Answer:
The second blank, which is the exponent on the y there, is -3.
[tex]3y^5 \cdot 4y^{-3}[/tex]
Step-by-step explanation:
[tex]3(?)=12[/tex] implies [tex]?=4[/tex] since 3(4)=12.
[tex]y^5 \cdot y^{?}=y^{5+?}=y^{2}[/tex] implies [tex]5+?=2[/tex].
[tex]5+?=2[/tex] can be solved by subtracting 5 on both sides. This gives you:
[tex]?=2-5[/tex]
[tex]?=-3[/tex]
Let's put it altogether:
[tex]3y^5 \cdot 4y^{-3}[/tex]
[tex]12y^{5+(-3)}[/tex]
[tex]12y^{2}[/tex]
So the first blank is 4.
The second blank, which is the exponent on the y there, is -3.
