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- Definitions: FSc Part 1 (Mathematics): PTB @fsc-part1-ptb
- Punjab Textbook Board (PTB) Lahore, Pakistan are given. We are very thankful to **Muhammad Waqas Sulaima... contribution. Definitions by Mr. Aurang Zaib are given at [[:fsc-part1-ptb:definitions-aurang-zaib]] =... ruth Table:** A table to drives truth values of a given compound statement in terms of its component part... Groupoid:** A non-empty set which is closed under given Binary Operation ‘*' is called Groupoid. * *
- Question 5, Exercise 1.3 @math-11-kpk:sol:unit01
- the equation ${{z}^{2}}+z+3=0$. ====Solution==== Given: $${{z}^{2}}+z+3=0.$$ According to the quadratic ... qrt{11}}{2}i\end{align} Thus the solutions of the given equation are $-\dfrac{1}{2}\pm\dfrac{\sqrt{11}}{2... the equation ${{z}^{2}}-1=z$.\\ ====Solution==== Given: $${{z}^{2}}-1=z$$ $$\implies {{z}^{2}}-z-1=0$$ A... sqrt{5}}{2}.\end{align} Thus the solutions of the given equations are $\dfrac{1\pm\sqrt{5}}{2}$. =====Que
- Question 1 and 2 Exercise 4.1 @math-11-kpk:sol:unit04
- == Find first four terms of the sequence with the given general terms: $a_n=\dfrac{n(n+1)}{2}$ ====Solution==== Given: $$a_n=\dfrac{n(n+1)}{2}$$ For first term, put $n... == Find first four terms of the sequence with the given general terms: $a_n=(-1)^{n-1} 2^{n+1}$ ====Solution==== Given: $$a_n=(-1)^{n-1} 2^{n+1}$$ For first term, put
- Question 2 Exercise 4.3 @math-11-kpk:sol:unit04
- of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that is missing: $a_1=2, n=17, d=3$. GOOD ====Solution==== Given: $a_1=2, n=17, d=3$ \\ We need to find $a_{17}$ a... of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-40, S_{21}=210$. GOOD ====Solution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$
- About Us
- y putting some related keywords in the search box given in main menu. This site contains old papers, mo... l institute or board or university. The resources given on this site holds no official position in govern... e or board or university). The material/resources given on this site are open educational resources (OER)... ut type="info" icon="true"> While using resources given on this site you agree to a term that we (MathCit
- Unit 03: Vectors (Solutions) @math-11-kpk:sol
- * Find a unit vector in the direction of another given vector. * Find the position vector of a point w... divides the line segment joining two points in a given ratio. * Use vectors to prove simple theorems o... e by a constant force in moving an object along a given vector. * Define cross or vector product of two... ween two vectors. * Find the vector moment of a given force about a given point. * Define scalar trip
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align... \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align
- Question 2, Exercise 2.2 @math-11-kpk:sol:unit02
- & 2 & 0 \end{matrix}\right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 0 \\ 3 & 1... -1 & 3 \end{matrix} \right|=0$. ====Solution==== Given $$\left| \begin{matrix} 1 & 2 & 3 \\ -8 & ... 2 & 1 & 4 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix} 1 & 3 & -2 \\ 3 & ... 2 & 4 & 2 \end{matrix} \right|$. ====Solution==== Given $$\left| \begin{matrix}3 & 2 & 0 \\1 & 1 & -3 \
- Question 7 Exercise 3.5 @math-11-kpk:sol:unit03
- 3 \hat{i}+\hat{j}+c \hat{k}$ ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec... ign} which is required value of $c$ for which the given vectors become coplanar. =====Question 7(ii)====... \hat{i}+\hat{j}-c \hat{k}$. ====Solution==== The given vectors are coplanar, therefore \begin{align}\vec... which is the required value of $c$ for which the given vectors become coplanar. =====Question 7(iii)===
- Question 2 Exercise 4.5 @math-11-kpk:sol:unit04
- a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $a_1=1, \quad r=-... a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=\dfrac{1}{2}, ... {\prime \prime}]}{1-r},\end{align} becomes in the given case\\ \begin{align}\Rightarrow S_9&=\dfrac{2^8[1... a_n, n_2 r$ and $S_n$ of a geometric sequence are given. Find the ones that are missing $r=-2, S_n=-63, a
- Question 4 Exercise 6.4 @math-11-kpk:sol:unit06
- ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI
- Question 3 Exercise 7.2 @math-11-kpk:sol:unit07
- t $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align}T_{r+1}&=\dfrac{9 !}{(9-r) ! r !}(\dfrac{4... t $T_{r+1}$ be the term independent of $x$ in the given expansion. $T_{r+1}$ of the given expansion is: \begin{align} T_{r+1}&=\dfrac{10 !}{(10-r) ! r !}(x)^{1
- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align... \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is... arrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align
- Formatting Syntax @wiki
- {{wiki:dokuwiki-128.png}} Resize to given width: {{wiki:dokuwiki-128.png?50}} Resize to given width and height((when the aspect ratio of the given width and height doesn't match that of the image, i... {{wiki:dokuwiki-128.png}} Resize to given width: {{wiki:dokuwiki-128.png?50}}
- Question 6, Exercise 1.3 @math-11-kpk:sol:unit01
- of the equation ${{z}^{3}}=-8$. ====Solution==== Given: $$z^3=-8.$$ This gives \begin{align} & z^3+2^3=0... \left( z-1 \right)}^{3}}=-1$.\\ ====Solution==== Given: $$(z-1)^3=-1.$$ Since we have $$(a-b)^3=a^3-b^3... \sqrt{3}i}{2}.\end{align} Hence solutions of the given equation are $0$, $\dfrac{3}{2}\pm \dfrac{\sqrt{3... the equation ${{z}^{3}}=1$. \\ ====Solution==== Given $$z^3=1.$$ Thus, we have \begin{align}&z^3-1^3=0\