How much would $300 invested at 9% interest compounded monthly be worth after 3 years?

F=P(1+(r/n))^(nt)

F=future amount

P=present amount

r=rate in decimal form

n=number of times per year it is compounded

t=time in years

F=P(1+(r/n))^(nt)

F=future amount

P=300

r=0.09

n=1

t=3

F=300(1+(0.09/1))^(1(3))

F=300(1+0.09)^(3)

F=300(1.09)^(3)

F=388.509

it would be worth $388.51 (rounded)

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: