You invest $3000 in an account at 3.5% per year simple interest. how much will you have in the account at the beginning of the 7th year? round your answer to the nearest whole dollar.

$3735

Step-by-step explanation:

The formula for simple interest is I = Prt, where I is the interest earned, P is the initial investment, r is the interest rate in decimal form, and t is the time in years.  We have everything we need to find the interest, which is the amount your investment earned while it sat there for 7 years.  Once we find that interest amount, we will add it to the intial investment to find the total amount after 7 years that your money has grown to.

I = 3000(.035)(7) so

I = 735

3000 + 735 = 3735

$3,630

Step-by-step explanation:

You invest $3,000 in an account at 3.5% per year simple interest.

We have to calculate the amount in the account at the beginning of the 7th year. This means we have to calculate the interest for completed 6 years.

Formula for simple interest

A = P(1+rt)

A = Amount after maturity

P = Principal amount ( 3,000)

r = rate of interest in decimal ( 0.035)

t = time in years ( 6 )

Now we put the values in to formula

A = 3,000(1 + 0.035 × 6)

A = 3,000 ( 1 + 0.021 )

A = 3,000 × 1.21

A = $3,630

The amount would be $3,630 at the beginning of the 7th year.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

the bolded word " farsighted" could be replaced with visionary.