using (a) substitution; (b) master table.T(n)={T(1) n=1 or 7T(n/4)+18n^2 n>1 }?
963851
The question doesn't make sense: what is T(n/4) supposed to mean when n = 2 or 3 or 5 ...??
211777
n=3\nl=1\nm sub l =1,0, or -1\nm sub s = + or - 1/2
107302
If you mean N=(mn)/(m-n)\nThen m=[mn/N]+n\n\nIf you mean N=[mn/m]-n\nThen m=(mn)/(N+n)
786690
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
66380
You have to use the formula -\nm1s1(t3-t1)=m2s2(t2-t3)\n=>191(4.184)(30.5-25)=38s2(192-30.5)\n=>s2=191(4.184)(5.5)/((38)(161.5)\n=>s2=0.716 J/g.k
808688
the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
549518
How about...\n\n( 3 + (3/7) ) * 7 = 24\n\nThat would be like (21/7 + 3/7)... which is (24/7)... times 7 = 24.
559200
By order of precedence of operations:\n\ny = 2^1/2  is the same as\ny = (2^1)/2  which is just\ny = 2/2 which yields\n\ny = 1 <<<<<<<<<<<<<<<<<<<<<<
527619
use ln\n\nln(5^t) = ln(7^(t-1))\n(t)*ln(5) = (t-1)*ln(7)\n(t)*ln(5) = (t)*ln(7)-ln(7)\n(t)*ln(5)-(t)*ln(7) = -ln(7)\nt*(ln(5)-ln(7)) = -ln(7)\nt*(ln (5/7)) = -ln(7)\nt = -ln(7) / ln(5/7)\nt = ln(7) / ln(7/5)\n\nt=~5.78
973961
a) i +(1/3)(j-i)\n\n  b)  \n1-12 =>1-4 & 5-12\n\n... 1-4=>1 & 3-4\n.......3-4=>3 & 4\n\n...5-12=> 5-6 & 7-12\n......5-6=> 5 & 6\n......7-12=>7-8 & 9-12\n..........7-8=>7 & 8\n..........9-12=> 9 & 10-12\n...............10-12=>10 & 11-12\n....................11-12=>11 & 12\n\n c) 1/3 of the computations lead to a subset of the previous set\nwhich is 1/2 the size of the other subset which is lead to twice\nas often. So it averages out to the same number of\ncomputations as a conventional binary search.
122227
T1=O(f)\n=> T1=c1*f for input>a\nT2=O(g)\n=> T2=c2*g for input>b\n\nlet d is maximum of a and b. and c=c1*c2\n\nso T1*t2=c*f*g for input>d\n\nSo T1*T2=O( f*g)
1095736
T=.00139x^2+.0304x=1.25 \n\nmaybe you meant\n\nT=.00139x^2+.0304x-1.25 \nor\nT=.00139x^2+.0304x+1.25 \n\n?????
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
440158
Since a = 9-t,\nThen 4(9-t) = 8t-12.\nThen 36-4t = 8t-12.\nThen 12t = 48.\nSo t = 4.\n\nThus a = 9-t = 9-4 = 5.
721887
1)t= -12    2) x= -8    3) x= 7/5 or 1 2/5