Mathematics Questions and Answers

Volume of a cube with side a cm = a\(^3\)

a\(^3\) = 64 \(\implies\) a = 4cm

Base area of a cube = a\(^2\)

= 4\(^2\)

= 16 cm\(^2\)

\(\implies\) Base area of the cuboid = 16 cm\(^2\)

Volume of cuboid = Base area x height

80 = 16 x h

h = \(\frac{80}{16}\)

= 5 cm

From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ² = OM² + MQ²

5² = 2² + MQ²

25 = 4 + MQ²

25 - 4 = MQ²

21 - MQ²

MQ² = 21

MQ² = \(\sqrt{21}\)

Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm

p^{\(\frac{1}{3}\)} = \(\frac{3\sqrt{q}}{r}\)(cross multiply)

3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply)

3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side

q = 3\(\sqrt{r}\) x p

q = r^{\(\frac{1}{3}\)}p = pr^{\(\frac{1}{3}\)}

\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)

\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2)

x + 4y = 6......(iii)

3x - 4y = 2.....(iv) add (iii) and (iv)

4x = 8, x = \(\frac{8}{4}\) = 2

substitute x = 2 into equation (iii)

x + 4y = 6

2 + 4y = 6

4y = 6 - 2

4y = 4

y = \(\frac{4}{4}\)

= 1(x + y)

2 + 1 = 3

\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)

substitute x = 64 and y = 27

\(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\)

= \(\frac{8 - 3}{27 - 16}\)

= \(\frac{5}{11}\)