They have the same symbolic representation, in mirror-image.
No, the primary (P) waves arrive first.
I believe it is the P Wave. A good way to remember is all of the Waves are in alphabetical order. P Wave, Q-R-S Waves and the T Wave
P or primary waves
yes
yes
The relational operators: ==, !=, =.p == q; // evaluates true if the value of p and q are equal, false otherwise.p != q; // evaluates true of the value of p and q are not equal, false otherwise.p < q; // evaluates true if the value of p is less than q, false otherwise.p q; // evaluates true if the value of p is greater than q, false otherwise.p >= q; // evaluates true of the value of p is greater than or equal to q, false otherwiseNote that all of these expressions can be expressed logically in terms of the less than operator alone:p == q is the same as NOT (p < q) AND NOT (q < p)p != q is the same as (p < q) OR (q < p)p < q is the same as p < q (obviously)p q is the same as (q < p)p >= q is the same as NOT (p < q)
Not sure I can do a table here but: P True, Q True then P -> Q True P True, Q False then P -> Q False P False, Q True then P -> Q True P False, Q False then P -> Q True It is the same as not(P) OR Q
p q and p q cannot be a useful part of a system of simultaneous equations since they are the same!
this is the waves of p waves that have same direction of waves
Think of 'not' as being an inverse. Not 1 = 0. Not 0 = 1. Using boolean algebra we can look at your question. 'and' is a test. It wants to know if BOTH P and Q are the same and if they are 1 (true). If they are not the same, or they are both 0, then the result is false or 0. not P and Q is rewritten like so: (P and Q)' = X not P and not Q is rewritten like: P' and Q' = X (the apostrophe is used for not) We will construct a truth table for each and compare the output. If the output is the same, then you have found your equivalency. Otherwise, they are not equivalent. P and Q are the inputs and X is the output. P Q | X P Q | X ------ 0 0 | 1 0 0 | 1 0 1 | 1 0 1 | 0 1 0 | 1 1 0 | 0 1 1 | 0 1 1 | 0 Since the truth tables are not equal, not P and Q is not equivalent to not P and not Q. Perhaps you meant "Is NOT(P AND Q) equivalent to NOT(P) AND NOT(Q)?" NOT(P AND Q) can be thought of intuitively as "Not both P and Q." Which if you think about, you can see that it would be true if something were P but not Q, Q but not P, and neither P nor Q-- so long as they're not both true at the same time. Now, "NOT(P) AND NOT(Q)" is clearly _only_ true when BOTH P and Q are false. So there are cases where NOT(P AND Q) is true but NOT(P) AND NOT(Q) is false (an example would be True(P) and False(Q)). NOT(P AND Q) does have an equivalence however, according to De Morgan's Law. The NOT can be distributed, but in doing so we have to change the "AND" to an "OR". NOT(P AND Q) is equivalent to NOT(P) OR NOT(Q)
No, the primary (P) waves arrive first.
no, there are p-waves, s-waves, surface waves
I believe it is the P Wave. A good way to remember is all of the Waves are in alphabetical order. P Wave, Q-R-S Waves and the T Wave
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
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.
The sum of p and q means (p+q). The difference of p and q means (p-q).
q + p