Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
combining like terms in a polynomial is adding or subtracting together terms that have the same degree in the exponents.
example. 3x^3 + 5x^6 + 2x^3 +7x + 9x^6
= 7x + 5x^3 + 14x^6
Answer:
I don't think any are a correct graph of the function. But here are the ordered pairs. (1,0),(0,-2),(-1,-4)
Step-by-step explanation:
Take three values for x and plug them in for your y's. For instance x values of 1,0,-1. Plug each in to find the y value. 2(1) - 2 =y so (1,0)
Answer:
A
Step-by-step explanation:
you don’t make money for working 0 hours
