help-0

[modern sedge]

now if we want to find t(4x-5), what we do is replace y with 4x-5 in the previous equation

[tacit iron]

Yeah

[modern sedge]

it doesnt matter if you use y or x in definition of the function, I just changed it to y to make the substitution less confusing, as now I can refer to it as "replace y with 4x-5" instead of "replace x with 4x-5"

[tacit iron]

Oh right yeah I get you

Because it's being replaced anyway

[modern sedge]

mhm

so what do you get when you replace it

[tacit iron]

T(4x-5)?

[modern sedge]

yes, thats the left hand side

[tacit iron]

Ok

[modern sedge]

now do the same on right hand side

[tacit iron]

That's what's tripping me up a bit but I'll try

[modern sedge]

^ of this equation

[modern sedge]

just replace every occurence of y with (4x-5)

dont overthink it

just do the string manipulation, treat it like a text

[tacit iron]

So I'd be t(4x-5)= 4x-5-5(4x-5) +1?

[cinder tundra]

No

[modern sedge]

almost

[tacit iron]

Shoot

[modern sedge]

but there was y^2

[tacit iron]

I kind of get you but I thought I was right with 4x-5 for the y^2

Oh ther was a two

[cinder tundra]

Imagine you want to find t(2), what would you do? So this is the same

[modern sedge]

yeah, in part a) you probably just replaced x with 6, so you got 6^2 - 5*6 + 1

[tacit iron]

So I'd be
2(4x-5)?

[modern sedge]

do the same here

[modern sedge]

no

[tacit iron]

Dam it

[modern sedge]

y^2 isnt same as 2y

[modern sedge]

^ this is what you did in part a) right? You essentially just replaced X with 6

do the same here, but replace it with 4x-5

[tacit iron]

Just trying to think
So if I reply slow that's why

Ok yeah so part a is yeah
6 squared-5(6) + 1 which gives you whatever
So then part b is
t(4x-(5))?

I'm making this more complicated for myself 🗿

[modern sedge]

in part a, you computed t(6) = (6)^2 - 5(6) + 1

now you're finding t(4x-5)

[tacit iron]

Ok

[modern sedge]

do the same thing as you did with t(6)

[tacit iron]

Oh wait so your literally doing the same thing

[modern sedge]

just replace every y with (4x-5) in t(y) = y^2 -5y + 1

[tacit iron]

Expect it's with 4x-5

[modern sedge]

Yes

[modern sedge]

Exactly

completely the same

just replacement

[tacit iron]

So why do they say 16 squared beside part b, is that what your meant to get or is it irrelevant?

[modern sedge]

it's 16 x^2, it will be relevant later, but not right now

now just do the replacement

[tacit iron]

Yeah

I'll do it rn

[modern sedge]

(4x-5)^2 - 5(4x-5) + 1 is what you should get

[tacit iron]

Ok I got 16 squared minus 60x plus 51 by
Subbing in (4x-5) into t(4x-5) or just the x
Equation:
(4x -5) squared - 5(4x-5)+1?

[modern sedge]

yes, that's it

[tacit iron]

Yesssss!!!

Thanks

[modern sedge]

and when you expand it you get the answer

[tacit iron]

Yeah

[modern sedge]

which ig you already did

[tacit iron]

I've a thingy that makes questions like that harder to do

Dam reading is tough sometimes 💀

Thanks for being patient w me @modern sedge

[modern sedge]

np

[tacit iron]

Yk for simultaneous equations
5x+6y=4
7x+8y=8 gets multipleyed by to make 35 cause of LCD
Is the bottom equation with 7 multiplied with a minus because it's on the bottom?

Oh nvm it's to get rid of the x

So I'd be done to make 35x and a -35x to make them disappear

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[azure zealot]

Hey am having a hard time solving this, could u help me?

[alpine sable]

notice that both are linear equations
so rearrange both equations to the form of y = mx+c
and for 2 linear equations to have an infinite number of solutions, they must be perpendicular to each other, i.e. same the product of their slopes is -1 (m1 • m2 = -1), and their y-intercept (c) must be the same

[azure zealot]

How do I rearrange them?

[alpine sable]

move all y terms to the LHS and other terms to the RHS, then divide both sides by the coefficient of y and simplify the RHS in case it's not 1
for example the second equation:
2y + 2bx = 4
2y = -2bx + 4 (divide both sides by 2)
y = (-2bx + 4)/2
y = (-2bx/2) + (4/2)
y = -bx + 2

[alpine sable]

after doing this you'll get 2 equations in terms of a and b which you can simply just solve

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[azure zealot]

Thanks

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[humble sigil]

Can someone help part c

[humble sigil]

<@&286206848099549185>

[junior vigil]

what is z? what is 1/z?

you can write 1/z as a power of z

[ocean seal]

w

w

[main junco]

and if you multiply those

you get -1/4

[humble sigil]

Can’t tell if u just found the exact values off the Internet but this made realized how to find them off previous part so thanks

[junior vigil]

it may help to find the roots of the quadratic. then you use the product of roots property.

[main junco]

i remembered the first one

[versed hound]

You can derive the value of cos 18 quite easily

[main junco]

i had to search up the second one becoz i forgot 😦

[versed hound]

Cos 5*18 =0

[humble sigil]

Yeah that’s not how ur meant to do the question…

[main junco]

oops 😳

[humble sigil]

The hence means u need to use the previous part

[main junco]

my bad

[humble sigil]

Finding the roots for part b gets u those tho

[versed hound]

Derive cos 18 quite easy

[humble sigil]

Yeah but “hence”

[main junco]

i have a list of useless trig identitties

https://artofproblemsolving.com/community/c7h3296089

@humble sigil

found that a long time ago

[humble sigil]

Yeah they are useless cos I will need to prove any of them if I wished to use them in an exam

.close

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[humble sigil]

.reopen

[lone heart]

✅

[humble sigil]

Acc

Can someone help with a and c

[humble sigil]

Here is part a and b for the question on the top of that page

@main junco can u help

[main junco]

e

[main junco]

u can do the same thing u did witht the last one

[humble sigil]

?

[main junco]

u know how to find form of cos(2pi/5) or cos(4pi/5)

so just use same process

[humble sigil]

I don’t think that’s how it works…

Acc

I found out how to do it

[main junco]

nice

[humble sigil]

But do 3a

[alpine sable]

cos⁴(theta) = (cos²(theta))²
use the formula:
cos²(theta) = (1/2)(1-cos(2•theta))
sub that in

[humble sigil]

ok thanks

[alpine sable]

urwelcome

.close

are the one who can close help threads only the OP?

[humble sigil]

prob

.close

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[alpine sable]

Me help for every question

Give me steps to the answer for each question

[empty moth]

I'll help with 1 question

what can we decompose this into

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[vast harbor]

/rotate

[vast harbor]

ABC is a traingle

AEDC and AFGB are squares

I in the middle of BC

FC = EB , and they are perpendicular

The question is prve that IOÒ are isoceles right traingle

[limpid turret]

,rotate 65

[vast harbor]

<@&286206848099549185>

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@vast harbor Has your question been resolved?

[vast harbor]

<@&286206848099549185> what do you think

Or u don't know geometry

It's kinda hard I'm feeling y'a

[exotic belfry]

is this given or is this to prove?

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[vast harbor]

Given

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[tight beacon]

I'm working on practice questions but I need help understanding this, I was wondering if someone is willing to draw this out.

[tight beacon]

Here is the full question if more context is needed, but the top picture should be enough.

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@tight beacon Has your question been resolved?

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[lone heart]

@vapid steppe Has your question been resolved?

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[quasi sand]

I don't how to start

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[viscid rune]

hey, i just did this question and i got pi x [128 - 32sqrt(2)]/3 units^3. i can't find the posted answers and im not too sure about what i put, so can anyone double check?

[hollow wren]

uh

i got 2pi

[viscid rune]

welp

[hollow wren]

wait

WAIT

NVM

lolol

i found the area instead of volume

gimme a min

im tweaking today

[viscid rune]

alright

[ocean seal]

Potatomonke

[hollow wren]

if i did it right

[viscid rune]

huh

i'll try to do it once more and show my working

[hollow wren]

ill try to as well

ok wait i got a diff ans

but i think its acc the correct one this time

wait u mightve been correct

lemme check

ok i got my final ans

$\pi \frac{128- 64 \sqrt{2}}{3}$

[ocean seal]

Potatomonke

[viscid rune]

how come you got 64sqrt(2) and not 32sqrt(2)?

[hollow wren]

idk

prob messed up somewhere idrk

also i plugged it into desmos and it got the same value (roughly ~39.25978)

@viscid rune lmk if i did smth wrong

[viscid rune]

hold on

for the integrals you got, wouldn't you need to subtract the bottom line from the top line? @hollow wren

[hollow wren]

which lines?

[hollow wren]

like where in here

[viscid rune]

hold on i might be tripping

[viscid rune]

OH

LOL

[hollow wren]

the first integral solves for the area rotated under that purple line

second one over

[viscid rune]

why did i think i needed to subtract the top line from the bottom line when it's in respect to the y axis, not the x axis 😭

[hollow wren]

ah lol

[hollow wren]

common mistake

dw

[viscid rune]

i thought "oh, well y = x is above y = sqrt(16 - x^2), so i'll have to subtract them" instead of not realising it didn't matter

that makes so much sense

tysm

[hollow wren]

np

[viscid rune]

that's all settled then

.close

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[floral sail]

I'm reading about the proof of the Completeness theorem and don't understand why Henkin Theories are complete. If H is some Henkin Theory, then by the first incompleteness theorem, there should exist sentences which are not provable from H. But H is complete, meaning for every single sentence in its language, H can prove either the sentence or the sentence's negation. How can this be?

[candid hamlet]

can i use this channel

[floral sail]

there are three others available

four now

ok nvm we back to three

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@floral sail Has your question been resolved?

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[rancid fractal]

im confused about what this theorem is saying

like P and Q are my change of bases matriceis but

like is Q the change of basis from A to A' or from A' to A?
what i mean by that if i had v expressedin A: $[v]A$
and v expressed in A': $[v]{A'}$

is Q the matrix such that
$$ Q[v]{A'} = [v]{A} $$
or
$$ [v]{A'} = Q[v]{A} $$

same question for P

[ocean seal]

abraxas

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[lunar anvil]

https://www.youtube.com/watch?v=xcIb91wleuM&ab_channel=XanderGouws
this video goes through the derivation of an epicycloid, a circular shape. within this derivation, 2 angles are used for 2 circles where they touch and have one rolling around the other. But it mentions that the side length produced by both angles must be the same. can anyone help and tell why they must be the same?

secondly, the triangle that consists of alpha and theta purely says that it must be noticed, but why it is as so is not mentioned?

[lone heart]

@lunar anvil Has your question been resolved?

[wide tartan]

what is the timestamp for this

[lunar anvil]

allow me to provide

for the first one 1:04, and the second one is 2:08

i believe one look into the full video may give an idea into what is being found

[wide tartan]

for the first one that’s just because of the “rolling without slipping” condition, you can think of it as “if i roll an object on another and the arc lengths are not the same, one of the circles has slipped a bit”

think about rolling a wheel on the ground for example

when the wheel rolls around one time, the distance it traveled better be 2pi, or else you have a defective wheel or the ground is slippery

[lunar anvil]

that may be so, but then wouldn't the arc length on the smaller circle be a bigger proportion of the circle's circumference?

[wide tartan]

yes that’s right

the video is not asserting that the angles are the same

only that the length of the arcs are the same

so in fact you’re right that the length will take up a bigger proportion of the smaller circle

[lunar anvil]

so then it's not, "the distance of the arc length formed by the α and θ by the rolling and static circle respectively, is the same." it's "the α and θ, are formed by the distance of the arc lengths being the same between the rolling and static circles"

[lunar anvil]

ofc, hence the separate variables

[wide tartan]

i think it’s just a perspective thing, but the way you put it is fine

we are only outside observers of this process, so we are not determining anything really, we only observe that the arc lengths are the same, and see that this means that the angles must be different, but i would just remain neutral on whether one thing causes the other to happen, or whether one forms the other

[lunar anvil]

hmmmm, that's annoying because this is for a mathematics assignment, and so after getting it reviewed, i've been told there is no proof of such

stating that one of these happens due to another may create understanding for this

[wide tartan]

okay, well just put it as “the arc lengths being the same causes the angles to be different” because the assumption that we started with was that we roll without slipping, and that’s equivalent to the arc lengths being the same

which is a fine take to have for now

[lunar anvil]

alright easy, i'll denote it as such

pls help w the second part too

[wide tartan]

sure let me take a look

wait what are you asking specifically

are you asking why he mentioned the triangle in the first place?

[lunar anvil]

no, why is the angle of the triangle alpha plus theta

[wide tartan]

oh okay

oh let me draw something rq

that’ll make it clear

[lunar anvil]

alright bet

[wide tartan]

so

here we have both circles

one mark is theta

two marks is alpha

the right triangle

is this

[lunar anvil]

okay maybe this will help

[wide tartan]

nice drawing. now also we can draw a horizontal line through the center of the small circle

and notice that it is parallel to the radius of the big circle that was already drawn

[lunar anvil]

yep

[wide tartan]

okay so that means that the opposite angles are congruent

[lunar anvil]

congruent? or alternate interior angles?

[wide tartan]

the alternate interior angles have the same measure

[lunar anvil]

yep

[wide tartan]

okay so now the right triangle has an angle of theta plus alpha

because it’s literally just the theta angle stacked next to the alpha angle

[lunar anvil]

OHHHHHHH

OKAY, OKAY, YEP

that makes sense now

[wide tartan]

nice

okay time to sleep for me i think

it’s 6 am lol

[lunar anvil]

easy, thanks, i needed your representation, i had the drawing i sent, but i didn't see that this is why it is as so

[wide tartan]

gl on the project

[lunar anvil]

thanks, imma need it

.close

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[candid surge]

yo so

i got the eigen values

i have to get the eigen vectors

from what i know the eigen vectors are the null of a+eigenvalues*In

i tried solving that and i couldnt reach a conclusion

[tardy mango]

what have you tried?

and what eigenvalues did you obtain?

[candid surge]

eigen values are 2 and -1

then i gotta solve the null

equalling it to zero

but for 2 i got the three elements of the vecotor equal zero

[tardy mango]

yup your approach is right

can you maybe show your work?

[candid surge]

how do i like

type matricies on here

[tardy mango]

a picture will do just fine

or a screenshot

[candid surge]

i just scribbled on it

cuz it was wrong so

L

lemme write it again

this is what i gotta solve

[manic terrace]

what did you get after row reducing

well actually

there is a somewhat easier way

to find this

looking at the row sum will clue you in on one of the eigenspaces

and since the matrix is symmetric you can also find out the others quite easily

[tardy mango]

ahh i think you made a tiny mistake in constructing the matrix
Null (A- \lambda I) = Null (A + 1 I)

so not Null (A - 1 I)

[manic terrace]

yes

[candid surge]

what do you mean

[tardy mango]

sorry i wrote it wrong myself 😭

[candid surge]

so am i correct ?

[manic terrace]

the matrix should be all ones

[candid surge]

how so

the lamda is -1

did i write the rule wrong

[tardy mango]

it's the null space of matrix A - lambda

[candid surge]

oh

gotcha

[tardy mango]

not matrix A + lambda

[manic terrace]

yess

[candid surge]

ok ok

so now i gotta put it in reduced echelon form

correct ?

[manic terrace]

yes

[tardy mango]

yupp correct

[candid surge]

but the zeroes will never change that way though

like

[manic terrace]

true

[candid surge]

everything will equal everythign

[manic terrace]

youll get 1s in the first row of the coefficient matrix

[candid surge]

so free variables ur saying

[manic terrace]

yes

[candid surge]

but how

[manic terrace]

and plug them back into the variables

[candid surge]

if it's in reduced

there shouldnt be

[manic terrace]

why not?

[candid surge]

reduced echelon form

like

[manic terrace]

yes

[candid surge]

above anud below each one there should be a zero

[manic terrace]

not neccesarily

it's still in rref there just isn't a pivot in every column

so when you put the variables back in you'll get something like

x1 = -x2 - x3

x2 = x2

x3 = x3

since the second column and third column correspond to two free variables

[candid surge]

ok

[manic terrace]

if the eigenspace exists, there should always be at least one free variable in the matrix $(A-\lambda I)$

[ocean seal]

tim

[manic terrace]

have you covered the spectral theorem yet?

[candid surge]

nope

[manic terrace]

oh okay

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[indigo mica]

Very quick questioni:
This problem should be solved with mathematical induction. I solved some mathematical induction problems already so I some what know what i should get as a result.
What the does the line between 5 and 2 mean in this kind of problems?

[prime badge]

it means "divides"

so divisible but backwards

[indigo mica]

how should i understand divisible but backwards? how does it look on a very simple example?

[prime badge]

5 | 15

5 | 13

[indigo mica]

ohhh, okay okay, so it means that the expression in the brackets is divisible by 5?

[prime badge]

15 = 0 mod 5

yes

[indigo mica]

okay okay, thank you for clearing that up, what should i do with it now? most of the times when i was solving mathematical induction it equaled something and you were looking for the same thing. This doesn't equal anything, so how should i interpret that or get it into such form?

.close

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[nocturne sage]

I was told that this problem is incorrect and Im not sure why. The only thing I could think of is that I have to flip the problem.

[arctic lintel]

why did u flip the sign?

[gusty shale]

you only flip the sign when dividing or multiplying by a negative

[nocturne sage]

I didnt. I was gave this problem by a teacher and have to correct it. However I was confused. Would the correct answer be U > -3 ??? @arctic lintel @gusty shale \

[gusty shale]

you have flipped the less than sign to be a greater than sign, but you only do that when you are multiplying or dividing by a negative, the correct answer is -3 < U or U > -3

[nocturne sage]

Ahhh gotcha. Thank you really appreciate that!

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[upbeat pilot]

I didn't understand what I have to do to find the right answer

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[vernal linden]

Hello

[hot pewter]

why did you open 2 mh channels at the same time

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@vernal linden Has your question been resolved?

[vernal linden]

Sorry

[hot pewter]

nah its fine

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[lone heart]

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

[thorny parcel]

3

[main junco]

what answer

[thorny parcel]

I got answer as D

it should be B

[main junco]

did you

rearrange to

$x^2 < 4$

[ocean seal]

w

[thorny parcel]

yes

i did

idk after that

its confusing

[main junco]

u didnt know what to do after?

[thorny parcel]

yea idk

i thought it will be

-2 and 2

thus D

[hollow wren]

well

D is basically just saying x < 2

which isnt correct

[stark scaffold]

how can x be -2

[thorny parcel]

oh

[main junco]

cant you just square root both sides

[thorny parcel]

exactly

why not

[main junco]

trivial inequality, x^2 > 0

[thorny parcel]

is my ques

[stark scaffold]

ok got it

[thick lynx]

It can't, if you mean the image, it's probably just written like that to indicate the bounds

[stark scaffold]

yeah gotcha

[main junco]

x < 2 (principal root) and x > -2

so -2 < x < 2

[stark scaffold]

yeah

[main junco]

so whats the confusion-

[drowsy sun]

I did not include 2 and -2, I identified the critical points

[thorny parcel]

so do u solve it like quadratic?

like uh

(a+b)(a-b)=

[drowsy sun]

yep

[thorny parcel]

(a-b)^2

[drowsy sun]

no

[drowsy sun]

no sorry

difference of two squares

[stark scaffold]

a^2 - b^2

[thorny parcel]

i dont understand tho

[main junco]

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

[ocean seal]

w

[thorny parcel]

(x+2)(x-2)<0

after this how can u say for sure that
x=2 and x=-2

[main junco]

noo i should have put (a+b) first 😦

[stark scaffold]

x aint -2

[thorny parcel]

but this-

[main junco]

strictly < and strictly >

[stark scaffold]

oo

[main junco]

he just identified critical points

[thorny parcel]

ooo

[thick lynx]

What does (x+2)(x-2) represent, geometrically/graphically?

[main junco]

*in addition, only one of (x+2) or (x-2) has to be negative for inequality to hold

rest is obvious

[thorny parcel]

ooo

[thick lynx]

A parabola with zeros at x = -2 and at x = 2. In particular, it is opened upwards

[thorny parcel]

alr

[thick lynx]

From this, it should be clear that (x+2)(x-2) < 0 is fulfilled for values between these two zeros

[thorny parcel]

alright thanks

yall

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[bold silo]

how would i solve this? i tried subbing in 2k & k which got me -16/9k^2 but idk if thats relevant

[echo socket]

Just start by integrating

[bold silo]

bro im actually dumb, i subbed 2k&k in without intergrating

so i got 2/(3k)

thanks

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[eternal crest]

I got this equation I came up with and I’m trying to figure out how to get x by itself, it’s been a while since I’ve done this math.

[eternal crest]

And I’m also on vacation so I don’t have access to a pencil lol

Also you can ping me

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[eternal crest]

<@&286206848099549185>

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[eternal crest]

<@&286206848099549185>

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@eternal crest Has your question been resolved?

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@eternal crest Has your question been resolved?

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[worthy valley]

I get everything except the parallel vector idea

[worthy valley]

Can someone explain it for me

[sour verge]

The derivative has value 4 at this point, i.e. the tangent line there has slope 4.

That means you want a unit vector with a "slope of 4", in other words, it goes up four times as much as it goes right. But as they argue, (1,4) is such a vector, (in fact, given any value a, (1,a) will have slope a).

[worthy valley]

makes sense thanks!

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[visual ivy]

ping me when you respond

thanks!

[ocean holly]

ok... what is your status for these two questions?

!status

[lone heart]

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

[visual ivy]

1

don't know where to begin

:/

[ocean holly]

ok. that's fine

for a), you probably want to make sure that the function is in the format that is useful. Specifically, you probably want it to be in a standard form. Do you know what is standard form?

[visual ivy]

yes

[ocean holly]

a hint for you is that if you have a standard form for this parabola and know the shape of it (which is hinted by the problem that f(x) > 1 for all x), you can determine where the vertex should be

it will look a bit messy but that is fine. Overall, what do you care for this problem is y-k and x-h term

can you try to turn that into the standard form?

[visual ivy]

oh ok

sure thanks

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[lyric bay]

can anyone help me with this

[raw stream]

nice

[ocean holly]

there you go. But is this 8?

or is it B?

[lyric bay]

no 13

lol

13

[ocean holly]

oh ...

my bad

[lyric bay]

no my bad

had to paint it in paint

[ocean holly]

ok. There is a systemic way to solve this. Can you solve the system of equations or is it too advanced for you?

[lyric bay]

english is not my first language and i really suck at math bro so i think its too advanced

like i really really suck

[raw stream]

what grade are you in?

tell me your real age

it will help me explain

you better

[lyric bay]

14

8th grade i think in english

[tribal field]

Or a general approximation

Nvm

[raw stream]

You know about forming equations and solving system of linear equations?

[lyric bay]

lemme translate this

na i dont think i can i dont think yall understand how bad i am in math loll

[raw stream]

where you from?

[lyric bay]

netherlands

[tribal field]

Ok, different approach

[raw stream]

ok

[tribal field]

Top right plus bottom right is 8
Top right plus top left is 8
What does that tell you about the relationship between top left and bottom right

?

[lyric bay]

sorry but i have no idea

my english is not that great

[tribal field]

A+B=8
B+D=8
how are A and D related?

[lyric bay]

like they both have something to do with 8 ????? no idea bro

im so sorry bro but like i said im just really bad and my english is not that good

[tribal field]

Ok
Ok
Ok
Completely different approach
Guess and check
At this level of math, you probably aren't working with anything other than integers anyways
At this point just start by plugging in one corner and solving for the other 3 to see if you're right

[raw stream]

there are no integer solutions

you need to form equation and solve

[tribal field]

Sorry, just solved it, you're right

[tribal field]

But I think
That might be rather difficult for them

@lyric bay did you choose the numbers yourself? Or is this a problem someone gave you?

[lyric bay]

my teacher gave me the problem

she said

[runic aurora]

Hmm, this one is gotta be quite frustrating to solve
Skip it

[lyric bay]

there was a little trick to do it

but

everyone in my class is clueless

[tribal field]

I mean, there is a trick

[tribal field]

I really don't know how to hint at it more than this

[lyric bay]

she will reveal it at the end of the day but like i cant work without knowing the answer of this🤣

[runic aurora]

But that still leaves a lot to solve

Well I guess it does reduce to 3 var

Augh boring calculation problem anyway

[tribal field]

True, but if you're looking for a trick to get you started, that's it

[ocean holly]

just a heads up, I think the answer is not integer so it might be hard to solve

[lyric bay]

do yall know the answer

[tribal field]

Yeah

It isn't an integer

[lyric bay]

tell me bro i need them free chocolate

[tribal field]

If you can't get it from here
Your teacher will never believe you got it on your own

[lyric bay]

wait

[tribal field]

E.g.
Give up on the free chocolate and leave it be

[runic aurora]

Chocolate is overrated anyway

[ocean holly]

it is a trick question at your level

[lyric bay]

if yall give me the answer i will send all of you 60% of my chocolate

[tribal field]

The shipping will be worth more than the chocolate

At this point
Buy your own chocolate

[lyric bay]

i really cant work if i dont know the answer fam

[tribal field]

There are times in life where you just gotta get over it

This one of them

You'll live without chocolate

[ocean holly]

A hint for you if you really, really, really need to solve this is to guess the finer detail. Instead of 1,2,3. Try 1.01, 1.02,...

otherwise, I would say give up.

[lyric bay]

im on my hands an knees for u guys to tell me the answer

please

[ocean holly]

The problem is not the answer. The problem is that it is hard for us (and thus, for you) to explain why this is the answer to claim the chocolate.

[tribal field]

you will live till the end of class without knowing

[lyric bay]

trust me i will get it

pls tell me the answer

some guy told me it had to with like 4/15 shit

I BEG U GUYS!!!

[tribal field]

trust me, we've given you enough hints, you won't get it

[lyric bay]

can u guys just tell the answer

pls

i will pay 10000 euros

i wont get in trouble with my teacher

pls

[ocean holly]

Damn, I would be glad to accept that money but still, I would not tell you the answer.

[lyric bay]

plsssssssssssssssssssss

[ocean holly]

The point for this question is to learn how to solve not just what is the answer

[tribal field]

nope

[lyric bay]

no i need to know the answer bro

[tribal field]

no you don't
no you really don't

[lyric bay]

i have to

can yall just maybe say

2 numbers

and then i can work from that

[tribal field]

nah

[lyric bay]

1 number

i really wanna get it

[tribal field]

also, if you're representative of the average student in your class,
there's no guarantee that your teacher actually has chocolate

[lyric bay]

no she showed it

my teacher said its actually so easy

and when i get the answer i will find myself dumb

i have no clue

[ocean holly]

Have you learned about fraction or decimal?

Like, the operation and such

If not, then this is a lie.

Your teacher just lie to you

[tribal field]

it is in fact easy for the rest of us
you? not so much

that choclate exists to taunt you
no one's taking that home except your teacher

[ocean holly]

This, however, does not imply that you are bad at math. It is just that this question is quite beyond your scope of feasibility a bit. I might as well label this question as a high school level or, heck, linear algebra problem in college.

[runic aurora]

Yeah LA problem in disguise

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[prime glacier]

I'd like to confirm my answer if it's right. I got 20.

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@prime glacier Has your question been resolved?

[lyric bay]

i got it

[lone heart]

@prime glacier Has your question been resolved?

[prime glacier]

u got 20 too?

[lone heart]

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[rose sigil]

,w expand (x+x^2+x^3)^4

[ocean seal]

Results provided by WolframAlpha

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

Requested by slayyla

[rose sigil]

i did not get 20

i got 4 + 16 + 4

[prime glacier]

what's your process

[wet harness]

Possible prime numbers 3,5,7,11

when 1×1 matrix only , 1 possible case

When 1×2 matrix :

3 = 1+2

5 = 3+2

7 and 11 arent possible using 1,2,3 summation

So 4 more ways

So total ways is 4+1 = 5 no ?

[wet harness]

Multinomial expansion

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[alpine sable]

Hello, I'm relearning algrebraic expression, like writing them and theres this one question, "the difference of the product of 2 and a number and 15" I dont know when to use ( ) at all, I wrote the problem as 2(x-15) but in the answer sheet it says 2x - 15? Isnt that the same thing? or is there something im missing

[boreal jetty]

2(x-15) is not the same as 2x - 15

[alpine sable]

How do you know when to use ( )?

[shell token]

@alpine sable if u need to when we are making at least two nested groups for calculation

did u understand it

[boreal jetty]

in your exemple: 2(x-15) can be simplified. You have to 'times' 2 with every term in the paranthesis.
so: 2(x-15) -> 2x + 2(-15)

[modern sedge]

"the difference of the product of 2 and a number and 15"
Is in some way analogous to
"the box with package of eggs and with an apple", which obviously refers to something like the image below:

In the same way, "the difference of the product of 2 and a number and 15" refers to (2 * x) - 15, note that the 2 * x is packaged in parenthesis

[alpine sable]

Ohhh okay, thank you

[shell token]

your welcome

[boreal jetty]

Please start this course if you want to learn more
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:combine-like-terms/v/combining-like-terms

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[prime glacier]

That's not what 1 less than Or equal to i, j less than Or equal to 2 means

[prime glacier]

It means that thr matrix is a 2x2

a11, a12, a21, a22

Are the only possible values, aka a 2x2 matrix

On top of this since it's an intersection from open 0 to 4

1, 2, 3 are the only possible values

Meaning 4 Is the minimum value and 12 is the max

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[prime breach]

Hi, how do I do this question?

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[woven turret]

how do i go on about doing this difficult problem

[placid zinc]

This is a double integral. The region is that triangle, the integrand is that function

[woven turret]

yea ik its a double integral but are there 3 points

[placid zinc]

Yeah they make a triangle. This is your region that you'll be integrating over

[woven turret]

but how am i going to do 2 integrals with 3 points

[placid zinc]

Well, I'd start by drawing a picture of the region

Then I'd try to describe the region with inequalities

The region can't be covered with a single integration. You might want to split it into two, or use a change of variable

[woven turret]

how can i do that

[amber elm]

Plot the points on a graph

And then draw a triangle connecting the points

[woven turret]

ok

any idea how to plot pi,pi on a graph?

[amber elm]

Go to the right π units and go up π units

[hushed locust]

it should be a little bit farther up and right of 3,3

[woven turret]

can i use 3.14 will it be the same

[amber elm]

Pretty much

[woven turret]

ok

[amber elm]

Since its just a sketch

[woven turret]

ok i drew the triangle

now what

@placid zinc

anyone

@amber elm

.close

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[summer pumice]

need

[summer pumice]

.close

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[robust bronze]

totally lost cuz i forgot how to do this 🧍 a walkthrough would be helpful since i need to know how to do this on finals

[simple wigeon]

set both equations equal to 90 degrees since they add to a right angle and find x
after you find x just plug it in for the angle bdc

[robust bronze]

thanks!

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[simple wigeon]

ofc ofc

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[cerulean garnet]

i dont get it, how does limit with x approaching 1 from the right of g(h(x)) equal to -2

shouldn't it be 0?

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[lone heart]

Please don't occupy multiple help channels.

[winter olive]

Can someone explain what I did wrong in the above problem where I’m multiplying complex numbers?

[hollow willow]

it's not multiplication

[winter olive]

YOURE RIGHT

ugh i realize im bad at math bc i keep messing up these little details

thank you, let me do it again really quick and see if i get the correct answer now that i know..

[hollow willow]

it happens to everyone

[winter olive]

you re right, once i just did the subtraction i got it correct. thank you again

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[crimson mesa]

how i do this? $\begin{bmatrix}
-2 & 2 \
2 & -2
\end{bmatrix}_{m2}$

[ocean seal]

Lil

[crimson mesa]

$\left[ \begin{pmatrix}
-2 & 2 \
2 & -2
\end{pmatrix}\right]_{m2}$

[ocean seal]

Lil

[crimson mesa]

figured out, it was easy

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[devout jetty]

If anyone could tell me if I did good or not(pretty sure I did badly) and point out which ones wrong that would be much appreciated

[devout jetty]

Does that mean 3 are wrong or number 3 is wrong

Ohh

Ok take ur time thx for the help

[pseudo ice]

3 and 9 so far are wrong

[devout jetty]

So the rest r good so far right ;-;

[devout jetty]

Anything else?

[pseudo ice]

And 13

[devout jetty]

Anything else

Or that's all that's wrong?

[pseudo ice]

4 and 5 (thanks Garlic )

[devout jetty]

OK so 3, 4,5, 9, 13

And the rest are right

Ok ill redo them

Thanks for the help

[pseudo ice]

Well some of them I haven't checked yet still

[devout jetty]

Well if u find more please tell me;-;

[pseudo ice]

I'd also check 11: remember that f^{-1} means the inverse function of f

Other than that, seems fine to me

[devout jetty]

Ok thanks

Anyways I THINK I did it good now

If you can check it that would be nice (unless you're busy you were an already big help anyways)

[pseudo ice]

3 and 13 are alright now

And 4 as well actually

For 9, remember you don't want either of the denominators being zero

[devout jetty]

Ok

[devout jetty]

Alr

So what's left is 9 5 13 alr

Wait no

9 5 11

[pseudo ice]

just 5 and 9 need to be checked

[devout jetty]

Ok thanks and 11

[pseudo ice]

Oh yea, that too forgot about 11, but yea

[devout jetty]

Ok xD

These emojies look so funny

@pseudo ice (if you don't like pings tell me and Ill stop) I think it's good

[pseudo ice]

well, my bio does say to ping while replying
And you didn't answer 11, the other two are still no

[devout jetty]

Oh oops meant to put 2

[pseudo ice]

For (9), notice that x^2 - 1 has roots 1 and -1, but x^2 - x - 2 has roots 2 and -1, so you can't have -1, 1, or 2, hence it was (b)

For (5), you may e.g. want to complete the square or use the quadratic formula for it

[devout jetty]

Ohhh now I understand

[devout jetty]

Ok

How about 11for 2?

[pseudo ice]

Hmm, I wouldn't say it's 2, but then now those options look kinda sus

Unless I'm fried, it's none of them

[devout jetty]

So 5 is d

And 9 is b

[devout jetty]

Lol-

Thank you so much for the help

So pretty much everything is right but 11

[pseudo ice]

Yeeaaa, unless there's something I'm missing, $x = f^{-1}(2)$ if and only if $f(x) = 2$, so you want to solve the latter for $x$

[devout jetty]

Ok

[ocean seal]

@pseudo ice

[devout jetty]

I think I'll close it now then

So I pretty much get 14/15 if I don't solve 11

Right

[pseudo ice]

Yea, though I'd remove 11 from the list tbf

[devout jetty]

XD

Thank you so much f for the help chart

I'll close it now.

.close

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[sonic wharf]

Consider a random walk where steps are decided based on coin tosses. A head is a positive step and a tail is a negative step. After 10 steps, the random walker finds himself at the same position as where he started. In how many ways, can this be done?

[tribal field]

Well, if you end where you've started, you've taken the same amount of steps forward as you have backwards

[sonic wharf]

I understand, but how do i go on from there?

[tribal field]

Well then, how many forwards and backwards steps are there?

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[eager iron]

Hi sorry last minute question before exam

[eager iron]

In linear algebra if we have a quadratic Matrix nxn, if the determinant isn’t 0 then the Matrix is injective, surjective and bijective correct?

And Quick check injective means that if f(u) = v and f(w) = v then u=w for it to be injective, aka we can only map one input vector to one output vector Yeah?

And surjective means that we must have a mapping to every possible output in the dimension we’re going to?

Well if my understanding is correct so far, does that mean that the function F(x) = Ax = Y is linear when A’s determinant isn’t 0? Assuming A is a nxn Matrix?

Aka there exists an inverse to F and by extension A?

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[eager iron]

<@&286206848099549185>

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[ocean seal]

kisnar

kisnar

[livid sage]

you can use a trig identity to solve this integral

actually sorry that's not necessary*

does the integral of (secx)^2 look familiar to you?

do you have a list of derivatives?

i assume it should be on there

so true

that's interesting

yeah that's pretty much the idea

no

i was just wondering what you had

normally this is info given to students

hm

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[minor pawn]

Why is U{a}= a

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[valid merlin]

I'm just guessing because I can only see like half the page, but I assume that its saying that the union of a is a??

[minor pawn]

Ya

But what is with the bracket

[valid merlin]

just to show that thats all they are taking the union of

ig, so you dont get mixed yp

[minor pawn]

Wait

U(a) =a,
I have somewhere read that to remove the brackets you use union

[valid merlin]

okay, im not really following but if you are then nice 🙂

[minor pawn]

Ok

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[indigo anchor]

Hew how do I do A?

I assume I have to get the left side to look like the right side, and I know how to equate coefficients

Im just not sure how to get it to look like that

[earnest saddle]

Did you try simplifying?

[indigo anchor]

i tried simplifying right side

and it didnt match the left

[earnest saddle]

what did you get?

[indigo anchor]

sorry expanding

not simplifying

[quasi vector]

you can simplify the left side instead

[earnest saddle]

x^3 terms cancel out ig

[quasi vector]

or expand the right side it doesnt matter

[indigo anchor]

i expanded right side and got 8x^3 - 8a^3 + b

i think i did it wrong

[quasi vector]

nope

$(a+b)^3\ne a^3+b^3$

[ocean seal]

kheerii

[earnest saddle]

Yes you did

[indigo anchor]

yeah

so like

(a + b)(a + b)(a + b)

ah got it

[quasi vector]

you've never encountered a problem where you've had to expand $(a+b)^3$?

[ocean seal]

kheerii

[indigo anchor]

i have lol, just had a brain forgetting moment

[ocean seal]

Pro_Hecker

[indigo anchor]

sometimes this server gives a bit of a refresh

i will be back and let you know if I get matching sides

[indigo anchor]

could I just plug it into this?

[indigo anchor]

this

[quasi vector]

yes

[indigo anchor]

because i am expanding and am getting x^2

and ax

im not getting x^3

[quasi vector]

how

[indigo anchor]

now i think about it i have onyl ever expanded brackets squared

so I am not sure how to do this

[indigo anchor]

idk

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[bold silo]

how would i do this question? i have no clue where to start, the set notation is throwing me off

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[bold silo]

<@&286206848099549185>

[junior vigil]

$fg=f \circ g$

[ocean seal]

🫎MooseyMooseMooser 🫎

[junior vigil]

top line is saying 'f maps x to 3x+ln(x), for x>0, x in real numbers', so essentially another way of saying f(x)=3x+ln(x) for x in positive reals

[bold silo]

so how do i solve it?

[bold silo]

<@&286206848099549185>

[royal grail]

are you confused about the part where $fg(x) = x^2 + 3e^{x^2}$, or the part about differentiating $fg(x)$?

[ocean seal]

lgkoo

[bold silo]

didnt they already differenciate it here?

[royal grail]

I don't think so

Try differentiating fg(x) yourself

[bold silo]

what is this?

[royal grail]

just differentiate fg(x), and equal them and solve it

[bold silo]

so 2x+6xe^(x^2) = x(xe^(x^2) + 2)

[royal grail]

yes

now simplify the equation and see what you can spot

[bold silo]

x=6?

[royal grail]

there should be 2 soln I think

since you have a quadratic

[bold silo]

i did divide by x

[royal grail]

yh, never do that

always best to write as x(x-6) = 0

so x = 0 or x = 6

[bold silo]

ight thanks for the help

[royal grail]

np