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\(\begin{vmatrix} 2 & -5 & 3 \\ x & 1 & 4 \\ 0 & 3 & 2 \end{vmatrix} = 132\) \(\implies 2 \begin{vmatrix} 1 & 4 \\ 3 & 2 \end{vmatrix} - (-5) \begin{vmatrix} x & 4 \\ 0 & 2 \end{vmatrix} + 3 \begin{vmatrix} x & 1 \\ 0 & 3 \end{vmatrix} = 132\) \(2(2 - 12) + 5(2x) + 3(3x) = 132\) \(-20 + 10x + 9x = 132\) \(19x = 152\) \(x = 8\)

\(A = \begin{vmatrix} 3 & -2 \\ 1 & 6 \end{vmatrix}\) |A| = (3 x 6) - (-2 x 1) = 18 + 2 = 20. A\(^{-1}\) = \(\frac{1}{20} \begin{vmatrix} 6 & 2 \\ -1 & 3 \end{vmatrix}\) = \(\begin{vmatrix} \frac{6}{20} & \frac{2}{20} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\) = \(\begin{vmatrix} \frac{3}{10} & \frac{1}{10} \\ \frac{-1}{20} & \frac{3}{20} \end{vmatrix}\)

y = 8x\(^3\) - 3x\(^2\) + 7x - 1 \(\frac{\mathrm d^2 y}{\mathrm d x^2} = \frac{\mathrm d}{\mathrm d x} (\frac{\mathrm d y}{\mathrm d x})\) = \(\frac{\mathrm d}{\mathrm d x} (24x^2 - 6x + 7)\) \(\frac{\mathrm d^2 y}{\mathrm d x^2} = 48x - 6\)

No expanded working has been attached for this question yet. The verified answer is Option D.

= \(\sqrt{(5 - 2)^2 + (6 - 2)^2}\) = \(\sqrt{3^2 + 4^2}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 units

Equation: 4y = 7x + 3 \(\implies y = \frac{7}{4} x + \frac{3}{4}\) Slope = coefficient of x = \(\frac{7}{4}\) Slope of perpendicular line = \(\frac{-1}{\frac{7}{4}}\) = \(\frac{-4}{7}\) The perpendicular line passes (-3, 1) \(\therefore\) Using the equation of line \(y = mx + b\) m = slope and b = intercept. \(y = \frac{-4}{7} x + b\) To find the intercept, substitute y = 1 and x = -3 in the equation. \(1 = \frac{-4}{7} (-3) + b\) \(1 = \frac{12}{7} + b\) \(b = \frac{-5}{7}\) \(\therefore y = \frac{-4}{7} x - \frac{5}{7}\) \(7y + 4x + 5 = 0\)

Marks (x) 1 2 3 4 5 Frequency (f) 2y - 2 y - 1 3y - 4 3 - y 6 - 2y 3y + 2 fx 2y - 2 2y - 2 9y - 12 12 - 4y 30 - 10y 26 - y Mean = \(\frac{\sum fx}{\sum f}\) \(3 = \frac{26 - y}{3y + 2}\) \(3(3y + 2) = 26 - y\) \(9y + 6 = 26 - y\) \(9y + y = 26 - 6\) \(10y = 20 \implies y = 2\)

Length of arc = \(\frac{\theta}{360^\circ} \times 2 \pi r\) = \(\frac{57}{360} \times 2 \times \frac{22}{7} \times 7\) = 6.97 cm

|AC| = |DF| = 13 cm Using Pythagoras theorem, |AC|\(^2\) = |AB|\(^2\) + |BC|\(^2\) 13\(^2\) = 12\(^2\) + |BC|\(^2\) |BC|\(^2\) = 169 - 144 = 25 |BC| = \(\sqrt{25}\) = 5 cm Volume of triangular prism = \(\frac{1}{2} \times base \times length \times height\) = \(\frac{1}{2} \times 5 \times 12 \times 18\) = 540 cm\(^3\)

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\) 5\(^2\) = 4\(^2\) + adj\(^2\) adj\(^2\) = 25 - 16 = 9 adj = \(\sqrt{9}\) = 3 \(\tan x = \frac{4}{3}\) \(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\) \(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\) \(\csc x = \frac{1}{\sin x}\) = \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\) \(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\) \(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\) = \(\frac{25}{16} \div \frac{9}{16}\) = \(\frac{25}{9}\)

\(\frac{2\sin 30 + 5\tan 60}{\sin 60}\) \(\sin 30 = \frac{1}{2}; \tan 60 = \sqrt{3}; \sin 60 = \frac{\sqrt{3}}{2}\) \(\therefore \frac{2\sin 30 + 5\tan 60}{\sin 60} = \frac{2(\frac{1}{2}) + 5(\sqrt{3})}{\frac{\sqrt{3}}{2}}\)\) = \(\frac{1 + 5\sqrt{3}}{\frac{\sqrt{3}}{2}}\) = \(\frac{2(1 + 5\sqrt{3})}{\sqrt{3}}\) = \(\frac{2 + 10\sqrt{3}}{\sqrt{3}}\) Rationalizing, we get = \(\frac{2\sqrt{3} + 30}{3}\) = \(\frac{2}{3} \sqrt{3} + 10\)

Flowering plants having both the male and female flowers on the same flower are called Monoecious plants

Secreting irritating fluid is a behavioral adaptation because it involves an action or response performed by an organism—such as a toad or certain insects—specifically to deter or repel a predator once it is threatened. This behavior serves as a direct defense mechanism to increase the chances of survival during an encounter with a predator.

Aestivation is a period of no activity seen in some animals during period of heat or drought.

When an insect traps movement of leaves, it is a behavioural adaptation for obtaining food.