How do you simplify ${(y^{9})}^{\frac{1}{2}} {(p^{9})}^{0}$ $(y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}$ ?

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Answer 1 · 2018-01-30T17:03:28

$y$ $y$ ^(9÷2)#

Consider

$y^{{(9)}^{(\frac{1}{2})}}$ $y^((9)^((1/2))$

By the law of indices

${(a^{m})}^{n} = a^{m n}$ $(a^m)^n = a^(mn)$

$y^{{(9)}^{\frac{1}{2}}} = y^{9 \times \frac{1}{2}}$ $y^((9)^(1/2))=y^(9xx1/2)$

$= y^{\frac{9}{2}}$ $=y^(9/2)$

Consider

${(p^{9})}^{0}$ $(p^9)^0$

Anything raised to the power zero is $1$ $1$ . Hence

$y^{{(9)}^{\frac{1}{2}}} \times {(p^{9})}^{0} = y^{\frac{9}{2}} \times 1$ $y^((9)^(1/2)) xx (p^9)^0 = y^(9/2) xx 1$

The answer is

$y^{\frac{9}{2}}$ $y^(9/2)$

Answer 2 · 2018-01-30T17:07:52

$y^{3}$ $y ^3$

${(y^{9})}^{\frac{1}{2}} {(p^{9})}^{0}$ $(y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}$

According to rules of exponents, any number with an exponent zero gives 1, so ${(p^{9})}^{0} = 1$ $( p ^ { 9} ) ^ { 0} = 1$ .

$\Rightarrow {(y^{9})}^{\frac{1}{2}} \cdot (1)$ $=> (y ^ { 9} ) ^ { \frac { 1} { 2} } cdot( 1 )$

$\Rightarrow {(y^{{(3)}^{2}})}^{\frac{1}{2}}$ $=> (y ^ ((3)^2))^ { \frac { 1} { 2} }$

$\Rightarrow {(y^{3})}^{\frac{2}{2}}$ $=> (y ^ (3))^{ \frac { 2} { 2} }$

$\Rightarrow y^{3}$ $=> y ^3$

How do you simplify (y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}(y9)12(p9)0(y ^ { 9} ) ^ { \frac { 1} { 2} } ( p ^ { 9} ) ^ { 0}?