A person invests 4500 dollars in a bank. The bank pays 4.75% interest compounded semi-annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 5900 dollars?

It should take 5.8 years to reach his goal.

Step-by-step explanation:

In order to solve this question we must use the compounded interest's formula shown below:

M = C*(1 + r/n)^(t*n)

Where M is the final amount, C is the initial amount, r is the interest rate, n is the compound rate and t is the time elapsed in years. Applying the data from the problem we have:

5900 = 4500*(1 + 0.0475/2)^(2*t)

5900 = 4500*(1 + 0.02375)^(2*t)

5900 = 4500*(1.02375)^(2*t)

4500*(1.02375)^(2*t) = 5900

(1.02375)^(2*t) = 5900/4500

(1.02375)^(2*t) = 59/45

log[1.02375^(2*t)] = log(59/45)

2*t*log(1.02375) = log(59/45)

t = log(59/45)/[2*log(1.02375)] = 5.77008

It should take 5.8 years to reach his goal.

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake