Answer:
The second one:
L∪J={i,j,k,l,m,n,o}
Step-by-step explanation:
The union is the elements listed in either set.
So since l,m,n, and o are elements of set L, they will also be elements of whatever it is "unioned" with.
Since i,j,k,l and m are elements of set J, they will also be elements of whatever it is "unioned" with.
When you write the union, just be sure to include each element that occurs in either set once.
So the union of L and J is {i,j,k,l,m,n,o}.
The answer is the second one.
The intersection would actually be that upside down U thing, the ∩ symbol. The intersection of two sets is a list of elements that both sets include. So here the intersection would just consist of the elements l amd m.
Answer:
All numbers less than 2 or
or 
Step-by-step explanation:
Given that:
Number is added to itself is less than the number subtracted from 6.
To find:
All such numbers.
Solution:
Let the number be
.
Here, an inequality will be made.
When solved, it might give more than one answers.
As per the question statement, let us write the inequality.
Number added to itself 
Number subtracted from six = 
As per question:

So, the answer is:
All numbers less than 2 or
or
.
Answer:
what are we actually solving for?....
anyways I had a suggestion ion know if it's right or wrong....
<em>Isolate</em><em> </em><em>the</em><em> </em><em>variable</em><em> </em><em>by</em><em> </em><em>diving</em><em> </em><em>each</em><em> </em><em>side</em><em> </em><em>by</em><em> </em><em>factor</em><em> </em><em>that</em><em> </em><em>don't</em><em> </em><em>contain</em><em> </em><em>the</em><em> </em><em>vari</em><em>able</em><em>.</em>
<em>Therefore</em><em> </em><em>x</em><em>=</em><em>7</em><em>.</em><em>2</em>