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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
- 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 @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}^{
- 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
- 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}
- Question 9 & 10 Exercise 4.3 @math-11-kpk:sol:unit04
- arrow 9 n&=396 \\ \Rightarrow n&=44.\end{align} $\therefore$ Required sum is:\\ \begin{align} S_{44}&=\df... = The total money for distribution $S_4=1000$, \\ therefore we have $n=4$\\ \begin{align}\text{Let the fi... \ \begin{align}S_n&=\dfrac{n}{2}[2 a+(n-1) d] \\ \therefore S_4&=\dfrac{4}{2}[2 a+3(-20)] . \\ \Rightarro
- Question 13 & 14 Exercise 4.3 @math-11-kpk:sol:unit04
- in the third row and so forth. How many seats are there in the theater? ====Solution==== \begin{align}\te... ver this is including $1$ and $50$ as terms,\\ so therefore there would need to be $16$ terms between $1$ and $50$. ====Go To==== <text align="left"><btn type=
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- hat\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \dfrac{1}{16}&=8(\dfrac{1}{2})^{n-1} or\\ (\d... We know that\\ \begin{align}a_n&=a_1 r^{n-1}, \\ \therefore 2^{10}&=2^4(2)^{n-1} \text { or } 2^{n-1}=\df... ign}r&=\dfrac{-1}{\dfrac{8}{5}}=-\dfrac{5}{8} \\ \therefore S_{\infty}&=\dfrac{a_1}{1-r}=\dfrac{\dfrac{8}
- Question 11 & 12 Exercise 4.5 @math-11-kpk:sol:unit04
- ==== The general term of G.P $a_n=a_1 r^{n-1}$.\\ Therefore, $a_p=a_1 r^{p-1}=a \quad a_q=a_1 r^{q-1}=b$ ... frac{a_1}{1-r}$, but we are given $S_{\infty}=6$. Therefore,\\ $$\dfrac{a_1}{1-r}=6 \text { or } 6 a_1=1-... four times the sum of all the terms following it, therefore\\ \begin{align}a_1&=4(a_1 r+a_1 r^2+a_1 r^3+\
- Question 5, Exercise 10.1 @math-11-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 1 Exercise 3.4 @math-11-kpk:sol:unit03
- ec{a}&=2 \hat{i}-3 \hat{j}\\ \vec{b}&=\hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{c... {k}\\ \vec{b}&=6 \hat{i}+2 \hat{j}-3 \hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{c
- Question 5 Exercise 3.4 @math-11-kpk:sol:unit03
- htarrow{P Q} \times \overrightarrow{P R}|&=30 \\ \therefore \text { Area of triangle }& =\dfrac{1}{2}|\ov... {P Q} \times \overrightarrow{P R}|&=\sqrt{76} \\ \therefore \text { Area of triangle }&=\dfrac{1}{2} | \v
- Question 4 & 5 Exercise 4.4 @math-11-kpk:sol:unit04
- akistan. =====Question 4===== How many terms are there in a geometric sequence in which the first and th... a three terms geometric sequence,\\ \begin{align}\therefore \dfrac{a_2}{a_1}&=\dfrac{a_3}{a_2} \\ \Righta