Explanation

(a) Let the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) be less than 7 by z.

Then z = \(7 - (3\frac{2}{3} + 2\frac{1}{5})\)

= \(7 - (\frac{11}{3} + \frac{11}{5})\)

= \(7 - \frac{88}{15}\)

= \(\frac{17}{15} = 1\frac{2}{15}\)

Hence, the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) is less than 7 by \(1\frac{2}{15}\).

(b) \(h = 6 + 4\cos (15p)°\) (Given)

(i) When p = 4,

\(h = 6 + 4\cos (15 \times 4)°\)

\(h = 6 + 4 \cos 60°\)

= \(6 + 4 \times 0.5\)

= \(6 + 2 = 8 m\)

(ii) When h = 9 m, the given equation becomes

\(9 = 6 + 4\cos (15p)°\)

\(9 - 6 = 4 \cos (15p)°\)

\(\cos (15p)° = \frac{3}{4}\)

\((15p)° = \cos^{-1} (0.75)\)

\((15p)° = 41.4°\)

\(p° = \frac{41.4}{15}\)

\(p° = 2.76°\)

\(p \approxeq 2.8°\) (2 significant figures).

Explanation

a. a = (10t - 4t\(^2\))m/s\(^2\), t = t seconds, u = 0

from first equation of motion,

v = u + at

v = 0 + (10t - 4t\(^2\))t

v = (10t\(^2\) - 4t\(^3\))m/s\(^2\)

ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4

Δa4 = 10(t4-t3) - 4(t4-t3)\(^2\)

Δa4 = 10(4-1) - 4(4-3)\(^2\)

Δa4 = 10(1) - 4(12)

Δa4 = (10-4)m/s\(^2\)

Δa4 = 6ms-2

b. R = m(g+a)

When the lift moves with a constant velocity, a = 0

R = mg

R = 120 × 10

R = 1200N

ii. when the lift moves upward with an acceleration of 3m/s\(^2\)

R = m(g+a)

R = 120(10+3)

R = 120 × 13

R = 1560N

Correct Answer: B. 75

Explanation

No. of ways = 3 x 5 x 5 = 75

Explanation

WHILE Baadi is the SUJUD you make after SALAM when you mistakenly ADD during solat. When you noticed you've shortened or omitted something in ur salaat then compensate with QABLI. It is compensated by two sujuud after the final Tashahhud before final Tasleem.

Correct Answer: A. delimits the cytoplasm

Correct Answer: A. N'dama and muturu

Correct Answer: C. mail order firms

Correct Answer: A. Advertising is paid for while publicity is not

Explanation

(a)To demonstrate the effect of auxin on growth of a plant shoot, cut off the growing tip of coleoptile or shoot of three potted maize seedlings A,B, and C. Get three agar blocks A.B.C. Place one of the cut tips on an agar block A, so that the auxins in the tip can diffuse into the agar block. Then put three drops of IAA or Synthetic auxin on agar blockB. Put agar blocks A and B on one side or on top of the cut tip of the coleoptile or shoot A and B respectively. Then place the untreated agar block C on the side or tip of third cut coleoptile or shoot tip C as control experiment. Leave the three coleoptiles or shoots in unilateral light for 12 to 36 hours. The Coleoptiles or Shoots carrying agar block A and B show growth curvature while C shows no growth curvature. This shows that auxin promotes growth

(b)The use of auxin in agriculture are:

(i) Auxin helps to induce ripening of fruits.

(ii) It also helps to produce lateral bud development in stem

(iii) It helps in controlling weeds as well as controlling flowering and fruiting in plants.

(v) It assists in breaking seed dormancy.

(vi) It helps to initiate adventitious roots in stem cuttings and for parthenocarpic fruiting.

(c) Mechanism of transmission of impulses through a nerve fibre is characterised by change in electrical, potential and ionic charge. When at rest, the outside of the nerve is positively charged while the inside is negatively charged. Chemically, the nerve fibres possess potassium ions which are concentrated inside the fibre, while sodium ions are concentrated more on the outside. The change in permeability of the cell membrane is brought about when the fibre is stimulated and sodium ions enter the nerve while potassium ions move to the outside. The change in permeability along the nerve or change in electrical potential and ionic change along the fibre cause the movement of the electrical current along the fibre. In this way, the impulse is propagated or transmitted while the original resting stage of the nerve is restored as soon as the impulse passes.

Correct Answer: A. mark each traverse station with a peg