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- 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
- 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 6 Exercise 8.2 @math-11-nbf:sol:unit08
- for the expression: $2 \cos ^{2}\left(\frac{\pi}{12}\right)-1$ ** Solution. ** We have a double-ang... alpha -1=\cos 2\alpha.$$ Put $\alpha= \dfrac{\pi}{12}$, we have \begin{align*} 2\cos^2 \left(\frac{\pi}{12}\right)-1&=\cos 2\left(\frac{\pi}{12}\right)\\ &=\cos \left(\frac{\pi}{6}\right)\\ \end{align*} \begi
- Question 9, Exercise 2.2 @math-11-nbf:sol:unit02
- \begin{align*} A &= \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31}... end{pmatrix} \\ B &= \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31}... } A + B &= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & a_{13} + b_{13} \\ a_{21} + b_{21} & a_{22} + b_{22} & a_{23} + b_{23} \\ a_{31} + b_{3
- Question 14, Exercise 4.5 @math-11-nbf:sol:unit04
- al notation for the infinite geometric series; $0.12121212 \ldots$ ** Solution. ** We can express the decimal as $$0.121212 \ldots = 0.12 + 0.0012 + 0.00001
- Unit 04: Sequences and Seeries @math-11-nbf:sol
- * [[math-11-nbf:sol:unit04:ex4-1-p6|Question 11 & 12]] * [[math-11-nbf:sol:unit04:ex4-1-p7|Question ... * [[math-11-nbf:sol:unit04:ex4-2-p7|Question 11 &12]] * [[math-11-nbf:sol:unit04:ex4-2-p8|Question ... * [[math-11-nbf:sol:unit04:ex4-3-p6|Question 11 &12]] * [[math-11-nbf:sol:unit04:ex4-3-p7|Question ... on 23 & 24 ]] * [[math-11-nbf:sol:unit04:ex4-3-p12|Question 25 & 26]] </panel> <panel type="default
- Question 5, 6 and 7, Exercise 4.4 @math-11-nbf:sol:unit04
- -2)= -6 \\ & a_{3}=a_{1} r^{2}=(3)(-2)^{2}=3 (4)= 12 \\ & a_{4}=a_{1} r^{3}=(3)(-2)^{3}=3 (-8) = -24 \end{align*} Hence $a_1=3$, $a_2=-6$, $a_3=12$, $a_4=-24$. GOOD =====Question 6===== Find the ... our terms of the geometric sequence. $\quad a_{1}=12, r=\frac{1}{2}$ ** Solution. ** Given $a_{1}=12$ and $r=\frac{1}{2}$. Use the formula $$a_{n}=a_{1
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- 3 \\ 2 & 4 \end{array} \right| = 4 - 6 = -2\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 3 \\ ... array}{cc} 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{a... t| = 5 - 6 = -1\\ A&= \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} &... \\ 2 & -5 \end{array} \right| = 15 - 4 = 11\\ A_{12} &= (-1)^{1+2} \left| \begin{array}{cc} 2 & 2 \\
- Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
- $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\... }$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an identity: $$\co... t{1+\tan^2\beta} \\ &=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\ &=-\sqrt{1+\dfrac{25}{144}} \\ & =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} \end{align*} \begin{align*} \Rightarrow \quad
- Question 13, Exercise 8.1 @math-11-nbf:sol:unit08
- Express in the form of $r \sin (\theta+\phi)$: $12 \sin \theta-5 \cos \theta$ ** Solution. ** Let $12=r\cos \varphi $ and $-5=r\sin \varphi$.\\ Squaring and adding \begin{align*} & (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \varphi \\... }=13 \end{align*} Also \begin{align*} & \frac{-5}{12}=\frac{r\sin \varphi }{r\cos \varphi } \\ \implie
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- (5+6i)\\ \implies &(2i+4) z + (6i-4) \omega = 10i-12 \\ \implies &(4+2i) z + (-4+6i) \omega = -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \implies &(3-14i)\omega=15-11i\\ \end{align... ts(2) \end{align} Multiply (1) by $2$ and (2) by $12$, we have \begin{align}-6i z - (12 + 4i) \omega =
- Question 25 and 26, Exercise 4.7 @math-11-nbf:sol:unit04
- {6 \cdot 7^n}\\ &=\frac{7}{6} + \frac{7(7^n - 1)}{12 \times7^n } - \frac{1 + 3n}{6 \times 7^n}\\ &=\frac{2\times 7^{n+1}+7^{n+1}-7-2-6n}{12\times 7^n} \\ &=\frac{3 \times7^{n+1}-9-6n}{12\times 7^n}\\ &=\frac{3(7^{n+1}-3-2n)}{12 \times7^n}\\ &=\frac{7^{n+1}-3-2n}{4\times 7^n}\\ \end{align
- Question 2, Exercise 5.3 @math-11-nbf:sol:unit05
- ld during the match can be modeled by $t(x)=x^{3}-12 x^{2}+48 x+74$, where $x$ is the number of games ... ket season. ** Solution. ** Given: $$t(x)=x^{3}-12 x^{2}+48 x+74.$$ When $t=12$ \begin{align*} t(12)&=(12)^3-12(12)^2+48(12)+74 \\ &=650. \end{align*} During the 12th game 650 tick
- Question 5 and 6, Exercise 4.5 @math-11-nbf:sol:unit04
- ==== Find the sum of the geometric series. $a_{1}=12, a_{5}=972, r=-3$ ** Solution. ** Given $a_1 = 12$, $a_5 = 972$ and $r = -3$.\\ Now, we can use \(r... $$a_n = a_1 \cdot r^{n-1}$$ \begin{align*} 972 &= 12 (-3)^{n-1}\\ 81 &= (-3)^{n-1}\\ (-3)^{n-1} &= 3^{... uad r \neq 1.$$ Thus, \begin{align*} S_5 &= \frac{12 \left(1 - (-3)^{5}\right)}{1 - (-3)} \\ &= \frac{