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- Question 1, Exercise 1.3 @math-11-kpk:sol:unit01
- hat \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ \mathop+\limits_{-}2z&\mathop+\limits_{-
- Question 1 Exercise 5.2 @math-11-kpk:sol:unit05
- n=1.2^2+2.2^3+3.2^4+4.2^5+\ldots +n \cdot 2^n....(ii)\end{align} Suburacting the (ii) from (i), we get \begin{align} (1-2) S_n&=1 \cdot 2+(2-1) 2^2+(3-2) ... S_n& =2+(n-1) 2^{n+1}\end{align} =====Question 1(ii)===== Sum up to $n$ terms the series $1+4 x+7 x^2... =x+4 x^2+7 x^3+10 x^4+\ldots +(3 n-2) x^{4 t}....(ii)\end{align} Subtracting the (ii) from (i), we get
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- os \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \right)$. ====Solut... ght)&=\frac{33}{65}.\end{align} =====Question 4(ii)===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact valu
- Question 5(iii) & 5(iv) Exercise 3.5 @math-11-kpk:sol:unit03
- ====== Question 5(iii) & 5(iv) Exercise 3.5 ====== Solutions of Question 5(iii) & 5(iv) of Exercise 3.5 of Unit 03: Vectors. Th... TB or KPTBB) Peshawar, Pakistan. =====Question 5(iii)===== Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \... olution==== We have already calculated L.H.S in (ii) that\\ \begin{align}|\vec{a} \times \vec{b}|^2&=
- Question 5 & 6 Exercise 4.5 @math-11-kpk:sol:unit04
- _{2 n}&=\dfrac{a_1(r^{2 n}-1)}{r-1} \ldots . . . (ii)\\ \text { and } S_{3 n}&=\dfrac{a_1(r^{3 n}-1)}{r-1}...(iii)\end{align}\\ Puting (i),(ii) and (iii) in L.H.S of the given, we get \begin{align}S_n(S_{3 n}-S_{2 n})&=\dfrac{a_1(r^n-1)}{r-1}[\
- Question 1, Exercise 3.2 @math-11-kpk:sol:unit03
- \\ &=-\hat{i}+\hat{j}\end{align} =====Question.1(ii)===== If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b}... &=13\hat{i}-21\hat{j}\end{align} =====Question.1(iii)===== If $\vec{a}=3\hat{i}-5\hat{j}$ and $\vec{b... } …(i)\\ |\vec{b}|&=\sqrt(-2)^2+(3)^2=\sqrt{13} …(ii)\end{align} Subtracting (i) from (ii). We get $$|\hat{a}|-|\hat{b}|=\sqrt{34}-\sqrt{13}$$ =====Quest
- Question 5(i) & 5(ii) Exercise 3.5 @math-11-kpk:sol:unit03
- ====== Question 5(i) & 5(ii) Exercise 3.5 ====== Solutions of Question 5(i) & 5(ii) of Exercise 3.5 of Unit 03: Vectors. This is uni... a} \times \vec{b}\perp \vec{b}$. =====Question 5(ii)===== Let $\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \h... "><btn type="success">[[math-11-kpk:sol:unit03:ex3-5-p4|Question 5(iii) & 5(iv) >]]</btn></text>
- Question 1 and 2 Exercise 4.1 @math-11-kpk:sol:unit04
- quence whose last term is $50 $. =====Question 1(ii)===== Classify into finite and infinite sequences... in this sequence, we don't know. =====Question 1(iii)===== Classify into finite and infinite sequence... of the sequence are $1,3,6, 10$. =====Question 2(ii)==== Find first four terms of the sequence with t... the sequence are $4,-8,16,-32$. =====Question 2(iii)==== Find first four terms of the sequence with
- Question 6 & 7 Exercise 4.4 @math-11-kpk:sol:unit04
- ^{\mathbf{A 5}}=n\end{align} Multiplying (i) and (iii), we get\\ \begin{align}a_{10} \cdot a_{16}&=\ln... \quad \ln &=m^2 \because m=a_1 r^{12} \text { by (ii) }\end{align} Hence showed that $\ln =m^2$. ====... , \dfrac{1}{a_1 r^{n-1}}$,\\ General term of the (ii) sequence is:\\ \begin{align}a_n&=\dfrac{1}{a_1 r... endent of $n$, \\ which implies that sequence in (ii) is also geometric sequence with common ratio $\d
- Question 4 Exercise 4.5 @math-11-kpk:sol:unit04
- . \overline{8}=\dfrac{8}{9}$$.\\ =====Question 4(ii)===== Convert each decimal to common fraction $1 ... 7}{11} \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\dfrac{7}{11}=\dfrac{18}{11} \text {. }$$\\ =====Question 4(ii)===== Convert each decimal to common fraction $2
- Question 4 & 5 Exercise 5.2 @math-11-kpk:sol:unit05
- ac{7}{3^2}+\dfrac{9}{3^2}+\dfrac{11}{3^3}+\ldots.(ii) \end{align} Subtracting the (ii) from (i), we get \begin{align} & \dfrac{2}{3} S_{\infty}=5+\dfrac{2... $$r S_{\infty}=3 r+5 r^2+7 r^3+\ldots \infty....(ii)$$ Subtracsing the (ii) from (i), we get \begin{align} (1-r) S_{\infty}&=3+2 r+2 r^2+2 r^3+\ldots \
- Question 1 and 2 Exercise 6.1 @math-11-kpk:sol:unit06
- .5}{3.2 .1}\\ &=4200 \end{align} =====Question 1(ii)===== Evaluate the $\dfrac{3 !+4 !}{5 !-4 !}$ ===... )}\\ &=\dfrac{5}{16} \end{align} =====Question 1(iii)===== Evaluate the $\dfrac{(n-1) !}{(n+1) !}$ ==... =\dfrac{19 !}{13 !} \end{align} =====Question 2(ii)===== Write $2.4.6 .8 .10 .12$ in term of factori... t 6\\ &=2^6 \cdot 6 !\end{align} =====Question 2(iii)===== Write $n(n^2-1)$ in term of factorial. ===
- Question 1 and 2 Exercise 6.2 @math-11-kpk:sol:unit06
- {(6-6) !}\\ &=6 !=720\end{align} =====Question 1(ii)===== Evaluate $^{20} P_2$ ====Solution==== \begi... }\\ &=20 \times 19=380\end{align} =====Question 1(iii)===== Evaluate $^{16} P_3$ ====Solution==== \beg... can not be negative, so $n=11$. =====Question 2(ii)===== Solve $^n P_5=9(^{n-1} P_4)$ for $n.$ ====S... \\ \Rightarrow n&=9 \end{align} =====Question 2(iii)===== Solve $n^2 P_2=600$ for $n$ ====Solution==
- Question 1 Review Exercise 7 @math-11-kpk:sol:unit07
- id="a1" collapsed="true">(a): $2520$</collapse> ii. How many two digits odd numbers can be formed fo... e id="a2" collapsed="true">(c): $28$ </collapse> iii. How many six digits number can be formed from t... (b) $120 x y^2$ (c) $5000 x y^3$ (d) $6 x^2 y^2$ (ii) What is the coefficient of the term $$ \left(x^3... ^2\right)^7 ? $$ (a) 84 (b) -280 (c) 560 (d) 448 (iii) The expansion of $\left(x+\sqrt{x^2-1}\right)^5
- Question 1, Exercise 3.3 @math-11-kpk:sol:unit03
- times 3)\\ & =3-4-3=-4\end{align}. =====Question(ii)===== If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, $... times-5)\\ & =6+4+5=15\end{align}. =====Question(iii)===== If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$, ... t(\vec{b}+\vec{c})&=11\end{align}. =====Question(iii)===== If $\vec{a}=3 \hat{i}+4 \hat{j}-\hat{k}$,