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3 years ago

How can we get Equation B from Equation A? A. 5 = -2 (x - 1) B. 5 = -2x + 2 How can we get Equation B from Equation A?

Mathematics

1 answer:

lesya [120]3 years ago

7 0

Answer:

Please check the explanation.

Step-by-step explanation:

Determining equation B from Equation A

Given the equation A

$5\:=\:-2\:\left(x\:-\:1\right)$

Apply distributive law $a\left(b-c\right)=ab-ac$

$5= -2x+2$ which is equation B

Hence, equation B can be obtained by implementing the distributive law operation on equation A.

Determining equation A from Equation B

Given the equation B

$5 = -2x + 2$

Factor out common term -2 on the right-hand side of the equation

$5=-2\left(x-1\right)$ which is the equation A

Hence, equation A can be obtained by factoring out the common term -2 on the right-hand side of equation B.

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Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

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If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

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If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

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