Complete the equations bellow.

The following integral requires a preliminary step such as long division or a change of variables before using the method of par

shtirl [24]

Division yields

$\dfrac{x^4+7}{x^3+2x} = x-\dfrac{2x^2-7}{x^3+2x}$

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

$\dfrac{2x^2-7}{x(x^2+2)} = \dfrac ax + \dfrac{bx+c}{x^2+2}$

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

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

$\dfrac{2x^2-7}{x(x^2+2)} = -\dfrac7{2x} + \dfrac{11x}{2(x^2+2)}$

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

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

7 0

2 years ago

The perimeter of a rectangular swimming pool is 160 feet . The length of the swimming pool is four times its width. Find the dim

Murrr4er [49]

Formula for Perimeter of Rectangle:

P = 2(L + W)

Plug in 160:

160 = 2(L + W)

L = 4W
So we can plug in '4W' for 'L' in the first equation.
160 = 2(L + W)

160 = 2(4W + W)

Combine like terms:

160 = 2(5W)

160 = 10W

Divide 10 to both sides:

W = 16

Now we can plug this back into any of the two equations to find the length.

L = 4W

L = 4(16)

L = 64

So the width is 16, and the length is 64.

6 0

3 years ago

Read 2 more answers

PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

andrew-mc [135]

Step-by-step explanation:

The sum of all angles in a triangle is always 180.

So we can set up an equation

3/2x+1/2x+x= 180

3/2+1/2x+2/2x=180

3x+1x+2x=360

6x=360

x=60

4 0

3 years ago

3. For the equation-4y = &r, what is the constant of variation? (1 point) ООО 02

alexandr1967 [171]

Answer:

3).

$- 4y = 8x \\ \frac{y}{x} = \frac{8}{ - 4} \\ \frac{y}{x} = - 2 \\ y = - 2x \\ hence \: constant \: is \: - 2$

4).

$y \alpha x \\ y = kx \\ where \: k \: is \: the \: constant \: of \: proportionality \\ when \: y = 24 \: and \: x = 8 \\ substitute \: in \: y = kx \\ 24 = 8k \\ k = \frac{24}{8} \\ k = 3 \\ value \: of \: constant = 3$

If it was by 10, k = 3 by 10

k = 30

5).

$general \: equation \: of \: line \\ y = mx + c \\ but \: m = \frac{2}{3} \: and \: c = 9 \\ substitute \: for \: m \: and \: c \: in \: y = mx + c \\ y = \frac{2}{3} x + 9 \\ alternatively \\ 3y = 2x + 27$

7 0

3 years ago

Peppermint candies per jar

Harlamova29_29 [7]

Leaf #2 because 27 and 23 are closest to 30

4 0

3 years ago