Precalculus Archive: Questions from May 15, 2023
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\( \begin{array}{ll}\sin \left(\arccos \left(\frac{12}{19}\right)\right)=( & \varnothing^{\infty} \frac{\sqrt{217}}{19} \\ \tan \left(\arccos \left(\frac{12}{19}\right)\right)=( & \sigma^{\infty} \fra2 answers -
6) \[ \left\{\begin{array}{l} x y-x^{2}=-20 \\ x-2 y=3 \end{array}\right. \] A) \( x=5, y=1 ; x=-\frac{11}{2}, y=-8 \) B) \( x=-5, y=-1 ; x=\frac{11}{2}, y=8 \) C) \( x=-5, y=-1 ; x=8, y=\frac{11}{2}2 answers -
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cos 2u = cos²u - sin²u = 2 cos²u - 1 = 1 - 2 sin²u tan 2u = 2 tan u 1 - tan²u Given that cot u = √14; π
\[ \begin{aligned} \cos 2 u & =\cos ^{2} u-\sin ^{2} u \\ & =2 \cos ^{2} u-1 \\ & =1-2 \sin ^{2} u \\ \tan 2 u & =\frac{2 \tan u}{1-\tan ^{2} u} \end{aligned} \] Given that \( \cot u=\sqrt{14} ; \pi2 answers -
9x + y - z = 8 8x - Y = 1 X - Z= 9
\( \left\{\begin{aligned} 9 x+y-z & =8 \\ 8 x-y & =1 \\ x-z & =9\end{aligned}\right. \)0 answers