A solid metal sphere of radius 7 cm is melted and recast into smaller spheres of radius 1.4 cm. a) Find the number of smaller spheres obtained. b) If the metal used costs Ksh 450 per cubic centimetre, find the total cost of the metal in the original sphere. (Take pi = 22/7)

The radius of the large sphere, $R = 7$ cm.
The volume of a sphere is given by the formula $V = \frac{4}{3}\pi R^3$.

$$V_L = \frac{4}{3}\pi (7)^3$$

The radius of a small sphere, $r = 1.4$ cm.

$$V_S = \frac{4}{3}\pi (1.4)^3$$

The number of smaller spheres ($N$) is the total volume of the large sphere divided by the volume of one small sphere.

$$N = \frac{V_L}{V_S} = \frac{\frac{4}{3}\pi (7)^3}{\frac{4}{3}\pi (1.4)^3}$$

Cancel out $\frac{4}{3}\pi$:

$$N = \frac{(7)^3}{(1.4)^3} = \left(\frac{7}{1.4}\right)^3$$ $$N = (5)^3 = 125$$

The number of smaller spheres obtained is $\boxed{125}$.

b) If the metal used costs Ksh 450 per cubic centimetre, find the total cost of the metal in the original sphere.

The radius of the original sphere, $R = 7$ cm.

$$V_L = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (7)^3$$ $$V_L = \frac{4}{3} \times \frac{22}{7} \times 343$$ $$V_L = \frac{4 \times 22 \times 49}{3} = \frac{4312}{3} \text{ cm}^3$$

The cost per cubic centimetre is Ksh 450.
Total cost = Volume of original sphere $\times$ Cost per cubic centimetre.

$$\text{Total Cost} = \frac{4312}{3} \times 450$$ $$\text{Total Cost} = 4312 \times 150 = 646800$$

The total cost of the metal in the original sphere is $\boxed{\text{Ksh 646,800}}$.

5. a) Form an equation in $x$ and solve it.

The exterior angle of a regular polygon is $x^\circ$.
The interior angle of the regular polygon is $5x^\circ$.
The sum of an interior angle and its corresponding exterior angle is always $180^\circ$.

$$x + 5x = 180$$

Combine like terms:

$$6x = 180$$

Divide by 6:

$$x = \frac{180}{6}$$ $$x = 30$$

The equation is $x + 5x = 180$ and the value of $x$ is $\boxed{30}$.