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- Exercise 2.4 (Solutions) @matric:9th_science:unit_02
- {\frac{-1}{5}}}{\sqrt(196)^{-1}}$ * (ii) $\left(2x^5y^{-4}\right)\left(-8x^{-3}y^2\right)$ * (iii) $\left(\frac{x^{-2}y^{-1}z^{-4}}{x^4y^{-3}z^0}\right)^{-3}$ * (iv) $\frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\ri
- Exercise 6.3 @matric:9th_science
- +9y^6}{9x^4-24x^2y^2+16y^4},(x \neq 0)$\\ (vi) $\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)$\\ (vii) $\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12,(x \neq 0)$\\ (viii) $(x^2+3x+2)(x^2+4x
- Review exercise @matric:9th_science
- 2+b^2}$\\ **Answer:**\\ $a$\\ (xiv) Simplify $\left(\frac{2x+y}{x+y}-1\right)\div \left(1-\frac{x}{x+y}\right)$= ---\\ (a) $\frac{x}{x+y}$ (b) $\fr... root of $x^4+\frac{1}{x^4}+2$ is ---\\ (a) $\pm \left(x+\frac{1}{x}\right)$ (b) $ \left(x^2-\frac{1}{x^2}\right)$\\ (c) $\pm \left(x-\frac{1}{x}\right
- Exercise 6.2 @matric:9th_science
- rac{x+4}{x+3}\end{align}$ ====Question 2:==== $\left[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\... 4x}{x^4-1}$\\ **Solution:**\\ $\begin{align} \left[\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right]+\frac{4x}{x^4-1}&=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)-\frac{4x}{x^2+1}+\frac{4x}{x^4-1}\\&= \left(\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}\right)-\frac{4
- Exercise 2.2 (Solutions) @matric:9th_science:unit_02
- = \sqrt{24}$ ... ... ... * (ii) $\frac{-2}{3} \left( 5 + \frac{7}{2}\right) = \left(\frac{-2}{3}\right){5} + \left(\frac{-2}{3}\right)\left(\frac{7}{2}\right)$ ... ... * (iii) $\pi + \left(-\pi\right) = 0$ ... ...
- Exercise 2.3 (Solutions) @matric:9th_science:unit_02
- {array}{cl} \sqrt[4]{32} &= \sqrt[4]{{2}^5}\\ &= \left(2^{4}\times{2}\right)^\frac{1}{4}\\ &= \left(2^{4})^\frac{1}{4}\right)\times\sqrt[4]{2}\\ &= 2\sqrt[4]{... ) $$\begin{array}{cl} \sqrt[5]{\frac{3}{32}} &= \left(\frac{3}{{2^5}}\right)^\frac{1}{5}\\ &= \frac{\sq... n{array}{cl} \sqrt[3]{\frac{-8}{27}} &= \sqrt[3]{\left(\frac{-2^3}{3^3}\right)}\\ &= \left(\frac{-2}{3}\
- Unit 08: Linear Graph and their Application @matric:9th_science
- t angles at the point $O$ . * Identify origin $\left( O \right)$ and coordinate axes (horizontal and v... e rectangular plane. * Locate an ordered pair $\left( a,b \right)$ as a point in the rectangular plane
- Exercise 2.5 (Solutions) @matric:9th_science:unit_02
- (iv) $\left(-i\right)^8$ (v) $\left(-i\right)^5$ (vi) $i^{27}$ **
- Mathematics 9 (Science Group)
- number $a$ denoted by $|a|$ is defined as $$ |a|=\left\{\begin{matrix} a & \text{if } a\geq 0, \\ -a