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MrRissso [65]
3 years ago
8

Which expression is equivalent to

Mathematics
2 answers:
Mkey [24]3 years ago
8 0
I believe A is the correct answer
AURORKA [14]3 years ago
7 0
<h3>The answer is option A</h3>

Step-by-step explanation:

<h3>\frac{ {x}^{ - a} }{ {x}^{ - b} {x}^{c}  }</h3>

Using the rules of indices

Since the bases are the same and are dividing we subtract the exponents

That's

<h3>\frac{ {x}^{d} }{ {x}^{e} }  =  {x}^{d - e}</h3>

So we have

<h3>\frac{ {x}^{ - a} }{ {x}^{ - b} {x}^{c}  }  =  {x}^{ - a -  - b - c}</h3>

We have the final answer as

<h2>{x}^{ (- a + b  - c)}</h2>

Hope this helps you

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Answer:

\boxed{\pink{\tt I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C}}

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\  dx

  • Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
  • Now , Rewrite using du and u .

\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I =  \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C }}}}}

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Step-by-step explanation:

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