The square root of 4 is 2
The square root of 16 is 4
The square root of 4 plus the square root of 16 is 4+2 = 6
6 squared is 36 so 6 is not the square root of 20.
2006-09-05 16:53:07
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answer #1
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answered by Demiurge42 7
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No...the square root of 4 (2) + the square root of 16 (4) is 6 and that is the square root of 36 not 20. It helps to write it out like i just did.
2006-09-05 17:01:50
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answer #2
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answered by hazeleyedbambi 2
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No.
That would be 2 + 4 = 6... but the square root of 20 has to be less than 5 since 5*5 would be 25.
Aloha
2006-09-05 16:52:37
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answer #3
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answered by Anonymous
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The square root of 20 is 4.4721359... The square root of 4 is 2, and the square root of 16 is 4. 2+4=6,and that isn't 4.47...!
2006-09-05 17:04:03
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answer #4
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answered by miyuki & kyojin 7
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What is the squaare root of 4....2? What is the sq rt of 16? 4. So is 2+4 = Sq root of 20? What is the sq root of 20? Not a real number and 2+ 4 = 6. so no.
2006-09-05 17:00:36
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answer #5
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answered by blacklicorice 2
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nope because square root of 4 is 2 and square root of 16 is 4 so add them together and you get 6. Sqare root of 20 is 4.234567
2006-09-05 16:52:40
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answer #6
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answered by Anonymous
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No it doesnt 2+4 = 6 square root 0f 20 isnt 6 its 4,47 something
2006-09-05 16:55:40
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answer #7
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answered by sam (joe thornton) pro 3
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No it is the square root of 36
2006-09-05 16:53:06
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answer #8
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answered by Anonymous
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No.
square root of 4=2
square root of 16=4
2+4=6
square of 6 =36
Hence,no.
Square root of 20 is 4.4721359
2006-09-05 23:36:11
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answer #9
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answered by Firdaus 3
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2 issues to bear in techniques for Diff of two squares (DOTS): DOTS constantly components as (squareroot of 1st term - squareroot of 2d term)(squareroot of 1st term + squareroot of 2d term) or basically: a^2 - b^2 = (a - b)(a + b) Exponents that are even numbers are squares. To take squareroot of even exponent, halve the exponent: squarert of a^8 = a^4 a million. sixteen - a^sixteen = (4 - a^8)(4 + a^8) notice: 1st element is DOTS (4 is a sq., exponent is even , and this is a distinction (subtraction) --> element returned: (2 - a^4) (2 + a^4)(4 + a^8) won't have the ability to be factored added b/c 2 isn't a sq., the different 2 components are actually not variations (-) 2. Rewrite to coach the version: (x^8 - a million/80 one) = (x^4 - a million/9)(x^4 + a million/9) = (x^2 - a million/3)(x^2 + a million/3)(x^4 + a million/9) squarert of fraction is squarert of the two numerator and denominator purely the element with "-" is eligible to element returned. this might't be factored added b/c of the three attempt the others your self now. a splash for #4 is 256 is sixteen^2.
2016-10-14 09:06:28
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answer #10
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answered by ? 4
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