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3 years ago

14

Which methods could you use to calculate the y-coordinate of the midpoint of a vertical line segment with endpoints at (0,0) and

(0,15)

1 answer:

konstantin123 [22]3 years ago

5 0

The mid point of a line with end points (x₁,y₁) and (x₂,y₂) is given by the formula

$[\frac{x1+x2}{2} , \frac{y1+y2}{2}]$

Here we only need the y coordinate hence lets calculate the y coordinate alone

= $\frac{0+15}{2}$

=7.5

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What is the measure of an interior angle in a regular octagon?

andrew11 [14]

To find the interior angle of any regular polygon you can use the formula $\frac{(n-2)*180}{n}$ where n is the number of sides
So, $\frac{(8-2)*180}{8} =135$
The measure of an interior angle in a regular octagon is 135

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3 years ago

What is the GCF of 20, 36 and 48

nlexa [21]

20= 1,2,4,5,10,20
36=1,2,3,4,6,9,12,18,36
48=1,2,3,4,6,8,12,16,24,48

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3 years ago

If y varies directly as x, and y=8 when x=3, find y when x=15.

Crank

Answer:

y=40

Step-by-step explanation:

The formula for Direct Variation is y=kx or k=y/x. In this case I would use k=y/x. If you're y is 8 and x is 3, this means that K=8/3. Using this, we know that X=15. We have to find a Y value so that the fraction with an x of 15 simplified is 8/3. To do this you would write 8/3 and y/15. Now cross multiply to get 3y=120. Divide by 3 to get y=40. View my attachment for the work!

8 0

3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

Simplify Simplify 6 - 23 + (-9 + 5) · 2.

Vlad [161]

Answer:

The answer is −42

7 0

3 years ago