Answer:
1st piece = Piece A = 5 inches
2nd piece = Piece B = 10 inches
Third Piece = Piece C = 27 inches
Step-by-step explanation:
A 42-inch piece of steel is cut into 3 pieces
Let the lengths be represented as:
1st piece = Piece A
2nd piece = Piece B
Third Piece = Piece C
Hence:
A + B + C = 42
The 2nd piece is twice as long as the 1st piece
B = 2A
The 3rd piece is 2 inches more than 5 times the length of the first piece.
C = 5A + 2
Hence: we substitute
A + B + C = 42 inches
A + 2A + 5A + 2 = 42 inches
8A = 42 - 2
8A = 40 inches
A = 40 inches/8
A = 5 inches
Solving for B
B = 2A
B = 2 × 5 inches
B = 10 inches
Solving for C
C = 5A + 2
C = 5 × 5 + 2
C = 25 + 2
C = 27 inches
Therefore, the lengths of the 3 pieces =
1st piece = Piece A = 5 inches
2nd piece = Piece B = 10 inches
Third Piece = Piece C = 27 inches
Answer:
197
Step-by-step explanation:
Assuming the question is asking for the sum of 14 squared and 1 cubed, 14 squared is 196 and 1 cubed is 1. Adding these together is 196+1 which gives 197.
It would be A x=1/2 i hope this helped
If b is in the first position then c can be in any 1 of the remaining 6 positions.
If we start with ab then the letter c can be in any one of 5 positions and if we have aab there are 4 possible positions for c and so on.
So the total number of possible sequences where b comes first = 6+5+4+3+2+1 = 21.
The same argument applies when c comes before b so that gives us 21 ways also.
So the answer is 2 *21 = 42 different sequences.
A more direct way of doing this is to use factorials:-
answer = 7! / 5! = 7 * 6 = 42.
( We divide by 5! because of the 5 a's.)
