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Users vs Mods (& Admin)

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DJ Enetro
37
Why is Meah not helping :(
Achromalia
38

*shrug*

well I'm currently doing homework
DJ Enetro
I have a couple hours per day sadly (of homework)

39
Achromalia
usually it takes anywhere between 2-7 hours for me.

40
DJ Enetro
Well rip

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Achromalia
42

ye I get distracted easily and sometimes homework can be unusually time-consuming
DJ Enetro
distractions add to time consumption tho
43
Achromalia
potentially. even pure focus can take up around 4 or 5 hours of time. i get distracted and add on a couple more, dependent on what im distracted by (like osu forums), or how tired I am of solving THAT ONE VAGUE AS SHIT MATH PROBLEM

also enetro your siggy is pretty weird. What's the naive part in knowing things arent fine, but acting as if they are? To expect that people will think they're okay? Not exactly sure how its naive. Sometimes that can work perfectly well. "progression into Societal acceptance" is a pretty iffy phrase, but sure why not, being vague seems cool nowadays.

44 :D
DJ Enetro
45

math huh? Please show the problem to me.
After all math is pretty simple for me compared to words
Achromalia
46.

(5.2.1.) #67:

"What are the possible side lengths for side ML in the triangle /(at the right)/? Show how you know."

Given information: MS = 10", SL = 4". Angle MSL can be observed to have an obtuse angle.

Edit: All I've been able to assume is that ML could be 12.
DJ Enetro
i can see the answer already, but i don’t want to give you the answer straightaway. May i proceed?
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Husa
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Achromalia
sure. at least explain how it works, and i'll be okay.

49 (husa. unsuccessful ninja, youre 48.)

pretty long explanation to have to last approx. 10 15 minutes, especially for something you already know. did something happen?

hopefully youre not writing an essay trying explain how to do this xD
DJ Enetro

Xuequinox wrote:

(5.2.1.) #67:

"What are the possible side lengths for side ML in the triangle /(at the right)/? Show how you know."

Given information: MS = 10", SL = 4". Angle MSL can be observed to have an obtuse angle.

Edit: All I've been able to assume is that ML could be 12.
I'm going to explain this in detail, forgive me if this is annoying.

a^2 + b^2 = c^2
where a, b, and c are sides of a triangle
(a, b and c are all greater than 0 because they are side lengths)
That's the Pythagorean Theorem, which is the core for solving problems like these.
For example, a triangle with sides of 3, 4, and 5 can be considered a right triangle:
3^2+4^2=5^2
9+16=25
25=25
If we take this one step further, we can find out that:

1. if a^2 + b^2 < c^2, then the triangle is obtuse. generally you have to use the longest side length for c to make this work.
2. if a^2 + b^2 > c^2, then the triangle is acute.

Great, i'm going off-topic.

Let's look at the given first:

  1. MS = 10"
  2. SL = 4"
  3. Angle MSL can be observed to have an obtuse angle

Because MSL has an obtuse angle, (MS)^2+(SL)^2 < (ML)^2.
So let's plug in the values for MS and SL:

10^2 + 4^2 < ML^2
100 + 16 < ML^2
116 < ML^2
(taking the square root of both sides)

sqrt(116) < ML
which approximates to 10.7703296

That's only half of it though.

We need to find a maximum value for ML as well, because think about it:
we can put MS and SL into a single line, can we not? Since a straight line cannot form a triangle anyway, we can take the addition of the two known sides to find the maximum. (sorry, idek if this part will make sense)

So to find our maximum value of ML, we simply add the other two side lengths we know.
ML < MS + SL
ML < 10 + 4
ML < 14

So if we put these two together, our final answer will be:

SPOILER
sqrt(116) < ML < 14

p.s.: i forgot to mention that there's a thing called the Triangle Inequality Theorem.
It's important to know if you want to solve problems like this in the future.

50
Achromalia
aaaaaaaaaaaaaaaaaaaaand you wrote... well, its not exactly an essay.

thanks though, that's actually relatively easy to digest. (seems like I had half of the answer already. I wrote down 'ML < 14"' as my answer, but I didn't know what to do with the minimum.)

51.
DJ Enetro
That seems like geometry-level stuff isn’t it? Hit me up anytime you need help with that.
Well, I gotta sleep.
26*2.
Achromalia
I'm only one grade advanced in math, so IM2 is something I'm working on for my second semester in 9th grade.

I think it was mostly confusing because no one ever explained this, especially not the teacher. :/

53.
DJ Enetro
54
you have to put the minimum because they specifically said obtuse angle.
Achromalia

DJ Enetro wrote:

54
you have to put the minimum because they specifically said obtuse angle.
they didn't, I mentioned that "it can be observed", as in it wasn't directly said.

55.
Husa
58
Achromalia
why am I awake

57 (husa again that's supposed to be 56 ;w;)
Husa
56 then
vinnicci
59
[Lunatic Elf]
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Husa
61
DJ Enetro
62
(Crap, sqrt(116) can be simplified to 2sqrt(29))
Husa
uwu what are you talking about 63
DJ Enetro
Look in the previous page
2^6
Husa
65
Achromalia
(200 / 3) - 2/3

(edit: shit I forgot the 1/3 wqas supposed to be 2/3)
DJ Enetro
More like (199/3)-1/3

67
Achromalia
1`2`3
,
1`2`3`2
DJ Enetro
What is that

69
If it’s some math tell me about it
Achromalia

DJ Enetro wrote:

What is that

If it’s some math tell me about it
` = "and", or +.

I just go from 1-9 and back unless the number goes in between, in which case I stop and go back, and I can end before 1 or 9 if I need to. You can bounce back up if you need to.

for example, 9 = 1`2`3`2`1.

, = is simply a seperate digit down.

. = is a decimal, i messed that up, will fix.

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vinnicci
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Husa
72(7)pp
DJ Enetro
73
Lmao
Husa
74
vinnicci
75
Husa
76
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