Question #75cc2

2 Answers
Apr 14, 2017

Answer:

#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#

Apr 14, 2017

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#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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.