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

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Answer 1 · 2018-01-30T17:03:28

$y$ ^(9÷2)#

Consider

$y^((9)^((1/2))$

By the law of indices

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

$y^((9)^(1/2))=y^(9xx1/2)$

$=y^(9/2)$

Consider

$(p^9)^0$

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

$y^((9)^(1/2)) xx (p^9)^0 = y^(9/2) xx 1$

The answer is

$y^(9/2)$

Answer 2 · 2018-01-30T17:07:52

$y ^3$

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

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

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

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

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

$=> y ^3$

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