Question 281927
to solve these u must understand how multiplication and division properties work
in these proplems  like {{{8y^5/2y^2=4y^3}}} because 8/2=4 and the 5th power of y divided by the 2nd power of y is really yyyyy/yy=yyy or powers 5-2=3
{{{(8*10^5)/(2*10^2)=a}}}8/2 and powers 5-2
{{{4*10^3=a}}}


{{{(6*10^7)/(3*10^9)=b}}} 6/3 and powers 7-9
{{{2*10^(-2)=b}}}

{{{(2*10^4)/(5*10^2)=c}}}2/5 and powers 4-2
{{{(2/5)*10^2}}}