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If r per annum is the rate at which the principal A is compounded annually, then at the end of k years, the money due is
Q = A (1 + r)k
Suppose compounding is done continuously. i.e. at every instant the principal A is compounded at R per annum. Then,
Q = A eRk
The relationship between R and r is given by the following reasoning:
A (1 + r)k = A eRk
Example
If R = 5.25%, then ln(1 + r) = 5.25% or r = 5.39%
Suppose Rs.100 is being compounded annually at the rate of 10% per annum. What is the future value of Rs.100 at the end of the third year? What is the effective continuously compounded rate of interest? What is the future value of Rs.100 at the end of the third year, using this interest rate?
FV(Rs.100) = 100 x (1.10)3 = 133.1
If r = 0.1, then the continuously compounded rate of interest R is given by
R = ln(1 + 0.1) = 0.0953
FV(Rs.100) = 100 e0.0953 x 3 = 100 x 1.331 = 133.1
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Necessity of holistic marketing or importance of holistic marketing
Example of Fractional Equations: Example: Solve the fractional equation (3x +8)/x +5 =0 Solution: Multiply both sides of the equation by the LCD (x). (x) ((3x
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In an equilateral triangle 3 coins of radius 1cm each are kept along such that they touch each other and also the side of the triangle. Determine the side and area of the triangle.
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We will be looking at solutions to the differential equation, in this section ay′′ + by′ + cy = 0 Wherein roots of the characteristic equation, ar 2 + br + c = 0 Those
we know that log1 to any base =0 take antilog threfore a 0 =1
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