2. Determine between which two consecutive integers does $$ \sqrt{11} $$ lie 3. Write 0,7 as a common fraction. Clearly show all your working. QUESTION 2 [ 6 Marks] 1. Determine the product of the following and simplify fully: $$ (x-2)\left(x^{2}+5 x-1\right) $$ 2. Factorise $$ y^{2}(y-2)+x^{2}(2-y) $$ fully

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UpStudy AI · Accepted Answer

Numbers by following steps:
- step0: Solution:
  $$0.7$$
Calculate or simplify the expression $$ \sqrt(11) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{11}$$
Factor the expression $$ y^2*(y-2)+x^2*(2-y) $$.
Factor the expression by following steps:
- step0: Factor:
  $$y^{2}\left(y-2\right)+x^{2}\left(2-y\right)$$
- step1: Rewrite the expression:
  $$y^{2}\left(y-2\right)-x^{2}\left(y-2\right)$$
- step2: Factor the expression:
  $$\left(y^{2}-x^{2}\right)\left(y-2\right)$$
- step3: Factor the expression:
  $$\left(y+x\right)\left(y-x\right)\left(y-2\right)$$
Expand the expression $$ (x-2)(x^2+5*x-1) $$
Simplify the expression by following steps:
- step0: Multiply the terms:
  $$\left(x-2\right)\left(x^{2}+5x-1\right)$$
- step1: Apply the distributive property:
  $$x\times x^{2}+x\times 5x-x\times 1-2x^{2}-2\times 5x-\left(-2\times 1\right)$$
- step2: Multiply the terms:
  $$x^{3}+5x^{2}-x-2x^{2}-10x-\left(-2\right)$$
- step3: Remove the parentheses:
  $$x^{3}+5x^{2}-x-2x^{2}-10x+2$$
- step4: Subtract the terms:
  $$x^{3}+3x^{2}-11x+2$$
Let's solve each part of your problem step by step.

### Question 2: Determine between which two consecutive integers does $$ \sqrt{11} $$ lie

1. We know that $$ \sqrt{11} $$ is between two perfect squares. The perfect squares around 11 are $$ 3^2 = 9 $$ and $$ 4^2 = 16 $$.
2. Therefore, $$ 3 < \sqrt{11} < 4 $$.
3. To find a more precise value, we can calculate $$ \sqrt{11} $$ approximately, which is $$ \sqrt{11} \approx 3.316625 $$.
4. Thus, $$ \sqrt{11} $$ lies between the consecutive integers 3 and 4.

### Question 3: Write 0.7 as a common fraction

1. The decimal 0.7 can be expressed as a fraction. 
2. We can write $$ 0.7 = \frac{7}{10} $$.
3. This fraction is already in its simplest form.

### Question 2 (continued): Determine the product of $$ (x-2)(x^2+5x-1) $$ and simplify fully

1. We expand the expression:
   $$
   (x-2)(x^2+5x-1) = x(x^2+5x-1) - 2(x^2+5x-1)
   $$
2. This results in:
   $$
   = x^3 + 5x^2 - x - 2x^2 - 10x + 2
   $$
3. Combining like terms gives:
   $$
   = x^3 + (5x^2 - 2x^2) + (-x - 10x) + 2 = x^3 + 3x^2 - 11x + 2
   $$

### Question 2 (continued): Factorise $$ y^2(y-2) + x^2(2-y) $$ fully

1. We can rewrite the expression:
   $$
   y^2(y-2) + x^2(2-y) = y^2y - 2y^2 + 2x^2 - x^2y
   $$
2. Rearranging gives:
   $$
   = -x^2y + y^2y - 2y^2 + 2x^2
   $$
3. Factoring out common terms:
   $$
   = (y+x)(y-x)(y-2)
   $$

### Summary of Results
1. $$ \sqrt{11} $$ lies between the integers **3 and 4**.
2. $$ 0.7 $$ as a fraction is **$$\frac{7}{10}$$**.
3. The product $$ (x-2)(x^2+5x-1) $$ simplifies to **$$ x^3 + 3x^2 - 11x + 2 $$**.
4. The factorization of $$ y^2(y-2) + x^2(2-y) $$ is **$$ (y+x)(y-x)(y-2) $$**.
1. $$ \sqrt{11} $$ lies between 3 and 4. 2. $$ 0.7 = \frac{7}{10} $$. 3. $$ (x-2)(x^2+5x-1) = x^3 + 3x^2 - 11x + 2 $$. 4. $$ y^2(y-2) + x^2(2-y) = (y+x)(y-x)(y-2) $$.

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