1)Is 5 to the negitve second power positive or negitive number?\nExplain.\n\n2) Two numbers are their own reciprocals. What are they?\n\n3) What number has no reciprocal?
128197
The negative second power is the reciprocal of the the number to the second power.  Any positive number to the power of a positive power is always positive.  Any positive number to a negative power is also always positive.\n\n1 = 1/1 and -1 = 1/-1\n\n0 has no reciprocal (otherwise 1/0 would be a number)
187382
I know it has already been answered but I"ll take a chance with something different. Ok 3 to the third power plus two would be 3 to the fifth power which is 243.\n :)
271512
1. E\nIf you plug in 1, you get 0 >= 2, which is false.  This eliminates Choice A and D because 1 is a positive integer.\nIf you plug in 4 for S, you get 9 >= 8, which is true.\nAlso if you plug in 6, you get 25 >= 10, which is also true.\nSince B and C are both true the correct answer is E.\n\n2. B\nThe only number that makes the inequality false out of the possible choices is 2.
318519
1. no\n2. 1\n3. 12\n4. 6\n5. no\n6. 25\n7, 2\n8. 90\n9. 9\n10. none\n11. don't know
568639
1) 1nm=10 to the power of minus 1\n589nm=10 to the power of minus*5.89*10 to the power of 2\n           =5.89*10 to the power of 1 or 5.89*10m\nfor second question mail me at soodrachit@yahoo.com cos it is not possible to show you the working here.seriously
470169
1. the number before 10 is 9\nmultiply by two\n9 *2 = 18\nsubtract 1\n18 - 1 = 17\n\nthen for 17\n16*2 = 32\n32 - 1 = 31\n\nthen for 31\n30 * 2 = 60\n60-1 = 58\n\nthen for 58\n58 * 2 = 116\n116-1=115\n\nthat is the first five numbers (including 10):\n10, 17, 61, 58, 116\n\n2. multiply two by three\n2 * 3 = 6\nadd one\n6 + 1 = 7\n\nthen for 7\n7*3 = 21\n21+1 = 22\n\nthen for 22\n22*3 = 66\n66+1=67\n\nthen for 67\n67*3 = 201\n201+1=202\n the first 5 numbers are 2, 7, 22, 67, 202
518366
Anything raised to the 0th power = 1.  Anything raise to the -1 power is its reciprocal.\n\ntherefore the first question is multiplying by 1.  Taking that, the solution becomes:\n\n(2x^-1)^3 * 1 = 1/(8x^3)\n\nThe second one is the reciprocal squared, so it's...\n\n(4/3)^2 = 16/9\n\nThe third one is as follows:\n\ny + 2 = 3/8(x-16)\ny + 2 = (3/8)x - 6\ny = (3/8)x - 8\n\n(hence m = 3/8 and b = -8)
448514
1) 2.01 * 10^2\n\n2) 1.995 * 10^23\n\n3) 5.35 * 10^-4\n\n4) 3.8418 * 10^-7
482844
1) Yup. a_1 - 5 = a_2, so a_1 must be 3.\n2) What does the common ratio actually mean? Its any term divided by the term before it. So in this case, the common ratio is just (-3)/(-2) = 1.5. Now, how would you get the first term? To go from the first term to the second you'd multiply by 1.5; thus to go from the second to the first, divide by 1.5: -2/1.5 = -4/3 is the first term.
498549
If John had a 1 then Sheila's number would need to be 2\n\nIf John had a 5 then Sheila's number would need to be 4\n\nSo when John says "I don't know your number", he has just effectively told Sheila he does not have 1 or 5\n\nThe same logic dictates Sheila cannot have 1 or 5 and also lets you say:-\n\nIf Shiela had a 2, she would know John had a 3 (Since she knows he does not have a 1)\n\nIf Shiela had a 4, she would know John had a 3 (Since she knows he does not have a 5)\n\nSo when Sheila says "I don't know your number" she is effectively saying, "I don't have 2 or 4"\n\nSo Sheila's number must be 3
522381
As the hint says, every number is some power of two multiplied by an odd number, or 2^j * s, where s is odd.\nNow, lets say you have written every number in the set T like that. What possible values for s have you come up with? They all have to be <= 2n-1, since the whole number is <= 2n. That means there are precisely n possibilities: 1, 3, .., 2n-1.\nHowever, you have chosen n+1 numbers. So two of those numbers must have the same 's' part.\nBut then if the numbers were 2^a * s and 2^b * s, one of those numbers is a multiple of the other - assuming a>b, (2^a * s)/(2^b * s) = 2^(a-b) (and similarly for a<=b).
458284
1. 2(n-1) -1= x, where n starts as 10 then n-1 will be equal to 9 and then becomes your answer, or x. 2(9)-1=17, then n is 17 the second time you use this equation....the first 5 numbers are 17, 31, 61, 121, 241. each answer is one less than one less twice the number before it, confirming our equation.\n\n2. 3n+1=x, where n starts as 2. The answers are 7, 22, 67, 202, 607. each number is triple the number before it, plus one.......\n\nsorry if my answers are wrong.......\n\nbut hope this is correct because it is very familiar to me.......it is like an IQ test...
575972
Assuming by "plus" and "minus", you are referring to "positive" and "negative"... they are fundamental rules for mathematics.  To understand them, you need to understand the definition of a "negative number", where subtraction of a positive number is the equivalent of addition of the number's negative form:\n\n5 - 3 = 5 + (-3) = 2\n\nYou can think of negative values as the "explicit absence of" a value or number of items.  With this definition, it should be clear why the multiplicative rules are as you stated.\n\n(positive) * (positive) = (positive) ...is just the obvious, since any positive number of items added multiple times will always result in a positive number of items.  Add 3 apples to a basket 4 times, and you will have 3*4=12 apples.  Mathematically:\n\n3 + 3 + 3 + 3 = 12\n\n(positive) * (negative) = (negative) ...you can use the "explicit absence of" idea I stated above to visualize this one.  If you have 2 holes into which a ball would fit and consider that number to be -2 (negative 2), and you had three rows of them, you will have 3 * (-2) = -6 holes, into which 6 balls would fit.  Mathematically:\n\n(-2) + (-2) + (-2) = -6\n\n(negative) * (negative) = (positive)  ...is derived directly from the 2nd rule above.  If a single negative number changes the sign of the product, then two of them should change the sign right back.  You can also use the "explicit absence of" idea.  If you were counting pairs of holes as negative values (-2), but you were to fill up 4 of these pairs with balls (the filling of which would be considered as "negating the hole", or creating a filled hole), then you would have (-2) * (-4) = 8 holes filled with balls.  Mathematically:\n\n-[ (-2) + (-2) + (-2) + (-2) ] = -(-8) = 8\n\nI hope this makes some sense, and that I didn't just create even more confusion.  ^.^
646202
subtracting a negative is the same as adding a positive\n\nmultiplying/dividing like signs (+,+ ; -,-) will give a positive product/quotient\n\nmultiplying/dividing opposite signs will give a negative answers
384646
First, start listing the prime #'s starting with the number 2.\nAnd let's only go up to 50, because 2 x 50 = 100 and we need 3 #'s.\n\nPRIME #'S LESS THAN 50:\n\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53 ........oops!  STOP, too far.\n\nOkay.  Now we want the product of the 3 #'s needs to be 100.\n\n100 is an even #, so one of the numbers must be even.  \nThat eliminates, quite a lot.\n\nActually, that eliminates all of the #'s but the number 2.\nSo, 2 is the 1st prime number less than 100.\n\nYa, know. . . .\n \nI gave it a try.\n\nAnd I can't seem to find the other two numbers.\n\nYou got me stumped.\n\nPerhaps I read the question wrong.\n\nDid you mean:\nWhat three different prime numbers, multiplied together, have a product of 100?\n\nIf so, you stumped the professor!  :o)
540058
Consecutive numbers differ by 1.\nIf the first one is n, then the second one is n+1\nThat means the third one is the second number plus one (n+1)+1 = n+2\nThe fourth one is (n+2)+1 = n+3\nand the fifth one is (n+3)+1 = n+4\nAdd them altogether\nn + n+1 + n+2 + n+3 + n+4 = 5n + 10
370799
There was actually a very gradual development in attitudes toward \nnegative numbers; people didn't just suddenly start to use them at \nthat time. You can read some of this history here:\n\n   Millennium and the Year 0\n   http://mathforum.org/library/drmath/view/52565.html   \n\nThis refers to the Kevin Brown's Negative Numbers page:\n\n   http://www.mathpages.com/home/kmath298.htm   \n\nAnother page I have found with more details is\n\n   Some elements of history of the negative numbers - Anne Boye\n   http://nti.educa.rcanaria.es/penelope/uk_confboye.htm   \n\nwhich, unfortunately, seems to be a poor translation into English, but \nshows negative attitudes toward negative numbers persisting almost to \nthe present.