Points A,B, and C are collinear. Points M and N are the midpoints of segments AB and AC. Prove that BC = 2MN
2 answers:
Look at the picture.
1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
Hello ! (<span>I attached a picture of the drawing)</span>
<span>First, we draw a line with 3 points B,A,C in this order.Then </span><span>we will note the middle of AB with M and also the middle of AC with N.
</span>
<span></span><span>We can notice that BC = AB + AC
</span>
<span>M - the middle of AB ⇒ AM=MB=AB/2</span>
<span>N - the middle of AC ⇒ AN=NC=AC/2
</span>
<span>We can write MN as AM+AN , which means AB/2 + AC/2 . We can also write AB as BC-AC , so we will have :
</span>
<span>MN = AB/2 + AC/2</span>
<span>MN = (BC-AC)/2 + AC/2</span>
<span>MN = BC/2 - AC/2 + AC/2 ( AC/2 with -AC/2 are reduced)</span>
MN = BC/2 or BC = 2·MN
<span>First, we draw a line with 3 points B,A,C in this order.Then </span><span>we will note the middle of AB with M and also the middle of AC with N.
</span>
<span></span><span>We can notice that BC = AB + AC
</span>
<span>M - the middle of AB ⇒ AM=MB=AB/2</span>
<span>N - the middle of AC ⇒ AN=NC=AC/2
</span>
<span>We can write MN as AM+AN , which means AB/2 + AC/2 . We can also write AB as BC-AC , so we will have :
</span>
<span>MN = AB/2 + AC/2</span>
<span>MN = (BC-AC)/2 + AC/2</span>
<span>MN = BC/2 - AC/2 + AC/2 ( AC/2 with -AC/2 are reduced)</span>
MN = BC/2 or BC = 2·MN
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Answer:
Two zero pair
Step-by-step explanation:
We are given the following quadratic equation:
Since, this a quadratic equation we would have two zero pairs for this model.
Factored form of polynomial:
is the required factored form of the given polynomial.