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-BARSIC- [3]
3 years ago
14

Solve the equation: 2 (x + 2x) - 36 A) X=-18 B) x = 18 cx = -6 D) x = 6

Mathematics
1 answer:
bazaltina [42]3 years ago
5 0
<h2><u>Answer:</u></h2><h2>\boxed{x=6}<u></u></h2><h2><u></u></h2><h2><u>Solution Steps:</u></h2>

______________________________

<h3>1.) Divide both sides by 2:</h3>
  • 2 ÷ 2= Cancels Out
  • 36 ÷ 2=18

<em>  - We do this first because we need the get rid of the parenthesis in order to solve this like a regular problem. In this case the only way to do so was to divide by 2 to separate the Number/Variable from the parenthesis. </em>

<u>Equation at the end of Step 1:</u>

  • <u />x+2x=18<u />

<h3>2.) Combine x and 2x:</h3>
  • x+2x=3x

<em>  - We do this because we can't have 2 variables at the end of the equation so it's best to just combine them. </em>

<u>Equation at the end of Step 2:</u>

  • <u />3x=18<u />
<h3></h3><h3>3.) Divide both sides by 3:</h3>
  • 3x ÷ 3=x
  • 18 ÷ 3=6

<em>  - We divided for the last step since their was 1 variable connected by a number and another number on the opposite side of the equal. </em>

<em></em>

______________________________

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Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of −3
RSB [31]

Answer:

\Large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Step-by-step explanation:

Hello,

Based on the indication, we can write this polynomial as below, k being a real number that we will have to identify (degree = 3 and we have three zeroes -3, -1, and 2).

   \Large \boxed{k(x+3)(x+1)(x-2)}

We know that the point (1,10) is on the graph of this function, so we can say.

k(1+3)(1+1)(1-2)=10}\\\\4*2*(-1)*k=10\\\\-8k=10\\\\k=\dfrac{10}{-8}=-\dfrac{5}{4}

Then the solution is:

\large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

8 0
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A
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Answer:

D. Zacpszcap. is probably the answer

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