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3 years ago

Suppose that a loan of $6000 is given at an interest rate of 9% compounded each year.

Mathematics

2 answers:

kirza4 [7]3 years ago

5 0

Answer: a) $6649

b) $7128.6

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 6000

r = 9% = 9/100 = 0.09

n = 1 because it was compounded once in a year.

t = 1 year

Therefore,

A = 6000(1 + 0.09/1)^1 × 1

A = 6000(1.09)

A = $6649

b) t = 2 years

A = 6000(1.09)^2

A = $7128.6

lara [203]3 years ago

3 0

Answer:

$6540.00
$7128.60

Step-by-step explanation:

Each year, the balance is multiplied by (1+r) where r is the annual interest rate.

(a) Balance at the end of year 1:

$6000×1.09 = $6540

(b) Balance at the end of year 2:

$6540×1.09 = $7128.60

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The solution of the equation can be determined using the quadratic formula.

The quadratic formula is given by

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Hence, from the equation, we have,

$a=-1,\:b=5,\:c=-3$

Substituting these values in the quadratic formula, we get,

$x=\frac{-5\pm \sqrt{5^2-4\left(-1\right)\left(-3\right)}}{2\left(-1\right)}$

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$x=\frac{-5\pm \sqrt{25-12}}{-2}$

Simplifying the terms within the root, we get,

$x=\frac{-5\pm \sqrt{13}}{-2}$

Thus, the roots of the equation are $x=\frac{-5+ \sqrt{13}}{-2}$ and $x=\frac{-5- \sqrt{13}}{-2}$

Taking out the negative sign, we get,

$x=\frac{-(5- \sqrt{13})}{-2}$ and $x=\frac{-(5+ \sqrt{13})}{-2}$

Cancelling the negative sign, we get,

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