m = 2

243 = 3 x 3 x 3 x 3 x 3
In the above factorisation 3 x 3 remains after grouping the 3’s in triplets. Therefore, 243 is not a perfect cube.

(a) 729
(c) 1728
(e) 6859

4.05 \(\times\) 10⁶

It cannot be always true. So, the statement is false.
e.g. Let m = -1
Then, m² =(-1)² =1
But m³ = (-1)³ =-1
So, 1>-1
Hence, the above statement m² < m³ is false, when m = -1.

Yes

We have, 0.0000000000085
=\(\frac{85}{10^{13}}=\frac{8.5\times 10}{10^{13}}=8.5\times 10 \times 10^{-13}\)      \([\because \frac{1}{a^{m}}=a^{-m}]\)
= \(8.5\times 10^{(1-13)}=8.5\times 10^{-12} \quad [\because a^{m}\times a^{n}=a^{m+n}]\)
which is the required standard form.

3² \(\times\) 3^-5 \(\times\) 3⁶ = 3^2+(-5)+6 = 3^8-5 = 3³

269 is an odd number, so its square is also odd.

Given number is 1234.
Unit's digit of 1234 = 4
So, square of unit's digit = (4)²= 16
Hence, unit's digit of the square of unit's digit of the given number is 6.

100

736000

Given, the volume of cube = 512 cm³
We know that, the volume of a cube = (Side)³
=(Side)³
Let the length of each side be a.
So, (a)³ = 512
a = \(\sqrt [ 3 ]{ 512 } \)
a = \(\sqrt [ 3 ]{ \underline { 2\times 2\times 2 } \times \underline { 2\times 2\times 2 } \times \underline { 2\times 2\times 2 } } \)
= 2 x 2 x 2
=8 cm
Hence, length of the each side of a cube is 8 cm.

4.42 X10²

2 x 2 = 4

If a number has 4 or 6 in the unit's place, then its square ends in 6.

0.07 x 10¹⁰= 700000000.

\((-2)^{-2}={1\over (-2)^2}={1\over 4}\)

\(\therefore\)Number of zeroes at the end of the cube = 6
\(\therefore\)Number of zeroes at the end of the cube root = \(\frac{6}{2}\)= 2