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Darya [45]
3 years ago
11

HELP PLZZ. Select the correct answer Which number best represents the slope of the graphed line?​

Mathematics
2 answers:
bija089 [108]3 years ago
7 0

Answer: A. -5

Step-by-step explanation:

i just took this

Gre4nikov [31]3 years ago
6 0
The answer is negative five so a
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What is the value of x? A) 10 B) 12 C) 15 D) 17
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Answer:

c) 15

Step-by-step explanation:

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3 years ago
1) find the equation of the line parallel to
In-s [12.5K]

<u>Answer:</u>

1) The equation of the line parallel to  x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is  2y = 5x -12

<u>Solution:</u>

<u><em>1) find the equation of the line parallel to  x-5y=6 and through (4,-2).</em></u>

Given, line equation is x – 5y = 6  

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }

y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14

Hence, the line equation is 5y = x -14

<u><em> 2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)</em></u>

\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line \times slope of given line = -1

slope of required line = -1 \times \frac{5}{-2}=\frac{5}{2}

And it passes through (2, -1)

Now, using point slope form

\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}

Hence, the line equation is 2y = 5x -12

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2x+4y=22 \\&#10;2x-2y=-8 \ \ \ |\times (-1) \\ \\&#10;2x+4y=22 \\&#10;\underline{-2x+2y=8} \\&#10;2x-2x+4y+2y=22+8 \\&#10;6y=30 \ \ \ |\div 6 \\&#10;y=5 \\ \\&#10;2x-2y=-8 \\&#10;2x-2 \times 5=-8 \\&#10;2x-10=-8 \ \ \ |+10 \\&#10;2x=2 \ \ \ |\div 2 \\&#10;x=1 \\ \\&#10;\boxed{(x,y)=(1,5)}
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