Answer five questions only in this part. 6. (a) Express the sum of 112012three and 21121three in base six. (b) If (8x16x/28x44) = 100, find the value of x.

The volume of water is the internal volume of the cylinder.

$$V_{water} = \pi r_i^2 h$$

Substitute the values:

$$V_{water} = 3.14 \times (5.0 \text{ cm})^2 \times (350 \text{ cm})$$ $$V_{water} = 3.14 \times 25 \text{ cm}^2 \times 350 \text{ cm}$$ $$V_{water} = 3.14 \times 8750 \text{ cm}^3$$ $$V_{water} = 27475 \text{ cm}^3$$

Convert the volume from cm³ to litres ($1 \text{ L} = 1000 \text{ cm}^3$):

$$V_{water} = \frac{27475}{1000} \text{ L}$$ $$V_{water} = \boxed{\text{27.475 L}}$$

The volume of metal is the difference between the external volume and the internal volume of the pipe.

$$V_{metal} = \pi r_e^2 h - \pi r_i^2 h = \pi h (r_e^2 - r_i^2)$$

Substitute the values:

$$V_{metal} = 3.14 \times 350 \text{ cm} \times ((7.5 \text{ cm})^2 - (5.0 \text{ cm})^2)$$ $$V_{metal} = 3.14 \times 350 \text{ cm} \times (56.25 \text{ cm}^2 - 25 \text{ cm}^2)$$ $$V_{metal} = 3.14 \times 350 \text{ cm} \times 31.25 \text{ cm}^2$$ $$V_{metal} = 3.14 \times 10937.5 \text{ cm}^3$$ $$V_{metal} = \boxed{\text{34361.25 cm}^3}$$

8. (b) A sector of a circle with radius 12 cm and angle 240° is used to form a cone. Calculate the:

• Given:
  • Radius of sector ($R$) = $12 \text{ cm}$
  • Angle of sector ($\theta$) = $240^\circ$

When a sector is formed into a cone, the radius of the sector ($R$) becomes the slant height ($l$) of the cone, and the arc length of the sector becomes the circumference of the base of the cone.
Arc length of sector = $\frac{\theta}{360^\circ} \times 2\pi R$
Circumference of cone base = $2\pi r$, where $r$ is the radius of the cone.
Equate the two:

$$2\pi r = \frac{240^\circ}{360^\circ} \times 2\pi (12 \text{ cm})$$ $$2\pi r = \frac{2}{3} \times 24\pi \text{ cm}$$ $$2\pi r = 16\pi \text{ cm}$$

Divide both sides by $2\pi$:

$$r = \frac{16\pi}{2\pi} \text{ cm}$$ $$r = \boxed{\text{8 cm}}$$

Let the vertical angle of the cone be $\alpha$. We have the radius of the cone ($r = 8 \text{ cm}$) and the slant height ($l = 12 \text{ cm}$).
Consider the right-angled triangle formed by the cone's height, radius, and slant height. The angle at the apex of this triangle is $\frac{\alpha}{2}$.

$$\sin\left(\frac{\alpha}{2}\right) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{r}{l}$$

Substitute the values:

$$\sin\left(\frac{\alpha}{2}\right) = \frac{8 \text{ cm}}{12 \text{ cm}}$$ $$\sin\left(\frac{\alpha}{2}\right) = \frac{2}{3}$$

Now, find $\frac{\alpha}{2}$:

$$\frac{\alpha}{2} = \arcsin\left(\frac{2}{3}\right)$$ $$\frac{\alpha}{2} \approx 41.8103^\circ$$

Multiply by 2 to find $\alpha$:

$$\alpha \approx 2 \times 41.8103^\circ$$ $$\alpha \approx 83.6206^\circ$$

Correct to two decimal places:

$$\alpha = \boxed{\text{83.62}^\circ}$$