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For all positive numbers x, Δx is defined as the cube root of x, and ∇

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For all positive numbers x, Δx is defined as the cube root of x, and ∇  [#permalink]

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New post 22 Aug 2018, 04:12
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A
B
C
D
E

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  15% (low)

Question Stats:

79% (01:29) correct 21% (01:54) wrong based on 66 sessions

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For all positive numbers x, Δx is defined as the cube root of x, and ∇  [#permalink]

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New post 22 Aug 2018, 04:18
1
Bunuel wrote:
For all positive numbers x, Δx is defined as the cube root of x, and ∇x is defined as the square root of x. If ∇(Δk)=m^2 , then k =


A. m^(12/5)

B. m^6

C. m^12

D. m^36

E. m^64


Best way to go about doing this problem is to substitute a value for \(m\). Let \(m = 2\)

Therefore, we have to find the value of ∇(Δk) = \(m^2\) or 4

\(4(2^2)\) is the cube root of \(4*4*4 = 64(2^6)\), which the square root of \(64*64 = 4096(2^{12})\)

Substituting the values in the answer options, we arrive at the value of k which is \(m^{12}\) (Option C)
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For all positive numbers x, Δx is defined as the cube root of x, and ∇  [#permalink]

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New post 22 Aug 2018, 04:23
1
Bunuel wrote:
For all positive numbers x, Δx is defined as the cube root of x, and ∇x is defined as the square root of x. If ∇(Δk)=m^2 , then k =


A. m^(12/5)

B. m^6

C. m^12

D. m^36

E. m^64


\(∇(Δk)=m^2\)
Or,\(∇(k^{\frac{1}{3}})=m^2\)
Or, \((k^{\frac{1}{3}})^{\frac{1}{2}}=m^2\)
Or, \(k^{\frac{1}{2}*\frac{1}{3}}=m^2\)
Or, \(k^{\frac{1}{6}}=m^2\)
Or,\(k=(m^2)^6=m^{12}\)

Ans. (C)
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For all positive numbers x, Δx is defined as the cube root of x, and ∇   [#permalink] 22 Aug 2018, 04:23
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