Melanie said:
Every angle bisector in a triangle bisects the opposite side perpendicularly.
A 'counterexample' would show an angle bisector in a triangle that DOESN'T
bisect the opposite side perpendicularly.
See my attached drawing of a counterexample.
Both of the triangles that Melanie examined have
equal sides on both sides
of the angle bisector. That's the only way that the angle bisector can bisect
the opposite side perpendicularly. Melanie didn't examine enough different
triangles.
Answer:
n=5/4 or n=-1/4
Step-by-step explanation:
16n^2 - 10n-5=6n
16n^2-16n-5=0
(4n)^2-2.4n.2+2^2 = 2^2 + 5
(4n- 2)^2=9
4n-2 = 3 or 4n-2=-3
4n=5 or 4n=-1
n=5/4 or n = - 1/4
<h3>
Answer: x-2</h3>
Explanation:
If x > 2, then x-2 > 0 after subtracting 2 from both sides.
Since x-2 is always positive when x > 2, this means the absolute value bars around the x-2 aren't needed. The results of |x-2| and x-2 are perfectly identical.
For example, if we tried something like x = 5, then
- x-2 = 5-2 = 3
- |x-2| = |5-2| = |3| = 3
Both outcomes are 3. I'll let you try other x inputs.
So because |x-2| and x-2 are identical, this means |x-2| = x-2 for all x > 2.
In short, we just erase the absolute value bars.
2 and 1/5
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3rd option