DanNr
Answered

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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.

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 class=
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 class=
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 class=

Sagot :

940172

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