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3 years ago

The population of a town increased from 5,600 to 6,300 people. What was the percent of the increase?

2 answers:

8 0

To get the percentage, first, we must subtract 5,600 from 6,300.
6,300 - 5,600 = 700
.
Second, we must divide 700 by 5,600.
700 / 5,600 = 0.125
.
Then, move the decimal place two places to the right and add a percent sign (%).
0.125 = 12.5%
.
Therefore, the percent of increase of the population of the town is 12.5%.
:)

Mademuasel [1]3 years ago

4 0

X=6300×100/5600=112,5
112,5-100=12,5
<span>The percent of the increase is 12,5</span>

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Step-by-step explanation:

Given electric charge is dristributed over the disk,

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

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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