1. The problem asks to find $x$ given the determinant equation: $$\begin{vmatrix} 2 & 6 \\ 3 & x \end{vmatrix} = 0.$$ 2. Calculate the determinant using the formula for a 2x2 matrix: $$2 \cdot x - 6 \cdot 3 = 0.$$ 3. Simplify: $$2x - 18 = 0.$$ 4. Solve for $x$: $$2x = 18 \implies x = \frac{18}{2} = 9.$$ --- 1. Find value of $$\left(\frac{25}{16}\right)^{-\frac{1}{2}}.$$ 2. Recall that $$a^{-b} = \frac{1}{a^b}$$ and $$a^{\frac{1}{2}} = \sqrt{a}.$$ 3. So, $$\left(\frac{25}{16}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{\frac{25}{16}}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}.$$ --- 1. The logarithm of unity to any base is asked: $$\log_a 1.$$ 2. Recall: The logarithm of 1 is always 0 because $$a^0 = 1$$ for any valid base $a$. --- 1. Simplify $$\frac{a^2 - b^2}{a + b}$$. 2. Note that $$a^2 - b^2 = (a - b)(a + b)$$ (difference of squares). 3. Substitute: $$\frac{(a - b)(a + b)}{a + b} = a - b$$ assuming $$a + b \neq 0.$$ --- 1. Complete the square for the quadratic expression $$x^2 + 4x + m$$. 2. Formula for a complete square trinomial: $$x^2 + 2bx + b^2 = (x + b)^2.$$ 3. Here, coefficient of $x$ is 4, so $$2b = 4 \implies b = 2.$$ 4. Then, $$m = b^2 = 2^2 = 4.$$ --- 1. Simplify $$\frac{a}{9a^2 - b^2} + \frac{1}{3a - b}.$$ 2. Factor denominator: $$9a^2 - b^2 = (3a - b)(3a + b).$$ 3. Rewrite first fraction: $$\frac{a}{(3a - b)(3a + b)}.$$ 4. Make common denominator: $$\frac{a}{(3a - b)(3a + b)} + \frac{1}{3a - b} = \frac{a}{(3a - b)(3a + b)} + \frac{3a + b}{(3a - b)(3a + b)}.$$ 5. Add numerators: $$\frac{a + (3a + b)}{(3a - b)(3a + b)} = \frac{4a + b}{9a^2 - b^2}.$$ --- 1. Identify if $$x = \frac{3}{2}$$ satisfies $$-2 < x < \frac{3}{2}.$$ 2. Since $$x = \frac{3}{2}$$ equals the upper bound but the inequality is strict ($<$), $x=\frac{3}{2}$ is not a solution. 3. So correct solution from choices given is $$\frac{3}{2}$$ is not included, but the question wants an option matching the solution inside the interval; if question means value lying between, verify options. --- 1. If $$(x, 0) = (0, y)$$, equate coordinates: $$x = 0$$ and $$0 = y.$$ 2. Hence, $$x = 0$$ and $$y = 0$$ so the point is $$(0, 0).$$ --- 1. Find distance between points $$(0, 0)$$ and $$(1, 1)$$. 2. Use distance formula: $$\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1 + 1} = \sqrt{2}.$$ --- 1. Identify symbol for (1 - 1) correspondence between two triangles; it is congruence symbol: $$\equiv.$$ --- 1. Medians of a triangle are known to be: Concurrent. --- 1. A point equidistant from endpoints of a line segment lies on the perpendicular bisector of that segment. --- 1. Ratio is a dimensionless quantity, so it has no unit. --- 1. A triangle with two congruent sides is called an isosceles triangle. --- 1. Area of a rectangle is base times height: $$6 \mathrm{cm} \times 3 \mathrm{cm} = 18 \mathrm{cm}^2.$$