55 dogs come upon 1 dog respectively. They adjectives enjoy a huge spectator sport of frisbee next to thier doggy brothers, how heaps dogs?
Question:
Answers:
54
55?
I hate math!!
I know the answer if it be for cats, but dogs. I'm not sure.
2 dogs if theyre all appointment the same one
110, the doggies enjoy girlfriends too.
110
Is it 55?
If they have adjectives met the same dog later there are 56. If they enjoy met a seperate dog each next there should be 110...
Whichever it is...that's one heck of plentifully of dogs in one place at one and the same time!
220 hows that
56
110 dogs, but I think they be chasing cats not the frisbee.
if they all met duplicate dog then 56. if they adjectives met seperate dogs then 110.
56
I reckon 56, becouse they may of met duplicate dog! ?
55
There's not enough info to find a single answer. It could be 55 (and probably is intended to be), assuming that the imaginative 55 were adjectives strangers and they all met another inwardly that group. But it could be anywhere up to 110 (55x2) depending on how many of these dogs met next to a stranger outside the group. That number is not specified anywhere in the query. The frisbee game can be determined to be all-inclusive, but we are disappeared to assume whether or not the meeting occur within or outside the artistic group, so since all possible assumptions enjoy to be considered, the solution is any number between (and including) 55 and 110.
110.
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