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aniked [119]
3 years ago
5

Can a segment have a more than one bisector

Mathematics
2 answers:
dmitriy555 [2]3 years ago
8 0
No. All segments have one perpendicular bisector.
mel-nik [20]3 years ago
4 0
For every line segment there is one perpendicular bisector that passes through the midpoint.there are infinitely many bisectors but only one perpendicular bisector for any segment.
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Find the center and radius of the circle with equation (x + 8)2 + (y + 3)2 = 49.
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Center is at (h,k) in standard form:
(x-h)2 + (y-k)2 = r2
Center: (-8, -3)
Radius (r): sr49 = 7
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Directions: Complete all 3 questions. Make sure you include a graph, work and conclusion.
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Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.

I think in these problems the first step is to express each side as a vector.  A vector is the difference between points.  When two sides have the same vector (or negatives) it means they're parallel and congruent.  So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL.  That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.

Let's calculate these vectors:

IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)

LK = K - L = (7, 6) - (3, 3) = (4, 3)

IJ = LK, so far so good

(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)

JK = K - J = (7,6) - (2,6) = (5,0)

IL = L - I = (3, 3) - (-2, 3)  = (5, 0)

Those are the same too.    

Now we have to show IJ ≅ JK

The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK

We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.

8 0
3 years ago
Log 64 = x. Find X<br> ⁴
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Change 64 to index notation

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