geometry-and-trigonometry

[dull egret]

i mean, this looks so much like there d be some trick with pythagoras

[foggy oxide]

hint : try some geometric transformation

almost like that

(by geometric transformation i mean, reflection, rotation)

[dull egret]

yeah yeah

[foggy oxide]

This one is very geometric and i remember this solution because there was another problem with the same case (AP^2 = BP^2 + CP^2) where it wanted the total area of the triangle

[dull egret]

so well, by some property of the circle the name of which i forgot P must be on a circle around A right?

or wait

[foggy oxide]

idk 😂 , let me try to remember it

[dull egret]

if there is a singular solution, then a valid point to use is the vertex B(actuall C too)

then the angle doesnt exist

great

[foggy oxide]

😂 dafuq

[dull egret]

but in the limit it is gonna be 180-30 degrees

so 150

my guess

[foggy oxide]

You're right xD

i have no idea how you came to this sol

but yeah , that's the answer 😂

[dull egret]

well, get the point P arbituarily close to B which is a possible position for P

then the angle PBC is gonna be half of ABC

and cause that is 60 in an equilateral it is 30

[foggy oxide]

did you liked this one ?

one of my favorites too

[dull egret]

similarly it is gonna be 0 for PCB

yeah just that i have a more algebraic solution again xDDD

[foggy oxide]

there is a more algebraic sol too

i won't send it

😂

[dull egret]

xD

i m prob the only person on this planet who d solve it this way, but it works

[foggy oxide]

xDDDDDD, i would not say that, but i can almost have certainty that you're the only one who solved it like that 😂

[dull egret]

hahaha

[foggy oxide]

well bro, i'll watch something or practice M U S I C see you later xD

[dull egret]

alright :P

[upper karma]

can someone help with math

[dull egret]

probably

[crude kraken]

Sorry, this is a French Culture discussions server

[upper karma]

All points in a grid lie in the same plane. Each point is 1 unit from its nearest neighbors. How many circles in the plane have a radius 1 unit and pass through at least 2 of the 4 points?

Can someone help?

[crude kraken]

4

[upper karma]

lmao I drew more than 4

I just wasn't sure if I'm right

[crude kraken]

I read 4 points sorry xP

[upper karma]

Oh np

[crude kraken]

Oh wait

9?

[upper karma]

mhm

[crude kraken]

13?

13 and then my mental picture fades but I think 17

[upper karma]

lol

[crude kraken]

:p

[upper karma]

Wait is 13 the final answer?

[foggy oxide]

a good way to prove Cauchy-schwarz inequality

[upper karma]

Someone ;-;

[crude kraken]

Never heard of it, can you explain? @foggy oxide

[foggy oxide]

ax +by < or = sqrt {a^2 + b^2} * sqrt(x^2 + y^2}

that's the inequality

[crude kraken]

And skarz I think yea

Lemme get paper

[foggy oxide]

You can see that (y+a)(x+b) - ba - xy = the parallelogram area

(ax + by)

and it's also b * sqrt{x^2 + y^2}

ax + by = b sqrt{x^2+y^2} < or = sqrt{x^2 + y^2} * sqrt{a^2 + b^2} (supposing that the parallelogram area it's minor or equal to the rectangle area)

<=>

b < or = sqrt{a^2 + b^2} (squaring both sides)

b^2 < or = a^2 + b^2

[crude kraken]

Skarz, 9

And interesting

[foggy oxide]

i have no idea where this inequality is supposed to be used

never had to use it

[upper karma]

@crude kraken ok thanks

Wait what's the 9th one?

I got 8

[crude kraken]

Yea just realized my mistake, only 8

[upper karma]

ok

Thanks for the help :)

[oblique raft]

This is Khan academy's explanation for the initial question I had. We had to derive that the two lines were parallel using the left triangle

Here is the pic of the triangles btw

@dark sparrow

[dark sparrow]

oh

right

yeah

you do need that left triangle

to verify that those two lines are parallel

sorry

[oblique raft]

😃

No problem

[marsh wing]

if a rectangle and a parallellogram (a tipped one like this one: https://www.mathplanet.com/Oldsite/media/44000/parallelogram_499x300.jpg) has the same area

[dark sparrow]

then...?

[marsh wing]

which of them has the

how do you say it

hmm

least amount of perimeter

does not sound right

[dark sparrow]

which of them has the shorter perimeter?

[marsh wing]

yea yea

[dark sparrow]

hmm

not sure if there's a conclusive answer to that

[marsh wing]

hmm

if anybody knows let me know!

[dark sparrow]

you can make a rectangle's perimeter arbitrarily big

just take an x by 1/x rectangle

its area will be 1 but its perimeter will be 2x + 2/x

and that can be made as large as you wish

lol

it's always at least 4 tho

while say the parallelogram (0,0)-(1,0)-(2,1)-(1,1) also has area 1

but perimeter 2 + 2 sqrt(2)

so the answer to your question is

not enough info

[marsh wing]

hmm

ok

it has let's say 77cm^2

will the parallellogram have the shorter perimeter compared to the rectangle

[dark sparrow]

as i said

there is no way to tell

[marsh wing]

alright

[heavy path]

Each of the tubes that make up the Holland Tunnel's inbound and outbound lanes are about the same length. The north tube is 8558 ft, while the south tube is 8371 ft long. The diameter of each tube, which were initially cylindrical, is roughly 20 feet.

a)Based on the above information, what is the volume of dirt that had to be removed from the ground to create these tubes? Express your answer in cubic meters. My answer to this was 150600.38m^3 which is correct however for part B i dont get, B) Using your answer for the previous question, what would the height of a giant cube be that was made out of this much dirt?

[dark sparrow]

what is the edge length of a cube whose volume is 150600 m^3?

[heavy path]

no information is given for part A they want you to find amount of dirt removed, Part B is asking hieght of a giant cube that was made of all that dirt

[dark sparrow]

do you know how to calculate the volume of a cube given its edge length?

[heavy path]

it has the same legth width and hieght so V=a^3

[dark sparrow]

yes

so if a^3 = 150600, what is a?

[heavy path]

so i just square it to the third , sorry dont know how to write it in computer

[dark sparrow]

raise, not square

150600^(1/3) is what you tried to refer to

[heavy path]

yea

[dark sparrow]

that's called taking the cube root

[heavy path]

oh

[dark sparrow]

or raising to the power of 1/3

== 150600^(1/3)

[charred spear]

53.20367811

[heavy path]

yes

[dark sparrow]

...that's your answer

[heavy path]

for part c they are asking this If they wanted to dig the tunnels in no more than 91 days, how many kilograms per minute of dirt would have to be removed? Assume a density of 1826 kg/m3 for the wet dirt, clay and rock that would be found under the Hudson, and that they could work around the clock.

oh thats a neat bot

thank you

[dark sparrow]

do you want me to walk you through that one?

[heavy path]

Yea i dont know which formulas they might be asking me fore

[dark sparrow]

...

okay, so

you want to remove 150600 m^3 of dirt

1 m^3 of dirt weighs 1826 kg

how many kilograms of dirt do you want to remove?

[heavy path]

isnt the question asking if for new values? do you still have to use the answer for part a?

[dark sparrow]

is this all part of the same question?

[heavy path]

yea its all one problem

[dark sparrow]

it's reasonable to assume that since you had to reuse in part b the values you got in part a, you should also use them here

you want to remove 150600 m^3 of dirt
1 m^3 of dirt weighs 1826 kg
how many kilograms of dirt do you want to remove?

[heavy path]

Kg/min

[dark sparrow]

we'll get to that in a moment

answer my question, please

i want to guide you through it step by step

[heavy path]

yes, im reading the question to see if i can get a hint to the answer

150600/1826

[dark sparrow]

no

if one bottle of wine cost $25, would 3 bottles of wine cost $25/3?

[heavy path]

oh snap lol

if only

so multiply

[dark sparrow]

yes

[heavy path]

thats a big number..

[dark sparrow]

== 150600 * 1826

[charred spear]

274995600

[dark sparrow]

now

how many minutes are there in a day?

[heavy path]

3600? is it?

[dark sparrow]

no

[heavy path]

606060

[dark sparrow]

no

3600 is 60^2, not 60^3

[heavy path]

lol i thought i was using my calculator

[dark sparrow]

3600 is how many seconds there are in an hour

how many hours are there in a day?

[heavy path]

24

[dark sparrow]

how many minutes are in an hour?

[heavy path]

60

[umbral rivet]

[dark sparrow]

how many minutes are there in a day?

[heavy path]

1440?

[dark sparrow]

yes

[umbral rivet]

👍

[rare iron]

== 60*60 *24

[charred spear]

86400

[dark sparrow]

how many minutes are there in 91 days?

@heavy path

[heavy path]

..... okay give me a sec

i got this

[dark sparrow]

you can use a calculator

[heavy path]

yes, 131040

[dark sparrow]

== 91 * 1440

[charred spear]

131040

[dark sparrow]

yup

so now

you need to remove 274,995,600 kg of dirt
and you have 131,040 minutes to do it
how many kilograms will you have to remove every minute?

[heavy path]

we just devide

[dark sparrow]

divide, but yes

[heavy path]

== 274,995,600 / 131,040

[charred spear]

40

[heavy path]

40 kg per min?

[dark sparrow]

...the bot doesn't respond well to numbers with thousand separators

== 274995600/131040

[charred spear]

2098.56227106

[dark sparrow]

2098 kg/min

2.1 tons

[heavy path]

oh, okay, learning to use the bots

[umbral rivet]

Wouldn't you round it to 2099? 🤔

[dark sparrow]

2100

is the most reasonable figure to round to

[umbral rivet]

:3

[heavy path]

so to summarize you have to first muliply the density times the amount then just figure out the time it takes in the amount given

[dark sparrow]

mass = density * volume

and quantity = rate * time

really, that's it

those are the two "formulae" you could use

[umbral rivet]

Formulae :D

[dark sparrow]

?

[heavy path]

thats neat, i was not always good with real life applications hehe, but i feel like this was super easy to understand thanks alot

[dark sparrow]

:+1:

[umbral rivet]

I was unaware that was the plural word for formula :p

[dark sparrow]

formulas and formulae are both fine according to OED, iirc

i prefer the Latin-derived plural

[umbral rivet]

X3

[crude kraken]

Formulas is what I say, formulae is what I wish I could get used to saying because it sounds fancy :3

[umbral rivet]

Toast

🍞

[crude kraken]

🍞

[umbral rivet]

that's bread you ding dong

[crude kraken]

That's not geometry you ding dong

[umbral rivet]

:(

Let's make a geometry problem with bread

[crude kraken]

Okay

[umbral rivet]

"How many loaves of bread will it take to fill a 20 inch by 10 inch by 10 inch box?"

We need to find the average measurements for a piece of bread first

And we're not using end pieces >:T

[crude kraken]

nope, too simple

I'll make it

[umbral rivet]

K

[crude kraken]

If a loaf of bread is baked in an empty 3d space, in the end it is a perfect 3d oval with a width of 5 inches and a length of 16 inches. When the loaf of bread is instead baked on a sheet, because the sheet presses onto it, the bottom 2 inches from the perspective of width are cut off. What is the volume of the loaf of bread?

[umbral rivet]

wat

How does it

Turn into an oval

[crude kraken]

shh

it's math

[umbral rivet]

What kind of bread are we using

Okay

I have no idea where to start

[crude kraken]

I do 😏

but that's only cuz I made the problem and I probably asked it wrong ;3;

[umbral rivet]

;3;

;w;

[crude kraken]

=tex \text{Area of oval } = \frac{length}{2} \frac{width}{2} \pi

[crude kraken]

2.5 * 8 * pi = 62.8

Not sure how to turn area of oval into volume of elipse

[umbral rivet]

uwu

[crude kraken]

=tex V = \frac{4 \pi r^2}{3}

[crude kraken]

for sphere

but idk how that works for ovals

but I do know the area of the oval is 62.8, and I know how to calculate the area of a segment of a circle

I don't know how to apply circle maths to ovals though. 🤔

[umbral rivet]

😡🖐

[crude kraken]

so wait

pi(r)^2 = area of circle

so it's pirr

oval is just pi(l)(w)

so therefore any time I see r squared I just replace with length * width

so

[umbral rivet]

😡🖐

[crude kraken]

=tex V = \frac{4 \pi lw}{3}

[umbral rivet]

🛑

[crude kraken]

whatchuwant

[umbral rivet]

👀

:3

[crude kraken]

so we know the volume of the ovaloid is 4pi(5)(16)/3 = 1004.8/3 = 334.93

Now spherical cap is

=tex V= \pi h^2(R-(h/3)) = \frac{1}{6} \pi h(h^2 + 3r^2)

[crude kraken]

We know to replace r^2 with lw, and that h is 2

[foggy oxide]

sup guys

[crude kraken]

so if we work this out, 3(80) or 240 + 4

sup

[foggy oxide]

Woah, what is this formula

[crude kraken]

2(244) = 488

488pi/6 = 255.39

[foggy oxide]

this is wrong

it's r^3

[crude kraken]

So the solution is 334.93 - 255.39 and because that makes absolutely no sense I know I dun messed up

and oh, that changes a lot

[foggy oxide]

idk, depends

[crude kraken]

So wait

how to you calculate volume of a 3d oval

[foggy oxide]

What are you guys discussing ?

[crude kraken]

If a loaf of bread is baked in an empty 3d space, in the end it is a perfect 3d oval with a width of 5 inches and a length of 16 inches. When the loaf of bread is instead baked on a sheet, because the sheet presses onto it, the bottom 2 inches from the perspective of width are cut off. What is the volume of the loaf of bread?

[foggy oxide]

Oh god

[crude kraken]

shrug

[foggy oxide]

there are so many words that idk

like loaf

and loaf

and baked

[crude kraken]

xD

[foggy oxide]

and baked too

huayuhuhaa

[crude kraken]

mk so basically

the same problem in 2d is: you have an oval of length 16 and width 5, you take off a segment of height 2 from the width-side and you calculate the area remaining when the segment is gone

but in 3d

[dull egret]

well, just integrate that right?

[crude kraken]

I've been trying to figure out what an integral is for the longest time

and I don't even know

so I'm trying to use simpler formulas made for spheres and apply them to ovals and it's not working and I'm sad ;3;

[dull egret]

well, so, i mean we know a circle is defined by r^2=x^2+y^2. r is 5 and the whole thing is moved down two.

wait diameter

so 6.25=(x+1/2)^2+y^2

[crude kraken]

area of oval is pi * length/2 * width/2

but idk volume of ovalloid

and yea

[dull egret]

now find the volume of the bottom section essentially

[crude kraken]

Mhm

h is 2 where a and c are both 2.5 and b is 8

Trying to find the area of not-blue

*volume

that's this problem in a nutshell

[dull egret]

yeah i got that

[crude kraken]

mk

[dull egret]

i m just contemplating what is the easiest way to go about things kinda

i mean. let s just be stratic about it. lets first find the area of a circle which has the bottom section disregarded i guess.

and then integrate that formula along the x axis while cahnging r depending on x

so with that we re back to 2.5=sqrt((x+1/2)^2+y^2)

positive sqrt

solving for y

sqrt(6.25-(x+1/2)^2)=y

wait

this is gonna have problems cause it s two sides

hmmmmm

sqrt(6.25-(x-1/2)^2)=y

better

so... integrate that

let s be lazy

should have shifted in y not x

[charred spear]

Query made by @dull egret
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sqrt(6.25-(x-1%2F2)^2)%3Dy
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[dull egret]

6.25=x^2+(y-1/2)^2

therefore

sqrt(6.25-x^2)+1/2=y

=wolf integrate sqrt(6.25-x^2)+1/2=y

[charred spear]

Timeouts
Integral, GlobalExtrema, Inequality2D, Geometry, Inequality, Simplification, ImplicitDifferentiation, FredholmIntegralEquation

Query made by @dull egret
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=integrate+sqrt(6.25-x^2)%2B1%2F2%3Dy
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[dull egret]

should have shifted it in x too, now stuff is partially negative

do you get where i m going?

[crude kraken]

So therefore, the volume is 209.44 - 73.72 = 135.72

Is what you're trying to say

[dull egret]

where did you suddenly get that from?

[crude kraken]

I found a formula for the volume of an ellipsoid cap because I don't know the first thing about integrals 🤔

Still interesting solution tho

[dull egret]

not a full solution yet xD

[crude kraken]

Still interesting semi-solution tho

I don't get it but it looks fancy 😮

that but without transparent background

[dull egret]

essentially i would have gotten the caps area for each slice, subtract that off the circles area and split the ovaloid into infinitecimally small slices of cylinders

[crude kraken]

Ohhh

Yea I've seen that analogy before, interesting way of looking at it

and I do know the formula for taking off a segment of a cylinder so I can definitely see it working

I just don't know the "syntax" (if that's a math term) for it

[dull egret]

generally works quite well for geometry problems.

[crude kraken]

But isn't it calculus?

[dull egret]

what is the way you are expected to solve this if you dont know calculus

cause yes, it is calculus

[crude kraken]

Well uhh

I didn't exactly know it was calculus 🤷

I just thought of changing the formulas for segments of a sphere to match ovaloids

and hope that works

*also learned that ellipsoid is the proper term

[dull egret]

whatever

is this some school task or so?

like, how are you mean to solve it?

what are people expecting you to do with it?

[crude kraken]

If you scroll to right before I posted the problem at first you would see that me and @umbral rivet were trying to come up with a geometry problem about bread

[dull egret]

oh lol

[crude kraken]

and I came up with a perfect ellipsoid loaf volume being an interesting problem

because loaves are often in that shape but flat at the bottom so

y'know

[dull egret]

getting yourself into complicated stuff, great...

[crude kraken]

I dunno if I'll need it but will note when I learn calculus that it can do this

[dull egret]

trivial task with calculus tbh with you. bit tedious tho

just need to move stuff around correctly :P

[crude kraken]

turns out I need to know what a "derivative" is to know how to integrate, so guess I'll just self-teach myself derivatives sometime

[dull egret]

lol

[crude kraken]

and hope I don't have yet another prerequisite for that, so see ya around I guess 😛

[dull egret]

well, good knowledge of algebra then it s all very simple really

[crude kraken]

Searches derivatives I have to know limits, integrals are a lot more complex than I thought ;3;

[dull egret]

limits are stupidly complex if done rigorously, but simple af if done with theorems

[crude kraken]

I semi-get limits though

sorta

not really but I know that the bottom thing means "as x approaches y"

[dull egret]

lol

[crude kraken]

but I don't know what the thing on the right is, and I also don't think that I should ask what that is in #geometry-and-trigonometry xP

[umbral rivet]

:3

[dull egret]

on the right side?

[crude kraken]

Yea

Whenever I see limit there's something on the right

idk what that means :i

[dull egret]

the actual formula?

[umbral rivet]

:o

[crude kraken]

:o

[dull egret]

tell me you know calculus @umbral rivet

[crude kraken]

Angel is a grade level below me

[umbral rivet]

I have no knowledge of calculus :3

[crude kraken]

=tex \lim_{x \rightarrow 0}3x+4

[crude kraken]

What is the 3x+4

[dull egret]

she s as old as i am and i learned it myself too

the formula

xD

[crude kraken]

...

[dull egret]

this limit would go to 4 for instance

[umbral rivet]

Who? 👀

[crude kraken]

Then what does the limit mean about the formula >:o

[umbral rivet]

:u

[dull egret]

cause you can just plug in 0 cause it s continous everywhere

[crude kraken]

okay so as x approaches 4, 3x?

[dull egret]

....

wait

[crude kraken]

=tex \lim_{x \rightarrow 4}3x?

[dull egret]

https://www.youtube.com/playlist?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr

[crude kraken]

ooo

[dull egret]

wrong one

[crude kraken]

uuu

[umbral rivet]

ggg

[dull egret]

fixed

[crude kraken]

ooo

[umbral rivet]

uuu

[dull egret]

...

[umbral rivet]

!!!

[dull egret]

we should stop

[crude kraken]

We should.

[dull egret]

otherwise sqrt 2 is gonna make us go bakc to the bin again

[umbral rivet]

I don't wanna go back to the bin >:(

Hmph

[dull egret]

:P

[umbral rivet]

:P

[crude kraken]

:P

[umbral rivet]

:3

:T

[crude kraken]

:T

[dull egret]

stfu

[umbral rivet]

Begone

:)

[crude kraken]

re-eats

[dull egret]

😛

[crude kraken]

You're now -eatenjelly

[umbral rivet]

uneatenjelly?

👀

[crude kraken]

no

if you eat something already eaten

it's now negative

so he's -jelly

because 3 - 3 = 0

0 - 3 = -3

[umbral rivet]

[crude kraken]

so jelly + eaten = eatenjelly, eatenjelly + eaten = -jelly

simple

[umbral rivet]

#prealg-and-algebra 👈

[pale crow]

http://prntscr.com/ggq6wy

[umbral rivet]

Need help with that? @pale crow

[crude kraken]

2.5x+x = 5x-3

After simplification: 3.5x = 5x-3 (Combine like terms)

3.5x - 5x = 5x - 3 - 5x

-1.5x = -3

[umbral rivet]

ugh

[crude kraken]

x = 2

[umbral rivet]

I was solving it

:(

[crude kraken]

sorry ;c

[thorn talon]

Rip

[umbral rivet]

;c

[crude kraken]

You can carry on c:

[umbral rivet]

aghhh

I got the same answer

u3u

[crude kraken]

❤

but that's not the answer

[umbral rivet]

tru

[crude kraken]

that's just a declaration that x=2

[umbral rivet]

Need to substitute that in

[crude kraken]

2.5(2) = 5

5 = answer

c:

[umbral rivet]

@pale crow do you understand how we solved this problem?

(also I'd like to answer more question if you have them) :3

[pale crow]

uh

@umbral rivet i dont get it

[umbral rivet]

Hm okay

[pale crow]

hold on one minute

[umbral rivet]

http://prntscr.com/ggq6wy

[pale crow]

im gonna redo it

[umbral rivet]

Do you understand that MN + NR = 5x - 3?

okie :)

[pale crow]

yes

i understand

[umbral rivet]

^3^

[pale crow]

keep explaining please 😂

@umbral rivet

[umbral rivet]

Okie :)

So here

All we need to do is substiture all those letters for the numbers and variables

so...

MN + NR = 5x - 3

Would turn into...

2.5x + x = 5x - 3

Does this make sense?

[pale crow]

I GOT 2

OH

yes thank u everyone

[umbral rivet]

:)

Any more questions, don't hesitate to ask 💕

[pale crow]

ookk

[umbral rivet]

:D

[pale crow]

this makes much more sense

[umbral rivet]

:DDD !

[pale crow]

http://prntscr.com/ggqp8v I could solve this (find the answer, but if anyone could explain how. it would be nice

[umbral rivet]

16 or 17?

[pale crow]

17

school started this week and hes making us do these for homework before we do the lesson

so im kinda lost

[umbral rivet]

:(

School started for me recently as well

Let me see...

ah okay

The keyword in this problem is "midpoint"

Whenever you see the word midpoint, it always means that the line is split into two equal parts

Because these two parts are equal, we can, well... Set them equal to each other

So...

2x + 4 is equal to 3x -1

Once again, this is because we know E is the midpoint between them

Does that make sense?

[pale crow]

i got x=5

[umbral rivet]

I got...

Let me do the problem

yup, x = 5

Nice :D

[pale crow]

2x+4=3x-1

ya

[umbral rivet]

^3^

[pale crow]

http://prntscr.com/ggqt12 what does bisect mean??

divide into 2 parts?

so like half??

[scarlet quail]

divides into two, yes

[pale crow]

ook

[umbral rivet]

Yup :3

[pale crow]

So its like the midpoint

[thorn talon]

kind of yes

[scarlet quail]

Yes it is

[umbral rivet]

yee

[pale crow]

ok so i know that

i drew a line

Q is in the middle

but idkhow to find QR

[umbral rivet]

I got chu

okay so

Since we know that PQ = QR

It means that QR = 3y

Right? :D

so...

6y = 42

:3

Does that make sense? :3

[pale crow]

yes so PQ = QR is 42

???

the whole thing equals 42??

[umbral rivet]

Yes

Do you know why? :o

Here's my bad picture explaining it :D

[pale crow]

YAHHH

THATS WHAT MINE LOOKS LIKE

ok so i understand the first step

now hwat

oh

just divid

ook

[mint axle]

Beth we have the same math problem :O

[umbral rivet]

:3

hm

I mean

both sides add up to 6y

And the whole segment is 42

6y = 42

:P

[mint axle]

How are you so good at math

[pale crow]

how did u get 6

[thorn talon]

studying and practicing

and understanding

[umbral rivet]

I'm not xP

[thorn talon]

is how i assume he is good at math

/she

[umbral rivet]

I'm bad at math x3

[thorn talon]

idk

same

[umbral rivet]

I'm a guy ^.^

So

I got 6y because

If you remember earlier...

Q is the bisector of PR

[pale crow]

so it doubles??

[umbral rivet]

Which means that both sides are equal

if one side equals 3y, the other side must equal 3y as well

Pretty much, yeah Beth

[pale crow]

im slowly understading

[umbral rivet]

:)

[thorn talon]

btw my brain is dead

can somebody help me?

[pale crow]

y=7

[thorn talon]

How do I do b?

[umbral rivet]

ooh

pythagorean theorm it looks like

[thorn talon]

I tried to do Pythagoras and got nowhere

[umbral rivet]

yepper

alright so

[dusty timber]

Oh, no. It's similar triangles

[umbral rivet]

It is? o.o

oh god

[dusty timber]

The ratio between the big and the small triangles will remain constant. Get the ratio and then find the missing leg of the small triangle

[mint axle]

I know how to get the top triangle does this mean I'm smart

[umbral rivet]

👍

Smart to me

I don't think similar triangles and ratios are used here

Let me solve the problem real quick

one min

[thorn talon]

25/37 = x/x+10 ?

[dusty timber]

Yes

[thorn talon]

Alright thanks

I forgot similarity was a thing

[umbral rivet]

[dusty timber]

It happens. You knew how to set it up teal quick tho

[thorn talon]

Wait do I have to prove it?

[umbral rivet]

I forgot everything about similar triangles

D:

[dusty timber]

Uh I don't think so. It's a proven theorem that a line parallel to any leg cutting through a triangle creates a similar smaller triangle

[thorn talon]

Alright thanks

[umbral rivet]

owo

I love this type of geometry <3

Nice and easy for me to understand uwu

[dusty timber]

I never liked geometry lol

How old are y'all?

[umbral rivet]

15 :3

[thorn talon]

16

geometry is my least favourite topic currently

[dusty timber]

I'm 17. I hated my geometry class. My teacher had a terrible monotone voice

And guess what... he's my calculus teacher now lolol

[thorn talon]

we didn't have a single geometry class

they assigned the chapter as holiday work

[umbral rivet]

w t f

[crude kraken]

df

[dusty timber]

That's unusual. Geometry is usually where proofs are first introduced. It usually ends up diving into probability and such

[umbral rivet]

^

Hopefully you don't do 2 column proofs

If you do though

🙏

They were the hardest thing for me to learn

I always failed that part on every single test

[thorn talon]

what's that?

[dusty timber]

I always hated those. They give a bad name to actual proofs.

[umbral rivet]

^^^^

there are two types of proofs:

1. 2 column proofs

2. paragraph (written proofs)

2 column proofs are basically like

You have one column with a statement, and another column with reasoning

[thorn talon]

that sounds like what we did

[umbral rivet]

So like an example would be

"A is parallel to B"

And in the other column you would write the reason or the theorm that proves it

It's awful

[thorn talon]

sounds like what we did

[dusty timber]

It's horrible. It's just a series of steps and you are given all the pieces, which in turn gives no understanding at all

[umbral rivet]

Once my teacher introduced proofs to us, my grade in geometry went downhill

I ended with a C- or a D at the end of the year I think :\

[thorn talon]

better than me

[umbral rivet]

That was an honors class

Now I'm in an academic algebra 2

Which is actually good :3

<3

[thorn talon]

i have no idea what i'm doing

[umbral rivet]

Slower pace <3

o.o

What do you mean? D:

[thorn talon]

i have to study basic arithmetic to quadratic functinos

[dusty timber]

Learn your algebra well. Very well. Without that you're gonna have a hard time with everything else

[thorn talon]

basic arithmetic, i think i'm fine

algebra and surds, i think is fine

equations is fine

geometry, i should probably work on

same with functions and graphs

and trig

linear functions is fine for me i think

intro to calculus i think is alright

quadratics is not too bad

[umbral rivet]

me + graphing = 💥

[thorn talon]

so i need to study at least 3

and probably revise calc

cause i don't remember a thing from calc

[dusty timber]

What's the most advanced math you've taken?

[thorn talon]

quadratics functinos apparently

[umbral rivet]

👀

[dusty timber]

I took a trig class in a junior college and honesty I don't remember half of it

[umbral rivet]

hehe

No idea what I'm looking at <3

Ty though

[thorn talon]

diagonal asymptotes?

hmm

haven't done that yet

[dusty timber]

Is it really useful to learn this stuff? I mean, I understand it but is it really worth it to learn all these rules despite having graphing calculators or is it just for the fun of it? @upper karma

[thorn talon]

we don't get graphing calculators

so i guess i need to learn it

hmm

i understand only the basic graphs

i guess i'll need to learn

[umbral rivet]

o3o

[thorn talon]

i got my exam on friday

[dusty timber]

Yeah, you're right, but I guess a little rigor would be ideal instead of juat learning the algorithm

[thorn talon]

hopefully enough time

my brain still feels stuffed from english

yeah

i can do that

ehh

for what types?

[dusty timber]

I thought it'd be easier to graph things like hyperbolas and such. I figured there would be a way to graph them quickly just by knowing the coordinates of the foci and stufd

Stuff*

I see. Sines/cosines are kinda easier because they cycle

[blissful hill]

Imagine you are in 7th grade and you are given this task, how would you solve it

[dull egret]

trying to remember what you knew then is the hardest part

[blissful hill]

Yea

I know solution which using tan(15)

But why the hell i was given this task in the 7th grade

[dull egret]

what did you know then?

[blissful hill]

Maybe piphagorean theorem

I wrote it wrong right

[dull egret]

pythagorean

[umbral rivet]

x3

[blissful hill]

Thx

[dull egret]

i wouldnt have complained actually, but oh well, you asked for it

[blissful hill]

it is good for me to learn more and more in english

[dull egret]

were you maybe meant to just solve this by measurement?

where are you from?

russia?

[blissful hill]

Russia

[dull egret]

but yeah, only thing i cna think of is measuring things

[blissful hill]

For ?

I need to find angle a

A

And this is possible to do with trygonometry

But i think there must be another way

[dull egret]

trigonometry

[blissful hill]

Oh

[dusty timber]

There is another way

They are similar triangles. The ratio between the 2 is... 2. The angle is 30

I mean, if there's any other way, it's similarity

[topaz valley]

Wait, do you know that that triangle is solvable?

Because at first glance it doesn't look so.

No nvm you're right it's 30.

No, it's not 30

angle = atan(x/2x)

[dusty timber]

TanA= 1/2

Tan^-1 (1/2)= angle A

[topaz valley]

26.6 degrees

[dusty timber]

Yup

I was almost right lol

[topaz valley]

Lol that second triangle really threw me off

[blissful hill]

Hm

Similarity

2 similar triangles have same angles

They are not similar

Answer is 30 but i don't know how to prove it without trigonometry

Btw A is 30, B is 75 and C is 75, its an isosceles triangle

Let H be the point of height

2CH=AB

[upper karma]

I'm having a hell of a time trying to prove the volume formula for a frustrum. I still just can't figure out what I'm doing wrong

https://cdn.discordapp.com/attachments/222063343798583297/354378596615323658/JPEG_20170904_133344.jpg

[surreal bolt]

still there?

[upper karma]

always and forever

Was just torturing myself physically since I've mentally beat everything out of myself

the original question.

[surreal bolt]

Hmm is it assumed the pyramid has an equilateral triangular or square base?

I'll do this for a square base, but the answer works for other shapes as well.

(knowing that it works for lots of shapes means we can just find it for one ... and that is how I am cheating a bit here.)

[upper karma]

I believe square, since the first was square

[surreal bolt]

So I think h is supposed to be the height of the fulcrum. If the whole pyramid were intact, what would be the height of the pyramid (not fulcrum) with base s1?

It ends up being a ratio. Let H (capital) be the height of the nonexistent pyramid. (H-h)/H = s2/s1

Hs1 - hs1 = Hs2 --> H(s1 - s2) = hs1 --> H = hs1/(s1 - s2)

The pyramid with base length s1 has height H --> V1 = 1/3 (H)(s1)^2
The pyramid with base length s2 has height H-h --> V2 = 1/3 (H-h)(s2)^2

V = V1 - V2 = (1/3)(H(s1)^2) - (1/3)(H(s2)^2) + (1/3)(h(s2)^2)
= (1/3) (H ((s1)^2-(s2)^2) + h (s2)^2) ... Sub hs1/(s1 - s2) for H
= (1/3) ((hs1/(s1 - s2))((s1)^2-(s2)^2) + h (s2)^2)
= (1/3) h (s1(s1 + s2) + (s2)^2)
= (1/3) h ((s1)^2 + s1s2 + (s2)^2)

And you are right I wasn't paying attention. It was definitely square.

It says two ways ... have a preference?

anyway that is one way. Hope that helps.

[upper karma]

sorry about that

was talking to my daughter

so where does the h (not H) come from?

[surreal bolt]

um from the original problem 😃

[upper karma]

alright, alternative wording, where does H come from?

it says (in original problem) that h = height from base to top

[surreal bolt]

Yes. But note that the "top" in the problem is also a base.

[upper karma]

right, so it isn't the entire distance of a pyramid

[surreal bolt]

Yeah. Just making sure .... you have a picture of a fulcrum in your books or notes, no?

[upper karma]

ja

so from s1 to s2 = h

[surreal bolt]

yes

from s1 extending past s2 to a point ... H

[upper karma]

Thanks, sorry bunch of household mess getting handled on top of me working on this as well

[surreal bolt]

yup yup. We are here to support your math -- something which can distract and be distracted by life.

[upper karma]

far too true

[upper karma]

I'm having trouble here "= (1/3) ((hs1/(s1 - s2))((s1)^2-(s2)^2) + h (s2)^2)
= (1/3) h (s1(s1 + s2) + (s2)^2)"
I can post what I have, but I feel like I'm straying further from the path

[surreal bolt]

you are basically there if you understand up to: (1/3) h (s1(s1 + s2) + (s2)^2)

[upper karma]

I doon't understand how you got rid of the 1/(s1-s2)

[surreal bolt]

oh cause s1^2 - s2^2 is a diff of 2 squares.

[upper karma]

OH!

bloody hell

[surreal bolt]

hehe

[upper karma]

helps if you look in the right place

[surreal bolt]

definitely.

I'd share a story I heard, but it is a bit lengthy.

[upper karma]

I've got nothing but pencil lead, ink, paper, and time

[surreal bolt]

oh I found it online.

From a hopefully reputable / accurate website:

Nikola Tesla visited Henry Ford at his factory, which was having some kind of difficulty. Ford asked Tesla if he could help identify the problem area. Tesla walked up to a wall of boilerplate and made a small X in chalk on one of the plates. Ford was thrilled, and told him to send an invoice. The bill arrived, for $10,000. Ford asked for a breakdown. Tesla sent another invoice, indicating a $1 charge for marking the wall with an X, and $9,999 for knowing where to put it.

(not as long and dramatic as the version I heard but it might be more accurate)

anyway ... about the "right place" 😃

[upper karma]

I'm still messing up/missing something

[surreal bolt]

oh last small mistake. s1^2 - s2^2 = (s1 + s2)(s1 - s2) ... which one cancels away and which one stays?

[upper karma]

the (s1+s2) stays

the one that doesn't match

[surreal bolt]

s1(s1 + s2) + s2^2

[upper karma]

where is that other s1 coming from?

[surreal bolt]

you dropped it between the second to last and last line 😃

oh

I see.

H = hs1/(s1 - s2)

H has an extra s1 term

[upper karma]

ahh yes, it does

thank you

that makes sense

[surreal bolt]

yay!

nice going.

you are welcome.

[upper karma]

and so that sqrt in the middle is fabricated?

because now it matches save for the sqrt(s1^2 s2^2)

[surreal bolt]

I think they were worried you didn't have enough math to do 😃

[upper karma]

so it's just made up?

[surreal bolt]

they don't like conserving ink 😃

I dunno 😦

[upper karma]

I don't blame you. Thanks at least!

much better/closer than I've been in days

[surreal bolt]

good good. progress is usually touted here.

[upper karma]

Now to determine how to prove it once more. Last time I just did 6 of these is a cube, but with this. I don't think it will work as nicely

[thorn nymph]

3√-4 = -3√4 or no?

[vale raven]

==3*sqrt(-4)

[charred spear]

(3.6739404e-16)+6i

[vale raven]

They're not equal

[upper karma]

Due to being a creature of habit. Multiplying it by 6 would give me s^3 - s2^2(H-h)/3, right?

[blissful hill]

@thorn nymph you can't take square roots of negative numbers in real numbers

[thorn nymph]

@blissful hill yeah, I misstyped it, supposed to be cbrt(-4x)

[blissful hill]

Then you are right

[supple knoll]

So I know that Euclid described in his axioms that we could "describe any circle with any center and radius r", but how exactly does he give meaning to the notion of circumference of a circle? If I wanted to define pi geometrically, I would say that it's the ratio of this circumference to 2r, but that doesn't mean much to me. Without using any calculus to describe it based on arc length etc., how was the circumference of a circle defined?

Probably sounds like a retarded question but I'm being serious

[foggy oxide]

two times pi, times the radius @supple knoll

that's exactly how we think about "creating" a circuference

well, idk euclid, now we can think in another way about circuference

but i think about it as the set of points that are equidistant, in a certain distance (radius) to a single point, (center)

and that's how the circle is defined in a equation

{x - xP}^2 + {y- yP}^2 = r^2

where, (xp, yp) are the center

and r the radius

this is a consequence from pytaghorean theorem, and distance of two points formula (both are basically the same)

but, in the time of euclid they didn't have rigorous definitions for this, i think xD...

[supple knoll]

Isn't that circular? Pi is defined as the ratio of the circumference to the diameter, so saying that the circumference is defined in terms pi doesn't make sense

[foggy oxide]

so

take the equation as a definition of the circuference

it does not involve pi

[supple knoll]

I was looking for something that didn't involve coordinates and could just be described using the postulates

but it seems that, going through the Elements, he just defines a line (not necessarily a straight one) as some length, and defines a circumference to be a line

and talks about ratios somewhere else

[foggy oxide]

well, that's probably what he did

[supple knoll]

So I guess we're restricted to talking about ratios as equivalence classes, of which pi is one such equiv. class

[foggy oxide]

i heard that he avoided using the postulates that he wasn't confident in the definition

[supple knoll]

I think you're referring to the last postulate (5), which he thought could be proven as a proposition but no one ever managed to

the one about parallel lines

There weren't really any concerns about the others

[foggy oxide]

i'm back, well

i think it was about parallel lines or "planes"

something like that

[supple knoll]

Yeah, he specifically tried to avoid using the last postulate, but that doesn't really have anything to do with the construction of circles

[foggy oxide]

yeah, i was just commenting hehe

[glacial grail]

Guys how can I find b?

a I found by using pythagorean theorem

@upper karma How?

Using 30

@upper karma So it will be b = sin60/8 ?

8 * Pi/3 = b

Thanks alot

[foggy oxide]

hm

my brain it's kinda messed up today

this problem asked up

to find the pink angle

how do i prove the P P' C angle its right

i got the correct answer, (i'm probably missing something elementar)

i just rotated 90 degree the APB clockwise

wait

if PP'

it's 2sqrt2

OH

(2sqrt2)^2 + 1^2 = 3^2 .-.

so it's a right triangle

[past mantle]

I'm having trouble with algebraic proofs 😭

[upper karma]

I need help with this one problem

[past mantle]

Here

[upper karma]

??

@upper karma Can you help me?

[past mantle]

Yes

Do I need to use name the property's?

K

[upper karma]

😅

ok

Ok

Thanks 👍

Got 60

Well you helped

So thanks

[pale crow]

Does anyone know the formula i can use to solve these???

[dark sparrow]

i take it that you don't know how to find the midpoint of a segment

[pale crow]

Uh

[dark sparrow]

(everyone else: no spoilers)

[dull egret]

awwww

[pale crow]

Wahat

[dark sparrow]

i would like you to answer a few questions for me, so that you can derive the formula yourself

[pale crow]

Ok

[dark sparrow]

what is the midpoint of (0,0) and (2,0)?

[dull egret]

there is nothing special to this, what is there to be derived?

[dark sparrow]

i don't want to give the formula as a black box

[pale crow]

Um

[dark sparrow]

can you mark the points (0,0) and (2,0) on the xy plane?

[pale crow]

Yes

Just put it on a graph

[dark sparrow]

can you identify their midpoint?

can you find the point that is halfway between (0,0) and (2,0)?

[pale crow]

1???

[upper karma]

👏

[dark sparrow]

not 1

(1,0)

this is important

now
can you find the midpoint of (2,1) and (3,5)?

[pale crow]

(1,4)??

[dark sparrow]

no

[pale crow]

Oh

[dark sparrow]

how did you get that?

[pale crow]

well i kinda guessed

[dark sparrow]

...

[upper karma]

LOL

[pale crow]

Im not good at this.

Im sorry.

[upper karma]

no

don't apologize

not everyone is good at math some people just need a little help

[pale crow]

its ok.

[dark sparrow]

okay, so
let's consider only the x coordinates of my two points

what number is halfway between 2 and 3?

[pale crow]

Theres no point in helping me

Im fine

[upper karma]

no beth

[dark sparrow]

don't give up now

[upper karma]

keep trying

[pale crow]

Its ok

[upper karma]

you're getting it i promise u r