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Misha Larkins [42]
3 years ago
5

BRAINLIEST + 30PTS Use the form x^m • x^n = x^m+n to solve for (x2y)(x3y4) = x^5y^5

Mathematics
1 answer:
Dimas [21]3 years ago
7 0

Step-by-step explanation:

  • (x²y)(x³y⁴)
  • x²y×x³y⁴
  • x²×x³×y×y⁴
  • x^2+3×y^1+4
  • x⁵×y⁵
  • x⁵y⁵
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What is the equation of the line that is parallel to the line 2x+3y=-8 and passes through the point (2,-2)?
Leya [2.2K]

Equation of line passing through (2, -2) and parallel to 2x+3y = -8 is y=\frac{-2 x}{3}+\frac{-2}{3}

<h3><u>Solution:</u></h3>

Need to write equation of line parallel to 2x+3y=-8 and passes through the point (2, -2)

Generic slope intercept form of a line is given by y = mx + c

where "m" = slope of the line and "c" is the y - intercept

Let’s first find slope intercept form of 2x+3y=-8 to get slope of line

\begin{array}{l}{2 x+3 y=-8} \\\\ {=>y=\frac{-2 x-8}{3}} \\\\ {\Rightarrow y=-\frac{2}{3} x-\frac{8}{3}}\end{array}

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c,

\text {for line } 2 x+3 y=-8, \text { slope } m=-\frac{2}{3}

We know that slopes of parallel lines are always equal

So the slope of line passing through (2, -2) is also m=-\frac{2}{3}

Equation of line passing through (x_1 , y_1) and having slope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=2 \text { and } y_{1}=-2

Substituting the values in equation of line we get

(y-(-2))=-\frac{2}{3}(x-2)

\begin{array}{l}{\Rightarrow y+2=\frac{-2 x+4}{3}} \\\\ {=>3(y+2)=-2 x+4} \\\\ {=>3 y+6=-2 x+4} \\\\ {3 y=-2 x-2}\end{array}

y=\frac{-2 x}{3}+\frac{-2}{3}

Hence equation of line passing through (2 , -2) and parallel to 2x + 3y = -8 is given as y=\frac{-2 x}{3}+\frac{-2}{3}

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