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Juliette [100K]
3 years ago
12

A principal of $3800 is invested at 6.75% interest, compound annually. How much will the investment worth after 9 years?

Mathematics
1 answer:
frozen [14]3 years ago
6 0

Answer:

$6840.60

Step-by-step explanation:

formula od compound interest A= P (1 =r/n)ⁿ⁹

where n=1 when it is compunded annually  and t= 9 years

A= 3800(1+6.75/100)⁹

A= 3800 ( 1+0.0675)⁹

A= 3800(1.0675)⁹

A=3800(1.800)

A= 6,840.60 $

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Add and Subtract Rational Expressions with a Common Denominator
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Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

7 0
2 years ago
The function f(x) = x2. The graph of g(x) is f(x) translated to the left 6 units and down 5 units. What is the function rule for
Y_Kistochka [10]

Answer:

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Step-by-step explanation:

To translate f(x) to the left by 6 units, replace x with x+6.

To translate f(x) down 5 units, replace f(x) with f(x)-5.

Together, these replacements give you ...

... g(x) = f(x+6) -5

... g(x) = (x +6)^2 -5

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3 years ago
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ludmilkaskok [199]
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Answer:

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Step-by-step explanation:

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