This looks like a science lab you have to do on your own
Monthly payments, P = {R/12*A}/{1- (1+R/12)^-12n}
Where R = APR = 4.4% = 0.044, A = Amount borrowed = $60,000, n = Time the loan will be repaid
For 20 years, n = 20 years
P1 = {0.044/12*60000}/{1- (1+0.044/12)^-12*20} = $376.36
Total amount to be paid in 20 years, A1 = 376.36*20*12 = $90,326.30
For 3 years early, n = 17 year
P2 = {0.044/12*60,000}/{1-(1+0.044/12)^-12*17} = $418.22
Total amount to be paid in 17 years, A2 = 418.22*17*12 = $85,316.98
The saving when the loan is paid off 3 year early = A1-A2 = 90,326.30 - 85,316.98 = $5,009.32
Therefore, the approximate amount of savings is A. $4,516.32. This value is lower than the one calculated since the time of repaying the loan does not change. After 17 years, the borrower only clears the remaining amount of the principle amount.
Perhaps we can treat this as a "distance to horizon" problem, the observer is 7 mi above the earth
One formula is: d+=+sqrt%28%283h%29%2F2%29 where d is in miles and h is in feet
Change 7 mi; 5280*7 = 36960 ft
so we have
d = sqrt%28%283960%2A3%29%2F2%29
d ~ 235.5 mi
looking in opposite directions a total of 2*235 = 471 mi can be viewed
Find the circumference of the earth
C = pi%2A8000
C = 25133 mi
:
The arc: 471%2F25133 * 360 = 6.7 degrees
:
This should be pretty close, because the earth is so large compared to the altitude. would appreciate if it you would let me know. C
U have to get two 5s for a 10 for top numbers
