how do you reduce 4 to the 5th power divided by 4 squared root?
140600
sqrt 4 is the same as 4^(1/2), so the question becomes 4^5/4^(1/2)\n\nWhen dividing, if the bases (in this case 4) are the same, then you can subtract the exponents.  5 - 1/2 = 4 1/2, or 9/2.  So now you have 4^(9/2).\n\nAlso, in a fractional exponent, the denominator is a root, and the numerator is the power it is raised to, so 4^(9/2) is like saying the 2nd root of 4, raised to the 9th power.  This is written as (sqrt 4)^9
800980
1. Move the constant to the right side of the equation:\n4x^2 + 2x = 1\n\n2. Divide out the x^2 coefficient:\nx^2 + 1/2x = 1/4\n\n3. Divide the coefficient of x (in this case, 1/16) by 2 and square it, then add it to both sides:\nx^2 + 1/2x + 1/16 = 5/16\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x + 1/4)^2 = 5/16\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx + 1/4 = +/-sqrt(5/16)\n\n6. Move the constant on the left to the right and simplify:\nx = -1/4 +/-sqrt(5/16) = -1/4 +/-(sqrt 5)/4 = (-1 +/- sqrt 5)/4
141530
Steps:\n1) Divide by 2\nx^2=32\n2)Find square root\nx=sq.rt.32\n3)Simplify Radical\nx=sq.rt.(2)(16)\nANSWER:\nx=4+sq.rt.2\n\n*=times
216863
((4 * 4) / 4) - 4 + 4 = 4
766787
If this is a problem in a math class, the answer provided by abcde12345 is correct. He shows you the technique to simplify it, which is to multiply numerator and denominator by the "conjugate" of the denominator (which just means you change the sign of the square root part, so you get (3 - sqrt(5)).\n\nIf this is a problem in physics or some other class, and all you need is a value, then follow the suggestion of some others who recommend using a calculator.
207303
We could change the second + to 4 by putting a vertical small line\n\n5 + 5 + 5 = 550\n5 + 5 4 5 = 550
896906
The strategy is to reduce the number of square roots by repeatedly squaring and rearranging:\n\nSquare\n\n     11 - 2 √11√10 + 10 < 6 - 2 √6√5 + 5\n\nRearrange\n\n     5 <  √11√10 - √6√5\n\nSquare\n\n     25 < 110 - 2 √11√10 √6√5 + 30\n\nRearrange\n\n     2 √11√10 √6√5 < 115\n\nSimplify\n\n     4 √3√11 < 23\n\nSquare\n\n     4² × 3 × 11 < 23²\n\n     528 < 529\n\nwhich is true.\n\nNote that all steps are reversible because both sides of the "<" are positive. For example, √11 - √10 > 0. Else you get into trouble "proving" that 1 < -2 from 1² < (-2)² !!!
775457
multiplying by x\nx^2-5=4x\nx^2-4x-5=0\n(x-5)(x+1)=0\nx=5 or -1
1144250
2) new-old divided by old\n3) find the square root first\n4) i think its 184... not sure\n5) Distribute 3 to 4x and 4 to get 12x+12 and the 2 tho 5x and 9 to get: 12x+12+10x+18-12 Then simplify (I'm not gonna do all ur work 4 u!)
767712
First you find the square root of both the numerator and the denominator which would end up being √5/√8. This would be a fine answer, but in my math classes teachers have had rules against keeping a radical in the denominator. so, to get rid of that multiply √5/√8 by √8/√8, which gives you √40/√64, or √40/8.
812516
Concentrate on the ^2, because that will have the most significant impact on the value of the expression being raised to the second power:\n\nFirst I tried just raising the 4 to the second power.  4^2 = 16, so "n" / 16 = 6, therefore n = 96.  If you can group the other numbers to reach 96, you have an answer.  But working that out for a while got me nowhere, so I started working leftwards to bring the next number and operator into the square:\n\n(6/4)^2 = 2.25\n\nThat is not an integer so it probably won't work.\n\nEven bringing the next number and operator in, which is 2 -, I still end up with a non-integer:\n2 - (6/4)^2 = -0.25\n\nBut look -- the next number and operator are "8 x " so lets work that out:\n\n8 x (2 - (6/4)^2) = 8 x (2 - 2.25) = 8 x (-0.25) = -2\n\nso that might work if we can come up with a number which works out for ___ + (- 2) = 6\n\nWhat plus -2 is =6?  It is 8 of course.  Now we just need to group 3 x 5 - 3 + 2 to equal 8:\n\n(3 x (5-3)) + 2\n(3 x (2)) + 2\n(3 x 2) + 2\n6 + 2\n8\n\nAha!  We have it.\n\nSo then, or grouping is:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n\nTechnically some of those parenthesis are extras because the order of operations says multiplications and divisions are always done first, you don't need to explicitly show that they are (but it helps keep things clear).  So this would be a simpler form:\n3 x (5 - 3) + 2 + 8 x (2 - (6 / 4) ^2) = 6\n\nBut anyway, putting it back in the more complex form and proving it true works like this:\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x (5-3)) + 2 + (8 x (2 - ((6 / 4) ^2))) = 6\n(3 x ( 2  )) + 2 + (8 x (2 - (( 1.5 ) ^2))) = 6\n(3 x   2   ) + 2 + (8 x (2 - (    2.25   ))) = 6\n(3 x   2   ) + 2 + (8 x (      -0.25        )) = 6\n(6) + 2 + ( -2) = 6\n6 + 2 - 2 = 6\n6 = 6 <<<<< TRUE!\n\n\nHere is another grouping that also works out:\n3 x (5-3) + 2 + 8 x ((2 - 6) / 4^2) = 6\n3 x    2    + 2 + 8 x     (-4 / 16) = 6\n3 x    2    + 2 + 8 x (-0.25) = 6\n3 x    2    + 2 - 2 = 6\n6 + 2 - 2 = 6\n\nand another, this one involves only integers in the proof so might be more what you are looking for:\n3 x (5-3) + 2 + (8 x (2 - 6)) / 4^2 = 6\n3 x 2 + 2 + (8 x (-4)) / 16 = 6\n3 x 2 + 2 + (-32) / 16 = 6\n3 x 2 + 2 + (-2) = 6\n3 x 2 + 2 - 2 = 6\n6 + 2 - 2 = 6\n6 = 6
907035
You have \n100 +4X^2 -16y -40x\n\nFirst let's factor out a 4, since 4 is a factor of each term, that gives\n\n4*(x^2 -10X +25) -16Y\n\nOf course I rearranged the terms so the highest powers of X are\non the left.  I didn't factor 4 out of -16Y because I want the \nquadratic (X^2-10X+25), which I know I can factor.\n\nsince (X^2-10X+25) = (X-5)*(X-5) = (X-5)^2, we have\n\n4*(X-5)^2 -16Y\n\nThe secret is being able to recognize that \n(x^2 -10X +25) = (X-5)^2
162281
The fourth root of 625 is 5 (5 x 5 x 5 x 5 = 625).\nThe square root 0f 121 is 11.\nThe square root of 169 is 13.
309645
Here : \n\n4^2x3-64 reads 4 squared times the cube root of 64.\n\n= 4x4 x (64)^1/3 = 4 x 4 x 4 = 4^3 = 64\n\nAND \n\n16^7/2 x 2^0 = ((4)^2)^7/2 x 1 = (4)^ (2 x 7/2) = 4^7\n\nCOMPRENDE` ?
413999
In step 4 you divided by (a-b) which is zero.\n\nIf you could do this you could just say that 5 times 0 = 8 times 0 means that 5 = 8, by dividing both sides by 0 and canceling it.
282386
First, I suggest that when you encounter the words "square root" you immediately translate to "the one half power." Now your equation is f(x)= 5*[(5x - 5)^(1/2)] -6; write that as 5*[(5x - 5)^(1/2)] -6 = y, so that f(x) = y. Now exchange the variables, x for y, and y for x. The new equation becomes 5*[(5y - 5)^(1/2)] -6 = x. Solve this equation for y and you now have the equation for the inverse of the function f(x), specifically f'(x). Step by step: Add 6 to each side, then divide both sides by 5; you get [(5y - 5)^(1/2)] = (x+6)/5. Now square both sides: (5y - 5) = [(x+6)/5]^2. Add 5 to each side: 5y = [(x+6)/5]^2 + 5; then divide both sides by 5: y = {[(x+6)/5]^2 + 5}/5 = {[(x+6)/5]^2 }/ 5 + 1. Further manipulation if you wish, yields  {[(x+6)^2]/25}/5 + 1, and a final form of [(x+6)^2/125] + 1 = y = f'(x), the inverse function. All you do is exchange the x's and y's and solve for y in the "exchanged" equation.
243752
Well, the definition of a square root has been explained pretty well.  The square root of 4 is 2.  sqrt(9)=3.  sqrt(2)=sqrt(2).  This is in simplist form becuase you cannot write sqrt(2) in a finite decimal expansion.\nHow to Simplify:\n\nConsider sqrt(99).\n99=11*9.\nso, sqrt(99)=sqrt(11*9)=sqrt(11)*sqrt(9)=sqrt(11)*3\nThe steps to follow are as described:\n1) try to break down the number being considered into its prime factors.  In this case: 11,3,3\n2)  look at those factors.  Are any of them perfect squares?  In our case, 9 is 3*3.\n3)  Use this property: sqrt(a*b)=sqrt(a)*sqrt(b) to break thye original number up into an irreducible number and a perfect square if possible.\n\nSome notes:\nThis is not possible if the number does not reduced to perfect squares.  For example: 143=11*13 neither of which are perfect squares so we are stuck here and sqrt(143) is in simplist form.  Hope this helps.