Hi, I need help with classwork. I am only really great at geometry, so I would appreciate it if someone lend a hand, thank you.



Answer:
1: x=16
2: I'm not sure
3: I'm not sure
4a: Yes (Diagonals Bisect)
4b: Yes (Opposite Angles Congruent)
4c: Yes (Opposite Sides Congruent and Parallel)
4d: Yes (Opposite Sides Parallel/Opposite Angles Congruent)
5: (7,9)
6a: AB=4/3 CD= 4/3
6b: AB=5 CD= 10
6c: No, it doesn't any of the characteristics of a parallelogram
Explanation:
1: Consecutive angles are supplementary because it is a proven parallelogram
[tex]3x+5+9x-17=180\\12x-12=180\\12x=192\\x=16[/tex]
2: I haven't learned this yet
3: I haven't learned this yet
4a: See Above
4b: See Above
4c: See Above
4d: See Above
5: [tex]mp(x)= \frac{x_{1}+x_{2}}{2} \\ mp(y)= \frac{y_{1}+y_{2}}{2} \\[/tex] [tex]4=\frac{1+x}{2} \\8=1+x\\x=7[/tex] [tex]6=\frac{3+y}{2}\\12=3+y\\y=9[/tex]
6a:[tex]\frac{7-3}{0-(-3)}=\frac{4}{3} \\\frac{-1-7}{2-8}=\frac{-8}{-6} =\frac{4}{3}[/tex]
6b: [tex]d=\sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}\\d=\sqrt{(0-(-3))^{2}+ (7-3)^{2}}\\d=\sqrt{(3)^{2}+ (4)^{2}}\\d=\sqrt{9+ 16}\\d=\sqrt{25}\\d(AB)=5[/tex] [tex]d=\sqrt{(x_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}}\\d=\sqrt{(-1-7)^{2}+ (2-8)^{2}}\\d=\sqrt{(-8)^{2}+ (-6)^{2}}\\d=\sqrt{64+ 36}\\d=\sqrt{100}\\d(AB)=10[/tex]
6c: See Above