geometry-and-trigonometry

[sick veldt]

what?

[upper karma]

If you have to ask here to verify the answer, you don't understand it @upper karma so stop

Ask only if you get the wrong answer and don't understand why

you should verify it yourself and make sure it's right on your own

not guess then ask here if it's right

im only 12

im trying to learn geometry

none of this is for my benefit

if it's not for your benefit then don't learn it lmao

it is for my benefit

[upper karma]

i just don't gain anything but knowledge

[sick veldt]

Is this proof correct?

[upper karma]

you just said it wasn't

[sick veldt]

I believe the last box should be HL

but I'm not so sure

[upper karma]

you highly underestimate the value of knowledge lmao

…..?

good luck being a redneck who don't know whether the earth is flat or nah

[sick veldt]

what are you talking about

lol

[upper karma]

wow..

he's being rude to me because im asking help

you didn't ask for help lmfao

you just asked someone to verify your blind guess

it wasn't a blind guess

then answer me, why would that answer not be right?

ive been stuck on the same problem for 35 minutes

because their not congruent

but that makes no sense because length wise, they look like they are

[sick veldt]

whats not congruent

[upper karma]

SNK AND RNL

length is irrelevant

[sick veldt]

yeah...

the angles look corresponding

[upper karma]

that's what i meant

[sick veldt]

parallel lines that is

[upper karma]

the angles look congruent/corresponding

just inverted

[sick veldt]

idk dude haha

I just need to know if my answers are correct

next time

Explain what you did like I did

after sending a question

<@&286206848099549185>

help pls

I need help with this

15 min already

Number one might be a rectangle

and the last box might be SAS

[tacit sparrow]

Yeah that’s not the definition of a parallelogram

Parallelograms don’t have right angles

[sick veldt]

right

And for the last one?

it would be HL

cause it does have a rectangle

with right angles

[tacit sparrow]

Yeah that’s HL I think

HL is just specialized SAS anyway

Where the angle is right

[sick veldt]

@tacit sparrow Sorry to bother you again, but could you also confirm this one?

[tacit sparrow]

Looks fine to me
I don’t rly remember all of the terms from geometry but

[upper karma]

https://gyazo.com/f6da20d89c0f84b22df4b3b0687075cd

how do I go abouts solving this

[tacit sparrow]

Well BC and DE being parallel would make ABC and ADE similar

How would you contradict that

[sick veldt]

Looks like the second option

[tacit sparrow]

Yeah

[sick veldt]

remember

always add the ratios

you're gonna use that later on

with partitions

[merry siren]

How do i get AE

BE is 20.8

[upper karma]

?help

!help

[merry siren]

!help

[sick veldt]

<@&286206848099549185> Three people here need assistance

[upper karma]

yeah i’m kinda busy now doing work too

Lol

[jaunty horizon]

Perhaps you should use a fraction

[merry siren]

Do it sir

[dire rampart]

oof

@sick veldt thats correct

[sick veldt]

thanks

[spring apex]

@sick veldt still need help?

[sick veldt]

yeah

i got another one

[spring apex]

If a line is perpendicular to another line

What does this mean?

[sick veldt]

No I answered that already

This one I'm working

on

I haven't graphed it yet

but heres what I have written down so far

[spring apex]

Oh you pinged helpers for someone else’s question

Big mistake

[sick veldt]

No lol

I pinged for mines

and the other guy

i included myself

[spring apex]

Well idk about that graph thing so 🤷

But I can tell you about the perpendicular line

[sick veldt]

sigh

FINALLY

done

Is this graphed correctly?

[upper moat]

The point at (4, 3) should be at (4, 4), since we’re doubling the distance from (-2, -2) to (1, 1)

[tawdry cape]

guys, if i want to find the x intercepts of 2sinx + 1 = 0, the first one is -1/2 and second is 7pi/6. but how is the second one found?

[somber coyote]

Yue:

[serene field]

The sine inverse function yields the quadrant 1 or quadrant 4 solution.

In order to find the other solution, we take a look at the unit circle.

[bitter stratus]

what is the derivative of

[gritty siren]

=pup derivative sin^2(x)-2cos^2(x)

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@gritty siren

[hard gale]

proof by wa

[serene field]

🤔

I'm getting a different solution

Wait, it's the same thing.

[gritty siren]

It's got a nice expression surprisingly

[bitter stratus]

i dont get it though

how does it become that

this is how i did it ;-;

[hard gale]

🤕 my chain rule hurts......

[upper karma]

My calc 2 professor made this one of our standard derivatives actually

[hard gale]

so you can write $\sin^2$ as $sq\circ\sin$, with $sq : x \mapsto x^2 \$ therefore $\left(sq \circ\sin\right)^{'} = \sin^{'} * (sq^{'} \circ \sin)$, ie $$ \boxed{\left(\sin^2\right)' = 2\sin\cos}$$

almost the same thing for cos^2

wat

[upper karma]

The fuck did you do lol

[hard gale]

fack tex

[somber coyote]

emeric75:

[bitter stratus]

is this an identity?

[hard gale]

if you want

[bitter stratus]

what

im lost

product rule?

[hard gale]

chain rule demit

i've said it at the beginning

[bitter stratus]

wait

oh

it makes sense

thank you

sorry

[upper karma]

“Demit”

I’m stealing dis

K bye

[hard gale]

so concise lel

[steady sleet]

[queen socket]

So if the slope of a line is the tangent does that mean the hyptonese of a triangle is equal to the tangent?

Because tan = a^2+b^2= c^2 or sin/cos =tan

[spring apex]

The slope of a line is it’s gradient

Is it’s rate of change

I’m not sure exactly what you mean by the slope of a line is the tangent

If you put it into context

Suppose I have a car which drives at 3m/s east and 4m/s north

Then the car is driving 5m/s at an angle of x = tan^-1(4/3)

But yeah could you elaborate more on your question?

@queen socket

[queen socket]

Like in a unit circle your hypo is equal to one and your x = cos and y = sin and using the trig identity x2 +y2 =1^2

So your hypo is also the length of the radius of the unit circle

[spring apex]

For me I would say you are overthinking it a bit, the tangent to something is the straight line or plane which touches a curved line or surface at one and only one point

To call the hypotenuse of a triangle the ‘tangent’ is a bit strange to me

[queen socket]

Oh okay

[spring apex]

But I would also ask someone else because I could be completely off

sorry I couldn’t give you the answer you may have been looking for

[queen socket]

I just looked it up, it says the following in calculus, the slope of a function at a given point is the slope of the tangent line. However, not every line with the same slope is tangent.

[spring apex]

No I know what a tangent is with calculus

I meant I could be off about the trig stuff

Is the tangent the hypotenuse, maybe 🤷 I’ve never heard of it personally

[queen socket]

Thats all good, have you been unbanned from piano yet

[spring apex]

Lol nope

[queen socket]

Unlucky

[woeful flame]

why my answer is incorrect?

:\

[boreal warren]

derivative of 2/(x^2+2x) is wrong

[woeful flame]

why?

[boreal warren]

try using the full quotient rule instead of the chain rule

d/dx (u/v) = (u'v-v'u)/v^2

[woeful flame]

😮

sorry but i didnt understand

[boreal warren]

idk how to use the LaTeX bots so can't write it out clearer 😄

do u know the quotient rule? or have u used it before?

[woeful flame]

i did:
(2')(x^2+2x) - (-2)((x^2+2x) ') / (x^2+2x)^2

that should be ok

[boreal warren]

seems correct

now simplify

[woeful flame]

I got -4x-4 / (x^2+2x)^2

[boreal warren]

o.O how?

2' = 0 so the first term is gone completely

(x^2+2x)' = 2x+2

[woeful flame]

Fixed

[boreal warren]

ah yes

thats correct 😄

[woeful flame]

but that what I sent

[boreal warren]

oh yea

well derivative calculators are giving me the answer in a different form

they factorise the denominator into x^2 (x+2)^2

[woeful flame]

i dont understand

[boreal warren]

sorry u were correct all along haha

[woeful flame]

my english is very bad

[boreal warren]

i think

hmm i can't think why it is marked wrong#

try just expanding the denominator into the form x^2*(x+2)^2, even though it should be equivalent the computer may only recognise it in a particular form depending on how the question is coded

i worked for my uni's maths department writing online quiz questions... sometimes it can be difficult to code such things to recognise all inputs

[spring apex]

=pup derive \frac{1}{x+2} + \frac{2}{x^{2}+2x} - \frac{1}{x} + 3

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@spring apex

[spring apex]

=pup derivative \frac{1}{x+2} + \frac{2}{x^{2}+2x} - \frac{1}{x} + 3

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@spring apex

[spring apex]

?!

[boreal warren]

well i didn't anticipate that

if its a half decent program it should recognise that it simplifies to zero on its own

[spring apex]

=pup derive \frac{1}{x+2}

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@spring apex

[spring apex]

=pup derive \frac{2}{x^{2}+2x}

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@spring apex

[spring apex]

There’s where you went wrong

Did you write that?

[boreal warren]

ye he got that

[spring apex]

He did?

I thought it was something different

[boreal warren]

denominator was written differently

in the form (x^2+2x)^2

[spring apex]

Try writing the minus out of the fraction

So instead of -1/2

It’s -(1/2)

I know it’s the same but computer tests are retarded

[boreal warren]

they are... very xD

[glacial hazel]

Hello

[steady sleet]

no

[elfin mica]

need help here. i did a method but i think i messed up somewhere

is there a formula for sinθcosθ again?

[warm pine]

Split it in to 12 triangles

sinxcosx = (1/2)sin2x

[elfin mica]

ok thanks

im rechecking my work then ill ask for help again

ahh ok i found my mistake

thanks for the assistance btw

[valid pike]

Does anyone know how I can find the legnth of x

[serene field]

Make a triangle with x as the hypotenuse

Relate the side lengths of the outer and inner triangles

Pythagorean Theorem to solve for x

[valid pike]

Thanks

[elfin moon]

Not sure where this goes but what is this problem called?

Given a polygon P with holes (obstacles inside P) and a point q, find q' that is furthest away from q (Euclidian distance) and not on a boundary of P

[upper karma]

what are the conditios for 2 segments to intersect?

defined each segment with 2 points

[keen aspen]

The slopes have to be different

[upper karma]

something else?

[kind mist]

https://cdn.discordapp.com/attachments/367528357614845952/523970434614558730/IMG_20181216_131106.jpg

Stuck on #9 pls help

Ive got the solutions pi/6 and pi/3 but not the other two

[keen aspen]

There are 2 solutions in (0, 2pi]

Look on the unit circle to see where sine is positive and has the same reference angle

[upper karma]

I Need help

#smart peaple plz help meh

[chrome fiber]

[upper karma]

Lead has a density of 11.35g/cm"3. Calculate the maximum number of lead spheres of radius 1.5mm that can be made from 1kg of gold

@chrome fiber if you can help be that would be awesome!

😦

[upper karma]

guys help

[plucky marlin]

[upper karma]

we have three points: A(-1,3), B(1,4), C(3,y)

they are collinear

we wanna find y

how can i do that ?

[plucky marlin]

the line connecting those points would have the same slope

or u could connect those points and say that the area = 0

and then solve for y

[upper karma]

let me try that first one then i'll come back with the results

ok solved thanks

y = 5

hi can i also post a question here? or does it have to be in the questions channel

it's about analytical geometry

[hard gale]

you can post it here

[upper karma]

thank you 😃

Q: if the line segment connecting (x, 3) and (5, y) is bisected by the point (7, -5) , find the value of x and y

i tried to come up with equations however still haven't found the answer

[hard gale]

bisected, ie the point in the middle of the line segment is (7, -5)? (my geo vocab in english is pretty bad)

[upper karma]

bisected means it divides the line in two specifically at the middle

oh wait hahaha

i think i get it now lol

i was thinking point 7,-5 is not the one in the middle

[hard gale]

oof

[upper karma]

omg you're a genius octo!!! thank you!!!!

i can solve this now

[hard gale]

👌

[upper karma]

Lead has a density of 11.35g/cm"3. Calculate the maximum number of lead spheres of radius 1.5mm that can be made from 1kg of gold

pls help meh

[hard gale]

"Calculate the maximum number of lead spheres of radius 1.5mm that can be made from 1kg of gold" dafuq?

[crisp hornet]

Density*volume=mass

Convert the g/cm”3 to kg/m”3

And find the volume of each lead sphere(4/3 * pi*r^3)

Multiply both to get mass of one lead sphere

Then equate the masses to find the no. Of spheres required

[civic monolith]

can someone help me?

i’m doing geometry

kinda understand nothing

[dire rampart]

sure, ask your question

[civic monolith]

so this question says “(3,-7) and 3x-4y=12 write an equation for a line perpendicular”

i literally know nothing of what i’m doing

[serene field]

What is the slope of a line that is perpendicular to another?

More specifically, what slope will be perpendicular to the given linear equation?

[upper karma]

the relation is given by (slope1)*(slope2) = -1

lets say the slope of a line is 5, then the slope of a line perp to it is -1/5

[upper karma]

Lead has a density of 11.35g/cm"3. Calculate the maximum number of lead spheres of radius 1.5mm that can be made from 1kg of gold

plz help

bk thx but i dont think that worked

[keen tangle]

how do you even make lead spheres of gold?

reverse philosopher's stone

[dire rampart]

atom bender

[civic monolith]

sorry i never responded back, but i understand my question now so thanks

[upper karma]

i need halp plz

XD

[trail helm]

I need help with factoring

[sly belfry]

@trail helm go ahead

[frozen saffron]

Hellp

I need help

So I have 10th grade math right now and I need some help about similar triangles.

Like SOH CAH TOA

[trail helm]

IM in 8th grade geometrey rn and my teacher is useless

[frozen saffron]

maybe I can help you

[trail helm]

I need help with factoring

[frozen saffron]

what kind of factoring?

binomials?

[trail helm]

10x^2 + 27x + 5

[frozen saffron]

so basically u need to transform that on a factored from

gimme a minute

imma solve it for you

[trail helm]

I just need help cuz I have a semester test tmmrw

And im doing the study guide

[frozen saffron]

do you know the criss cross method?

[trail helm]

No

I knno the greatest common factor thingy

[frozen saffron]

do that to find the factored form of that equation

imma show the procedures to you aight

[trail helm]

2x^2 + 27x + x

[trail helm]

What about the exponent

Ill try that

[frozen saffron]

the exponent will be gone

[trail helm]

Ok

How would I use that on 10x^2 + 27x + 5

Im confused

[frozen saffron]

multiply any number that can result to 10

like 5 x 2

[trail helm]

Oh ok

How about the 5

[somber coyote]

Yue:

[serene field]

We can then factor by grouping.

[frozen saffron]

that procedure is way too confusing

[serene field]

Product of the first and last

Add up to the middle

[trail helm]

@frozen saffron Thank you

That makes sense now

I got (5x+1)(2x+5)

I think thats right, right?

[frozen saffron]

yup

remember always check the middle

[trail helm]

Ok

What do you mean

How do I check the middle

[frozen saffron]

I mean you need to have the same result like in the middle

if you cross multiply

So can someone help me about my similar triangles thing?

I need a help about SOH CAH TOA

[upper karma]

Does anyone know how to find the missing angle

[frozen saffron]

80 i think

because the other line is parallel to that one

[serene field]

Just ask, Chris

Many of us know enough trig to help out. 🤔

[frozen saffron]

Can someone help me out?

[serene field]

,rotate 90

[serene field]

Draw out your diagram

[frozen saffron]

kk

Is that correct?

[serene field]

Looks right.

[frozen saffron]

how do I find the height tho?

the total height

[burnt relic]

im a grade 10 but ill try to help

add up the angles on the left 37 and 58

or whatever they were

then use tan(angle on left)=x/9.8

tan because the angle in question is adj and u are trying to find the opp so tan uses that

to solve just do tan of the angle then multiply by 9.8(cross multiply)

^^^^

not sure if correct but pretty sure

[woven wigeon]

yeah its fine

[serene field]

It only implies the angle of elevation from the anchor point to the top of the tower is 58 degrees from the base, not 58 degrees on top of the 37 degree angle

If it were 58 degrees on the 37, that would be > 90 and there would be an obtuse angle.

[woven wigeon]

owo, so its not a right triangle 🤔

[burnt relic]

oh ur right

[serene field]

We have a triangle with two known angles and one side. We can easily find the third angle by the sum of angles of a triangle. Law of sines should quickly find the solution.

[burnt relic]

lol rip sry

[woven wigeon]

damn big f

[serene field]

This question threw me off too. 🤔

Don't really need the 37 degree thing, as far as I can see it.

[burnt relic]

Anyways, can i get some help on this?

Maria needs to load cars onto a transport
truck. She is planning to drive up a ramp, onto the truck bed. The truck bed is 1.5 m high, and the maximum angle of the slope of the ramp is 35 degrees .

How long should the ramp be? Round your
answer to one decimal place.

[serene field]

Draw your diagram

[woven wigeon]

so basically sin(35)=1.5/x.... hopefully lo im not a math person

[serene field]

You know you have a right triangle and can find all angles via sum of angles. Law of sines looks like it'd work here as well.

[burnt relic]

ya i was just gonna ask about that

idk about that studd yet

stuff

[woven wigeon]

same

[serene field]

Then work with SOHCAHTOA

[burnt relic]

we have a test on cos,tan,sin and sim triangles

tom

[serene field]

Have you created your diagram?

[burnt relic]

i ahve but cant send a pic

but angle on left is 35

and height (or opp to angle) is 1.5

m

[serene field]

You have multiple ways to go about this, but your goal is basically to find the length of the ramp.

Tbh, based on the question, I'm not sure whether they want the second leg or the hypotenuse.

[woven wigeon]

the hypo

[burnt relic]

thats what i thought

but i did both

and did not get either of them to what the ans was

[serene field]

Well then, look at your trigonometric ratios.

Only sine and cosine have hypotenuse, so those will get us to the hypotenuse.

[woven wigeon]

owo

[serene field]

With respect to the 35 degree angle, for example, we would need to use sine.

As we don't have the length of the adjacent.

With respect to the 55 degree angle, cosine would be needed, as we don't have the length of the opposite to that angle.

Either way, you can solve for hypotenuse.

[burnt relic]

okay thanks for that. I will try it out

[summer barn]

is this question possible without a graphic calculator? i seriously have nowhere to start with solving this

[merry bough]

@summer barn what formulas do you have that relate cosine and sine?

[summer barn]

cos^2x+sin^2x=1

[merry bough]

what else?

something that has adding in the brackets like cos(x+b) = sin??

[burnt relic]

Peggy’s Cove Lighthouse in Nova Scotia is possibly the most photographed lighthouse in the world. The observation deck is about 20 m above sea level. From the observation deck, the angle of depression to a boat is 6°. How far from the boat is the lighthouse, to the nearest metre?

^^^^^^^^^^^

can i get some help w/ that please

[mystic shadow]

did you try a picture?

[burnt relic]

sry for bad quality but top right is 6

[timber hinge]

They cram so much useless shit into those questions

[burnt relic]

left*

[mystic shadow]

okay which side is the side youre trying tosolve for

[burnt relic]

opposite to the angle right?

[timber hinge]

Hypotenuse

[mystic shadow]

pretty sure they mean the bottom side

that is the distance from the lighthouse to the boat

[burnt relic]

k so if i were to find the bottom side,

i would do tan 6=x/20

but that leaves me with 2.1

which is not the answer

and i fi were to solve for hypotenuse,

i would do cos 6 = 20/x

[mystic shadow]

it asks for the nearest metre

try inputting 2 m

[burnt relic]

ohh i think i got it

my problem was with doing the tan not rounding

but thanks for your help!

[mystic shadow]

np

[burnt relic]

I actually need a bit more help lol

If you were in a hot air balloon at a vertical height of 500 m above the ground, at what angle of depression would you look at a point on the ground 800 m horizontally on the ground from directly under the balloon?

[mystic shadow]

you need to use an inverse trig function on your calculator

otherwise it's solved the same as the last problem

try drawing the picture first

[glad ocean]

is it arctan (800/500)?

[mystic shadow]

seems right

[upper karma]

Boi

[mystic shadow]

@upper karma ?

[upper karma]

whats up?

Is it fine that I forget triggernometry cuz I moved on xD

[summer barn]

@merry bough cos(x+3/2π) right?

sorry i was away just now

[upper karma]

@glad ocean yesh that should be rigth

right

[glad ocean]

one does not simply move on from trigonometry

[serene field]

I mean, I haven't done trig stuff since calc 3-ish

[burnt relic]

can i get some help pls

its pretty easy im just not understanding it rn

[summer barn]

@burnt relic find CB using pythag

[burnt relic]

ive done that

[summer barn]

find the scale factor

[upper karma]

^

[serene field]

I would find angle BAC and use tangent, but that works too 👀

[upper karma]

Then you get DA

Then you minus BA

I guess

[summer barn]

^

[timber hinge]

tru

[upper karma]

Idk Im just like 13 so :/

Oops wrong

[summer barn]

@serene field yeah thats prolly better lol

[burnt relic]

for the scale factor

is it 2.5/4.7=db/3

[timber hinge]

CB = 2, 2.5/2 = 1.25, 1.25 * 3 - 3 = 2.5

wait

hmm looks right

mine

Idk what you're doing

[burnt relic]

lol same

[serene field]

<BAC: cos(t) = 3/sqrt(13)
Once you find t:
tan(t) = 2.5/x
Once you find x:
x - 3 cm = final answer

[timber hinge]

no trig needed

[burnt relic]

this has to be done without trig

[summer barn]

@burnt relic have you done sin cos tan yet

[serene field]

I like my trig

[burnt relic]

liek sin cos tan

[summer barn]

or do you know what they mean

[burnt relic]

i know

[summer barn]

oh ok use his method then

[timber hinge]

similar 📐

[upper karma]

I forgot triggernometry since I did it last year XD

It's okay right?

[burnt relic]

its just that our teacher wanted us to do it with scale factor

[upper karma]

To forget u h

[summer barn]

@serene field but that requires a calculator right

[serene field]

If you're good enough with ratios, you don't need a calculator 🤔

[summer barn]

cuz you need to calculate t=arccos(3/sqrt13)

[timber hinge]

gl with that

[burnt relic]

Ok i got one more

A plane takes off in a straight line and travels
along this line for 10 s, when it reaches a height
of 300 m. If the plane is travelling at 60 m/s, at
what angle is the plane ascending?

would i divide 300 by 10 to get the m/s to make all units the same?

[mystic shadow]

it's still the same problem

however you should see how far it travels by using its speed

$$\frac{60\text{ m}}{\text{s}}\cdot\frac{10\text{ s}}{}$$

[somber coyote]

MiracleMan:

[mystic shadow]

be sure to write units so you can cancel them correctly

[burnt relic]

Surveyors need to determine the height of a hill. They set up a laser measuring device on a pole that is 1m tall and shine the laser toward the top of a second pole, which is 1.6m tall. Then they adjust the distance between the two poles until the laser hits the top of the longer pole and the top of the hill. The 1.6m pole is 415m from the centre of the hill. The 2 poles are 12m apart. Find the height of the hill.

final q for the day if any1 can help

[mystic shadow]

draw the picture

[burnt relic]

solved

thanks for responding

[mystic shadow]

then draw a triangle eith the base starting from the top of the first pole

ok

similar triangles and all that stuff

[river marsh]

i propose we delete trigs from the entirety of human knowledge

because they are hard

[last imp]

@burnt relic

[dark osprey]

,tex just use pyt similarty and find AD and then DB

[somber coyote]

ISIPINI:

[last imp]

Hey wut ?

[dark osprey]

the triangles are similar for obvious reasons 😄

similarty implies taht AB/AD = BC / DE

[last imp]

That's what I do

[dark osprey]

so its correct ...

[last imp]

Just proved it with trig

[dark osprey]

aah u were the one helping ;D

[last imp]

😟

[upper karma]

Does anyone understand Proofs?

[mystic shadow]

pretty sure someone exists that does

[upper karma]

But can you prove that?

i just need someone to explain ASA AAS and all the other ones

Those are generally considered postulates in elementary geometry, no?

I believe so but im in high school

That's what I meant. Between algebra last year and algebra ii/trig next year, right?

algebra 2 next year yes

OK, so in your level of geometry, I'm pretty sure SSS, ASA, and so on are considered postulates, no?

[astral hornet]

you can prove similar triangle, then use prop of similar triangles

[upper karma]

Can someone explain to me how I go about solving question seven from above?

[limpid basin]

Have you learned transversals yet?

[upper karma]

Yes

@limpid basin

[kind wagon]

can anyone help me discover the third coterminal angle of angle 67?

[split laurel]

What do you have so far @kind wagon

[kind wagon]

@split laurel 427, 787

[split laurel]

why not just add 360?

1147

Or subtract 360

-293

@kind wagon

[kind wagon]

hmm cheers @split laurel

[split laurel]

anytime

[copper valve]

what

[limpid basin]

wat

[steady sleet]

ew geometry

[pliant cloak]

No geometry is fun

and combinatorial geometry is masterrace

[upper karma]

no geometry is fun. that's true, no part of geometry is fun

[warm pine]

And so began the great geometry war of discord

[upper karma]

Can someone help me with this question please

The graph of y=aSinb theta has a maximum value of 5 and a period of 45 degrees What are the values of a and B

since sin x has a maximum value of 1

max(a sin bx) = 5 implies that a = 5

the period is 45

45 degrees is pi/4

and regular sine has period 2pi

sin(kx) has period 2pi/k

so you can figure out b from that

Oh i see thanks

[sacred cradle]

@upper karma

[upper karma]

how do i prove EM=DM and EP=DP

[wild hamlet]

prove that ADB and BEA are similar triangles

[upper karma]

ADF?

[wild hamlet]

ADB sorry

[upper karma]

are they even similar

<A!=<B

i found that <EAH=<DBH 😄

<HCD=<HBM

<ECH=<HAM

[lapis geyser]

what do i need to research to solve these 3 problems ? https://www.dropbox.com/s/3bwgn4iln0zify5/wordpad_2018-12-21_14-52-26.png?dl=0

[keen aspen]

@lapis geyser sin function is odd and cos is even

[lapis geyser]

can you recommend a quick trig course online?

i need a refresher again

[keen aspen]

Hmm you can use Khan Academy

[lapis geyser]

ok thanks

[sharp nymph]

@lapis geyser Paul's notes

http://tutorial.math.lamar.edu/Classes/CalcI/ReviewIntro.aspx

Very useful trig

[lapis geyser]

oooooh that looks like gold, thanks @sharp nymph

[wild hamlet]

Avoid posting in multiple channels

[upper karma]

how would i let this rotate

[mystic shadow]

could you be more specific?

[upper karma]

like lets say i would want rotate this "T" about 50 degree how would i do that

[mystic shadow]

by the use of a rotation matrix

where the new coordinates (x',y') are given by multiplying the rotation matrix on the right by the original coordinates (x,y)

[lapis geyser]

can someone tell me yes, no if x = 4

[hard gale]

Ya x=4cm

[lapis geyser]

thanks

[upper karma]

@mystic shadow when doing only cos for x and sin for y it also works

[mystic shadow]

no it doesn't

if you can't do matrix multiplication, the full equations are:
x'=xcos(theta)-ysin(theta)
y'=xsin(theta)+ycos(theta)

[upper karma]

well i am doing

y = length * sin(rad(deg))

and it works fine

[mystic shadow]

you're just breaking the vector into its x and y components

not rotating it like your original request was

hope you figure it out bbl

[young barn]

what could be the graph of a two variable linear equation

like

y = m1x1 + m2x2 + c

[serene field]

That's a plane.

i.e.

[sick veldt]

https://cdn.discordapp.com/attachments/377497721583697930/526581492793475072/Screen_Shot_2018-12-23_at_9.06.43_PM.png

is this correct?

[keen aspen]

Yep

[upper karma]

if i know a position vector OA and another position vector OB

how do i know AB?

do i just add them

[dire rampart]

OB-OA

@upper karma

[upper karma]

ah i see

[lapis geyser]

can someone link me info i need to solve this please

https://www.dropbox.com/s/29h9i8z4vf66kng/firefox_2018-12-24_12-17-59.png?dl=0

[umbral snow]

Two things give most of these questions away
sin(-x) = -sin(x) (sin is odd)
cos(-x) = -cos(x) (cos is even)

@lapis geyser

[mystic shadow]

cos(-x)=cos(x)

[lapis geyser]

i never learned Trigonometry, learning it from scratch, and homework is due today, totally lost atm lol

any Khan series that explains the picture i linked. Pressed for time

after homework is due im going to go back over it again and study it further

[mystic shadow]

http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/TrigIntro.aspx

[dire rampart]

https://www.youtube.com/watch?v=5p8hokJ3Cqo

[lapis geyser]

thank you very much both of you

[upper karma]

dudesmen

if the triangle with vertices X(3, 5) , Y(4, 2), Z(-5, a)

is right angled at Y

how can we find 'a'

?

[hard gale]

pythagoras will certainly do the trick

[upper karma]

how ?

[hard gale]

if our triangle is rectangle at Y, then

$$XY^2 + YZ^2 = XZ^2$$

[somber coyote]

emeric75:

[upper karma]

right

then ?

so i get the distance between the sides ?

and work my way backwards with the distance "algorithm"

?

[hard gale]

ya, and you'll get an equation for a

[upper karma]

the distance rule is as i know it something like this:

$$sqrroot((x2-x1)^2+(y2-y1)^2)$$

[somber coyote]

Sentoss:

[hard gale]

yea get the lengths of the sides using dat formula

and plug them inside that pythagoras thing

[upper karma]

let me try and get back to you

bro

all i got is

XY = sqr10

XZ = sqr(64+(a-5)^2)

YZ = sqr(81+(a-2)^2)

where can i go from here

[hard gale]

yea seems right

ok so let's plug those things inside that pythagora's equation (we need those distances to verify pythagoras if we want XYZ to be a right triangle)

[upper karma]

XZ= sqr(YZ^2 + XY^2)

[hard gale]

$$XY^2 + YZ^2 = XZ^2 \iff 10 + 81 + (a-2)^2 = 64 + (a-5)^2$$

(tbh just leave the squares)

[upper karma]

there is no "(a-1)^2"

[hard gale]

typo'd

[somber coyote]

emeric75:

[upper karma]

"(a-2)^2" is the correct one

np

so i should move the 64 to the left and the (a-2)^2 to the right

[hard gale]

ya

[upper karma]

ok let me try

my bad math skills gave me:

-27 = -29

😂 😂 😂 😂

[hard gale]

oof

1sec brb

[upper karma]

ok

[hard gale]

ok so

moving our variables on one side and the constants on the other one

$$91-64 = (a-5)^2 - (a-2)^2$$

[somber coyote]

emeric75:

[upper karma]

mine is like

$$64-81-10 = (a-2)^2 - (a-5)^2$$

[somber coyote]

Sentoss:

[hard gale]

yea exactly the same thing

(just the signs are reversed)

[upper karma]

ok

so the prob is with the next step

[hard gale]

expanding those squares :/

[upper karma]

for me it turned out like

:

$$-27 = a^2 - 4 - a^2 -25$$

[somber coyote]

Sentoss:

[hard gale]

my eyes bleed

(a-b)^2 does not equal a^2 - b^2 in most cases

[upper karma]

wow!!!

[hard gale]

$$(a-b)^2 = (a-b)(a-b) = aa -ab - ba +bb = \boxed{a^2 -2ab+b^2}$$

[somber coyote]

emeric75:

[upper karma]

factorization ?

again!!!!!!!

no!!!

[hard gale]

that's distributing

[upper karma]

really ?

how do you do these things ?

[hard gale]

(that's what i just did?)

[upper karma]

so what should i do again ?

[hard gale]

1. memorize it /or/ 2. remember that (a-b)^2 = (a-b)(a-b) by definition of squaring and expand

[upper karma]

hmm

gimme a sec

so basically it'll be

$$-27 = (a-2)(a-1) - (a-5)(a-1)$$

[somber coyote]

Sentoss:

[upper karma]

right ?

[hard gale]

why a-1?

$$-27 = (a-2)(a-2) - (a-5)(a-5)$$

[somber coyote]

emeric75:

[upper karma]

oh sorry i was still stuck in the factorization thing

then ?

[hard gale]

then expand those

[upper karma]

how ?

if i did that i'll end up again in:

$$a^2 - 4 - a^2 - 25$$

[somber coyote]

Sentoss:

[hard gale]

didn't think i'd be here for basic algebra but i am dommed to help you now

[upper karma]

bro sorry but i'm bad at math

like really bad as you can see

ughhhhhh

[hard gale]

$$(a+b)\cdot(c+d) = a(c+d) + b(c+d) = ac + ad + bc +bd$$

[somber coyote]

emeric75:

[upper karma]

bro ?

[hard gale]

so for example if we take $(a-2)(a-2)$

[somber coyote]

emeric75:

[hard gale]

$(a-2)(a-2) = a(a-2) - 2(a-2) = a^2 - 2a -2a + 4 = a^2 -4a +4$

[somber coyote]

emeric75:

[upper karma]

jaw drop really hard

how did you do that ?

[upper moat]

Distributive property

a * (b + c) = ab + ac
In this case, he expanded it into:
(a + b) * (c + d) = a * (c + d) + b * (c + d)

[upper karma]

aha

so in short....

black magic

[upper moat]

Well, here's an example:

4 * (5 + 6) = 4 * 11 = 44, yes?

Well, with the distributive property, this becomes

[upper karma]

yea i think so

[upper moat]

4 * (5 + 6) = 4 * 5 + 4 * 6

= 20 + 24

= 44

You basically multiply the stuff outside the brackets with everything inside

[upper karma]

but here we have 2 brackets and there is no outside

[upper moat]

Yes

But

(a + b) * (c + d)

Pretend (a + b) is ONE THING

[upper karma]

so basically

i think it'll be something like

$$-27 = a(a-2) - 2(a-2) - a(a-5) - 5(a-5)$$

[somber coyote]

Sentoss:

[upper moat]

What was the original?

[upper karma]

$$-27 = (a-2) (a-2) - (a-5) (a-5)$$

[somber coyote]

Sentoss:

[upper karma]

and the original to this is :

$$-27 = (a-2)^2 - (a-5)^2$$

[somber coyote]

Sentoss:

[upper moat]

$$-27=a(a-2)-2(a-2)-a(a-5)+5(a-5)$$

[somber coyote]

Hello World:

[upper moat]

since -(a - 5)^2 = -1(a - 5)(a - 5)

[upper karma]

oh that's why the +

[upper moat]

yep

[upper karma]

but still why not the same with the 2(a-2) ?

[upper moat]

Because it's (a - 2)^2, not -(a - 2)^2

[upper karma]

oh ok

now how can i solve this "equation" ?

[upper moat]

Continue expanding

For example

a(a - 2)

Seem familiar?

a * (b + c)?

[upper karma]

NOOOOOOOOO!!!!

i can't expand more than that

like this is the best i can do

so...

a(a-2) will be a^2*-2a ?

[upper moat]

Yes

Gj

Now do that expansion like 3 more times

for the other 3 parts

[upper karma]

well it turned out to be at the end

that $$ a = -0.6$$

[somber coyote]

Sentoss:

[upper karma]

is that right ?

ughhhh

[plucky marlin]

=pup -27 = (a-2)^2 - (a-5)^2

[charred spear]

Do more with Wolfram|Alpha pro

Query made by

@plucky marlin

[plucky marlin]

did u mess up the signs or something ?

[upper moat]

-27 = (a - 2)(a - 2) - (a - 5)(a - 5)
-27 = a(a - 2) - 2(a - 2) - a(a - 5) + 5(a - 5)
-27 = a^2 - 2a - 2a + 4 - a^2 + 5a + 5a - 25
-27 = 6a -21
-27 + 21 = 6a
-6 = 6a
-1 = a

Was your process anywhere close to that?

[bright nebula]

-27=(a - 2 + a - 5)(a - 2 - a + 5)
-27=(2a - 7)3
-9=2a - 7
-2=2a
-1=a
shorter version

[upper moat]

That works, but i was talking to him about the distributive property above, so it was only fair to solve it the same way

[dusky yarrow]

absolutly aweful diagram here but this is the problem i have been presented with and i have no clue how to find the length of x or if it is even possible

[umbral heron]

try making triangles with the center of the circles

[dusky yarrow]

o

thats genius

[umbral heron]

lol

[dusky yarrow]

ty

[umbral heron]

np

[dusky yarrow]

ok now im stuck again

how would i find the hypotenuse

[thorn talon]

which triangle did you make?

[dusky yarrow]

lemme make a new better diagram

[thorn talon]

i assume the right triangle?

with the hypotenuse through the two circles?

[dusky yarrow]

not better but yeah that triangle

[thorn talon]

there's a better triangle you can draw i think

that vertical side

how about you extend it further?

so it goes down to the centre of the bottom circle

[thorn talon]

yep

like that

do you know the length of each side now?

[dusky yarrow]

no

[thorn talon]

well the vertical one is 1 + x right?

[dusky yarrow]

yes

[thorn talon]

top horizontal is 1 right?

[dusky yarrow]

yes

[thorn talon]

and the hypotenuse

it's equal to 2 radii length right?

[dusky yarrow]

im not sure what radii is

[thorn talon]

2 radius length

[dusky yarrow]

oh yeah

[thorn talon]

so 2 * 1 = 2 right?

[dusky yarrow]

yes

[thorn talon]

and then you can just pythagorean theorem from here

[dusky yarrow]

alright thank you

[lusty maple]

Would I be right in saying that given 2D vectors A and B, if (A + B)'s x and y components are both not zero, A * q added to B * p can be equal to any 2D vector by changing values of q and p (allowing q and p to be negative, zero or positive)? If that makes sense to you.

[hard gale]

what if A and B are collinear ?

[lusty maple]

if (A + B) has non-zero x and y components then wouldn't they inherently be not collinear?

or am I understanding you wrong

[hard gale]

A=(1,1) / B=(2,2) they're collinear right?

[lusty maple]

ah I see, I only thought of them pointing in opposite directions. Good call

So would A + B and A - B and B - A all have to have non-zero x and y components?

[hard gale]

well you'd need A != kB for any k in R

(which is much more cases than you can write in a "A+B A-B" way)

[lusty maple]

All right, makes sense.

And in 3D I'd need 3 vectors?

[hard gale]

non colinear one another again but ye

[lusty maple]

That's great, thank you. I'm working on a programming project and want to balance a mass on thrusters automatically, or at least attempt to get as close to it as possible.

Maybe a physics server might be a better place for assistance with the rest of it though

[hard gale]

oof good lucks : if you wanna go further about this vectory idea just take any lin alg book

yeah physics would be better ^^

[lusty maple]

Thanks once again, I appreciate the help 😃 I'll keep at it

[hard gale]

[lusty maple]

[umbral snow]

You want to look into control theory for holding a state

Feel free to ask anything about control here or in the physics server!

[lusty maple]

Currently I'm at the point where I have the torque that each thruster provides and can also get the change in angular velocity for firing it for one physics update (1/50s), my approach is to give each thruster a factor from 0 to 1 based on how (inversely-)"useful" it is to correcting the error. So the more torque a thruster has (if it is acting in the correct direction) the more useful it is therefore it needs to be at a lesser throttle % (factor) than a weaker thruster to generate the same torque on the centre of mass.
If I'm thinking it through right, assigning each thruster a factor based on the amount and direction of torque produced so that the torque * factor are equal for all thrusters should produce a balanced system.

My problem is calculating the usefulness properly. I tried a couple approaches and some worked a bit but not fully. Best I had is a stable balancing system but it would constantly switch between two engines firing at full power and two being off, so whenever it would lean any amount to the left of centre the right engines would turn off and vice versa. I tried fixing it by making the error-based factor smaller the closer to the correct orientation the object was, so if you were 45 degrees off the engines would try hard to correct it but as you approached the correct orientation the factor would fade towards one which simply has all the thrusters balanced proportional to how much torque they produced. I messed up somewhere and couldn't get it right though and anyway it felt like an improper cheaty solution.

How would you approach a problem like this:
vectors t0 through tn are known (torque from thrusters)
coefficients c0 through cn are unknown (factors or throttle, but for the sake of this they are not constrained to 0.0-1.0, they can be any real number)
error to correct e is known

find values of c0 to cn where (c0t0 + c1t1 + c2t2 + .. + cntn) is as close to e as is possible

there would be multiple answers I assume and that makes it even harder. Is there a better approach to something like that? Where would I find info on this?

[serene field]

Something like

[somber coyote]

Yue:

[serene field]

That?

[lusty maple]

yes, although that wouldn't cover edge cases where they can not be equal to e and instead you'd have to get the closest approach

Would they need to be broken down into simultaneous linear equations recursively? Ignoring the edge cases for now

[serene field]

Honestly, I have no idea. A linear system is probably the way I'd attempt to solve this. Maybe one of the helpers can solve this better than I know how. 🤔

[lusty maple]

That's grand, I'll look into the different methods. Thanks 😃

[lusty maple]

A lad on a computer science discord said that it's a least squares problem but I can't see any way to use the method for my problem although I do see some similarities

[lusty maple]

Now that I think about it, I'm pretty sure I can't break it down into linear equations because even if it was 2(c0) + 3(c1) = 4, there are infinitely many values for c0 and c1 that would work. The only values I actually want are where |c0-c1| is the smallest it can be.

Maybe I should think of it as a line on a n+1 dimensional graph

If that would get me anywhere closer to an answer? Probably not but worth to think on

e.g. say error correction value e is the y axis, c0 is x axis and c1 is z axis. Using the previous example, y = 4 so equation is 2x + 3z = 4.

[umbral snow]

Okay, so what is this thing? Are we in 2D? A plank with two forces acting on each side?

[lusty maple]

My ideal result would be where x and z are the closest and I know it's easy af to calculate that but it's late and I can't put my finger on it lol

It's 3D

Imagine a hovercraft with engines

[umbral snow]

Four of them?

[lusty maple]

for example 1 each corner so 4 engines

yea

And ideal scenario would be all 4 at 100%

but a possibility would be 2 corner engines at 100 and 2 others at 0

and it would still technically be stable

so I guess total % should be max if I'm thinking straight

either that or all %s should be as similar as possible

I'm leaning towards the latter actually

Also important to note is that in this example the engines are all pointing straight down so it's more 2D-ish but that may not always be the case

[umbral snow]

I imagine you're trying to control two things, the height of the hovercraft and the angle at which it hovers

3D manouvering is tough, lol.

[lusty maple]

The height would be easier I assume

get the angle right then adjust total thrust

so getting the angle stable is the main challenge

[umbral snow]

You can do both at once with an ideal system

[lusty maple]

How is that?

[umbral snow]

Well, how much do you know? Understand calculus?

[lusty maple]

Not really just the overall idea of it, not in uni yet so just doing this as a hobby out of interest

I have the vague idea of getting the slope of a curve by being more and more precise with an average over a time, or however you would put it. But in-depth not at all

[umbral snow]

In physics, velocity is the slope of displacement, and acceleration is the slope of velocity

You also have F = ma, so you can control "a" directly with thrusters.

So you have full control over acceleration, but you want control over displacement.

[lusty maple]

So I imagine there'd be a most efficient way of changing s by controlling a, since if you just go full on until you overshoot you're gonna have to backtrack etc?

[umbral snow]

Exactly, you're calling it. And, it's not a crazy simple problem

[lusty maple]

And how does that play into angle?

[umbral snow]

All the while, you have to control the angle

You have Newton's second law of rotation.
τ = Iα

τ, or torque, is controlled by the thrusters, so you can control acceleration of angle, or α

Once again, you want to control the angle, but you have control of the second derivative

[lusty maple]

ah, I see

[umbral snow]

So I've said a lot but haven't said much for results

You should have some I from the physics simulation, so you should know how fast it starts to rotate, given a thruster is on

[lusty maple]

Yeah, I can get the angular acceleration (if that's the word) for firing it for 1/50th of a second (one physics frame), I assume that dealing in discrete units of time simplifies things as well compared to a real-world simulation.

I could be wrong through since that may only be the case if you are doing compounding calculations (or however they're called)

[umbral snow]

Run one thruster for two "cycles", how much does the angle change in each one?

[lusty maple]

angle or angular velocity?

[lusty maple]

Angular velocity changes almost same amount as predicted by the code, so for x axis it's about 7.5 deg/s per cycle. Angle itself is more difficult because the whole thing flies off at an awkward rotation and direction so the axes get messed up, since I know ang vel around local axes but the rotation itself is shown about the global axes. I use unity and the physics tend to be a bit wonky but realistically implemented. My money is on local x angle being 0 at beginning, then 1/50 of 7.5, then another 7.5 is added to vel so 1/50 of 15 added to angle so total after 2nd update would be 7.5/50 + 15/50 = 22.5/50
Phys frame 0: angle 0 vel 0
1: angle 7.5 vel 7.5
2: angle 22.5 vel 15

I'd have to do a more in depth test for 100% accurate results, I through that one together sloppily since it's 3am so I'll be back at it tomorrow. Thanks a lot for the insight btw, sorry about running off partway through but I shouldn't be messing up my sleep more than already lol. I'll check back the next day

[lusty maple]

Before I forget though, if I wasn't dealing with vectors but instead scalars, having an equation e = c0t0 + c1t1 etc. Wouldn't be hard to get optimal values for. I believe it would be:
optimal coefficient = error correction amount * (1 - (torque/total torque))
including signs and so on, not absolute values. Ofc if total torque cancelled out by default then it would be dividing by zero, so I'd have to have a failsafe for that and just set them all to one. That's besides the point though. My problem is that in 1D, A = kB no matter what, for real numbers. In 3D though, it's much more complex. Also torque can be in the wrong direction and in 3D that is usually not solveable by scaling, hence why my initial question where it was deducted that you'd need 3 vectors which would not be collinear to have full control over where they point if scaled independently and summed.

[surreal bolt]

Hello!

Question 1: So you have an 8 inch x 8 inch standard issue chessboard. You are also provided with unlimited 2 inch x 1 inch rectangles (dominoes).

Can you tile the chessboard with the rectangles with no overlap and no extra rectangles nor parts of rectangles hanging off the playable part of the chessboard?

SPOILER

Turns out the answer is yes. Just tile each row with four 2 x 1 rectangles end to end.

The following a famous problem ...

Question 2: So you have an 8 inch x 8 inch standard issue chessboard AND you remove 1 pair of diagonally opposite corners. You are also provided with unlimited 2 inch x 1 inch rectangles (dominoes).

Now, can you tile THIS chessboard (with corners removed) with the rectangles with no overlap and no extra rectangles nor parts of rectangles hanging off the playable part of the chessboard?

SPOILER

It turns out ... no :(

But, of course, in math you need proof.

And this turns out to be a tricky proof to come up with, but very simple to understand.

Each rectangle MUST cover two squares: one square of each color.

Since the two removed squares are the same color (WLOG black) you have 32 white and 30 black squares left over. This cannot be tiled by 2 x 1 rectangles.

But then one has to ask ... Question 3: What about other shapes? Maybe Tetris shapes?

Of course you can't tile an 8 x 8 chessboard with two corners removed and have it be tiled by pieces with an area of 4. That's 62 for the leftover area and 4 doesn't divide 62. 😃

Question 4: But what if I said you can remove Any 4 squares of the same color? Or any 4 squares where 3 are of one color and 1 is of the other? We solved the problem of not having the whole area being divisible by the area of the piece.

It turns out that all of these Tetris pieces:

++  ++      ++  +++ +
++ ++  ++++  ++   + +++

... must also cover exactly 2 white and exactly 2 black squares.

This means that not just any one of the above Tetris pieces but Any Combination of pieces cannot tile any board with unequal white and black squares. The only odd piece out is the T-shaped piece. But I'll leave that for further study.

[weary ingot]

Hey guys what is the best book for trig, a book.that explains concepts, and gives examples, easy to understand. Overall a great book to learn trig

[upper karma]

how do you sketch this graph with just the function and A values? thanks 😃

[waxen gorge]

Take the derivative

[vital frost]

So do I have just have to draw it and explain right?

[glad ocean]

yeah I guess

for AA postulate, say that:

1. the sum of angle measures in a triangle is 180 degrees
2. say there are three angles in two triangles: in triangle ABC, There are angles 1 2 and 3 and in triangle DEF there are angles 4 5 and 6
3. given angle 1 is congruent to angle 4 and angle 2 is congruent to angle 5, angle 3 is congruent to angle 6 using the difference of 180 - (angle 1 + angle 2) = 180 - (angle 4 + angle 5)
4. since this is AAA, the ratios of the corresponding sides in each triangle
5. in conclusion, AA postulate is an extension of AAA, proving that two triangles are similar

[azure tundra]

a. since ADC is a right angle at D, ACD (angle) = angle B (since they both equal 90-A) therefore (by AAA similarity) ADC is similar to ABC, same thing for CBD but this time it's angle DCB=90-B (aka same as A)

hopefully this helps (sorry for the amateur proof writing)

[weary ingot]

Standard form of a circle is (x-h)^2+(y-k)^2=r^2,let's say center of the circle is at 3,-2. Now its (x-3)^2+(y+2)^2=81, for what type of question would they ask me to plus in x and y Into there and what does that give me?

[vital frost]

I'm so confused what

[glad ocean]

X + y = 10

oh wait 81 should be the sum of two squares

[vital frost]

Are y'all talking to me?

@keen aspen since you really helped me.last time can you help me with the picture above

[glad ocean]

can someone Mathbot this for me please?

[copper valve]

no

[glad ocean]

all the lines that you can make out

[copper valve]

no

[surreal bolt]

use excel --> save delimited --> get rid of all commas ....

[vital frost]

Help...

[umbral snow]

Which are you having trouble with? Any definitions you're not good on?