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- Question 9 Exercise 6.3 @math-11-kpk:sol:unit06
- n==== Total men are $7$ and total women are $6.$ Therefore, Total number of persons $=7+6=13$ Committee... . In how many ways can the committee be chosen if there must be at least two men? ====Solution==== Total men are $7$ and total women are $6$. Therefore, Total number of persons $=7+6=13$ There must be at least two men The number of committees that wil
- MTH321: Real Analysis I (Spring 2023) @atiq
- erges to $s$. - For each irrational number $x$, there exists a sequence $\left\{ {{r}_{n}} \right\}$ of... and only if for any real number $\varepsilon >0$, there exists a positive integer ${{n}_{0}}$ such that $... r $\lambda $ that lies between $f(a)$ and $f(b)$, there exist a point $c\in (a,b)$ with $f(c)=\lambda $. ... us on $[a,b]$and differentiable on $(a,b)$. Then there exists a point $c\in (a,b)$ such that $\frac{f(b
- MTH322: Real Analysis II (Spring 2023) @atiq
- {\,\infty }{f(x)\,dx}$ converges if, and only if, there exists a constant $M>0$ such tha $\int\limits_{a}... or every $b\ge a$ and for every $\varepsilon >0\,$there exists a $B>0\,$ such that \[\left| \,\int_{b}^{... or every $\varepsilon>0$ and for all $x\in[a,b]$, there exist an integer $N$ such that $$\left|f_{n+p}(x)... converge uniformly (and absolutely) on $[a,b]$ if there exists a convergent series $\sum M_n$ of positive
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- 1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutat... 3$ are $E$. $m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations a... $ are $E$, $m_2=2$ are $L$ $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutat... are $E$, $m_2=2$ are $L$ and $m_3=1$ are $C$. $\therefore$ \begin{align}\text{Number of permutations a
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- extbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. ===== Question 1(i... a \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22... ha \cos \beta +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ }}\cos {{53}^... ha \cos \beta -\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{19}^{\circ }}\cos {{5}^{
- Question 1, Exercise 10.1 @math-11-kpk:sol:unit10
- extbook Board (KPTB or KPTBB) Peshawar, Pakistan. There are four parts in Question 1. ===== Question 1(i... a \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22... ha \cos \beta +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ }}\cos {{53}^... ha \cos \beta -\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{19}^{\circ }}\cos {{5}^{
- History of Mathematics by Muhammad Usman Hamid @notes
- political and military events are always chaotic; there is no way to predict the rise of a Genghis Khan, ... chaotically—changed in the subsequent millennia, there are elements of greatness in all the fashions. Si... ble, or repulsive. But only among the sciences is there true progress; only there is the record one of continuous advance toward ever greater heights. And yet,
- Question 3 Exercise 6.4 @math-11-kpk:sol:unit06
- ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ay i.e. $${ }^8 C_8=\dfrac{8 !}{(8-8) ! 8 !}=1$$ Therefore probability to $8$ answers are correct is: $$... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc
- Definitions: FSc Part 1 (Mathematics): PTB by Aurang Zaib @fsc-part1-ptb
- rt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). Therefore, \( |z| = 5 \). =====Chapter 02: Set and Oper... well-defined. ====Methods to describe a set==== There are 3 method to describe a set ====Descriptive Me... ====Equivalent Set==== Two sets are equivalent if there exists a one-to-one correspondence between their ... forms a group because for every integer \( a \), there exists its additive inverse \( -a \), and additio
- Definitions: FSc Part 1 (Mathematics): PTB @fsc-part1-ptb
- ct collection of distinct object is called set.\\ There are three ways to describe a set.\\ * **Descri... two numbers $a$ and $b$. If $a,G,b$ are in $G.P$. Therefore $\frac{G}{a}=\frac{b}{G} \longrightarrow G^2... 1}{5},\frac{1}{7}$ are in harmonic sequence since there reciprocals $1,3,5,7$ are in $A.P$. * **Harmon
- Formatting Syntax @wiki
- es_do_not_work|Mozilla Knowledge Base]]. However, there will still be a JavaScript warning about trying t... be added to the [[doku>entities|pattern file]]. There are three exceptions which do not come from that ... xbasic xml xojo xorg_conf xpp yaml z80 zxbasic// There are additional [[doku>syntax_highlighting|advance
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\... minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\... minal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\
- Question 12 & 13, Exercise 3.3 @math-11-kpk:sol:unit03
- are the midpoints of sides shown\\ \begin{align}\therefore \quad \overrightarrow{O D}&=\dfrac{\vec{b}+\v... htarrow{O D} \perp \overrightarrow{B C} \quad \\ \therefore \quad \overrightarrow{O D} \cdot \overrightar... rightarrow{O E} \perp \overrightarrow{C A} \quad \therefore \quad \overrightarrow{O E} \cdot \overrightar
- Question 7 Exercise 3.5 @math-11-kpk:sol:unit03
- ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \ve... ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \vec{v} \times \ve... olution==== Since the given vectors are coplanar, therefore \begin{align}\vec{u} \cdot \bar{v} \times \ve
- Question 5 & 6 Exercise 4.3 @math-11-kpk:sol:unit04
- $Condition-2$\\ The sum of their square is $120$, therefore\\ \begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2... $ and $d=-1$ then the numbers are\\ \begin{align}\therefore x_1+(x_1+6 d)+(x_1+9 d)&=-6 \\ \Rightarrow 3 ... text { and } \\ x_{22}&=x_1+21 d=3+21(-1)=-18 \\ \therefore \quad x_3+x_8+x_{22}&=1-4-18=-21\end{align}