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Leokris [45]
3 years ago
10

Simplify -3(y+5)+3(2y+6)

Mathematics
2 answers:
likoan [24]3 years ago
5 0
-3(y + 5) + 3(2y + 6)
-3y - 15 + 6y + 18
3y + 3

Answer:
3y + 3

Hope this helps.
Fantom [35]3 years ago
3 0
Hey there! 

In order to find the answer to this question you can do the distributive property to solve this particular problem

-3(y) = -3
\\ \\ -3(5) = 15 
\\ \\ 
-3y + (-15)  
\\ \\ \\ \\  
3(2y) = 6y 
\\ \\ 
3(6) = 18 
\\ \\ \\ \\ \\ 
-3y + -15 + 6y + 18

Once when we do the distributive property we move onto the like terms 

Our Like terms: -3y + 6y \\  and \\ -15 + 18

So, lets solve our like terms 

-3y + 6y = 3y  \\ \\ \\ -15 + 18 = 3

Finally!  we came to our answer

Answer: 3y + 3

Good luck on your assignment and enjoy your day 

~
LoveYourselfFirst:)
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-------------------------------------
-------------------------------------

Work Shown:

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Can you find the limits of this ​
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Answer:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Constant]:                                                                                             \displaystyle \lim_{x \to c} b = b

Limit Rule [Variable Direct Substitution]:                                                             \displaystyle \lim_{x \to c} x = c

Limit Property [Addition/Subtraction]:                                                                   \displaystyle \lim_{x \to c} [f(x) \pm g(x)] =  \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}

Let's substitute in <em>x</em> = -2 using the limit rule:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}

Evaluating this, we arrive at an indeterminate form:

\displaystyle  \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}

Substitute in <em>x</em> = -2 using the limit rule:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{3(-2)^2}{4(-2)^3}

Evaluating this, we get:

\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit:  Limits

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