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2007-08-16 11:05:53 · 9 answers · asked by coolkid 1 in Science & Mathematics Mathematics

9 answers

= (8 x^6 y^4)^(1/2)
= 2 √2 x ³ y ²

2007-08-20 10:49:25 · answer #1 · answered by Como 7 · 0 0

sqrt(8 x^6 y^4)

memorize this rule:
sqrt(ab) = sqrt(a) * sqrt(b)

sqrt(8) * sqrt (x^6) * sqrt(y^4)

prime factor
sqrt(2 * 2 * 2) * sqrt (x^3 * x^3) * sqrt(y^2 * y^2)

2sqrt(2) * x^3 * y^2

2 x^3 y^2 sqrt(2)

assume all variables are positive. If not the correct answer is:
2 l x^3 l y^2 sqrt(2)

2007-08-16 11:11:28 · answer #2 · answered by      7 · 0 1

2(x^3)(y^2) and inside the square root (2)

2007-08-16 11:11:11 · answer #3 · answered by icemoon 2 · 0 0

sqrt ( 8 x^6 y^4 ) =
= + or - 2 sqrt(2) x^3 y^2

2007-08-16 11:09:21 · answer #4 · answered by vlee1225 6 · 0 0

Remember these rules for radicals and exponents:
√(abc) = √a√b√c
a^(bc) = (a^(b))^c

So here we have
√[ 8 x^6 y^4 ]
√[ 4*2 (x^3)^2 (y^2)^2 ]
2√[ 2 (x^3)^2 (y^2)^2 ]
2 x^3 √[ 2 (y^2)^2 ]
2 x^3 y^2 √[ 2 ]
(2√2) x^3 y^2

2007-08-16 11:10:27 · answer #5 · answered by Anonymous · 0 0

a million. The sq. root of fifty 4=9 X squarert. of 6 2. The sq. rt. of 10/32 (fraction)=(sq. rt. 10) / (sq. rt. 32) =(sq. rt. 10) / (4 X sq. rt. 2) multiply by (sq. rt. 2) / (sq. rt. 2) to eliminate the novel interior the denominator =(sq. rt. 20) / 8 =(2 X sq. rt. 5) /8 sq. rt. is my abbreviation for the sq. root image in case you have hardship simplifying sq. roots, basically touch me (i admire math :] ).

2016-12-15 17:19:10 · answer #6 · answered by ? 4 · 0 0

√(8x^6y^4)
= 2√(2) x^2|x|y^2
-----------
Ideas: Since √(8x^6y^4) cannot be negative, when you get x^3 out, you can use absolute sign "| |"

2007-08-16 11:09:26 · answer #7 · answered by sahsjing 7 · 0 1

sqrt((8x^6)(y^4))
=sqrt(8x^6) sqrt(y^4)
=(2x^3)(y^2)sqrt(2)

2007-08-16 11:11:24 · answer #8 · answered by Larry C 3 · 0 0

sq rt (8 x^6y^4)
=sq rt{2.2^2.(x^3)^2(y^2)^2}
=sq rt{2(2x^3y^2)^2}
=+-2x^3y^2 sq rt(2)

2007-08-16 11:12:50 · answer #9 · answered by MAHAANIM07 4 · 1 0

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