Answer:
the norm of V = =11.357816691601..
Step-by-step explanation:




If you took

of a pie then only ate

of that slice, you ate

of

.

Simplify.

So, you ate 1/6 of the pie.
Answer: 1/10 is greater; niether its equal; 2 1/2, .25, -.2, -1/2, -4/5
Step-by-step explanation:
1/10 is a tenth and .09 is a hunderedth
I know these are all the factors of 12
1,2,3,4,6,12
<em> </em><em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>0</em><em>.</em><em>7</em><em>5</em><em>.</em><em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>