Mathematics, 20.06.2019 18:04 salejandro68

James purchased a $205,000 home with a 30-year mortgage at 5.78%. if makes a $1500 monthly mortgage payment, how many months early will he pay off his mortgage? a. 180 months b. 224 months c. 127 months d. 136 months

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Answer from: dayantapia2007

James will pay off his mortgage 136 months early.

Step-by-step explanation:

Given : Amount to be paid(A)= $205000

Interest rate per annum = 5.78% = 0.0578

monthly interest rate(i) = $\frac{0.0578}{12}$ = 0.004817 (approx)

Monthly payment (P) = $1500.

we have to find how many months early will he pay off his mortgage

We know, $P=\frac{A(i)}{1-(1+i)^(-t)}$

Substitute values , we get,

$1500=\frac{205000(0.004817)}{1-(1+0.004817)^(-t})$

$1500=\frac{205000(0.004817)}{1-(1.004817)^(-t)}$

$1500=\frac{987.416653}{(1-(1.004817)^(-t)}$

Solving for t ,

${(1-(1.004817)^t)}=\frac{987.416653}{1500}\\\\\\{(1-(1.004817)^(-t))}=0.65828\\\\\(1.004817)^(-t)=0.341722$

Applying ln both sides,we get,

$\ln \left(1.004817^(-t)\right)=\ln \left(0.341722\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$-t\ln \left(1.004817\right)=\ln \left(0.341722\right)$

$\mathrm{Divide\:both\:sides\:by\:}\ln \left(1.004817\right)$

$\frac{-t\ln \left(1.004817\right)}{\ln \left(1.004817\right)}=\frac{\ln \left(0.341722\right)}{\ln \left(1.004817\right)}$

$-t=\frac{\ln \left(0.341722\right)}{\ln \left(1.004817\right)}$

$t=223.44650\dots$

Thus, t = 224 months.

1 year = 12 month thus, 30 years = 360 months.

Difference = 360 - 224 = 136 months

Thus, James will pay off his mortgage (360-224) 136 months early.

Answer from: Quest

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).