If AE is x+50, then the whole thing must be x=50.
if a portion of that, CE, is x+32, then the rest of it, AC, must be x+18.
Because no matter what X is, 32+18=50.
Answer:
=−36p+52x+1324
Step-by-step explanation:
7x−7+35−34p+45x−2p+64
=7x+−7+35+−34p+45x+−2p+1296
Combine Like Terms:
=7x+−7+35+−34p+45x+−2p+1296
=(−34p+−2p)+(7x+45x)+(−7+35+1296)
=−36p+52x+1324
Answer and Explanation:
Solution: The operation of concatenation for a set of string on p. and the set is
AB = {XY | X ∈ A and y ∈ B}.
We need to satisfy all these following properties to find out the standard set is closed under concatenation.
1- Union of two standard sets also belongs to the classic collection. For example, A and B are regular. AUB also belongs to a regular group.
2- Compliment of two standards set A and B are A’ and B’ also belonging to the standard set.
3- Intersection of two standards set A and B is A∩B is also a regular set member.
4- The difference between two regular sets is also standard. For example, the difference between A and B is A-B is also a standard set.
The closure of the regular set is also standard, and the concatenation of traditional sets is regular.
Answer:
20
Step-by-step explanation:
If you use PEMDAS(Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) you would take care of what's in the parentheses, so you would at 5 and 3 to get 8. Then you would multiply 6 and 4 to get 24, so far your equation looks like this 24 - 8 ÷ 2. You then divide 8 by 2 to get 4 then subtract 24 from 4 to get 20.
