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
P=s r t , so, s= P/(st)
American Dreams - 2002 R-E-S-P-E-C-T 2-2 was released on: USA: 5 October 2003
The formula for logarithmic growth is ( y = a \cdot \log(x) + b ), where ( y ) is the output, ( a ) is a growth factor, ( x ) is the input, and ( b ) is a constant. The logarithmic function grows slowly at first but then accelerates as the input increases, often used to model growth that levels off over time.
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 \\'