You can calculate it using the law of cosines: c^2=a^2+b^2-2*a*b*cos(C)
your triangle is
CD=15=a
CE=?=b
DE=CE+3=b+3=c
and C=90°
-> insert those values, with c substituted with b+3 to remove c
c^2=a^2+b^2-2*a*b*cos(C)
(b+3)^2=15^2+b^2-2*15*b*cos(90)
cos(90)=0->
(b+3)^2=15^2+b^2
b^2+2*3*b+3^2=225+b^2
6b+9=225
6b=216
b=36=CE
DE=CE+3=36+3=39
Answer:
8
Step-by-step explanation:
13x - 6y = 22
x - y = 6
6(x-y) = 6*6
6x - 6y = 36
subtract both the equations
13x - 6x -6y + 6y = -14
7x = -14
x = -2
13*-2 -6y=22
-26 -6y=22
-6y = 48
y = -8
answer: x = -2
y = -8
Hello.
JK (or any other combination of letters) doesn't stand for anything in geometry. The purpose of these letters is to reference certain angles and sides of a figure without having to describe the angles/sides in relation to other angles/sides to know exactly which angle or side is being referenced.
Hope this helps!
