In a rectangle ABCD, AB=3cm, BC=4cm, and AD=13cm. Calculate: a) CD b) tan (angle ADC) c) AC. The table shows the cost of 5000 shares of company X. The shares are traded on the Nairobi Stock Exchange.

Since $\angle ABC = 90^\circ$, we have:

$$AC^2 = AB^2 + BC^2$$ $$AC^2 = (3 \text{ cm})^2 + (4 \text{ cm})^2$$ $$AC^2 = 9 \text{ cm}^2 + 16 \text{ cm}^2$$ $$AC^2 = 25 \text{ cm}^2$$ $$AC = \sqrt{25 \text{ cm}^2}$$ $$AC = 5 \text{ cm}$$

Since $\angle ACD = 90^\circ$, we have:

$$AD^2 = AC^2 + CD^2$$ $$(13 \text{ cm})^2 = (5 \text{ cm})^2 + CD^2$$ $$169 \text{ cm}^2 = 25 \text{ cm}^2 + CD^2$$ $$CD^2 = 169 \text{ cm}^2 - 25 \text{ cm}^2$$ $$CD^2 = 144 \text{ cm}^2$$ $$CD = \sqrt{144 \text{ cm}^2}$$ $$CD = 12 \text{ cm}$$

In $\triangle ACD$, $\angle ACD = 90^\circ$.
The side opposite to $\angle ADC$ is $AC$.
The side adjacent to $\angle ADC$ is $CD$.

$$\tan(\angle ADC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{CD}$$ $$\tan(\angle ADC) = \frac{5 \text{ cm}}{12 \text{ cm}}$$ $$\tan(\angle ADC) = \frac{5}{12}$$

Depreciation = 15% of GH¢ 6,500.00

$$\text{Depreciation} = 0.15 \times \text{GH¢ } 6,500.00$$ $$\text{Depreciation} = \text{GH¢ } 975.00$$

Value after one year = Original value - Depreciation

$$\text{Value after one year} = \text{GH¢ } 6,500.00 - \text{GH¢ } 975.00$$ $$\text{Value after one year} = \text{GH¢ } 5,525.00$$

The value of the car after one year is $\boxed{\text{GH¢ } 5,525.00}$.

Given $a = \frac{1}{3}$ and $b = -\frac{1}{2}$.

$$2a - 3b = 2\left(\frac{1}{3}\right) - 3\left(-\frac{1}{2}\right)$$

$$2a - 3b = \frac{2}{3} - \left(-\frac{3}{2}\right)$$ $$2a - 3b = \frac{2}{3} + \frac{3}{2}$$

The least common multiple of 3 and 2 is 6.

$$2a - 3b = \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3}$$ $$2a - 3b = \frac{4}{6} + \frac{9}{6}$$ $$2a - 3b = \frac{4+9}{6}$$ $$2a - 3b = \frac{13}{6}$$

The value of the expression is $\boxed{\frac{13}{6}}$.