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Doss [256]
2 years ago
12

Could you explain how to solve this problem? 10n = 40

Mathematics
2 answers:
kondor19780726 [428]2 years ago
8 0

Answer:

n=4

Step-by-step explanation:

n is 4 because 10(4)=40 the n represents a number and because there is nothing in between we know to multiply them, because we know that do the opposite of the question which is 40/10=4 meaning that n=4

irina1246 [14]2 years ago
7 0
Since were trying to solve for n, you have to isolate it so you would divide both sides by 10, and that will give you n = 4
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If a=2, b= -1, c= -3 and d= -2 find :<br> 1/2bc-6ac+2ab<br> 1/4(ac)² +1/6ac
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34

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Step-by-step explanation:

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7 0
3 years ago
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.

3 0
3 years ago
What is 0 divided by 6
Pepsi [2]
0 / 6 is equal to 0 because anything that divides 0 will always be zero but never if the problem is 6/0
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