Absolute inequalities??
447443
Ok...you have to solve two equations\n\nFIRST ONE\n\nif (8-4x) < 5\n\nthen (8-4x-5) < 0 (by subtracting 5 from both sides of the equation)\n\nthus (3-4x) < 0 (obviously)\n\nso 3 < 4x (by adding 4x to both sides of the equation)\n\nso 4x > 3\n\nso......... x > 3/4\n\nSECOND ONE\n\nif -(8-4x) < 5\n\nthen -8+4x < 5 (obviously...see that -*- = +)\n\nthen -8+4x-5 < 0 (by subtracting 5 from both sides of the equation)\n\nso -13+4x < 0 (obviously)\n\nand 4x < 13 (by adding 13 to both sides of the equation)\n\nso......x < 13/4\n\nFINAL ANSWER IS.......(3/4 < x < 13/4)
780452
Solve it just like a regular equality:\n\nx-6 ≤ 7\n\nAdd 6 to both sides, to get:\n\nx ≤ 13  ← this is your answer\n\nWhen working with inequalities, you can add and subtract from both sides just as you would in a regular equality.  You can also multiply and divide both sides, but the trick to keep in mind is that if you multiply or divide by a negative number, you need to change the sign!  For example, if you had:\n\n-2x ≤ 10\n\nDivide both sides by -2 AND REVERSE THE INEQUALITY to get:\n\nx ≥ -5\n\nHope that helped!\n\n~ ♥ ~\n\nEDITED TO ADD:  A "regular equality" would be where you have an equal sign rather than a < or > sign.  So x-6 = 7 is an equality.\n\nGood luck kid ... sounds like you will need it!
433611
Read the book Savage Inequalities by Jonothan Kozol, and you shall have your answers.  There is much more to this issue than access to honors classes.
973577
given x < y\n\nnot sure if this would be using mathmatical induction but this is how i would do it\n\ntake\nx < 1/2 (x+y) < y\n2x< x + y < 2y\nx < y < 2y\n\nx<y is stated to be true and 2y is certainly bigger than y\ntherefore x < 1/2 (x+y) < y
754121
Given:\n7<=3-2p<11\n\nYou want to isolate p, and as long, we only add and subtract its OK as long we do it in every term in the inequality.\n\nOf cause, we want to add -3 in the term in the middle, and therefore in rest of the expression, we get:\n7-3<=3-3-2p<11-3 <=>\n4<-2p<8\nNow we want to divide with 2 and we can do so, as long, we do it in all term, but we also want to multiply with -1, but it's also OK if we remember to reverse the inequality sign. \nThis use to be the point of confusion in calculation with inequalities\nHowever, it is easy to se that it is true\nEx 7<8 multiply with -1 gives\n-7<-8 but that not true therefore we have to reverse the inequality sign, if we multiply with a negative number.\n\nBack on track:\n\n4<=-2p<8<=>\n-4/2=> p >-8/2\n.......NB!!.NB!!!\n -2=>p>-4 \nOr if you better like the result in adequate form\n\nx={ x in R l (-4<p<= -2)}\n================
907446
Absolute increase is (1.00 - 0.92) units\nThe relative increase = (0.08) units / 0.92 units = 0.087 = 8.7%\nThe relative increase = percental increase = 8.7%. That's it.\n\nTh
880673
hint: mean value theorem
388704
it cannot be f(x) = a for all x, unless a=0; otherwise you would be pretending that a = a+a with a non-0 !! This doesn't happen in R.
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
787480
if x is positive, then I x I is always equal to the value, but not less than it.\nif x is negative, then I x I will be a positive value which is never less than a negative value.\nso, the inequality holds good only for positive values of x.
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
110904
No. you may count yourself lucky to have these conditions because with these you are able to get an unique solution ψ.\nBut there are 4 main conditions:\n\nNorm: |ψ| := 1\n\nContinuity of ψ\n\nContinuity of ψ'    (only for finite potentials)\n\nLimit: ψ(x) -> 0 for x -> ∞
288230
It should work on any norm, whether the norm is the infinity norm, |[a,b]| = max(a,b) or the 2 norm, |[a,b]| = sqrt(a^2 + b^2).  In the former case you have a rectangle.  In the latter case you have a disk.  All that is necessary is that the norm satisfies the properties of being a norm, which is a generalization of a distance function:\n\nPositive definiteness: |[a,b]| > 0 for all [a,b] != [0,0]\nLinearity: |[ka,kb]| = k|[a,b]| for all [a,b] and real k.   \nTriangle inequality: |[a,b]| + |[b,c]| >= |[a,c]|.
871092
I guess you probably use the Epsilon definition for convergence? Then, you know:\nFor every Epsilon greater than 0, there exists an N such that n>N implies /xn - a/ < E'.\nYou want to show that this also implies:\n/xn^(1/3) - a^(1/3)/ < E.\n\nI think that you can probably do something with:\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\nmaking x = a in this equation, the proof's a = b.\nThen, you can find an N so that /[xn^(1/3) - a^(1/3)][xn^2/3 + xn^(1/3)a^(1/3) + a^(2/3)]/ < E'. Perhaps this is easier to relate to the first one?\n\nGood luck!
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
902313
Since the constant term is negative, the signs must be opposites. A little trial and error shows the answer to \nbe (x + 6)(x - 1)
7873
well in the demostrantion of the Real numbers...\n\nall the numbers and relations depend of the existence of 1... suck that there are some rules..\n\nso we even demostrate that 1 exists... and that 1 > 0... after that we can prove the existence of any other Real number..