NSC Mathematics Grade 12 Exam Information Sheet

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$\begin{align*}ax^{2}+bx+c&=0\\x&=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{align*}$

$A=P(1+ni)$

$A=P(1-ni)$

$A=P(1+i)^{n}$

$A=P(1-i)^{n}$

$T_{n}=a+(n-1)d)$

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{n}=\frac{a\left ( r^{n} -1\right )}{r-1}\;\;;r\neq 1$

$T_{n}=ar^{n-1}$

$S_{n}=\frac{a\left ( r^{n} -1\right )}{r-1}\;\;;r\neq 1$

$S_{\infty }=\frac{a}{1-r}\;\;;-1< r< 1$

$F=\frac{x\left [ \left ( 1+i\right )^{n}-1 \right ]}{i}$

$P=\frac{x\left [ 1-\left ( 1+i\right )^{-n} \right ]}{i}$

$f'\left ( x \right )=\lim_{h\rightarrow 0}\frac{f\left ( x +h\right )-f(x)}{h}$

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\textrm{M}=\left ( \frac{x_{1}+x_{2}}{2};\frac{y_{1}+y_{2}}{2} \right )$

$y=mx+c$

$y-y_{1}=m(x-x_{1})$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$m=\tan \theta$

$(x-a)^{2}+(y-b)^{2}=r^{2}$

$\begin{align*} \textrm{In }\bigtriangleup ABC:\frac{a}{\sin A}&=\frac{b}{\sin B}=\frac{c}{\sin C}\\\\a^{2}&=b^{2}+c^{2}-2bc.\cos A\\\\\textrm{area}\bigtriangleup ABC&=\frac{1}{2}ab.\sin C \end{align*}$

$\sin(\alpha +\beta )=\sin\alpha.\cos\beta+\cos\alpha.\sin\beta$

$\cos(\alpha+\beta)=\cos\alpha.cos\beta-\sin\alpha.\sin\beta$

$\sin(\alpha-\beta)=\sin\alpha.\cos\beta-\cos\alpha.\sin\beta$

$\cos(\alpha-\beta)=\cos\alpha.\cos\beta+\sin\alpha.\sin\beta$

$\cos2\alpha=\begin{cases} \cos^{2}\alpha-\sin^{2}\alpha \\1-2\sin^{2}\alpha \\ 2\cos^{2}\alpha-1\\ \end{cases}$

$\sin2\alpha=2\sin\alpha.\cos\alpha$

$\bar{x}=\frac{\sum_{}x}{n}$

$\sigma ^{2}=\frac{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}{n}$

$P(A)=\frac{n(A)}{n(S)}$

$P(A\,\,\mbox{or}\,\,B)=P(A)+P(B)-P(A\,\,\mbox{and}\,\,B)$

$\hat{y}=a+bx$

$b=\frac{\sum (x-\bar{x})(y-\bar{y})}{\sum (x-\bar{x})^{2}}$