An investment of $1,000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the $1,000 increased to $4,000 by earning interest. In how many years after the initial investment was made would the $1,000 have increased to $8,000 by earning interest at that rate

it will take 18.4 years

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 = $1000

A = $4000

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

t = 12 years

Therefore,.

4000 = 1000(1 + r/1)^1 × 12

4000/1000 = (1 + r)^12

4 = (1 + r)^12

Log 4 = 12 log (1 + r)

0.602/12 = log (1 + r)

0.0501 = log (1 + r)

Taking inverse log of both sides, it becomes

10^0.0501 = 10^log(1 + r)

1.122 = 1 + r

r = 1.122 - 1 = 0.122

At r = 0.122 and A = 8000

Therefore,

8000/1000 = (1 + r)^1 × t

8 = (1 + 0.122)^t

8 = (1.122)^t

Log 8 = tlog 1.122

0.903 = 0.049 × t = 0.049t

t = 0.903/0.049

t = 18.4 years

11.2

step-by-step explanation:

when doing tangent the tangent of the angle is equal to the opposite side from the angle over the adjacent side to the angle.

tan43 = x / 12

move 12 over by multiplying each side by 12

12tan43 = x

x = about 11.2