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AVprozaik [17]
3 years ago
15

Is it possible to have a bisector that is not perpendicular to the segment that it bisects?

Mathematics
1 answer:
Usimov [2.4K]3 years ago
7 0
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Line segments JK and JL in the xy-coordinate plan both have a common endpoint J(-4,11) and midpoints at M1 (2,16) and M2 (-3,5),
OleMash [197]
The first part of the question is not needed 
we only need to know M1 and M2
(y2-y1)^2+(x2-x1)^2=(your answer)^2
(5-16)^2+(-3-2)^2=(your answer)^2
(-11)^2+(-5)^2=(your answer)^2
121+25=(your answer)^2
146=(your answer)^2
√146=12.0830459735945721
nearest tenth=12.1
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Step-by-step explanation:

7 0
3 years ago
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Find the length indicated. find CE
torisob [31]
Okay, to find length CE, your going to know the value of <em>x</em>. Length BC + CE = BD + DE.
3x+47+x+26=27+x+10
Simplify the equation to get
4x+73=37+x
you can choose one of four ways to continue, but I will choose to subtract x
3x+73=37
Subtract 73 from both sides of the equal sign
3x=-36
divide by 3 on both sides of the equal sign to get the value of x
x=-12

Now, plug in -12 for x in length CE to get -12+26=14
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3 years ago
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lesya692 [45]

Answer:

the third one is the correct answer

Step-by-step explanation:

hope that helps

7 0
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Help find the value of x
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Answer: the leg x would equal 76.8, or 77 rounded. The actual number would be 76.8374908492.
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