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sammy [17]
3 years ago
14

Find the length of the missing side

Mathematics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

c = 17

Step-by-step explanation:

Since this is a right angle triangle we can use the Pythagoras theorem that states that c^{2} = a^{2} + b^{2} (where c is the Solve for  hypotenuse and "a" and "b" are the legs of the right angle triangle). From here we know that the answer to this question is....

c^{2} = a^{2} + b^{2}\\c^{2} = 15^{2} + 8^{2} \\c^{2} = 225 + 64\\c^{2} = 289\\c = \sqrt{289} \\c = 17

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On Thursday afternoon at camp Alice played basketball and went swimming before dinner she spent 45 minutes playing basketball an
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Answer:

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4 years ago
The area of a rectangle is 40 meters. One of the sides is 10 meters. What are the measurements, in meters of the
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Step-by-step explanation:

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Trigonometric question, 30 points, will give brainliest.
zheka24 [161]

hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.

\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]

\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill

\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill

3 0
4 years ago
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