powers and roots

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powers and roots

by klaud » Sun Feb 12, 2012 3:35 pm
Question 1

All of the following are equal except:

A)2^20
B)2^19+2^19
C)4^10
D)4^5+4^5
E)16^5


Question 2

2√3+3√2
-------- =
√2

A)2+2√3
B)2+√6
C)2+3√6
D)3+2√3
E)3+√6

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by luiscarlos59 » Sun Feb 12, 2012 3:40 pm
Q1= B, you cant add exponents when you are adding

Q2=E , use the reciprocal to cancel out 2 sqrt 3 / sqrt 2 which equals sqrt 6. And 3sqrt2 / sqrt 2 cancels out to 3. So 3 + sqrt 6

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by pemdas » Sun Feb 12, 2012 4:05 pm
luiscarlos59 wrote:Q1= B, you cant add exponents when you are adding
algebraic simplification is required.
B) 2^19+2^19=2*(2^19)=2^(19+1)=2^20, this isn't correct answer.

the correct choice is d

4^5+4^5=2*4^5=2*2^10=2^(10+1)=2^11
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by Anurag@Gurome » Sun Feb 12, 2012 8:01 pm
klaud wrote:Question 1

All of the following are equal except:

A)2^20
B)2^19+2^19
C)4^10
D)4^5+4^5
E)16^5
A)2^20
B)2^19 + 2^19 = 2^19(1 + 1) = 2^19 * 2 = 2^20
C)4^10 = 2^20
D)4^5+4^5 = 2^10 + 2^10 = 2^10(1 + 1) = 2^11
E)16^5 = (2^4)^5 = 2^20

Only [spoiler](D)[/spoiler] is not equal to other answer choices.

The correct answer is D.
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by Anurag@Gurome » Sun Feb 12, 2012 8:08 pm
Question 2

2√3+3√2
-------- =
√2

A)2+2√3
B)2+√6
C)2+3√6
D)3+2√3
E)3+√6
Multiply the numerator and denominator of given fraction by √2,
(2√3 + 3√2)/√2 = √2(2√3 + 3√2)/(√2)(√2) = [2√6 + 3(√2)(√2)]/2 = [2√6 + 6]/2 = √6 + 3

The correct answer is E.
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