4.036 divided by 0.04 gives :

Let the numbers be x and x + 2.

Then, (x + 2)² - x² = 84

⟹ 4x + 4 = 84

⟹ 4x = 80

⟹ x = 20.

∴ The required sum = x + (x + 2) = 2x + 2 = 42.

Let the number be x.

Then, x + 17 =	60
	x

⟹ x² + 17x - 60 = 0

⟹ (x + 20)(x - 3) = 0

⟹ x = 3.

Let the numbers be a, b and c.

Then, a² + b² + c² = 138 and (ab + bc + ca) = 131.

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

⟹ (a + b + c) = √400 = 20.

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

⟹ 9x = 36

⟹ x = 4.

∴ Required difference = (2x + x) - (2x - x) = 2x = 8.

Let the numbers be x and y.

Then, xy = 120 and x² + y² = 289.

∴ (x + y)² = x² + y² + 2xy = 289 + (2 x 120) = 529

∴ x + y = √529 = 23.

Given expression	= (0.96)3 - (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= ❨ a3 - b3 ❩ a2 + ab + b2
	= (a - b) = (0.96 - 0.1) = 0.86

3.14 x 10⁶ = 3.14 x 1000000 = 3140000.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.

Given Expression =	a2 - b2	=	(a + b)(a - b)	= (a + b) = (2.39 + 1.61) = 4.
	a - b		(a - b)

Let 3889 + 12.952 - x = 3854.002.

Then x = (3889 + 12.952) - 3854.002

   = 3901.952 - 3854.002

   = 47.95.