X equal 7 so 49=x7 that is what I would do at least
Answer:
According to Angle Bisector Theorem, if Ray YZ is an angle bisector, then MZ and NZ are congruent.
EH= 23. It is shown that ray FH bisects EFG, EF and EH are perpendicular, FG and HG are perpendicular. Then EH and GH are congruent. Therefore since GH= 23, then EH= 23
To determine the number of possible arrangements for 6 out of 8, we should use combinations. That is
ₐC₆ = 8!/(6!2!)
Answer: b. Combination
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.
D. 5 5/6
Volume = l * b * h
= 2 1/2 * 3 1/2 * 2/3
= 5/2 * 7/2 * 2/3
= 35/6
= 5 5/6
