Find the 13th term of the geometric sequemce with first term 10 and common ratio 2.
202611
10 * 2^(13-1) =\n\n10 * 2^12 =\n\n40,960
447193
its a square with 13 feet sde.\n\nIf a proof is to be sumbitted ask her her to find are at different side lenghts,\n\n1,12\n2,11\n......\n......\n.......\n13,13\n\n14,2 etc and hence conclude.\n\nI i dont know a better proof for fifth grader. otherwise i will nedd to use maxima minima or a quadratic equation atleast to prove that\n\n\nsorry i cant help much\n\n\n\nbythe way you dont adjust anything  like  13.00001 or 12.9999999\n\nsquare is a special case of rectangle with length=breadth
222447
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
370524
[k+2]=13\nk+2-2=13-2\n       k=11
224920
Here are the answers:\n\n9= F\n10= d\n11= h\n12= e\n13= A\n14= g\n15= c\n16= b
6373
Geometric -- multiplying the previous term by 4
102453
D.\n13 + 12=25\n25 + 12=37\n37 + 12=49\n49 + 12=61
581128
First term = x\nSecond term = 2x\nthird = 2(2x) = 4x\nfourth = 8x\nfifth = 16x\nsixth = 32x\nseventh = 64x\neighth = 128x\n\n8th/5th = 128x/16x = 128/16 = 8/1
432476
c.  It's the 7th term\n\nHere is the entire sequence:\n3/16, 3/8, 3/4, 3/2, 3/1, 3/(1/2), 3/(1/4).  Each time you take half of the denominator.  3/(1/4) = 3*4=12.
244869
The Fundamental Counting Principle gives us 12 combinations in this case, so if the first 12 are wearing different things, then the 13th must be wearing one of the other 12's combos. The answer to your second question is 25.
345994
use arithmatic progression formula n(2a+(n-1)d)/2\nwhere n=no. of terms. a=value of first term and d=common difference between each consecutive term\nsay fro your first eg. no. of black ball = 1+3+5+...\nn=10 a=1 d=2\nso sum = 10(2+9*2)/2 = 100\nsolve the rest by changing values of n, a and other variables respectively
282203
You asked for the nth term, so here it is\n\nFigure: n\n\nSquares: n(n+1)/2\n\nDots:  (n^2 + 5n + 2)/2\n\neg. n = 5\n\nsquares = 5 x 6/2 = 15\n\ndots: (25 + 25 + 2)/2 = 52/2 = 26
544230
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
403909
1)  13/107 is already in the simplest form\n\n2)  Notice that 28 = 7*4.  Therefore multiply 32 by 4 to get the denominator that goes under 28.\n\n3)  Notice that 6 = 1.5 * 4.  Therefore multiply 20 by 1.5 to get the denominator that goes under 6.
426917
the first one is (x+10)(x+8)\nand the second one is your second answer
517609
I agree\n\n1) 1x2 = 2\n2+10 = 12\n12/2 = 6\n6-1=5\n\netc
78234
phi(n)=10 ... n=11\nphi(n)=12 ... n=13, 21\n\n14=2x7=phi(3)xphi(n), where phi(n)=7. But phi(n)=7 has no solution.