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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- $ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}{5} \righ... 25}}\\ &= \sqrt{\frac{16}{25}} = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 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 =
- Exercise 6.1 @matric:9th_science
- )$, $54(27x^4-x)$ **Solution:**\\ (i) $\begin{align} x^2+5x+6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3-27 &=x
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- $\alpha=180^{\circ}$, $\beta=60^{\circ}$. \begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \be... ht) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies... ght) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*} \begin{align*} \sin (\alpha + \beta) & = \sin
- Quotes for the March @quote-of-the-day
- "#A11400" title="Quote by Mathematician"> <TEXT align="right">"نئی ریاضی" کی اہمیت بنیادی طور پر اس حقی... d the circle. [Isaac Todhunter (1820-1884)] <TEXT align="right" type="danger" size="small">(Courtesy: Mac... "#A11400" title="Quote by Mathematician"> <TEXT align="right">ریاضی کی تعلیم آپ کی توقع سے کہیں زیادہ پ... han you expected. [Edward Begle(1914-1978)] <TEXT align="right" type="danger" size="small">(Courtesy: Mac
- Quotes for the May @quote-of-the-day
- ="#A11400" title="Quote by Mathematician"> <TEXT align="right">مختصراً، پوری دنیا خلا اور وقت میں اشیا ک... d machine. --- **Morris Kline (1908-1992)** <TEXT align="right" type="danger" size="small">(Courtesy: Mac... ="#A11400" title="Quote by Mathematician"> <TEXT align="right">دنیا کی ہم آہنگی شکل اور تعداد میں ظاہر ہ... beauty. --- **D'Arcy Thompson (1860-1948)** <TEXT align="right" type="danger" size="small">(Courtesy: Mac
- Quotes for the April @quote-of-the-day
- "#A11400" title="Quote by Mathematician"> <TEXT align="right">کاسمولوجسٹ اکثر غلط ہوتے ہیں، لیکن کبھی ش... er in doubt. --- **Lev Landau (1908-1968)** <TEXT align="right" type="danger" size="small">(Courtesy: Mac... "#A11400" title="Quote by Mathematician"> <TEXT align="right">ہمیں یاد رکھنا چاہیے کہ ایک زمانے میں فطر... y aspect. --- **Hannes Alfvén (1908-1995)** <TEXT align="right" type="danger" size="small">(Courtesy: Mac
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- II, therefore $\cos 2\theta$ is negative. \begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}\\ &=-... \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-... ta$ lies in QI and $\sin\theta > 0$. Thus \begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\theta}{2}} ... ac{\frac{32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \fr
- Quotes for the February @quote-of-the-day
- avoid them [Werner Heisenberg (1901-1976)] <TEXT align="right">ایک ماہر وہ ہوتا ہے جو اپنے مضمون میں ہون... جانتا ہو [ورنر ہیزنبرگ (1901-1976)]</TEXT> <TEXT align="right" type="danger" size="small">(Courtesy: Mac... live teacher [Bertrand Russell (1872-1970)] <TEXT align="right">ایک اچھی علامت میں باریک بینی اور تجویز ہ... رح لگتی ہے [برٹرینڈ رسل (1872-1970)]</TEXT> <TEXT align="right" type="danger" size="small">(Courtesy: Mac
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- z-(2+5 i) \omega=2+3 i$. ** Solution. ** \begin{align} &(1-i) z+(1+i) \omega=3 \quad \cdots(1)\\ &2 z-(2+5 i) \omega=2+3i \quad\cdots(2) \end{align} Multiplying Eq. (1) by $2$: \begin{align} &(2-2i)z+(2+2i) \omega=6 \quad \cdots (3) \end{align} Multiplying Eq. (2) by $(1-i)$: \begin{align} &2
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- \overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (1) \end{align} Now \begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies |-z_1| &= \sqrt{(3)^2 + (-2)^2} \\ &= \sqrt{9 + 4} = \sqrt{13} \,\, -- (2) \end{align} Also \begin{align} \overline{z_1} &= -3 - 2i
- Question 6 Exercise 4.1 @math-11-kpk:sol:unit04
- xt{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1}
- Exercise 6.3 @matric:9th_science
- $4x^2-12xy +9y^2$\\ **Solution:**\\ $\begin{align}4x^2-12xy +9y^2\\&=4x^2-6xy-6xy +9y^2\\&= 2x(2x-3... )-3y(2x-3y)\\&= (2x-3y)(2x-3y)\\&= (2x-3y)^2 \end{align}$\\ $\begin{align} \sqrt{4x^2-12xy +9y^2}&= \pm (2x-3y) \end{align}$ (ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$\\ **Solution
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- nt, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Righ... sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Ri... \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- nt, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Righ... sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Ri... \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{
- Question 11 and 12, Exercise 4.8 @math-11-nbf:sol:unit04
- present the $k$th term of the series. Then \begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $k(k+2)$, we get \beg