1 Find the integers between which sqrt(33) lies. 1.2 Convert the recurring decimal 0.45 to a common fraction. 2.1 Simplify ((3x)2 (-2xy)1)/(2x3 y4).

Question 1

1.1 To find the integers between which $\sqrt{33}$ lies, we find the perfect squares closest to 33.

$$5^2 = 25$$ $$6^2 = 36$$

Since $25 < 33 < 36$, we can take the square root of all parts:

$$\sqrt{25} < \sqrt{33} < \sqrt{36}$$ $$5 < \sqrt{33} < 6$$

The value of $\sqrt{33}$ lies between the integers 5 and 6.
The integers are $\boxed{\text{5 and 6}}$.

1.2 Let $x = 0.4\dot{5}$. This means $x = 0.4555...$

Step 1: Multiply by 10 to move the non-repeating part to the left of the decimal.

$$10x = 4.555... \quad (1)$$

Step 2: Multiply by 100 to move one full repeating block to the left of the decimal.

$$100x = 45.555... \quad (2)$$

Step 3: Subtract equation (1) from equation (2).

$$100x - 10x = 45.555... - 4.555...$$ $$90x = 41$$

Step 4: Solve for $x$.

$$x = \frac{41}{90}$$

The common fraction in its simplest form is $\boxed{\frac{41}{90}}$.

Question 2

2.1 Simplify: $\frac{(3x)^2 (-2xy)^1}{2x^3 y^4}$

Step 1: Expand the terms in the numerator.

$$\frac{(3^2 x^2) (-2xy)}{2x^3 y^4}$$ $$\frac{9x^2 (-2xy)}{2x^3 y^4}$$

Step 2: Multiply the terms in the numerator.

$$\frac{-18x^{2+1}y}{2x^3 y^4}$$ $$\frac{-18x^3y}{2x^3 y^4}$$

Step 3: Simplify the coefficients and variables.

$$\frac{-18}{2} \cdot \frac{x^3}{x^3} \cdot \frac{y}{y^4}$$ $$-9 \cdot x^{3-3} \cdot y^{1-4}$$ $$-9 \cdot x^0 \cdot y^{-3}$$ $$-9 \cdot 1 \cdot \frac{1}{y^3}$$ $$-\frac{9}{y^3}$$

The simplified expression is $\boxed{-\frac{9}{y^3}}$.

2.2.1 Factorise completely: $6p + 40 - p^2$

Step 1: Rearrange the terms in standard quadratic form.

$$-p^2 + 6p + 40$$

Step 2: Factor out $-1$.

$$-(p^2 - 6p - 40)$$

Step 3: Factor the quadratic expression $p^2 - 6p - 40$. We need two numbers that multiply to $-40$ and add to $-6$. These numbers are $-10$ and 4.

$$-(p-10)(p+4)$$

Alternatively, distribute the negative sign into one of the factors:

$$(10-p)(p+4)$$

The factorised expression is $\boxed{(10-p)(p+4)}$.

2.2.2 Factorise completely: $-xy - (y-x)b + b^2$

Step 1: Distribute the negative sign and $b$ in the middle term.

$$-xy - yb + xb + b^2$$

Step 2: Group terms with common factors.

$$(-xy - yb) + (xb + b^2)$$

Step 3: Factor out common terms from each group.

$$-y(x+b) + b(x+b)$$

Step 4: Factor out the common binomial $(x+b)$.

$$(x+b)(-y+b)$$ $$(x+b)(b-y)$$

The factorised expression is $\boxed{(x+b)(b-y)}$.

Question 3

3.1 Solve for x: $\frac{x+2}{x^2-3x-4} = \frac{3}{x-4} - \frac{1}{2+2x}$

Step 1: Factor the denominators.

$$x^2-3x-4 = (x-4)(x+1)$$ $$2+2x = 2(1+x)$$

The equation becomes:

$$\frac{x+2}{(x-4)(x+1)} = \frac{3}{x-4} - \frac{1}{2(x+1)}$$

Step 2: Identify restrictions on $x$. The denominators cannot be zero, so $x \neq 4$ and $x \neq -1$.

Step 3: Find the Lowest Common Denominator (LCD), which is $2(x-4)(x+1)$.

Step 4: Multiply every term by the LCD.

$$2(x-4)(x+1) \cdot \frac{x+2}{(x-4)(x+1)} = 2(x-4)(x+1) \cdot \frac{3}{x-4} - 2(x-4)(x+1) \cdot \frac{1}{2(x+1)}$$

Step 5: Simplify and solve for $x$.

$$2(x+2) = 2 \cdot 3 \cdot (x+1) - (x-4) \cdot 1$$ $$2x+4 = 6(x+1) - (x-4)$$ $$2x+4 = 6x+6 - x+4$$ $$2x+4 = 5x+10$$ $$4-10 = 5x-2x$$ $$-6 = 3x$$ $$x = \frac{-6}{3}$$ $$$$