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Slav-nsk [51]
2 years ago
14

Find the value of k so that (x + 3) a factor of H(x) = 3x ^ 3 - 2x ^ 2 + kx - 3

Mathematics
1 answer:
gavmur [86]2 years ago
4 0

Step-by-step explanation:

By Factor Theorem, (x + 3) is a factor of H(x) if and only if H(-3) = 0.

3(-3)³ - 2(-3)² + (-3)k - 3 = 0

-81 - 18 - 3k - 3 = 0

3k = -102

k = -34.

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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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p(x)

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Which are the figures shown below.

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