There are six letters between M and T. They are n, o, p, q, r, and s. The midway point is in the middle between P and Q, so there is no one letter midway between M and T. If you must have an answer, it would be both P and Q.
1.a 2.i 3.c 4.u 5.u 6.m 7.h 8.b 9.q 10.m 11.b 12.a 13.i 14.t 15.p 16.t 17.c 18.h 19.t
p=t*C, where C is a constant value (in this case, the person's rate of pay). To determine C, solve for C for the p and t given. 10800=16C C=10800/16 C=800 Therefore, the equation is: p=800t
That equation is, q(Joules) = mass * specific heat ( symbol is C ) * (delta, a triangle) change in temperature That is to say delta means, Temperature Final - Temperature Initial q is a constant and not subject to change as temperature is
There is no simple way. A polynomial of the form f(x) = ax4 + bx3 + cx2 + dx + e may have four real factors: it may have none. Binomial factors will be of the form px + q, where p is one of the factors of a and q is one of the factors of e. In general, p and q can be positive or negative. That gives a very large number of possible binomial factors of the polynomial. Evaluate f(x) for x = -q/p, that is, substitute x = -q/p in the polynomial and calculate its value. If f(-q/p) = 0 then (x + q/p) = (px + q) is a factor. It may be possible to find the zeros of the quadratic by numerical or graphical methods. If x = z if a root then (x + z) is a factor. If the four factors are (x - s), (x - t), (x - u) and (x - v) then s+t+u+v = b/a st +su+ sv + tu + tv +uv = c/a stu + stv + suv + tuv = d/a and stuv = e/a One option is to solve these equations simultaneously for s, t, u and v.
6 quadrants on a trivial pursuit counter
p --> q and q --> p are not equivalent p --> q and q --> (not)p are equivalent The truth table shows this. pq p --> q q -->(not)p f f t t f t t t t f f f t t t t
p > q~qTherefore, ~p| p | q | p > q | ~q | ~p || t | t | t | f | f || t | f | t | t | f || f | t | t | f | t || f | f | t | t | t |
P Q (/P or /Q) T T F T F T F T T F F T
75 per cent is three quartersSy031109
h t t p s : / / w w w . y o u t u b e . c o m / w a t c h ? v = H k Q 7 _ o W q K p c
P | T T F F Q | T F T F Q' | F T F T P + Q' | F T F F The layout is the best I could do with this software. Hope it is OK.
Cave
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.
Pointy
T
it is true when at least one statement is true p q p v q t t t t f t f t t f t f You sure about this?