With 5 elements in A={20,1,6,10,11}, there are 2^5=32 possible subsets, including the null set, and A itself. Any subset that is identical to A is NOT a proper subset. Therefore there are 31 proper subsets, plus the subset {20,1,6,10,11}.
The subsets are: null set {} (has no elements) ........total 1 {20},{1},{6},{10},{11}.......................total 5 {20,1},{20,6}...{10,11}.....................total 10 {20,1,6},{20,1,10},...{6,10,11}.........total 10 {20,1,6,10}...{1,6,10,11}.................total 5 {20,1,6,10,11}.................................total 1 Altogether 32 subsets.
6(2) -2(-5) Substitute the values given into the expression and multiply. 12+10 ... 1. 2x-(3-4y), when x = 5 and y=-3. 215)-(3-4(-3)). 10-(3+12). 10-15. 2. ... + (3 - 3 x + 1 +. 6-3x = 4x + 4. Bebetto. 2=7x -. - 4-2=34+12. - ý 2=2y+12. + ... y = 2x-1. Simplify. Subtract 2x from each side. Multiply both sides by -1. 2+3 m.
Step-by-step explanation:
If 2x + y = 7 and x + 2y = 5, then (x + y)/3 = (A) 1 (B) 4/3 (C) 17/5 (D) ... Divide both sides by 3 to get: x + y = 4 ... 2*(x+2y = 5) equals 2x+4y=10 ... or subtracting one from the other and it likely happens 90% times that ... If a question involves a system of two equations, and we're asked to find the value of ONE