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- Exercise 2.4 (Solutions) @matric:9th_science:unit_02
- \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) $\frac{\left(81\right)^n.3^5-\left(3\right)^{4n-1}\left(243\right)}{\left(9^2n\right)\le
- Exercise 6.3 @matric:9th_science
- +16y^4},(x \neq 0)$\\ (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) $(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)$ \\ (ix) $(
- Review exercise @matric:9th_science
- = Chose the correct answers. <quizlib id="quiz" rightanswers="[['a1'],['a0']]" submit="Check Answers"> ... $a$\\ (xiv) Simplify $\left(\frac{2x+y}{x+y}-1\right)\div \left(1-\frac{x}{x+y}\right)$= ---\\ (a) $\frac{x}{x+y}$ (b) $\frac{y}{x+y}$\\ (c) $\frac{... 1}{x^4}+2$ is ---\\ (a) $\pm \left(x+\frac{1}{x}\right)$ (b) $ \left(x^2-\frac{1}{x^2}\right)$\\
- Exercise 6.2 @matric:9th_science
- [\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right]+\frac{4x}{x^4-1}$\\ **Solution:**\\ $\begin{... [\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{4x}{x^2+1}+\frac{4x}{x^4-1}\\&= \frac{x^2+
- Exercise 2.2 (Solutions) @matric:9th_science:unit_02
- v) $ab = ba$ ... ... ... * (vi) $a + c = b + c \Rightarrow a = b$ ... ... ... * (vii) $5 + (-5) = 0$ ... mes \frac{1}{7} = 1$ ... ... ... * (ix) $a > b \Rightarrow ac > bc? (c >0)$ ... ... ... **Solution**\\... ve w.r.t multiplicative) * (vi) $a + c = b + c \Rightarrow a = b$ ... ... ... (Cancellation property o... . ... (Multiplicative inverse) * (ix) $a > b \Rightarrow ac > bc?(c >0)$ ... ... ... (Multiplicative
- Mathematics 9 (Science Group)
- a\geq 0, \\ -a & \text{if } a<0. \end{matrix}\right. $$ e.g. $|6|=6$, $|0|=0$, $|-3|=3$. **Defining... al triangle. * an isosceles triangle. * a right angled triangle. * a scalene triangle. * Us... t. * Prove that if in the correspondence of two right-angled triangles, the hypotenuse and one side of ... will be able to: * prove that any point on the right bisector of a line segment is equidistant from it
- Exercise 2.3 (Solutions) @matric:9th_science:unit_02
- {32} &= \sqrt[4]{{2}^5}\\ &= \left(2^{4}\times{2}\right)^\frac{1}{4}\\ &= \left(2^{4})^\frac{1}{4}\right)\times\sqrt[4]{2}\\ &= 2\sqrt[4]{2} \end{array}$$ (ii... } \sqrt[5]{\frac{3}{32}} &= \left(\frac{3}{{2^5}}\right)^\frac{1}{5}\\ &= \frac{\sqrt[5]{3}}{2} \end{arra... \frac{-8}{27}} &= \sqrt[3]{\left(\frac{-2^3}{3^3}\right)}\\ &= \left(\frac{-2}{3}\right)^{3\times\frac{1}
- Unit 08: Linear Graph and their Application @matric:9th_science
- ane consisting of two number lines interesting at right angles at the point $O$ . * Identify origin $\left( O \right)$ and coordinate axes (horizontal and vertical ax... ar plane. * Locate an ordered pair $\left( a,b \right)$ as a point in the rectangular plane and recogni
- Exercise 11.1 (Solutions) @matric:9th_science:unit11
- m\angle C$ (Opposite angles of parallelogram)\\ $\Rightarrow \,\,\,\,m\angle C=50^\circ$ </panel> </col><... \begin{align} & m\angle B =m\angle D=40^\circ \\ \Rightarrow \,\,\,\,\,\,\,\,\, & m\angle D=40^\circ \en... } Also \begin{align} & m\angle C=m\angle DAB \\ \Rightarrow \,\,\,\,\,\,\,\,\, & m\angle C=140^\circ \e
- Exercise 2.1 (Solutions) @matric:9th_science:unit_02
- nd 1 into 3 equal parts. - (ii) Take 2 parts on right of 0 - (iii) Point M represents $\frac{2}{3}$ ... 0 and -1 into 5 equal parts. - Take 4 parts on right of 0 - Point M represents $-\frac{4}{5}$ on th
- Exercise 2.5 (Solutions) @matric:9th_science:unit_02
- (iv) $\left(-i\right)^8$ (v) $\left(-i\right)^5$ (vi) $i^{27}$ **Solution**