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Nady [450]
3 years ago
14

Part 1 (Unit 2): Subtract Polynomials:

Mathematics
1 answer:
Alinara [238K]3 years ago
8 0
Part 1
 We have the following polynomials:
 (3-6n5-8n4)
 (-6n4-3n-8n5)
 Subtracting the polynomials we have:
 (3-6n5-8n4) - (- 6n4-3n-8n5)
 n5 (-6 + 8) + n4 (-8 + 6) + 3n + 3
 Rewriting:
 2n5 - 2n4 + 3n + 3

 
Part 2
 
For this case we have the following polynomial:
 4x2-2x-5 = 0
 Using resolver we have:
 x = (- b +/- root (b ^ 2 - 4 * a * c)) / (2 * a)
 x = (- (- 2) +/- root ((- 2) ^ 2 - 4 * 4 * (- 5))) / (2 * 4)
 x = (2 +/- root (4 + 80)) / (8)
 x = (2 +/- root (84)) / (8)
 x = (2 +/- root (4 * 21)) / (8)
 x = (2 +/- 2raiz (21)) / (8)
 x = (1 +/- root (21)) / (4)
 The roots are:
 
x1 = (1 + root (21)) / (4)
 
x2 = (1 - root (21)) / (4)
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Given:

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Time = 11 years.

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The accumulated value of the given investment.

Solution:

Formula for amount or accumulated value after compound interest is:

A=P\left(1+\dfrac{r}{n}\right)^{nt}

Where, P is the principal values, r is the rate of interest in decimal, n is the number of times interest compounded in an year and t is the number of years.

Compounded semiannually means interest compounded 2 times in an years.

Putting P=14000,r=0.10,n=2,t=11 in the above formula, we get

A=14000\left(1+\dfrac{0.10}{2}\right)^{2(11)}

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A=14000\left(1.05\right)^{22}

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