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2 years ago

If you deposit $25,000 now at 6.5% compounded quarterly, how much will you have in 30 years?

Mathematics

1 answer:

Mice21 [21]2 years ago

4 0

Answer:

$172,984.44

Step-by-step explanation:

We can use the formula

$A = P(1 + \frac{r}{n})^{nt}$ to compute the final amount

Here P is the principal amount, the original deposit = $25,000

r is the annual interest rate = 6.5% = 0.065 in decimal

n is the number of times the compounding takes place. Here it is quarterly so it is 4 times a year

t is the number of time periods ie 30 years

A is the accrued amount ie principal + interest

Computing different components,

$nt = 4 \times 30 = 120$

$r/n = 0.065/4 = .01625$

$1 + r/n = 1.01625$

$(1 + r/n)^{nt} = 1.01625^{120}$

Therefore

$A = 25000 \times 1.0625^{120} = 172,984.44$

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Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y

professor190 [17]

Answer:

Required total charge is $256\pi$ coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

$x^2=y^2\leq 16$ so that the charge density at (x,y) is,

$\rho (x,y)=2x+2y+2x^2+2y^2$

To find total charge on the disk let Q be the total charge and $x=r\cos\theta,y=r\sin\theta$ so that,

$Q={\int\int}_Q\rho(x,y) dA$ where A is the surface of disk.

$=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA$

$=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta$

$=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta$

$=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta$

$=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta$

$=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi$

$=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi$

$=256\pi$

Hence total charge is $256\pi$ coulombs per square meter.

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eimsori [14]

Answer:

Step-by-step explanation:

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