Find the approximate perimeter of AABC plotted below. A(2,7) C(6,7) B(-4,-3)

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

8 0

Answer:

29.8 units

Step-by-step explanation:

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If x = a sin α, cos β, y = b sin α.sin β and z = c cos α then (x²/a²) + (y²/b²) + (z²/c²) = ?

Oduvanchick [21]

$\large\underline{\sf{Solution-}}$

Given:

$\rm \longmapsto x = a \sin \alpha \cos \beta$

$\rm \longmapsto y = b \sin \alpha \sin \beta$

$\rm \longmapsto z = c\cos \alpha$

Therefore:

$\rm \longmapsto \dfrac{x}{a} = \sin \alpha \cos \beta$

$\rm \longmapsto \dfrac{y}{b} = \sin \alpha \sin \beta$

$\rm \longmapsto \dfrac{z}{c} = \cos \alpha$

Now:

$\rm = \dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }$

$\rm = { \sin}^{2} \alpha \cos^{2} \beta + { \sin}^{2} \alpha \sin^{2} \beta + { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha (\cos^{2} \beta + \sin^{2} \beta )+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha \cdot1+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha + { \cos}^{2} \alpha$

$\rm = 1$

Therefore:

$\rm \longmapsto\dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

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3 years ago

Find the equation of a line that goes through the points (-3,5) and (5,-11)

Rasek [7]

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-11}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-11}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{(-3)}}}\implies \cfrac{-16}{5+3}\implies \cfrac{-16}{8}\implies -2$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-2}(x-\stackrel{x_1}{(-3)}) \\\\\\ y-5-2(x+3)\implies y-5=-2x-6\implies y=-2x-1$