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Inessa05 [86]
3 years ago
12

PLEASE HELP!!! Identify the side or angle that is common to ∆ABC and ∆CDA.

Mathematics
1 answer:
Sedbober [7]3 years ago
6 0

Answer:

AC

Step-by-step explanation:

both triangles contain this side

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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
Two angles are complementary if the sum of their measures is 90°. Find two complementary angles such that one of the angles is 1
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Bro that's a lot, i don't know what the heck you are talking about.
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