Aye the Answer is D ~hope that helps :)
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
We will need like a graph for that
Answer:
y = (-1/2)x + 5
Step-by-step explanation:
As we move from the y-intercept (0, 5) to the x-intercept (10,0), x increases by 10 units while y decreases by 5 units. Thus, the slope of this line is
m = rise / run = -5 / 10 = -1/2.
Since we know both the slope and the y-intercept of this line, let's use the slope-intercept form of the equation of a straight line: y = mx + b.
Substituting -1/2 for m and 5 for b, we get:
y = (-1/2)x + 5
Solve the "f" function with substitute 4 and solve the "g" function with what we get for the "f" function.
f(4) = 2(8) + 3
f(4) = 16 + 3
f(4) = 19
g(19) = 4(19) - 1
g(19) = 76 - 1
g(19) = 75
Best of Luck!
