8 balls = 16
1 ball = 2
32 balls = 64
Answer:
All of them.
Step-by-step explanation:
For rational functions, the domain is all real numbers <em>except</em> for the zeros of the denominator.
Therefore, to find the x-values that are not in the domain, we need to solve for the zeros of the denominator. Therefore, set the denominator to zero:
Zero Product Property:
Solve for the x in each of the three equations. The first one is already solved. Thus:
Thus, the values that <em>cannot</em> be in the domain of the rational function is:
Click all the options.
I think it would be A.) but at the same time i think it's wrong
Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
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You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
