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nadezda [96]
3 years ago
6

A polygon has the vertices P(-2,6), Q(4,6), R(4,1) and S(-2,1). What is the length of side RS?

Mathematics
1 answer:
inessss [21]3 years ago
8 0
It is 6 units because you find the absolute value of the two different coordinates, in this case it is 4 and -2. Since they belong in different quadrants (one x or y value is positive and the other is negative) you add them. If they are both in the same quadrant, you subtract them. 
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Simplify. Consider all cases. |14−(x−6)|
yanalaym [24]

Answer:

20-x; -20+x; -20-x; 20+x

Step-by-step explanation:

|14-x+6|

|20-x|

20-x; -20+x; -20-x; 20+x

4 0
3 years ago
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Using a sheet of graph paper, solve the following system of equations graphically. Be sure to show any work, use a straight edge
madam [21]

Answer:

<h2> (-1,1)</h2>

Step-by-step explanation:

Given system is

\left \{ {{x+y=0} \atop {x-y+2=0}} \right.

If we graph, the solution would be the interception point between these two lines, because each linear equation represents a line. The lines are attached.

In the graph, you can observe that the solution is the point (-1,1), that is x = -1 and y = 1.

Now, if we want to solve this system by another method, we can just sum both equations and solve for <em>x</em>

\left \{ {{x+y=0} \atop {x-y+2=0}} \right.

2x+2=0\\x=\frac{-2}{2}=-1

Then, we replace this value in one equation to find the other value

x+y=0\\-1+y=0\\y=1

Therefore, the solution is (-1,1)

4 0
3 years ago
Line ef is tangent to circle g at point h. segment gh is a radius of circle g. what can be concluded about triangle fhg?
solong [7]
We know that

A line tangent to a circle is perpendicular to the radius to the point of tangency.
so
<span>Line ef is </span>perpendicular to the segment gh

hence
Triangle FHG is a right triangle
3 0
3 years ago
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Expand to write an equivalent expression: 4(-8x+6y)
Naddika [18.5K]

Answer:

-32x + 24y

Step-by-step explanation:

4(-8x + 6y) ⇒ -32x + 24y

4 0
3 years ago
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M=-4/3, point (0,-12)
postnew [5]
Y - y1 = m(x - x1)
slope(m) = -4/3
(0,-12)....x1 = 0 and y1 = -12
sub
y - (-12) = -4/3(x - 0) =
y + 12 = -4/3(x - 0) <=== point slope form

y + 12 = -4/3x 
y = -4/3x - 12 <=== slope intercept form

y = -4/3x - 12
4/3x + y = -12
4x + 3y = -36 <=== standard form
3 0
3 years ago
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