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3 years ago

Lots of points if answer pls

Mathematics

2 answers:

mario62 [17]3 years ago

4 0

Answer:

The answer is D.

Lelu [443]3 years ago

3 0

Answer:

i think d

Step-by-step explanation:

surry if im wrong

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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

3 years ago

Pie graphs should be used when the data is represented as a percentage. True False

horrorfan [7]

Answer: True

Hope this helps!

3 0

3 years ago

Bobby has already ran 1 mile on his own, and he expects to run 1 mile during each track practice. How many track practices would

german

Answer:

In order to have ran 33 miles, Bobby would have to attend 32 track practices.

Step-by-step explanation:

Solving this problem entails of uncovering the amount of track practices Bobby must attend in order to have ran 33 miles. Start by reading the problem carefully to break down the information provided.

You can see that Bobby has already ran one mile on his own. This is important to remember for later. The problem also states that he expects to run one mile at every track practice.

Setting up an equation will help us solve. Here is how we could set up the equation:

(amount of miles already ran = 1) + (number of track practices = x) = (total miles to run = 33)

1 + x = 33

The equation is now in place. You can solve this, or isolate 'x', by using the subtraction property of equality. This means we will subtract one from both sides of the equation, thus isolating the variable.

1 + x = 33

1 - 1 + x = 33 - 1

x = 32

The variable is the only term left on the left side of the equation. This means Bobby must attend track practice 32 times in order to have ran 33 miles.

8 0

4 years ago

How much wrapping paper would you need to cover a rectangular box that is 18.25 inches by 12 inches by 3 inches if you need 10%

Reika [66]

Answer:

681 square inches

Step-by-step explanation:

The surface area of a rectangular prism of dimensions L, W, and H is given by ...

A = 2(LW +H(L+W))

The surface area of the given box is ...

A = 2(12·3 +18.25(12 +3)) = 2(36 +273.75) = 619.5 . . . square inches

The amount of wrapping paper required is 10% more than this value, so is ...

paper required = (1 +.10)(619.5) = 681.45 ≈ 681 . . . square inches

8 0

3 years ago

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the

ipn [44]

For each $x$ in the interval $0\le x\le2$ , the corresponding shell has radius $x+9$ (horizontal distance from $x$ to the rotation axis) and height $8-x^3$ (vertical distance between $y=8$ and $y=x^3$ ). A shell of radius $r$ and height $h$ has area $2\pi rh$ , so the volume is

$\displaystyle2\pi\int_0^2(x+9)(8-x^3)\,\mathrm dx=2\pi\int_0^272+8x-9x^3-x^4\,\mathrm dx=\boxed{\frac{1176\pi}5}$

8 0

3 years ago