6n<u>^2</u>-5n-6 (btw <u>"6n^2"</u> means its 6n squared )

Step-by-step explanation:

Use the box Method. (Picture shows the box method)

once your done the simplified answer should be 6n^2-5n-6

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Use stokes' theorem to evaluate c f · dr where c is oriented counterclockwise as viewed from above. f(x, y, z = xyi + 5zj + 7yk,

Helga [31]

The intersection can be parameterized by

$C:=\mathbf r(t)=\begin{cases}x(t)=6\cos t\\y(t)=6\sin t\\z(t)=5-6\cos t\end{cases}$

with $0\le t$ .

By Stoke's theorem, the integral of $\mathbf f(x,y,z)=xy\,\mathbf i+5z\,\mathbf j+7y\,\mathbf k$ along $C$ is equivalent to

$\displaystyle\int_C\mathbf f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r(t)=\iint_S\nabla\times\mathbf f\,\mathrm dS$

where $S$ is the region bounded by $C$ . The line integral reduces to

$\displaystyle\int_0^{2\pi}(36\sin t\cos t\,\mathbf i+(25-30\cos t)\,\mathbf j+42\sin t\,\mathbf k)\cdot(-6\sin t\,\mathbf i+6\cos t\,\mathbf j+6\sin t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(54(\cos3t-\cos t)-30(3\cos2t-5\cos t+3)+(126-126\cos2t)\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}(36+96\cos t-216\cos2t+54\cos3t)\,\mathrm dt$
$=72\pi$