Olivia has taken out a $13,100 unsubsidized stafford loan to pay for her college education. she plans to graduate in four years. the loan has a duration of ten years and an interest rate of 7.6%, compounded monthly. by the time olivia graduates, how much greater will the amount of interest capitalized be than the minimum amount that olivia could pay to prevent interest capitalization? round all dollar values to the nearest cent. a. $654.45 b. $477.27 c. $995.60 d. $354.22

If Olivia wants to prevent interest capitalization, she must pay the accrued interest each month. That amount is

... I = Prt = $13,100×0.076×(1/12) = $82.97

Over 4 years (48 months), these payments total $82.97×48 = $3982.56.

If no payments are made, the loan balance grows by the multiplier (1 +r/12) each month. Then the amount of interest that will be capitalized at the end of 48 months is ...

... $13,100×((1 +0.076/12)^48 -1) = $4637.01

The difference in these amounts is ...

... $4637.01 -3982.56 = $654.45 . . . . . matches selection a.

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(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3