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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{alig... \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- lign*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ &= \sqrt{1-\frac{9}{25}}\\ &= \sqrt{\frac{... 2 \sin\theta \cos\theta \\ &= 2\left(\frac{4}{5} \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}.... & = 1-2\sin^2\theta \\ &= 1-2\left(\frac{4}{5} \right)^2\\ &= 1-2\left(\frac{16}{25} \right)\\ \end{ali
- Question 9, Exercise 1.2 @math-11-nbf:sol:unit01
- and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following for... \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 y_2 - x_2 y_1}{x_1^2 + y
- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right. \right] \\ \underset{\sim}{R} & \left[\begin{matrix} 1 & 4 & 14 \\ 0 & 5 & 6 \\ -1 & 2 & 3 \end{matri... & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} \right. \right]\text{ by }R_1+3R_3\text{ and }R_2+2R_3 \\ \underset{\sim}{R}&\left[\begin{matrix} 1 & 4 & 14 \
- Question 8 & 9, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- he identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2... in \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\s
- Question 8 & 9, Review Exercise 10 @math-11-kpk:sol:unit10
- he identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2... in \alpha \sin \beta =\cos \left( \alpha -\beta \right)-\cos \left( \alpha +\beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\s
- Quotes for the March @quote-of-the-day
- 0" title="Quote by Mathematician"> <TEXT align="right">"نئی ریاضی" کی اہمیت بنیادی طور پر اس حقیقت میں ... ircle. [Isaac Todhunter (1820-1884)] <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor) ... 0" title="Quote by Mathematician"> <TEXT align="right">ریاضی کی تعلیم آپ کی توقع سے کہیں زیادہ پیچیدہ ہ... expected. [Edward Begle(1914-1978)] <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor)
- Quotes for the May @quote-of-the-day
- 00" title="Quote by Mathematician"> <TEXT align="right">مختصراً، پوری دنیا خلا اور وقت میں اشیا کی ریاضی... ne. --- **Morris Kline (1908-1992)** <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor) ... 00" title="Quote by Mathematician"> <TEXT align="right">دنیا کی ہم آہنگی شکل اور تعداد میں ظاہر ہوتی ہے،... --- **D'Arcy Thompson (1860-1948)** <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor)
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- -z\in \mathbb{C} \mbox{ such that}\; z+\left( -z\right) =0 \\ \mbox{In fact if } z=a+bi, \mbox{ then } -... ber\] - Associative Law for Addition \[\left( z+w\right) +v= z +\left( w+v\right)\nonumber \] </panel> <panel> **Question 2** Verify the multiplication proper... $\quad (2,6)\div (3,7)=\dfrac{(2,6)}{\left( 3,7 \right)}$ $=\dfrac{2+6i}{3+7i}=\dfrac{2+6i}{3+7i}\times
- Quotes for the April @quote-of-the-day
- 0" title="Quote by Mathematician"> <TEXT align="right">کاسمولوجسٹ اکثر غلط ہوتے ہیں، لیکن کبھی شک میں ن... oubt. --- **Lev Landau (1908-1968)** <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor) ... 0" title="Quote by Mathematician"> <TEXT align="right">ہمیں یاد رکھنا چاہیے کہ ایک زمانے میں فطری فلسفہ... t. --- **Hannes Alfvén (1908-1995)** <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor)
- Quotes for the February @quote-of-the-day
- them [Werner Heisenberg (1901-1976)] <TEXT align="right">ایک ماہر وہ ہوتا ہے جو اپنے مضمون میں ہونے والی ... ہو [ورنر ہیزنبرگ (1901-1976)]</TEXT> <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor) ... acher [Bertrand Russell (1872-1970)] <TEXT align="right">ایک اچھی علامت میں باریک بینی اور تجویز ہوتی ہے ... ہے [برٹرینڈ رسل (1872-1970)]</TEXT> <TEXT align="right" type="danger" size="small">(Courtesy: MacTutor)
- Exercise 2.4 (Solutions) @matric:9th_science:unit_02
- \sqrt(196)^{-1}}$ * (ii) $\left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right)$ * (iii) $\left(\frac{x^{-2}y^{-1}z^{-4}}{x^4y^{-3}z^0}\right)^{-3}$ * (iv) $\frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\right)}{\left(9^2n\right)\le
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- ix}x\\y\\z\end{bmatrix}$. Verify that $\left( AB \right)C=A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmat... \left[ x( ax+hy+gz )+y( hx+by+fz )+z( gx+fy+cz ) \right] \\ &=\left[ a{{x}^{2}}+hxy+gxz+hxy+b{{y}^{2}}+fyz+gxz+fyz+c{{z}^{2}} \right] \\ &=\left[ a{{x}^{2}}+2hxy+2gxz+b{{y}^{2}}+2fyz
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \... in \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \... in \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \right)\left( \frac{1}{\sqrt{2}} \right)-\left( \frac{1}{2} \right)\left( \frac{1}{\sqrt{2}} \right) \\ & =\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2