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# For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?

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Joined: 02 Sep 2009
Posts: 58340
For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?  [#permalink]

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12 Feb 2019, 01:27
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Difficulty:

35% (medium)

Question Stats:

67% (01:02) correct 33% (01:26) wrong based on 49 sessions

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For all a ≠ 0 and b ≠ 0, $$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$?

(A) $$8a^6b^8$$

(B) $$8a^6b^4$$

(C) $$8a^4b^7$$

(D) $$2a^6b^8$$

(E) $$2a^4b^7$$

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For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?  [#permalink]

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12 Feb 2019, 01:40
1
Bunuel wrote:
For all a ≠ 0 and b ≠ 0, $$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$?

(A) $$8a^6b^8$$

(B) $$8a^6b^4$$

(C) $$8a^4b^7$$

(D) $$2a^6b^8$$

(E) $$2a^4b^7$$

$$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$

Becomes

$$\frac{(16a^8b^6)}{2a^2b^{-2}}$$

$$8a^6b^8$$
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Re: For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?  [#permalink]

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12 Feb 2019, 05:07
Bunuel wrote:
For all a ≠ 0 and b ≠ 0, $$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$?

(A) $$8a^6b^8$$

(B) $$8a^6b^4$$

(C) $$8a^4b^7$$

(D) $$2a^6b^8$$

(E) $$2a^4b^7$$

$$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$

would be

$$8a^6b^8$$
IMO A
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Re: For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?  [#permalink]

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13 Feb 2019, 19:52
Bunuel wrote:
For all a ≠ 0 and b ≠ 0, $$\frac{(4a^4b^3)^2}{2a^2b^{-2}}$$?

(A) $$8a^6b^8$$

(B) $$8a^6b^4$$

(C) $$8a^4b^7$$

(D) $$2a^6b^8$$

(E) $$2a^4b^7$$

Simplifying we have:

(16a^8 x b^6)/(2a^2 x b^(-2))

We move b^(-2) to the numerator, where it becomes b^2:

(16a^8 x b^6 x b^2)/(2a^2)

(16a^8 x b^8)/(2a^2)

8a^6 x b^8

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Re: For all a ≠ 0 and b ≠ 0, (4a^4b^3)^2/(2a^2b^(-2))?   [#permalink] 13 Feb 2019, 19:52
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