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Flura [38]
3 years ago
9

Please help me with this.

Mathematics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

3

Step-by-step explanation:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

3^4 +9^2÷3^3-9^2

81+81÷27-81

81+3-81

3

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Answer:

12

Step-by-step explanation:

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3 years ago
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 <em>a</em>, <em>b</em>, and <em>c</em> such that

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

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which gives <em>a</em> + <em>b</em> = 2, <em>c</em> = 0, and 2<em>a</em> = -7, so that <em>a</em> = -7/2 and <em>b</em> = 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 <em>y</em> = <em>x</em> ² + 2 and d<em>y</em> = 2<em>x</em> d<em>x</em> 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
How to Simplify. (–21 × 3) + (15 ÷ –3)
vladimir2022 [97]

Answer:

-68 is your answer

Step-by-step explanation:

Remember to follow PEMDAS (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction)

First, solve each parenthesis

(-21 x 3) = -63

(15/-3) = -5

Next, add

(-63) + (-5) = -63 - 5 = -68

-68 is your answer

~

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Andy's soccer team scored 5 goals in each of 7 games and 6 goals in another game.How many goals did Andy's team score?Write an e
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Answer:

Step-by-step explanation: 41

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If you have a 5 letters A,B,C,D,E and you can put them in any order how many ways can you place them
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The first letter can be any one of 5.  For each of those . . .
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