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
The nearest whole percent would just be 62%
Answer:
so yea what was the answer again
Step-by-step explanation:
Answer: +/- 7
Step-by-step explanation:
3x² = 147
To solve for x divide through by three first.
3x² = 147
x² = 49, now we take the square root of both side by trying to apply laws of indices.
√x² = √49
The square root will neutralize the effect of the square because
√a = a¹/² so (x²)¹/², and (x²)¹/² =
x²×¹/² = x, therefore the solution is
x = +/- 7.
Answer: y=7x
Step-by-step explanation:
The way I did it was, I looked at the chart. The highest it goes is 30 and 210. So for your answer, you have to multiply. 7x30 gives you 210 and there y=7x
