how do i rationalize a number?
454725
If you have a radical in the denominator of a fraction, you have to multiply that fraction by that radical (or some form of it) in order to rationalize it.  If you had 12/sqrt(2), you'd have to multiply that fraction by sqrt(2)/sqrt(2), which will give you 12sqrt(2)/2, or 6sqrt(2).  (You're multiplying by sqrt(2)/sqrt(2) because that reduces to 1 - you have to do this, or else you change the value of the fraction.)\n\nNow if you're fraction has a more complicated denominator, you still have to multiply by some form of the radical in order to rationalize.  Take 12/(1-sqrt(2)).  You can't just multiply by sqrt(2)/sqrt(2) like we did for the last one because your denominator will still have a radical: sqrt(2)[1-sqrt(2)] = sqrt(2)-2.  So instead of just using the sqrt(2), use the conjugate of the denominator.  The conjugate is basically the same denominator, but changing the sign of the second term.  Here, your denominator is 1-sqrt(2), so your conjugate will be 1+sqrt(2).  So multiplying your original expression by 1+sqrt(2)/1+sqrt(2) will give you:\n\n[12/(1-sqrt(2))] * [1+sqrt(2)/1+sqrt(2)] = 12(1+sqrt(2))/(1-sqrt(2))(1+sqrt(2))\n\nUsing FOIL on the denominator gives you\n=12(1+sqrt(2))/[1+sqrt(2)-sqrt(2)+(sqrt(2))^2]\n=12(1+sqrt(2))/(1+2)\n=12(1+sqrt(2))/3\n=4sqrt(2)
398683
Use log you will find\n\nor put\n\nx=3.597285024\n\nIts done
921908
1\n\njust do a little bit of trial you will see that the smaller number you use it nears 1.
942101
start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
1025493
It's rational and it's 2.\n\nI think you answered your own question, 1.9999... is a repeating decimal, so it is a rational. \n\nI think some of the confusion comes from the fact that some repeating decimals, like .33333..., come out of repeated long division. But there isn't any long division that will give you 1.99999...\n\nThe thing to remember is that either you accept that 1.9999... represents a number, or you don't. And if you accept that it represents a number, then 2 is the only value it can take that won't break all of your rules of arithmetic.
685997
u should write it like\n7.52*10^-4
286367
Take the last number off, double it, and subtract it from the rest of the number.  If that is divisible by 7, then the original number is.\n\nFor example 2457\n\n2457 is 245 & 7; 7*2 = 14\n245-14=231\n\nNot sure?  do it again\n\n231 is 23 & 1; 1*2 = 2\n23 - 2 = 21; this is divisible by 7, so 2457 is also divisible by 7.
522815
Do you want to know why you get your phone number?\n\nOkay, so if your phone number was actualy one number as opposed to a series of digits and it was abc-defg then it is actualy \n\n(a*10^6) + (b*10^5) + (c*10^4) + (d*10^3) + (e*10^2) + (f*10^1) + (g*10^0)\n\nwhich means you start with (a*10^6) + (b*10^5) + (c*10^4)\n\nMultiplying by 80 gives\n(8a*10^7) + (8b*10^6) + (8c*10^5)\n\nAdding 1 gives\n(8a*10^7) + (8b*10^6) + (8c*10^5) + 1\n\nMultiplying by 250 gives\n(2000a*10^7) + (2000b*10^6) + (2000c*10^5) + 250\nwhich equals (2a*10^10) + (2b*10^9) + (2c*10^8) + 250\n\nAdding the last 4 digits is the same as adding\n(d*10^3) + (e*10^2) + (f*10^1) + (g*10^0) which gives\n\n(2a*10^10) + (2b*10^9) + (2c*10^8) + (d*10^3) + (e*10^2) + (f*10^1) + (g*10^0) + 250 \n\nAdding them again gives\n(2a*10^10) + (2b*10^9) + (2c*10^8) + (2d*10^3) + (2e*10^2) + (2f*10^1) + (2g*10^0) + 250 \n\nSubtracting 250 gives\n(2a*10^10) + (2b*10^9) + (2c*10^8) + (2d*10^3) + (2e*10^2) + (2f*10^1) + (2g*10^0)\n\nAnd so dividing by 2 gives\n(a*10^6) + (b*10^5) + (c*10^4) + (d*10^3) + (e*10^2) + (f*10^1) + (g*10^0)\n\nwhich is your phone number. This process is the same if you do it backwards. (Multipy where you would divide etc...) This is the same thing as multiplying a number by 4 then dividing it by 2 then dividiing it by 2 then adding 5 then subtracting (20/4)\n\nSorry if I took the magic out of it, but that's the math that explains why it works.
523467
yes, get a calculator
1051751
Use yor calculator....
954164
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
129253
.5024\n\nYou could just use a calculator.
1183043
Well, put it in a calculator. Heck, enter "34567890 multiplied by 9876543" into Google and it'll tell you.
482701
"mode"\n"display digits"\n"float 12" (or however many you want - 12 is the highest)
981666
A number is rational if and only if it can written as a ratio of integers, otherwise it is irrational.\n\nHere are two rational numbers between 5 and 7: \n6 and 6.5.\n(6 can be written as 6/1 and 6.5 as 13/2.)\n\nThat's easy.  I'll give you one irrational.  You'll need to think of one more.\nsquare root of 47 = 6.855654...
1080702
See, first let us see what exactly is meant by an irrational no. I rrational nos. are non-terminating non-repeating numbers such as 1.123456789637631...... Numbers such as 0.33333333..... are rational and not irrational. \n\nRational numbers are those numbers which can be expressed in the form of p/q where q not equal to 0.  0.33333333....  is rational since it can expressed as 1/3 . \n\nAccording to ur ques., i think 0.375932667 is a rational numbr since it can be represented as 375932667/1000000000.\n\nBut if it was 0.375932667.....  that wud become non-terminating non-repeating no. thus it wud be an irrational number.\n\n\nAlso, 0 is a rational as well as a real number.
1084833
Calculators can't tell the difference between a rational and an irrational number. The reason is that calculators only do approximations to some number of decimal places and approximations are not enough to determine rationality. \n\nWe say a number is iirational if it cannot be written as a ratio p/q where both p and q are integers. The usual way to show a number is irrational is via contradiction: assume that it *is* rational and show that something goes wrong. The details of this are different for each number. In general, it is a very difficult thing to show that a number is irrational. If you know the complete decimal expansion of your number, it is enough to show that this expansion never cycles. For example,\n\n.1101001000100001000001....\nis irrational where the number of zeroes increases by one between each 1. At the next level, sqrt(2), sqrt(3), and cbrt(4) are irrational and it isn't too difficult to show this. Also, pi, e, and ln(3)\nare irrational (even transcendental), but this is trickier to show. Finally, there are numbers where it is not known whether they are rational or irrational like zeta(7).