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Alex Ar [27]
3 years ago
13

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

of 169 immediately and another payment of 169 at the end of two years. (ii) A payment of 225 at the end of two years and another payment of 225 at the end of four years. Calculate K.
Mathematics
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

The present value of K is, K=251.35

Step-by-step explanation:

Hi

First of all, we need to construct an equation system, so

(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}

(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}

Then we equalize both of them so we can find i

(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}

To solve it we can multiply (3)*(1+i)^{4} to obtain (1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}), then we have 225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}.

This leads to a third-grade polynomial 169i^{3}+451i^{2}+395i-112=0, after computing this expression, we find only one real root i=0.2224.

Finally, we replace it in (1) or (2), let's do it in (1) K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35

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The zeros of the polynomial are -3, -2, 4 and 6 and the graph of the polynomial is graph (a)

<h3>How to factor the polynomial?</h3>

The polynomial is given as:

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Factor out x^2 - 10x + 24

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Expand each bracket

P(x) = (x^2 + 3x + 2x + 6)(x^2 - 4x - 6x + 24)

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P(x) = [x(x + 3) + 2(x + 3)][x(x - 6) - 4(x - 6)]

Factor out x + 3 and x - 6

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Set to 0

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Read more about polynomials at:

brainly.com/question/20896994

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