Afamily has a $93,411, 20-year mortgage at 5.4% compounded monthly. find the monthly payment. also find the unpaid balance after 5 years and after 10 years the monthly payment is so. (round to two decimal places.)

Step-by-step explanation:

1. The relevant formula for computing the monthly payment A from principal P and interest rate r for loan of t years is ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

Filling in the numbers and doing the arithmetic, we get ...

  A = $93,411(0.054/12)/(1 -(1 +0.054.12)^-(12·20)) ≈ $637.30

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2. The relevant formula for computing the remaining balance after n payments of amount p on principal P at interest rate r is ...

  A = P(1 +r/12)^n -p((1 +r/12)^n -1)/(r/12)

Filling in the given values and doing the arithmetic, we get ...

  A = $93,411(1.0045^60) -637.30(1.0045^60 -1)/(0.0045) ≈ $78,505.48

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3. The same formula with n=120 gives ...

  A = $93,411(1.0045^120) -637.30(1.0045^120 -1)/(0.0045) ≈ $58,991.59

1

step-by-step explanation:

you are throwing one dart

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}