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- Question 18, Exercise 2.2 @math-11-kpk:sol:unit02
- atrix.\\ $$A=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right]$$ $$|A|=a_{11}a_{22}-a_{12}a_{21}$$ $$AdjA=\left[ \begin{matrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \\ \end{matrix} \right]... 1}{|A|}AdjA$$ $$A^{-1}=\dfrac{1}{a_{11}}a_{22}-a_{12}a_{21}\left[ \begin{matrix} a_{22} & -a_{12}
- Chapter 12: Application of Trigonometry @fsc:fsc_part_1_solutions
- ====== Chapter 12: Application of Trigonometry ====== {{ :fsc:fsc_part_1_solutions:fsc-1-chap-12-ptb.jpg?nolink|Chapter 12: Application of Trigonometry}} Notes (Solutions) of Chapter 12: Application of Trigonometry, Text Book of Algebr
- Ch 12: Application of Trigonometry: Mathematics FSc Part 1 @fsc:fsc_part_1_solutions:ch12
- ====== Ch 12: Application of Trigonometry: Mathematics FSc Part 1 ====== Notes (Solutions) of Chapter 12: Application of Trigonometry, Text Book of Algebr... ====Here is the list of all exercises of Chapter 12==== * [[vfsc1>ch12:view&cp=01&p=01&ch=12&fp=Ex-12-1-FSC-part1-ver1|Exercise 12.1]] * [[vfsc1>ch
- View Online (Solutions of Chapter 12) @fsc:fsc_part_1_solutions:ch12
- ====== View Online (Solutions of Chapter 12) ====== Notes (Solutions) of Chapter 12: Application of Trigonometry, Text Book of Algebra and Trigon... ====Here is the list of all exercises of Chapter 12==== * [[mdoku>fsc:fsc_part_1_solutions:ch12:viewer&cp=1&p=1&ch=12&fp=Ex_12_1_FSC_part1|Exercise 12.1]] * [[mdoku>
- Exercise 6.3 @matric:9th_science
- e root of the following expressions.\\ (i) $4x^2-12xy +9y^2$\\ (ii) $x^2-1+\frac{1}{4x^2}, (x\neq 0)$\\ (iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$\\ (iv) $4(a+b)^2-12(a^2-b^2)+9(a-b)^2$\\ (v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4-24x^2y^2+16y^4},(x \neq 0)$\\ (vi) $\left(
- Exercise 2.6 (Solutions) @matric:9th_science:unit_02
- \begin{array}{cl} 2(5+4i)-3(7+4i) &= 10+8i-21-12i\\ &= 10-21+8i-12i\\ &= -11-4i \end{array}$$ (iii) $$\begin{array}{cl} -(-3+5i)-(4+9i) ... 9i-6i^2\\ &= 6-5i-6(-1)\\ &= 6+6-5i\\ &= 12-5i \end{array}$$ ====Question 4==== * Simplify ... c{2(4+i)+3i(4+i)}{(4-i)(4+i)}\\ &= \frac{8+2i+12i+3i^2}{16-i^2}\\ &= \frac{8+14i-3}{16+1}\\
- Formatting Syntax @wiki
- w) like this: [[http://php.net|{{wiki:dokuwiki-128.png}}]] [[http://php.net|{{wiki:dokuwiki-128.png}}]] Please note: The image formatting is the only... Real size: {{wiki:dokuwiki-128.png}} Resize to given width: {{wiki:dokuwiki-128.png?50}} Resize to given width and height((when
- Question 8, Exercise 2.2 @math-11-nbf:sol:unit02
- \begin{align*} A &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}\\ B &= \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{bmatrix}\\ AB &= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{bma
- Khuram Ali Khan
- , Advances in Inequalities and Applications, 1 (2012), No. 1, 12-32, (Published on July 9, 2012). ([[http://scik.org/index.php/aia/article/view/347|Link]]) - Laszlo... ights, Proc. A. Razmadze Math. Inst., vol. 158 (2012), 33-56. ([[http://www.rmi.ge/proceedings/volumes
- Question 13, Exercise 2.1 @math-11-kpk:sol:unit02
- tion==== $$A=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33... )$$ $$A+A^t=\left[ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ... gin{matrix} a_{11} & a_{21} & a_{31} \\ a_{12} & a_{22} & a_{32} \\ a_{13} & a_{23} & a_{33
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- n \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in ... lpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an ide... quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}... \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- n \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in ... lpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an ide... quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}... \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{
- Question 2, Review Exercise @math-11-nbf:sol:unit08
- 2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\ph... This implies $\cos\pi >0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\p... sin \phi\\ &=\left(\frac{3}{5}\right) \left(\frac{12}{13}\right)-\left(-\frac{4}{5}\right) \left(\frac... 2 \theta&=\frac{144}{169} \\ \cos \phi&=\pm\frac{12}{13}\\ \end{align*} As $\phi$ is acute, so $\ph
- Question 2, Exercise 2.3 @math-11-nbf:sol:unit02
- The elements of \(R_1\) are \(a_{11} = 3\), \(a_{12} = 2\), and \(a_{13} = 3\). Now we find their cor... } (5 \cdot 0 - 1 \cdot 1) = (1) (-1) = -1 \\ & A_{12} = (-1)^{1+2} \left|\begin{array}{cc} 4 & 1 \\ 2 ... ant: \begin{align*} \det(A) &= a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} \\ &= 3(-1) + 2(2) + 3(-6) \\ &= -3 + 4 - 18 \\ &= -17 \end{align*} Thus, the
- Question 3 & 4 Exercise 4.3 @math-11-kpk:sol:unit04
- rac{66}{2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question 4===== The sum of ... a+d)&=36 \\ \Rightarrow 3 a&=36 \\ \Rightarrow a&=12\end{align} Now by the second condition, the sum o... \Rightarrow 3 a^3+6 a d^2&=6336 \\ \Rightarrow 3(12)^3+6(12) d^2&=6336 \text { as } a=12 \\ \Rightarrow 3(1728)+72 d^2&=6336 \\ \Rightarrow 72 d^2&=6336-
- 4th International Conference on "Recent Developments in Fluid Mechanics", QAU, Islamabad (09-11 August 2010) @conferences
- Chapter 12: Graph of Trigonometric and Inverse Trigonometric Functions and Solutions of Trigonometric Equations @fsc:kpk_fsc_part_1