If you can do your homework at a rate of 5.3 minutes for every problem, how many problems could be done after 9.01 hours? Round
1 answer:
You might be interested in
The complete question in the attached figure
we know that
t<span>he centroid is always 2/3 the distance of each median to each vertex
TW=30
TW=TY+YW
TY=(2/3)*30---------> TY=20
then
YW=TW-TY-----------> YW=30-20-----------> YW=10
the answer is
YW=10</span>
we know that
t<span>he centroid is always 2/3 the distance of each median to each vertex
TW=30
TW=TY+YW
TY=(2/3)*30---------> TY=20
then
YW=TW-TY-----------> YW=30-20-----------> YW=10
the answer is
YW=10</span>
Method One
Your calculator might be able to do this. Mine does it like this.
6
nCr
2
=
15
Method 2
You could simply set up 6C2
This gives you
Method Three
You only have to do this a couple of times to see how the cancellation works.
After all the cancellation takes place you have
6*5/2 = 15
Your calculator might be able to do this. Mine does it like this.
6
nCr
2
=
15
Method 2
You could simply set up 6C2
This gives you
Method Three
You only have to do this a couple of times to see how the cancellation works.
After all the cancellation takes place you have
6*5/2 = 15