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Serjik [45]
3 years ago
9

Can u guys please help me with this question??!!! 10 points and brainy will be awarded PLEASE!?

Mathematics
1 answer:
Over [174]3 years ago
8 0

Answer:

The answer is the last option.

Step-by-step explanation:

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I trust those links^
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3x+6y = 27<br> x + 2y = 11<br> Solve by elimination
AlladinOne [14]

Answer:go0gle

gle come in clutch

Step-by-step explanation:

Next, let's solve 3x+2y = 10 for the variable y. Move the 3x to the right hand side by subtracting 3x from both sides, like this: 3x - 3x = 0. The answer is 2y. The answer is 10-3x.

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Find -sqrt(36<br> O t6<br> O -6<br> O 6<br> O 18
gavmur [86]
-sqrt(36) = -6

Hope this helps :)
8 0
3 years ago
Please help me with this
GREYUIT [131]

Ths phrase that represents the algebraic expression (3p + 6)/(7p - 9) will be D. the sum of three times a number and six divided by the difference of seven times the number and nine.

<h3>How to illustrate the algebra?</h3>

It should be noted that an algebra is simply used to show the relationship between variables.

Here, the phrase that represents the algebraic expression (3p + 6)/(7p - 9) will be the sum of three times a number and six divided by the difference of seven times the number and nine.

In conclusion, the correct option is D.

Learn more about algebra on:

brainly.com/question/723406

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7 0
2 years ago
Integration of ∫(cos3x+3sinx)dx ​
Murljashka [212]

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 }}}}}

6 0
3 years ago
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