Jennifer invested $400 at 6% interest compounded continuously. Write a model m(t) that represents the money in Jennifer's account in t years. A. B. How much money is in Jennifer's account after 5 years? Round to the nearest cent. C. Approximately when will Jennifer have $800 in her account? Round to the nearest tenth of a year.

11.5 years

Step-by-step explanation:

Given data

Principal= $400

Rate= 6%

For the compound interest at time t, the expression is given as

A= P(1+r)^t

Substitute

A= 400(1+0.06)^t

A=400(1.06)^t

B. How much money is in Jennifer's account after 5 years

put t= 5

A=400(1.06)^5

A=400*1.338

A=$535.2

C. Approximately when will Jennifer have $800 in her account

A=$800

P=$400

r=6%

t= ln(A/P)/r

t= ln(800/400)/0.06

t= ln(2)/0.06

t= 0.6931/0.06

t=11.551

Hence the time is about 11.5 years

answer are

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.