help-10

[grave crypt]

if we looked at it from the right angle

op / hyp

wait

[azure anchor]

sin is the y axis of the unit circle

radius of unit circle is 1

[grave crypt]

i was never showed a unit circle sorry lol

but continue

[grave crypt]

so -1 here?

[azure anchor]

yes

1/-1 =

[grave crypt]

oh alr

so applying that for e and f of 4

[azure anchor]

positive angle only

for what positive angle is cot = -1

[grave crypt]

do you mean like cast rule

[azure anchor]

?

[grave crypt]

sry idont get this

[azure anchor]

cot is identity for what

[grave crypt]

1/tan

[azure anchor]

and tan is the identity for what

[grave crypt]

sin x cos?

[azure anchor]

tan = sin/cos

1/tan = cos/sin

so there's an angle where cos/sin = -1

when does cos/sin = +-1?

what special triangles have both adjacent and opposite sides the same?

[grave crypt]

45-45-90

[azure anchor]

tan = opposite/adjacent

[grave crypt]

yeah

so

1/1

[azure anchor]

but we need negatives

so what quadrants do you think that would happen

[grave crypt]

2 and 4

[azure anchor]

what angles would give us 45 45 90 triangles in those quadrants?

[grave crypt]

180-45

so

135

and

360-45

315

[azure anchor]

looks like you got it

[grave crypt]

okay last one

please

and thank you

sin = 1

[azure anchor]

sohcahtoa
sin = opposite/hypotenuse

what type of triangle you think we need for sin to be 1?

[grave crypt]

dont none of them give that

[azure anchor]

opposite/hypotenuse = 1 means opposite = hypotenuse. so we can still use a 45 45 90.

[grave crypt]

waitcant i just look at a singe function

sin at 90 is - 1

=1

sine

[azure anchor]

sin 90 = 1, right. like i said, sin is the y-axis of the unit circle/triangle.

[grave crypt]

hm okay

but i still wanna hear your way

[grave crypt]

how do i use the 45 45 90 in this case

[azure anchor]

imagine theta is our angle, ad we know that in this case sin theta = 1

crap hold on

its been so long since i had to use triangles

[grave crypt]

we are moving to radians tmrw

is it going to be easier then

[azure anchor]

i actually cant really think of a way to visualize sin theta = 1 with triangles and the opp/hyp definition

sorry

[grave crypt]

that’s okay

i’m going to skip this and move to radians

is it easier for you to explain with radians then?

[azure anchor]

if i can use the unit circle, then yes.

https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html maybe this might help? it at least shows how the triangles and the circle are related.

[grave crypt]

are you meant to memorize the unit circle

this is confusing

[tardy epoch]

Only your teacher knows the answer to that

[grave crypt]

could you try explaining the values

[tardy epoch]

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/v/unit-circle-definition-of-trig-functions-1

[obtuse pebble]

@grave crypt Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silent otter]

i dont rly understand what i have to prove here apart from commutativity of addition

[silent otter]

heres def 5.18

[silent otter]

i mean ii) and iii) here already imply that this algebra is a ring iff the addition is commutative, no?

[obtuse pebble]

@silent otter Has your question been resolved?

[obtuse pebble]

@silent otter Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[honest flicker]

I have this so far idk what to do

[final hornet]

so thats the volume of the sphere

it has a density of 0.9 kg/m^3

multiply that by the volume and you should get the mass

[honest flicker]

Uhhh 2748

Idk😭

I don't think that's the weight of the air

Oh wait kilograms

Idk how much that is

[final hornet]

idk

[honest flicker]

Is that the answer

@final hornet

[final hornet]

in theory yes

[final hornet]

cant you submit it here

[honest flicker]

I'm scared

Omg I got it right

Ty

.close

[obtuse pebble]

Channel closed

Closed by @honest flicker

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thin pendant]

how do I solve this? (sorry for rotated photo)

[narrow vault]

,rccw

[lethal breach]

first you'll need to find the angle AOB (considering O is the center of the circle)
well, technically thats the answer

in radians

[thin pendant]

ok thank you

.close

[obtuse pebble]

Channel closed

Closed by @thin pendant

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[compact gate]

.close

[obtuse pebble]

Channel closed

Closed by @compact gate

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dawn vessel]

How to do 22b) and 22c) ??

[dawn vessel]

Also 23b)iii)

[lethal breach]

how did you find angle BEC?

[dawn vessel]

BEC = CFB

[lethal breach]

the answer to this is a null set
{} or phi

[dawn vessel]

Thx

[lethal breach]

how do you know that?

[dawn vessel]

Do you know this?

[lethal breach]

they haven't provided any info

[dawn vessel]

Same way AEB = BDA we learnt this in 9th

[dawn vessel]

2 triangles sharing the same side have same corresponding angle in a circle

[dawn vessel]

this <@&286206848099549185>

[lethal breach]

noooo. its not true

who told you that?

there isnt any info given. you cant just assume

[dawn vessel]

brother its property of circle

[lethal breach]

nope

trust

[dawn vessel]

do you see the 2 triangles in EADB

[lethal breach]

i'll show you one example. one sec

[dawn vessel]

you can see 23 degrees

AEB = ADB

they share 2 same points

[dawn vessel]

what grade are you in?

[lethal breach]

Tell me how are C and D angles same?

@dawn vessel

they have the same side AB

or am i trippin

[dawn vessel]

Ok, first of all that is not a proper circle, second the lines arent straight so when you use protractor you would get different results, wait lemme google it for you

[lethal breach]

i am

yes

i am trippin

fuck

youre right

okay i get it

im sorry

[dawn vessel]

https://www.brightstorm.com/math/geometry/similarity/similar-triangles-in-circles-and-right-triangles/. see this

[dawn vessel]

nice

[lethal breach]

nono. i get it

the last time i did circles was 3yrs ago

its hard to remember

now that i do

yes

mb

im sorry

so yes

[dawn vessel]

good

[dawn vessel]

its fine, happens

[lethal breach]

so. angle EAB is 90
cuz in triangle EAB, you know E and B. so A is 180 - those 2 angles = 90

since EAB is 90, ECB is also 90

thats c) done

@dawn vessel

now for b)
you can use the quadrilateral ABCE. you know E, A and C. you can find B

does this help?

[dawn vessel]

one sec lemme see

you are correct for c)

so b) will be 138 right? @lethal breach

[lethal breach]

correct

[dawn vessel]

ight thx

[dawn vessel]

thx so much

.close

[obtuse pebble]

Channel closed

Closed by @dawn vessel

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[uneven horizon]

why is r=4?

[fierce lagoon]

Because it's a circular arc centered at the origin. It's pretty clear that r goes from 2 to 4

[uneven horizon]

actually im not very clear on polar coordinates

[fierce lagoon]

It's legit on the graph

Basically a circle.

x^2 + y^2 = 4

[uneven horizon]

so is r supposed to be a radius

[fierce lagoon]

Yeah

[uneven horizon]

oh

🗿

well that was very stupid

then why isnt the lower limit 2 if we apply the same logic

my prof did this

[fierce lagoon]

Oh they're tryna find the pink line

The equation of the pink line

x = 2

rcos(θ) = 2

Since x = rcos(θ)

[uneven horizon]

oh is it cause now we do polar coords so we cant take x

[fierce lagoon]

Mmhm

[uneven horizon]

but if thats the case

then why wouldnt the upper limit be rcos(θ) = 4

[fierce lagoon]

There's no vertical line at x = 4

[uneven horizon]

then 4secθ

[fierce lagoon]

The region is bounded by a circular arc (the red curve) on the right sode

It's not a vertical line

[uneven horizon]

so i take the arc instead?

[fierce lagoon]

Well yeah

[uneven horizon]

but isnt 4 the radius

its not the arc right

[fierce lagoon]

r = 4 on polar coordinates is the arc

Well

The full circle

But the arc in this context as a bounding curve

Here hold on

[uneven horizon]

okie

[uneven horizon]

i think i kinda get it now

[fierce lagoon]

Yeah I made the full bound. Desmos is being a bitch and won't let me graph the stupid horizontal line but there you can see the region

[uneven horizon]

yes

and what about finding the theta limits?

[fierce lagoon]

Hm

Gonna dig deep into my brain for this because I kinda willfully forgot how to do some of this lmao after a month of pure mental torture

θ obv starts at 0

[uneven horizon]

is this always the case

0 and something pi

[fierce lagoon]

Well, in most nice cases

Until you get into cones and shit, no

They can be ass

[uneven horizon]

like triple?

integrals

[fierce lagoon]

Yeah

Well actually

No you can just use pythag

The upper bound is arccos(2/4)

[uneven horizon]

how did you know the h is 4

[fierce lagoon]

The red curve is a circular arc of radius 4 to any distance from the origin to any point on the red curve has to be 4

[uneven horizon]

oh rightttt

okay i got it now

thank you so much!

.close

[obtuse pebble]

Channel closed

Closed by @uneven horizon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[undone plume]

HOW DO YOU FIND THETHA

[undone plume]

😢💔

[pulsar quarry]

when is value of sin equal to root 3/2

[undone plume]

huh

[shut lagoon]

They are just known identities. Look at this .

[undone plume]

how in the world am i supposed to memorize that for a quiz

[timid silo]

hello no need to fear macro is here

so with this it is simple trigonometry and equations

[warm shale]

Mr. Macro

[timid silo]

we can use this formula to solve for theta by rearranging the formula, but be for so we must substitute the given values

$Tan(\theta)=\frac{(\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}$

[warm shale]

Mr. Macro
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[timid silo]

oo

[undone plume]

how do you know you need to use tan

[timid silo]

this

dam i had the solution but it was to much for the bot

[undone plume]

LOL

no but why tan

why not sin or cos

[ember frost]

$\tan{\theta}=\frac{\frac{\sqrt3}2}{\frac12}$

[undone plume]

silly bot

[timid silo]

ok have you learned basic trig?

[undone plume]

yes

i know how to find them

but i dont know how to find the missing angle

using the sin cos tan thing

[timid silo]

ok wait i have to fix the bot

[undone plume]

okay

[timid silo]

$Tan(\theta)=\frac{(\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}}$

[warm shale]

Mr. Macro

$Tan(\theta)=\frac{(\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}}$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.52 ...rac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}}$
                                                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[timid silo]

GRRRRR

why

[lilac pebble]

u learn the unit circle before sohcahtoa??

[undone plume]

yes 😢

my teacher is a bit silly

he thinks we remember everything from last year

but i forgot

maybe im the silly one

[lilac pebble]

T=o/

t=o/a

[warm shale]

FungusDesu

[ember frost]

there

[timid silo]

yessssss

WOOOAA

[undone plume]

but why tan

[timid silo]

we get the value oppisit toit

and the value adjacent

[undone plume]

what about sin

[timid silo]

given tan = opposite \ adjacent

[undone plume]

that has oppisite

[timid silo]

you will still get the same

since all sides are given

[undone plume]

how does it know which angle

this is so hard 😢

why 😢

[timid silo]

$Tan(\theta)=\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}$

[warm shale]

Mr. Macro

[timid silo]

you could use any

ok now the next step

we want theta alone so we must remove tan

[undone plume]

OHHH

[timid silo]

when we divide tan we must inverse it

[undone plume]

i thought that was the only step

LOL

so divide the other side with tan?

how do you do sin inverse

[timid silo]

$\theta=Tan^{-1} (\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})})$

[warm shale]

Mr. Macro

[undone plume]

ooooo

[timid silo]

ok i shall now do it the sine way

well i since

all the sides are given

we could use cos, sin or tan

[undone plume]

wait let me try for cos

[timid silo]

ok

[undone plume]

yay it works

[timid silo]

😁

i got the same answer with tan and sin

[undone plume]

if one of the sides was negitive would i have to change the side to a positive

or does that make no sense

[timid silo]

ohhh like if the measurement of one side is forsay -5

[undone plume]

yes

[timid silo]

yes you would have to use -5

[undone plume]

okay

also what is this asking

[timid silo]

unit circle

[undone plume]

whats a terminal arm

[undone plume]

😟

[timid silo]

also use this

[undone plume]

do i just do the oppisite

[timid silo]

oh remember the saying Soh Cah Toa

[undone plume]

yes

[timid silo]

and the rule how opposite angles are equal

[undone plume]

what

how is it equal

equal to what

[timid silo]

no no not opp

opposite angles on a parallel

When any two straight lines intersect each other, there are different pairs of angles that are formed. The angles that are directly opposite to each other

[timid silo]

this

[undone plume]

ohhhh

oh yeah

oops

so 180 + 60

OH

I GET IT

[undone plume]

ITS 240

😍

[timid silo]

wait which are you solving

[undone plume]

d

wait a sec

240° is not a coordinate

💔

[timid silo]

💀 ive never learnt unit circles

[undone plume]

me too

i was sick on that lesson

wait wont it just be the same coordinates but oppisite signs

[timid silo]

ima quickly learn it and try to help

[undone plume]

this is the answer

[timid silo]

nice

[undone plume]

but how

[timid silo]

ohhhhh

think of a number plane

what is 4 - 10

[undone plume]

what is a number plane

[timid silo]

-4

[undone plume]

-6

what

[timid silo]

wait

i mean

what is the opp +4

-4

[undone plume]

yes

[timid silo]

its flipped like a mirror

[undone plume]

i see

so x is opposite sign

yay

🎉

[timid silo]

🎉

wait wait

[undone plume]

wait is that wrong

OH NO!

wait no

its right

yay 🎉

[timid silo]

it depends on which you are looking for

here is a diagram for all sides

[undone plume]

the question says pi - thetha

180 - 60

120

so the one with the smiley face

help

[timid silo]

Idk <@&286206848099549185>

[undone plume]

😢

[obtuse pebble]

@undone plume Has your question been resolved?

[timid silo]

Oh dear

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[long bluff]

hello

[long bluff]

i made a script in matlab for a predator-prey model

i just need something explained to me

so these are 3 graphs I made from 3 sets of inputs

the first being 4000 prey, 80 predators

second: 5000, 100

third: 200, 50

The second and the third graphs make sense, the first absolutely makes no sense to me

why did that happen?

[lone echo]

without the script itself we cannot really know much

[long bluff]

sure 1 min

[wooden cipher]

Seems like a fixed point kinda thing

[candid rock]

The second graph is just a weird way to plot a limit cycle

[wooden cipher]

Im not the best at differential equation, but it has to do with that kinda stuff

[candid rock]

It makes sense if you think about it

[lone echo]

the cycle seems pretty weird actually. The populations should tend to the stability point, not oscillate so wildly

[candid rock]

Do you have some kind of time delay

[long bluff]

well its a very basic model with simple differential equations

[lone echo]

yeah but they dont look like it would be correct at the most basic level

[long bluff]

this is the model

[lone echo]

Originally, let's assume you have a (virtually unlimited) amount of prey and 1 predator

predator has free reign so it multiplies

[long bluff]

i built the script using eulers method

[lone echo]

until the prey is depleted enough, where you have k prey, n predators, and the n predators start competing

then the growth starts being slower, until it stops growing

[long bluff]

@lone echo that seems to happen clearly in images 2 and 3 but i dont understand how 1 happened

[lone echo]

at that point, you have a stable k2/n2 relation between prey and predators

[long bluff]

just from arbitrary values of 4000 and 80

[lone echo]

no, because in your case the predators keep multiplying even after competing way too much for survival, so both get almost extinct. That doesnt happen.

[long bluff]

in image 3 yes?

[lone echo]

your oscillations are constant tho

oscillations should converge to 0 as the population gets closer to the stability point

Something along these lines

[long bluff]

true, this model is intentionally unrealistic

from what ive learned in class to make it realistic we add another factor to the diff eqs

to achieve that image

[lone echo]

fair enough

[long bluff]

but for right now in this assignment I have to explain the failures of this model i showed you

so far ive got what you said, but there is something else related to that first image

I dont understand why it occured that way

[lone echo]

okay, your problem is with the graph that represents a straight line, correct?

[long bluff]

yeah

which is 4000 prey, 80 pred

if I change it to 4100, 80 or 4000, 70 it looks more normal with oscillations

but for some reason at specifically 4000,80 it produces those straight lines

[lone echo]

can you represent, on the pop vs time graphs, the predator multiplied by 1000?

[long bluff]

like 4000, 8000?

[lone echo]

(or by whatever factor makes both pops be in the same order of magnitude)

[long bluff]

okay sure ill test that

[lone echo]

because on these kind of models, the important factor is not "how many" you have of each, the important factor is the proportion

okay, that was a misunderstandingç

i'm not asking you to introduce more predators

[long bluff]

i think that apparent rise in predators at the beginning is misleading btw, because the graph starts at t=1

[lone echo]

im asking you to represent, instead of the number of predators, 1000 times the predators that there are

[long bluff]

im confused

[lone echo]

let's say oyou have 4k prey and 40 predators

represent the 4k prey, and tell it to represent 100 times whatever predators it has

in other words, change the scale for the pred graph

or represent them in different graphs

so you only have prey vs time in one, and predator vs time in another one

[obtuse pebble]

@long bluff Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sly matrix]

can someone check this through? idk if this is the correct method

[obtuse pebble]

@sly matrix Has your question been resolved?

[sly matrix]

<@&286206848099549185>

[obtuse pebble]

@sly matrix Has your question been resolved?

[obtuse pebble]

@sly matrix Has your question been resolved?

Channel closed

Closed by @sly matrix

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[olive heath]

Could someone please explain this to me?

[drifting wraith]

i don't understand what it's saying

it shouldn't be true though

the function that relates angle to x-coordinate is cos, not sin

maybe it starts at 12 o'clock

still doesn't work

it works if it starts at 6 o'clock

[viral blade]

anticlockwise

Wait nvm

You're right

[polar fossil]

anticlockwise = positive

[viral blade]

I read anti-clockwise and thought ah yes the negative direction

[olive heath]

um

so it's false?

[drifting wraith]

there's not enough information, it's mostly false

[polar fossil]

think about whether it's correct at ø = 0

[olive heath]

hm

[past sand]

"with respect to an angle ø"... they don't specify where the angle is

[olive heath]

hm

i think that's the angle

[past sand]

I suppose they want you to assume it's the angle between the origin-ball vector and the x axis, but it's a bit of a stretch honestly

[olive heath]

from starting point x = 0

[viral blade]

Presumably the standard position then

[olive heath]

but it's still going anti-clockwise

[past sand]

x = 0 is the y axis

[olive heath]

yeah

but then it moves to the left

but that would mean negative x

but sine x goes positive first

is that good reasoning?

[past sand]

Not really

[drifting wraith]

it is

[past sand]

If the ball starts at (0,-3) (and ø starts at 0) then 3sin(ø) would describe its x coordinate

[olive heath]

yeah

[olive heath]

but inverted, no?

[past sand]

No

[drifting wraith]

0,−3, bottom of the circle, 6:00

[olive heath]

oh, -3

right

but i doubt that's the case

do you know if you go anti-clockwise from the x axis (on the cartesian plane) to measure angle?

[drifting wraith]

the chance that it's true because it's like "i didn't say it doesn't start there" exists, but it's small enough to ignore

[olive heath]

hm

[past sand]

The standard way we think of these things is with a unit circle described by (sin t, cos t), where t starts at 0 and ends at 2pi

At t=0, (sin t, cos t) = (1, 0) so it is on the x axis, to the right

t is the angle here, just like ø in your problem, but your problem never specified what ø represents exactly

Here, animate l in this: https://www.desmos.com/calculator/fuetzowbqe

[olive heath]

oh

i was just

doing that lol

[past sand]

Yeah but this goes clockwise

Do -cos n to go anticlockwise

[olive heath]

ah

[autumn adder]

Well people doing Desmos and not telling me

[obtuse pebble]

@olive heath Has your question been resolved?

Channel closed

Closed by @olive heath

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hushed seal]

okay so im doing heat transfer and i seem to be getting the calculations wrong, asking chat gpt to give me questions to practice and when i do the calculations on my own, they seem to be different from the answers i ask it to give.

[tardy epoch]

Don't use chat gpt for math

[hushed seal]

LMAO

lemme show you

[tardy epoch]

No thanks

[hushed seal]

can i at least show you the question and the answer i got?

really, i just need someone to tell me if im getting these right or not

[obtuse pebble]

@hushed seal Has your question been resolved?

[neon vector]

You'll have more luck finding such tasks by searching on the internet rather than asking ChatGPT anything other than factual questions.

[hushed seal]

ahh, okay
thank you

[obtuse pebble]

Channel closed

Closed by @hushed seal

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

@nimble gulch Has your question been resolved?

[nimble gulch]

<@&286206848099549185>

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[nimble gulch]

.close

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tawny kite]

for the case yRx I do not understand how transitivity plays its role here?

I understand all the rough work

[frosty spoke]

I didn't read the whole thing, but I suspect that suppose you have a singular element and a maximal element of the set of size n

then if the singular element is "bigger" than the maximal element of the size-n set

then you need to show that the singular element is bigger than all of the elements of the size-n set

and you use transitivity to show this

[tawny kite]

say case yRx, the solution says that suppose z in B, x<z, so y < z, This is to show how if the maximal x is smaller than z then y is also smallet than z because y<x. But z can't be in B if x is maximal

but then if you are already claiming x to be maximal why do you have to say z not in B{x} by maximality at y?

[frosty spoke]

okay yeah this solution kinda makes it a little confusing because it proves it by contradiction

it supposes that there is a z in B that is bigger than x

and then uses transitivity to show that z is bigger than y, which contradicts the maximality of y (because y is alleged to be maximal in B and z is alleged to be in B)

I guess you have to do it this way, because it's a partial order

[tawny kite]

Ahh okok I dont want to close this channel cause I still wanna continue this question, but I have 2 and a half hours of lecture after

can I just keep this channel

[frosty spoke]

I think it'll autoclose after a while

[obtuse pebble]

@tawny kite Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[split pecan]

I need immediate help please

[split pecan]

<@&286206848099549185>

[regal tundra]

Do you see some similarity between this and series of ln(1-x)

[obtuse pebble]

@split pecan Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crystal hull]

Hi. I am trying to understand the Möbius inversion formula properly. I've digested the proof of the version, where the arithmetic function f and its summator F are being inverted via the Möbius function mu. However, it is not clear why this proof should hold for the case, when f and F are not arithmetic functions, but just mappings from natural numbers N to any abelian group G.

[crystal hull]

How can we be multiplying here, if it's an additive abelian group for example?

<@&286206848099549185>

[kind hawk]

you are multiplying by an integer

which is basically repeated addition

which is fine in an abelian group

[crystal hull]

Ok, i was just wondering if that's ok, i need to just write down the justification properly. Thank you. More in #advanced-number-theory

.close

[obtuse pebble]

Channel closed

Closed by @crystal hull

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

Solve for BD

$AC^2= \frac{BD^2+BD^2 \times Tan(\Theta)^2 \times Tan(\alpha)^2 }{ Tan(\theta)^2 \times Tan(\alpha)^2}$

[warm shale]

Mr. Macro

[timid silo]

wait

.close

[obtuse pebble]

Channel closed

Closed by @glossy sky

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[low ravine]

help please

[obtuse pebble]

Please don't occupy multiple help channels.

[low ravine]

hello?

[pulsar quarry]

!1c

[obtuse pebble]

Please stick to your channel.

[timid silo]

hm

find the current

then subtract that thing from both the downstream and upstream time

[obtuse pebble]

@low ravine Has your question been resolved?

[obtuse pebble]

@low ravine Has your question been resolved?

[obtuse pebble]

@low ravine Has your question been resolved?

[teal turret]

.close

[obtuse pebble]

Channel closed

Closed by @teal turret

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[alpine oar]

Hello, I am stuck trying to alter the form of the equation to the general form of linear differential equations of the first order and the first degree

[obtuse pebble]

@alpine oar Has your question been resolved?

[alpine oar]

.close

[obtuse pebble]

Channel closed

Closed by @alpine oar

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dark vector]

have you drawn the figure yet?

[obtuse pebble]

@mossy thorn Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silk maple]

is there a quicker way to solve this than partial integration?

[forest sinew]

power reduction?

a few times

[silk maple]

does that work?

i just simplified it to this

[forest sinew]

hm this may lead to tabular

but might be fast

[silk maple]

this integral is just 0

but it took me forever to figure it out

[forest sinew]

yea... write...

[silk maple]

ohh you can probably say that the integral from 0 to 2pi of any sin or cos to some uneven power is always 0

[forest sinew]

$\cos x \cos 2x = \cos(2x-x) - \sin2x \sin x$

[warm shale]

jan Niku

[silk maple]

ty anyway

.close

[obtuse pebble]

Channel closed

Closed by @silk maple

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[smoky dove]

what is the asnwer to this question

[obtuse pebble]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[smoky dove]

I got D but my friend says A

@versed wolf

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[oak marlin]

How did you get D

[smoky dove]

ok so if z is cis-2pi/3 then the conjugate must be cis2pi/3 so if we square z by 4 then we add cis-2pi/3 4 times to get 2pi/3

[runic void]

How do you represent z ?

[smoky dove]

rcistheta

[runic void]

I mean z= e^-2pi/3 right?

[oak marlin]

cis(-4*2pi/3) = cis(-8pi/3) = cis(-8pi/3 + 6pi/3) = cis(-2pi/3)

[smoky dove]

the modulus in this question is a bit iffy since it should not remain constant

[runic void]

$z* z’= |z|^2$

[warm shale]

.doc

[runic void]

since you are in unit circle

$z* z’= 1$

[warm shale]

.doc

[runic void]

or z’= 1/z

[smoky dove]

is it B then?

[oak marlin]

Yes

[smoky dove]

ah I see my mistake

I added the arguments inscorrectly

thank you guys 👍

[runic void]

This just means, the conjugate of your complex number is the multiplicative inverse

[oak marlin]

👍

[runic void]

This particularly forms a group under complex multiplication

[smoky dove]

.close

[obtuse pebble]

Channel closed

Closed by @smoky dove

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[low matrix]

i'm kinda confused by the leaking and pumping at the same time, help would be appreciated!

it's related rates btw

[obtuse pebble]

@low matrix Has your question been resolved?

[mighty lava]

Are you online rn?

[low matrix]

yes!

[obtuse pebble]

@low matrix Has your question been resolved?

[static sandal]

if ur still here, just use u for the combined pumpage/leakage ignore that part for now

if this hjelps

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[forest sinew]

you know i got this one sorted, btw

[forest sinew]

just rewrite $\cos ^3 x + 1 = \cos x (1-\sin ^2 x) + 1$

[warm shale]

jan Niku

[forest sinew]

this leads to a very natural and easy u sub integral as the "hardest part"

$\int \cos x - \int \sin ^2 x \cos x + \int$

[warm shale]

jan Niku

[forest sinew]

clearly only $\int _0 ^{2\pi} \dd x$ is the only one with value over 0 to 2 pi

[warm shale]

jan Niku

[forest sinew]

just figured id say

.close

[obtuse pebble]

Channel closed

Closed by @forest sinew

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[high vortex]

Hello. I am having a world of problems trying to understand how to work out part b in this problem. I'm really not seeing the big picture. Any poiinters??

Laplace's equation in two dimensions is the PDE for $u(x, y)$
$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 .
$$

Such $u$ are "steady-state" (i.e., time-independent) solutions to the 2-dimensional heat equation. Consider this on $\left[0, L_1\right] \times\left[0, L_2\right]$, with vanishing Neumann boundary conditions on the left, right, and bottom sides:
$$
\frac{\partial u}{\partial x}(0, y)=0=\frac{\partial u}{\partial x}\left(L_1, y\right), \frac{\partial u}{\partial y}(x, 0)=0 .
$$

We'll analyze that boundary value problem (with vanishing Neumann conditions along 3 of the 4 sides).
(a) Show that the nonzero separated BVP solutions $X(x) Y(y)$ are the functions
$$
c_n \cos \left(\left(n \pi / L_1\right) x\right)\left(e^{\left(n \pi / L_1\right) y}+e^{-\left(n \pi / L_1\right) y}\right)
$$
for $n \geq 0$ with scalar $c_n \neq 0$ (don't overlook $n=0$, related to constant solutions to an ODE).
(b) Consider solutions $u(x, y)$ to the BVP subject to the further Dirichlet-style boundary condition $u\left(x, L_2\right)=f(x)$ along the top side, for a given function $f:\left[0, L_1\right] \rightarrow \mathbf{R}$ vanishing at the endpoints. Writing $u(x, y)$ as an infinite series
$$
\sum_{n=0}^{\infty} c_n \cos \left(\left(n \pi / L_1\right) x\right)\left(e^{\left(n \pi / L_1\right) y}+e^{-\left(n \pi / L_1\right) y}\right)
$$
in separated solutions, express the $c_n$ 's in terms of Fourier-type integrals involving $f(x)$.
Hint for (b): the $2 L_1$-periodic series $\sum_{k=0}^{\infty} a_k \cos \left(\left(k \pi / L_1\right) x\right)$ computing the even extension $f_{\text {even }}:\left[-L_1, L_1\right] \rightarrow \mathbf{R}$ of $f(x)$ has coefficients given in Proposition 21.1.8(i) with $L=2 L_1$.

[warm shale]

True

[obtuse pebble]

@high vortex Has your question been resolved?

[silk maple]

Thanks :)

[obtuse pebble]

@high vortex Has your question been resolved?

[tardy epoch]

Probably should show the proposition mentioned

[obtuse pebble]

@high vortex Has your question been resolved?

[high vortex]

sorry, one sec

Proposition 21.1.8. Let $f(x)$ be an $L$-periodic function, and define $\omega=2 \pi / L$.
(i) For even $f(x)$, the associated Fourier series is $a_0+\sum_{k=1}^{\infty} a_k \cos (k w x)$ ("cosine series") with $a_0=\frac{1}{L / 2} \int_0^{L / 2} f(x) d x, \quad a_k=\frac{2}{L / 2} \int_0^{L / 2} f(x) \cos (k \omega x) d x$ for $k \geq 1$.
(ii) For odd $f(x)$, the associated Fourier series is $\sum_{k=1}^{\infty} b_k \sin (k \omega x)$ ("sine series") with
$$
b_k=\frac{2}{L / 2} \int_0^{L / 2} f(x) \sin (k \omega x) d x \text { for } k \geq 1 \text {. }
$$

[warm shale]

True

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pale drift]

can a linearly dependent set of vectors be orthogonal?

[sage geode]

Perhaps if there are two vectors and one of them is the zero vector

Can't see any other possibility

Or if it's just {0}

[obtuse pebble]

@pale drift Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[slate saddle]

How would I go about solving this?

[slate saddle]

Don't tell me the answer but what identities would I use?

okay

[limber quartz]

angle addition formula for cosine

[slate saddle]

hmm

Thank you I will try it out.

[limber quartz]

^

[limber quartz]

are you curious how I knew that

(eventually)

[slate saddle]

yeah

I am

[limber quartz]

because the angle addition identies follow a pattern

for cosine, it's cos(A +/- B) = cos(A)cos(B) -/+ sin(A)sin(B)

for sine, it's sin(A +/- B) = sin(A)cos(B) +/- cos(B)sin(A)

[slate saddle]

hmm

What about the product to sum identities?

Could that be used in this scenario?

[limber quartz]

I don't know, but the angle addition/subtraction identities are more well-known

you should use the angle addition formula for cosine.

get 1 by itself on RHS, factor out the 2 on the LHS.

[slate saddle]

okay I am trying it out

So

2cos(-3x)cos(2x) = 1 + 2sin(-2x)sin(2x) can be translated as cos(-3x +/- 2x) = cos(-3x)cos(2x) +/- sin(-3x)sin(2x)?

@limber quartz

thats for the LHS

[limber quartz]

no

you messed it up or typed it wrong not quite: you kept my notation for the signs when you should just stick to what you were given

rearrange what you were given.

get the 1 by itself.

on the left hand side, factor out the 2.

[slate saddle]

oh

[limber quartz]

then divide both sides by 2

the left hand side will be in the same form as the angle addition formula for cosine.

translate that

[limber quartz]

oh I see

you are kind of right

but

pick which sign you have

you have subtraction on the right, which is associated with addition on the left (that's why it's the angle ADDITION formula for cosine)

[slate saddle]

okay

This is what i have

[limber quartz]

don't worry about my formulas

[slate saddle]

:

[limber quartz]

yet

page 2 under "Sum and Difference Formulas"
https://tutorial.math.lamar.edu/pdf/trig_cheat_sheet.pdf

[slate saddle]

2cos(-3x)cos(2x) = 1 + 2sin(-2x)sin(2x) can be translated as cos(-3x)cos(2x) - sin(-3x)sin(2x) = 1/2?

[limber quartz]

yes

good job

[slate saddle]

yes finally

then

Thank god

I finally see it

Imma solve it then send the answer (if you have time to review it)

[limber quartz]

did you translate it yet?

[slate saddle]

not yet

am doing it now

[limber quartz]

use the formula sheet

[slate saddle]

yessir

[limber quartz]

$$\cos{(A \pm B)} = \cos{(A)}\cos{(B)} \mp \sin{(A)}\sin{(B)}$$

[warm shale]

Disorganized

[slate saddle]

cos(-3x + 2x) = cos(-3x)cos(2x) - sin(-3x)sin(2x)

Is what I have so far.

[limber quartz]

good

[slate saddle]

What happens to the 1/2?

[limber quartz]

tell me.

[slate saddle]

does that appear afterward

?

[limber quartz]

it's happening right now

!

😛

transitive property

"if x = y and y = z, then x = z"

[slate saddle]

got it

so

cos(-3x + 2x) = 1/2

[limber quartz]

yes

keep going

[slate saddle]

cos(-3x + 2x) has the result of:

cos(-x)?

[limber quartz]

yeah

what else do we know about the cosine function that will simplify that even more

"cosine is an (blank) function"

[slate saddle]

umm

let me think

[limber quartz]

what does this mean:
f(-x) = f(x)

[slate saddle]

odd

[limber quartz]

no

[slate saddle]

an odd function

whoops

oh

[limber quartz]

odd is
f(-x) = -f(x)

[slate saddle]

even then

since

[limber quartz]

yeah

we can see it graphically as y-symmetry

[slate saddle]

the "-" has to equal "+" for it to be even

[limber quartz]

which cosine has

[slate saddle]

hmm

[limber quartz]

phrasing is a little imprecise

[slate saddle]

rather

[limber quartz]

cosine is an even function because you can fold the graph over the y-axis onto itself.

[slate saddle]

yep

okay

[limber quartz]

algebraically (or graphically, if you plot it), for any input x, cos(-x) = cos(x)

[slate saddle]

so the final answer (without finding the solutions in the interval) is cos(x)?

[limber quartz]

no

that's not the final answer

this is the next logical step

use the fact that cosine is an even function with transitive property again

[slate saddle]

x = (2k + 1) pi/2?

[limber quartz]

"cos(-x) = 1/2
and cos(-x) = cos(x)
therefore,
cos(x) = 1/2"

[slate saddle]

oh

[limber quartz]

wouldn't be that

because when k=0, x = pi/2, which isn't going to work

use what I just wrote

(typed)

the last line

to draw a right triangle with angle x

[slate saddle]

doing it now

[limber quartz]

or, you know

you don't need to draw the picture if you have the unit circle memorized

what angle theta (in our case, x) has a cosine value of 1/2?

[slate saddle]

pi/3

and

[limber quartz]

cosine

not sine

pi/3 --> 60 degrees

oh!

you are right actually

[slate saddle]

so pi/3 and 5pi/3

[limber quartz]

yes.

in the interval [0,2pi)

very good job

[slate saddle]

thank you

would

or rather

is x already in the interval [0,2pi) ?

Or would I need to solve that as well?

wait a min

[limber quartz]

it is in the interval.

you made it so

[slate saddle]

aha

I see

[limber quartz]

you found the most accute angle that had that trig ratio

or as we would say, the reference angle/triangle

[slate saddle]

So there are only two solutions within the interval [0,2pi)?

As those are the only ones we found.

[limber quartz]

I wish I could draw you the picture

you would draw the reference triangle in Q1

and that angle in Q1 would have its terminal side as the hypotenuse

I hope that is obvious

but then, since cosine is also positive in Q2 (positive is desired because since we had cos(x) = +1/2)

[slate saddle]

ah

[limber quartz]

we lay the same reference triangle in Q2 on the -x axis

and the new angle terminates at that hypotenuse

but notice that would just be 180 degrees - x

or pi - x

that is,

pi/3, pi - pi/3 = 2pi/3

oh

you got the other angle wrong

I think

oh cheese and crackers

I got it wrong

I'm thinking of sine again!

[slate saddle]

lol

[limber quartz]

don't orient the second triangle along the -x axis

flip it OVER the +x axis

then the angle you desire is 2pi - pi/3

6pi/3 - pi/3 = 5pi/3

so you are correct

yes

sorry

[slate saddle]

Ah

So

The 1st triangle is in Q1 and I would flip it into Q2?

[limber quartz]

nah

[slate saddle]

Q2 being the +x axis?

[limber quartz]

flip it into Q4

because cosine has positve values in Q1 and Q4

and that is because the horizontal components of the right triangles are positive in Q1 and Q4

[slate saddle]

yep

WHAT AM I THINKING Q2

I meant Q4

yes

got it

Thank you very much for your assitance and help!

This is a very through conversation.

[limber quartz]

hang on

[slate saddle]

oh dear there's more?

[limber quartz]

Just a picture

[slate saddle]

ah

[limber quartz]

in the unit circle

the hypotenues is always "1 unit"

any triangle can have all its sides scaled up or down by the same amount to make this so.

[slate saddle]

got it

[limber quartz]

it's only the x and y-components of the triangle that have a sign (positive or negative). The hypotenuse is always positive.

so the rules for flipping the reference triangles can be visualized picorally by just noticing where the sine and cosine values are the same result for congruent triangles in different quadrants

the signs of the sides are different in different quadrants, obviously

[limber quartz]

you're welcome

[slate saddle]

hmmm

[limber quartz]

I added the quadrant labels and the signs of cos, sin, tan (in that order) for each quadrant

[slate saddle]

Thank you very much!

[limber quartz]

no problem

[slate saddle]

I wish you a great rest of your night or day (wherever you are in the world).

Cheers!

.close

[obtuse pebble]

Channel closed

Closed by @slate saddle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

anyone?

[timid silo]

Calculate the length of segment AD, if |OA| = 50 cm, |OB| = 28 cm, and [OCI] = 7 cm.

into english ^^

[alpine bison]

just use the intercept theorem

[obtuse pebble]

@timid silo Has your question been resolved?

[robust raven]

to jest dosyć łatwe, pomyśl o tw. Talesa 🙂

najpierw obliczasz OD, a potem sumujesz

[timid silo]

a juz kumam

A to

[alpine bison]

wow you know a lot of things

[timid silo]

“• W troikacie ABC o bokach drugosci: | AB| = 18 cm, | BC| = 24 cm, | AC| = 20 cm poprowadzono dwusieczna kata ABC, która preciela bok AC w punkcie D. Oblicz dugosci odcinkow AD i CD.”

Strasznie nie kumam o co tu chodzi

[robust raven]

oznacz sobie szukane boki jako x i y

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[robust raven]

.reopen

wteyd masz x + y = 20

następnie:

[robust raven]

smiling

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

??

[timid silo]

Tak

[robust raven]

jestem , potem piszesz x / y = x / (20-x) = 18/24 = 3/4

obliczasz 4x = 3(20-x) <=> 7x = 60 itd

x = 60/7 , y = 80/7,

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @gusty latch

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ocean arrow]

hi

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crude coral]

why is my answer in negatives? it should be 10/9

the question is $$\int^1_0 \frac{3x}{\sqrt{4-3x}}$$

[warm shale]

JustToPro