Question 212274: 2 SQRT over6^2 + 6 SQRT over 2^2 = c^2
My friend said that this was the answer..
12 + 12 = c^2
c=SQRT over 24
But I don't know where she got it or if its right.
I'm really not good with Square roots.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your friend is right.
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next time, write it as sqrt(6^2)
this would be the way to do it in algebra.com
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your formula is, as best i can figure out, the following:
2 * sqrt(6^2) + 6 * sqrt(2^2) = c^2
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sqrt(x^2) = x for any number.
take the number 6
square it to get 36
take the square root of 36 and you get 6 because 6 * 6 = 36
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since sqrt(6^2) = 6 and since sqrt(2^2) = 2, your equation of:
2 * sqrt(6^2) + 6 * sqrt(2^2) = c^2 becomes:
(2 * 6) + (6 * 2) = c^2
this becomes:
12 + 12 = c^2 which becomes:
24 = c^2
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to solve for c, you take the square root of both sides of this equation to get:
sqrt(24) = sqrt(c^2) which becomes:
sqrt(24) = c
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you can take any number and square it and then take the square root of the result and you will get the number.
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6^2 = 6*6 = 36
square root of 36 = square root of 6*6 = 6
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similarly, you can take any number and get the square root of it and then square the result to get the number.
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it works both ways.
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10 squared equals 100
square root of 100 = 10
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square root of 25 equals 5
5 squared equals 25
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roots and exponents are related.
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square root of x = x^(1/2)
cube root of x = x^(1/3)
nth root of x = x^(1/n)
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6^2 = 36
36^(1/2) = 6
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if you are having trouble, then you need to review the basics.
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try these websites
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut26_exp.htm
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut37_radical.htm
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http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut38_ratexp.htm
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just copy these website addresses and paste them into your browser address bar and hit the return.
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the main website for these tutorials is:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm
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