I believe it would be 6. the first number and exponent are usually your quadratic terms
C = 2m^2 + m
D = 2 - 6m + 2m^2
2C = 2(2m^2 + m) = 4m^2+2m
2D = 2(2-6m+2m^2) = 4-12m+4m^2
2C - 2D =
4m^2+2m-(4-12m+4m^2) =
4m^2+2m-4+12m-4m^2 =
0m^2 + 14m -4 =
14m - 4
Answer: Yes, a triangle can have sides of those lengths.
Step-by-step explanation:
Answer:
0.0000001 (one ten millionth)
Step-by-step explanation:
first number and second number and third number = Total
78 possibilities x 78 possibilities x 78 possibilities = 234
Since "order" matters, this is a permutation.
So, this can be calculated using: ₇₈P₃
"Permutation lock" would be a more appropriate name.
