Solve the following equations -
a1+a2=8 b1-b2=6 a1+b1=13 a2+b2=8
a1 = 3.5, b1 = 9.5, a2 =4.5 and b2 = 3.5
a1 + a2 = 8 and a2 + b2 = 8. So a1 + a2 = a2 + b2 which gives a1 = b2. Now take b1 - b2 = 6 and using a1 = b2 we get b1 - a1 = 6. we have a1 + b1 = 13, So adding these two equations we get 2b1 = 19, So b1 = 9.5 It is easy to get remaining values from the remaining equations.
Calculate the number of ordered triples (A1, A2, A3) such that:
A1 ∪ A2 ∪ A3 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A1 ∩ A2 ∩ A3 = ∅
Write your answer as 2a3b5c7d where a, b, c, d are non-negative integers.
Let a1 and a2 be the roots or the equation (x- a)(x - b) = c and c != 0, then what are the roots of the equation (x - a1)(x - a2) + c = 0?
If 1,4,3=13; 4,5,4=9; 5,6,5=11 and 7,7,6=7 then 9,8,7=?
