The upper pair of angles 60 degrees and (2x-y) degrees are supplementary angles. This is because of the parallel lines. Note how they are same side interior angles. Therefore, (2x-y) and 60 combine to 180 degrees like so

(2x-y)+60 = 180

2x-y = 180-60 ... subtract 60 from both sides

2x-y = 120 ... call this equation 1

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Similarly, (2x+y) and 40 also combine to 180

(2x+y) + 40 = 180

2x+y = 180-40 ... subtract 40 from both sides

2x+y = 140 ... call this equation 2

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Line up equation 1 and equation 2. Then add straight down

$\ \ \ 2x-y = 120\\+ 2x+y = 140\\\hrule\\\\\\ \ \ \ 4x+0y = 260\\\\$

That becomes 4x = 260 which solves to x = 65 when you divide both sides by 4.

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If x = 65, then,

2x-y = 120

2(65)-y = 120

130 - y = 120

-y = 120-130

-y = -10

y = 10

2x+y = 140

2(65)+y = 140

130+y = 140

y = 140-130

y = 10

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Either way end up with x = 65 and y = 10

4 0

3 years ago

In two or more complete sentences, explain the theorem used in solving for the range of possible lengths of the third side, AB o

Luden [163]

You can use the Pythagorean Theorem to find the length of the third side AB (Identified as "x" in the figure attached in the problem), which says that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:
a² = b²+c²
As we can see the figure, the triangle does not have an angle of 90°, but it can be divided into two equal parts, leaving two triangles with a right angle. We already have the values of the hypotenuse and a leg in triangle "A" , so we can find the value of the other leg:
b = √(a²-c²) b = √(10²-4²) b = 9.16
With these values, we can find the hypotenuse in the triangle "B": x = √b²+c² x = √(9.16)²+(4)² x = 10

3 0

3 years ago

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