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. (a) IV, V and VIl (b) VI

. (i) Axile placentation (ii) Marginal placentation

(a) Monometer (b) It is used to measure chloroplast which is the main centre for photosynthesis (c) Mercury (d) From 1

(a) (i) It posses chloroplast which is the main centre for photosynthesis (ii) it has air spaces in which carbon (iv) oxide accumulates and is used to carry out photosynthesis (b) in order to remove the chlorophyll

(i) Mechanical support (ii) rigidity to plant (iii) flexibility to plant

Average speed = \(\frac{total Distance}{Total Time}\) from Calabar to Enugu in time t1, hence t1 = \(\frac{P}{U}\) also from Enugu to Benin t2 \(\frac{q}{w}\) Av. speed = \(\frac{p + q}{t_1 + t_2} = p + q * \frac{p}{u} + \frac{q}{w}\) = p + q x \(\frac{uw}{pw + qu}\) = \(\frac{uw(p + q)}{pw + qu}\)

P = \(\pi\)r + \(\pi\)r + R = \(\frac{22}{7}\) x 7 + \(\frac{22}{7}\) x \(\frac{7}{2}\) + 7 P = 22 + 11 + 7 = 40cm

\(\frac{27}{5^1}\)(3)-3 x \(\frac{(1)^{-1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\) = \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\) = \(\frac{1}{25}\)

I = \(\frac{PRT}{100}\) = \(\frac{10 \times 2 \times 4}{100}\) = \(\frac{4}{5}\) = 0.8 Total amount = N10.80 He pays N8.00 Remainder = 10.80 - 8.00 = N2.80

W \(\alpha\) \(\frac{\frac{1}{uv}}{u + v}\) ∴ w = \(\frac{\frac{k}{uv}}{u + v}\) = \(\frac{k(u + v)}{uv}\) w = \(\frac{k(u + v)}{uv}\) w = 8, u = 2 and v = 6 8 = \(\frac{k(2 + 6)}{2(6)}\) = \(\frac{k(8)}{12}\) k = 12 i.e 12 ( u + v) = uwv

The region is denoted by 4x + 3y \(\geq\) 6

Potential Energy = Heat Energy mgh = Mc\(\Delta \theta = \frac{gh}{c} = \frac{10 \times 240}{4.20 \times 10^3}\) = 1.0oC

(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms\(^{-2}\). \(a = \frac{v - u}{t}\); a - uniform acceleration, v - final velocity, u - initial velocity, t - time. Graphically, acceleration is the same acceleration at x is different every point from that at Y. Note - acceleration is given by slope in both graphs. (b) SEE DIAGRAM ABOVE. V\(_1\) = u + at = 0 + 8 x 30 = 240m/s V\(_2\) = V\(_1\) + at = 240 + 50 x 20 = 340m/s (i) Maximum speed = 340 m/s. (ii) \(a = \frac{v - u}{t} = \frac{0 - 340}{20} = -17 ms^{-2}\) (iii) Area of A = \(\frac{1}{2} \times 30 \times 240 = 3600m\) Area of B = \(\frac{1}{2}\) x ( 240 + 340) x 20 = 5600m Total distance in the first 50s: 3600 + 5800 = 9400m (iv) Average speed during the first 50s = \(\frac{9400m}{50s} = 188 ms^{-1}\)

An induction coil is a device used to produce high emf from a low-voltage direct current (DC) supply.