The three counters shown in the table are used for a board game. if the counters are tossed , how many ways can at least one counter with side a occur?

Total ways = {AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB}

Ways at least one counter with side A = {AAA, AAB, ABA, BAA, ABB, BAB, BBA}

Hence, the number of ways at least one counter with side A = 7.

i really dont know

step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions