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square root of 4489?

sqare root of 4489?

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Square Root of 4489 -

What is Square Root of 4489 ? The square root of 4489 is an odd, 2 digit prime number 67. It is the 19 th prime number.
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What is the square root of 4489 -

The square root of a number is the factor which when multiplied by itself is equal to that number. i.e. 2 X 2 = 4, so 2 is the square root of 4.
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Square Root Of 4489 Simplified? - Math Question [SOLVED]

Simplify square root of 4489 in radical form? What is the square root for the number 4489. Learn how to simplify square roots into radical forms.
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Find the square root of 4489 by factorization method ...

Find the square root of 4489 by factorization method Purchase. 011-40705070 or . Call me . Select ... find square root by long division method. 16 ...
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Square root of 4489 using prime factorisation -

Square root of 4489 using prime factorisation - 1662422. 1. Log in Join now Katie; a few seconds ago; Hi there! Have questions about your homework?
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Square And Square Roots Class Eight Mathematics CBSE Maths

Square And Square Roots. Exercise 4. 1. Find the square root of each of the following numbers by Division method. (i) 2304 (ii) 4489 (iii) 529 (iv) 3249
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Square root of 4500 - Conversion Calculators

Find the square root of 4500. ''Square Root Calculator' with step-by-step solution using the Babylonian Method or Hero's Method. Square Root Table.
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Shortcut to Find Square Root of a Number - Testbook Blog

Learn shortcut to find square root of a number. You can learn this quick easy trick in 2 minutes and save upto 5 minutes in bank exams.
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list of perfect squares - Square Root Calculator

Perfect Squares and their Square Roots Perfect Square: Taking a positive integer and squaring it (multiplying it by itself) equals a perfect square.
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Suggested Questions And Answer :

complete the square to find the x intercepts of the function

x^2 +14x+78 1 wae tu look at it...(x+7)^2 +29 or yuze quadratik equashun & get roots= -7+-5.385164807i  (komplex variabels)
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how do I solve x2 -6x +6 = 0

x2- 6x  +6 =0 compare with ax2 + bx + c = 0 here a = 1; b = -6 ; c = 6; D = b2 -4ac    = (-6)(-6) - 4 * 1 * 6   = 36 - 24  = 12 [ It is not a perfect square, so, you cannot get roots what u imagined] Use below method; x = -b+ square root of D/2a  or -b - square root of D/2a ​x = - (-6) + square root of 12/ 2*1 or - ( -6) - square root of 12/2*1 x = 6 + 2*square root of 3/2 or 6 - 2* square root of 3/2 x = 3 + square root of 3 or 3 - square root of 3
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zeros of 2x^6+5x^4-x²

All the powers of 2 are even so this 6-degree expression can be reduced to a cubic if we put y=x^2: 2y^3+5y^2-y. Putting this equal to zero to find the y zeroes: 2y^3+5y^2-y=0. This factorises: y(2y^2+5y-1)=0. Take the quadratic: 2y^2+5y-1=0 and rewrite it as: y^2+5y/2=1/2 by dividing through by 2 and moving the constant across. Now we can complete the square by dividing the y term by 2 and squaring its coefficient: y^2+5y/2+25/16=1/2+25/16, and you can see we've added 25/16 to both sides of the equation to make it balance. We now have a perfect square on the left: (y+5/4)^2=1/2+25/16=(8+25)/16=33/16, so we can take the square root of each side:  (y+5/4)=+sqrt(33)/4. Therefore, y=-5/4+sqrt(33)/4. This is the same result as we would have found using the formula for solving a quadratic equation. We need to calculate what these solutions are: 0.18614 and -2.68614. But we want x, not y, so we need to find the square root of these values of y. Assuming we don't want complex solutions (involving the imaginary square root of -1), we can only use the positive solution 0.18614 and take the square root of that, which is +0.43144 approx. so the real solution is x=0.43144 or -0.43144. (The complex solution is +1.63894i, where i=sqrt(-1).) But we're not finished yet, because y=0 was also a solution, and that means x=0 is a solution, so we have three possible real zeroes for x: 0, 0.43144 and -0.43144 (and two complex zeroes).
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show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!
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how to solve this quadratic equation x^2-2^x-13=0

I think you may have typed this in wrongly because the term 2^x is not a permitted term in a quadratic equation. As it stands the solution to the equation is about -3.6168. Let's say you made a mistake in typing and the middle term is 2*x or 2x, then the solution exists but it is irrational (involves square roots). The easiest way to solve x^2-2x-13=0 is to complete the square: (x^2-2x+1)-1-13=0, which is (x-1)^2-14=0, so (x-1)^2=14. Take square roots of each side: x-1=+sqrt(14)=+3.7417 approx. A square root always has two solutions, one positive and one negative, but the same magnitude number. So x has two possible values: x=1+3.7417=4.7417 or x=1-3.7417=-2.7417. If you meant the question to be x^2-12x-13=0, then the solution is (x-13)(x+1)=0 and x=13 or -1. To work this out we ask: what are the factors of 1 (x^2 coefficient) and 13 (the constant term). The factors of 1 are 1 and 1 because only 1 times 1 make 1; and the factors of 13 are only 1 and 13. The coefficient of the x^2 term tells is how many x's go in each bracket. That's 1x or just x in each bracket. And the factors of 13 tell us what. Numbers to write in each bracket, so that's 1 and 13. So we have (x 1)(x 13). What about the signs between them? We look at the sign in front of 13 in the quadratic. It's minus, and that means there will be a plus in one bracket and a minus in the other. But which way round? Well, there's one more test: we take the factors of 13 and subtract them because the minus sign in front of 13 tells us we need to subtract. If it had been plus, we would have added the factors. 13-1=12. If 12 is the coefficient of the x term then the quadratic can be solved. (If the number had not been 12 we could not have solved the quadratic this way.) The sign in front of 12x is the sign that goes in front of the larger number in the brackets, so minus goes in front of 13. So we have (x+1)(x-13)=0. One or other of these factors is zero, so x+1 or x-13 is zero. x+1=0 means x=-1 and x-13=0 means x=13. These are the solutions or roots.
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What is the y-coordinate of the endpoint in quadrant III.

The point (1,6) is one endpoint of a line segement that has a length of square root of 80 units. The other endpoint is in quadrant III and has an x-coordinate of -3. We can draw a right triangle using these two points, and use the Pythagorean theorem to find what we are looking for. The length of the x leg is (1 - -3), which is 4. The length of the y leg is (6 - y). We are given the length of the hypotenuse, the square root of 80. The formula is x^2 + y^2 = c^2  (I used x and y in place of a and b) 4^2 + (6 - y)^2 = 80  (square root of 80, squared) 16 + 36 - 12y + y^2 = 80 Subtract 80 from both sides. 16 + 36 - 12y + y^2 - 80 = 80 - 80 y^2 - 12y - 80 + 36 + 16 = 0 y^2 - 12y - 28 = 0 Factoring, we get the following: (y - 14) * (y + 2) = 0 Because we are multiplying and getting a zero, one of the two factors has to be zero. y - 14 = 0 y = 14 Or... y + 2 = 0 y = -2 The problem statement says the endpoint we are interested in is in quadrant III, which means the y co-ordinate has to be negative, therefore y = -2 The point is (-3, -2)
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solve : x^2+kx+m=0

Simplest solution: we know that x=1 is a solution so x-1 is a factor. In a quadratic the constant term is the product of the roots so m must be the other root. The x term is minus the sum of the roots, so k=-(1+m). 1st alternative solution: Divide the quadratic by x-1 or use synthetic division applying the root 1 and you get the result x-m. 2nd alternative solution: We can apply the quadratic formula or complete the square. Let's apply the formula: x=(-k±sqrt(k^2-4m))/2. We know that one root is x=1, so (-k±sqrt(k^2-4m))/2=1. Let's assume that we take the positive square root: (sqrt(k^2-4m)-k)=2 or sqrt(k^2-4m)=k+2. Squaring both sides we get: k^2-4m=k^2+4k+4 from which m=-(k+1) or k=-(m+1). But that means we can calculate the square root in the quadratic formula and get the other root: x=(-k-(k+2))/2=(-2k-2)/2=-(k+1)=m. The quadratic becomes: (x-m)(x-1)=x^2-x+mx-m=0: x^2-x(m+1)+m=0.
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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how do i solve the quadratic equation 16x^2=25

This is a perfect square.  First set it equal to 0 by subtracting 25 from both sides 16x^2-25=0 Take the Square root of both terms Square root of 16x^2 is 4x Square root of -25 is 5 and -5 The factored form is (4x+5) (4x-5) If you set each of these equal to 0 and solve for x you get 4x+5=0 subtract 5 4x=-5 divide by 4 -5/4=x Do the same with the next factor 4x-5=0 add 5 4x=5 divide by 4 x=5/4 the two roots for this are 5/4 and -5/4
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How to Find Square Root

98=49*2, so sqrt(98)=sqrt(49)*sqrt(2)=7sqrt(2)=7*1.4142=9.8994 approx. There's another way using the binomial theorem. 98=100-2=100(1-0.02). sqrt(100)=10 so sqrt(98)=10(1-0.02)^(1/2) because square root is the same as power 1/2. (1+x)^n expands to 1+nx+(n(n-1)/1*2)x^2+(n(n-1)(n-2)/1*2*3)x^3+... Putting n=1/2 and  x=-0.02, we get sqrt(98)=10(1-0.02)^(1/2)=10[1-(1/2)0.02+((1/2)(-1/2)/2)0.0004+...]. This gives us: 10(1-0.01-0.00005+...)=10*0.98995=9.8995. A third method is to use an iterative process, which means you keep repeating the same action over and over again. Look at this: x=10-(2/(10+x)). If we solve for x we get x=sqrt(98); but we're going to find x in an iterative way. Start with x=0 and work out the right hand side: 10-2/10=9.8. This gives us a new value for x, 9.8, which we feed back into the right hand side: 10-(2/(10+9.8))=10-2/19.8=9.8989..., giving us another value for x, 9.8989... which we feed back into the right hand side: 10-(2/(10+9.8989...))=9.89949..., giving us yet another value for x and so on. Very quickly we build up accuracy with each x. You can do this on a calculator, a basic one that doesn't even have square roots, using the memory to hold values for you. Here's a very simple program, where STO means store in memory (if your calculator doesn't have STO use MC (memory clear) followed by M+ (add to memory)); MR means read memory (the steps show what calculator keys to press in order; / may be ÷ on your calculator): 0= +10=STO 10-2/MR= GO TO STEP 2 OR STOP (display should show the answer for sqrt(98)) Note: In STEP 3 the division must be carried out before subtracting from 10, otherwise you get the wrong answer. If your calculator doesn't do this you need to replace STEP 3 with: 0-2=/MR=+10= You should only have to go round the loop a few times before you get a really accurate result. To find the square root of 2 directly the iteration equation is x=1+1/(1+x) and the program is: 0= +1=STO 1/MR+1= GO TO STEP 2 OR STOP STEP 3 should work on all calculators.
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