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olga_2 [115]
3 years ago
13

The product of three consecutive odd integers is 75.

Mathematics
2 answers:
Zigmanuir [339]3 years ago
4 0
1). 3X+6=75
75-6=69=23
Answer 23,25,27

Inessa [10]3 years ago
3 0

Answer:it’s A

Step-by-step explanation: I jus got it right

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If sin theta = 2/5 and theta is in quadrant I, determine the following.
anzhelika [568]

\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm\sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a\implies \stackrel{I~Quadrant}{+\sqrt{21}=a}


recall that cosine is positive on the I Quadrant, so though we get a ± valid roots, only the positive one applies.


\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}


now, for tangent and secant, let's rationalize the denominator.


\bf tan(\theta)\implies \cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{21} \\\\\\ sec(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{5\sqrt{21}}{21}

5 0
3 years ago
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