[ ] is bracket .eg [2.4]= 2
174019
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
186589
(0, 1)\n(1, 1.1)\n(2, 1.21)\n(3, 1.331)\n(4, 1.4641)
389844
☺☺☺☺☺☻☻☻☻☻♥♥♥♥♥Alt+1=☺ Alt+2=☻Alt+3=♥♥♥♥\nand you can do a bunch more.
306002
A = {1,2}, B = {2,3}, C = {1,3} (A U B) = (A U C) = {1,2,3} \n\nNone of the 3 statements is true.
395312
3x + 4 < x + 8\n2x < 4\nx < 2\n\n{x | x < 2}
553937
basically, think of a function as a machine:\n\ntake this an an example: f(x) = 2x +3\nsee the x in brackets?\n\nif x = 1, then f(1) = 2(1) + 3 = 5\nif x = 2, then f(2) = 2(2) + 3 = 7\nif x = 3, then f(3) = 2(3) + 3 = 9\n\nsee whats happening? the answer DEPENDS on number inputted in the brackets.\n\nso therefore, the definition you quoted "a consistent, predictable, relationship between the independant and dependant variable" is true.
549429
[i]  = I\n[d]s = does (without the s it would be do)\n[t]  = to or the\n[o] = of\n\nAnyway I think all the [ ] are suppose to be abbreviations and these are the most likely.◘
45590
First u has to do the operation in side the bracket. (BODMAS rule.)\nSo 0.25-4.99= -4.74\nNow multiply with 0.85 \n-4.74*0.85 = -4.029\n\nround off to -4.03
147645
Since [L : K] = [L : E][E : K] and [L : K] is prime, \neither [L : E] = 1 or [E : K] = 1.\n\nThus either E = L or E = K
648447
d#  A   G   Bm  \nE  6   5   3    2\nB  8   5   3    3\nG  8   6   4    4\nD  8   7   5    4\nA  6   7   5    2\nE  x   5    3    x
426620
Use the definition\nie\n|a| = a for a ≥ 0\n= -a for a < 0
436382
2 zeroes, one in the range -3<z<-1, and the other in the range 1<z<3.\n\nSteve
565431
2x<1 , x<0.5\n -1<x\n\n     -1< x<0.5
276533
Correct.\n\n(1350+99)*1.06
607140
for the first one it's '1'. this is because of the AB'+1. since anything or'ed with 1 gives u 1. the whole thing turns out to be 1. same thing happens in the other bracket. so when 1 n 1 are ANDed u get 1.\n\nfor the second one remember ther are two terms. one is DE' other is (A+F) in the second bracket the (A+F) term has been complemented. when DE' is ANDed by itself u get DE' again. When DE' is Anded with (A+F) and it's complement u are able to actually factorize it so u can take DE' out and u get the or'ing of a term an it's complement inside. that is 1. therfore it evaluates to DE'. then (A+F) and it's complement gives a 0 when anded. oring everything left u get ED' as the final answer. if u need the working pls msg me
247208
you have to use instinct, there is no clear rules that states what the expressions must look like (except if the problem insist you on giving the answer in a specific form). \n\nPerhaps one of the guidelines is make variables that have same indices grouped into one. So x + 3x becomes 4x, group them out don't leave them straying.
544599
Easy, type $A$10+B8. The $ before A will anchor the column whereas the one before 10 will anchor the row.