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.