Answer:
For the code we have 3 selections.
The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}
So we have 5 options.
The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)
So here we have 27 - 5 = 22 options
The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.
Here we have 22 - 1 = 21 options.
The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:
C = 5*22*21 = 2310.
There are 2310 different possible codes
The most simplified answer is the greatest factor being six. Divide both by six and six is on the outside of parentheses to be equivalent making the answer 6(2s+18r) or -6(-2s-18r) if you used negatives as I would jot down negative and positive
A polynomial with roots a and b is (x - a)(x - b).
(x - 2)(x - (-1))(x - 4) = 0
(x - 2)(x + 1)(x - 4) = 0
has roots 2, -1, and 4.
Answer:
A. {x,y}={-2,-3}
// Solve equation [2] for the variable x
[2] x = 2y + 4
// Plug this in for variable x in equation [1]
[1] (2y+4) - y = 1
[1] y = -3
// Solve equation [1] for the variable y
[1] y = - 3
// By now we know this much :
x = 2y+4
y = -3
// Use the y value to solve for x
x = 2(-3)+4 = -2
B. [1] 3x=3y-6
[2] y=x+2
Equations Simplified or Rearranged :
[1] 3x - 3y = -6
[2] -x + y = 2
Solve by Substitution :
// Solve equation [2] for the variable y
[2] y = x + 2
// Plug this in for variable y in equation [1]
[1] 3x - 3•(x +2) = -6
[1] 0 = 0 => Infinitely many solutions
C.Step by Step Solution
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System of Linear Equations entered :
[1] 4x - y = 2
[2] 8x - 2y = 4
Solve by Substitution :
// Solve equation [1] for the variable y
[1] y = 4x - 2
// Plug this in for variable y in equation [2]
[2] 8x - 2•(4x-2) = 4
[2] 0 = 0 => Infinitely many solutions
