The expression "SSD 2 mod 1" seems to contain a misunderstanding or typo, as "SSD" is not a standard mathematical term. However, if you're asking for the result of "2 mod 1," the answer is 0, since when you divide 2 by 1, there is no remainder. If you meant something else by "SSD," please clarify for a more accurate response.
Don't. You're in the ARMY I'm sure they'll do something to you if you do.
NO CHEATING! -Your teacher
NO CHEATING! -Your teacher
SSD 3 Module 2 typically refers to a specific module in the Social Security Disability (SSD) training program or course. The answers or content would generally cover topics related to social security benefits, eligibility criteria, and the application process. For detailed answers, it's best to refer to the course materials or resources provided by the training program.
what type of insurgent approach involves a few leaders and a militant cadre or activist party seizing control of government structures
1 + 1 = 0 in binary. Why does this happen?Note: Adding binary numbers is related to modulo 2 arithmetic.Let's review mod and modular arithmetic with addition.modulus 2 is the mathematical term that is the remainder from the quotient of any term and 2. For instance, if we have 3 mod 2, then we have 3 / 2 = 1 + ½. The remainder is 1. So 3 ≡ 1 mod 2.What if we want to add moduli?The general form is a mod n + b mod n ≡ (a + b) mod n.Now, for the given problem, 1 mod 2 + 1 mod 2 ≡ 2 mod 2. Then, 2 mod 2 ≡ 0 mod 2.Therefore, 1 + 1 = 0 in binary.
what type of insurgent approach involves a few leaders and a militant cadre or activist party seizing control of government structures
1 mod 3 means the remainder after 1 has been divided by 3. Which is 1. 2 mod 3 = 2, 3 mod 3 = 0
You can find answers to the SSD 2 course through your unit's training library, on the Army Training Network (ATN) website, or by speaking with your unit's training administrator. Make sure to follow proper procedures to access the answers in a legitimate and ethical manner.
There are many possible answers. Some are A: Un = (2n3 - 15n2 + 34n - 18)/3 B: Un = 2 - mod(n, 2) C: Un = 1.5 + 0.5*(-1)n
1. D 2. B 3. A 4. A 5. C 6. B 7. D 8. D 9. A 10. C 11. A 12. B 13c 14c 15a 16d 17c 18a 19 d 20. A 21. A 22. D 23. D 24. B 25. B 26. D 27. C
No because it is impossible. Let mod(x.3) denote the remainder when x is divided by 3. Let n be any integer. Then mod(n,3) = 0,1 or 2. When mod(n,3) = 0, mod(n2,3) = 0 When mod(n,3) = 1, mod(n2,3) = 1 When mod(n,3) = 2, mod(n2,3) = 4 and, equivalently mod(n2,3) = 1. So, there are no integers whose squar leaves a remainder of 2 when divided by 3.