Deja had $25 to spend at the fair. if the admission to the fair is $4 and the rides cost $1.25 each, what is the greatest number of rides daniel can go on?

It depends. Did you write it correctly?There is two names Deja and Daniel. If its two people going into the fair and the admission fair is $4 each , then for both the admission would be $8. If that is the case then subtract 8 from $25 and you are left with $17. We will asume Deja, as well will get on the rides not Just Daniel, but the money is for both. So I just kept multiplying a number which represents the rides by $1.25. For example, 1.25×13= $16.25, 1.25×14=17.50 it is too high, 1.25×12=$15, so if both get on then it has to be even the rides, So dived the rides which the greatest was 12 , then you will be left with 6 rides each. So the greatest number daniel go on was 6. However, if you just wrote the name wrong and it is only one person we are talking about, then it is still the same process just instead we will subtract $4 from $25, and Daniel is left with $21 for the rides. Just multiply 1.25 by a number until you get close but not past the amount of $21. example,
1.25×16=$20, 1.25×17=$21.25. The answer is 16. The greatest number Daniel rides is 16.

Yeah she/ he needs look how long that is man