geometry-and-trigonometry

[bleak rivet]

does BAQ = RPQ

since RP is parallel to BA

[fresh quiver]

Yes, as they are corresponding angles

[bleak rivet]

right

so

BAQ isnt = to angle x?

[fresh quiver]

No

[bleak rivet]

But

ABQ is = to Angle X

right?

[fresh quiver]

From the position, I think it's ABR

[bleak rivet]

Well ABR is angle X

But what Im saying is

Is ABQ corresponding to angle x

so is the answer 87?

[fresh quiver]

No, it isn't

What is angle X for you?

[bleak rivet]

Angle X is RBA

Im trying to find angle X

Im trying to figure out

what angle corresponds with angle X (RBA)

Im so confused

I understand that the answer is either 87 or 46 but I dont know which one

[fresh quiver]

It's neither

RBA doesn't correspond to anything.

[bleak rivet]

So this is what I got

[fresh quiver]

Yeah, QR is a straight line, so angle RBA + angle QBA = 180

Hence, angle X = 180 - angle QBA

[bleak rivet]

=93

thanks

[fresh quiver]

Exactly

No problem

[bleak rivet]

Hey another quick question

I have a few questions in a row that ask something like this

in this context what would supplement mean?

[dark sparrow]

supplementary angles add up to 180°

[fresh quiver]

This refers to supplementary angles, two angles whose sum is 180

[bleak rivet]

so easy then

168

[fresh quiver]

Base 20?

[bleak rivet]

What?

lol

[dark sparrow]

ignore that

[bleak rivet]

No thats for my Digital Engineering classes

but 168 would be right?

[dark sparrow]

yes

[dark sparrow]

nope

it isn't

it'd be 8A

[bleak rivet]

I would be done like 10 questions ago but some jerk in my class thought it would be funny to ask for 70 questions

soooo

here I am

wasting more hours

so if supplement is 180

[dark sparrow]

oh wait

[bleak rivet]

would complement be 360

[dark sparrow]

that's a 10 subscript there >.>

and uh

no, complementary angles add up to 90°

iirc

bc like

cosine = complementary sine

these are all

fancy terms not really worth memorizing tbh

[bleak rivet]

aight

thought so

[cobalt folio]

How did you check if two lines in R3 are parallel? Was that cross/dot product = 0 with the Director Vector of each one? Can't find it on Google

[fresh quiver]

I think it's cross product

of the director vectors

[dark sparrow]

yes

[cobalt folio]

ok thanks

[bleak rivet]

Ummm

None of these answers are right... are they

unless he meant to put 112 for D

2x + 2y

46 x 2 = 92 + 10 x 2 = 20

92+20 = 112

OH

I misread the question lol

sorry

[dark sparrow]

what have you tried?

[bleak rivet]

Well im not sure

cause there isnt a third measurement

[dark sparrow]

if you had a third measurement you'd have a 4-dimensional thing

you're given the base area and the height

V = Sh

[bleak rivet]

isnt is base x width x height?

[dark sparrow]

if you have a box, its volume is the product of length, width and height

yknow

V = lwh

[bleak rivet]

yea

[dark sparrow]

yeah

but that solid can be a cylinder or something

[bleak rivet]

but this one is a rectangle

which should be lwh

wait

so it would be A x Height

yeah

so it just be 24 * 8.25

[bleak rivet]

whats a right circular cylinder?

[dark sparrow]

a cylinder whose base is a circle such that the line connecting the bases' centers is perpendicular to the planes of the bases

it's what you probably think of when seeing the word cylinder

[upper karma]

A robot is programmed to do the following moves in a coordinate system: start at the origin facing in the positive y direction and move 1 unit. For each subsequent move, it is to turn 90 degrees and go forward 1 unit farther than it went it the previos move. What is the sum of the coordinates of the robot's posistion after it moved 14 units in one directions?

I'm confused

What do I do lmao

@crude kraken can you help?

[crude kraken]

Ahah

Do you know what summation is?

[upper karma]

No???

[crude kraken]

Okay

Let's say that the number of times it has moved is r, for repetitions.

The short answer is that this can be written as

=tex \sum_{n=1}^{r}n

[charred spear]

Support the bot on Patreon: https://www.patreon.com/dxsmiley

[upper karma]

Idk this stuff yet ;-;

[crude kraken]

It's actually really simple

[upper karma]

o

[crude kraken]

So, first let's look at the scary part

The thing under the sygma is your starting number, and r is the amount you go to.

Actually, images will explain this best

[upper karma]

ok

[crude kraken]

=tex \sum_{n=1}^{3}n = 3 + 2 + 1 = 6

[crude kraken]

oh wait I'm figuring out the coordinates of the bot this'll be harder than I thought 🤔

But, summation is how you can figure out the total distance travelled, so that's cool I guess

Basically, for the bottom until it reaches the top, you're going to add the function on the right that many times.

=tex \sum_{n=0}^{5}f(x) = f(0)+f(1)+f(2)+f(3)+f(4)+f(5)

[upper karma]

ok

[crude kraken]

Anyways, now we figure out the hard part, sum of coordinates.

Easiest method is probably to figure out the x and y formulas separately

[upper karma]

Wait does it mean just adding x and y

[crude kraken]

Not really

[upper karma]

o

[crude kraken]

So we start at 0 and move 1, then we turn left and move 1 (Our x,y is 1,1), then we turn left and move 2 (-1,1), then we turn left and move 3 (-1,-2)

(By the way, the numbers are the fibonacci sequence)

[upper karma]

oh lol

[crude kraken]

So how can we define f(x) and g(x) so that we get the y coordinate?

*y and x

(Just so you know, i'm winging it, which is why I'm explaining so much)

(So that anyone who knows this can be like "Bro u dun messed up")

[upper karma]

???

[crude kraken]

So, first is y coordinate. if R is the fibonacci sequence, then r1 = 1, r2 = 1, r3 = 2, r4 = 3, r5 = 5.

[upper karma]

oooooo

[crude kraken]

If we look at only the y coordinate, we find that we go up r2, down r4, up r6, down r8

If we look at the x coordinate, we go right r1, left r3, right r5, left r7

so on

[upper karma]

ok

[crude kraken]

So, how can we write a formula to get the x and y (and add them

I don't actually know notation to represent the fibonacci sequence so gimme a minute to figure out how to manually have a function that will calculate it

[upper karma]

k

r is the bot right?

Sorry my mind is really messed up today

[crude kraken]

Thing is I could write a program to do this in a few seconds but I can't write a formula to do it

[upper karma]

I can barely think

[crude kraken]

And, it's fine.

[upper karma]

Oh

[crude kraken]

Is it allowed if you write things like

[upper karma]

All I need is the work and the answer

[crude kraken]

=tex x = r_1 - r_3 + r_5 - r_7...

[upper karma]

No because the teacher hasn't taught that yet

:/

[crude kraken]

Gosh darnit

Okay, I don't see another way so forgive me

=tex F_n = F_{n-1} + F_{n-2}

[upper karma]

It's fine

[crude kraken]

You should write next to it that this is the fibonacci sequence

your geo teacher will be proud

;p

[upper karma]

xD

[crude kraken]

GLORY TO ARZTOCHKA

[upper karma]

shgidf;'hf

[crude kraken]

I'm trying to find a formula that you can use to define specifically Fn

[upper karma]

I'm gonna google it and see if google has an answer hehe

Ok

Thanks

[crude kraken]

Aha!

[upper karma]

You got it?

[crude kraken]

=tex F(n) = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)

[crude kraken]

Or if you would prefer to write this instead:

=tex F(n) = \frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}

[upper karma]

Simplier is better for me lol

[crude kraken]

Whichever you prefer

[upper karma]

second one

[crude kraken]

k

Would your teacher be mad if you didn't rationalize the denominator?

Watch this, someone is gonna come in like "Oh it's just this easy thing and you're done"

and here we are with op formulas

[upper karma]

Lol

[crude kraken]

So now we have the idea that for x<=number of jumps we make

[upper karma]

Maybe not

[crude kraken]

then

[upper karma]

Well I gtg now sorry

[crude kraken]

oh rip

good luck 😮

[upper karma]

Tomorrow

Thanks xD

[crude kraken]

The thing is I know exactly what I'm trying to say

but I can't represent it with notation

:I

[upper karma]

It's due in like 2 weeks idc lol

[crude kraken]

oh gg

[upper karma]

Just wanna do it early

Cya tomorrow

[rough juniper]

I need help specifically with 6-3 to 6-5

the rest I can do alone

[crude kraken]

Is <ABC a right angle?

And, what does it say about <XBC?

For 6-3 though, the definition of a bisector is something that cuts something into two equal parts.

So, it is liable to make that claim because BX bisects ABC and leaves you with two equal parts, ABX and CBX

For 4 and 5, this is only true if ABC is a right angle (which I assume is)

And again, definition of a bisector is something that cuts something into two equal parts.

So therefore, by the above rule that the two are equal, both must be the same, and 90/2 is 45.

If <ABC is not 90 degrees, then 4 and 5 are both false.

[rough juniper]

(sorry, just came back from halfway eating dinner, will be back soon, or I'll be sleeping. Apologizes)

[raw tree]

hey guys what methods do you use to solve b ? 🤔

[dark sparrow]

CB and AC are perpendicular

what do you know about perpendicular lines?

[raw tree]

ow is it that easy

f me

thanks lol

[spare sapphire]

Can someone help me real quick

im trying to write biconditional statements and are having issues.

so for the biconditional would i write it as a converse with if and only if or would i write it as a conditional with if and only if

?

pls

@ me if you get the answer

[crude kraken]

I can't find a formula for this, anyone have the formula for center of mass (xyz coords) of an elliptical cone with height d and width j

Oh, and the weight to count it as when comparing to rest of object would be helpful

(in the graph y=-x^2 + 9, d would be 5 and j would be 6, but a cone is 3d)

[umbral rivet]

👀

[crude kraken]

wolfram had me covered :p

[trim belfry]

is this the right place to ask about the mathematics of rotation?

[dark sparrow]

sure

[trim belfry]

okie dokie

(btw I am just stating my reasonings first and then I will ask the question)

so I think I figured out a way to determine that the specifying of any given orientation in n-dimensional space needs at most (n^2-n)/2 dimensions:

we have n n-vectors representing an n by n matrix transform which computes our rotation on a given n-vector

for each axis (represented by one of the n-vectors in the matrix) it must be orthogonal to all other axes for the overall transformation to be a valid rotation (note that this allows reflections), and it must have unit length (to exclude scaling)

the condition of unit length reduces the dimension of scaling for each axis by 1, so the overall dimension is reduced by n

meanwhile, the condition of orthogonality must by nature be separate from the condition of unit length as orthogonality is about *orientation and not scale* and meanwhile unit length is about *scale not orientation*

I will recursively figure out what dimensions are restricted by the condition of orthogonality, starting with n=2

so for n=3, the third axis is restricted in respect to the first and second axis, meaning that it is restricted in two axes (minus 2 dimensions)

then for n=4, the fourth axis can't rotate towards or away from the first, second or third axes (minus 3 dimensions)

n^2 - (n-1) - (n-2) ... - (2) - (1)  - (n)
    \_____________   _____________/  \_  _/
                  \_/                  \/
    condition of orthogonality   condition of no scaling

n + (n-1) + (n-2) + ... + 2 + 1 = (n^2+n)/2

n^2 - (n^2+n)/2 = (2*n^2 - n^2 - n)/2 = (n^2 - n)/2

Therefore, to specify any given orientation in n dimensions where scaled axes and non-orthogonal axes are rules out, you need (n^2 - n)/2 dimensions. Note that this includes reflections, however these reflections when removed as a possibility do not reduce the dimensionality (in order to specify one number out of all positive real numbers you need one dimension and to specify one number out of all real numbers you also need one dimension).

the block text is my reasoning

but according to this http://www.euclideanspace.com/maths/geometry/rotations/theory/nDimensions/ compound rotations in n dimensions are 2^n/2-1 dimensional? is this correct?

I don't understand how that's possible, surely an n by n matrix transform can specify every possible compound rotation because an n by n matrix has each column represent where the new 'axis' is? surely that covers all possible orientations? I don't understand how it couldn't cover all possible rotations

(sorry I am a noob in maths)

[ruby swallow]

@spare sapphire both but only one iff

[spare sapphire]

i figured it out but thx

[opal rock]

https://math.stackexchange.com/questions/129351/extremely-hard-geometric-problem

Wondering how he got the 3rd equation?

(3)

Help pleease

[fathom seal]

If you want, you could make a stackexchcange account and ask him directly.

[opal rock]

Dw, I got it

[versed falcon]

quick question does cos have to do with 2pi and sin with pi? in any sort

[opal rock]

@versed falcon if you take the unit circle definition of trigonometry and if you use radians then yeah

2pi is the distance along the x-axis sin or cos goes through in 1 period

In radians

Meaning, cos(0) is the same as cos(2pi) in radians

In fact, cos(n) is the same as cos(n + 2pi) and also the same as cos(n + 4pi) ...so on

[versed falcon]

I have another problem with that phrase above coming from the link: https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/solving-sinusoidal-models-precalc/a/trigonometric-equations-review . I dont understand how he comes up with the identity you need to use for the problem

[thin hound]

<@&268886789983436800>

[thorn talon]

What do you need help with?

[thin hound]

21

The image on the 2nd page is what we"re looking at

How would i det this up with law of sines

[thorn talon]

There is enough information to determine every angle

[thin hound]

Yea nvm im good

[quasi lion]

@thin hound Why did you called mods?

[upper karma]

He thinks mods have it as a duty to help

[thin hound]

I was just asking fot some assistance

[dark sparrow]

@help

is a thing

[thin hound]

Im innocent😱

Sorry

[upper karma]

hehe

[trail pollen]

Who can help me to resolve the problem? i don't all understand if you can help me in private ^^

[spare sapphire]

what's the angle addition postulate?

[prime junco]

I need help anyone?

So I'm doing a project about if then statements (easy) but what confuses me is include theorem.

<@&286206848099549185>

What is theorem?

[umbral rivet]

What's a theorem?

A theorem is like

Something that is true and can be supported with evidence in the form of proofs

[prime junco]

So the theorem will be related to the conditional or no?

Basically my if then statements is about not studying, failing a test, and falling the class.

@umbral rivet ?

[umbral rivet]

uwu

I don't know :'(

[prime junco]

Oh lol

How does a theorem looks like

[novel matrix]

hey guys just wondering if there's anyone who can help me with this geometry problem 😃

[crude kraken]

Given two points, I know slope, but how do I find y intercept?

[dark sparrow]

as in, of the line passing through those points?

[crude kraken]

Mhm

[dark sparrow]

y = kx + b

you know k, and you can replace x and y with the coordinates of either point

and then just solve for b

[crude kraken]

Okay

Thanks.

[tropic island]

I need help!

Plaese!

Is anyone ON!

[crude kraken]

Don't ask to ask, ask.

[raven vessel]

what is you're problem ?

[hot grail]

Quick question :
What is the normalized version of a vector called ?
Like if I have two vectors A and B and B is normalized A, then B is the ...???.... of A
Is there such a word ?

ah alright, thanks

[finite nexus]

i checked the method for it and it says that tanSTC = 400/300

how is that possible?

oops

there

guys, why is tanSTC = 400/300???

<@&268886789983436800>

[topaz valley]

<@&286206848099549185>

[finite nexus]

o yea, i see it now. thanks

[tropic island]

Graph the following lines and write the equation in slope-intercept form.
Through the point (2,−4) with y-intercept of −2.

Graph the following lines and write the equation in slope-intercept form.
Through the point (2,−4) with y-intercept of −2.

Can someone help me!

[dark sparrow]

@tropic island

[upper karma]

i have an assignment due in a few hours and ive finished most of the questions, having trouble with the remaining ones, if anyone could assist, this is one of them

https://i.imgur.com/QZlQt8y.png

figured id start with this one first since its requires more time spent to read it

[dark sparrow]

k so you know AC + CB = AB + 4

or well, AC + 10 = AB + 4

so AB = AC + 6

and you know that the angle at C is 120°

law of cosines

[upper karma]

how'd you get 4, ive missed the past few lectures to focus on my other classes and now im behind in math, thought id be able to catch up

[dark sparrow]

The trip would've been 4 km shorter...

[upper karma]

nvm, i thought you did something fancy

yeh i saw

with knowing that angle C is 120 and CB is 10, whats the next step from what you wrote there, AB = AC + 6

[dark sparrow]

you know the law of cosines, right?

[upper karma]

a^2 = b^2+c^2 - 2bc cos A?

[dark sparrow]

well, yes, but with different letters in this case

AB^2 = AC^2 + BC^2 - 2 AC * BC * cos(C)

replace AB with (AC+6)

replace BC with 10

solve for AC

[upper karma]

ugh i am lost

[dark sparrow]

at what point did you get lost?

[upper karma]

im watching the lecture recordings, probably doesnt help that its 4am

well ive never been great at trigonometry, but

we know cb is 10, and acb = ab + 4

we only know 1 length and im not sure how to get the 2nd, let alone the 3rd at this point

[dark sparrow]

one moment

[upper karma]

my knowledge of the cosine rules isnt great

*rule

[dark sparrow]

cos(120°) = -1/2, so just replace it with that

and then you have an equation in x which you can clean up and solve

do you need help doing that?

[upper karma]

yeh

[dark sparrow]

okay first off

do you understand where the equation came from?

[upper karma]

well, we had the cosine rule of AB^2 = AC^2 + CB^2

the first part of it where you just substituted the values in

the order you wrote the equation in confuses me a bit too

[dark sparrow]

AB^2 = AC^2 + CB^2 -2 * AC * CB * cos(C)

[upper karma]

yep

[dark sparrow]

and i mean

i just wrote it on two lines

=tex x^2 = 10^2 + (x-6)^2 - 2 \cdot 10 \cdot (x-6) \cdot \cos(120^\circ)

[upper karma]

actually pretty clear now

thanks alot for taking the time to explain it, not just giving me the answer 😃

is there a particular reason you wrote it in a different order, obv a + b = b + a, but i noticed you did AB^2 = CB^2 + AC^2, was it because of the quadratic?

or am i just reading too far into it

[dark sparrow]

you're reading too far into it

sometimes i write things in a certain order for clarity's sake, but not always

[upper karma]

i have two more questions left on my assignment, while i have you, if you could assist me here that'd be fantastic too please

https://i.imgur.com/lGW4nTt.png

[dark sparrow]

okay so for number 4

let's say that the population at t = 0 has size 1

it doesn't have to be like, 1 bacterium, just 1 arbitrary unit

then the size of your population at time t is

P(t) = 1.03^t

make sense?

[upper karma]

unfortunately these last 2 questions i am in a similiar boat to the earlier one, where i am behind and dont really understand much of it

[dark sparrow]

ok so

so let's say that at the start, we have 1 unit of bacteria

then after 1 hour, we'll have 1.03 units

is that clear?

@upper karma

[upper karma]

yep

sorry, have lecture on other monitor, got distracted

[dark sparrow]

aight

so every hour, the size of the population gets multiplied by 1.03

so P(t) = 1.03^t

is that clear?

[upper karma]

yep

[dark sparrow]

aight

so now

we started with 1 unit

and we want to find when the size of our population reaches 2 units

[upper karma]

ok

[dark sparrow]

so we want to solve the equation 1.03^t = 2

=tex t = \log_{1.03}(2)

[dark sparrow]

@upper karma

[upper karma]

hmm

i dont know, but im reading about it, so in this

t is the power we raise 1.03 to to get..2

going through my notes, log_a b = log b/log a

so i did ln(2)/ln(1.03) got 23.44

not sure if underscore is the propery way to write it here

*proper

so 1.03^23.44 = 2

well, 1.9994

[tulip spruce]

1.03^[ln2/ln1.03]= 2

[upper karma]

is that the correct way of writing it

[dark sparrow]

== log(2)/log(1.03)

[charred spear]

23.44977225

[tulip spruce]

well you'll see a pattern if you write it this way

[dark sparrow]

23.45

[tulip spruce]

but your precise answer will be ln(2)/ln(1.03)

[dark sparrow]

exact*

but yes

[upper karma]

i see

so in 23.45 days or ln(2)/ln(1.03), it will have doubled

[dark sparrow]

yes

uh

[tulip spruce]

iirc your t was in hours

[upper karma]

or hour

[dark sparrow]

hours, not days

yeah

[tulip spruce]

so 23.45 hours

[upper karma]

i see

[dark sparrow]

the other one is actually a lot simpler

every 1000 years the amount of your substance gets multiplied by 0.98

== 200 * 0.98^13

[charred spear]

153.80447785

[tulip spruce]

should be subtracted from 200

but ye

[dark sparrow]

no

[tulip spruce]

wait

imafuckingret

🤔

[upper karma]

could you go into a bit more detail about how you got the 200 * 0.98^13, i understand you just flipped the question

really want to understand what im doing

[dark sparrow]

okay so

every thousand years, the amount of your substance decreases by 2%

in other words, it gets multiplied by 0.98

[upper karma]

yep

[dark sparrow]

so

to find out what happens after 13000 yeasr

just do that 13 times

multiply the amount of your substance by 0.98, 13 times

[upper karma]

oh

[upper karma]

thanks alot for your help @dark sparrow and you too @tulip spruce

hi can anyone help me with a comp sci project?

i'm struggling with the geometry part

i'm trying to simulate a trajectory of a ball inside a billard pool, it has no acceleration or any effect whatsoever, basically we dont need to care about physics

what i'm trying to do is first :

an x and y coordinate are given as parameters, also an angle, the pool will always be a rectangle

how do i know which side it will hit first?

then whenever it's meeting with a wall, how to predict where it will go?

that 2nd part will probably be easier but i have no idea how to complete the first part pls help 😦 !!!

[umbral snow]

@upper karma Obviously, there are only two walls the ball can hit, depending on the angle. So, you need to consider between the two.

I would assign a coordinate system to the table. One wall is y = 0, the other perpendicular to it is x = 5 or something

Find the distance required to hit either wall, and go with the shorter one

[coarse lagoon]

just learned about the triangle inequality theorem

how come i never learned this in school

[dark sparrow]

the one that says detours can't be shortcuts?

[tropic island]

Hey!

[dark sparrow]

hi

[upper karma]

how do i find the distance? im shit at geometry 😦

[dark sparrow]

the distance between what two things?

@lam#9958

...

[umbral snow]

@upper karma You can make a line for the pool ball. If we start at (a, b) with an angle w, then:
y = arctan(w)(x - a) + b

Find any points of intersection. Use Pythagorean theorem for distances

[mossy oriole]

Hi

I have this problem on my HW and I was having some trouble with it

it states:

"Noel and Kirk are building a new roof. They wanted a roof with two sloping planes that meet along a curved arch. Is this possible?

"

I just drew an arc basically

Its like /curve\

[charred spear]

2

Failed to parse equation: Invalid token at position 11

log_1.03(2)\
           ^

Failed to parse equation: Invalid syntax at position 1

log_1.03(2)
 ^

[vale raven]

Statements 3 and 4 are similar to statements 5 and 6

[dark sparrow]

two col proofs of trivialities 😒

[midnight oak]

Hot

[upper karma]

is this the right place where I can ask questions about simple interest etc?

[dark sparrow]

that sounds more like an #prealg-and-algebra thing

[upper karma]

A question:

Two circles, both in 1st quadrant, are tangent to x-axis and y-axis. Both the circles pass through (9,11). Find the product of their radii

I got a method, but it's too complicated, and I am looking for an elegant solution

[surreal bolt]

Is the answer necessarily a constant answer or do they want an expression?

actually it cannot be a constant unless there is also tangency of the circles and even then maybe not.

nope 😃

(9, 11) can be the right most and top most point.

(of the two circles respectively)

Diameter of circle A (9, 11) and (0, 11) while diameter of circle B (9, 11) and (9, 0)

But that isn't the only way.

um yes ...

oh nvm

I still think there is more than one solution.

I just have to show the circle tangent to the y-axis has multiple possibilities.

Or show that the circle tangent to the x-axis has multiple possibilities.

BTW

Diameter of circle A (9, 11) and (0, 11) while diameter of circle B (9, 11) and (9, 0)

is a possible answer as both circles are in the first quadrant.

Circle A has center (4.5, 11) and perpendicular diameter (4.5, 6.5) to (4.5, 15.5)
Circle B has center (9, 5.5) and perpendicular diameter (3.5, 5.5) to (14.5, 5.5)

anyway ...

[upper karma]

The answer is constant

It's 202

[upper karma]

area of a trapezoid anyone?

[molten ingot]

just split it up into triangles and a rectangle

[crude kraken]

=tex (\pi r^2 - \pi(r-4)^2) mm^2 \ (\pi r^2 - \pi r^2 - 4r - 4r + 16) mm^2 \ (-8r + 16) mm^2

[crude kraken]

I would assume A at this point as it nearly matches the format of that answer, but I don't know why we don't take out the pi

@bleak rivet

Oh wait nvm

=tex (\pi r^2 - \pi(r-4)^2) mm^2 \ (\pi r^2 - \pi(r^2 - 4r - 4r + 16)) mm^2 \ (\pi r^2 - \pi r^2 - 4\pi r - 4 \pi r + 16\pi) mm^2 \ (-8 \pi r + 16\pi) mm^2

[crude kraken]

But that's not a choice, if anyone knows what I did wrong it'd be appreciated, would still assume A at this point.

[umbral snow]

You didn't distribute the negative after line 2

@crude kraken

The answer is B

@bleak rivet

[crude kraken]

^

[bleak rivet]

Thank you guys.

[upper karma]

A square region with perimeter 60 inches is made with square inch tiles. Bob removes one tile from the square and rearranges the remaining tiles without any overlap to make a rectangular region with minimum perimeter. How many inches are in the perimeter?

ANYONE?

60?

Sorry for the delay

@sour briar

U there still

Lol sorry

Ok

That was simple

Thanks

And I have one more question

It was thanks

Ok

I got 15 for this one

The sum

Like if it's (1,1)

The sum is 2

Ok

Yeah sorry for bothering you

gn

nvm got it

It's 15

:)

Ok last one

The Prevu Theater keeps a set of five counterweights to help stagehands move heavy props. The weights can balance exactly any load that is a multiple of 10 kg, from 10 kg, up to the total of the five weights. Give the masses of weights which meet these conditions and allow the stagehands to counterbalance the maximum possible load under the conditions below.

The weights must all go on the same side.

The weights may be used on one or both sides

I'm confused with this one

[lost jetty]

https://gyazo.com/cf3747f2ab6d8db0a5918115a9725057

why are all three of these angles θ?

well, we start with θ_1, so i guess why are the second and third angles equal to θ?

<@&286206848099549185>

[crude kraken]

Do you have the context of the problem? All I can say is that theta is an angle of the three triangles

[upper karma]

:/

[lost jetty]

well usually it's like

"object on an incline of angle θ"

and then the triangle with θ_2 is the mg vector with its parallel and perpendicular components

and θ_3 triangle is the normal force with its parallel and perpendicular components

[lost jetty]

https://gyazo.com/6d67a05e888a7e2b313943c2814778a4

here's an example

@crude kraken

[umbral snow]

@lost jetty I remember being confused about this in early physics. They're all provable using geometry facts like "the Z rule" or whatever, but there's a few different cases. You gain an intuition about which angles are equal through practice

If two lines intersect at some angle, then their normal lines will also intersect at that same angle, I'm pretty sure.

[upper karma]

are pyth triples unique?

[crude kraken]

https://www.youtube.com/watch?v=QJYmyhnaaek To an extent, video will demonstrate better than me but if you have a particular question then ask away.

[umbral rivet]

👀

[crude kraken]

If I'm trying to find the length of a line in 3d space between point (a,b,c) and point (x,y,z), do I just use pythagorean theorem to find the 2d length or 'shadow' that the line would make, and then just pythag that with c/z, or is there a more efficient method?

=tex L=\sqrt{\sqrt{(a-x)^2+(b-y)^2}^2+(c-z)^2} = \sqrt{(a-x)^2+(b-y)^2+(c-z)^2}

[charred spear]

The renderer took too long to respond.

Something went wrong while handling the message:

=tex L=\sqrt{\sqrt{(a-x)^2+(b-y)^2}^2+(c-z)^2} = \sqrt{(a-x)^2+(b-y)^2+(c-z)^2}

If this error keeps recurring, you should report it to DXsmiley on the official MathBot server: https://discord.gg/JbJbRZS

[crude kraken]

:i

=tex \sqrt{\sum_{a=1}^{n}\left(x_i-y_i)^2}

[charred spear]

The server sent back the following error:

Can be used only in preamble.
leading text: \begin{document}
Extra }, or forgotten \right.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Missing } inserted.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Extra }, or forgotten \right.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Missing } inserted.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Extra }, or forgotten \right.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Missing } inserted.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Extra }, or forgotten \right.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Missing } inserted.
leading text: ...m_{a=1}^{n}\left(x_i-y_i)^2}  \end{gather*}
Extra }, or forgotten \right.

[crude kraken]

sigh...

[simple reef]

lol

[crude kraken]

Figured that out :p

[wide portal]

=tex If the area of a circle is π cm2, how long is the radius

[wide portal]

great answer

[upper karma]

is that a question? xd

[visual field]

@wide portal Not sure what you're looking for with that, but tex is a command used to render the message sent using LaTeX, a markup language that is really great at typesetting math. It can't answer any questions, just make the words and numbers you put in it into fancy and easy to understand numbers and symbols. If you actually need help with the question, remember that the area of a circle is π*radius^2, so you can set the area you have equal to that formula and then solve for r

[wide portal]

ok thanks

[lilac kindle]

Also the answer is one @wide portal

[wide portal]

i know i got it right

[lilac kindle]

Oh nvm I didn't see someone already said

[night lotus]

Idk how to do most of these

[dark sparrow]

this picture is too low-res to see the text

[night lotus]

How about now

[dark sparrow]

okay that's better

[night lotus]

i forgot how to find x

if i can find x then i can do them

[dark sparrow]

do the words "pythagorean theorem" sound familiar to you?

[night lotus]

yes

[dark sparrow]

so

for example, for that first triangle

x^2 + 8^2 = 10^2

[night lotus]

so i use pythagrean theorem to find x?

[dark sparrow]

well

yes

[night lotus]

ok thanx

[feral fractal]

Ideas?

[rare talon]

Noticce triangle BEH and ADH is similiar

[dense holly]

namely because AD and BC are parallel

[upper karma]

you could also state that the potential line BD would have distance 8 too since it bisects both <ADC and <ABC

oh actually

you can set up a ratio

ill put it in english and u can translate to math

"3 is to 5 as to x is to 8+x"

this may be incorrect idk

[feral quiver]

https://cdn.discordapp.com/attachments/359052604149465088/369282108977315840/image.jpg Can anyone help me with this?

[upper karma]

i cant access this pic sorry

copy the img itself and paste here

[upper karma]

how te heck do you do this using trigonometry?

[rustic talon]

@upper karma tangent i guess

[upper karma]

what 2 sides are you given (in respect to the angle that you're given)

hypotenuse ? adjacent ? opposite ?

[rustic talon]

opp and adj

u have the angle though

[upper karma]

you should see that @upper karma you have opposite and adjacent, mind telling me what each of those are ?

[rustic talon]

sure

[upper karma]

@upper karma u here

well i'll explain here anyways

you are given side ? (we'll call it x ) and 2100 cm

2100 cm is your adjacent side to the angle of 29 degrees

that makes x your opposite side

the trig function that relates these two together is tangent (since tan = opposite/adjacent)

so you'd set it up this way

=tex tan (29^\circ) = \frac{x}{2100} \
tan (29^\circ) \cdot 2100 = x

==tand(29) * 2100

[charred spear]

1164.04900805

[upper karma]

1164.05 should be your answer

[dark sparrow]

=tex \tan(29^\circ)

[dark sparrow]

@upper karma

[upper karma]

oo

tand is right for the calc right ?

[upper karma]

1164cm?

yes

do you understand the process we just took

?

[dark sparrow]

yeah, tand(29) is tan(29°)

[upper karma]

Ye

So 11.6 Meters

This seems odd

how so

[dark sparrow]

the base was 21 meters

[upper karma]

This is to calculate my school building

There are 3 floors in the building

if you look at the triangle's angles, the angle that is related to the hyp is 90 degrees, so the hyp would have to be the largest side.
the angle that is related to the opposite side we just found is 29 degrees (the smallest), so the side that we found should be the smallest one.
the last angle is (180-119 = 61) is in the middle of the other two, so its measurement is neither the smallest nor the largest

if we were to have gotten a value of x larger than 2100 we should've been concerned since it would've been incorrect

[dark sparrow]

or to put it another way

29° < 45°, so tan(29°) < tan(45°) = 1

so if we'd gotten something above 2100 cm that'd mean that our angle was above 45°

[upper karma]

oly shit

thats good thinking lol

well i should head to do my own hw now

see ya

Used a clinometer tho

And a meter

Well actually it was a stick cause I did not have a meter

And then calculated the length of the stick

[feral fractal]

Looking at similarities, how would I equalize them if there are two variables?

[tulip spruce]

you can find the height of the smaller triangle

then you can find y

then you can find x

[upper karma]

pyth theorem

to find the height of the smaller one

[feral fractal]

lemme try

would then a similarity 45:28=73:y be correct?

[tulip spruce]

uh

it would be 45:(45+28) to something

=wolf sqrt(53^2-45^2)

[charred spear]

Query made by @tulip spruce
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=sqrt(53^2-45^2)
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[tulip spruce]

oh nvm

yeah you're fine

they're both 28 lol

[feral fractal]

Yeah, I got confused for a sec. How do I determine how similarities are set tho. Is there some rule/law or something for it? If you get what i'm asking?

Like what is put into the num1, and why isn't it in num2,3,4 part? (num1:num2=num3:num4)

[upper karma]

um idk the name

but its the ratio between sides that you just described

i.e the hyp of one / hyp of the larger one = leg of one / leg of the larger one

and so on

[feral fractal]

wait, so it's not leg1/leg2=big leg/big leg2(y)? I'm confused now

[umbral snow]

The ratio of the legs stays the same. That is,
53 / 45 = (53 + x) / (45 + 28)

You don't have enough info to find y based on similarity. So you need to use pythag

[cloud shell]

@Bager#9612 The rule is called Basic Proportionality Theorem

"If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio"

https://cdn.discordapp.com/attachments/326138757474680852/369560641314226176/unknown.png

so... 53 / x = 45 / 28

then fine y using pytho-mama

[mossy oriole]

hey

I have this problem on my homework

it states:

The measures of the angles in a triangle have a ratio 1:4:7. Find the angles

[thorn talon]

Well it totals to 12

So what's 1/12 of 180?

[mossy oriole]

=calc 1/12(180)

[charred spear]

Error: 12 is not a function

[mossy oriole]

=calc 1/12 * 180

[charred spear]

15

[mossy oriole]

Its 15

[thorn talon]

Yes

So that's 1

[mossy oriole]

done all in my head totally

[thorn talon]

What's 4 times 15?

[mossy oriole]

60

Oh

[thorn talon]

And 7 times?

[mossy oriole]

and 7*15 = uhh

=calc 7*15

[charred spear]

105

[mossy oriole]

105

[thorn talon]

So

[mossy oriole]

so those are the angles

[thorn talon]

15:60:105

[mossy oriole]

15. 60, 105

[thorn talon]

Yep

[mossy oriole]

thank you!

[glass canopy]

Have a kind of a problem. Let's say there's a car (at X: 30 and Y: 80). It's going to X: 130 and Y: 10. The total time of the trip is 500 milliseconds. At what coordinates will it be after 200ms?

I tried with calculating the angle then velocity but I got lost... Can anyone help me?

[umbral rivet]

👀

[upper karma]

seems like you are describing something of a position/time function (x being position and t being time?)

if so then i can help if not no lol

if not it'd clear up if you stated what your x and y represented lol

[glass canopy]

Yeah

x y are coordinates of the car

[upper karma]

huh

shit i was hoping for one of them to be time lol

yes really it confuses me without one of them being time sadly

sry

[dark sparrow]

@glass canopy how familiar are you with vectors?

[glass canopy]

not much, unfortunately

started to play with them yesterday or so

so I'd look for another solution

[dark sparrow]

mh

okay

[upper karma]

its probably related to it then if u were shown the concept yesterday

[dark sparrow]

you can do the x and y coordinates separately

[upper karma]

i'd go for that xd

[dark sparrow]

you start from (30, 80) and you're going to (130, 10)

now let's measure time in 500-millisecond increments

so the x coordinate of your car is

x(t) = 30 + 100t

make sense?

t here is measured in half-second units; e.g. t = 1 corresponds to 500 milliseconds, t = 0.2 to 100 milliseconds, etc

@glass canopy

[glass canopy]

yeah

I get it

[dark sparrow]

likewise y(t) = 80 - 70t

so now, you just need to find those coordinates when t = 0.4

since that's what 200 ms corresponds to, with the way we made it

[glass canopy]

oh, now gonna finish it on a paper

could have this idea before

🤔

[umbral rivet]

Ann 💕

[glass canopy]

thank you

[dark sparrow]

💕 @umbral rivet

[bitter moon]

How would you reflect a point over any line (in f(x) format)?

[umbral rivet]

Hm

Just draw it on the opposite side of the x axis :3

[gritty light]

Hey guys, could i get some help with my trig hw>

?

[upper karma]

yeh

DON'T ASK TO ASK. JUST ASK

[gritty light]

Could we pM?

PM?

[upper karma]

yeh sure

[gritty light]

anyone that can help me, with this

please DM me!

A mini-triathlon requires the competitors to run a 230m on a bearing of 072T from point A to point B. From point B they swim 380m across to point C on a bearing of N24E. They then cycle from point c DIRECTLY back to the starting point A.
2a) Draw a diagram showing the above information
2b) Find the total distance the competitiros had to cover correct to 1 decimal place.
2c) Find the bearing of point C from A to the nearest degree and minute

[umbral snow]

@gritty light 072T is the angle 72 degrees counterclockwise from north?

[feral fractal]

On 9.

What would be x?

[upper karma]

Is segment AD parallel to EB?

Becuase if so, you can use the Alternate Interior Angles theorem to figure that angle AD_missing point, which is what I think reads to be 105 degrees.

Other angle is 40 degrees, which makes x=35 degrees. Though this does take the assumption that those two lines are parallel, it does seem to make sense in the given diagram.

@Bager#9612

[feral fractal]

It is paralel.

Why is it 35 degrees tho?

[upper karma]

AIA, makes other one 105 too

Angles in triangle add to 180

180-(105+40)=35

[feral fractal]

And that one is 40 degrees because of the aia theorem? I got 105 because of 180-75-30

What would be that theorem simplified tho?

[upper karma]

What do you mean @feral fractal

[quartz igloo]

0

[gritty light]

@umbral snow yeah

It isn't counter-clockwise though, as it's stated that it rotates clockwise

so 072T Degrees

[umbral snow]

@gritty light The first step is to break all the vectors into their rectangular components.

A handy way to do this is to find the vector's angle away from east which I will call t, then rsin(t) is the north component, and rcos(t) is the east component

[nocturne bane]

im going to find an equation for the coloured area in the circle

with AB being r and AC being 2r

pls help

The drawing shows how many half circles. The diameter in the smallest half circle is AB = r. I the second lowest half circle the diameter is AC = 2r. The diameter in the big circle is AD = 6r. Find an equation for the areal of the coloured area.

[rustic talon]

@nocturne bane I think I can do it

Give me 5min I go to a book

Then I solve it

[nocturne bane]

where u atttt

[rustic talon]

I'm here ,just arrive at the my desk ,hang on

I'll try it

Right now

[rustic talon]

@nocturne bane im into it now,

@nocturne bane done

[nocturne bane]

we got the right answer nice

thank you so much dude 😄

[rustic talon]

hahah you welcome

more?

[nocturne bane]

nah was the hardest one

but anyways ty!

[rustic talon]

yo welcome,im free,i spend most free times ,doing math hahahahaha

@nocturne bane was it hard though?

[nocturne bane]

@rustic talon it was kinda difficult since i wasnt into it then, but i figured it out

[rustic talon]

THATS good cheesey cheese

[nocturne bane]

@rustic talon if u can solve that task u could most likely get the highest grade

[rustic talon]

Haha sure

Have another? Don't hesitate

[nocturne bane]

School ends 15.15 here in norway

[rustic talon]

Its 17:07

Here

[nocturne bane]

But i can give u another one tomorrow though if i fins one

[rustic talon]

OK bro/sis

Its bro though I guess

[nocturne bane]

Oh lol i didnt put on the ass lol

And yeah its bro

[rustic talon]

Hahah lol

Have a nice evening

[feral fractal]

Is there a way to construct a trapeze with only given all 4 lenghts of sides?

When given a=9, b=5, c=2 and d=6 i tried to calculate the height via s=(a+c)/2 and then h=2s, although I'm getting wierd numbers that don,t work out for me, e.g h=11.

oh, nvm im just stupid lmao

[rustic talon]

lol@feral fractal

did u solve it

[feral fractal]

Yeah, i did it

I made a triangle first and then a paralelogram

[dense crater]

I'm unsure of how to go ahead on this one

..Surely the third "pair" is orthogonal by symmetry?

[dark sparrow]

what exactly do you mean by "symmetry"?

you know PQ · RS = 0 and PR · QS = 0

[lusty sparrow]

Can a perp slope have a 0 or does it have to be simplified?

Like the negative reciprocal of 0/4 is -4/0, is that the final answer or does it have to be -4?

[tulip spruce]

perpindicular to what

if its to a line of the form y=k then the slope is undefined

[upper karma]

@tulip spruce I found the book. I got the pdf of it for free

[summer canopy]

Anyone here

I need some help

[topaz valley]

@summer canopy it's usually best to just ask

[summer canopy]

its just proofs

most people ive talked to dont remember it

[dark sparrow]

@lusty sparrow a. division by zero is not allowed
b. the perpendicular to a horizontal line is a vertical line, which cannot be put in the form y = kx + b

[summer canopy]

@dark sparrow please help

I started trying to do it

[dark sparrow]

two-col proofs? 😒

[summer canopy]

Yep

[dark sparrow]

they are obfuscatory and useless and i really don't want to do anything pertaining to them

[summer canopy]

Neither do I but its a quiz grade that I really need

[slim plover]

I already forgot how two column proofs work lol sorry

[summer canopy]

And im not cheating to be clear, she said you could look anything up you could or get any help you could

but im currently sitting at an 89% and this would push me over to an A if I did well on it

[royal dew]

one column is a statement and the other is the proof

[summer canopy]

I got that much haha

I just dont know what reasoning to put down is the main issue

[royal dew]

have you done khan academy's geometry? I am pretty sure those proofs are pretty sufficient if you were to use that style

[dark sparrow]

i mean honestly

the main idea is simple as hell

[summer canopy]

I understood the ones in class, but dont know how to do it with supplementary angles

just proving something is a paralellogram

[dark sparrow]

A + B = C + B (= 180°)

so A = C

[summer canopy]

Yeah i understand the concept, just dont understood how to write that out properly

[dark sparrow]

ugh

fine...

[royal dew]

One is a worded statement and the other is the math equation

[dark sparrow]

starting from what you wrote:
mB + mC = 180 / definition of supplementary angle
mA + mB = mB + mC / transitive property of equality
mA = mC / subtraction property of equality
A congruent to C / definition of congruence

bleh

[summer canopy]

Wait is what I wrote so far completely correct

[dark sparrow]

seven lines to prove something that can be proved in 2 if not 1

and yes

[summer canopy]

Yeah i think its completely unneccesary as well

[dark sparrow]

two column proofs are horrible

[summer canopy]

but unfortunately the curriculum thinks it matters

[summer canopy]

@dark sparrow mind if I ask one more question

Not 2 column proof

I know that the answer I have selected is incorrect

[rare talon]

Who will use two column proof in real math proof problem lmao

[dark sparrow]

@summer canopy the first one

the ratio between those triangles is 2

[summer canopy]

So sides must just be proportionate to be similar?

[rare talon]

Or, the angles of the pair of the triangles are the same (not in any choice)

[dark sparrow]

well, yes

[summer canopy]

Alright

Thank you for the help

[rustic talon]

The angles,

Will be equal

Are they said to be corresponding really?

[upper karma]

a pyramid has a quadrate base with width 2p and length 5p. Every 2 triangles face of the pyramid has its surface 2p(3p+2), and the other triangles left have the surface 5p(p-3)
A: Find a formula for the surface of the pyramid.
B: Factor the formula to find A excersise

I'd apperciate who helped me on this

[topaz valley]

what have you tried?

[leaden kindle]

I cant seem to figure out the proof why D2 = F2

[tropic stirrup]

I'm assuming DAF = DBF

Then you can see that A, B, F, d are on a circle

And therefore ADB = AFB

[tulip spruce]

<A = <B

and E1=E2

they're both triangles so you can only have a sum of 180 for the angles

[tropic stirrup]

Oh

I thought it too hard

[leaden kindle]

I have that problem too sometimes

Thinking too hard

[slim basin]

Should just take a protractor and measure the angles

[hidden geode]

has anyone worked with or spent any time with Geometry: A Comprehensive Introduction (Dan Pedoe)? looks interesting, wondering if it's suitable for someone with a solid calculus background & some matrix math

[visual jacinth]

sorry, didn't.

[civic sigil]

Hey guys, so I'm sorry if this question sounds super stupid, but what does it mean when a question asks you to calculate the slope of a rectangle? Or the midpoint of the rectangle? What does "slope of a rectangle" and "midpoint of a rectangle" means?

[thorn talon]

slope of the rectangle

or of the lines of the rectangle?

[civic sigil]

Slope

Read where it says "Assignment"

Or Whats in red

[topaz valley]

I think he just wants us to read it

[rustic talon]

Oh I read and not getting anything

Maybe I review

[eager pendant]

ah my apologies

didnt see the message above which detailed the problem

[vale raven]

👀

[upper karma]

am i not allowed to say that

[vale raven]

Well

[dark sparrow]

no shitposting

[vale raven]

there's a channel for you to say that

[upper karma]

freedom of speech !

[vale raven]

¯_(ツ)_/¯

Seriously though

#chill

[upper karma]

alright man

[vale raven]

Okie dokie

[upper karma]

big apology!

[dark sparrow]

https://xkcd.com/1357/

xkcd: Free Speech

I can't remember where I heard this, but someone once said that defending a position by citing free speech is sort of the ultimate concession; you're saying that the most compelling thing you can say for your position is that it's not literally illegal to express.

freedom of speech does not apply

we are not a government

[upper karma]

Yes, but we are a community.

[vale raven]

the whole "freedom of speech" thing doesn't work on internet forums or chat groups

[dark sparrow]

yeah, what kao said

[vale raven]

there are rules to prevent disruption

[upper karma]

understood

big mistake, big apology!

[dark sparrow]

k

[vale raven]

Just go to #chill xD

[upper karma]

A cube has the sides 1cm, calculate the space diagonals (direct translation) length

[dark sparrow]

what have you tried?

[upper karma]

Idk, my teacher isn't too good at explaining this

[dark sparrow]

do the words "pythagorean theorem" ring a bell to you?

[upper karma]

I know that I am supposed to take involvement of pythagorean theorem

I was typing when you said that, haha

[dark sparrow]

do you understand why the face diagonals are equal to me? 😛

[upper karma]

I just know that the top part is 1^2 + 1^2

[dark sparrow]

square root of that

leg^2 + leg^2 = hyp**^2** after all

[upper karma]

sqrt 2^2?

[dark sparrow]

no

sqrt(2)

[upper karma]

oh

[dark sparrow]

1^2 = 1, after all

[upper karma]

Yes

oh

so 3

[dark sparrow]

no

sqrt(3) is the space diagonal

[upper karma]

Yeah

since I only took 2 sides

[upper karma]

can someone explain me, I don't get trigonometry at all

[slender socket]

e?

[upper karma]

me*

my bad

[slender socket]

Ah

A good trick for trigonometry is the soh cah toa triangles

Have you heard of them?

[upper karma]

ye I did

[slender socket]

Sweet