DragonLord

We know that Bill covers $120$ feet at $3$ feet per sec.

Thus, the time taken to cover the walkway will be $\frac{120}{3} \to 40$ seconds. 

Since his default walking speed is $6,$ then the amount he can cover in $40$ seconds is $6 \cdot 40 \to240.$

Now he has $60$ feet left to cover, which will take him $\frac{60}{3} \to 20$ seconds.

Since the Average speed=The Total distance/TheTotal time, we get $\frac{300}{20+40} \to \frac{300}{60}=\boxed{5}$ feet per second is his average rate. 

I got the center of the hyperbola is $(-2, -2)$, but Idk what you want.

We know that there are $\frac{15}{2}$ yards of ribbon. So, to find the amount of bows we can divide.

Thus, $\frac{15}{2} \div \frac{3}{5}$ will become $\frac{15}{2} \cdot \frac{5}{3} \to \frac{5}{2} \cdot 5 \to \frac{25}{2} \to \boxed{12.5}$ 

Thank you for explaining me what the flaw is! Thank you!!! 

I was experimenting with numbers myself, to come apon this! It's quite interesting. 

Oh, I see. Thanks for the response, Melody.

The process repeats, I'm pretty sure all of them will be divisible by 9.

That's a smart way to think about it. Thank you for the help, geno!!! 

Since the common ratio between cats and dogs is 4:5. We want to know cat and dogs in the ratio x:20. Thus, we can set up the equation as $\frac{4}{5}=\frac{x}{20}$. Cross multiplication gets us $80=5x.$ Solving for $x$ we get that $\frac{80}{5}=x \to 16=x.$ Now, we know that the number of cats there are is $\boxed{16}.$

I would appreciated if you put the latex into the latex translator.