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3 years ago

Simplify the expression (8x^3y^2/4a^2b^4)^-2

1 answer:

3 0

Simplify the following:
((8 x^3 y^2 a^2 b^4)/4)^(-2)
8/4 = (4×2)/4 = 2:
(2 x^3 y^2 a^2 b^4)^(-2)
Multiply each exponent in 2 x^3 y^2 a^2 b^4 by -2:
(x^(-2×3) y^(-2×2) a^(-2×2) b^(-2×4))/(2^2)
-2×3 = -6:
(x^(-6) y^(-2×2) a^(-2×2) b^(-2×4))/2^2
-2×2 = -4:
(y^(-4) a^(-2×2) b^(-2×4))/(2^2 x^6)
-2×2 = -4:
(a^(-4) b^(-2×4))/(2^2 x^6 y^4)
-2×4 = -8:
b^(-8)/(2^2 x^6 y^4 a^4)
2^2 = 4:
Answer: 1/(4 x^6 y^4 a^4 b^8) thus the answer is A

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General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

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$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
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[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

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