How do you divide square roots by ordinary numbers, i.e (square rt of 12)/3 ?
138815
Using your example:\n\n  sqrt(12) / 3\n\nThe square root (sqrt(x)) and the exponential of the second power (x^2) are inverse functions, then you can write:\n\n  sqrt(12) / sqrt(3^2)\n= sqrt(12) / sqrt(9)\n\nThis is equivalent to:\n\n= sqrt(12/9)\n= sqrt(4/3)\n= sqrt(4) / sqrt(3)\n= 2 / sqrt(3)\n\nMultiply the numerator and denominator by sqrt(3) each:\n\n= 2 * sqrt(3) / [sqrt(3) * sqrt(3)]\n= 2 * sqrt(3) / 3\n= (2/3) * sqrt(3)\n\nIf that's that kind of division you were looking for, that's your answer.
353568
sqrt= square root or radical\n\n3sqrt(125) = 15sqrt(5)\n2sqrt(45) = 6sqrt(5)\n3sqrt(20) = 6sqrt(5)\n15sqrt(5) - 6sqrt(5) + 6sqrt(5) = 15sqrt(5)
510280
Consider your order of operations. Does it look like sqrt(7/100) or [sqrt(7)]/100.  If it is the first one then your answer in decimal form is .264 and if the second then the answer is .0264.  Otherwise 7 is a prime number, so you really cant take the square root of it, unless it is put in decimal form.
637727
Just plug in 1 for x.\n√(1-1) = √0 = 0
161364
You can multiply it out like this:\n\n(2 + 3 times square root of x) times (4 - 5 times square root of x)\n\nequals (2 times (4 - 5 times square root of x)) + ((3 times square root of x) times ((4 - 5 times square root of x))\n\nequals (2 times 4) - (10 times square root of x) + (12 times square root of x) - ((3 times square root of x) times (5 times square root of x))\n\nequals 8 + (2 times square root of x) - (15 times x)
178578
multiply the numerators (numbers on top) to get a final numerator..then multiply the denominator (numbers on the bottom) to get the final denominator...then simplify if/when nesscessary\n\ntherefore\n-4      X     -3    =       12    =      3\n 5             8            40          10
19718
i = cos(pi/2) + isin(pi/2)\n\nsqrt(i) = cos(pi/4) + i sin(pi/4) = sqrt(2)/2 + i sqrt(2)/2\n\nif z = r(cos(t) + i sin(t), to take the square root, you take the square root of r (in your case 1) and divide t by 2.
537004
assuming you mean:\n4 sqrt(x-3) =13\nthen\nsqrt(x-3) = 13/4\nx-3 = 169/16\nx= 3 + 169/16\n~
155335
http://www.nist.gov/dads/HTML/squareRoot.html\nhttp://www.math.com/students/calculators/source/square-root.htm\n\nI hope it will help
452804
If you meant take the square root, then your answer would be (5y times the square root of y) divided by (x squared). You just take the square root of each term. \n\nFor variables, if you are trying to find the square root of x^5, then just think about 5 divided by 2. Since that's 2, you would have x squared outside of the radical and x inside the radical (square root sign). I hope that helps!
169546
Evaluation for it square root\n\nAs we know x^(y*2)=(x^y)^2\ntherefore 81^(1/4)=81^((1/2)*(1/2))=(81^1/2)^1/2\nhence 81^1/2=9\nand 9^1/2=3\nso answer is 3
430363
sq rt2 x sq rt49\n\nsq rt49 = 7\n\nanswer is 7 x sqrt2\n\nwritten 7sqrt2
370130
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!
506075
The answer is:  4.47, and that's all there is to it!\n\nOk Vanessa, all you have to do is pull up your calculator (Start>Accessories).\nI am not suggesting you don't work it out in your head first: it's just basic. Check the answer on the calc, or, if you get stuck....go back.... to POWERS OF HIGH HOLY ROOTISM   SQUARE YOU KNOW, 1x1+(2x2)+2+(10-5)/1=       Simple:\n1+4+2+5/1=12
455236
Saw your edit, see below...\n\nSo if I understand correctly, you start with a decimal number that has a finite number of decimals, like 0.41 or 0.87293, and you want to know if you can build an irrational number that begins with those same decimals, expressing it as a function of integers only.\n\nIn your example, this can be done with 0.41, because sqrt(2) = 1.41421... (the left side of the decimal doesn't matter to you).\n\nSure then.  There are infinitely many ways to do this, but here's just one:\n\nGiven the number "0,a" having n decimals.  So for 0.41, a=41 and n=2.  Then the given number can be expressed as a/10^n.\n\nIf you take sqrt(2)=1.41421... and divide it by 10 often enough, you'll get the number 0.000...0141421... which is irrational.  All you need to do then is add this to the original number.  Just make sure the first 1 appears on the (n+1)th decimal so as not to perturb the first n decimals.  This means dividing by 10^(n+1).\n\nSo the number a/10^n + sqrt(2)/10^(n+1) satisfies the conditions.  Everything is there is integers.\n\nIn the 0.41 case, this would yield the solution 41/100 + sqrt(2)/1000 = 0.41141421356...\n\nEdit: So you don't like it because it uses a.  OK, how about this:\n\nDefine "0,x" as the square of "0,a".  By adding an extra decimal to "0,x" and taking its square root, you will get "0,a" plus an irrational component.  This can be expressed as \n\nsqrt(x/10^n+1/10^(n+1))\n\nwhere n is the number of decimals for "0,x".  In the 0.41 example, this yields the number sqrt(1681/10000 + 1/100000) = sqrt(0.16811) = 0.410012194...
289094
Here's how to work it out:\n\n(Sorry, because I can't use special symbols, I'm using _/ for the square root symbol and <= for the less than or equal to symbol.)\n\n_/3x + 6 + 4 <= 7\n_/3x + 10 <= 7\n3x + 10 <= 7 x 7\n3x + 10 <= 49\n3x + 10 - 10 <= 49 - 10\n3x <= 39\n3x/3 <= 39/3\nx <= 13
528164
It's the same thing as cube root (3a^2 / 9a) which simplifies to:\ncube root (a / 3).\n\nYou don't want a radical in the denominator, so multiply top and bottom by 9.\ncube root (9a / 27)\n\nThis comes out to cube root (9a) / cube root (27) or:\ncube root (9a) / 3.