help-10

[left mesa]

is that wrong?

[coarse ibex]

that lets you "cancel the powers"

here we "cancelled the base"

i.e. we went from 3^(3x+1) = 3^(2x) to 3x+1 = 2x

[left mesa]

so it's one to one because? Sorry I'm pretty slow

[coarse ibex]

I'm saying that we can do the cancelling because f(x) = 3^x is one-to-one

[left mesa]

Ohhh

wait but why 😭 how do you consider something one-to-one again? Sorry it's just our teacher focused on solving more and she didn't tackle on this one that much

[coarse ibex]

one-to-one means f(x_1) = f(x_2) implies x_1 = x_2
we use that here with x_1 = 3x+1 and x_2 = 2x

[left mesa]

ohh

Ohhhhhh Thanks!

I at least understood it now I just hope I remember... Considering how our gen bio that consists of 1000 things to remember I'll probably forget abt this 😭

Anyways thank u so much ((: Have a wonderful day!

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[thorn marsh]

Proof by induction: 2^n>n^2 for all integer n greater than 4

[thorn marsh]

This is the inductive step of the proof

I'm having trouble understanding why (n-1)^2 is used. And what exactly are the first and second inequalities here?

[marsh geyser]

I think you have a typo

[thorn marsh]

oh

[ember frost]

2n > n^2 is false

[thorn marsh]

yeah

sorry i changed it

i just took it from stackexchange

still having trouble with understanding what they did

[gilded needle]

what is actually being proved here

can you show the problem statement

[thorn marsh]

Proof by induction: 2^n>n^2 for all integer n greater than 4

[gilded needle]

ah got it

[thorn marsh]

that is the problem statement

[gilded needle]

i suspect that when putting the proof together, they started with 2n^2 > (n+1)^2 and worked their way backward, and found that it reduced to (n-1)^2 > 2

that's usually what happened when someone pulls a fact out of the air like that

[thorn marsh]

can you explain this

this is what i pieced so far for what they meant for inequality 2

[gilded needle]

in the highlighted line, they claimed that 2n^2 > (n+1)^2 for n >= 5

since it's not obvious why that is true, they need to provide a proof

they probably started with 2n^2 > (n+1)^2 and rearranged it to get (n-1)^2 > 2

and said aha, but that is true

then when they wrote the proof, they did it in the reverse order

[thorn marsh]

i'll probably take a minute to process this and then reply to you

ok now i am confused as to where did 2*n^2 come from because
I am looking at it from this inductive step

and how did he get that whole inequality to equal 2^(n+1)

is he assuming that 2* 2^n> 2*n^2 as 2^n >n^2?

In that case it does make sense

but why is there an equal sign

that makes no sense

oh

nvm

i think i got it now

[gilded needle]

yea that should be the least controversial part haha

[thorn marsh]

it's very confusing to wrap my head around this

[gilded needle]

it's not a very clearly written proof

[thorn marsh]

can you show me how they reduced it to to (n-1)^2>2 or provide some insight

like how did they come up with (n-1)

[gilded needle]

just reverse these steps:

i.e. start with the last line and do the opposite manipulations

all four lines are equivalent

[obtuse pebble]

@thorn marsh Has your question been resolved?

[thorn marsh]

yes

[obtuse pebble]

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[sinful falcon]

Is this correct

[sinful falcon]

i jiust need to find derivative of

$\ln(xe^{-2x})$

[warm shale]

Light

[sinful falcon]

$x* -2e^{-2x} + e^{-2x}*1$ isnt that what i did

[warm shale]

Light

[sinful falcon]

can i remove e^-2x

i dont think i can

so thats just it right

oh ok

,w diff of ln(xe^(-2x)

,w diff of ln(xe^(-2x))

[warm shale]

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[sinful falcon]

ty

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[sinful falcon]

.reopen

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✅

[sinful falcon]

wait

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[sinful falcon]

is derivative of that first part

my last part

is that righ

im nots rue

obut i think im right

[junior granite]

derivative of arscosine is negative of 1/sqrt{1-{thing}^2}

[sinful falcon]

thought it was this

[junior granite]

yeah

[sinful falcon]

hea so i took derivative which is 1/2sqrtx

and then over that

which is sqrt 1- sqrtx^2

[junior granite]

yeah
other part is alright

just that negative was off

[sinful falcon]

oh i see

i forgto nebatyive

negative

.close

thnk

[obtuse pebble]

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[sinful falcon]

thank

.reopen

[obtuse pebble]

✅

[sinful falcon]

is this right

[timid silo]

https://tenor.com/view/tank-m60t-turkish-tank-tsk-syria-gif-23298742

[sinful falcon]

find derivative of $e^{sin^-1(z^2)}$

[warm shale]

Light

[sinful falcon]

am i righ

[junior granite]

how the heck that extra "e" came to existence?

[sinful falcon]

ihg yeah

oh

i do not know

goof[y brain or soemthing bruh

[junior granite]

the way you factored out that "e" is scary tbh'

[sinful falcon]

it would just be 2e right

here

i wil show

[junior granite]

huh?

[sinful falcon]

no?

because i take dydx of sin^-1

right

[junior granite]

$\frac{d(e^{f(x)})}{dx}=f'(x)e^{f(x)}$

[sinful falcon]

got it

.close

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[sinful falcon]

thank u

[obtuse pebble]

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[quiet plover]

I need help figuring out how to solve these

[quiet plover]

If anyone helps can you @me?

[slow prawn]

Give a number or letter you wanna solve first

[quiet plover]

2

[obtuse pebble]

@quiet plover Has your question been resolved?

[eternal moth]

Read about parallel line angle theorems, ive attached a page that discusses some which I assume are all that are needed for your questions. https://thirdspacelearning.com/us/math-resources/topic-guides/geometry/corresponding-angles/

[obtuse pebble]

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[hidden river]

This type of questions, here to proceed i squared both of the sides( sinalpha + sinbeta + singamma )^2 = 0 then i'm stuck, How do i use the x= cosalpha + i sinalpha

[restive acorn]

Which part are you workkng on?

[hidden river]

First

cos2alpha + cos 2beta + cos2gamma=0

[restive acorn]

Did you try double angle identity?

[hidden river]

cos^x-sin^x one?

OH MB

^2*

[restive acorn]

Yea that would do

That's the first thing I would try

[hidden river]

Yes and then i squared the given equations, but how do i proceed

we do the sin^2 terms and cos^2 terms, but 2(sinalphasinbeta+ ... )

[restive acorn]

Can you do me a favor and write it expanded from the identity

Don't do any additional algebra just invoke the double angle one

[hidden river]

You mean like
cos^2 alpha - sin^2 alpha + cos^2 beta - sin^2 beta + cos^2gamma - sin^2 gamma?

[restive acorn]

Yea

[hidden river]

I'm using laptop is it okay if i just type?

[restive acorn]

It's fine I'll try to see it in a sec

[hidden river]

okayy

[restive acorn]

I'm out rn hard for me to think about it

[hidden river]

Oh

( sin alpha + sin beta + sin gamma)^2 = 0 (1)
sin^2 alpha + sin^2 beta + sin^2 gamma + 2(sinalphasingamma + sinalphasinbeta + sinbetasingamma) = 0

( cos alpha + cos beta + cos gamma)^2 = 0 - (2)
cos^2 alpha + cos^2 beta + cos^2 gamma + 2(cosalphacosgamma +cosalphacosbeta + cosbetacosgamma) = 0

subtracting 2 from 1

cos^2 alpha - sin^2alpha + cos^2beta - sin^2beta + cos^2 gamma - sin^2gamma +2 ( cosalphacosgamma +cosalphacosbeta + cosbetacosgamma - ( sinalphasingamma + sinalphasinbeta + sinbetasingamma ) = 0

[obtuse pebble]

@hidden river Has your question been resolved?

[obtuse pebble]

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[final light]

i know the volume is 2970meters cubed, just dont understand the second part

[final light]

the trapezium 80 percent full is 2376

wait nvm

im stupid

[candid dawn]

demn

[final light]

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[dark agate]

Help

[obtuse pebble]

Please don't occupy multiple help channels.

[dark agate]

my ans

[wary badger]

$\log_2(5) - 3\log_2(a) + \frac{7}{3}\log_2(9)$

[frigid burrow]

Is this the question?

[wary badger]

it’s supposed to have 5

she stole the question from someone else

https://discord.com/channels/268882317391429632/903481106819612714

[frigid burrow]

Ohk

[frigid burrow]

What is your answer?

[warm shale]

knief

[dark agate]

Log2(-5xa^3x9^(7/3))

[wary badger]

please don’t use x for multiplication

[frigid burrow]

Use * or •

[wary badger]

and no

why -5

[frigid burrow]

Yeah

[dark agate]

log2(-10935*a^3)

[frigid burrow]

No minus

[brazen viper]

No

[frigid burrow]

Log of negative numbers isn't defined

[wary badger]

$\log(a) - \log(b) = \log(\frac{a}{b})$

[warm shale]

knief

[dark agate]

ohhhh

[frigid burrow]

Similarly for addition it's *

[dark agate]

Log2(5/(a^3)*2187)

am I right?

[wary badger]

,w 9^(7/3)

[warm shale]

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[wary badger]

2187?

[frigid burrow]

,w log_3 2187

[wary badger]

,w 3^7

[warm shale]

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[frigid burrow]

Nice

[wary badger]

Nice

[wary badger]

no

[dark agate]

How do I do it??????

[frigid burrow]

7/3 log_2(9) becomes?

[dark agate]

9^{7/3}

ohh that's the mistake

[wary badger]

yup

[dark agate]

ty guyss

[wary badger]

what’s your final answer

[dark agate]

168.49*x^3

[obtuse pebble]

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[cunning parrot]

e

[cunning parrot]

Is the theorem able to have wlog because proving for y=2k is the same thing as

proving for x right

[wary badger]

oh you’re asking for confirmation that that’s the reason why

yea

[cunning parrot]

gotcha

[cunning parrot]

ty

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[near saddle]

$V = \frac{\pi}{3} \cdot \frac{h^3}{16}$

Then following the product rule

[wary badger]

you don’t need the product rule

r can be written in terms of h

[near saddle]

ooo

[wary badger]

similar triangles

[near saddle]

that saves so much effort

h = 2r

[wary badger]

h = 2d

not r

[near saddle]

ohh

so its

[wary badger]

but d = 2r

[near saddle]

h = 4r

[wary badger]

so r = ..

[near saddle]

r = h/4

r^2 = h^2/16

[wary badger]

put that here

[near saddle]

or h^3 / 16

[wary badger]

/48

$V = \frac{\pi}{48} h^3$

[near saddle]

ooh

3*16

[warm shale]

knief

[near saddle]

that's simpler

[wary badger]

or

you could just leave it

because power rule gets rid of the 3

🤷🏼‍♂️

[near saddle]

either way is fine

$dV = \frac{3\pi}{48} \cdot h^2 \cdot dh$

[warm shale]

smeagol

[near saddle]

dV = 20 ft^3/min

h = 15

$\frac{64}{45\pi} = dh$

[warm shale]

smeagol

[near saddle]

Yay that's what I was supposed to get :)

ty!

somehow in my previous attempt I got
16/45pi

[wary badger]

dv/dt and dh/dt*

[wary badger]

you’re welcome

just make sure to write dv/dt and dh/dt

not just dv and dh

[near saddle]

gotcha

[near saddle]

(because I did 2r = h instead of 4r = h)

.close

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[south star]

Can anybody help me please 🙏

Anyone???

[final steppe]

what do you need help with

[south star]

This question

[final steppe]

what have you solved so far

[south star]

Nothing 🙂

[final steppe]

well what do you get

what can you interpret

[south star]

I can't solve this problem so i need help

[final steppe]

ok what don't you get

like did you get stuck somewhere?

[south star]

Yes all the problem

[final steppe]

ok well

when you divide a fraction by another number

that dividend comes up and gets multipled

so in the case of 544 you can bring it up to multiply with (6/11+2)

from there onwards just use pemdas/bodmas/bidmas whatever it is you use

[final steppe]

sorry i mean divisor

[final steppe]

you can convert 2 into 22/11

and then add

then multiply the 544

in the numerator

and then multiply the 10/816 to the -6/11

add (28/11)544 to the product

and then you'll get a new denominator

btw do this with the numerator you already have

and then multiply the new denominator to the numerator you already have

im sure you could also just simplify

[south star]

Thanks bro 🙏🙏

[final steppe]

np

[south star]

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[fierce vale]

$A,B,C \subset E \$Prove $A\Delta B=A\Delta C \iff B=C$

[simple ermine]

what does delta mean here ?

[warm shale]

Adam Ch.

[simple ermine]

what does the sign delta mean in the question

scalar product ?

[simple ermine]

oh a logic problem

[obtuse pebble]

@fierce vale Has your question been resolved?

[obtuse pebble]

@fierce vale Has your question been resolved?

[obtuse pebble]

@fierce vale Has your question been resolved?

[fierce vale]

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[arctic sage]

My brain is fried, simple thing tho

4 whole numbers must equal 140

• numbers 1-8, one time use

(multiply only)

[void quartz]

hint: 140 = 20 x 7

[arctic sage]

ty

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[bronze flame]

halp i have no idea how to do this

[obtuse pebble]

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[fleet dock]

Let f(x)=x^3+x+1. Suppose g is a cubic polynomial such that g(0)=-1, and the roots of g are the squares of the roots of f. Find g(4)

[fleet dock]

I can't find a good approach to this problem

I think that if I figure out the first step I can do the rest, any ideas?

[obtuse pebble]

@fleet dock Has your question been resolved?

[fleet dock]

<@&286206848099549185>

[obtuse pebble]

@fleet dock Has your question been resolved?

[obtuse pebble]

@fleet dock Has your question been resolved?

[obtuse pebble]

@fleet dock Has your question been resolved?

[obtuse pebble]

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[vital verge]

can i have some help with 20b i keep on getting y=7

[vital verge]

.close

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[lime venture]

How can I get better at 3D trigonometry just the basic stuff with the sine and cosine rules etc.

[lime venture]

<@&286206848099549185>

[timid panther]

3d and trigonometry?

Anyway what I can tell you is simple. Noone can make you better at anything, only you can do that. Just keep practicing that's the cure.

[lime venture]

Yeah I mean where do I practice

[timid panther]

You must have questions in your books right?

[lime venture]

I'm relatively good at maths compared to my year group anyway I just feel like I want extra practise

[lime venture]

yeah I was thinking something more on the digital side

[timid panther]

You always need practice dude dont think you are good at maths. Maths is never ending.

[timid panther]

Use books

[lime venture]

Yeah that's why I'm here

[lime venture]

Well alright then

[warm dome]

#book-recommendations

[lime venture]

Thanks

[obtuse pebble]

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[obtuse pebble]

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[wintry lark]

found the domain is [-4,2) union (2,4]

but how would i determine the rnage of it

little confused

[timid silo]

you can differentiate the function to find the maximum and minimum

that is one way

another way is to do it intuitively

another way is to do it graphically by plotting it and seeing where the value is maximum and minimum

,w plot y = (sqrt{16 - x^2} / (x-2))

[warm shale]

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[timid silo]

it doesn;'t seem like this function has any specific maximum or minimum

[trim portal]

look at the points of discontinuity ig

[wintry lark]

wait is my domain answer right

[wintry lark]

this

[trim portal]

yes, it is

sorry i read it wrong

[timid silo]

yup yup

[wintry lark]

alright

[trim portal]

at -4 the value of the function is 0, and it will decrease after it

because sqrt(16-x^2) is gonna be always positive

[wintry lark]

yep that's how i got my domain

[trim portal]

and x - 2 is gonna be negative for x around -4

[wintry lark]

range looks like il just have to do it intuitively

[trim portal]

yeah

next point of discontinuity is 2

at that point, the function jumps from -infinity to infinity

and then the last one is 4

where it's 0 again

so it's gonna look approximately like this

very approximately

[junior granite]

(you could also make a trig sub to figure out the range, imo it'll be a bit easier to transform this function into something that know the range of than analysing the function in its base form)

[obtuse pebble]

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[fierce ice]

I'm not sure if this is the right channel to ask in, but I wanted to ask if someone would be willing to explain to me what the XOR operation does and how it's defined inside of Set notation? I'm a little lost

[fierce ice]

In bitwise arithmetic, I understand what XOR does, but I have no idea if that carries over to the topic of sets.

[last pilot]

like if you perform S OR T, it returns all elements in S or in T but not both

[warm shale]

ren

[last pilot]

XOR returns all elements such that they are in either S or T but not both

[fierce ice]

ahhh

okay so it is equivalent to the arithmetic definition

thank you!

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[violet yarrow]

i got 14x-1 for 15, which sounds right but this is such a weird question

[lunar nymph]

no

check again

[violet yarrow]

i ended up getting 19x+2

after redoing it

still wrong

[frigid burrow]

Yup wrong

Can you name the figure

[violet yarrow]

it's just an irregular one

[frigid burrow]

And then share each side length?

[violet yarrow]

i'm going to be honest, i haven't done this in awhile

[frigid burrow]

Like ABCDE

[violet yarrow]

wait

i could split it into 2

but it doesn't give the top

[violet yarrow]

no

it's asking for perimeter

[frigid burrow]

[violet yarrow]

i have no clue 😞

[frigid burrow]

I am saying something just like this

[violet yarrow]

ohh 😞

no i've never done that

[frigid burrow]

(tu ru ru tu ru ru)

[violet yarrow]

not that i can remember

[frigid burrow]

Do you not know how to name polygons?

Just write a letter beside the corners

[frigid burrow]

@violet yarrow ?

[violet yarrow]

i do

[violet yarrow]

this just made so much more sense 😭

[lunar nymph]

glad to hear that

[violet yarrow]

you would add 5x-3 + 3x, giving you 8x-3,no?

[lunar nymph]

Yes!

[violet yarrow]

yay!

[lunar nymph]

YAY!

and now what do you get

[violet yarrow]

8x-3+3x+2x+4x+2

[lunar nymph]

there's one more side left

[violet yarrow]

for now

[violet yarrow]

it would be another 4x+2

i split it

ignore the 3x, i erased it

[lunar nymph]

Yes!!

[violet yarrow]

so then just add the other 4x+2 and 2x?

[lunar nymph]

Correct!!

[violet yarrow]

okay!

[violet yarrow]

23x+1

wait

[lunar nymph]

oh you also forget the 5x-3 side

[violet yarrow]

it would 28x-2

[lunar nymph]

Yes!

you get it

[violet yarrow]

okay! so area is a=w•l

[lunar nymph]

yes, this is the equation for rectangle

[violet yarrow]

for the perimeter what would be the width?

[lunar nymph]

try spliting the shape to small rectangles first

[violet yarrow]

so 6x+6x+2x+2x

[lunar nymph]

i don't understand

For example, you could split it like this

[obtuse pebble]

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[obtuse pebble]

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[hard imp]

anyone got any pointers on how to solve this?

[hard imp]

i'd start by converting 0,2 and 0,04 to fractions

if not, where should i start?

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[left lotus]

So I have a type of questions saying
[Translation] find a generic (/general) term of numerical sequence {x_k}, which is defined
x_1 = A
x_(k+1) = B*x_k+C
(A B C depend on the problem but they're integer constants)
And calculate their sum, s_k, of terms from 1, to k.
I don't really know how to go about solving it
I didn't really do well in a school so I'm trying again but it's really hard

[obtuse pebble]

@left lotus Has your question been resolved?

[left lotus]

<@&286206848099549185> hii

[fallen elm]

hi

done u got help

[left lotus]

What does that even mean

I didn't specify A B and C but the answer turns out to be a constant?

[fallen elm]

.reopen

[little jewel]

Try summing the equation on both the sides from k=1 to k=n

[left lotus]

Wait, whats about the first question of finding a generic term?

[little jewel]

Oh ok I misunderstood and thought that was a translation problem

But yes you'd need that too to solve the s_k

It means we have to find x_k in terms of A,B and C (and k)

[left lotus]

Oh translation means I translated it from my native language

[little jewel]

Like x_1 = A

x_2 = BA + C

And so on

What would be x_k

[left lotus]

x_1 is A,
x_2 is BA + C
x_3 is B(BA + C) + C
= BBA + BC + C
x_4 is B(BBA + BC + C) + C
= BBBA + BBC + BC + C

So, um let me think

Do we use Σ? Like
k-1
Σ B^n
n=0
Then multiply that by C, and add B^k * A?
Umm but it mismatches if I put k=4

[little jewel]

Yes we use summation

Separate the first term from the rest and then see if you can see any pattern

First term is easy to guess for x_k I think

Do you know sum of a geometric series?

Like C + CB + CB^2 + CB^3 + ... CB^(n-1) = ?

[left lotus]

are you asking if I can write it with sigma?

[little jewel]

Hmm like do you know the sum of that series for n terms?

The formula

[left lotus]

Hmm?

[little jewel]

Ok do you have any idea of

[left lotus]

There is a shorter way to write it?

[little jewel]

What would be the sum of 1 + 3 + 9 + 27 + ... upto 10 terms?

[left lotus]

I think I got it, let me write it down

[little jewel]

Yes, it solves out to a single short formula

[left lotus]

In base 3, it would be
1, 10, 100, 1000
Then adding them results in
1111
Add one more 0 to the biggest term
10000 then subtract 1
So it's 2222, divide by 2 and it's the answer

It's embarrassing if it's wrong :p

For base b, sum up to k is
(b^(k+1) - 1)/(b-1)

[little jewel]

Oh thats a rather cute / unique way to do it XD

But yes!

That formula is right

Except

For 1 + b + b^2 + ... b^(k-1) = (b^(k) - 1)/(b-1)

[left lotus]

Ok!!

[little jewel]

ok now do you have any idea for how you can simplify this?

[left lotus]

For k = 4, I can separate to BBBA and BBC + BC + C, then bring C to the back (BB + B + 1)*C
Then it's just what I did earlier
The biggest exponent is 2 (= k-2)
(B^(k-2+1)-1)/(B-1)*C
The first part is A*(B^(k-1))
So the sum is (B^(k-1)-1)/(B-1)*C + (B^(k-1))*A ?

[little jewel]

Yesss

Except I don't get why you wrote it like (B^(k-2+1)-1)/(B-1)*C

There are n = k-1 terms and the formula puts the n as the exponent so u can directly write it like k-1

But maybe you meant it like the highest term exponent (B^2 = 2) + 1 = 3

[left lotus]

I just used this one
Here k is the biggest exponent

[little jewel]

That's wrong tho

[little jewel]

Notice how there's k here

I'm talking about 1 + b + b^2 + .. b^(k-1) aka k terms

Maybe you're talking about k + 1 terms like 1+b+b^2+...+b^k ??

[little jewel]

Because your final answer here is correct

[left lotus]

Oh sorry I named the variable k too
For k terms highest exponent is k-1 so your k-1 corresponds to my k

[little jewel]

Oki

[left lotus]

But I think I got the main idea!

[little jewel]

Yupp

So you already found the x_k

Now try summing the equation on both the sides for k=1 to k=n and you'll find s_n

[left lotus]

It's getting late so I'll do it in tomorrow (actually todays) morning
Nice help <3 I appreciate your kindness
But I'm just happy I've made some progress

.close

[obtuse pebble]

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[bleak chasm]

hello
I have a question about vectors.

Assume we have two vectors:
u = (1;1;1)
and vector v

I want v to be parrallel and equal to vector u but not on the same line.

What do I need to have?

[simple ermine]

wdym not on the same line ?

also for vectors we do not say parallel we say colinear

[bleak chasm]

it means the points of the vector are not on the same,
like AB and CD are two vectors that have to be colinear and equal

[simple ermine]

not on the same what ?

[bleak chasm]

Suppose we have a line AB,
and we have another line CD

They need to be equal and colinear

[simple ermine]

well

if you want to find a parallel vector that is not equal

you can multiply the three coordinates by a same number

for ex if you have v(1,1,1) ; u(2,2,2) is colinear but with unequal coordinates

idk if that is what you are asking for

[bleak chasm]

but their length is not equal

they need to be equal
vector AB has to be equal to vector CD

[simple ermine]

wdyouwant then

[simple ermine]

then they are the same vector

[bleak chasm]

I need to find a vector AB = CD

and also |AB| = |CD|

[simple ermine]

if vector AB = vector CD then the vector is the same

a vector is not static

[bleak chasm]

so how do I find it?

[simple ermine]

wdym hw do you find it

it is the samer

same*

[bleak chasm]

I have vector u(1,1,1)
how do I find vector v that's equal and colinear to vector u?

[simple ermine]

it can only be u then lol

[bleak chasm]

<@&286206848099549185>

[simple ermine]

v has the same coordinates than u

bruh i'm answering your question

[simple ermine]

listen to me

if the vectors have the same coordinates then they are the same vector

[bleak chasm]

@hot hazel

[hot hazel]

Please do not ping individual users for math help

The response you've already gotten is correct in any case

Formally vectors have no "placement" on the plane; by convention we often position them at the origin

But we could place them anywhere

And they'd be the same vector

[bleak chasm]

but say I want a vector that is equal in length and colinear to another vector
how would I find it?

[hot hazel]

It's the same vector

Idk what else to say

Those two conditions imply the vectors are equal

[bleak chasm]

there's only one vector on a plane that has the same length and is colinear to a vector?

[hot hazel]

Oh I guess you could have a collinear vector in the opposite direction

Like (1, 1) vs (-1, -1)

[bleak chasm]

ok

thank you

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[opaque mesa]

There were 48 students from different classes. 6 of them had 1 friend from their class, 9 of them had 2 friends from their class, 4 of them had 3 friends from their class and the rest didn't have any friends from their classes. How many classes were there?

[haughty stump]

is this an actual problem or did you like find this from tiktok and want an answer

[zenith raft]

[obtuse swan]

I mean it is a problem in combinatorics techincally

Let me solve it

So if 6 of them had a friend form their class that means there was 3 pairs of people from same class

[haughty stump]

/shrug

[obtuse swan]

?

9 means there were 3 triplets

And 4 means there was one 4

[glass dagger]

I agree with this

[obtuse swan]

The rest i.e. 48-6-9-4=29 were solo

So 29+4+3+1=37 classes

But looks like some tiktok fr

Or instagram reels

Answer is sus

[obtuse pebble]

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[fossil raven]

Could anyone check if those derivatives are correct please ? 🙏

[obtuse pebble]

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[flat shell]

I was finding the gradient of F to get the normal vector and everything was right except the Z part of the gradient, which apparently is -1. Can someone explain why that is? I checked the answer and apparently we subtract z in the original equation?

[obtuse pebble]

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[plain tiger]

how do you do this one?

[tough reef]

Do you know the formula for the derivative of the inverse function at point x?

[plain tiger]

(f-1)'(x) = 1/f'(f-1(x))

[versed pier]

perf

juse use that here

[plain tiger]

yeah i was having some trouble finding the inverse of f

i couldnt isolate for y

[versed pier]

well is there even an inverse?

[dry anchor]

I think the inverse of this function is a little annoying though right?

[plain tiger]

yeah

i couldnt isolate for y

so i got stuck

[versed pier]

you don't even need the inverse

if you have a point (x,y) on the graph f

[ripe jay]

dont find the inverse

[versed pier]

f^-1(y) = ?

[plain tiger]

oh

[plain tiger]

x

so f(x) = 90

idk what to do guys.......

[obtuse pebble]

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[idle patio]

I don’t need an answer I just wanna know what type of math is this and does anyone recognize it

[empty musk]

It's just the graph properties of x^n and -x^n

[idle patio]

How does it work?

[empty musk]

So you know if you square a number it's always positive, yes?

[idle patio]

Yea

[empty musk]

That works for every even power

[idle patio]

And odd is negative?

[empty musk]

thats why it says that for x^n (even) the graph is only above Ox (the result is always positive)

[empty musk]

Odd powers dont change the sign

- * - = +

and if you multiply it again with a minus you get a minus back

It's just a game of signs

[idle patio]

Hm ok thanks for the help

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[bitter terrace]

Why is 17 false?

[bitter terrace]

it makes intuitive sense

because if they have the same vector field

then at each point on that vector field the derivates must equal each other

meaning that the two systems would need to be equal

and 18 would be true since it does take into account the magnitude?

[obtuse pebble]

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[eternal monolith]

quick question on rationalizing denominators

[eternal monolith]

lets say for example i have a cube root in the denominator, could i multiply it by that expression squared to rationalize the denominator

[glass dagger]

Sounds like it could work. Have you tried it?

[eternal monolith]

not yet

[glass dagger]

Maybe you could try it for yourself :)

[glass dagger]

,rotate

[eternal monolith]

Does that seem right?

[eternal monolith]

hold up

thats wrong my bad

im brainfarting rn

[glass dagger]

I was gonna show that they're not equal by plugging in t=0 but they have the same value at zero 😂

[glass dagger]

,rotate

[glass dagger]

t=7 gives 2/6 from the first expression
t=7 gives 4/6 from the second expression

so something's wrong

[eternal monolith]

How to I rationalize the cube roots?

[glass dagger]

Your issue is that a cube root would need to be multiplied by itself three times to give back what's inside

so cbrt(t+1) * cbrt(t+1) is not sufficient to give t+1

[eternal monolith]

ohh

so how would i do that?

a cube root multiplied by itself 3 times ?

[glass dagger]

yess

well

[eternal monolith]

i understand but how do i multiply a cube root by itself 3 times

[glass dagger]

you could do what you did here but do it twice

[eternal monolith]

why twice?

[glass dagger]

at first there's one lonely little root

then you give the root a friend

then you give it another friend

[eternal monolith]

so if it was an expression raised to 1/4 and i wanted to rationalize the denominator i would multiply it by itself 3 times?

[glass dagger]

sounds like it could work

There's also a shorthand for multiplying it three times

you can multiply by the third power of whatever you want to multiply three times

[obtuse pebble]

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[bold junco]

hi

[latent walrus]

allo

[bold junco]

i need help with differintiation

[latent walrus]

are t and n both variables?

[bold junco]

yes

[latent walrus]

alrighty, what have you tried

[bold junco]

i tried the quotient rule

[latent walrus]

you can do that though i dont think its the most efficient

[bold junco]

and i also tried splitting the nominator and demominator then using product rule on the bottom

[bold junco]

whats the most effecient way?

[latent walrus]

I would write it as (t^4 * n)^(-3)

then use the chain rule and product rule

[bold junco]

how can you use them both?

[latent walrus]

well, youre first using the chain rule
but youll have a (t^4 * n)' which needs the product rule

[bold junco]

shouldnt you distribute the power to whats inside the brackets then use power rule for each one?

[latent walrus]

even if you distribute the exponent it still needs the product rule

t^(-12) * n^(-3)

[bold junco]

every single time i have a variable times a variable i MUST use product rule?

[latent walrus]

yeah, essentially

[bold junco]

t^(-12) * n^(-3) so i just use product rule on this and thats it?

[latent walrus]

yup

[bold junco]

in that case chain rule wasnt necessary right?

[latent walrus]

youre still technically going to be using it

since we're diff with respect to x

so eg n^(-3) goes to -3n^(-4) dn/dx

chain rule

[bold junco]

these technichalities confuse me

i just use product rule right?

whatever technicly comes let it come

[latent walrus]

we usually call this implicit differentiation, but it is just the chain rule

[latent walrus]

yeah

oops wrong message

one below

[bold junco]

how is this implicit when implicit is when we differentiate with respect to a variable thats inside the finction and we just add a derivative multiplier of it next to its normal version

[latent walrus]

this is implicit differentiation, and implicit differentiation is just the chain rule

eg if i have y^2, my outer function is u^2, my inner is v=y

so i do v' * u' which is dy/dx * 2u = dy/dx * 2y

its the chain rule, youre just not thinking of it as doing the chain rule

[bold junco]

alright

well like i said lets keep it uncomplicated

i basicly just

remove the division by making the power 3 a negative

then distributing it to remove the brackets

then use the product rule

all done?

[latent walrus]

yeah, its a valid approach

[bold junco]

if i write that in the final exam it will be correct?

[latent walrus]

nothing wrong with it at all

you would be fine

[bold junco]

thanks

u need any financial help tell me

.close

[obtuse pebble]

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[timid silo]

Given the equation of the line 𝑥+𝑦+8=0. Determine the equation of the line equivalent to this
in a coordinate system with the origin at (3,4).

[timid silo]

what is about to do with it, what kind of formula does exist to it

what does it means

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[final ledge]

I need help making this function so that it outputs the exact same value but so that h is defined as any integer >=1,the only issue is h=1 as it creates a singularity, I would preferably like it to be a "discrete" function per se ie,. it doesnt have any domain restrictions and doesnt have any other external definitions

[final ledge]

also, both x and h are restricted to positive integers >=1

<@&286206848099549185>

[obtuse pebble]

@final ledge Has your question been resolved?

[obtuse pebble]

@final ledge Has your question been resolved?

[final ledge]

bruh

an hour and a half is crazy

[obtuse pebble]

@final ledge Has your question been resolved?

[final ledge]

ion think its gonna get resolved

[odd sage]

can you not just make it piecewise

[final ledge]

no, not for this certain problem as this was originally intended to be apart of a much larger function

[drifting wraith]

i don't understand what you expect instead

[final ledge]

a way to rewrite or a different form of the function that still has the same output values at each integer input >=1 that doesnt have a singularity at h=1

[drifting wraith]

but what should it be instead at h=1

[final ledge]

if it would be helpful i can provide a table of values if reconstructing a whole new function would work

[final ledge]

any integer >=1 works

[drifting wraith]

what does that mean, you want a horizontal line?

[final ledge]

no, I want when you plug in 1 for h into the function to not get an undefined term, however dividing by 0 gives that

[odd sage]

make it lim g(x)?

[final ledge]

i had already considered a limit but as the function only takes integer inputs, unless this would work

ill provide a table of outputs to the function in case that might be helpful to construct a new one

[odd sage]

wait why wouldnt a limit work

im confused

[final ledge]

im not sure how a limit could be applied to this but if it can im well open to this

[odd sage]

just do like lim g(x) as h-> c and replace all the hs with c in the equation

[final ledge]

if this were applied to a sum for example, this would still give discrete integer values

[drifting wraith]

do this

[final ledge]

I also need to prevent it from going to infinity at h=1

[drifting wraith]

replace 1 with sign of 1-h

wait that's not what i meant

[final ledge]

?

[drifting wraith]

i meant this

[final ledge]

ah

this is giving me non integer outputs

[drifting wraith]

no like this

[final ledge]

still giving non integer outputs, but its closer than the latter

[drifting wraith]

no it works

you have it inside mod, don;t forget that one

[final ledge]

what do you mean mod inside?

[drifting wraith]

you have x−1 / h−1 twice in the formula

do it both times

[final ledge]

well dang

thats alot messier of a function than i originally intended but it does what i need it to

i guess then

is there any way to simplify it?

if not thats also fine, just wondering

im not familar enough with the modulo function to know if its possible

[drifting wraith]

@final ledge

[final ledge]

much thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

is this correct?

10^2^2^2
times
10^2^2^2
equals
10^4^4^4

[timid silo]

no

aw

how do i do it then

[fathom light]

Simplify
10^2^2^2 as 10^x then multiply 10^x with 10^x or square them
Same thing

[timid silo]

cant

[fathom light]

a^b^c = a^(bc) afaik

[timid silo]

i cant do that

[fathom light]

Oh yeah ig

That won't work mb

[timid silo]

i need something that works for any amount of exponents

[fathom light]

Got it

So it's better you paste your original question

[timid silo]

so like 10^10^10 times 10^25

[timid silo]

my original question is a little too much-

but i guess

my question, how to represent a number?

[timid silo]

where you may represent ludicrously large numbers

[obtuse pebble]

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[timid silo]

.close

[obtuse pebble]

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[tardy kestrel]

Not sure how to answer this using cosine rule. Do I have to rearrange?

[graceful rapids]

cosine rule solves for the side opposite angle
so in this case you have a, c, y and are looking for b

= yes rearrange

[formal island]

you can calculate all through sine rule which for me is easier, but whatevs > takes 1 more step

[tardy kestrel]

ok i did:

let unknown side = x

19^2 = 12^2 + x^2 - 2 * 12 * x * cos70

and struggling to rearrange for x

[tardy kestrel]

do you mind showing me how you would do it using sine rule?

just so i have an alternate way to view the problem

[formal island]

shakes sine rule* $\frac{a}{sin(\alpha)}=\frac{b}{sin(\beta)}=\frac{c}{sin(\gamma)}$

[safe haven]

whats the problem with that

[formal island]

just simply look for the angle opposing 12cm, and then find the other angle, and do another sine rule

[warm shale]

zzz0nnn

[safe haven]

looks to me like a quadratic

[formal island]

a little bit cursed of a quadratic, but yeah

[graceful rapids]

not really

you can evaluate cos70

[tardy kestrel]

this is what im basing it off

[safe haven]

uh huh

[graceful rapids]

yeah this is good

quadratic formula is the way to go

[tardy kestrel]

redid equation using sine rule, found it longer but easier then struggling to rearrange cosine rule

[graceful rapids]

or that i guess

that works

[tardy kestrel]

thank you all for your assistance

.close

[obtuse pebble]

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[timid silo]

So.. Another off topic no math subject. I don't know what the hell am I doing in research..

[timid panther]

What?

[brazen gorge]

just ask

[timid silo]

It turns out they didn't do anything yet

[obtuse pebble]

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[timid silo]

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[floral bramble]

how to get access to topology channel?

[floral bramble]

#point-set-topology to be exact

[kind hawk]

#get-advanced-access

[floral bramble]

thnx buddy

[obtuse pebble]

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[timid silo]

what have u done alrdy ?

[heady turtle]

sta1 as I said

[timid silo]

ye but idk what stat1 mean

[heady turtle]

basically no clue

[timid silo]

ok

[frigid burrow]

!status

[obtuse pebble]

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

[frigid burrow]

this 1 i think

[heady turtle]

ye

this question is just so damn hard I can't really make any progress for hours

[timid silo]

agree

just to know if i am maybe able to do it, what grade are you ?

[heady turtle]

doesn't matter

[timid silo]

if u are like 4 grade above me, i dont think i can solve so we will both loose time on it

even just 1 grade above

[obtuse pebble]

@heady turtle Has your question been resolved?

[timid silo]

are u allowed to do the limit in the integral, and after, calculate the integral ?

[heady turtle]

wait im gonna delete and ask another q cuz I feel that no one can help this q for now 😂

@timid silo ty maybe I'll ask later

[obtuse pebble]

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as this action is irreversible, and this channel may abruptly lock.

[timid silo]

wait im trying rn

[obtuse pebble]

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[lapis spire]

Hi! I need a little help studying for an exam, when dealing with transformations like stretches, and you want to find the mew coordinates of a specific point, what would you do?

[lapis spire]

Would you like sub in the point in x and multiply for the streches and add for translations ect?

[crimson pike]

Simply yes
but mostly people mess up in rotations and reflections

[obtuse pebble]

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[raven kettle]

can anyone help me

[sonic rover]

You want to get the equation in the form ax^2+bx+c=0, with no fractions

[raven kettle]

yeah

this is the last question and its confusing me

[sonic rover]

Ok so following that, how can you get rid of the /5 and /4

[raven kettle]

lcd

oh so its 20

[sonic rover]

You don't necessarily need to do that but I guess you could

[raven kettle]

multiply both sides its uh

nvm im not sure anymore

[sonic rover]

Nope you're good

[raven kettle]

oh ait

I get it

[sonic rover]

By the other denominator

[raven kettle]

i divide

then miltiply

thx

holdup let me check if ingot it righ

how do i get the bx?

[sonic rover]

bx is whatever the 5x turns into after undoing the fractions

[raven kettle]

oh

so 20 divided by 4 is 5 x 5x

oh so its c

thanks

[hidden apex]

just times everything by 20

[sonic rover]

Yea

[raven kettle]

.close

[obtuse pebble]

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