The numbers in this problem are ordered pairs, which are points on a graph.
These are (10, 20), (-10, 20), (-10, -10), and (10, -10).
To find the area and perimeter of this shape, you must first find the distance between each point.
Distance between (10, 20) and (-10, 20):
Since the y-value remains the same here, we just have to find the difference in x-values.
This means 10 - (-10)
A negative being subtracted is the same as a positive being added.
That means 10 - (-10) is the same as 10 + 10.
10 + 10 = 20, so the distance between (10, 20) and (-10, 20) is 20 units.
Distance between (-10, 20) and (-10, -10):
The x-values are the same here so just find the difference between the y-values.
20 - (-10) = 20 + 10 = 30
The distance between the (-10, 20) and (-10, -10) is 30 units.
Distance between (-10, -10) and (10, -10):
The y-values are the same so just find the difference between the x-values.
10 - (-10) = 10 + 10 = 20
The distance between (-10, -10) and (10, -10) is 20 units.
Distance between (10, -10) and (10, 20):
The x-values are the same so find the difference between the y-values.
20 - (-10) = 20 + 10 = 30
The distance between (10, -10) and (10, 20) is 30 units.
So now we know the side lengths of the room are 20 units, 30 units, 20 units, and 30 units.
To find the perimeter, add all the side lengths together.
20 + 30 + 20 + 30 = 100
The perimeter of the room is 100 units.
To find the area, multiply the length by the width.
The length is 20 units and the width is 30 units.
20 • 30 = 600
The area of the room is 600 units.
Final answers:
Perimeter = 100
Area = 600
Hope this helps!
X=2 because first you have to combine like terms. Than you have to bring the Varible on one side. Finally you solve for the Varible and you get the answer2.
Answer:
A system of linear equation could only have 1 solution. This is because the straight lines will only have to meet, cross, or intersect each other once.
A system of linear equation could only have 1 solution. This is because the straight lines will only have to meet, cross, or intersect each other once.
There are many different methods in arriving to the final answer. However, errors cannot be perfectly avoided. One of these errors to mistakenly identify equations as linear. It is important that we know that the equations we are dealing with are of exact or correct characteristics.
Also, if she had used substitution method, she might have mistakenly taken the value of one variable for the other.
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