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The pyramid has 8 edges in itself while the cuboid has 12 edges. When merging the two shapes together, the edge of the base of the pyramid becomes same as the edges of the top of the cuboid. Hence, the new structure will have (12 + 8) - 4 = 16 edges.

The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\) Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) = \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\) = \(\frac{\pi}{24\sqrt{3}}\)

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

(x + y) is a factor 2x3 + 2x2y - xy + 3x - y^2

(25)\(^{x - 1}\) = 64(\(\frac{5}{2}\))\(^6\) (\(\frac{5}{2}\))\(^6\) = (\(\frac{5^6}{64}\)) 25\(^{x - 1}\) = 64 x (\(\frac{5^6}{64}\)) 5\(^{2x - 2}\) = 5\(^6\) 2x - 2 = 6 = \(\frac{8}{2}\) = 4

Assuming loge 4.4 = 1.4816 and loge 7.7 = 2.0142 then value of loge \(\frac{7}{4}\) = loge7.7 - loge \(\frac{7.7}{4.4}\) = loge\(\frac{7}{4}\) = 2.0142 -1.4816 = 0.5326

\(\frac{(a - \frac{1}{a}) (a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}})}{a^2 - \left(\frac{1}{a}\right)^2}\) \( = \frac{(a - \frac{1}{a}) \left(a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}}\right)}{\left(a - \frac{1}{a}\right)\left(a + \frac{1}{a}\right)} \) \(= \frac{a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}}}{a + \frac{1}{a}}\) \(= \frac{\frac{a^{2} + 1}{a^{\frac{2}{3}}}}{\frac{a^2 + 1}{a}}\) \(= \frac{1}{a^{\frac{2}{3}}} \cdot a\) \(= a^{1 - \frac{2}{3}} = a^{\frac{1}{3}}\) \(\boxed{a^{\frac{1}{3}}}\)

In < AFT: \(\frac{TF}{AF}\) = tan60º therefore, TF = xtan60º ..................1 In < BFT: \(\frac{TF}{BF}\) = tan45º TF = ( 2 + x )tan45º .......................2 equating equations 1 and 2 xtan60º = ( 2 + x )tan45º But tan60º = \(\sqrt3\) and tan45º = 1 x \(\sqrt3\) = 2 + x x \(\sqrt3\) - x = 2 x ( \(\sqrt3\) - 1) = 2 x = \(\frac{2}{(\sqrt3\)) - 1}\) rationalizing the denominator x = \(\frac{2(\sqrt3 + 2)}{3 - 1}\) = \(\frac{2(\sqrt3 + 1)}{2}\) = \(\sqrt3\) + 1

a = 2i - 3j - 3i + 6j = -i + 3j = \(\sqrt{5 \times 2}\) = \(\sqrt{10}\)

Let x be the no. \(\frac{3 + x}{

Possible multiples of 8 and 9 = 72, 144, 216, 288, 360, 432, 576, 648. \(\frac{387}{8}\) has remainder = 3; \(\frac{387}{9}\) has remainder 0. \(\frac{493}{8}\) has remainder 5; \(\frac{493}{9}\) has remainder 7. \(\frac{502}{8}\) has remainder 6; \(\frac{502}{9}\) has remainder 7. \(\frac{441}{8}\) has remainder 1; \(\frac{441}{9}\) has remainder 0. \(\frac{634}{8}\) has remainder 2; \(\frac{634}{9}\) has remainder 4. Hence 502 is the answer.