Q:

Only problem 6a. Can someone demonstrate how to do this so I can do the rest solo?

Accepted Solution

A:
as far as I can tell, is simply asking to write two more expressions, that are equivalent to the provided one, namely, grab the provided one and expand it, if you simplify the expanded version, you'd end up with the provided, for example[tex]\bf \boxed{6a.1}~\hfill Β \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill Β \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}}\\\\\\\boxed{6a.3}~\hfill \stackrel{ch anging}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}[/tex]so let's do 6a1, 6a3 and 6a5.[tex]\bf \boxed{6a.1}~\hfill \stackrel{changing}{\cfrac{29\cdot 3}{30\cdot 3}}\implies \stackrel{one}{\cfrac{87}{90}}~\hfill \stackrel{changing}{\cfrac{29\cdot 7}{30\cdot 7}}\implies \stackrel{t wo}{\cfrac{203}{210}} \\\\\\ \boxed{6a.3}~\hfill \stackrel{changing}{\cfrac{15\div 3}{30\div 3}}\implies \stackrel{one}{\cfrac{5}{10}}~\hfill \stackrel{changing}{\cfrac{15\div 5}{30\div 5}}\implies \stackrel{two}{\cfrac{3}{6}}[/tex][tex]\bf \boxed{6a.5}~\hfill \stackrel{changing}{(9\cdot 10)\div (2\cdot 10)}\implies \stackrel{one}{90\div 20}~\hfill \stackrel{changing}{(9\cdot 70)\div(2\cdot 70)}\implies \stackrel{two}{630\div 140}[/tex]