simplying indices

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Jan 17th 2009, 09:38 AM
Tweety

simplying indices

$(4y^3)^3 \div 2y^3$

$4y^9 \div 2y^3$

$4 \div 2 = 2$ $y^9 \div y^3 = y^6$

$2y^6$

is this working right? Because my book says the correct answer is $32y^6$ and I dont understand were 32 came from ?
Jan 17th 2009, 10:02 AM
ursa

Quote:

http://www.mathhelpforum.com/math-he...389952d9-1.gif

http://www.mathhelpforum.com/math-he...59e50af8-1.gif

http://www.mathhelpforum.com/math-he...214b540c-1.gif http://www.mathhelpforum.com/math-he...65ebae49-1.gif

http://www.mathhelpforum.com/math-he...4aab06fe-1.gif

is this working right? Because my book says the correct answer is http://www.mathhelpforum.com/math-he...90d25bd4-1.gif and I dont understand were 32 came from ?

hi
you forgot to make 4^3=64 also(Wink)
Jan 17th 2009, 10:04 AM
Chris L T521

Quote:

Originally Posted by Tweety

$(4y^3)^3 \div 2y^3$

$4y^9 \div 2y^3$

$4 \div 2 = 2$ $y^9 \div y^3 = y^6$

$2y^6$

is this working right? Because my book says the correct answer is $32y^6$ and I dont understand were 32 came from ?

Note that $\left(4y^3\right)^3=4^3y^3=\dots$
Jan 17th 2009, 10:13 AM
Tweety

Quote:

Originally Posted by ursa

hi
you forgot to make 4^3=64 also(Wink)

Hi thanks for answering but I was just wondering why in this expression $(3x^2)^3 \div x^4$ the number 3 is not 3^3 , ?

as the correct answer for this is $3x^2$ so how comes 4 has to be raised to the third power in the expression above?
Jan 17th 2009, 10:17 AM
hmmmm

the answer is wrong as it should be 3^3.

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