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2 years ago

An investment of 100,000 AED increases at a rate of 16% per year. What is

Mathematics

1 answer:

stepan [7]2 years ago

7 0

Answer:

1,946,075.95 AED

Step-by-step explanation:

Firstly, we write the general formula for an exponential increase as follows;

V(t) = A( 1 + r)^t

where V(t) is the value after some number of years t

A is the investment amount which is 100,000 AED

r is the rate of increase which is 16% = 16/100 = 0.16

t is the number of years which is 20

substituting these values;

V(20) = 100,000( 1 + 0.16)^20

V(20) = 100,000(1.16)^20

V(20) = 1,946,075.95 AED

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Read 2 more answers

What is the equation that represents the sequence in this problem ? Find the price after the 8th month

AVprozaik [17]

ANSWER

$\begin{gathered} a_n=ar^{n\text{ - 1}} \\ a_8\text{ = \$38.26} \end{gathered}$

EXPLANATION

The problem represents a geometric progression.

The general form of a geometric sequence is:

$a_n=ar^{n\text{ - 1}}$

where a = first term

r = common ratio

The first term from the table is the first price (for the first month). That is $80.00

To find the common ratio, we divide a term by its preceeding term.

Let us divide the price of the second month from the first.

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The price after the 8th month is the value of a(n) when n = 8

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$\begin{gathered} a_8\text{ = 80 }\cdot0.9^{(8\text{ - 1)}} \\ a_8\text{ = 80 }\cdot0.9^7 \\ a_8\text{ = \$38.26} \end{gathered}$

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11 months ago

Anyone know the answer to this algebra problem?

ikadub [295]

Answer: $\bold{a=1\qquad b=\dfrac{1}{16}\qquad c=\dfrac{1}{64}\qquad d=1\qquad e=\dfrac{4}{9}\qquad f=\dfrac{16}{81}}$

<u>Step-by-step explanation:</u>

$\begin{array}{c|l}\underline{\quad x\quad}&\underline{\quad 4^{-x}\qquad \qquad}\\-1&4^{-(-1)}=4^1=4\\\\0&4^{-(0)}=4^0=1\\\\2&4^{-(2)}=\dfrac{1}{4^2}=\dfrac{1}{16}\\\\4&4^{-(4)}=\dfrac{1}{4^4}=\dfrac{1}{64}\end{array}$

$\begin{array}{c|l}\underline{\quad \bigg x \quad}&\underline{\quad \bigg(\dfrac{2}{3}\bigg)^x\qquad \qquad}\\\\-1&\bigg(\dfrac{2}{3}\bigg)^{-1}=\dfrac{3}{2}\\\\0&\bigg(\dfrac{2}{3}\bigg)^{0}=1\\\\2&\bigg(\dfrac{2}{3}\bigg)^{2}=\dfrac{2^2}{3^2}=\dfrac{4}{9}\\\\4&\bigg(\dfrac{2}{3}\bigg)^{4}=\dfrac{2^4}{3^4}=\dfrac{16}{81}\end{array}$

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3 years ago