The answer is -2.21428571429
I dont know man you gotta ask a question
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.
Answer:

Step-by-step explanation:
we would like to solve the following equation for x:

to do so isolate
to right hand side and change its sign which yields:

simplify Substraction:

get rid of only x:

simplify addition of the left hand side:

divide both sides by q+p Which yields:

cross multiplication:

distribute:

isolate -pq to the left hand side and change its sign:

rearrange it to standard form:

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so
factor out x:

factor out q:

group:

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

cancel out p from the first equation and q from the second equation which yields:

and we are done!