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
Yes you are correct. Nice work.
(h,k) = (10,-4) is the center; r = 2 is the radius
(x-h)^2 + (y-k)^2 = r^2
(x-10)^2 + (y-(-4))^2 = 2^2
<h3>(x-10)^2 + (y+4)^2 = 4</h3>
Answer:
116
Step-by-step explanation:
8 x 7 = 56
12 - 7 = 5
5 x 12 = 60
60 + 56 = 116
The answer for the question is <span>0.03</span>
