I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169
with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0
insert into I:
12x=0
x=0
-> only intersection is at (0,0) = option B
Hey there!
A quadrilateral has 360 degrees. So add all the degrees together to 360. 9x+1+5x-12+8x+61+90=360. 90 degrees is in the bottom left corner. Combine like terms. 22x+140=360. Subtract 140 from each side. 22x=220. Divide 22 on each side to get x=10. Plug in 10 for x in angle D. 5(10)-12=38. The measure of angle D is 38 degrees.
I hope this helps!
the answer is that each side is 6x feet long
The answer is C. (True) did some research and that’s what all my sources say ;)
