Find the volume of a prism 20cm long and whose cross-section is an equilateral triangle of side 9cm.

The formula for the area of an equilateral triangle with side length $a$ is $A = \frac{\sqrt{3}}{4}a^2$.
Given $a = 9$ cm.

$$A = \frac{\sqrt{3}}{4}(9 \text{ cm})^2$$ $$A = \frac{\sqrt{3}}{4}(81 \text{ cm}^2)$$ $$A = \frac{81\sqrt{3}}{4} \text{ cm}^2$$

The formula for the volume of a prism is $V = \text{Area of cross-section} \times \text{length}$.
Given length $= 20$ cm.

$$V = \frac{81\sqrt{3}}{4} \text{ cm}^2 \times 20 \text{ cm}$$ $$V = 81\sqrt{3} \times 5 \text{ cm}^3$$ $$V = 405\sqrt{3} \text{ cm}^3$$

Using $\sqrt{3} \approx 1.732$:

$$V \approx 405 \times 1.732 \text{ cm}^3$$ $$V \approx 701.46 \text{ cm}^3$$

The volume of the prism is approximately $\boxed{701.46 \text{ cm}^3}$ (or exactly $\boxed{405\sqrt{3} \text{ cm}^3}$).

12. The equation of a line L1 is $4y - 3x + 8 = 0$. Find the equation of line L2 which is perpendicular to L1 and passes through (2, 7).

Rearrange the equation $4y - 3x + 8 = 0$ into the slope-intercept form $y = mx + c$.

$$4y = 3x - 8$$ $$y = \frac{3}{4}x - 2$$

The slope of L1 is $m_1 = \frac{3}{4}$.

Since L2 is perpendicular to L1, its slope $m_2$ is the negative reciprocal of $m_1$.

$$m_2 = -\frac{1}{m_1}$$ $$m_2 = -\frac{1}{3/4}$$ $$m_2 = -\frac{4}{3}$$

Use the point-slope form $y - y_1 = m(x - x_1)$ with $m_2 = -\frac{4}{3}$ and the point $(x_1, y_1) = (2, 7)$.

$$y - 7 = -\frac{4}{3}(x - 2)$$

Multiply both sides by 3 to eliminate the fraction:

$$3(y - 7) = -4(x - 2)$$ $$3y - 21 = -4x + 8$$

Rearrange into the general form $Ax + By + C = 0$:

$$4x + 3y - 21 - 8 = 0$$ $$4x + 3y - 29 = 0$$

The equation of line L2 is $\boxed{4x + 3y - 29 = 0}$.

13. Solve for x in $3^{2x+1} + 5(3^x) - 2 = 0$.

$$3^{2x+1} = 3^{2x} \cdot 3^1 = (3^x)^2 \cdot 3$$

Substitute this back into the equation:

$$3(3^x)^2 + 5(3^x) - 2 = 0$$

Let $y = 3^x$. Since $3^x$ is always positive, $y > 0$.

$$3y^2 + 5y - 2 = 0$$

Factor the quadratic equation:

$$(3y - 1)(y + 2) = 0$$

This gives two possible solutions for $y$:

$$3y - 1 = 0 \implies 3y = 1 \implies y = \frac{1}{3}$$ $$y + 2 = 0 \implies y = -2$$

Since $y$ must be positive, we discard $y = -2$.
Using $y = \frac{1}{3}$:

$$3^x = \frac{1}{3}$$ $$3^x = 3^{-1}$$

Therefore,

$$x = -1$$

The solution for $x$ is $\boxed{-1}$.

14. Express $\frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{4-\sqrt{3}}{2-\sqrt{3}}$ in the form $a+b\sqrt{c}$.

The common denominator is $(2+\sqrt{3})(2-\sqrt{3})$.
Using the difference of squares formula $(A+B)(A-B) = A^2 - B^2$:

$$(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$$

$$\frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{4-\sqrt{3}}{2-\sqrt{3}} = \frac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} + \frac{(4-\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$$

Since the common denominator is 1, the expression simplifies to the sum of the new numerators:

$$= (2-\sqrt{3})^2 + (4-\sqrt{3})(2+\sqrt{3})$$

For the first term, $(2-\sqrt{3})^2$:

$$(2-\sqrt{3})^2 = 2^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$$

For the second term, $(4-\sqrt{3})(2+\sqrt{3})$:

$$(4-\sqrt{3})(2+\sqrt{3}) = 4(2) + 4(\sqrt{3}) - \sqrt{3}(2) - (\sqrt{3})(\sqrt{3})$$ $$= 8 + 4\sqrt{3} - 2\sqrt{3} - 3$$ $$= 5 + 2\sqrt{3}$$

$$(7 - 4\sqrt{3}) + (5 + 2\sqrt{3})$$ $$= 7 + 5 - 4\sqrt{3} + 2\sqrt{3}$$ $$= 12 - 2\sqrt{3}$$

The expression in the form $a+b\sqrt{c}$ is $\boxed{12 - 2\sqrt{3}}$.