SOLUTION: (square root of 7 with ab under the radical) to the 14th power notice the paranthesis
square root of 5 with 320x^7y^20 under the radical
square root of three with 48a^10b^
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Question 370590: (square root of 7 with ab under the radical) to the 14th power notice the paranthesis
square root of 5 with 320x^7y^20 under the radical
square root of three with 48a^10b^5 under the radical DIVIDED by square root of the 3 with 6a^8b^4 under the radical
normal square root... with 107-w under the radical then -8=3
Any help is greatly appreciated
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(square root of 7 with ab under the radical) to the 14th power notice the paranthesis
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Comment: Your problem is "the 7th root of (ab)^14"
Answer: (ab)^2
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square root of 5 with 320x^7y^20 under the radical
Comment: Your problem is "the 5th root of (320x^7y^20)
Answer: (32x^5y^20)^(1/5)*(10x^2)^(1/5)
= 2xy^4*(10x)^(2/5)
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square root of three with 48a^10b^5 under the radical DIVIDED by square root of the 3 with 6a^8b^4 under the radical
Comment: Your problem is "the cube root of 48a^10b^5"
Ans: (8a^9b^3)^(1/3)*(6ab^2)^(1/3)
= 2a^3b*(6ab)^(1/3)
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normal square root... with 107-w under the radical then -8=3
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sqrt[107-w) - 8 = 3
---
sqrt(107-w) = 11
---
Square both sides to get:
107-w = 121
---
w = -14
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Cheers,
Stan H.
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