Let's denote the cost of a geometry box as x. Since the cost of a bag is three times the cost of a geometry box, the cost of a bag is 3x.
Given that the cost of 2 bags and 3 geometry boxes is Rs.3,960, we can set up the equation:
2(3x) + 3(x) = 3960
6x + 3x = 3960
9x = 3960
Divide both sides by 9:
x = 3960 / 9
x = 440
Given a + b + c = 9 and ab + bc + ca = 20
We know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Substituting the given values:
(9)² = a² + b² + c² + 2(20)
81 = a² + b² + c² + 40
Now, find a² + b² + c²:
a² + b² + c² = 81 - 40
= 41
Let's factor the area:
2x^2 - 7x - 4 = (2x + 1)(x - 4)
Assuming these are the length and width:
Length (L) = 2x + 1
Width (W) = x - 4
Perimeter = 2(L + W)
= 2(2x + 1 + x - 4)
= 2(3x - 3)
= 6x - 6
Given:
A1 = 28 (first term)
d = -4 (common difference)
n = 7 (number of terms)
To find the nth term (an), use the formula:
an = a + (n-1)d
an = 28 + (7-1)(-4)
= 28 + 6(-4)
= 28 - 24
= 4
Given that 1 - i is a root of the equation x² + ax + b = 0, and the coefficients are real, the complex conjugate 1 + i must also be a root.
The sum of the roots = (1 - i) + (1 + i) = 2 = -a
So, a = -2
The product of the roots = (1 - i)(1 + i) = 1² - i² = 1 + 1 = 2 = b
So, b = 2
Now, a - b = -2 - 2 = -4
The limit lim x→∞ 2/x is equal to 0.
As x approaches infinity, the denominator x becomes very large, and the numerator 2 remains constant.
Therefore, the ratio 2/x approaches 0.
