1.1 Siliptify thy followhig expreations fally: \[ 1.1 .1(x-2)\left(x^{2}+6 x-7\right) \] \( 1.12 \frac{27 p^{3}+64}{p^{2}-12 p+16} \div \frac{9 p^{2}-16}{3 p-4} \) (3) (5) 1.2 Factofise the following expressions fully: \[ 1.2 .7 \quad r-3 r^{2}+4 r-12 \] (2) 1.3 Determine, without the use or a calculator; between which two corsecutive integers \( \sqrt{125} \) lies. 1.4 Prove that \( 0.9=1 \). (2) ESTION2 (4) [19] Solporgo ? \[ 211 \times(3 x-4)=0 \] \[ 21.2\left(x^{2}-3 x+6=0\right. \] \[ 2.1 .3 \quad x+2-1=0 \] 2.1.4 \( 4<-3 x-1 \leq 5 \) (Indicate yguy olution on a number line) (3) Solve for \( x \) and \( y \) simultaneoursiy from the given equations: \[ 6 x+10 y-16=0 \] and \[ 2 x=4 y-2 \] \( [14] \)

Let's delve into some cool aspects of algebra and numbers! Imagine using algebra not just on paper but to design algorithms! Programmers often use polynomial expressions to create efficient code that sorts or organizes data. For instance, the divide-and-conquer technique in sorting utilizes these principles for better performance—much like simplifying those expressions can make complex problems easier to solve! Now, consider another fascinating twist: the proof that \(0.9 repeating\) equals \(1\)! It's a neat trick that often astonishes students. You can set \(x = 0.999...\), multiply by \(10\) to get \(10x = 9.999...\), and subtract your original equation to uncover that \(9x = 9\), leading you right back to \(x = 1\), proving that \(0.9 repeating\) is simply another way to express \(1\)!