<h2>Hello!</h2>
The answer is:
The domain of the function is all the real numbers except the number 13:
Domain: (-∞,13)∪(13,∞)
<h2>Why?</h2>
This is a composite function problem. To solve it, we need to remember how to composite a function. Composing a function consists of evaluating a function into another function.
Composite function is equal to:
So, the given functions are:
Then, composing the functions, we have:
Therefore, we must remember that the domain are all those possible inputs where the function can exists, most of the functions can exists along the real numbers with no rectrictions, however, for this case, there is a restriction that must be applied to the resultant composite function.
If we evaluate "x" equal to 13, the denominator will tend to 0, and create an indetermination since there is no result in the real numbers for a real number divided by 0.
So, the domain of the function is all the real numbers except the number 13:
Domain: (-∞,13)∪(13,∞)
Have a nice day!
Step-by-step explanation:
Enter a problem...
Algebra Examples
Popular Problems
Algebra
Expand using the Binomial Theorem (3x+2)^4
(3x+2)4(3x+2)4
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=n∑k=0nCk⋅(an−kbk)(a+b)n=∑k=0nnCk⋅(an-kbk).
4∑k=04!(4−k)!k!⋅(3x)4−k⋅(2)k∑k=044!(4-k)!k!⋅(3x)4-k⋅(2)k
Expand the summation.
4!(4−0)!0!⋅(3x)4−0⋅(2)0+4!(4−1)!1!⋅(3x)4−1⋅(2)+4!(4−2)!2!⋅(3x)4−2⋅(2)2+4!(4−3)!3!⋅(3x)4−3⋅(2)3+4!(4−4)!4!⋅(3x)4−4⋅(2)44!(4-0)!0!⋅(3x)4-0⋅(2)0+4!(4-1)!1!⋅(3x)4-1⋅(2)+4!(4-2)!2!⋅(3x)4-2⋅(2)2+4!(4-3)!3!⋅(3x)4-3⋅(2)3+4!(4-4)!4!⋅(3x)4-4⋅(2)4
Simplify the exponents for each term of the expansion.
1⋅(3x)4⋅(2)0+4
Hope this helps!
That would be 4(16) - 8 - (4(0) - 8) = 64 - 9 - 0 + 8 = 63 Answer
If the hypothesis was wrong, all you simply have to do is state the hypothesis was wrong and why you think it was, and add what you can do to change it or to make it better so next time it will be right. Just state what you did and how you got your answer.
I hope that helps!