Answer:
Answer to fourth part is :
Angle ABC= Angle DBC
So for fifth part reason answer is ASA congruency.
Step-by-step explanation:
Here we are given that CB bisects angle ABD and angle ACD.
So we have ,
Angle ABC= Angle DBC
Answer to fourth part is :
Angle ABC= Angle DBC
Now here we have two angles and one side equal in two triangles.
So we can say that ASA congruency fits in the best here.
So for fifth part reason answer is ASA congruency.
This is essentially asking how many different ways to partition 6 into 3 segments.
I am assuming "no ball in a box" is not allowed.
6 can be partitioned as
(4,1,1), (3,2,1), and (2,2,2)
So, calculate each partition, we get
(6 choose 4) + (6 choose 3)*(3 choose 2) + (6 choose 2) * (4 choose 2)
= 15 + 20*3 + 15*6
=165
I think the answer to this is .6 or 1.7
Answer:
7x−3=7k,x=7k+3,k∈ℤ
Step-by-step explanation:
3x ≡ 2 (mod 7)
3 is a solution, since 3×3−2=7.
All solutions are 3+7ℤ
y=3x−2/7=3(x−3)+7/7=3(x−3)/7+1
7x−3=7k,x=7k+3,k∈ℤ
