Plot and connect the points A(-4,2), B(4,2), C(4,-2), and D(-4,-2). Then, find the perimeter of rectangle ABCD.

Mathematics

1 answer:

Anna71 [15]2 years ago

6 0

Answer: 456

Step-by-step explanation:

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Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Hitman42 [59]

$\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
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sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\$

$\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
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sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
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\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
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tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
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\textit{and now, let's rationalize the denominator}
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\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}
$

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3 years ago

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Montano1993 [528]

Answer:

Step-by-step explanation:

9.45 is rounded down to 9 because the number must reach half to round up. To round up to 10, it must be 9.5. hope this helps

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3 years ago

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