Calculate the yearly income of three workers (Aku, Brako, Dagadu), find the difference between Aku and Dagadu's income, and their total income. Illustrate info in a diagram.

To find the yearly income, multiply the monthly income by 12.

• Aku's monthly income = GH¢ 5,000.00 Yearly income for Aku = GH¢ $5,000.00 \times 12 = \text{GH¢ } 60,000.00$
• Brako's monthly income = GH¢ 6,500.00 Yearly income for Brako = GH¢ $6,500.00 \times 12 = \text{GH¢ } 78,000.00$
• Dagadu's monthly income = GH¢ 4,200.00 Yearly income for Dagadu = GH¢ $4,200.00 \times 12 = \text{GH¢ } 50,400.00$ The yearly incomes are: Aku: $\boxed{\text{GH¢ } 60,000.00}$, Brako: $\boxed{\text{GH¢ } 78,000.00}$, Dagadu: $\boxed{\text{GH¢ } 50,400.00}$.

Yearly income of Aku = GH¢ 60,000.00
Yearly income of Dagadu = GH¢ 50,400.00
Difference = GH¢ $60,000.00 - \text{GH¢ } 50,400.00 = \text{GH¢ } 9,600.00$
The yearly income difference is $\boxed{\text{GH¢ } 9,600.00}$.

Total yearly income = Aku's yearly income + Brako's yearly income + Dagadu's yearly income
Total yearly income = GH¢ $60,000.00 + \text{GH¢ } 78,000.00 + \text{GH¢ } 50,400.00 = \text{GH¢ } 188,400.00$
The total yearly income is $\boxed{\text{GH¢ } 188,400.00}$.

The situation forms a right-angled triangle where:

• The ladder is the hypotenuse.
• The distance from the foot of the ladder to the base of the building is the adjacent side (5 m).
• The height of the window is the opposite side.
• The angle between the ladder and the horizontal ground is $60^\circ$.

Where L is the length of the ladder and H is the height of the window.

We have the adjacent side (5 m) and the angle ($60^\circ$), and we need to find the hypotenuse (length of the ladder). We use the cosine function:

Given $\cos(60^\circ) = \frac{1}{2}$:

The length of the ladder is $\boxed{10.0 \text{ m}}$ (to one decimal place).

This is the height of the window (opposite side). We have the adjacent side (5 m) and the angle ($60^\circ$). We use the tangent function:

Given $\tan(60^\circ) = 1.732$:

To one decimal place, the distance is $\boxed{8.7 \text{ m}}$.

First, simplify each surd:

Substitute these into the expression:

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $3\sqrt{2} + 2\sqrt{3}$:

Numerator:

Denominator (difference of squares, $(A-B)(A+B) = A^2 - B^2$):

So the expression becomes:

The simplified expression is $\boxed{6 + 2\sqrt{6}}$.

First, distribute the $\frac{1}{2}$ on the left side:

To eliminate the fractions, find the least common multiple (LCM) of the denominators $2, 3, 10$, which is 30. Multiply every term by 30:

Subtract $10x$ from both sides:

Subtract 15 from both sides:

Divide by 20:

Simplify the fraction:

The solution is $\boxed{x \ge \frac{3}{5}}$.

The athlete runs 4 times around a circular track with radius $r = 70$ m.
The distance for one lap is the circumference of the circle, $C = 2\pi r$.

The total distance covered for 4 laps is:

The total distance covered is $\boxed{1760 \text{ m}}$.

First, list all the shoe sizes given:
6, 4, 7, 5, 5, 6
4, 5, 5, 4, 6, 7
5, 6, 4, 6, 7, 5
6, 4, 7, 5, 4, 6
5, 5, 4, 6, 7, 5

Count the frequency of each unique shoe size:

• Size 4: 7 times
• Size 5: 11 times
• Size 6: 9 times
• Size 7: 6 times

The size that should be purchased in large quantities is the one with the highest frequency (the mode).
From the table, shoe size 5 has the highest frequency of 11.

• 6(b)(i) Which size should be purchased in large quantities? $\boxed{\text{Size 5}}$
• 6(b)(ii) Give reason for the answer in (b)(i): $\boxed{\text{It is the most frequent shoe size among the learners.}}$

The size that will be purchased in less quantities is the one with the lowest frequency.
From the table, shoe size 7 has the lowest frequency of 6.

• 6(b)(iii) Which size will be purchased in less quantities? $\boxed{\text{Size 7}}$
• 6(b)(iv) Give reason for the answer in (b)(iii): $\boxed{\text{It is the least frequent shoe size among the