How long will it take for $3000 to grow to $11,700 at an interest rate of 6.3% if the interest is compounded continuously? Round the number of years to the nearest
hundredth.
A) 0.83
B) 2160.28
C) 21.60
D) 2.16

Step-by-step explanation:

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

Where,

P = principal amount (the initial amount you borrow or deposit), in this case $3000

r = annual rate of interest (as a decimal), in this case 0.06

t = number of years the amount is deposited or borrowed for, in this case we don't know it

A = amount of money accumulated after n years, including interest, in this case $21000

n = number of times the interest is compounded per year, in this case 12

     This gives us 21,000=3,000(1+0.06/12)^12t

    After simplification, 7=(1.005)^12t

By taking the logarithm of both sides,

                            Log7=Log1.005^12t

                               12t = Log 7/Log 1.005

                               12t =390.1545

          This gives us 32.5months

7

step-by-step explanation:

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation: