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2 years ago

In this rectangular box, EF = 16, FD = 5, and DB = 30. Find AF.

Mathematics

2 answers:

AVprozaik [17]2 years ago

6 0

Answer:

FA = 25.87

Step-by-step explanation:

Given: BD = 30 , DF = 5 , EF = 16

To find: FA

Rectangular box is a Cuboid.

Figure attached.

AB = FE = 16

So,

In ΔADB

using Pythagoras theorem,

$DB^2=AB^2+AD^2\\30^2=16^2+AD^2\\900=256+AD^2\\AD^2=900-256\\AD^2=644\\AD=\sqrt{644}\\AD=2\sqrt{161}$

Again using Pythagoras theorem in Δ ADF we get

$FA^2=DF^2+AD^2\\FA^2=5^2+(2\sqrt{161})^2\\FA^2=25+644\\FA^2=669\\FA=\sqrt{669}\\FA=25.87$

Therefore, FA = 25.87

lidiya [134]2 years ago

4 0

AD = sqrt(EF^2 + DB^2) = sqrt(16^2 + 30^2) = sqrt(256 + 900) = sqrt(1156) = 34

AF = sqrt(AD^2 + FD^2) = sqrt(34^2 + 5^2) = sqrt(1156 + 25) = sqrt(1181) = 34.4

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Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

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Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":