Given: x is the midpoint of RD, RM is parallel DN, Prove: MX is congruent MR?
813384
I drew the picture as you described it and it looks like a parallelogram with its diagonals drawn in.  MX is half of one diagonal and MR is one side of the parallelogram.  If this is indeed the case, there is no reason why MX is congruent to MR.  So, either you need a better description of what the figure looks like, or you left out a given, such as DN is congruent to XN or something like that.
766790
1. 5m^2 - 15mp - 2mr - 6pr <-- Impossible to factor by grouping. It needs to be either, exactly one of the signs needs to be reversed.\n\n2. 6-3x-2y+xy\n(6 - 3x) - (2y - xy)\n3(2 - x) - y(2 - x)\n(3 - y)(2 - x)
833002
Let u = x², so du = 2xdx\n\nSo, ∫ (x/1+x^4) dx =\n\n1/2 ∫ (1/1+u²) du =\n\n1/2 arctan(u) + C = \n\n1/2 arctan(x²) + C\n\n~ ♥ ~
338257
a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
840090
P1V1 = nRT = P2V2\nP1/P2 = V2/V1
467967
this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
871212
i2=i1R1/R2\ni1=R2i2/R1\n\nR1/R2=i2/i1\n\nR1/Rx = R2/R3\n(R1/R2)*R3=Rx\nRx=(i2/i1)*R3
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
543381
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
179545
Wow, that last person used way too much stuff.\n\nDraw in one diagonal of parallelogram ABCD, say BD.  You have congruent angles ABD and BDC, ADB and DBC because they are alternate interior angles and BD is congruent to itself.  The two triangles are congruent (ASA) which makes the parts you want congruent by CPCTC.
711618
You must show that E is reflexive, transitive and symmetric.\n\nxPx because P is reflexive, so xEx.\n\nSuppose xEy.  Then xPy and yPx, but that also implies yEx, so E is symmetric.\n\nSuppose xEy and yEz.  Then xPy, yPx, yPz, and zPy.  Because P is transitive, you have xPz and zPx, so xEz, therefore E is transitive.
844458
Let us suppose that when a is divided by b, the remainder is r. Then:\n\na=kb+r for some integer k.\n\nNow let us suppose that a and b both are divided evenly by some number, d. Then a=pd and b=qd for some integers p and q. So:\n\npd=kqd + r\n\nWhich implies that:\n\n(p-kq)d = r\n\nWhich means that r is also an exact multiple of d. Now suppose that d is the greatest common divisor of a and b. By the previous, it is clearly also a divisor of r. To show that it is the greatest common divisor of b and r, suppose b and r had a common divisor, d' > d. Then a = p'd' + q'd' = (p'+q')d', and so d' is a divisor of a, and thus a common divisor of both a and b greater than d, which is a contradiction. Therefore, the greatest common divisor of b and r is the same as the greates common divisor of a and b, which establishes the correctness of the recursive step of Euclid's algorithm. Q.E.D.
136546
here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
975361
d=nt2/p\npd=2nt\np=2nt/d
891869
A _______________ B\n... &#92;.................. ...............&#92;\n..... &#92;.................. ...............&#92;\n....... &#92;_______________&#92;\n...... D.................. ...............C\n\nLet ABCD be a quadrilateral with AB || DC and AB = DC\n\nDraw diagonal AC\n\nNow in Δs ABC, CDA\nAB = DC (Data)\n<BAC = <DCA ( Equal alternate angles, AB||DC)\nAC is common\nTherefore ΔABC ≡ ΔCDA (SAS property)\n\nTherefore < BCA = < DAC (Corresponding angles of congruent triangles are equal)\n\nTherefore BC||AD (alternate angles are equal)\n\nSo ABCD is a parallelogram as opposite sides are parallel
215876
Additive equivalence is a symmetric relation over the reals. See the source link.
406468
D = {x | x < -3 U x > 4 }