Let's consider two prime numbers where each is larger than 2
Say the primes 7 and 11. Adding them gets us 7+11 = 18. This counter example disproves the initial claim since 18 = 9*2 = 6*3 making 18 composite (not prime)
In general, if we let p and q be two primes such that q > p > 2 and q is the next prime after p, then p and q are both odd. If any of them were even then they wouldn't be prime (2 would be a factor)
Adding any two odd numbers together leads to an even number
Proof:
p = 2k+1 where k is some integer
q = 2m+1 where m is some integer
p+q = 2k+1+2m+1 = 2(k+m) + 2 = 2(k+m+1) which is in the form of an even number
That proof above shows us that adding any prime larger than 2 to its next prime up leads to an even number. This further shows us that the claim is false overall. It is only true if you restrict yourself to the primes 2 and 3, which add to 5. Otherwise, the claim is false.
Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra II</u>
- Distance Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
Point (-6, -5)
Point (2, 0)
<u>Step 2: Find distance </u><em><u>d</u></em>
Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>
- Substitute [DF]:
- Add:
- Exponents:
- Add:
Hello,
y=x²+2x+7
y=x+7
==>x²+2x+7=x+7
==>x²+x=0
==>x(x+1)=0
==>(x=0 and y=7) or (x=-1 and y=-1+7=6)
Sol={(0,7),(-1,6)}
Answer:
Answer in Explanation
Step-by-step explanation:
His reasoning is wrong. We can evaluate the validity of his reasoning by looking at the basics of what a polynomial is.
In the real sense of it, what makes a polynomial a polynomial is the power to which the variable is raised and not the integer attached to the variable.
In this case, we do not even have an integer attached.
What makes a polynomial a polynomial is that the variable is raised to an exponent or power which is a positive whole number ( integer)
In this case, while x^2 is a polynomial, x^-2 is not and also x^1/2 is not
Only x^2 is qualified as a polynomial because it is the only expression having its powers in positive integers
Hence; 0.5x^2 is a polynomial since it is raised to a positive whole number (integer)
"zeros" are solutions for x that make the equation equal to zero.
(x - 8)(x + 8)(x - 1) = 0