X = jane, y = jasmine, z = jocelyn
x + y + z = 56
x = 3z
y = 2z + 2
3z + 2z + 2 + z = 56
6z + 2 = 56
6z = 56 - 2
6z = 54
z = 54/6
z = 9 <==== jocelyn
x = 3z
x = 3(9)
x = 27 <=== jane
y = 2z + 2
y = 2(9) + 2
y = 18 + 2
y = 20 <=== jasmine
X = -6y - 12
4x + 5y =-39
4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3
x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6
(x, y) = (6, -3)
Answer:
-2x-10
Step-by-step explanation:
just distribute the -2 :)
In this question, we made a license that does not allow repeated digit and letter. That means the probability count of letter and digit will be decreased by 1 for every roll.
There is 10 kinds of digits (0-9) and 26 kinds of letters (a to z). So, 2 digits and 5 letter would be:
(10 * 9) * (26 * 25 * 24 * 23 * 22)= 710424000 kinds of license plates
