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3 years ago

R - 9S = 2 3R - 3S = -10

1 answer:

8 0

R-9s=2 add 9s to both sides

r=2+9s making 3r-3s=-10 become:

3(2+9s)-3s=-10 perform indicated multiplication on left side

6+27s-3s=-10 combine like terms on left side

6+24s=-10 subtract 6 from both sides

24s=-16 divide both sides by 24

s=-16/24

s=-2/3, making r-9s=2 become:

r-9(-2/3)=2 perform indicated multiplication on left side

r+6=2 subtract 6 from both sides

r=-4

So s= -2/3 and r= -4

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use Pythagoras $a^2 +b^2 = c^2$ where c is the longest side
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4 years ago

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4 years ago

What is an equation of the line that passes through the point ( 6 , − 2 ) (6,−2) and is perpendicular to the line 6 x + y = 2 6x

Diano4ka-milaya [45]

Answer:

An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

Step-by-step explanation:

We know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept.

Given the line

6x+y=2

Simplifying the equation to write into the slope-intercept form

y = -6x+2

So, the slope = -6

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

Thus, the slope of the perpendicular line will be: -1/-6 = 1/6

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 1/6 and the point (6, -2)

$y-\left(-2\right)=\frac{1}{6}\left(x-6\right)$

$y+2=\frac{1}{6}\left(x-6\right)$

subtract 2 from both sides

$y+2-2=\frac{1}{6}\left(x-6\right)-2$

$y=\frac{1}{6}x-3$

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

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3 years ago

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Dmitry [639]

Answer:

Step-by-step explanation:

62 is at least bigger than all the other numbers meaning that regardless of x or y axis it will always be much further away from all the other points by at lease 60.

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