Hi there
Payment per year
1,000×12months=12,000
total payment over the lifetime of the loan.
12,000×10years=120,000
Hope it helps
We have been given that a person places $6340 in an investment account earning an annual rate of 8.4%, compounded continuously. We are asked to find amount of money in the account after 2 years.
We will use continuous compounding formula to solve our given problem as:
, where
A = Final amount after t years,
P = Principal initially invested,
e = base of a natural logarithm,
r = Rate of interest in decimal form.
Upon substituting our given values in above formula, we will get:
Upon rounding to nearest cent, we will get:
Therefore, an amount of $7499.82 will be in account after 2 years.
According to all known laws
of aviation,
there is no way a bee
should be able to fly.
Its wings are too small to get
its fat little body off the ground.
The bee, of course, flies anyway
because bees don't care
what humans think is impossible.
Yellow, black. Yellow, black.
Yellow, black. Yellow, black.
Ooh, black and yellow!
Let's shake it up a little.
Check the picture below.
based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).
and that will give us two x-intercepts, at x = 3 and x = k.
since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.
now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.
let's find the y-intercept by setting x = 0 now,
![\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)](https://tex.z-dn.net/?f=%5Cbf%20y%3D0.5%28x-3%29%28x%2Bk%29%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28x-3%29%28x%2Bk%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsetting%20x%20%3D%200%7D%7D%7By%3D%5Ccfrac%7B1%7D%7B2%7D%280-3%29%280%2Bk%29%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28-3%29%28k%29%5Cimplies%20%5Cboxed%7By%3D-%5Ccfrac%7B3k%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20triangle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7Dbh%7D~~%20%5Cbegin%7Bcases%7D%20b%3D3%2Bk%5C%5C%20h%3Dy%5C%5C%20%5Cquad%20-%5Cfrac%7B3k%7D%7B2%7D%5C%5C%20A%3D1.5%5C%5C%20%5Cqquad%20%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B2%7D%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29)
now, we can plug those values on A = (1/2)bh,
![\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-2%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-2%29%5Cleft%28-%5Ccfrac%7B3%28-2%29%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%281%29%283%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-1%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-1%29%5Cleft%28-%5Ccfrac%7B3%28-1%29%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%282%29%5Cleft%28%20%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5)
