Question

Which value is equivalent to 7 multiplied by 3 multiplied by 2 whole over 7 multiplied by 5, the whole raised to the power of 2 multiplied by 7 to the power of 0 over 5 to the power of negative 3, whole to the power of 3 multiplied by 5 to the power of negative 9?

Answer 1

`=1(11)/(25)`

Explanation:

To understand this word problem and form an equation, we will go step by step.

A value:
`x`

is equivalent to 7 multiplied by 3 multiplied by 2:
`x = 7*3*5`

whole over 7 multiplied by 5:
`x=(7*3*2)/(7*5)`

the whole raised to the power of 2:
`x=((7*3*2)/(7*5))^2`

multiplied by 7 to the power of 0:
`x=((7*3*2)/(7*5))^2 * 7^0`

over 5 to the power of negative 3:
`((7*3*2)/(7*5) )^2*(7^0)/(5^(-3))`

multiplied by 5 to the power of negative 9:
`((7*3*2)/(7*5) )^2*(7^0)/(5^(-3))*5^(-9)`

So we have
`((7*3*2)/(7*5) )^2*(7^0)/(5^(-3))*5^(-9)`

`=((3*2)/(5) )^2*((1)/(5^(-3)))^3 *(1)/(5^(9))`

`=((3*2)/(5) )^2*5^9*(1)/(5^(9))`

`=((3*2)/(5) )^2`

`=((6)/(5) )^2`

`= (36)/(25)`

`=1(11)/(25)`

Answer 2

Given problem is 7 multiplied by 3 multiplied by 2 whole over 7 multiplied by 5, the whole raised to the power of 2 multiplied by 7 to the power of 0 over 5 to the power of negative 3, whole to the power of 3 multiplied by 5 to the power of negative 9. which can be written as:

`\left((7\cdot3\cdot2)/(7\cdot5)\right)^2\cdot\left((7^0)/(5^(-3))\right)^3\cdot5^(-9)`

Now we have to find the equivalent value for that so we will simplify it

`\left((7\cdot3\cdot2)/(7\cdot5)\right)^2\cdot\left((7^0)/(5^(-3))\right)^3\cdot5^(-9)`

`=\left((3\cdot2)/(5)\right)^2\cdot\left((7^0)/(5^(-3))\right)^3\cdot5^(-9)`

`=\left((3\cdot2)/(5)\right)^2\cdot\left((7^0)/(5^(-3))\right)^3\cdot5^(-9)`

`=\left((3\cdot2)/(5)\right)^2\cdot\left((1)/(5^(-3))\right)^3\cdot5^(-9)`

`=\left((3\cdot2)/(5)\right)^2\cdot\left(5^3\right)^3\cdot(1)/(5^9)`

`=\left((3\cdot2)/(5)\right)^2\cdot5^9\cdot(1)/(5^9)`

`=\left((3\cdot2)/(5)\right)^2\cdot1`

`=\left((3\cdot2)/(5)\right)^2`

`=\left((6)/(5)\right)^2`

`=(36)/(25)`

Hence final answer is
`(36)/(25)`.