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Sliva [168]
4 years ago
9

Irrational equations 64^1/3 how to simplify?

Mathematics
2 answers:
balandron [24]4 years ago
6 0
For indicies, the numerator is the normal power, and the denominator is a root as such. So 1/2 would be square root, 1/3 would be cube root, 1/4 would be 4th root and so on.

So, 64^(1/3) is the third root of 64, which is equal to 4.
Igoryamba4 years ago
3 0
My answer on pitcure :)

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\bf \begin{cases}&#10;A=\cfrac{1}{2}bh&#10;\\\\&#10;\textit{now, scaling "b" and "h" by 3}&#10;\\\\&#10;A=\cfrac{1}{2}\cdot 3b\cdot 3h\to \cfrac{1}{2}\cdot 9bh&#10;\\\\&#10;A=\cfrac{9}{2}bh\to 9\left(\cfrac{1}{2}bh  \right)\\&#10;--------------\\&#10;9\left(\cfrac{1}{2}bh  \right)\textit{ is really 9 times }\cfrac{1}{2}bh&#10;\\\\&#10;\textit{whatever the value of }\cfrac{1}{2}bh \textit{may be}&#10;\end{cases}

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\bf A=\cfrac{1}{2}bh\qquad &#10;\begin{cases}&#10;b=2\\&#10;h=3&#10;\end{cases}\implies A=\cfrac{1}{2}\cdot 2\cdot 3\to 3&#10;\\\\\\&#10;\textit{now, let us scale "b" and "h" by 3}&#10;\\\\&#10;A=\cfrac{1}{2}\cdot  3b\cdot  3h\qquad &#10;\begin{cases}&#10;b=2\\&#10;h=3&#10;\end{cases}\implies A=\cfrac{1}{2}\cdot (3\cdot 2)\cdot (3\cdot 3)&#10;\\\\\\&#10;A=\cfrac{1}{2}\cdot 6\cdot 9\to 3\cdot 9\to 27
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4 years ago
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