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3 years ago

Through (-4,3) , perpendicular to m=-2/3

Mathematics

1 answer:

stiv31 [10]3 years ago

5 0

Answer:

The equation of the line:

$y=\frac{3}{2}x+9$

Step-by-step explanation:

Given

The point (-4, 3)

m = -2/3

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -2/3

perpendicular to m = -1/m

$=-\frac{1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}$

Using the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 3/2 and the point (-4, 3)

$y-3=\frac{3}{2}\left(x-\left(-4\right)\right)$

$y-3=\frac{3}{2}\left(x+4\right)$

Add 3 to both sides

$y-3+3=\frac{3}{2}\left(x+4\right)+3$

$y=\frac{3}{2}x+9$

Thus, the equation of the line:

$y=\frac{3}{2}x+9$

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Now for partial fractions: you're looking for constants a, b, and c such that

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$\implies 2x^2 - 7 = a(x^2+2) + (bx+c)x = (a+b)x^2+cx + 2a$

which gives a + b = 2, c = 0, and 2a = -7, so that a = -7/2 and b = 11/2. Then

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Now, in the integral we get

$\displaystyle\int\frac{x^4+7}{x^3+2x}\,\mathrm dx = \int\left(x+\frac7{2x} - \frac{11x}{2(x^2+2)}\right)\,\mathrm dx$

The first two terms are trivial to integrate. For the third, substitute y = x ² + 2 and dy = 2x dx to get

$\displaystyle \int x\,\mathrm dx + \frac72\int\frac{\mathrm dx}x - \frac{11}4 \int\frac{\mathrm dy}y \\\\ =\displaystyle \frac{x^2}2+\frac72\ln|x|-\frac{11}4\ln|y| + C \\\\ =\displaystyle \boxed{\frac{x^2}2 + \frac72\ln|x| - \frac{11}4 \ln(x^2+2) + C}$