3. 27
11. 5
13. 2
20. 0
26. 1 (actually, it's a limit that approaches 1)
54. The cube of any negative number.
55. 4 * 4 = 16
2007-03-20 03:58:58
·
answer #1
·
answered by Tom :: Athier than Thou 6
·
0⤊
1⤋
3 No. is 2 7 cube root = 3 is sq.rt. of 2+7 i.e. 9
5 No. is 8 cube = 512 8 = 5+1+2
11 No. is 5 = 3 +2 and = 7 – 2
13 circumference = area of circle
2 pi à r = pi r^2 therefore radius r = 2
16 If the sum of divisors of number, excluding the number is equal to the number is perfect number.
Number is 28 = 3^3 + 1^3
Divisor are 1, 2, 4, 7, 14 & 28
Sum of divisors except 28 is 1+2+4+7+14 = 28
18 The smallest number is Only 2 out of these 50% i.e. one will have same birthday.
20
26 1/2 = 1 – 1/2, 1/4 = 1/2 + 1/2 1/8 = 1/4 + 1/4 and so on
when you add this all the term gets cancelled except 1
Therefore ans is 1
32 Triangular numbers are represented as n(n+1)/2
when n = 3, Triangular No. = 6, When n = 5 triangular No. is 15 and when n = 6 Triangular No. = 21 when n = 7 Triangular No. = 36
15 and 21 difference 21 – 15 = 6 and 15 + 21 = 36
Out of 15 and 21 , 15 is the smaller.
54 cube of any – ve number is less than square of that number
55 The number is 18 where L = 6 and B = 3
Area = 6Ã3 and perimeter = 2(6+3) =18
A square is also a rectangle therefore square with side 4 is having perimeter and area same.
2007-03-20 11:20:20
·
answer #2
·
answered by Pranil 7
·
0⤊
1⤋
3. 27 will work. 27^(1/3) = 3 = 9^(1/2) = (2+7)^(1/2).
5. 8. 512 = 5+1+2=8.
11. 5 is the only such prime: 5 = 3+2 = 7-2.
13. 2Ïr = Ïr², so r²=2r. The only positive
solution is r = 2.
16. Well, 28 = 3³ + 1³. I don't know if this is the
only such solution.
18. The answer is 23. See the reference below
http://en.wikipedia.org/wiki/Birthday_paradox
20. 0. x^0 = y ^0 = 1 for all x, y not 0.
26. This is an infinite geometric series which
converges to 1. The formula for the sum is 1/2 /(1-1/2) = 1.
32. 15. The larger number is 21.
21+ 15 = 36 and 21-15 = 6 are both triangular numbers.
54. This question has infinitely many answers.
Any positive integer cube other than 1 is less than its square.
I.e., The sixth power of n(n>1) is bigger than n³.
55. p = 2l+2w = lw
Certainly 16 will do if l and w are both 4.
2007-03-20 13:28:05
·
answer #3
·
answered by steiner1745 7
·
0⤊
0⤋
3. 27
5. 1
11. 5 (2+3) (7-2)
13. radius of 2, circumference would be 4pi area 4pi^2
16. 25 it is 5^2 and the sum of 3^2 and 4^2
18.
20.
26. 1
32.
54. the cube of any fraction will be less than its square
55. 16 (a square with a side length of 4)
2007-03-20 10:56:08
·
answer #4
·
answered by Jeffrey O 3
·
0⤊
1⤋
3. 27.....27^1/3 = (2 + 7)^1/2
5. 8.....8^3 = 512, 5 + 1 + 2 = 8
11. 5.....2 + 3 = 5, 7 - 5 = 2
13. 2.....PiR^2 = 2PiR, cancel PiR from both sides, R = 2
16. ???...I'm going to defer to Jeffrey for this one.
18. 23 see: http://en.wikipedia.org/wiki/Birthday_paradox
20. 0...everything to the zeroth power equals 1
26. approaches 1
32. 15...21 - 15 = 6, 21 + 15 = 36
54. Any number less than 1 will satisfy this (1/2, -3, etc.)
55. 16 and 18...sides 4x4 and 6x3
2007-03-20 10:52:50
·
answer #5
·
answered by gebobs 6
·
0⤊
1⤋
3) ???
5) ???
11) 5 (Sum primes is 3 and 2) (Difference between 7 and 2)
13) 2 (as pi r^2 = 2 pi r in this case)
16) ???
18) I think it is 183
20) n = 0 (as anything to the power zero = 1)
26) The series will end up being 1.
32) ???
54) Answer-: any number less than 1
55) ???
2007-03-20 10:56:27
·
answer #6
·
answered by Doctor Q 6
·
0⤊
1⤋
3. 27
5. 1
11. 5
13. 2
16.
18.25
20. 2
26. 1
32.
54. the cube of any fraction or negative no.will be less than its square
55. 16
2007-03-20 11:03:07
·
answer #7
·
answered by Aman G 1
·
0⤊
1⤋
3) 26
5) 8
11) 5 = 2+3 = 7-2
13) 2
16) 28 = 1^3 + 3^3
18) 23
20) 0
26) 1
54) (1/2)^3<(1/2)^2, ther are many...
55) 3,6
2007-03-20 11:21:51
·
answer #8
·
answered by a_ebnlhaitham 6
·
0⤊
1⤋
beats me!!!!!!!(not kidding..).
2007-03-20 10:49:14
·
answer #9
·
answered by coolest 1
·
0⤊
3⤋