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

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

The interaction between P. aurelia and P. caudatum is interspecific competition, specifically demonstrating competitive exclusion. In this scenario, P. aurelia outcompetes P. caudatum for resources, leading to a decline in the P. caudatum population while the P. aurelia population continues to grow. This is because P. aurelia is more efficient at obtaining food, demonstrating that two species cannot occupy the same ecological niche indefinitely.

From the diagram,we can infer that P aurelia is better adapted for obtaining food than P caudatum because as number of days increases the number of P aurelia increases in number as well while P caudatum decreases with increase in the number of days. This implies that P aurelia tend to adapt more despite the competition because instead of decreasing in number, they were increasing in number progressively.

From the diagram,we can infer that P aurelia is better adapted for obtaining food than P caudatum because as number of days increases the number of P aurelia increases in number as well while P caudatum decreases with increase in the number of days. This implies that P aurelia tend to adapt more despite the competition because instead of decreasing in number, they were increasing in number progressively.

x2 + 2x > 15 x2 + 2x - 15 > 0 (x2 + 5x) - (3x - 15) > 0 x(x + 5) - 3(x + 5) >0 (x - 3)(x + 5) > 0 therefore, x = 3 or -5 i.e. x< 3 or x > -5

Deliquescent substances are referred to known as those substances which absorbs moisture from the surrounding and dissolves themselves with the moisture or water to form a solution. Examples are sodium hydroxide(NaOH), calcium chloride(CaCl2), potassium hydroxide(KOH), ammonium chloride(NH4Cl) , etc

2q35 = 778 2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80 2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1 50 + 5q + 3 = 56 + 7 5q = 63 - 53 q = \(\frac{10}{5}\) q = 2

\(\frac{3\frac{2}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\) \(\frac{\frac{11}{3} \times \frac{5}{6} \times \frac{2}{3}}{\frac{11}{15} \times \frac{3}{4} \times \frac{2}{27}}\) \(\frac{110}{54} \div \frac{66}{1620}\) 50

S.I. = \(\frac{P \times R \times T}{100}\) If T = 9 months, it is equivalent to \(\frac{9}{12}\) years S.I. = \(\frac{5000 \times 4 \times 9}{100 \times 12}\) S.I. = N150

M:N:Q == 5:4:3 i.e M = 5, N = 4, Q = 3 Substituting values into equation, we have... \(\frac{2N - Q}{M}\) = \(\frac{2(4) - 3}{5}\) = \(\frac{8 - 3}{5}\) = \(\frac{5}{5}\) = 1

\((\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}\) \((\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}\) \((\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}\) \(\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}\) \(\frac{2}{3} \div \frac{4}{3}\) \(\frac{2}{3} \times \frac{3}{4}\) \(\frac{2}{4}\) \(\frac{1}{2}\)

log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3 log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = 3log33 log\(_{3}^{18}\) + log\(_{3}^{3}\) - log\(_{3}^{x}\) = log333 log3(\(\frac{18 \times 3}{X}\)) = log333 \(\frac{18 \times 3}{X}\) = 33 18 x 3 = 27 x X x = \(\frac{18 \times 3}{27}\) = 2

\(\frac{2 - \sqrt5}{3 - \sqrt5}\) x \(\frac{3 + \sqrt5}{3 + \sqrt5}\) \(\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}\) = \(\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}\) = \(\frac{6 - \sqrt5 - 5}{9 - 5}\) = \(\frac{1 - \sqrt5}{4}\)

(\(\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}\)) \(\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}\) = 2 - \(\frac {1}{3}\) = \(\frac {16 - 1}{3}\) = \(\frac{5}{3}\)