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a_sh-v [17]
2 years ago
5

Write the linear equation in slope-intercept form 2x - 3y = 6

Mathematics
1 answer:
noname [10]2 years ago
7 0
2x - 3y = 6
3y = -2x + 6

y = -\frac{2}{3}x + 2

m = -\frac{2}{3}
c  = 2

Linear Equation: 
y = -\frac{2}{3}x + 2
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H(x)=-2x-5, find h(-2)
Vedmedyk [2.9K]

Answer:

h(-2) = -1

Step-by-step explanation:

h(x)=-2x-5

Let x= -2

h(-2)=-2*-2-5

      = 4 -5

h(-2) = -1

5 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
Find the zeros of the polynomial p(x) 5-10x​
Charra [1.4K]

Answer:

Step-by-step explanation:

The given polynomial is :

p(x) = 5-10x​

We need to find the zeros of the above polynomial. To find it, put p(x) = 0

5-10x​ = 0

Subtract 5 from both sides

5-10x​-5=0-5

-10x=-5

or

10x=5

Divide both sides by 10.

x = 0.5 = 1/2

Hence, the zeros of the polynomial is 1/2.

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