Question

Question #75cc2

Answer 1

`8/r`

Explanation:

`4r^-3*2r^2=8/r`

`a^-3*a^4=a^(-3+4)=a^1=a`

`a^-1=1/a`

`:.4r^(-3+2)*2=8/r`

`:.4r^-1*2=8/r`

`:.4*1/r*2=8/r`

`:.8/r=8/r`

Answer 2

YOur working is correct until the very last step:

Multiply the numbers and add the indices of like bases.
Remember that the numbers and variables are independent of each other. You can think of it as:

`4r^-3 xx 2r^2`

`= (4 xx 2) xx (r^-3 xx r^2)`

`= 8 xx color(blue)(r^-1)" "larr` only the `r` is raised to a negative index

`= 8/color(blue)(r)`

There is a difference between `(8r)^-1 and 8r^-1`

`(8r)^-1 = 1/(8r) and 8r^-1 = 8/r`

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An alternative method is:

`4color(red)(r^-3) xx 2r^2`

`= 4/color(red)(r^3) xx 2r^2 = (8r^2)/r^3 = 8/r`

Hope this helps to explain your problem.