2 Times the Postman Always Rings
Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
T= Temperature P= Pulse R= Respiration
We will use the fact that if p prime, a divides p, then a = p or a = 1. Then if p + q = r, for primes p, q, r, then one of p,q,r is even, or all three are (consider mod 2). p = q = r = 2 clearly doesn't work, and p + q = 2 doesn't work for primes p,q >= 3. So without loss of generality p = 2, then r = q+2. r is also the difference of two primes, r = s - t. Again considering mod 2, knowing that r is odd, one of s or t is even (and so equal to 2). If s = 2 then r is negative, so t = 2, and we have q + 2 = r = t - 2, so t = q + 4. So we have q, r = q + 2 and t = q + 4 all prime. By considering q mod 3, one of them has a factor 3. If a prime has a factor 3, it is equal to 3. So q = 3, as q + 2 = 3 or q + 4 = 3 mean q is not prime. So, r = q + 2 = 5. Therefore, 5 is the only prime that can be represented as both the sum of two primes and the difference of two primes: 5 = 2 + 3 = 7 - 2. Since it is the only one, it is the greatest.
Codename Kids Next Door - 2002 Operation S-U-P-P-O-R-T- Operation T-A-P-I-O-C-A- 2-3 was released on: USA: 17 October 2003
The first half of the question yield two equations:1.I. p = r + (s - r) / 2II. t = r + (p - r) / 2The equation we are solving for, (s - t) / (t - r), does not have a p so we are going to Equation I for the pin Equation II. But first, to make things easier for us, let distribute the 2 in Equation II.2. 2t = 2r + p - r3. 2t = r + pSubstitute I4. 2t = r + r + (s - r) / 25. 2t = 2r + (s - r) / 2Again, lets distribute the 2, and combine the r's6. 4t = 4r + s - r7. 4t = 3r + sIn order to yield the desired quotient, we want a (s - t)on one side of the equation and (t - r) on the other. First let's get the (t - r) on the left.8. 4t - 3r = s9. t + 3t - 3r = s10. t + 3(t - r) = sNow move that extra t over to the right for the (s - t) we are looking for11. 3(t - r) = s - tNow divide both sides by (t - r) and we have our answer!12. 3 = (s - t) / (t - r)
u = p r t r = u / p t
American Dreams - 2002 R-E-S-P-E-C-T 2-2 was released on: USA: 5 October 2003
P=s r t , so, s= P/(st)
The answer depends on whether you are dealing with simple interest of compound interest. Suppose P = Principle R = Rate (in % per annum) T = Time (in years) I = Interest Then for simple interest: I = P*R*T/100 so that P = 100*I/(R*T) For compound interest P+I = P*(1+R/100)T so that P = I/[(1+R/100)T - 1]
The possible coordinates of the midpoint depend on the coordinates of A and T and these depend on what these two points are and how they are related.If A = (p,q) and T = (r,s ) then the midpoint of AT has coordinates [(p+r)/2, ((q+s)/2].
J. R. P. McKenzie has written: 'T. Mann \\' 'T. Mann \\'
P. T. R. Palanivel Rajan was born on 1932-02-27.