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Radius of curvature(r)=12cm focal length(F) = r \div 2 → 12 \div 2 = 6cm therefore F = 6cm magnification(M) = v \div u 1.5 = v \div u :v = 1.5u Using mirror formula 1/f =1/v + 1/u \(\frac{1}{6}\) =\(\frac{1}{1.5u}\) + \(\frac{1}{u}\) \(\frac{1}{6}\) = \(\frac{1 + 1.5}{1.5u}\) \(\frac{1}{6}\) = \(\frac{2.5}{1.5u}\) 1 * 1.5u = 2.5 * 6 u = \(\frac{15}{1.5}\) = 10 therefore object distance =10cm.

the power of a lens is defined as the reciprocal of the focal length ,expressed in metre.Mathematically, the power of a lens in dioptres is given by:(1)/f(in cm). thus for lens of power of 4.0 dioptres we have that: (1)/(f) = +4.0 f= (1)/(4.0) = 0.25cm

The quality or timbre of sound is that characteristic which enables us to distinguish one sound from another having the same pitch and loudness.

You will find convex lenses in telescopes, microscopes, magnifying glasses, cameras, slide and movie projectors, and other instruments that use light to create an image.

Fractional distillation of liquefied air, with Nitrogen distilling as a vapor at a lower boiling point of -196^\circC while O2 is left as a liquid. In the industrial preparation of nitrogen and oxygen from air, the process of fractional distillation is used. This process separates the components of air based on their boiling points. The boiling points of the main components of air are: - Nitrogen (N\(_2\)): -195.8^\circC - Oxygen (O\(_2\)): -182.96^\circC By cooling air to extremely low temperatures, the components can be separated based on their boiling points, allowing for the production of high-purity nitrogen and oxygen.

In option A, the oxidation number of Nitrogen increased from +2 to +4. In Option B, the oxidation number of Nitrogen decreased from +2 to +1 (Bearing in mind that the O.N of chlorine in hypochlorate, OCl\(^-\) is +1) In Option C, the oxidation number of Nitrogen decreased from +2 to +1 In Option D, the oxidation number of Nitrogen decreased from +2 to +1 (Bearing in mind that the O.N of bromine in hypobromate OBr\(^-\) is +1) Therefore, the answer is option A

In the electrolysis of a concentrated solution containing H\(^+\), Cu\(^{2+}\), OH\(^-\) and Cl\(^-\) ions using platinum electrodes, the Cu\(^{2+}\) ions will be discharged at the cathode. During electrolysis, ions migrate towards the oppositely charged electrode. Positive ions (cations) like H\(^+\) and Cu\(^{2+}\) will be attracted to the negative electrode (cathode). At the cathode, the ion that is easier to reduce (requires less energy to gain electrons) will be discharged first. In this case, Cu\(^{2+}\) ions are more easily reduced to Cu(s) than H\(^+\) ions are to H\(_2\)(g). The reaction at the cathode will be the reduction of Cu\(^{2+}\) ions: Cu\(^{2+}\)(aq) + 2e\(^-\) → Cu(s). Anode reaction: The reaction at the anode will be the oxidation of water: 2H\(_2\)O(l) → O\(_2\)(g) + 4H\(^+\)(aq) + 4e-

It also reduces potassium permanganate and dichromate to Mn2+ and Cr3+ respectively.

- Ethene is a plant hormone which causes fruit to ripen. - Ethene is used routinely within the food industry to provide controlled ripening during storage and transport or when fruit is displayed in shops, in sealed packages.

\(\frac{5}{\sqrt{3}}-\frac{3}{\sqrt{2}}=\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}}\\ =\frac{5\sqrt{2}-3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}Rationalize\\ \frac{\sqrt{6}(5\sqrt{2}-3\sqrt{3})}{6}\\ \frac{5\sqrt{12}-3\sqrt{18}}{6}=\frac{10\sqrt{3}-9\sqrt{2}}{6}\\ \frac{1}{6}(10\sqrt{3}-9\sqrt{2})\)

\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div\frac{33}{2}\\ =\frac{15}{2}-\left(\frac{5}{2}+\frac{3}{1}\right)\times\frac{2}{33}\\ =\frac{15}{2}-\left(\frac{5+6}{2}\right)\times \frac{2}{33}=\frac{15}{2}-\frac{11}{2}\times \frac{2}{33}=\frac{15}{2}-\frac{1}{3}\\ =\frac{45-2}{6}=\frac{43}{6}\) = 7.1666 ≈ 7

\(log_{10}18 - log_{10}2.88+log_{10}16\\ =log_{10}18 - log_{10}\left(\frac{288}{100}\right)+log_{10}16 = log_{10}\left(\frac{18\times 16}{1}\times \frac{100}{288}\right)\\ =log_{10}\left(\frac{288\times 100}{288}\right)=log_{10}100=log_{10}10^2=2log_{10}10=2\)

x2 (sum of roots)x + (product of roots) = 0 Sum of roots \(2+-3\frac{1}{2} = -1\frac{1}{2} = -\frac{1}{2}\) Product of roots \(= 2 \times - 3\frac{1}{2}\) = -7 x\(^2\) - \(\frac{-3}{2}\)x + (-7) = 0 ⇒ 2x\(^2\) + 3x - 14 = 0

The cost price of the whole mangoes = N5x The sold amount of the mangoes = 3x * 220 = N6.60x ( 660x converted to naira by dividing by hundred.) The gain made on mangoes = N6.60x - N5x = N8.00 => N1.60x = N8 => \(x=\frac{8}{1.60}=\frac{1}{0.2}=\frac{10}{2}=5\)

Using Pythagoras theorem \(|PR|^2=|PO|^2+|OR|^2\\ |PR|^2=5^2+12^2\\ |PR|=\sqrt{25+144}=\sqrt{169}=13km\)