Suppose that P dollars in principal is invested in an account earning 2.1% interest compounded continuously. At the end of 2 yr, the amount in the account has earned $193.03 in interest. A) Find the original principal.
B) Using the the original principal from part (a) and the model A = Pe", determine the time required for the investment to reach $6000.

See explanation

Step-by-step explanation:

Recall that

Amount = Principal + Interest

The formula is;

A = P(1 + r)^n

Where;

P= principal, n = time and r = rate

Hence;

P + I =  P(1 + r)^n

P + 193.03 = P(1.021)^2

P + 193.03 = 1.04P

P - 1.04P = -  193.03

-0.04 P =  -  193.03

P = 193.03/0.04

P = $ 4825.75

B) When A = $6000, n=?

From;

A=   P(1 + r)^n

6000 = 4825.75 (1.021)^n

6000/4825.75 = (1.021)^n

1.24 =  (1.021)^n

Taking logarithm of both sides;

log 1.24 = n log 1.021

n = log 1.24/log 1.021

n = 0.0934/0.009

n = 10 years

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c