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3, - 6, +12, -24 a = 3, r = -2 \(Sn = \frac{a(1-r^n)}{1-r}\\ ∴1-2^8 = \frac{3(1-(-2^{n}))}{1-(-2)}\\ 1-2^8 = \frac{3(1-(-2^{n}))}{3}\) 1 - 2\(^8\) = 1 - (-2)\(^n\) Since 1 - 2\(^8\) = 1 - (-2)\(^n\)

2p x 1 + 8 x 2 = 24 \(\to\) 4p = 24 - 8 = 16, p = 4 3 x 1 + -5q x 2 = -17 \(\to\) -10q = -17 - 3 -10q = -20 q = 2

2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4 2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1 2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1 \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

V = \sqrt2GM/R. =\sqrt[2 x 6.67 x 10\(^{-11}\) x 6.0 x 10\(^24\)] / 6.4 x 10\(^6\) = \sqrt12.506 x 10\(^7\) = 11,183.13 OR V =\sqrt2gR V=\sqrt[2 * 10 * 6.4 x 10\(^6\)] V = \sqrt128,000,000 11,313.708

16 days = 4 half-lives. The fraction left 16 days = after 4 half live, = (1/2)4 = (1)/16 Fraction that decayed = 1- \(\frac{1}{16}\) \(\frac{15}{16}\)

The correct answer is B. 2.0M of HCl. Here's why: - HCl (hydrochloric acid) is a strong acid, which means it completely dissociates in water to produce H\(^+\) ions and Cl\(^-\) ions. - The concentration of HCl is directly related to its ability to conduct electricity. A higher concentration of HCl means more ions are available to carry electrical charge. - Between options A and B, the 2.0M HCl solution has a higher concentration of ions, making it a better conductor of electricity. Ethanoic acid (CH\(_3\)COOH), on the other hand, is a weak acid, which means it only partially dissociates in water. As a result, it produces fewer ions and is a poorer conductor of electricity compared to HCl. So, even though option D has a higher concentration of ethanoic acid, it still won't conduct as much electricity as the 2.0M HCl solution.

During the electrolysis of an aqueous copper sulfate (CuSO\(_4\)) solution using platinum electrodes, copper ions (Cu\(^{2+}\)) are reduced to solid copper (Cu) at the cathode, and water (H\(_2\)O) is oxidized to oxygen gas (O\(_2\)) and hydrogen ions (H\(^+\)) at the anode. Here's a more detailed explanation: Cathode (Reduction): Copper ions (Cu\(^{2+}\)) from the solution migrate towards the cathode and gain two electrons (2e-) to form solid copper, which deposits on the electrode. Reaction: Cu\(^{2+}\)(aq) + 2e- → Cu(s) Anode (Oxidation): Water molecules (H\(_2\)O) are oxidized at the anode, losing electrons and forming oxygen gas (O\(_2\)) and hydrogen ions (H+). Reaction: 4OH\(^-\) + 4e\(^-\)(l) → 2H\(_2\)O(g) + O\(_2\)

For a reaction to be spontaneous, ∆G = - ve; ∆H = - ve; ∆S = + ve Where ∆G = ∆H - T∆S Given data; ∆G = ? ∆H = - 72KJ/mol, T = (25+273)K = 298K ∆S = -106J/mol or - 0.106KJ/mol/K : ∆G = -72 -(298 * - 0.106) = - 72 + 31.588 ∆G = - 40.412KJ/mol One of the condition for spontaneity of a reaction is ∆G to be negative.

Carbon exists in the combined state as petroleum, wood and natural gases.

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0. First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1. Then, a\(_0\) which is the coefficient void of x = 6. The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6. The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\). = \(\pm (1, 2, 3, 6)\). Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0. Therefore x = 1 is a root of the equation. Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6. x\(^2\) - x - 6 = (x - 3)(x + 2). x = -2, 3. \(\therefore\) The roots of the equation = 1, -2 and 3.

English = 52 students Health education = 36 students Total = 88 students = \(\frac{88}{160} \times 100%\) = 55%

(0.13)\(^3\) = 0.13 x 0.13 x 0.13 = 0.002197 = 0.00220 (3 s.f)

11011\(_2\) - 1101\(_2\) = 1110\(_2\)

\(\frac{25 \frac{2}{3} \div 25 \frac{1}{6}}{( \frac{1}{5})^{-\frac{7}{6}} \times ( \frac{1}{5})^{\frac{1}{6}}}\) = \(\frac{25^{4 - \frac{1}{6}}}{(\frac{1}{5})^{-7 + \frac{1}{6}}}\) = \(\frac{25^{\frac{1}{2}}}{(\frac{1}{5})^{-1}}\) = \(\frac{(5^2)^{\frac{1}{2}}}{(5^{-1})^{-1}}\) = \(\frac{5}{5}\) = 1

\(\frac{x - 4}{4} - \frac{x - 3}{6}\) = \(\frac{3(x - 4) - 2(x - 3)}{12}\) = \(\frac{3x -12 - 2x + 6}{12}\) = \(\frac{3x - 2x - 12 + 6}{12}\) = \(\frac{x - 6}{12}\)