Question

Renee is simplifying the expression (7) (StartFraction 13 over 29 EndFraction) (StartFraction 1 over 7 EndFraction). She recognizes that 7 and StartFraction 1 over 7 EndFraction are reciprocals, so she would like to find their product before she multiplies by StartFraction 13 over 29 EndFraction. Which property will allow Renee to do this without changing the value of the expression? associative property commutative property distributive property identity property

Answer 1

Answer:

The correct option is commutative property.

Step-by-step explanation:

i took the test and got it right thx to the person ontop of mehn give him credit :p

Answer 2

Answer:

The correct option is commutative property.

Step-by-step explanation:

The expression that Renee is simplifying is:

$(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})$

It is provided that, Renee recognizes that 7 and $\frac{1}{7}$ are reciprocals, so she would like to find their product before she multiplies by $\frac{13}{29}$.

The associative property of multiplication states that:

$a\times b\times c=(a\times b)\times c=a\times (b\times c)$

The commutative property of multiplication states that:

$a\times b\times c=a\times c\times b=c\times a\times b$

The distributive property of multiplication states that:

$a\cdot (b+c)=a\cdot b+a\cdot c$

The identity property of multiplication states that:

$a\times 1=a\\b\times 1=b$

So, Renee should use the commutative property of multiplication to find the product of 7 and $\frac{1}{7}$,

$(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})=(7\times\frac{1}{7})\times\frac{13}{29}=\frac{13}{29}$

Thus, the correct option is commutative property.