Question

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Drag Values To complete each equation

Answer 1

hi 1 and 17^9

Explanation:

Answer 2

Given:

Part A:
`(\left(17^(3)\right)^(6) \cdot 17^(-10))/(17^(8))`

Part B:
`\left(17^(6)\right)^(3) \cdot 17^(-9)`

To find:

The value of each expression.

Solution:

Part A:

Using exponent rule:
`(a^m)^n=a^(mn)`

`$(\left(17^(3)\right)^(6) \cdot 17^(-10))/(17^(8))=(\left(17\right)^(3* 6) \cdot 17^(-10))/(17^(8))`

`$=((17)^(18) \cdot 17^(-10))/(17^(8))`

Using exponent rule:
`a^m \cdot a^(n)= a^(m+n)`

`$=((17)^(18+(-10)))/(17^(8))`

`$=(17^(8))/(17^(8))`

Cancel the common factor, we get

= 1

`$(\left(17^(3)\right)^(6) \cdot 17^(-10))/(17^(8))=1`

Part B:

`\left(17^(6)\right)^(3) \cdot 17^(-9)`

Using exponent rule:
`(a^m)^n=a^(mn)`

`\left(17^(6)\right)^(3) \cdot 17^(-9)=\left(17\right)^(6* 3) \cdot 17^(-9)`

`=(17)^(18) \cdot 17^(-9)`

Using exponent rule:
`a^m \cdot a^(n)= a^(m+n)`

`=(17)^(18+(-9))`

`=17^9`

`\left(17^(6)\right)^(3) \cdot 17^(-9)=17^9`