Photo: Back-to-school dog math problem
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A back-to-school dog math problem:
There are 3 happy pups. Tucker has a lion. Jasper has a lemur. Lilah has a fox. How many seconds before Someone decides that Somebody Else has the better toy and tries to take it?
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hmmm if like my house 1 sec. Have a great day.
not more than 3 seconds… and I would place a bet the lemure is the object of desire… it has a special “must have” effect lol
I think it takes a “nano second” 🙂 Enjoy your Wednesday. Golden Woofs
It will happen as soon as you quit distracting their attention by taking photos. XO, Lily Olivia, Mauricio, Misty May, Giulietta, Fiona, Astrid, Lisbeth and Calista Jo
LOL Good one! At our house it would be about 2 seconds. Sort of the doggie version of the grass is always greener.
16.2 seconds…unless de trane goin EAST makez 7 mor stopz then de trane goin WEST… which iz runnin at 95 milez per hour but iz still 35 minits bee hind schedule…..
did we menshun we got expelled frum; we meen… we excelled in…. math class… at skewl ♥
Two seconds! Mr. N says having all the toys to himself is the best.
Never be satisfied with what you have, always want the other dog…our canine way of thinking.
Lol definitely 30 secs or less
Hmmmm…if woofies are like us kitties, we give it 2 seconds. 🙂
I’m an only cat so I would hog all three.
Wells, I wouldn’t knows since I’m an only pups, butts if I have a fursiblin’, I would say..a nanosecond!!! hehehe
Kisses,
Ruby ♥
I say the correct answer is C. Oh wait, I want to buy a vowel.
Hmmm… that’s a stumper.
MOL at Brian’s comment ^^
Wow, that’s a tough one! Word problems are hard! 🙂
Uhmm as soon as the camera clicked? Love Dolly
2.7 seconds!
I’m surprised you were able to get a picture before they decided someone else had the better toy!
In 3….2….1….I want that toy…
Let’s see, jealousy factor G for green multiplied by the square root of the mass of the combined stuffies equals 2-quick-to-count. 🙂
We give up! How long did it take?