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Maths Formulas

  • \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
  • \(\begin{aligned} A &= P\left(1 + ni\right) \\ A &= P\left(1 - ni\right) \\ A &= P\left(1 - i\right)^{n} \\ A &= P\left(1 + i\right)^{n}\end{aligned}\)
  • \(\begin{aligned} T_{n} &= a + \left(n - 1\right)d \\ S_{n} &= \dfrac{n}{2}\left[2a + \left(n - 1\right)d\right] \\ T_{n} &= ar^{n - 1} \\ S_{n} &= \dfrac{a\left(r^{n} - 1\right)}{r - 1} \: ; r \neq 1 \\ S_{\infty} &= \dfrac{a}{1 - r} \: ; -1 < r < 1 \end{aligned}\)
  • \(\begin{aligned} F &= \dfrac{x\left[\left(1 + i\right)^{n} - 1\right]}{i} \\ P &= \dfrac{x\left[1 - \left(1 + i\right)^{-n}\right]}{i} \end{aligned}\)
  • \(\begin{aligned} f^{\prime}(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} \end{aligned}\)
  • \(\begin{array}{c} d = \sqrt{\left(x_{2} - x_{1}\right)^{2} + \left(y_{2} - y_{1}\right)^{2}} \\ M\left(\dfrac{x_{1} + x_{2}}{2}; \dfrac{y_{1} + y_{2}}{2}\right) \\ y = mx + c \\ y - y_{1} = m(x - x_{1}) \\ m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m = \tan\theta \\ (x - a)^{2} + (y - b)^{2} = r^{2} \end{array}\)
  • \(\begin{array}{c} \text{In } \Delta ABC: \\ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\ a^{2} = b^{2} + c^{2} - 2bc{\cdot}\cos A \\ \text{area } \Delta ABC = \dfrac{1}{2}ab{\cdot}\sin C \end{array}\)
  • \(\begin{aligned} \sin(\alpha + \beta) &= \sin\alpha {\cdot} \cos\beta + \cos\alpha {\cdot} \sin\beta \\ \sin(\alpha - \beta) &= \sin\alpha {\cdot} \cos\beta - \cos\alpha {\cdot} \sin\beta \\ \cos(\alpha + \beta) &= \cos\alpha {\cdot} \cos\beta - \sin\alpha {\cdot} \sin\beta \\ \cos(\alpha - \beta) &= \cos\alpha {\cdot} \cos\beta + \sin\alpha {\cdot} \sin\beta \\ \cos 2\alpha &= \begin{cases} \cos^{2}\alpha - \sin^{2}\alpha \\ 1 - 2\sin^{2}\alpha \\ 2\cos^{2}\alpha - 1 \end{cases} \\ \sin 2\alpha &= 2\sin\alpha {\cdot} \cos\alpha \end{aligned}\)
  • \(\begin{aligned} \bar{x} &= \dfrac{\sum{x}}{n} \\ \sigma^{2} &= \dfrac{\sum_{i = 1}^{n}{\left(x_{i} - \bar{x}\right)^{2}}}{n} \\ P(A) &= \dfrac{n(A)}{n(S)} \\ P(A \text{ or } B) &= P(A) + P(B) - P(A \text{ and } B) \\ \hat{y} &= a + bx \\ b &= \dfrac{\sum {\left(x - \bar{x}\right)\left(y - \bar{y}\right)}}{\sum {\left(x - \bar{x}\right)^{2}}}\end{aligned}\)