Answer:
heyyyyyy
Step-by-step explanation:
So basically you just add the like terms to simplify the equation. You have 5q-p+p+1 you can cancel out the both p’s because one is negative and one is positive. That leaves you with 5q + 1 they are not like terms so you cannot simplify them so your simplified equation is 5q +1
<u>Answer</u><u>:</u>
<h2>
C. 9/12</h2>
<u>Explanation:</u>
<em>6</em><em>/</em><em>8</em><em> </em><em>simplifies </em><em>to </em><em>3</em><em>/</em><em>4</em>
<em>3</em><em> </em><em>×</em><em> </em><em>3</em><em> </em><em>=</em><em> </em><em>9</em>
<em>4</em><em> </em><em>×</em><em> </em><em>3</em><em> </em><em>=</em><em> </em><em>1</em><em>2</em>
<em>Since </em><em>they </em><em>are </em><em>both </em><em>multiples </em><em>of </em><em>3</em><em>,</em><em> </em><em>it's </em><em>e</em><em>quivalent</em><em> </em><em>to </em><em>6</em><em>/</em><em>8</em><em>.</em>
We have the expression:

When we have rational functions, where the denominator is a function of x, we have a restriction for the domain for any value of x that makes the denominator equal to 0.
That is because if the denominator is 0, then we have a function f(x) that is a division by zero and is undefined.
If we have a value that makes f(x) to be undefined, then this value of x does not belong to the domain of f(x).
Expression:

Answer: There is no restriction for x in the expression.