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- Exercise 2.4 (Solutions)
- \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 2.2 (Solutions)
- ... * (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) $\pi + \left(-\pi\right) = 0$ ... ... ... * (iv) $\sqr
- Exercise 2.3 (Solutions)
- {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}
- Exercise 2.1 (Solutions)
- 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)
- (iv) $\left(-i\right)^8$ (v) $\left(-i\right)^5$ (vi) $i^{27}$ **Solution**