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3 years ago

3. Write an equation of a line that is perpendicular to the line x – 2y = 8.

Mathematics

1 answer:

xenn [34]3 years ago

7 0

Answer:

y=0.5x+40

Step-by-step explanation:

Copy the equation.

x-2y=8

Subtract x from both sides.

-2y=-x-8

Divide both sides by -2.

y=0.5x+4

Now we know the slope is 0.5.

Any line with a slope of 0.5 will be perpendiculr to the original line.

One that you can use is y=0.5x+40.

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Drupady [299]

Answer:

Step-by-step explanation:

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3 years ago

Ou pick a card at random. Without putting the first card back, you pick a second card at random.

Arte-miy333 [17]

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3 years ago

Can someone pls pls help me ?

Yakvenalex [24]

9514 1404 393

Answer:

a = 9 meters

Step-by-step explanation:

The perimeter is the sum of the side lengths:

28 m = a + 2m + a + 8m

18 m = 2a . . . . . . . . . . . . . . . . subtract 10m, collect terms

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3 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$

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1 year ago