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Exercise 6.3 @matric:9th_science: 15 Hits, Last modified: 5 months ago; 2}, (x\neq 0)$\\ (iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$\\ (iv) $4(a+b)^2-12(a^2-b^2)+9(a-b)^2$\\ (v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4-2... {1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12,(x \neq 0)$\\ (viii) $(x^2+3x+2)(x^2+4x+3)(x^2+5... ) \end{align}$ (iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$\\ **Solution:**\\ $\beg
Exercise 2.6 (Solutions) @matric:9th_science:unit_02: 10 Hits, Last modified: 5 months ago; 9i-6i^2\\ &= 6-5i-6(-1)\\ &= 6+6-5i\\ &= 12-5i \end{array}$$ ====Question 4==== * Simplify ... =\frac{8-16i-6i+12i^2}{4-16(-1)}\\ &=\frac{8-22i+12(-1)}{4+16}\\ &=\frac{-4-22i}{20}\\ &=\frac{-4}{... -(\frac{11i}{10})^2\\ &= \frac{1}{25} - \frac{121i^2}{100}\\ &= \frac{1}{25} + \frac{121}{100}\\ &= \frac{4+121}{100}\\ &= \frac{125}{100}
Exercise 6.1 @matric:9th_science: 9 Hits, Last modified: 5 months ago; sions by factorization.\\ (i) $x^2+5x+6$, $x^2-4x-12$ \\ (ii) $x^3-27$, $x^2+6x-27$, $2x^2-18$ \\ (iii... &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x-6)\\ &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3... &= 2 \times 2 \times 3 (x+1)(x-1)(x-1)(x^2+1)\\ &=12 (x^4-1)(x-1)\end{align}$ ====Question 6==== For
Exercise 6.2 @matric:9th_science: 4 Hits, Last modified: 5 months ago; == $\frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}$\\ **Solution:**\\ $\begin{align} \frac{x^2-x-6}{x^2-9}&+\frac{x^2+2x-24}{x^2-x-12}\\ &=\frac{x^2-3x+2x-6}{(x)^2-(3)^2}+\frac{x^2+6x-4x-24}{x^2-4x+3x-12}\\&= \frac{x(x-3)+2(x-3)}{(x-3)(x+3)}+\frac{x(x+6... ac{x+3}{x(x-2)}\\&=1\end{align}$ ====Question 12:==== $\frac{2y^2+7y-4}{3y^2-13y+4} \div \frac{4y
Mathematics 9 (Science Group): 2 Hits, Last modified: 5 months ago; EW * [[matric:9th_science:unit11]] ===== Unit 12: Line Bisectors and Angle Bisectors ===== After ... Theorems | {{ :matric:9th_science:9th-science-ch-12-theorems.pdf |Download PDF}} ====== Unit 15: Pyt
Review exercise @matric:9th_science: 2 Hits, Last modified: 5 months ago; e H.C.F. of the following by factorization. $8x^4-128$ , $12x^3-96$\\ **Solution:**\\ $\begin{align}8x^4-128 &= 8(x^4-16)\\&=8[(x^2)^2-(4)^2]\\&= 2 \times 2\... (x^2+4)(x-2)(x+2)\end{align}$\\ $\begin{align}12 x^3-96&=12(x^3-8)\\&=2 \times 2\times 3 \times (x-2)(x^2+ 2 x+4)\end{align}$\\ $\begin{align} H.
Exercise 2.4 (Solutions) @matric:9th_science:unit_02: 2 Hits, Last modified: 5 months ago; ft(-3\right)}}\\ &=\frac{y^{-6}}{x^{-18} z^{-12}}\\ &=\frac{x^{18} z^{12}}{y^6} \end{array}$$ (iv) $$\begin{array}{cl} \frac{\left(81\right)^n.
Exercise 2.5 (Solutions) @matric:9th_science:unit_02: 2 Hits, Last modified: 5 months ago; (ii) $i^{50}$ (iii) $i^{12}$ ... &= -1 \end{array}$$ (iii) $$\begin{array}{cl} i^{12} &= (i^2 )^6\\ &= {-1}^6\\ &= 1 \en