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Vol. of cylinder = \(\pi r^2h\) = 1200cm2 Area of base = \(\pi(r^2)\) = 150cm2 h = \(\frac{\text{volume of cylinder}}{\text{base area}} = \frac{1200}{150}\) = 8.00cm

(a) < ZXY = 225^\circ - 135^\circ = 90^\circ. \(\therefore /ZY/ = 80^{2} + 60^{2}\) = \(6400 + 3600 = 10000\) \(/ZY/ = \sqrt{10000}\) = 100 km. (b) From \(\Delta\) XYZ, \(\tan \theta = \frac{opp}{adj}\) \(\tan \theta = \frac{80}{60} = 1.333\) \(\theta = \tan^{-1} (1.333)\) \(\theta = 53.13^\circ\) Bearing of Z from Y = \(360^\circ - (45^\circ + 53.13^\circ)\) = \(360^\circ - 98.13^\circ\) = \(261.87^\circ \approxeq 262^\circ\) (3 significant figures)

If the linear expansivity = ∝ => area expansivity \beta = 2∝. And cubic expansivity δ = 3∝. => 6.3 x 10\(^{-6}\) = 3∝ ∴ ∝ = \(\frac{6.3 \times 10^{-11}}{3}\) = 2.1 x 10\(^{-6}\) ∴ Area = 2∝ = 2 x 2.1x10\(^{-6}\) = 4.2 x 10\(^{-6}\)K\(^{-1}\)

Sodium chloride, which is obtained by neutralization of hydrochloric acid and sodium hydroxide, is a neutral salt. A neutral salt is formed when a strong acid reacts with a strong base. This is because the strong acid is completely neutralized by the strong base. When a strong acid and a strong base are combined in the right amounts, the result is a neutral solution with a pH of 7. The acid and base have neutralized each other, and the acidic and basic properties are no longer present. The salt formed from a strong acid and a strong base is neutral because the anion and cation are both weak. They are stable in water and do not react with water to form the original acid or base.

Elements in period 3 have 3 shells in their atomic structure. These include elements from sodium to argon.

This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils. Ethene reacts with hydrogen in the presence of a finely divided nickel catalyst at a temperature of about 150^\circC. Ethane is produced.

\((111_{two})^{2}=(111_{two})(111_{two})\\ =1 \times 2^{2}+1\times2^{1}+1\times 2^{0}=4+2+1=7_{_{ten}}\\ (111_{two})^{2}=7\times 7=49\\ \begin{matrix} 2 & 49\\ 2 & 24\hspace{1mm}R\hspace{1mm}1\\ 2 & 12\hspace{1mm}R\hspace{1mm}0\\ 2 & 6\hspace{1mm}R\hspace{1mm}0\\ 2 & 3\hspace{1mm}R\hspace{1mm}0\\ 2 & 1\hspace{1mm}R\hspace{1mm}1\\ & 0\hspace{1mm}R\hspace{1mm}1\hspace{1mm}\uparrow \end{matrix} \\ =110001_{2}\)

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

\((0.3\times 10^{5})\div (0.4\times 10^{7})\\ \frac{3}{10}\times 10^{5}\div \frac{4}{10}\times 10^{7}=3\times 10^{4}\div 4\times 10^{6}\\ \frac{3\times 10^{4}}{4\times 10^{7}}=\frac{3}{4}\times 10^{-2}=0.75\times 10^{-2}\\ 7.5\times 10^{-3}\)

\(p\propto \frac{1}{q} = q \propto \frac{1}{p}\)

\(3\sqrt{45}-12\sqrt{5}+16\sqrt{20}\) 3\(\sqrt{5\times 9}-12\sqrt{5}+16\sqrt{4\times 5}\) 3\(\times 3\sqrt{5}-12\sqrt{5}+16\times 2\sqrt{5}\) 9\(\sqrt{5}-12\sqrt{5}+32\sqrt{5}\) 29\(\sqrt{5}\)

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

\(\frac{3x^{3}}{(3x)^{3}}\\ \frac{3x^{3}}{3x\times 3x\times 3x}=\frac{3\times x\times x\times x}{3\times 3\times 3\times x\times x\times x}=\frac{1}{9}\)

\(Abu = 2x\\ Kayode = x\\ Uche = \frac{x}{2}\\ 2x+x+\frac{x}{2}=140,000\\ \frac{7x}{2}=140,000\\ 7x = 280,000\\ x=\frac{280,000}{7}\\ x=40,000\)Kayode receives N40,000

If I am x years old, then my brother will be (x + 3)years old then last year my brother was (x + 3) - 1 = (x + 2)years old