Willem Einthoven invented the ECG in 1903. The letters P Q R S T were assigned to the points on the ECG because it was thought that there may be more points in either direction to be found. The letters remained in use out of common practice even after it was determined there were no further points to be labelled.
The answer is Q.
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.
Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "If p q and q r then p r.?". There is no operator between the variables. Some operators are transitive, others are not. In the case of the operator "is not equal to", the answer is that it depends. In the case of "is the parent of" the answer is no.
Suppose the radius is r and the bearings of the two points, P and Q are p and q respectively. Then the coordinates of P are [r*cos(p), r*sin(p)] and the coordinates of Q are [r*cos(q), r*sin(q)]. The distance between these two points can be found, using Pythagoras: d2 = (xq - xp)2 + (yq - yp)2 where xp is the x-coordinate of P, etc.
P
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
Ifp < q and q < r, what is the relationship between the values p and r? ________________p
A rational number is a number of the form p/q where p and q are integers and q > 0.If p/q and r/s are two rational numbers thenp/q + r/s = (p*s + q*r) / (q*r)andp/q - r/s = (p*s - q*r) / (q*r)The answers may need simplification.
P=q/r* * * * *The correct answer is P = k*q/r where k is the constant of proportionality.
Two fractions are similar if they have the same denominator.So if p/r and q/r are two such fractions, then p/r + q/r = (p+q)/r.
tan x
On an ECG, p, q, r, s and t refer to the different spikes on the reading. P represents the depolarisation of the atria of the heart. Q, R and S represent the depolarisation of the ventricles. T represents the repolarisation of the ventricles.
The answer is Q.
p/q * r/s = (p*r)/(q*s)
Suppose you have the fractions p/q and r/s. Let the LCM of q and s be t.Then t is a multiple of q as well as of s so let t= q*u and t = s*v Then p/q = (p*u)/(q*u) = (p*u)/t and r/s = (r*v)/(s*v) = (r*v)/t have the same denominators.
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
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.