Evaluate: (23 * 32), [3]343, sqrt(58) to 3 decimal places, and 0.0000821 in standard form.

$$8^2 = 64$$

The value is $\boxed{\text{64}}$.

II. $\sqrt[3]{343}$

$$7^3 = 7 \times 7 \times 7 = 343$$

The value is $\boxed{\text{7}}$.

b) Find the value of $\sqrt{58}$, correct to 3 decimal places.

$$\sqrt{58} \approx 7.616$$

The value is $\boxed{\text{7.616}}$.

c) Express 0.0000821 in standard form.

Since it was moved 5 places to the right, the exponent of 10 is -5.

$$8.21 \times 10^{-5}$$

The standard form is $\boxed{8.21 \times 10^{-5}}$.

QUESTION 2

Express 0.00542 in standard form.

Since it was moved 3 places to the right, the exponent of 10 is -3.

$$5.42 \times 10^{-3}$$

The standard form is $\boxed{5.42 \times 10^{-3}}$.

Solve for $x: 3x^2+5x-2=0$, using the quadratic formula.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. For $3x^2+5x-2=0$, we have $a=3$, $b=5$, $c=-2$.

$$x_1 = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}$$ $$x_2 = \frac{-5 - 7}{6} = \frac{-12}{6} = -2$$

The solutions are $\boxed{x = \frac{1}{3} \text{ or } x = -2}$.

QUESTION 3

a) In triangle ABC, AB = 12 cm, AC = 9 cm, and angle BAC = 58°. Calculate the length of BC.
We use the Cosine Rule: $a^2 = b^2 + c^2 - 2bc \cos A$. Here, $a=BC$, $b=AC=9$, $c=AB=12$, and $A=58^\circ$.

$$BC^2 = 81 + 144 - 216 \cos(58^\circ)$$ $$BC^2 = 225 - 216 \cos(58^\circ)$$

$$BC^2 = 225 - 216(0.529919...)$$ $$BC^2 = 225 - 114.4625...$$ $$BC^2 = 110.5375...$$

$$BC = \sqrt{110.5375...} \approx 10.51368...$$ $$BC \approx 10.51 \text{ cm}$$

The length of BC is approximately $\boxed{10.51 \text{ cm}}$.

b) In triangle XYZ, angle X = 40°, angle Y = 65°, and side XY = 11 cm. Calculate the length of side YZ.
We use the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Here, we want to find side YZ (opposite angle X), and we know side XY (opposite angle Z).

$$\frac{YZ}{\sin X} = \frac{XY}{\sin Z}$$ $$\frac{YZ}{\sin 40^\circ} = \frac{11}{\sin 75^\circ}$$

$$YZ = \frac{11 \times 0.642788...}{0.965926...} = \frac{7.070668...}{0.965926...} \approx 7.3209...$$ $$YZ \approx 7.32 \text{ cm}$$

The length of YZ is approximately $\boxed{7.32 \text{ cm}}$.

QUESTION 4

a) Simplify $\log 45 + \log 8 - \log 6$, giving your answer as a single logarithm, then evaluate it.

$$\log 45 + \log 8 - \log 6 = \log(45 \times 8) - \log 6$$ $$= \log(360) - \log 6$$

$$\log(60) \approx 1.77815$$

The single logarithm is $\boxed{\log(60)}$ and its value is approximately $\boxed{1.778}$.

b) Solve for $x: \log_3(x) = 4$.

$$x = 3 \times 3 \times 3 \times 3 = 81$$

The solution is $\boxed{x = 81}$.

c) Simplify $5^3 \times 5^1 \div 5^2$, giving your answer as a single power of 5.

$$5^4 \div 5^2 = 5^{4-2} = 5^2$$

The simplified expression is $\boxed{5^2}$.

d) Solve for $x: 2^{x+1} = 32$.

$$x = 5 - 1$$ $$x = 4$$

The solution is $\boxed{x = 4}$.

QUESTION 5

a) Given that $(a - b)(a + b) = 40$ and $ab = 9$, determine the value of $(a^2 + b^2)$.

$$(a-b)(a+b) = a^2 - b^2$$

So, $a^2 - b^2 = 40$.

$$(a^2+b^2)^2 = 1600 + 4(81)$$ $$(a^2+b^2)^2 = 1600 + 324$$ $$(a^2+b^2)^2 = 1924$$

$$a^2+b^2 = \sqrt{1924} = \sqrt{4 \times 481} = 2\sqrt{481}$$

The value of $(a^2 + b^2)$ is $\boxed{2\sqrt{481}}$.

b) Solve the equation: $\frac{2x-1}{3} - \frac{x+2}{4} = 1$.

$$12 \left(\frac{2x-1}{3}\right) - 12 \left(\frac{x+2}{4}\right) = 12(1)$$ $$4(2x-1) - 3(x+2) = 12$$

$$8x - 4 - (3x + 6) = 12$$ $$8x - 4 - 3x - 6 = 12$$

$$5x = 12 + 10$$ $$5x = 22$$

$$x = \frac{22}{5}$$

The solution is $\boxed{x = \frac{22}{5} \text{ or } x = 4.4}$.

QUESTION 6

a) Calculate the length of an arc and the area of a sector of a circle which subtends an angle of 120° at the centre, if its radius is 9 cm.
Given: Angle $\theta = 120^\circ$, Radius $r = 9$ cm.
Formulas: Arc length $L = \frac{\theta}{360^\circ} \times 2\pi r$, Area of sector $A = \frac{\theta}{360^\circ} \times \pi r^2$.

$$L = \frac{120^\circ}{360^\circ} \times 2\pi (9)$$ $$L = \frac{1}{3} \times 18\pi$$ $$L = 6\pi \text{ cm}$$

$$A = \frac{120^\circ}{360^\circ} \times \pi (9)^2$$ $$A = \frac{1}{3} \times \pi (81)$$ $$A = 27\pi \text{ cm}^2$$

The length of the arc is $\boxed{6\pi \text{ cm}}$ (approximately $18.85 \text{ cm}$) and the area of the sector is $\boxed{27\pi \text{ cm}^2}$ (approximately $84.82 \text{ cm}^2$).