If u are using i as a variable the answer would be i= 3.
1 quarter note=2 eighth notes
1 eight note=2 sixteenth notes
so
1 quarter note=4 sixteenth note
so
times 3 both sides
3 quarter notes=12 sixteenth notes
he elements of the Klein <span>44</span>-group sitting inside <span><span>A4</span><span>A4</span></span> are precisely the identity, and all elements of <span><span>A4</span><span>A4</span></span>of the form <span><span>(ij)(kℓ)</span><span>(ij)(kℓ)</span></span> (the product of two disjoint transpositions).
Since conjugation in <span><span>Sn</span><span>Sn</span></span> (and therefore in <span><span>An</span><span>An</span></span>) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.
Has to be correct. I did it on paper and checked it. Then I checked it with a calculator
(3x-10) (3x-10)
Both of the 10's have to be negative because -10 x -10 = 100
and if you use foil -30 + -30 = -60
