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Two years and change ago, I discovered coffee at home as a doable, cost-saving thing. Sometimes I am a little late to the party of Obvious Things, but better late than never.

I ran some preliminary numbers on the cost to determine that yes, it was boatloads cheaper to make coffee at home. To that end, I have been making a concerted effort to do my coffee at home instead of buying it out and about. And I have better numbers now on how that is going.

Come and see! )
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So in the previous post, we discovered that logs 26" in diameter were like 40% bigger than logs 22" in diameter. Is there a point where a 4" increase in diameter makes the log 50% bigger? Heck, what is the relationship of 4" difference in trunk diameter as the overall size of the tree increases? How can we find out? (Yeah, it's maths. Look, this is not my fault. I just go where the current interests take me. Probably this all stems from the "half a dog" bullshit from badly-written homework at the beginning of the month. What can you do?)

Everyone wants to read about maths, right? )
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Mom shares her 7 yr old's homework problem, almost no-one can figure it out.

This is because people hate Al.

Al who?

Al-gebra. LOL.

Here's the problem:

There are 49 dogs signed up for the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?

This is a word problem. It's also DUMB AF but we're not there yet. Let's do the math first.

Step 1. Find out what they want to know.

Total number of small dogs.

Step 2. List what you know.

49 total dogs. There are an unknown number of large dogs and then there are 36 MORE small dogs than there are large dogs. The total number of small dogs is (36 + the unknown number of large dogs) and that's the thing we are supposed to find out.

Our friend Al says that we can represent the "unknown number of big dogs" as D. Let's do that because typing "unknown number of big dogs" is a pain in the ass.

49 = (36 + D) + D

Did you see how we got here? D is number of big dogs, 36 + D is number of small dogs. And there's 49 dogs total.

Now, let's move all the numbers to one side to isolate the Ds by themselves. If we minus 36 on the right, we have to minus 36 on the left to keep things balanced, like a see saw.

That gives us

13 = D+D

13 = 2D 2D is the same as D + D

Now, divide both sides by 2 so that we can have the value of just plain D. You gotta do both sides because balance like a see saw.

6.5 = D, the unknown number of big dogs.

Here's the dumbfuckery. Dogs do not come in "half a dog" amounts, particularly when you have them attending a dog show. A dog show presupposes real live dogs, not mathematical ideal dogs, trotting around in a circle on leashes and being stood up for a judge to examine them for conformation.

Even seven year olds know that dogs do not come in "half a dog" amounts for the purposes of "attending a dog show". Sanity check your fucking word problems, folks.

But we need "total number of small dogs" for the answer. Like, you're being GRADED on this even though it's dumb as fuck.

We know that there are 36 MORE small dogs than big dogs and we have previously represented "number of small dogs" as 36 + D and there are definitely 6.5 big dogs, like, the math don't lie... so... there are 42.5 small dogs.

Probably. I mean, if that were a thing but it's dumb as fuck because you can't have "half a dog" in any rational world. Is there some sort of vile dog butchery going on in this little kid homework?

Let's check the math part, anyway. Total dogs is 42.5 small dogs plus 6.5 big dogs equals 49 which is the total number of dogs. So yeah, it's probably correct. STUPID AS HELL but probably correct.

All done, thanks, Al! Too bad we can't have sensible answers in a world where all dogs come as whole numbers, a very difficult thing that would totally ruin the math problem that you could only make happen by (for example) setting the number of excessive small dogs to ANY ODD NUMBER FROM 1 to 47. Seriously.

1 more small dog than large dogs. 24 large dogs, 25 small dogs. No partial dogs.

47 more small dogs than large dogs. 48 small dogs. 1 large dog.

You only get half-a-dog issues if the number of excess small dogs is even. For the odd numbers from 1 to 47, the problem works just fine.

Anyway, there are 42.5 small dogs. At the dog show. Some lady in a nice pink twinset with heels is dragging half (cut longways like a side of beef) of a mini poodle around, smearing blood all over the artificial turf, while smiling rigidly and keeping her eyes on the judge. Everyone is pretending things are normal. Nobody says anything to the lady in the nice pink twinset about her HALF A DOG.

In the Big Dog classes, a man wearing an out of date suit is lugging half a Great Dane around, the front half, trailed by, well, the entrails. It's a bit of a mess. When it comes time for him to "stand the dog up" he tries his best but cannot balance the carcass on two front legs with the head and neck drooping despite his best efforts on the lead and collar.

This is the sort of nonsensical nightmare fuel that makes people hate math.

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