a - b = c -(a - b) = -c b - a = -c
2a. (a, b and c are all equal.)
A.
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
Yes.
61 degrees (180 degrees in a triangle)
a+b = 17a+c = 36 ----> c = 36 - ab+c = 25 ----> c = 25 - bSo since both are c, we say36 - a = 25 -bFrom this : 11 = a - bLook at 1st eqn: a + b = 17Now we have: a - b = 11 ----> a = 11 + bSo we can substitute that into a + b = 17 to make:(11 + b) + b = 1711 + 2b = 17So b = 3If b = 3, then a + 3 = 17 , so a = 14, 3 + c = 25, so c = 22So a=14, b=3, c=22, so a+b+c = 14+3+22 = 39
A triangle with side a: 5, side b: 8, and side c: 11 units has an area of 18.33 square units.
a - b = c -(a - b) = -c b - a = -c
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
2a. (a, b and c are all equal.)
And what is the question?
A.
A=B , A-B=B-B , A-B =0 B=C , B-B=C-B, 0=C-B So A-B=0 but also C-B=0 A-B=C-B ...add +b ...A-B+B=C-B+B , A=C
by transitive property
Yes.
2a + 2b = 26 2b + c = 22 ===> b = 11 - c/2 4c + 2b = 46 ===> b = 23 - 2c From the 2nd and 3rd equations: 11 - c/2 = 23 - 2c 22 - c = 46 - 4c 22 - 46 = -4c + c ===> -24 = -3c ===> c = 8 Substitute c=8 into the 3rd equation: b = 23 - 16 = 7 Substitute b=7 into the 1st equation: 2a + 14 = 26 2a = 12 ===> a = 6