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mestny [16]
2 years ago
5

What is the slope and y-Intercept of the equation 3(y – 2) + 6(x + 1) - 2 = 0?

Mathematics
2 answers:
aliina [53]2 years ago
5 0

⇒ 3(y - 2) + 6(x + 1) - 2 = 0

⇒ 3y - 6 + 6x + 6 - 2 = 0

⇒ 3y = -6x + 2

⇒ y = (-6x + 2)/3

⇒ y = -2x + ⅔

<h3><u>FINAL </u><u>ANSWER</u><u> </u><u>:</u></h3>

Slope = -2

Y - intercept = +⅔

<h3><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u> </u><u>:</u></h3>

Option (a) is the correct answer .

Ne4ueva [31]2 years ago
4 0

Answer:

  • A. Slope = -2, y-intercept = 2/3

Step-by-step explanation:

<u>Convert the given equation into slope-intercept form:</u>

  • 3(y - 2) + 6(x + 1) - 2 = 0
  • 3y - 6 + 6x + 6 - 2 = 0
  • 3y + 6x - 2 = 0
  • 3y = - 6x + 2
  • y = - 2x + 2/3

The slope is -2, the y-intercept is 2/3

Correct choice is A

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Can someone help me find the equivalent expressions to the picture below? I’m having trouble
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Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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