solve 81^(1/2) / 3^(-1)
am i correct????
or wrong .... lol
I can only understand the answer you marked there. And its wrong. See, 81 = 9^2 , right?
yes
Fine, so, let us first solve the numerator : \(\bf{81}^{\cfrac{1}{2}} = ? \)
We have 81 = 9^2 \(\left(9^2\right)^\cfrac{1}{2} = ? \)
=9 ??
Good work.
So, the numerator simplifies to 9. Now, the denominator : \(3^{-1} = ? \)
mmmhhh i actually dont know what would that be ?? i havent seen a negative exponent..
my best answer would be zero ????
No, it will not be zero. See, \(a^{-b} = \cfrac{1}{a^b} \)
So, \(3^{-1} = ? \)
81^(1/2) / 3^(-1) = 9/3^(-1) = 9 * 3 = 27 is the correct answer.
\[ \frac{ 1 }{ 3^1 }\]
=1/3
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@meln_n Yes, good work again. So, you have : \(\cfrac{9}{\cfrac{1}{3}}\)
Now, do you have any ideas to simplify this? @meln_n
mmmhhhh lol Fractions! -- i dont acutally ...
No problem. See, that 3 will jump to the numerator (he is promoted :P )
yes
So, you get : 9 *3 = ?
ohhh ok soooo simply 1/3 turns to 3 ...
9 times 3 = 27 n_n
Yes, basically , it happens like this : that 1/3 in denominator will be reversed (i.e. its reciprocal : 3/1) the reciprocal will be multiplied to the numerator (3/1 * 9 ) Thus, it is 27. For example : \(\cfrac{2}{\cfrac{3}{4}} \) Denominator is 3/4 Take reciprocal of it : 4/3 Multiply this reciprocal with the numerator : \(\cfrac{4}{3} \times 2 = \cfrac{8}{3}\)
ohhhhhhh omgoshhhh thank you - i understand n_n !! this @mathslover i really appreciate your help and i though this problem was simply wow i learn alot
:-) Good to hear! Have a great time ahead with mathematics.
thanksss
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