help-13

[foggy mica]

😔

[mental geyser]

How did you get 16.9

Like can you show me

[foggy mica]

10/0.5919

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[cedar kiln]

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[frigid dust]

is your calculator in radians or degrees?

[foggy mica]

.reopen

[cedar kiln]

✅

[foggy mica]

its supposed to be degrees right

?

[frigid dust]

yes

,tex .sohcahtoa

[wraith dagger]

Bonk

[frigid dust]

so lets look at this one

which two sides do we have

?

[foggy mica]

adjacent and opposite?

[frigid dust]

no

look at the angle

its angle A

and we are given BC and AB

BC=adjacent/opposite/hypotenuse?

[cerulean sail]

(Admittedly very poorly marked )

[frigid dust]

well, its definitely not angle CBA or ACB, so there is one angle left

[frigid dust]

@foggy mica

[foggy mica]

[frigid dust]

which one is it

[foggy mica]

oh

adjacent right? since its on the bottom

[frigid dust]

no

look at the angle thats given

its not the bottom right angle

its the top left angle

and is BC adjacent to the top left angle?

we can smash through these questions in 10 minutes if you work with me on this

[foggy mica]

im sorry. im trying, i rlly am, im just like

terrible at this stuff

[frigid dust]

its okay

[frigid dust]

notice here how the angle given is the bottom right angle?

[small yarrow]

Adjacent is the one next to the angle that isn't the hypotenuse

Always

[frigid dust]

and here its the top left angle?

[small yarrow]

As long as it's not that right angle

[foggy mica]

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[signal ibex]

How do I do this: Show that, for the complex numbers z_1, z_2, and z_3, if z_1+z_2+z_3=0 and |z_1|=|z_2|=|z_3|=1, then z_1, z_2, and z_3 form an equilateral triangle on the unit circle

[hard thistle]

rewriting the complex numbers as e^i*theta

might help

@signal ibex have you figured out more

[signal ibex]

I tried setting them equal to each other, for e.g. vec(z_1z_2)=vec(z_1z_3)cis(pi/3) but idk what to do

[hard thistle]

I cant think of a way without kinda assuming some sort of symmetry

cause since they sum to 0 you kinda can understand it to be symmetry around the origin

$e^{i\theta} + e^{i(\theta + x)} + e^{i(\theta - x)} = 0$

so you basically get this to represent z_1 z_2 and z_3

[wraith dagger]

Banana Steeler

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@safe island Has your question been resolved?

[safe island]

<@&286206848099549185>

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[wicked tulip]

just a general question but

when do we use the open and closed interveral for optimzation?

when working with fencing and such we use an open interval cause w and length have to be greater than 0?

and when working with anything else we use closed cuz they can be >= 0?

[cedar kiln]

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[wicked tulip]

<@&286206848099549185>

[dire geode]

Way too vague

Just give a concrete problem you're working on

[wicked tulip]

im just having a hard time understanding the concept of using a open and close interval

for question 1 the interval is open <
for question 12 the interval is closed <=

i can solve everything else but the domain is just confusing me

[cedar kiln]

@wicked tulip Has your question been resolved?

[cedar kiln]

@wicked tulip Has your question been resolved?

[cedar kiln]

@wicked tulip Has your question been resolved?

[barren canopy]

for the 1st question, set up a system of two equations

[barren canopy]

the first equation will hold information about the perimeter of and the 2nd about the area

can you do that now

assume x to be the length and y to be the width of the land

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[untold mortar]

when solving 2nd order ODEs, let's say those with a general solution of y = ae^mx + be^mx, solving for the y apparently requires an initial value of y and y'. however, i cant really work out why (as my teacher said) if we are given the initial value of y' and y'' this doesn't work?

[untold mortar]

surely they both give you some solvable simultaneous equations

or why not the fifth differential of y and the ninth? i don't see why i won't end up with simultaneous equations in any case

[fallen heath]

ae^{mx} + be^{mx} ?

[untold mortar]

yeah

[fallen heath]

Isn't that just (a + b)e^{mx} ?

[untold mortar]

uhh okay yeah i mean it's not m in both places mb, it's some m and some n

yk a general solution to y'' + by' + cy =0

[fallen heath]

Hmm got it. Also, it should work out just fine even if you're given y'(0) and say y''''(0)

[untold mortar]

really? why would he say otherwise? he's not normally wrong, ever.

he mentioned specifically to think about why having just initial values for y' and y'' wouldn't work, while y and y' would.

[fallen heath]

Hmm.

[cedar kiln]

@untold mortar Has your question been resolved?

[fallen heath]

y'(0) = 3(A - B) y"(0) = 9(A + B)
Equating, we get A - B = 1, A + B = 3, and A = 2, B = 1 subsequently```

I don't see the problem? Perhaps you missed some part of the lecture or his conversation and misunderstood the entire thing

[cedar kiln]

@untold mortar Has your question been resolved?

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[torn notch]

30 * cos(85° - Theta) = 35 * cos(60° - Theta) how would you proceed to get Theta?

[hollow kiln]

cos(x-y)=cosxcosy+sinxsiny

[torn notch]

thanks a bunch

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[oblique lynx]

hi, can somebody please help me understand, how would such sets look like?

i was thinking, would Z x Z be like squares?

here i tried to make a sketch

oh wait, it would be lines but points right

excuse the drawin haha

[livid swan]

Magnetic field out of page lol

[oblique lynx]

what

[livid swan]

Study physics you'll know

[oblique lynx]

i do physics so i know, but i dont know why you type it here when it doesnt contribute to my question at all

[livid swan]

You're right my vaf

Bad

[oblique lynx]

<@&286206848099549185>

[cedar kiln]

@oblique lynx Has your question been resolved?

[mint viper]

This is right. What have you tried for the other ones?

[oblique lynx]

R x Z would be straight lines parallel to R (x) axis, but what about the (R x Z) U (Z x R)

i was essentially thinking that it would go to (R x R) which would be whore R^2 plane

but im not really sure what is union of cartesian products

[mint viper]

You have R x Z right. Then Z x R would be that but rotated. Then (R x Z) U (Z x R) is both the previous ones on top of each other.

[mint viper]

It is nothing special. It is the union between two sets

[oblique lynx]

?

[mint viper]

Yeah

[oblique lynx]

oh cool, thank you

thnaks for the help

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[hot mulch]

Hi, can somebody help me with this missing Problem? i tried the "Correct option" and it doesnt lend me put it, meaning its wrong. Can somebody find the possible equation for B graph please it would help a LOT

[hot mulch]

<@&286206848099549185>

[idle gulch]

Do you have the extended graph ?

[steep badge]

once check the graph of D

verify those using desmos

[subtle harbor]

oh i misread

all of your other answers look right, what is the answer you got for B?

[steep badge]

-e^(-x + smth)

not passing through -2

[idle gulch]

-2+ln(1+x) kinda too

[steep badge]

that too

[subtle harbor]

you're all overthinking this 😭

[subtle harbor]

there is a form for B that matches the prompt though :3

[idle gulch]

How can a curve in the form k + e^(-mx) tend to 0 as x tends infinity!?

[subtle harbor]

constant multiple is allowed in front of e^mx

[idle gulch]

Oh ! I didn't think of that

Ty

[granite tundra]

-2e^(-x)?

[steep badge]

yep

[subtle harbor]

!noans

[cedar kiln]

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.

[subtle harbor]

ah wait

[granite tundra]

ok sry didnt know

[subtle harbor]

thats the wrong one but point still stands xd

its all good

[granite tundra]

ok lol

[cedar kiln]

@hot mulch Has your question been resolved?

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[hot mulch]

no

[hot mulch]

wym by that

i mean i cant put the answer -1-e^-x

that would be the "correct answer"

therefore theres another possibility

i cant find that one

[subtle harbor]

there is

[hot mulch]

give me a hint at least please 😭

like what prefix should i use

[subtle harbor]

the form you can put it in is a+be^(cx)

[hot mulch]

but i can only put the letter e and numbers

[subtle harbor]

a,b,c are constants

[hot mulch]

ik but i mean i can only put a limit of digits

i cant put the b constant

[subtle harbor]

think about these

yes i mean a,b,c are arbitrary numbers

[hot mulch]

i can only put this amount of digits tho

[subtle harbor]

how did you figure out c was 4+e^x?

[subtle harbor]

yes

this is very close

what is e^(0)

[hot mulch]

bc i did trial and error

[subtle harbor]

😭

[hot mulch]

😭

[hot mulch]

But i mean i cant put anything beyond this

[subtle harbor]

if i have f(x)=5e^(x), what is f(0)

[hot mulch]

Idk

im lost atm im sorry

[subtle harbor]

2^0, 3^0, anything to the 0 power is

[hot mulch]

0

[subtle harbor]

no

[hot mulch]

1

sorry

lmao

[subtle harbor]

yes

[subtle harbor]

so what is f(0)

[hot mulch]

1

[subtle harbor]

no

[hot mulch]

5

jkin

[subtle harbor]

yes

so

[subtle harbor]

what is B at x=0

[hot mulch]

-2

[subtle harbor]

so what do you think B should be

[hot mulch]

tysm

[subtle harbor]

np :)

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[hot mulch]

tyy

[cedar kiln]

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[torn breach]

Hello

[torn breach]

Can anyone help me

I need help with something simple

This

[slate lintel]

,tex .sohcahtoa

[wraith dagger]

hayley, who is not british

[torn breach]

Soo

Nvm I got it

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✅

[earnest quarry]

Hi

Hint: look at definition of sin and what they are asking u to solve for

The opposite number is not given

U have to solve the problem with adjacent and hypotenuse

So you use cos or tan

Pick one

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[shy tree]

Slightly embarrassing, I asked about the question “prove that in any non multiple Pythagorean triple, the difference b and c cannot be 16.” I was given the equation for the problem but I’m still completely stumped as I want to try and prove that no matter what non multiple Pythagorean triple I put into this equation, it will not result in a correct answer. The equation is as follows. a^2 + b^2 = (16+b)^2. Thank you in advance 🙏

[cedar kiln]

@shy tree Has your question been resolved?

[feral juniper]

maybe solve for a idk

[dull oxide]

or maybe b

[dire geode]

maybebebebebe

[shy tree]

I’m just not good at maths 😭

[blazing zephyr]

im pretty sure this works
send b^2 to the other side to get a^2=16(16+2b), take mod 4 and mod 4 to get rid of the factor of 16 (ill leave it to you on how to get that), and get k^2=16+2b, mod 4 would imply b is even, however, the "getting rid of the 16" part also proves a is even, however it cannot be a multiple of a pythagorean triple, contradiction

[shy tree]

Thank you

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[crimson sedge]

can someone help please

[pseudo pilot]

how can you express a_n using a_1 and r?

let me know if i'm not clear enough

[crimson sedge]

i dont understand

[oblique flare]

like an=?

[crimson sedge]

an=393216?

[oblique flare]

nah meant the formula for it

(it's in the question)

[crimson sedge]

i used the alternate formula

bc theres no n

[oblique flare]

an=ar^(n-1)

[crimson sedge]

how to find the n then

bc theres no a1

[oblique flare]

plug the numbers in the alternate formula and solve for a1

[crimson sedge]

thats what i did..

but i got the wrong answers everytime 😭

[oblique flare]

what ya get ?

[crimson sedge]

wait i'll try out this question then send it ok

[oblique flare]

it is ran-a1 not r(an-a1)

[crimson sedge]

ohhh

okok

so i should multiply r and an first

?

[oblique flare]

yea then subtract a1

[crimson sedge]

okok i get it now

no wonder why my answers are so off 😂

[oblique flare]

lol

solved it using a calculator cuz I'm sleepy

[crimson sedge]

yeah i gotchu

thanks dude

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[empty fable]

is this correct

-b not 6 btw

[kindred storm]

No.

It still needs the 1/2 power.

[digital cliff]

and a - sign went poof

[kindred storm]

Also, it's missing the -.

A negative exponent just exchanges the numerator and denominator. It won't change them other than that.

[empty fable]

omg i wrote the question wrong my bad

i meant

-1

not -1/2

oops

[kindred storm]

Oh, OK.

In that case, you still are missing the - sign.

[empty fable]

for 12?

[kindred storm]

No, 12 didn't have a negative sign when you started.

What had a negative sign when you started but doesn't now?

[empty fable]

-5 to 5

[kindred storm]

Right, a -1 power doesn't change that. It just switches the top and bottom, leaving them alone other than that.

(\qty(\frac{-5 - b}{12})^{-1} = \qty(\frac{12}{-5 - b}))

[wraith dagger]

Chai T. Rex

[kindred storm]

The old top is the same as the new bottom.

The old bottom is the same as the new top.

[empty fable]

ohh i see tysm

[kindred storm]

You're welcome.

[empty fable]

for part a

if i work out 2 lines which are parallel

and another 2 lines which parallel with a difference gradient

does that prove its a square?

[kindred storm]

You need three things.

• You need to prove that the opposite sides on each pair have the same gradient.
• You need to prove that the gradients of one pair are perpendicular to the other pair (m on one pair, -1/m on the other)
• You need to prove that all four sides are the same length.

[empty fable]

i drew a diagram

and i see that

CD and AB will have the same gradient

BC and AD will have a different but same gradient

[kindred storm]

OK, so that proves the first thing.

• You need to prove that the opposite sides on each pair have the same gradient.
• You need to prove that the gradients of one pair are perpendicular to the other pair (m on one pair, -1/m on the other)
• You need to prove that all four sides are the same length.

[empty fable]

i can do part 3 without doing part 2

[kindred storm]

Yes, that's true.

[empty fable]

so for 3 i can use the distance formula to see that CD is equal to AB

same with CB and DA

[kindred storm]

What about AB and BC?

You need AB = BC = CD = DA.

[empty fable]

oh yea

yes i see now

however for part 3

i can work out gradient of AB and BC then times them together to get -1

[kindred storm]

Yes, that will handle part 2.

[empty fable]

wouldnt that mean angle ABC = 90 degrees

[kindred storm]

Yes.

And you did part 1, so you know that all four angles are 90 degrees.

[empty fable]

ah yes

but the diagram

looks more of like a

parallelogram

which seems like there is an acute angle and obtuse angle thats why i was confused a little

[kindred storm]

Squares are also parallelograms. If you're using graph paper, make sure you plotted the points right and then rotate it so that one side is flat in your vision rather than rotated in your vision.

See if it looks like a square.

[empty fable]

i see

for part b itll just be length times hight

right

[kindred storm]

What's the area formula for squares?

[empty fable]

b x h

[kindred storm]

Well, that's for rectangles. It works for squares too, but the square formula is A = s^2.

Take any side length and square it to get the area.

[empty fable]

oh yea i forgot

itll give the same ans tho thats why i never differed between them

[kindred storm]

Oh, OK.

[empty fable]

but ty for making it clear

[kindred storm]

Squares are a kind of rectangle, so that'll work, but it can make you have to find two sides rather than just one.

[empty fable]

yea thats true

for part C i already kind of done it beacuse i know that CA is a diameter that goes through the centre

right?

[kindred storm]

It is.

[empty fable]

i see tysm this helped a lot

[kindred storm]

You're welcome.

[cedar kiln]

@empty fable Has your question been resolved?

[cedar kiln]

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[oblique flare]

secx is equal to what ?

Yep

And what is tanx equal to

Does the LHS equal the RHS

?

sinx(1/cosx)

Multiply

sinx/cosx

Nope

a(b/c)=ab/c

Distribute the tanx on the left side

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[olive thorn]

The black text is the question, I thoght the answer would be X is greator than or equal to 5 OR X is less than or equall to 5. The anwser key (blue text ) says I got the 1st part correct, but instead its X is less than or equall to -5 (Differnce is the -5) Can someone wxplain how?

[olive thorn]

(also pls ping or relpie if u have an answer so I get a notiication)

[humble karma]

I'm not sure which part of what you thought the answer was vs the blue bit doesn't agree?

[olive thorn]

wdym

I thoght it was X is greator than or equal to 5 OR X is less than or equall to 5

oops

I typed it wrong

pls check again

I thoght it was 5 not -5

ok i fixed it

[humble karma]

Okok now I see

[olive thorn]

👍

So i thohgt:

I had to flip the inequality and those were the 2 answers

but according to this I also have to have the negitive value?

[humble karma]

Algebraically, |x| can be treated piecewise.
One writes |x| = x if x >= 0 and |x| = -x if x < 0.

[olive thorn]

i kinna do get it

*dont

So can u explain what I have to do to solve it?

[humble karma]

You have to separate it into two cases.

If x >= 0, then this inequality says that x >= 5.
If x < 0, then this inequality says that -x >= 5, so x <= -5

[olive thorn]

oh cause its absolute value

[humble karma]

Yep

More geometrically, it says that x has to be at least 5 units away from 0

[olive thorn]

So to solve it if it is an OR:
X is greatoe than or equal to the number,
OR
X is less than or equall to the negitve vertion of that number

Correct?

these are the 2 answers?

@humble karma Thanks

DM me if this is not right cause im gonna continue

https://tenor.com/view/dad-gif-12845921214390797311

[humble karma]

For this specific type of problem yes.

[olive thorn]

okay

[humble karma]

|x| >= a iff x >= a or x <= -a

[olive thorn]

thx

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[tame blade]

can someone help with the domain and range for this in set builder notation:

[soft holly]

First- the domain

The smallest number that a sqrt can take is 0

So what value of x sets 1 - x = 0

And then which way can x go from there? (Hint: What happens when x increases/decreases)

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[solar tendon]

2. A uniform beam of length 4 m and weight 200 N is balanced on a pivot located 1 m from one end. What force must be applied at the far end of the beam to keep it in equilibrium?

[solar tendon]

so itried answering it and came up with 400

but i dont think thats right, what do?

[cedar kiln]

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@solar tendon Has your question been resolved?

[chrome quail]

what kind of drawing is that

You use a moment equilibrium. The mass of the beam can be represented as a point force of 200 N on the centre of the beam (so 1m left of the pivot). Thus it has a moment of 1m x 200N anticlockwise. To match it, you can either apply a 200Nm clockwise moment on the left end (upwards force), or a 200Nm clockwise moment on the right end (downwards force)

Either way, 400N is wrong, yeah.

[chrome quail]

.

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[faint hinge]

whats happening here?

[faint hinge]

they multiplied the second row by 1/3

which means it should be 3^-4 no?

[twilit escarp]

If they factor out yes, if they distribute 1/3 to the second line then its 3^-2

[faint hinge]

wdym distribute to the second line

[plucky owl]

$3^{-3} = 3^{-2} \cdot 3^{-1}$

[wraith dagger]

CaptainNova22

[plucky owl]

So they multiplied 3^-1 to the second row

[faint hinge]

ohhh

thanks

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[dreamy cedar]

i have a exam on functions in 2 days and i want to know how to study for it but its in french so im not sure you guys can help me, the photos i sent are an example question of whats coming up in the test, im fine with studying in english but i just need help knowing how to study and what i need to know to anwser these questions. its 12th grade math btw and im pretty sure you can use google to translate the photo directly

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<@&286206848099549185>

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[crimson sedge]

just got a quick question about like terms, if we have an expression like 3xyz + 2yxz - 7zyx, would this still have like terms or not?

[oblique flare]

yep bcz multiplication is commutative

[dawn junco]

terms are the parts inside a sum

and yeah you have a sum of 3 things 3xyz, etc...

they're all terms

[crimson sedge]

so basically even if those terms are rearranged in that manner, we can still add these up together (Ex. 3x + 2x - 7x) for this case

[dawn junco]

like 3xyz + 2yxz - 7zyx = -2xyz or something ?

yeah the number of terms depends on how you write it

on the left side there's 3 terms, the right side has only 1

and they're equal still

[crimson sedge]

i see, that makes sense

thanks for the assistance @dawn junco @oblique flare

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[empty walrus]

Hello, can anyone tell me whether these two equations are different in how you solve them: (x^1 + 5) -9(1+x) and 20 - (7m- 17)

[dull oxide]

$(x^1-5)$?

[wraith dagger]

SWR

[dull oxide]

By "solve", did you mean simplify?

[empty walrus]

yes

[dull oxide]

Basically same idea for both

[empty walrus]

could you show me how

if anybody knows how to simply these equations please feel free to show

[cedar kiln]

@empty walrus Has your question been resolved?

[peak vault]

Do u need something like a + b ?

Or like (a + b) ( a - b)

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[coarse dew]

Hey,
Im trying to understand ik solvers. I watched a video about it and im stuck at 3:38. How does he get the "possible side" when all the information he has, are two vectorslengths/sidelengths. This is the link to the video: https://www.youtube.com/watch?v=SGF3sE2fe_k
Any help is much appreciated!

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[coral jewel]

i'm so damn close, my idea should be correct but i can't quite put it into words

basically, call that linear map T, let e_1, ..., e_n be a basis of kerT and e'_1, ..., e'_m be a basis of imT

choose v_1, ..., v_m such that Tv_i = e'_i, then it's clear v_1, ..., v_m is linearly independent

what i want to prove is that e_1, ..., e_n, v_1, ..., v_m spans V

in fact, this is a basis of V, which makes sense because this is basically rank-nullity

my idea is to give a set of scalars

$T(a_1e_1 + \dots + a_ne_n + b_1v_1 + \dots + b_mv_m) = b_1e'_1 + \dots + b_me'_m$

[coral jewel]

the RHS is an arbitrary vector in imT

but i can't word it to make it clear that thus the vector on LHS is an arbitrary vector in V and thus this list is a basis

[cedar kiln]

@coral jewel Has your question been resolved?

[mental trail]

You're close

suppose v is in V

"e'_1, ..., e'_m be a basis of imT"

call f_1,...,f_m some antecedents

that way we can write instead T(f1),...,T(fm) basis of imT

[coral jewel]

antecedents?

[mental trail]

inverse images

they exist since e'_1,...,e'_m are in imT

(not unique)

now

T(v) = sum(a_i e'_i)

so what can you say about v - sum(a_i f_i)

I'll rewrite the same thing in latex to be more readable if you didn't get it

$e_1, ..., e_n$ basis of $Ker(T)$ and $e'_1, ..., e'_m$ be a basis of $Im(T)$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

$e'_i = T(f_i)$ for all $1\leq i \leq m$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

now take any $v\in V$: $T(v) = \sum_{i=1}^ma_iT(f_i)$ since $T(v)\in Im(T)$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

for some coefficients a_i

so what can you say about $v -\sum_{i=1}^ma_if_i$?

[wraith dagger]

rafilou is not not born in 2003

[coral jewel]

it's a vector that does not lie in span(f_1, ..., f_m) nor kerT?

[mental trail]

lies in neither?

[coral jewel]

yes

[mental trail]

you sure

?

[coral jewel]

actually, no

[mental trail]

how can you assess for example it's not in ker(T)

[coral jewel]

it lies in kerT

[mental trail]

there we go

[coral jewel]

yes, that's what i'm thinking

[mental trail]

well time to use linearity of T

[coral jewel]

yes, this vector is sent to 0 when applied T, thus it lies in kerT

but then what?

[mental trail]

so...

v - a vector in im T is a vector in ker T

so v is in Ker(T)+Im(T)

[mental trail]

soooo

[coral jewel]

wait

okk i get it now

because this thing lies in kerT, it can be expressed as a linear combo in terms of basis of kerT

[fallen heath]

Basically, you can write v - a = some linear combination of e vector

[coral jewel]

so this v, an arbitrary vector in v, can be expressed as the sum of linear combo of f's and e's

showing this spans V

i see now

[fallen heath]

So v is span {a_i, e_i}

[coral jewel]

yes

alright thank you

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[upbeat minnow]

can you help me with this?

[idle gulch]

ydx + xdy -dx + (y-2)dy = 0
d(xy) -dx + (y-2)dy = 0

[upbeat minnow]

what process did you do to get this

[idle gulch]

Seperation of variables

[upbeat minnow]

i tried it earlier and i got letter b. is that correct?

[blissful wolf]

exscuse me what is a differential equation?

[dusk goblet]

exact

[pastel vault]

you could try helping with #help-15

JEE just finished

a lot of people want DE help

[upbeat minnow]

(y-1)dx+(x+y-2)dy=0

[crimson sedge]

bro you indian ?

[dusk goblet]

bro you indian ?

[crimson sedge]

fortunately

[crimson sedge]

wby

[blissful wolf]

@dusk goblet I'm 15 years old and in my second year of high school, I was just curious and wanted to learn

[dusk goblet]

nice, it’s an equation relating a function to its derivatives

[upbeat minnow]

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[pastel vault]

you know how in algebra you have an equation relating a function of x to some other function, and you have to solve the equation to find what x is?
the answer will be a number, in other words

in differential equations, the answer will be a function

[pastel vault]

cause of this; the task will be "find the original function"

not "find the original number"

[blissful wolf]

thank you very much

[pastel vault]

so yes they're a special kind of functional equation like this

except differential equations specifically will have derivatives, can be a partial derivative, can have any number of derivatives, and so on

[pastel vault]

there's this form which you can make a differential equation by dividing everything by dx

so you get dy/dx, which is a derivative

[cedar kiln]

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[royal flint]

i have no idea how to solve this

[fallen heath]

f(x) = Asin(wt + C) has amplitude A

Think you can convert given expression in that form?

[royal flint]

the solution solved it like this and i have zero idea whats going on...

[royal flint]

what do you mean by w?

[fallen heath]

Uhh nevermind

[royal flint]

is that w from physics?

[fallen heath]

A function asin x + bcos x has amplitude √(a² + b²)

[deep sparrow]

Actually its not just from physics

[royal flint]

i thought that was angular velocity omega

[fallen heath]

Quick question, did you not read my message after that? Or you read and decided, for whatever purpose, to not respond to it

[fallen heath]

[royal flint]

i didnt get it

like what does that mean?

what do i do with that

i was googling that thing

[deep sparrow]

Google up: how to derive Asin(x)+Bcos(x)

[fallen heath]

You're given f(x) = √3 cos x + 3sin x

[deep sparrow]

lol

[royal flint]

so put that coefficient into that square root?

[fallen heath]

Yeah.

[fallen heath]

.

[royal flint]

and the answer is 2root3?

[fallen heath]

Did you get 2√3 ?

[royal flint]

yeah

[fallen heath]

Then why would it be π

[deep sparrow]

Yeah, and the whole solution is simply how to prove that formula

[royal flint]

?

[deep sparrow]

You can base on that to generalize how to derive Asin(x)+Bcos(x)

[royal flint]

the solution hella confused me

[fallen heath]

Of course if you followed the correct formula and the correct steps, you'll get the correct answer

Do you want to know how to convert f(x) in the form Asin(wx + C)?

Here A is Amplitude, w is frequency, C is phase shift

[deep sparrow]

It uses the fact that sin²(x)+cos²(x)=1 to convert the coefficients into a trigonemtry form so that a trig identity can be applicable

[royal flint]

so amplitude sin (frequency x + phase shift)? what does this do...?

[deep sparrow]

If you divide everything by √(A²+B²), the coefficient becomes A/√(A²+B²) and B/√(A²+B²). Sum of the square of those is exactly 1

[fallen heath]

That's how we work with waves. Do you wanna discuss waves?

You wanted Amplitude. I gave you that ._. and a formula to remember

[royal flint]

i think that's to much for my brain

[royal flint]

ok

[deep sparrow]

btw, harmonic motion originated entirely from math

[fallen heath]

I hope the next question doesn't ask for Frequency

[royal flint]

yeah

[fallen heath]

[royal flint]

anyways thanks yall for helping!

[deep sparrow]

!done

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[royal flint]

.close

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[undone scaffold]

I need help again #13

[dull oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[undone scaffold]

3

[dull oxide]

!show

[cedar kiln]

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

[undone scaffold]

I circled B and I found out it was wrong, the reason I circled B was because usually when a graph is shown, an equation like the ones above are shown.

But for this one all I see is a -5 a 5 for the x axis and a 5 for the y axis

[dull oxide]

Do you recall what the definition of "function" is?

[undone scaffold]

Something that has an input and output

[dull oxide]

loosely speaking, yea

Is there not an input and output for that graph?

[undone scaffold]

Ah I see now

So the answer is D and the answer says

“This graph is a horizontal linear function because it is a line with no rate of change.”

If this sounds dumb I apologize but my question is

What does it mean to have a line with no rate of change

Does it mean the equation is in a way that the line doesn’t change

Because I’ve never really heard of a line with no rate of change I’ve only heard about the line being linear or non linear

[dull oxide]

Rate of change is the measure of the change in y with respect to the change in x

That is, by how much will y change given a change in x

also known as the slope (or gradient depending on your country of origin)

In your graph, the y value is contant, meaning it never changes. So it has no rate of change.

[undone scaffold]

So everything is just balanced so it has no rate of change

Like if you put fruits on that weight machine in a store

And it moves but if the fruits are perfectly balanced they don’t move

[dull oxide]

"balanced" really isn't a great word. Constant (or even _static) is a better choice

[dull oxide]

Not quite.

It's like measuring the location of a car with respect to time

but, say the car is stopped

then its location will never change

so its rate of change is zero

[undone scaffold]

Oh ok that makes sense but how do I know if something has no rate of change

[dull oxide]

its graph will be a horizontal line

it will be of the form y=2, or y=0, or y= any constant value

any two y values on your graph are equal

[nocturne blaze]

You can tell by looking at it or you can use various different formulas

Like the slope formula or the differentiation forumla

If the slope is 0

Then it has no rate of change

[dull oxide]

slope, not slop

slop is what AI generates

[nocturne blaze]

Mb 😭🙏

[undone scaffold]

I don’t understand much so I’ll need to do my research on that but my question was answered and explained thank you @dull oxide and @nocturne blaze

[nocturne blaze]

Np

It might be helpful to just watch a few videos on the gradients of a function to really understand it

[dull oxide]

What do you know about functions, lines, graphs, and rate of change so far?

[undone scaffold]

For functions what I learned last time was how to intercept the y axis, which I need to practice on more, for lines, all I know is that there are linear, and non linear, for the rate of changes all the information that I understood was that a rate of change is a line that is horizontal or as you stated above, anything with y=0, y=2. Maybe y=5 but that’s a guess and not informative.

[dull oxide]

for the rate of changes all the information that I understood was that a rate of change is a line that is horizontal or as you stated above, anything with y=0, y=2. Maybe y=5 but that’s a guess and not informative.
This is for no rate of change

these are all examples of y not changing

They are fixed values

,w plot y=2

[wraith dagger]

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[dull oxide]

,w plot y=-5

[wraith dagger]

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[dull oxide]

Their graphs are always horizontal lines

You're familiar with rate of change if you've ever been in a car

[undone scaffold]

No rate of change is what I should’ve said

[dull oxide]

the speedometer is a measurement of your cars rate of change between position and time

Say you are driving in a var going at 40km/h

Let your position be $p$ in km, and your time be $t$ in hours. Then your position-vs-time graph can be plotted with the function $p=40t$

[wraith dagger]

SWR

[dull oxide]

your rate of change will be 40km/h

because, for every hour that goes by, your distance changes by 40 kilometers

Say you've driven for one hour. Then your car has traveled 40 km. Now say that you stop the car. Your speedometer is going to read 0km/h, because you are not moving. That is, your position is now constant

Your position is fixed at the value 40:
$$p=40$$

[wraith dagger]

SWR

[dull oxide]

Your rate of change is now 0, because p is constant

[undone scaffold]

Ok I get it now I was confused when reading because constant always means to keep going but that’s the point because you are constantly in that same position

Like water

[dull oxide]

I see your confusion

constant means the same, or unchanging

your idea of "keep going" is not too far off, but you were confusing constant position with constant speed

funny enough, when you think of your car going at 40km/h, your speed is constant

[undone scaffold]

How

Well speed determines if you go slow or fast

Which means it’s always changing

[dull oxide]

speed, loosely, defines how much your position is changing with respect to time

If, in 1 hour, your position changes by 10km, then your speed is 10km/h

if, in $3$ hours, your position changes by $60$ km, then your speed is
$$s=\frac{60\text{ km}}{3\text{ hr}}=\frac{60}{3}\frac{\text{km}}{\text{hr}}=20\text{ km/hr}$$

[wraith dagger]

SWR

[dull oxide]

But, now think of this...

If you get in your car, and you are travelling at 50km/hr, and your aren't changing your speed, then what can you say? Your speed is constant. Your position is changing, but your speed is always the same: 50km/hr

So, if you graphed position vs time, its rate of change would be 50km/hr. BUT.. if you graphed SPEED vs time, it would be a constant value of 50

Let me give you a graphical example

I'm gonna do a number smaller than 50, because it's graph is too steep

Let's say you are driving $4$ km/hr. One possible equation for your position with respect to time is $p=4t$

[wraith dagger]

SWR

[dull oxide]

Note that I labelled the axes here

the x axis is time in hours, and it determines how much time has passed since you started driving

and the y axis is position, it determines how far you have travelled since you started driving

Btw if I'm going too fast, let me know

[undone scaffold]

ok

My question is

[undone scaffold]

on the top right next to (4 x 2) earlier before I learned that x + x x2

are you doing the same thing with the hours?

or are you saying you duplicate 4km so each hour represents 4 km

which equals 8km

@dull oxide Are you still there?

[dull oxide]

Yeah basically

You're fine to think of it that way

[undone scaffold]

Okay thanks for all the help I have all of these saved so I can look back on it later

I have to move on to science now

[dull oxide]

[undone scaffold]

Is it ok if I add you incase I have more questions?

[cedar kiln]

@undone scaffold Has your question been resolved?

[dull oxide]

@undone scaffold, I don't help outside of math server, so I will ignore friend adds. But just post your questions in the server, and I'll try to help if I am around. I'll make a note of your name so I'm aware if I see you

[undone scaffold]

Okay thanks for the help

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[hybrid grove]

i need an answer check on this, is it ||16|| for part a) and ||2^3 * 3 * 5|| for b)?

the answer of b) may be wrong because i don't know combinatorics

[dull oxide]

!show

[cedar kiln]

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

[hybrid grove]

i'm not stuck

if you wanna do the problem and compare answers okay if not okay as well

[dull oxide]

I more want to know how you got to your answers

[hybrid grove]

i'm not here to give lessons tho

just to answer check

[dull oxide]

Because it's possible that you got the right answer for the wrong reasons

[dull oxide]

I think you are not understanding why I am asking you to explain how you got your answers

[hybrid grove]

you guys make everything so difficulty

i'll go back to study

[dull oxide]

I'm here to make sure you understand the material. I'm not a yes/no machine.

If it's that hard for you to explain your work, then I am sorry for you.

[hybrid grove]

yea i got more things to do

sorry

[dull oxide]

fair enough

but I'll just point out that, in all this time, if you just explained your reasoning, we'd have been done by now

[hybrid grove]

it's okay

[cedar kiln]

@hybrid grove Has your question been resolved?

[dire geode]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[hybrid grove]

learn how to read

[frigid dust]

can you show how you got your answer?

[dire geode]

help channels are for people who want help. not giving away answers

[dire geode]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[hybrid grove]

someone needs to take your helpful role off

[dire geode]

<@&268886789983436800>

[frigid dust]

!volunteers

[cedar kiln]

Helpers are just people volunteering their time to help you. Be polite and patient.

[limber marsh]

be respectful to other users.

[frigid dust]

what do you expect us to do? you want us to check your answer, but you dont even tell us how you got to your answer

[rocky harness]

dont be rude bruh. helpers want to see your work so they can check if your reasoning is correct, math just isnt about answers

[hybrid grove]

it's simple

you do the question and if the answer is equal you say yes

if not

we can discuss

about it

[frigid dust]

thats not how it works here

[dire geode]

!noans

[cedar kiln]

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.

[hybrid grove]

okay bro

[frigid dust]

you show how you got to your answer, and we can help you if you have questions

[rocky harness]

you can keep waiting till someone who agrees to do that comes here then

[frigid dust]

we are not here to check your solutions

[hybrid grove]

yea i'll do that

but the guy

keeps saying

@dire geode

[hybrid grove]

it's not homework

anyway

[dire geode]

doesn't matter what it is

[hybrid grove]

yea bro

[dire geode]

specifically that part

[hybrid grove]

you're annoying

[frigid dust]

can you show how you got your answer or not?

perhaps take a photo of your working out

[hybrid grove]

no

like the other guy said

[hybrid grove]

.

[frigid dust]

youre going to be waiting for a long time then, but sure

[hybrid grove]

it's okay

[limber marsh]

look this is all fine, but i’m just gonna recommend to you that you put your work out as well so that people are more inclined to help you.

[hybrid grove]

i just want an answer check of somebody who wants to try that problem, it's simple

[frigid dust]

your answer was 16 right?

[hybrid grove]

yes

for part a

[frigid dust]

okay, its wrong

[hybrid grove]

why it's wrong

now i can show you

what i did

now the guy disappear

this server is creppy

[frigid dust]

send a photo?

[hybrid grove]

what you found

[frigid dust]

no.

[hybrid grove]

see

lmao

enemies of mathematics

[frigid dust]

i show you my working out if you show yours, simple

[hybrid grove]

i'll just ignore you guys

too much bullshit for nothing

i need an answer check on this, is it ||16|| for part a) and ||2^3 * 3 * 5|| for b)?

[dull oxide]

!show

[cedar kiln]

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

[hybrid grove]

i need an answer check on this, is it ||16|| for part a) and ||2^3 * 3 * 5|| for b)?

[dull oxide]

they're both wrong

[hybrid grove]

okay thanks

[sacred grail]

both answers look right to me

[hybrid grove]

tysm

.close

[cedar kiln]

Channel closed

Closed by @hybrid grove

Use .reopen if this was a mistake.

[limber marsh]

you don’t have to be disrespectful, in the future don’t make these kinds of remarks about people who are trying to help you.

[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.

[plucky wind]

Please idk with this homework "Inside the right-angled isosceles triangle ABC with hypotenuse BC lies a point D such that AD is perpendicular to BD. In the half-plane determined by the line AD not containing the point B lies the square ADEF. Prove that the line EF passes through the point C."

[cedar kiln]

@plucky wind Has your question been resolved?

[cedar kiln]

@plucky wind Has your question been resolved?

[steel crest]

We want to prove ECBA are concylic

Consider AD intersects circle ABC at G

[cedar kiln]

Channel closed

Closed due to timeout

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[steel crest]

We want to show GD = DB

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[weary canyon]

Yo

[weary canyon]

Can someone help with 4

Geometric progression

[patent hearth]

english?

[weary canyon]

"for the geometric progression is given:"

That's all

I can send formulas if you'd like

[patent hearth]

so I am assuming they have provided 3 equations related to a GP right?

[weary canyon]

Gp?

Oh yeah

So

[quartz coral]

yes but it says something else

[patent hearth]

geometric progression

[quartz coral]

it tells you what to find

a1, n and something else

[weary canyon]

Yeah

[quartz coral]

cant see

[weary canyon]

A1 n q

[quartz coral]

i think q

[weary canyon]

If q<0

[quartz coral]

im assuming q is the common ratio?

for example

the common ratio is the r in $a_n = ar^{n-1}$

[wraith dagger]

alex <3

[quartz coral]

in your country i'm assuming it's $a_n = a_1q^{n-1}$?

[wraith dagger]

alex <3

[weary canyon]

Im done

Found it

?

Wdym

[patent hearth]

yep so
the first 2 equations you will have 2 unknows - a1 and q(common ratio)

solve for a1 and q and use them to solve for sum of first n terms in the 3rd question

[weary canyon]

Here

[patent hearth]

nah you are doing arithmetic progression

use the formulas of geometric progression

sorry my bad..

I assumed ur multiplication sign to be addition..let me check ur calculations

[weary canyon]

Tomorrow im getting a 2

[patent hearth]

yes you have done correctly! great work!

[weary canyon]

Thanks

[patent hearth]

now u have q as -2..use it to find a1 by plugging value of q in any of the equations

[weary canyon]

Idk how i will do the 6 questions in 40 mins

Yep

[patent hearth]

it will come with practice

[weary canyon]

Yeah

But still

I need to check like 3 methods in just one problem

[patent hearth]

areas in around the erstwhile soviets and eastern europe are known for their mathematical brilliance..so don't worry..u guys have it in ur genes

[weary canyon]

Bet

[patent hearth]

just practice a lot.. the speed will increase gradually

[weary canyon]

Not really much time for practice

So i will just believe im him

[patent hearth]

yep all the best
if you have nothing else to ask you can close this

[weary canyon]

How

[patent hearth]

add a "full-stop" and then write "close" without any space

[weary canyon]

Full-stopclose

[patent hearth]

nah

like .

[weary canyon]

Oooh

.full-stopclose

.close

[cedar kiln]

Channel closed

Closed by @weary canyon

Use .reopen if this was a mistake.

[cedar kiln]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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 badge]

if BD = xSin(a), how do i solve 7.2? Angle C doesnt even look like a right triangle so im completely lost

[forest badge]

If i take sin(theta) i get BC/xSin(a)

Not sure what to do with that

[velvet mortar]

Sine rule maybe?

[forest badge]

Ohhhhhh yeah forgot about those. Gonna do some revision then tackle this again. Thanks 🙌

.close

[cedar kiln]

Channel closed

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• 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.

Channel closed

Channel closed due to the original message being deleted.
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[dark cradle]

how do you do this

plpslsplsp

im stuck

[soft holly]

Send this in a new channel

[dusk finch]

this channel will soon close

and there is no way to reopen it

[dusk finch]

.

[dark cradle]

okay

/close

.close

[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!

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[cinder saddle]

how to solve for theta when cosine of theta=1/3

[noble flame]

what is the domain

[vagrant elbow]

if you're looking for an exact value you're gonna have to use a calculator

Otherwise it's just arccos(1/3)

• whatever periodicity

[quartz coral]

+-

[cinder saddle]

My answer sheet says its -pi/2 but i wanna know how to get there

[quartz coral]

cos(alpha) = x

alpha = arccos(x) + 2kpi, where k is an integer

AND

alpha = -arccos(x) + 2kpi, where k is an integer

[vagrant elbow]

that can't be right

!xy

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[quartz coral]

yeah can't be right

-pi/2 would be one of the solutions to cosx = 0

[cinder saddle]

The answrr sheet says the answer when cos(theta)=1/3 then sin(-pi/2) =-1

[quartz coral]

?

[quartz coral]

ahhh

[dire geode]

is this from a different chapter/section

[cinder saddle]

Maybe