A metal bar in the shape of a cuboid with a cross-sectional area of 6 cm2 and a length of 15 cm is melted down and recast into two identical cones. Each cone has a height of 3 cm. Show that the radius

The volume of a cuboid is given by the cross-sectional area multiplied by its length.

$$V_{cuboid} = \text{Cross-sectional area} \times \text{Length}$$ $$V_{cuboid} = 6 \text{ cm}^2 \times 15 \text{ cm}$$ $$V_{cuboid} = 90 \text{ cm}^3$$

The volume of a single cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^2 h$.
Given that the height $h = 3 \text{ cm}$.

$$V_{cone} = \frac{1}{3} \pi r^2 (3)$$ $$V_{cone} = \pi r^2 \text{ cm}^3$$

Since two identical cones are formed, the total volume of the cones is:

$$V_{total\_cones} = 2 \times V_{cone}$$ $$V_{total\_cones} = 2 \pi r^2 \text{ cm}^3$$

Since the metal bar is melted to form the cones without any waste, the volume of the cuboid is equal to the total volume of the two cones.

$$V_{cuboid} = V_{total\_cones}$$ $$90 = 2 \pi r^2$$

Now, we solve for $r^2$:

$$r^2 = \frac{90}{2\pi}$$ $$r^2 = \frac{45}{\pi}$$

To find $r$, we take the square root of both sides:

$$r = \sqrt{\frac{45}{\pi}}$$

We can simplify $\sqrt{45}$ as $\sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$.

$$r = \frac{3\sqrt{5}}{\sqrt{\pi}}$$ $$r = 3 \sqrt{\frac{5}{\pi}}$$

This matches the given expression.

Given $\pi = 3.14$.

$$r = 3 \sqrt{\frac{5}{3.14}}$$

To use logarithmic tables, we find $\log r$:

$$\log r = \log \left( 3 \times \left(\frac{5}{3.14}\right)^{\frac{1}{2}} \right)$$

Using logarithm properties, $\log(ab) = \log a + \log b$ and $\log(x^n) = n \log x$:

$$\log r = \log 3 + \frac{1}{2} \log \left(\frac{5}{3.14}\right)$$

Using another property, $\log(\frac{a}{b}) = \log a - \log b$:

$$\log r = \log 3 + \frac{1}{2} (\log 5 - \log 3.14)$$

From logarithmic tables:
$\log 3 \approx 0.4771$
$\log 5 \approx 0.6990$
$\log 3.14 \approx 0.4969$

Substitute these values into the equation for $\log r$:

$$\log r = 0.4771 + \frac{1}{2} (0.6990 - 0.4969)$$ $$\log r = 0.4771 + \frac{1}{2} (0.2021)$$ $$\log r = 0.4771 + 0.10105$$ $$\log r = 0.57815$$

Now, find the antilog of 0.57815 to get the value of $r$.
Using antilog tables, antilog of $0.57815 \approx 3.7855$.
Rounding to the first decimal place:

$$r \approx 3.8 \text{ cm}$$

The value of $r$ to the first decimal place is $\boxed{\text{3.8 cm}}$.