The amortization formua I'm familiar with assumes payments are made at the end of the period, so we'll use it for the part after the first payment has already been made.
.. A = 4,000
.. P = 500,000 -4000 = 496,000
.. i = 0.06
.. n = 12
.. t = to be determined
And the formula is
.. A = Pi/(n(1 -(1 +i/n)^(-nt))) . . . . . amortization formula with payments at the end of the period
.. 1 -(1 +i/n)^(-nt) = Pi/(An) . . . . . . rearrange to get "t" factor in numerator
.. 1 -Pi/(An) = (1 +i/n)^(-nt) . . . . . . get "t" factor by itself
.. log(1 -Pi/(An)) = -nt*log(1 +i/n) . . . . use logarithms to make the exponential equation into a linear equation
.. log(1 -Pi/(An))/(-n*log(1 +i/n)) = t . . . . divide by the coefficient of t
.. t = 16.1667 . . . . . years (after the first monthly withdrawal)
The plan will support withdrawals for 16 years and 3 months (195 payments).
Hello!
So you will use the Pythagorean Theron equation which is Asquare + bsquare=C square
Plug in what you are given to find the missing piece
A square+ (2.6)square= (6.2) square
Asquare+ 6.76= 38.44
(Subtract 6.76to both sides to isolate A)
Asquare= 31.68
(Take the square root of both sides to find out what A is)
A=5.628
Oyebolatoba Beginner
Let d = # digital cameras
Let m = # manual cameras
d + m = 240 equation 1
(d-82) = 3(m-26) equation 2
d - 82 = 3m - 78
d - 3m = 82 - 78
d - 3m = 16 equation 2 simplified
We have two equations with 2 unknowns:
d + m = 240
d - 3m = 16 subtract equation 1 from equation 2
---------------
-4m = -224
m = -224/-4
m = 56
d + m = 240
d = 240-m
d = 240-56
d = 184
He started with 184 digital cameras and 56 manual cameras
It is important so you can be more fluent or faster in knowing bigger problems
