I'm a brainlet so could you guys pls solve these questions and teach me how to do it...

I'm a brainlet so could you guys pls solve these questions and teach me how to do it, and pls use the simplest math you can. I'd really appreciate it, cause I'm trying to learn

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Hi senpai, can you send your mom over to my place for an hour or so and pls tell her to bring a female friend. I'd really appreciate it cause I'm trying to get my dick wet

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That's a very negative and immature attitude, broham, and you're not contributing all to helping me meet my goal. Please leave this thread unless you want to help out. I'm just trying to get my rocks off!

You really need to do your own homework but this should get you started on the first two, they're just geometric
Look up examples in your text book and do the first problem yourself, don't just copy

find area of square, divide by 4

I'll give you some hints. w is Width and h is Height.
B and C have same width. Only way they can have equal perimeters is if they have same height. Now we're down to 3 unknowns.
Dw=42
Bh+Ch+Dh=42
Aw=42-Cw
Ah=42-Dh
Aw+Ah = Dw+Dh = Bw+Bh = Cw+Ch

That should be more than enough. Start substituting.

Dude thanks, this really helps!

Dude thanks, this really helps!Oops forgot to add you too

I'll help you out on coin too, complete this graph with the limit on the right kept in mind, think about the constraints and the status of the system with each move

That looks great, but I'd imagine that there has to be a way to do it mathmatically

goddamn control your autism, don't reward this kind of low-effort homeworkposting

This is where it comes from, and I'm actually just trying to learn because I've always sucked at math and decided to come back to these after doing bad on them in high school because I felt bad about it

yeah whatever dude the absolute least you could have done is try to start working on these things and explain your process

I did the first 21 perfectly I think, it's just that these ones particularly #25 get really difficult that I don't know how to start

im literally a brainlet i cant do the first one

If you're talking about 13 I can help you if not there are the other anons who have posted

The solutions are here cemc.uwaterloo.ca/contests/past_contests/2017/2017CayleySolution.pdf

For 23) if it was rectangular x would be sqrt(10^2+17^2), since it is supposed to be obtuse (angle bigger than 90°) it has to be bigger than that. So the smallest possible value would be sqrt(10^2+17^2)+1. The highest possible value would be 10+17-1, because for 10+17 it wouldn't be a triangle but a line. So just sum up all numbers in between.

However I feel like this task is a bit weird written. In the examples the third side is always the biggest. Is this also supposed to be the case for x? The way this task is written I would say you also have to do the same thing again for the two cases that x is the 2nd or the 3rd longest side.

>sqrt(10^2+17^2)+1
Actually nvm that. I assumed the square root would be an integer, however it is 19.72, so the smallest x value would be 20. I hope they don't expect you to do this without a calculator, because then I have no idea how to solve this.

>x is the 2nd or the 3rd longest side.
Ok I also did a mistake here. There is only one other case and that is that 17 is the length of the side opposing the angle >90°.
>pic related
So in this case we take c=17, b=10 and a=x (b=x and a=10 isn't a different case but actually the same).
The highest possible value we get from the condition c>sqrt(a^2+b^2) so:
sqrt(c^2-b^2)>a
13.74>a
The biggest possible x value is therefore 13.
The smallest value we get from considering a>c-b (for a=c-b it wouldn't be a triangle but a line), so:
a>7
So the lowest x value would be 8.

Now the total result is the sum of all allowed x values for both cases:
20+21+22+23+24+25+26+8+9+10+11+12+13=224 (answer E)

Kms I just realized the full solution is already here Also how the fuck is anyone supposed to solve 25 tasks like that in 60 minutes, especially since this is made for high schoolers apparently.

The first 20 are easier, and this one specifically is for grade 10s