help-13

[unique valley]

great, now find the maximum of this function

[left bison]

is this right?

[unique valley]

it is

[left bison]

alrr

[unique valley]

by comparing f'(x)=0 first

you know how i hope

and write the sol. for x in terms of a

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[left bison]

.reopen

[cedar kiln]

✅

[left bison]

srry im

solving

[cedar kiln]

@left bison Has your question been resolved?

[left bison]

srry its taking me a while

getting distracted

wait

this is what I did

I based it on an example shown to me

but I honestly dont get the relation with all that solving

I mean I get some

but

not entirely

[cedar kiln]

@left bison Has your question been resolved?

Channel closed

Closed by @left bison

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.

[hardy swift]

So i understand math but i dont see a clear connection with the negative ones like i do with the ones that are positive

[hardy swift]

And i do understand how for example * and / are each others opposite though

[surreal hemlock]

uh can you phrase your question more clearly?

what is it you need help understanding?

[hardy swift]

how are -, / and sqrt connected. It doesnt seem to be as simple as how 2+2+2 = 23, and 22*2=2 to the power of 3

[silent steeple]

So you understand how do +, * and powers work, but you need an explanation on how do -, / and roots work? You need the explanation of how do opposite operations work?

[hardy swift]

I cant explain it.

.close

[cedar kiln]

Channel closed

Closed by @hardy swift

Use .reopen if this was a mistake.

[cedar kiln]

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.

[crimson sedge]

I need help with calc 1 work

[crimson sedge]

this is the problem ik the answer is supposed to be 1.5 but i having trouble trying to find that answer with the limit definition

and the f(x)=3/x with the lim = 0

if more is needed i can provide

<@&286206848099549185>

[patent badger]

The limit definition is calculating the derivative, but the table is just the function

[crimson sedge]

so would the answe be -3/4

shouldn't the derivative be close to the answer on the table which is 1.5

[patent badger]

No the table is just finding 3/x

You can plug in 2 there just fine

Ur mixing up derivatives and discontinuities i think

But yeah -3/4 sounds right for the derivative i think, dont quote me on that though

[crimson sedge]

oh okay than thank you

.close

[cedar kiln]

Channel closed

Closed by @quick tundra

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 dawn]

note that z* is the complex conjugate of z

and that z and w are complex numbers

I tried substituting but it I don’t work

[unique valley]

whats the question?

[high dawn]

solve for w and z

[slow jewel]

Could u show your work?

[unique valley]

did you try w = a+bi z=c+di?

[high dawn]

hmm, no

[inland ocean]

Is that 5 + 5i

[unique valley]

you just get 4 linear equation from that

[slow jewel]

Yes

[inland ocean]

And 2i

[unique valley]

for a,b,c,d

[slow jewel]

From 1st one you can get w*/z*= -2i and then sub in the second one

[high dawn]

I get it now

[unique valley]

i think both are fine

[high dawn]

.close

[cedar kiln]

Channel closed

Closed by @high dawn

Use .reopen if this was a mistake.

[cedar kiln]

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.

[pseudo merlin]

hi

[pseudo merlin]

what do these

symbols mean

[smoky idol]

All of them or only some of them?

[pseudo merlin]

all of them

i dont get it

whats the question asking

[smoky idol]

Well, those letters read a statement, and they‘re asking you to find a pair (x,y) such that the statement becomes false

[pseudo merlin]

does A mean

for all?

[smoky idol]

Yes

[pseudo merlin]

and E mean

there exists

[smoky idol]

And backwards E means "there exists"

[smoky idol]

Exactly

[livid hound]

upside down A
backwards E

[pseudo merlin]

how

is it

this chapter

[pseudo merlin]

is it saying for all x belongs to an integer and there exists y belongs to an integer

[smoky idol]

Such that 2xy = 24

[pseudo merlin]

can i just be like 2(3)(2) does not equal 24

[smoky idol]

It‘s fairly easy to disprove when you think about it. First you can divide both sides by 2 for convenience, but then, it‘s saying: for all integers X, you can find an integer y such that xy = 12. that‘s obviously not true, otherwise 12 would have an infinite amount of factors

[smoky idol]

Yeah pretty much

[pseudo merlin]

wtf

its that easy

[livid hound]

no

[pseudo merlin]

but what about the for all

and there exists

stuff

2(12)(1)?

wut

[livid hound]

choosing 3 for x
2 * 3 y = 24
there does indeed exist a integer value for y that results in 24

[smoky idol]

Oh I just read what they sent didn‘t compute. Oop

[pseudo merlin]

wait

do i have to prove

[livid hound]

for your counterexample, you'd want to find a value of x where you can't get an integer value for y that works

[pseudo merlin]

if its not an integer or something

for all

maybe lets say

if x=24

y has to be 1/2

wait im confused again

[livid hound]

yeh that works

[pseudo merlin]

how does it work cause then

isnt it

2(24)(1/2)

[livid hound]

because 1/2 isn't an integer

[pseudo merlin]

is there any

other way

i can solve it

to help me understand it better

[livid hound]

many many counterexamples

[smoky idol]

Because you just disproved the fact that for all integers X, there exists an integer Y such that xy = 12

[pseudo merlin]

so like

maybe

[pseudo merlin]

why such that xy=12

[livid hound]

2xy = 24 simplifies to xy = 12

[pseudo merlin]

oh

so i just

do whatveer = 12

and not an integer

for y

[livid hound]

for your counterexample, you'd want to find a value of x where you can't get an integer value for y that works

[pseudo merlin]

oh

i hate math

i rmb in this server i was asking how to prove similar triangles

and now im doing this

i hate life

.close

[cedar kiln]

Channel closed

Closed by @pseudo merlin

Use .reopen if this was a mistake.

[cedar kiln]

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.

[ivory finch]

Bob's tiktoks viralize with 10% chance. Model by X~Ber(0.10).
Note that Y represents the total number of viral Tiktoks.

[cedar kiln]

@ivory finch Has your question been resolved?

[ivory finch]

<@&286206848099549185>

[cedar kiln]

@ivory finch Has your question been resolved?

[azure horizon]

what have you tried

[cedar kiln]

@ivory finch Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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 pendant]

yo

[crystal pendant]

can i have some help

this is the question:

and I arrived at parametric equations

and a plane equation

however I am confused about how to check if they intersect because one equation is in t while the other involves three separate coordinates satisfying (x-1)-(y-0)-(z-1)=0

so how does checking if one point satisfies the equation going to prove that it doesnt intersect if there is no such t

would I just check a point on the plane other than the one i used to make the equation?

because i'm just confused as thats a single point

of infinite on the plane

and unlike with two lines or something

[azure horizon]

you can probably make the first equation in x, y, z also

instead of t

[crystal pendant]

like this is my work

[azure horizon]

yeah so do the circle equation in x, y, z instead

without t

[crystal pendant]

but thats not how we were told to do circles

[azure horizon]

well

this is a way that definitely works

i mean ok hmm

[crystal pendant]

shoot

[azure horizon]

ok yeah i'm being silly

you can just use t

[crystal pendant]

im sorry i can ask my prof i do appreciate the help

[azure horizon]

so take any t

[crystal pendant]

that was acc mean asf

[azure horizon]

then the point will be (5cos(t)+2), 2, 5sin(t)+2)

assuming you've worked that out right

so then you plug that into the plane equation

[crystal pendant]

yes

[azure horizon]

and see if it's ever true

[crystal pendant]

but how many ts should i test?

[azure horizon]

you don't test any specific t

you have an equation in t

you can solve it

(5cos(t)+2), 2, 5sin(t)+2) could be any point on the circle, depending on t

so if any of these points satisfy the plane equation, there's an intersection

otherwise there's no intersection

so if (5cos(t) + 2 - 1) - (2) + (5sin(t) + 2 - 1) = 0 has any solutions, there's an intersection

otherwise there isn't

do you see how i got there

[crystal pendant]

ohhhhhhhh

thank you

i appreciate it

i get it now

[cedar kiln]

@crystal pendant Has your question been resolved?

[cedar kiln]

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.

[nova snow]

can someoen walk me through how to solve this please

[dusky peak]

the way i would set it up is similar to the shell method

let's say f(x) looks like this

then we can slice the surface area into rings

rings with circumference 2 pi r, where r is f(x)

do you see what the area of such a small slice would be?

[nova snow]

can I put that into a software like desmos?

the graphinc calculator

[dusky peak]

the function, yes
the 3d shape, yes but i dont know how, maybe i can find it out

[nova snow]

what am i trying to find within the function

this is what i got

[dusky peak]

ok i got it to graph in geogebra

but it's really laggy

now to get the surface area, we can slice it into these rings

this red ring has a radus of f(x)

and the area can be approximated as it's circumference times width

we are assuming it to be a hollow cylinder pretty much

which is ok because the width will be very small

this approximation becomes reality when the width get's arbitrarily small

as to how to graph it, i followed this tutorial: https://www.youtube.com/watch?v=dkF7edNwt3M

[cedar kiln]

@nova snow Has your question been resolved?

[cedar kiln]

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.

[crimson sedge]

Yo what did I do wrong here…?

[kind snow]

look carefully at g(x)

[crimson sedge]

I missed the square…

I always make silly mistakes like this… idk how and why

With these types of questions… is there a way for me to confirm/check my answr afterwards by plugging my answr back in or something like that?

[kind snow]

not unless u use the computer

[crimson sedge]

So you can’t plug things back in to check?

I like to double check my answers while writing tests

[kind snow]

what do u wanna plug?

[crimson sedge]

Like idk..

[kind snow]

the same thing u manually calculated?

[crimson sedge]

I want to double check my answr

[kind snow]

use a computer

or wait till u correct the answer on the school

[crimson sedge]

Like you how with certain types of questions you can plug ur x val back in…. Can I do this here ?

[kind snow]

there is nothing bad about having things wrong

[kind snow]

what?

[crimson sedge]

Nvm

And this is answr key

Does it matter that my x^2 and x are in opposite places

.close

[cedar kiln]

Channel closed

Closed by @waxen hound

Use .reopen if this was a mistake.

[cedar kiln]

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.

[elder abyss]

Hi, anyone can explain me how to solve a matrix 3x3 using theorem of cramer?

[elder abyss]

i know when its a 2x2 you need to first calculate if you can do C-1, but if you can do it, how? (1 2 / 0 1) will be (1 1 / -2 -0) (i know 0 cant be - but only using it for the example) so how can i do that with a 2x2? Sorry for the hard explaining

<@&286206848099549185>

[cedar kiln]

@elder abyss Has your question been resolved?

[cedar kiln]

@elder abyss Has your question been resolved?

[cedar kiln]

@elder abyss Has your question been resolved?

[elder abyss]

any1?

[plush yarrow]

To solve a system of linear equations using Cramer's Theorem, you need to have a matrix equation of the form ( AX = B ), where ( A ) is the coefficient matrix, ( X ) is the column matrix of variables, and ( B ) is the column matrix of constants. Cramer's Theorem can only be applied if the coefficient matrix ( A ) is a square matrix (same number of rows and columns) and has a non-zero determinant.

[wraith dagger]

Flamey

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

Flamey

[cedar kiln]

@elder abyss Has your question been resolved?

[elder abyss]

the next step is to calculate the Inverse, how do you calculate that in a 3x3.

[tribal kite]

Extend the matrix

Like

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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.

[pseudo trellis]

How would I even start this problem? We certainly did not talk about this in class 😂

[dire geode]

Differentiate piecewise

[pseudo trellis]

Do I just use the power rule on basically all of them to get the answer?

so -2x-8 would be -2

but not sure what the domain has to do with any of them

[dire geode]

They stay unchanged

[pseudo trellis]

so like what's the point of them being there when i've never seen them before?

just to kinda confuse us? 😂

just to make sure

-2
-4x
-4
0

are the answers?

[dire geode]

No they're part of the function

[dire geode]

No you're missing the domain of the function

[pseudo trellis]

so I have to write

-2 -5 <= x <= -2
-4x -2<x<=1
-4 1<x<=2
0 2<x<=5

neater than that in LaTeX, but that's the gist?

[dire geode]

Yea

[pseudo trellis]

Sick, thank you!

[dire geode]

No

[dire geode]

Format should look like the function

[pseudo trellis]

oh

[dire geode]

Not individually

[pseudo trellis]

got it, so kinda copy what the original function is with v'(x)?

lemme write it up in LaTeX, hold up

[wraith dagger]

Someone

[spice kraken]

@pseudo trellis I think the derivative is not defined at x = -2 and 2

[pseudo trellis]

could you explain that a bit further? sorry

[spice kraken]

the left and right derivatives are not equal

so it's not defined

[cedar kiln]

@pseudo trellis Has your question been resolved?

[pseudo trellis]

wait, so which ones are wrong?

[spice kraken]

the domains should be -5<=x<-2 and 1<x<2 respectively

[pseudo trellis]

They are, no?

[wraith dagger]

Someone

[pseudo trellis]

I'm confused :P

[cedar kiln]

@pseudo trellis Has your question been resolved?

[crimson sedge]

hi

i need help

who is gona help

[cedar kiln]

@pseudo trellis Has your question been resolved?

[crimson sedge]

what the problem?

[cedar kiln]

@pseudo trellis Has your question been resolved?

[cedar kiln]

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.

[crimson sedge]

.close

[dapper mirage]

Can anyone help me out with the shape of these contours?

I don’t really see how each of these contour give a different shape,

[cedar kiln]

@dapper mirage Has your question been resolved?

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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.

[slow jewel]

I got |am|<=1 and |bn|<=1 but can't proceed after that

[wraith dagger]

Lorentz

[slow jewel]

This is the last option

.close

[cedar kiln]

Channel closed

Closed by @slow jewel

Use .reopen if this was a mistake.

[cedar kiln]

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 fiber]

who knows physics pls dm me

[rustic fable]

Help to calculate this integral

[cedar kiln]

@long fiber Has your question been resolved?

[cedar kiln]

@long fiber Has your question been resolved?

[cedar kiln]

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.

[pliant oxide]

How would I find the answer to this?

[pliant oxide]

I tried finding the derivative of x and using critical points to see where the intervals were increasing, but that didnt seem to work

i mean f(x) not x

[drifting marlin]

what's f(x)?

[pliant oxide]

sin(t-pi/2)

or f(t)

i mean

[drifting marlin]

but you're asked to find where g is increasing

why are you looking at the derivative of something that isn't g?

[pliant oxide]

do you even know how to help me, or are u just criticizing for no reason

lol

[drifting marlin]

i do know, and i'm not criticizing

you're asked to find where g is increasing

[pliant oxide]

so how do I find it

[drifting marlin]

the same way you find where any other continuous function is increasing

take the derivative of g and see where it's >0

[pliant oxide]

finding the derivative of g would be finding the derivative of the inside as well

and then finding critical points no?

[drifting marlin]

no it would not

recall the fundamental theorem of calculus

[minor crystal]

use ftc-1 to find the derivative of g

[drifting marlin]

thank you kanna but i'll be fine on my own

[pliant oxide]

sin(x-pi/2)

ohh then i find critical points

i see

[drifting marlin]

better

[pliant oxide]

i see

its pi/2 , 3pi/2

ok thanks

[minor crystal]

okay sorry

[cedar kiln]

@pliant oxide Has your question been resolved?

[cedar kiln]

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.

[crimson sedge]

I understand why they've turned this into a piecewise function

but how come the bottom piece has -4 included?

[weary thistle]

So we have an absolute value in our function right?

[crimson sedge]

yep

[weary thistle]

Well that's the one to determine the value of the function thats why we turn it into piece wise. So in one case the absolute value is positive and the other case its negative

The inside of the absolute value specifically

First case, x+4<0 so the absolute becomes -(x+4)

Second case, x+4≥0 so the absolute value is (x+4)

[crimson sedge]

but why

do they include

I'm confused why the first one didn't include -4

and the second did

[weary thistle]

Oh its the same thing if the problem you got doesn't specify where to put the equal

[crimson sedge]

lmao what

why

[weary thistle]

If the problem you got doesn't say where to put the equal you can either decide to put it for x<-4 or for x>-4

But are you sure it doesn't specify

[crimson sedge]

how are you allowed to do that

[weary thistle]

?

[crimson sedge]

no it doesn't

actually

let me take

a photo

[weary thistle]

Usually you put it for x>-4

For x greater

[weary thistle]

^

Thats what my teacher said well

Idk if you learn that any other way in uni

But usually my teacher said you put the equal for x greater

Wait, you want the function to be continuous?

[crimson sedge]

I'm checking if it's continuous

also there's nothing wrong with

two holes existing

i.e 2 unincluded values

it's just you can't (apparently) have two included values

in a piecewise

[weary thistle]

It depends where it came from

Pretty sure

Your initial function is a /

So |x+4|≠0

[crimson sedge]

yeah

so I think based on can't equal zero you can't have an equals

anywhere

[weary thistle]

Even if you simplify the excluded values should still be there

[weary thistle]

I didn't get this

[crimson sedge]

or maybe one should and one shouldn't

[crimson sedge]

so $x+4\neq0$

[wraith dagger]

wyldinwilliam

[crimson sedge]

$x\neq-4$

[wraith dagger]

wyldinwilliam

[weary thistle]

Yes

One excluded value

Due to the domain of the initial function

[crimson sedge]

there has to be an exclusion somewhere

which you achieved

but I'm curious why the flexibility

[weary thistle]

The function is also discontinuous at -4 btw

[weary thistle]

Wdym flexibility

[crimson sedge]

as in being able to set -(x+4) either less than or less than equal

to 0

[weary thistle]

≥ or > ?

The equal sign?

[crimson sedge]

yeah

[weary thistle]

Well it should be the same thing thats why

Graphing wise

The function will still be discontinuous.

It will still not be defined at x=-4

Graphing wise, same thing

[crimson sedge]

when i graph it

it seems it has an infinite discontinuity

at x=-4

[weary thistle]

You did graph your initial function

Im talking about the piecewise

You want to graph the piecewise, your problem is asking you thag

That*

[crimson sedge]

I think the proper way then to do the piecewise would be to not include any values because you would be saying that at a certain point either the left side or right side touches the asymptote

[weary thistle]

Yes, at x=-4 we have the asymptote

Well,

[crimson sedge]

so then how come the way this question is done they include -4?

[weary thistle]

Well I think a more proper way is to exclude -4 completely

If its not in the domain of the initial function

Then you should put it into the piecewise

[crimson sedge]

yeah, that's why I was skeptical when I saw the inclusion of it

[weary thistle]

Since it comes from this initial function

So a better thing would be to completely exclude -4

From the piecewise

[crimson sedge]

yup, going to ask my teacher about this

[weary thistle]

Due to the domain of the initial function

[crimson sedge]

weird they'd include it

[weary thistle]

👍

Think im gonna ask mine too, bcz im a little confused as well haha

But thats probably it

[crimson sedge]

mhm

just took graphing it

[weary thistle]

You keep the domain of the initial function even if it simplifies

I think rhats it

[crimson sedge]

that makes sense, but I do remember doing questions where

that isn't the case

so curious why it is here

[weary thistle]

Example?

[crimson sedge]

uhh can't really pull one up but I've done piecewise examples where you actually do include the x value

just not with an absolute value like this one

where you actually had to come up with a piecewise

[weary thistle]

If the initial function is defined at the value then yes you include it

Its not defined in a value usually if you have /, √ or ln

(or in an Interval of values)

Like if you had f(x)=x|x-3|

Then the function would be defined everywhere

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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.

[sand cradle]

Let $K$ be a field, $V$ a $K$-vector space and $\varphi: V \to V$ a linear mapping. Show that $\varphi(\varphi(v)) = \varphi(v) + v$ for all $v$ $\implies$ $\varphi$ bijective.

[sand cradle]

Is there anything I need to know about linear mappings to prove this?

I haven't read about them yet

This is similar to what we did before @crimson gulch

[crimson gulch]

There is maybe one theorem that can help you

[crimson gulch]

but i think you can do without

[sand cradle]

$\varphi$ is injective if $\operatorname{Kern}(\varphi) = {0_V}$. So here, we have $\varphi(v) = \varphi(\varphi(v)) - v$ and that is $0_V$ when $\varphi(\varphi(v)) = v$. That was my first idea

We need to show that that holds

[sand cradle]

Which one?

[crimson gulch]

bijective iff injective iff surjective since it s going from V to V

which has the same dimension

[sand cradle]

Oh

[crimson gulch]

so you only need to prove it is injective or surjective

[sand cradle]

So I can just show injective

[crimson gulch]

yeah

[sand cradle]

Does this look good

[crimson gulch]

if you prove one you are done

[flint plinth]

how do you know V is finite dimensional?

[crimson gulch]

oh yeah we dont

[sand cradle]

Ok, then we have to show both ig

[flint plinth]

for showing injectivity, try starting with
phi(v) + v = phi(u) + u
then apply phi to both sides
use both linearity of phi and the given condition

[sand cradle]

How did you come up with that

[flint plinth]

just playing around with it, tbh

[wraith dagger]

ThibaultF02

[sand cradle]

Yes

[keen nest]

I think there's an even simpler solution. Write the equation as $\phi^2 = \phi + Id \iff \phi^2 - \phi - Id = 0$ -- then you can say something very specific about $\phi$, I think

[wraith dagger]

spq_64_t

[keen nest]

Or maybe not, hmm

[sand cradle]

@crimson gulch But I mean, that doesn't lead to phi(v) + v = phi(u) + u?

[wraith dagger]

ThibaultF02

[sand cradle]

Oh

[flint plinth]

@sand cradle basic idea here is, show that phi is injective by showing that phi^2 is injective

[sand cradle]

$\varphi(\varphi(v)) + \varphi(v) = \varphi(\varphi(u)) + \varphi(u)$

$\varphi(v) + v + \varphi(v) = \varphi(u) + u + \varphi(u)$

$2\varphi(v) + v = 2\varphi(u) + u$

[flint plinth]

how did you get this first equation

[crimson gulch]

More like this $\varphi(\varphi(v))= \varphi(\varphi(u))$

[wraith dagger]

ThibaultF02

[flint plinth]

^

[sand cradle]

Applying phi to both sides of this

[sand cradle]

But he said

for showing injectivity, try starting with
phi(v) + v = phi(u) + u
then apply phi to both sides

That should be

No?

[sand cradle]

?

[flint plinth]

one second, let me actually write it down, i did it in my head earlier, might have screwed it up haha

ah it still works out but a bit more manipulation is needed

from there you get:

2phi(v) + v = 2phi(u) + u
as you said

then:

2(phi(v) + v) - v = 2(phi(u) + u) - u

rewrite phi(v) + v as phi(phi(v)) and similarly for the other one

i bet there's a simpler way, btw

[crimson gulch]

once you get there it's actually done

since you made the assumption that phi(v)=phi(u)

for injectivity

so v=u

[sand cradle]

What

Wdym

Maybe v \neq u

[flint plinth]

oh right, i'm overthinking it, i assumed phi(phi(v)) = phi(phi(u))

[keen nest]

If we write it as $\varphi^2 - \varphi = Id$. Then for injectivity, for instance, assume $\varphi(v_1) = \varphi(v_2)$. Then on one hand, $\varphi^2(v_1) - \varphi(v_1) = v_1$, on the other hand, $\varphi^2(v_1) - \varphi(v_1) = v_2$. ||For surjectivity, $\varphi \circ (\varphi - Id) = Id$ should do the job||

[wraith dagger]

spq_64_t

[flint plinth]

2phi(v) + v = 2phi(u) + u
if phi(v) = phi(u) then you can subtract 2phi(v) = 2phi(u) from both sides

[keen nest]

Ok, it does render spoilers as latex, my fault

[sand cradle]

Oh

[sand cradle]

Is this way valid?

[flint plinth]

(and if we're in characteristic 2, then 2phi(v) = 2phi(u) are already zero 😁 )

[sand cradle]

phi(u) = phi(v), then apply phi and do what you said

[flint plinth]

sure, which step do you have a concern with

[sand cradle]

Applying phi

[flint plinth]

well if x = y then phi(x) = phi(y)

so sure, it's valid

[sand cradle]

Alright, well then phi(u) = phi(v) => phi(v) + v = phi(u) + u (after applying phi) => v = u and we are done, right?

[flint plinth]

yea, just subtract phi(v) = phi(u) from both sides

[sand cradle]

Ok, then next, we need to show surjectivity

So all $v \in V$ can be written as $v = \varphi(z)$

For some z in V

[flint plinth]

yea, i'm still thinking about surjectivity haha

[sand cradle]

Maybe applying phi again

$\varphi(v) = \varphi(z) + z$

[crimson gulch]

i think you would have to find an expression of z

[sand cradle]

Yeah

[flint plinth]

ohh

it's actually easy

[sand cradle]

?

[flint plinth]

just rearrange phi(phi(v)) = phi(v) + v
as
phi(phi(v)) - phi(v) = v

use linearity on the LHS

[sand cradle]

phi(phi(v) - v) = v

[flint plinth]

yep

so if we call w = phi(v) - v

then phi maps w to v

v was arbitrary

so phi is surjective

[sand cradle]

Oh

[sand cradle]

How did you find this?

[flint plinth]

just noticed it for some reason

[sand cradle]

How, were you playing around with these on paper or just thinking about it?

[flint plinth]

i wrote down the original equation and was gonna start manipulating it when i just saw, wait, there's a v on its own, we can isolate it

and then use linearity

[sand cradle]

Ah, alright, thank you!

.close

[cedar kiln]

Channel closed

Closed by @sand cradle

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.

[velvet bone]

I need help understanding a few of these problems
vvvvvvvvvvv

[inner glacier]

you can find BD using pythagoras

assume ab=x and ab=d and form two equations to solve

[velvet bone]

How does BD contribute to AC?

[inner glacier]

because you can make two equations with it

[drowsy sage]

right triangles have well known ratios among all 3 sides

[inner glacier]

for two right angled triagles

[drowsy sage]

they are sine, cosine, and tangent

[velvet bone]

Haven't learned trig yet

[inner glacier]

trig is not needed

[drowsy sage]

you've learned a^2 + b^2 = c^2

[velvet bone]

Yeh

[drowsy sage]

They're the same thing 😉

[velvet bone]

Ik

[inner glacier]

ok, so which are all the right-angled triangles?

[velvet bone]

The 3 rights are ABC/ADB/BDC

[inner glacier]

correct

use pythagoras in bdc to get bd

assume ab=x and ad=y

[velvet bone]

Bc is the hypotenuse of the mid triangle and the leg

[inner glacier]

form an equation for the other two right angled triangles

[velvet bone]

So would it be 9/AC = AC/7?

[inner glacier]

no form two equations for the triangles using pythagoras theorem

[velvet bone]

9²=7²+BD²

[inner glacier]

yes

[velvet bone]

AC+7²=9²+BA²

Would I then solve these 2 for the answer?

[inner glacier]

how did you get this?

[velvet bone]

Large triangle

[inner glacier]

large triangle is AC^2=AB^2+BC^2

AC^2=AB^2+9^2

so you will need AB first

[velvet bone]

So after I solve ab I can get ac since it's js a normal theorem right

[inner glacier]

yes

[velvet bone]

Thank you imma try it

[inner glacier]

sure

[cedar kiln]

@velvet bone Has your question been resolved?

Channel closed

Closed by @velvet bone

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.

[rain oasis]

how to draw a trigonometry circle

[night swift]

You mean a unit circle?

[rain oasis]

yeah yeah that

[night swift]

I mean just google it. It will show all values. then if you have further quesitons I can answer them

Like what they represent

[rain oasis]

i tried but google is not googling

[night swift]

You googled "Unit circle"?

[rain oasis]

yeah

[night swift]

What didnt show up?

[rain oasis]

some vid bout teachers explining it

u see what i need is how to draw it with a ruler ad the stuff that draw circles that i dont know what its called in english

[night swift]

https://www.mathsisfun.com/geometry/unit-circle.html

[night swift]

Oh like the tool to make a circle?

Called a "compass" in english

[rain oasis]

idk english is not my first language

[night swift]

Yeah I got that

[vagrant pewter]

You may also want a protractor to determine the angles

[night swift]

Not cause its bad its very good english

[night swift]

Yeah I was just thinking that haha

[vagrant pewter]

But you could also use a 90 degree angle like on a piece of cardboard or something and at least get a start with that

You can also take a glass or a cup and use that the create the circle

You don’t need a compass explicitly

[rain oasis]

thats th point i cant use a protractor nor a marked ruler

[night swift]

I mean does it have to be a perfect representation of the unit circle? Or can you just lable a circle with like roughly the correct lines?

[rain oasis]

it have to be corret

[vagrant pewter]

Well once you have a circle you can always draw a diameter as long as you mark the centre

Then form that you can bisect it easily enough

That gives you right angles

Connecting points A and B gives you another line that you can bisect

So that can give you the 45 degree angles

One thing I’m not entirely sure about is how to get the 30 degree angles

I’m sure there’s a way

[rain oasis]

yeah thats what im stuck with the 30 and 60 degree angles

[vagrant pewter]

https://youtu.be/fiPtFoPMYPA?si=CgYkfesL5vl_FTmz

[drowsy sage]

Bisect a radius (B to O, for example)

[vagrant pewter]

I’ll be real i don’t have a proof for why this works

[drowsy sage]

The line from that bisected radius to a point perpendicular to your bisected radius gives you a 30/60/90 triangle

The reason this is the case is because cos 60 (your x value) is 1/2

mm.. not back to A, but back to a line parallel to B

[vagrant pewter]

Oh wait I just realized

The point at which the two circles intersect is exactly where you have a equilateral triangle

Since the radius is equal in all faces

And the angle measure of an equilateral triangle is 60

[drowsy sage]

... what second circle? is this from the video?

[vagrant pewter]

So by creating a circle at A then a circle with the same radius at B, the point where they intersect is the third point of an equilateral triangle

[rain oasis]

i dont know what that is but it surly workedd

[vagrant pewter]

Then the arc is 60 degrees

Sorry the second circle isn’t made from B

It’s made in the line where the first circle has its edge

But that gives a triangle with equal sides

So the interior angle is 60

Then with a 60 degree angle it’s easy enough to bisect it

That’s cool

[rain oasis]

ah well strnger on discord thank u lotssss

[drowsy sage]

took too long to sketch, but hopefully this better shows what I was trying to explain

cos 60 = .5

[rain oasis]

oh wait now i seeeeee

[drowsy sage]

But that's why you get an equilateral...

.. because both of these are the radius of the circle

[rain oasis]

yeah yeah i get ittt

[drowsy sage]

👍

[rain oasis]

.close

[cedar kiln]

Channel closed

Closed by @rain oasis

Use .reopen if this was a mistake.

[cedar kiln]

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 garden]

bruh im actually gonna ragequit

[slate garden]

vectors are so aids

i never know which direction to go in

How the hell does this guy know how to draw the triangle

and that he needs to go QP

and not PQ

I thought it was PQ + RP

[cerulean star]

Never heard this one

[slate garden]

for some reason i just drew a right angle triangle

i thought i could j straight up plot the points on a graph

and it would form the triangle

but no

[cerulean star]

You never know what direction to go in?

[slate garden]

like idk how the guy got the triangle in the first place

[cerulean star]

(I,j,k) <-> (x,y,z)

[slate garden]

how am i meant to know its meant to look like that

if i treat those letters

like x and y coordinates

they dont form a triangle at all

[cerulean star]

The number and direction of units in the cardinal directions

Cartesian coordinates

The direction is determined by the sign of the coefficients, - or +

[cerulean star]

Oh

Because you are supposed to draw both vectors from the same point

The starting point is (0,0)

However, the orientation and placement of the grid (not shown) is completely arbitrary for simple notes drawings like this

[slate garden]

does that mean

i cant plot

the points on a graph

[cerulean star]

It does not mean that

[slate garden]

cause im p sure thats what i did

[cerulean star]

The first point should Go right 3, up 5

[slate garden]

and i have PQ as (3,5) and PR as (13,-15)

yeah

but theres no way to connect these such that they form a triangle

[cerulean star]

Hmmm

Second point is right 13, down 15

[slate garden]

yeah exactly

looks nothing like the diagram in the picture tho

[cerulean star]

Weird drawing

[slate garden]

ohhh wait

ig its not a straight line up

?

yeah

ok

idek bruh

idk what other explenation it can be

cause thats looking like a rectangle otherwise

i think thats it

thanks

or i may be wrong

yeahhhhh

p sure i got it now

god i hate this shit

.close

[cedar kiln]

Channel closed

Closed by @slate garden

Use .reopen if this was a mistake.

[cerulean star]

To go from Q to R by following the existing vectors, you have to trace PQ backwards and proceed along PR

That is in fact
QR = QP + PR = -PQ + PR

[cedar kiln]

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.

[zenith wraith]

How come a cotangent function's graph starts from zero and is not centred around zero such as tangent?

[ancient lodge]

Are you asking why there’s an asymptote at $x=0$?

[wraith dagger]

Civil Service Pigeon

[zenith wraith]

well yes, because with tangent if the period is pi over 3 then the 2 endpoints are $(-\frac{\pi}{6}), (\frac{\pi}{6})$

[wraith dagger]

Tomi

[zenith wraith]

yes

[ancient lodge]

$\cot 0=\frac{1}{\tan 0}$

[wraith dagger]

Civil Service Pigeon

[ancient lodge]

What do you notice? (hint: ||what is tan 0||?)

[zenith wraith]

well $\tan 0 = 0$

[wraith dagger]

Tomi

[ashen shard]

and what happens when you divide by 0?

[zenith wraith]

its quite undefined.

[ashen shard]

And so you have an asymptote at that point

[zenith wraith]

but what is the difference between this and tangent? how come tangent has a real number at the centre?

[ancient lodge]

Because $\tan 0=0$ and $\cot 0$ is undefined

[wraith dagger]

Civil Service Pigeon

[zenith wraith]

ah yes because it is $\frac{0}{1} = 0$

[wraith dagger]

Tomi

[zenith wraith]

In terms of this, how would I graph it? Would I just add pi over 4 to my x axis so I would start there?

or would I change the period?

[ancient lodge]

Basically it’s the graph of y=tan(x) shifted pi/4 units to the left

[zenith wraith]

what would the period be?

simply 1 pi?

[ancient lodge]

Yup

[zenith wraith]

so an optimal interval would be $(\frac{-5\pi}{4})(\frac{3\pi}{4})$ being $\frac{-\pi}{4}$ at the middle.

[wraith dagger]

Tomi

[ancient lodge]

If you mean to show two cycles of the graph then sure

[zenith wraith]

hmm, isnt this a right shift?

[ancient lodge]

I mean a shift left pi/4 is the same as a shift right 3pi/4

b/c the period is pi

[zenith wraith]

makes sense

[cedar kiln]

@zenith wraith Has your question been resolved?

Channel closed

Closed by @zenith wraith

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.

[gaunt nebula]

I have 2 questions

[gaunt nebula]

where did 1/2 come from and why is it t^2

this is a defined integral, doesn't that mean that we don't just work it out (if we did then it would be t^(2)/2) but substitute whatever is in the integral with our defined borders?

[royal loom]

they integrated t dt

and they are substituting in the bounds of hte integral

see the vertical line

T on top 0 on bottom

@gaunt nebula

[gaunt nebula]

oh so we do normally integrate it first

thank you
ok second question now

sinusoidal signal

[royal loom]

that is how you do a definite integral

integrate, and then substitute the bounds

$\int_{a}^{b} f(x)dx=F(b)-F(a)$

[wraith dagger]

Austin

[gaunt nebula]

yeah i guess i forgot

exams nuked my brain

anyways

why do we add omega in a sinusoidal signal?

[royal loom]

I need more context to understand what you're asking about, but likely omega just represents the frequency of the sine wave

for example

,w plot y=sin(x), y=sin(5x)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by austinu

[royal loom]

if omega =1 or omega=5 the frequency changes

[gaunt nebula]

here

electricity

[gaunt nebula]

why did we add omega in the middle of the function then?

why not just in the beggining?

also is this only for a sinusoidal signal or?

[royal loom]

there's not enough context I don't know what your asking

[short blade]

beginning of what

[gaunt nebula]

calculating the mean of the signal
I think its RMS but im not sure

[gaunt nebula]

of the integral

we just seemingly randomly add it in step 2

[short blade]

how would you want to add the omega at the start

it’s not random, it represents the frequency of the signal

[gaunt nebula]

I don't see how the equation would change by just making omega appear earlier

[short blade]

can you show me what you mean

where do you want omega to appear

[gaunt nebula]

here

[short blade]

that makes no sense

[royal loom]

we don't change the bounds of integration before making our substitution

[short blade]

what are you trying to do with the omega there

[gaunt nebula]

wait why we do substitute T with omega * T

[short blade]

it’s a u-sub

[gaunt nebula]

Well they add it later on and it seems like the equation wouldn't change if we just add it in the beggining and its also less confusing

[short blade]

it does change it

[royal loom]

they don't just "add it"

[short blade]

i would suggest you go watch a video on definite integration, especially on definite integration using u-sub

it would clear up where the omega came from

[gaunt nebula]

hmm alright then i'll do that

Thanks for your help guys

.close

[cedar kiln]

Channel closed

Closed by @gaunt nebula

Use .reopen if this was a mistake.

[cedar kiln]

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.

[tulip silo]

Here, it says the formula for trapezoid is 1/2 (a + b)h but my teacher says the formula for trapezoid is base 1 + base 2 x height then divide by 2

[tulip silo]

Why one correct

[fair geyser]

you said the same thing twice

[tulip silo]

Huh

[slate lintel]

$\frac12(a+b)h = \frac{(a+b)h}2$

[fair geyser]

i don't understand either, what's different

[wraith dagger]

hayley!

[tulip silo]

What is a + b?

On base + base

Nvm

Ty

.close

[cedar kiln]

Channel closed

Closed by @tulip silo

Use .reopen if this was a mistake.

[cedar kiln]

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.

[sour horizon]

you solution is pretty poorly written. it's like you took cases where f(x) = x^2 and f(x) = -x^2 and showed the integral converges in both those cases

while i agree what you're ta wrote isn't what you tried to do, you're solution still isn't great

yea ok ig that's true...

other than the 'must' part

doesn't matter, just that's not the only option you have

yea i can't really argue with that lol

it is sufficient to do that, yes

it wasn't clear to me what you were doing before this discussion tho

[dapper raven]

Did you write anything after the statement both sides are convergent?

nvm i didn't read correctly

i don't get the ta

because it's clear to me that you're not trying to use squeeze theorem

[sour horizon]

are there some negative signs/inequality issues here?

[dapper raven]

forgot a minus sign

[sour horizon]

you wanted to bound it above and below right

no i think you misunderstand what i'm saying

[dapper raven]

i think what might've been more clear is if for the first case instead of that you wrote $-\int_{1}^{\infty}e^{-x} dx \leq -\int_{1}^{\infty} \frac{1}{\sqrt{1 + e^{2x}}} dx$

[sour horizon]

$\int_1^\infty \ldots dx \geq -\int_1^\infty e^{-x}dx$ is what you get

didn't wanna type all that out

[wraith dagger]

Katharine

soulgazer

[dapper raven]

in trying to prove it converges yes

as -infinity < that integral

:D

i think because they are both the same the ta thought it was supposed to be squeeze theorem

but idk

because to me there is no implication or statement that squeeze theorem is gonna be used

btw the graph of the function 1/sqrt(1+e^2x) is pretty

[sour horizon]

i don't see how the given solution is all that messy

[dapper raven]

,w int from -1 to infinity of 1/sqrt(1+e^2x)

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kathlijn

[dapper raven]

no not that

,w int from -1 to infinity of 1/sqrt(1+e^(2x))

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kathlijn

[dapper raven]

no

pi/2 is ~ 1.56

i think

,w pi/2

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kathlijn

[dapper raven]

it is a whole bunch of garbage plugged in

not a nice number :(

[crimson sedge]

:(

i was probably thinking of $\sqrt{3}$

[wraith dagger]

4C

[crimson sedge]

there actually is something interesting

the first two terms cancel

the integral from -1 to infinity of that function is $\sinh^{-1}(e)$

[wraith dagger]

4C

[crimson sedge]

nice

[cedar kiln]

@crimson sedge Has your question been resolved?

Channel closed

Closed by @grim kestrel

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.

[fair nymph]

I wanna know how to do c

[fair nymph]

Did I screw up on c

[cedar kiln]

Please don't occupy multiple help channels.

[ashen shard]

What is t-formulae?

[cedar kiln]

@fair nymph Has your question been resolved?

[wicked mantle]

@fair nymph

[fair nymph]

I wanna know c

[wicked mantle]

yea, it looks familiar right?

[fair nymph]

Wdym

[fair nymph]

Yh i did that

[fair nymph]

I wrote that it = tan theta

[wicked mantle]

Your confusion is about tan 11pi/6?

[fair nymph]

Yeah

It's wrong

[ashen shard]

exact value

you can simplify further

[fair nymph]

How

[wicked mantle]

11pi/6 = 330°

It’s special angle

[ashen shard]

you can find any angle thats a multiple of pi/6 or pi/4 (without half-angle)

[fair nymph]

Oh

[ashen shard]

(or you should know)

[fair nymph]

I didn't learn that 😭

Or I don't remember

[wicked mantle]

What a pity 🥲

[fair nymph]

The answer says - 1/root 3

Tho

[wicked mantle]

yep

Tangent on fourth dimension is negative

[fair nymph]

Wait but tan 330 ain't right

[wicked mantle]

and 330° = 360°-30°

[wicked mantle]

It is

But the question requires you to simplify it

[fair nymph]

Can you explain 😭

[wicked mantle]

I guess you’ll learn them after a few courses in class…

[fair nymph]

Isn't that pi radians

[ashen shard]

yea

if theres not deg sign you can (usually) assume that its in rad

[fair nymph]

Oh that makes more sense

But how does tan 330 equal - 1/root 3

[granite hare]

do you have a unit circle with you

here's one

[fair nymph]

Isn't this one for tan