help-10

[lyric ember]

multiply by x/x

[karmic hedge]

I dont really get what you mean by multiply by 1 or x/x, it will be the same no?

Like nothing will change

right?

[lyric ember]

well yes and no

[karmic hedge]

Ahh ok

So we just wanted to divide everything by x in the first place right? instead of x^2 like I did

[lyric ember]

yeah

its just so that you can easily read what the limit is

you dont want 0/0 or infinity/infinity scenarios

because then the limit can be anything

thats the main idea

[karmic hedge]

Ok I think I get it

[lyric ember]

if you take the limit here you would have infinity/infinity

[karmic hedge]

But its still kinda hard to read the limit, like is it just -1? Or is the sqare root 5 included? Or what am I supposed to do...

[lyric ember]

$\lim_{x\to \infty} \frac{1}{x} = 0$

[warm shale]

tobi.

[karmic hedge]

But the coefficient is the same on the top and bottom, no?

[lyric ember]

$\lim_{x\to \infty} \frac{a}{x} = a\lim_{x\to \infty} \frac{1}{x} = a\cdot 0$

[warm shale]

tobi.

[karmic hedge]

So its -1/(1 + √5) ?

[lyric ember]

yes and no

what is your limit

[karmic hedge]

infinity

[lyric ember]

then yes

[karmic hedge]

Its different for negative infinity?

[lyric ember]

yes

[karmic hedge]

Is the answer infinity (for limit negative infinity)

[lyric ember]

its not negative infinity

[karmic hedge]

https://cdn.discordapp.com/attachments/903430332324405288/1122459195363557406/IMG_20230625_113113_9852.png

Should I rearange this fraction or something to make it easier to read what the limit at minus infinity is?

But I dont really know what to change it to

[lyric ember]

oh I know the problem

sry

rookie mistake

$\sqrt{x^2} = \mid x \mid$

[warm shale]

tobi.

[karmic hedge]

Hmm

not really sure what that changes since the signs were positive already

[lyric ember]

well it changes something for the negative limit

[karmic hedge]

ahh

[lyric ember]

you would want to factor $\mid x \mid$

[warm shale]

tobi.

[karmic hedge]

Of this? Or the original...

[lyric ember]

$\frac{-x+4}{x+\sqrt{5x^2+9}} = \frac{-x+4}{x+\mid x \mid \cdot \sqrt{5+9/x^2}}$

[warm shale]

tobi.

[lyric ember]

now you can say let x>0 or let x > 0

to elimanate the amount function

you already did the positive part

[karmic hedge]

So it moves towards negative infinity?

[lyric ember]

if $x>0 \mid x \mid = x$ if $x<0 \mid x \mid = - x$

[warm shale]

tobi.

[lyric ember]

<@&268886789983436800>

[lyric ember]

no just make a case distinction

if x >0 or if x<0

[karmic hedge]

But do I not divide everything here by x again?

Like I did to find the limit at infinity

[lyric ember]

yeah you basicly do the same

if $x > 0 \quad \frac{-x+4}{x+\mid x \mid \cdot \sqrt{5+9/x^2}} = \frac{-x+4}{x + x \cdot \sqrt{5+9/x^2}}$

if $x < 0 \quad \frac{-x+4}{x+\mid x \mid \cdot \sqrt{5+9/x^2}} = \frac{-x+4}{x - x \cdot \sqrt{5+9/x^2}}$

[warm shale]

tobi.

tobi.

[karmic hedge]

if x < 0 its moving towards infinity

Because the negatives on the top and bottom cancel out

[lyric ember]

if x < 0 then the numerator is positive

[karmic hedge]

But what about the -x

beside the +4

[lyric ember]

do you remember your positive answer

as x goes to positive infinity

[karmic hedge]

-1/(1 + √5)

[lyric ember]

yes

maybe I should point out

$\lim{x\to -\infty} \quad \frac{1}{x} = 0$

[warm shale]

tobi.

[karmic hedge]

So its also -1/(1 + √5)?

[plain stag]

\lim_{x...

Need the _

[lyric ember]

same process as before

multiplying by $\frac{x^{-1}}{x^{-1}}$

[warm shale]

tobi.

[karmic hedge]

oooooooooooooooooooooooooooooooo

I finally get it

-1/(1 - √5)

[lyric ember]

yeah :D

[karmic hedge]

Omg

Im so sorry hahaha

[karmic hedge]

Thank you soooo much

[lyric ember]

this is worth to remember

[karmic hedge]

You are the most patient person I every met

Thank you hahahaha

Ok I will take a break from maths for a few mins 😅

Thanks again

❤️

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[timid silo]

Step 3, why did we raise everything to the nth power

[obtuse pebble]

@timid silo Has your question been resolved?

[devout fern]

the proof is in terms of a, x, and n and they dont have n yet.

to get n to where it needs to be, they introduce n at step 3 by using it as the power

they are trying to turn this

into this

[timid silo]

Ahhh

Thank you so much

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[granite axle]

ho to do

[proper kelp]

let u = (1-lnx)^n

[granite axle]

i meant ii

[proper kelp]

oh

prove P(1) is true

as a base case

and then P(k) using inductioin

and P(k+1) is true if P(k) is true.

so then you have your answer

[granite axle]

ohhh i see

thanks

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[toxic basin]

yo

[toxic basin]

i get pow(10,4/7)

but the book says there is no solutions

[timber island]

is this finding series convergence?

[toxic basin]

all it says is that "a" is a positive number not equal to 1

[toxic basin]

ye

first u put it into 1+log+log+etc=-7

then substract w 1

and u do 2logx/(1-2log x)=-8

[timber island]

nvm i dont think i can do this

[toxic basin]

and u get x=10^(4/7)

😭

[timber island]

$\prod_{n=1}^{\infty} a^{{2logx}^n}$

[timid silo]

infty

[warm shale]

itzkraken.

[rich plume]

how do you get -8

[toxic basin]

-7-1

[wintry fable]

no

basically geometric sequence

[toxic basin]

1/a^7 is a^-7

[wintry fable]

a^1+2logx+4log²x+8log³x + ... = a^-7

[toxic basin]

a^x=a^y, x=y

...

[wintry fable]

a = 1

not 2logx

i suppose youre using a/1-r

[toxic basin]

fisrst member is 2logx

i substract 1 from both sides

[wintry fable]

oh wait you set it as -8

yeah i didnt see that

[toxic basin]

it is the same eather way

[wintry fable]

um then it should be correct

[toxic basin]

but for some reason the book says the equation has no solutions

[rich plume]

I do get the solution 10^(4/7). probably the answer key is incorrect

[toxic basin]

it cannot be incorrect it is a state exam

[polar fossil]

state exams can be incorrect

[toxic basin]

it says a is positive number not equal to 1

maybe it is something there

[polar fossil]

is that log10?

[toxic basin]

ye

[polar fossil]

yeah i get no solutions either

[rich plume]

Ohh, the answer is indeed no solution

[polar fossil]

(if anything I get 10^(6/7) but that's not a converging ratio for a geom series)

[rich plume]

Oh yeah

you got x=10^(4/7)

But when you put that value of x into 2logx you get 8/7

so the common ratio is more than one

and hence the sum is diverging

*Talking about power of a on LHS

[polar fossil]

(ah right, it's 2logx not just logx; that explains my 6/7)

[toxic basin]

yeah but why does that matter

ohhhhhhh

[toxic basin]

q has to be from -1 to 1

[polar fossil]

idk... I get logx = 6/14

[rich plume]

That is incorrect recheck once

[toxic basin]

yeah, so when u get the x, u go bac kto check your conditions

log x has to be in interval of -1/2, 1/2

and 4/7 is over 1/2

[polar fossil]

$\frac1{1-2\log x} = 7$ \
$1 = 7 - 14\log x$ \
$14\log x = 6$ \
$\log x = 6/14$

[warm shale]

Hayley

[rich plume]

We had to because we assumed that 2logx was less than 1

[rich plume]

RHS should be (-7)

[polar fossil]

gragh it's in the denominator

I didn't even copy down that part of the problem

ok cool yeah makes sense

[toxic basin]

yep i get it now tnx

i also just found out they removed that from this years state exams lmao

[obtuse pebble]

@toxic basin Has your question been resolved?

[obtuse pebble]

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[shadow lava]

AXB = C to solve this, right?

A^-1(AXB) = A^-1(C)

[royal basin]

sure keep going

[shadow lava]

XB = A^-1(C)

(XB)B^-1 = A^-1(C)B^-1

X = A^-1(C)B^-1

this question will take some time to solve, I gotta use
[A | I] ~ [I | A^-1]
and
[B | I] ~ [I | B^-1]

to find inverse of A and B

[royal basin]

"gotta" as in "required by legally-binding contract" or "gotta" as in "i know no other way"?

[shadow lava]

it's the only way I know

to find inverse of 4x4

2x2 i can use shotcut

[royal basin]

right

[shadow lava]

but sometimes i will use [A | I] ~ [I | A^-1] on 2x2 as well

[royal basin]

well you might cut down on your steps somewhat by calculating $A^{-1}C$ with a single session of GE

[warm shale]

Ann

[royal basin]

[A | C] ~ [I | A^-1C]

B^-1 is easyish to find -- should take you only one row op

[shadow lava]

I don't understand this at all lol

we never learned it, it's OK i will just use my long method, keeping it mind it will take me a while to solve it

[royal basin]

when you apply row operations to the "double" matrix [A | I] to make its left half into the identity

what you are doing is multiplying this entire matrix by A^-1 from the left

without actually knowing what A^-1 is

but you achieve the same effect

do you understand this y/n

[shadow lava]

n

but it's OK

at least i can solve it

that's all i care about for final tomorow

no need for shortcut

ty

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[little lava]

I already found the constant coefficient to be 64, and 240 to be coefficient of x^6

but do not know how to use these results to find part b

[thick chasm]

Remember distributive property of multiplication

[little lava]

yeah

but still I am getting confused on what to do

the (-3/x^3) confuses me

[thick chasm]

Think what must be multiplied by 1/x^3 to get x^3

[little lava]

x^6

but then should i ignore

the first part?

i think i am understanding

[thick chasm]

No, what must be multiplied with x^3 to get x^3?

Constant multiple?

[little lava]

1

?

[thick chasm]

Yes

Now you have (x^3-3/x^3)(64+...+240x^6)

Can you figure it out now?

[little lava]

yeah so 64x^3 and then what -240x^3 right?

[thick chasm]

Yeah, now add them

[little lava]

-176

[thick chasm]

That is the answer

[little lava]

ok thanks

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[little lava]

Basically the answer i got was wrong. in the ms and wolfram, its -416 for part b)

but i do not know why or how

[balmy mortar]

if you could show us your working

or reasoning

[austere shard]

you only care about the x^3 so you should take in consideration just the terms x^3*(integer) and -2/x^3(coefficient of x^6)x^6.

you only have to find the coefficients with the binomial theorem

[little lava]

Yeah my reasoning is here

[little lava]

Like on the.

I dis that but i got the wrong answer

Wait i think i see my mistake

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[timid silo]

Trying to find out what 13 is as a percent of 20. it is 65%.

I figure this out by doing x * 20 = 13, divide by 20 each side and get x = 0.65, 0.65 * 100 = 65%...

But, when I do x * 20 = 13, divide by x each side and get 20 = 13/x, then multiply by 13 each side I get x = 260 which is not the right answer.

Very silly question but I'm just trying to understand why the first solution works and the 2nd one doesn't. I get why the 1st solution works, but why doesn't the second one please?

[royal basin]

20 = 13/x, then multiply by 13 each side I get x = 260
nope

$\frac{13}{x} \cdot 13 \neq x$

[warm shale]

Ann

[timid silo]

Oh, don't the 13's cancel out?

[royal basin]

no, they do not

$\frac{1}{x} \cdot 13 \cdot 13$

[warm shale]

Ann

[royal basin]

why would the thirteens cancel out

[timid silo]

This puts it better into perspective. I thought because I'm multiplying by 13 it would get rid of the numorator 13 from 13/x and I'd be left with just x.

Thank you! 🙏

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[forest sinew]

what happen

looks like u can get to $n(n-1)(n+1)(n^2-n+1)(n^2+n+1)$

[warm shale]

janniku

[forest sinew]

assuming thats where u were

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[fluid bough]

$n(n-1)(n^2+n+1)(n+1)(n^2-n+1)$

[warm shale]

cravingbanana

[fluid bough]

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[twilit cloud]

God

[obtuse pebble]

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[forest sinew]

right

[twilit cloud]

#latex-testing

[fluid bough]

I wasn't testing.

[obtuse pebble]

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[patent zealot]

Task: We have a graph G with the maximum degree of 2019. We need to prove that there exists a graph H that is 2019-regular and contains graph G as an induced subgraph.

My idea was to identify the node with the maximum degree in graph G and connect all its neighbors to each other to construct graph H. If there exists a node in graph G with the maximum degree of 2019, we can directly connect all of its 2019 neighbors to each other. This will result in a 2019-regular graph H that contains graph G as an induced subgraph. This construction satisfies the definition of an induced subgraph because all the nodes in G that are connected to each other are included in graph H.

[warm canopy]

But G may have more nodes than just the highest degree one and all its neighbours no?

[patent zealot]

Yeah, I didnt even think of that, facepalm

[obtuse pebble]

@patent zealot Has your question been resolved?

[obtuse pebble]

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[vivid creek]

How do I find T?

[teal turret]

Set v = 0

@vivid creek

And solve for t

To get T

[vivid creek]

Oh yea

Ty

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[timid silo]

https://youtu.be/1nXSO3POyZ8?t=260

[timid silo]

yo so I put this video at a specific time

I'm confused exactly what's going on at that step

Sal substitutes 1/2 and then that multiplied by 1/20 = 1/10?

[kind hawk]

1/20 = 1/2 * 1/10. put the 1/2 inside the integral and leave the 1/10 outside

[timid silo]

makes sense

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[timid silo]

thanks a lot

[obtuse pebble]

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[timid silo]

How do I start this problem?

[timid silo]

None of the properties of log or any results from them, seem to help here

[normal oracle]

the sign you changed from - to + is because...?

[timid silo]

I didn’t, the instructor did

He said it was a typo

[normal oracle]

so the exercise is with x+1 instead of x-1

[timid silo]

Yes

But I’d also like to learn what happens if it was not changed

[normal oracle]

Do you know what to do when you have x there?

like a^x

how do you make so the x is not there anymore

[timid silo]

Take log?

[normal oracle]

ok

do it

[timid silo]

left hand side would be

(x+1)(x+1)?

[normal oracle]

where is the log?

[timid silo]

Log_10 ((x+1)^Log_10(x+1))= log_10 (x+1)(x+1)?

[normal oracle]

no

[timid silo]

Why? Where did I go wrong?

[normal oracle]

LHS is correct

RHS is not

[timid silo]

Oh oh oh

log_10 (100x+100)

[normal oracle]

yes but not necessary

don't expand

sry, i did next step

i delete so you do yourself

[timid silo]

I did that to avoid multiple parentheses

[normal oracle]

parenthesis are good

now

LHS

what is ln(a^x)?

i know u have log, just asking a general question

[timid silo]

I Haven’t learnt natural log yet

[normal oracle]

no prob, rules are for logs

no matter the base

what is log (a^x)?

[timid silo]

Depends on base right?

[normal oracle]

do you know this?

[timid silo]

Yes

[normal oracle]

ok, so what can you do with your LHS

[timid silo]

I can treat Log_10(x+1) like n

[normal oracle]

yes you can

so what do you have left

[timid silo]

Log_10(x+1)Log_10(x+1)???

[normal oracle]

ok

so now you have this, right?

[timid silo]

Log_10(x+1)^2?

[normal oracle]

what is a * a

[timid silo]

a^2

[normal oracle]

let that separated for now

go for the RHS now

[timid silo]

2 + log_10(x+1)

???

[normal oracle]

ok

now what can u do?

[timid silo]

Take log_10 (x+1) to LHS

And take it common?

Wait no

[normal oracle]

you have to mistake to learn

if you think u have to do that do it

[timid silo]

Log_10(x+1) = 2

10^2 = x+1

x = 99

Is that correct?

[normal oracle]

yes, but I didn't understand your proccess of thought

[timid silo]

Log_10 (x+1)^2 - log_10 (x+1) = 2

Log_10 (x+1) = 2

[timid silo]

We apply the property of log there

[normal oracle]

why do you say Log_10 (x+1)^2 - log_10 (x+1) = log_10 (x+1)

[timid silo]

Log_a (m/n) = log_a(m) - log_a(n)

[normal oracle]

ok

now only one last thing

u skiped one step here

Log_10(x+1) = 2
10^2 = x+1

in the (now RHS)

you did 10^2 but didn't show the other side

you should put all the steps to know what you're doing

and to prove you know it

[timid silo]

Oh

I intended to directly convert it into exponential form

[normal oracle]

yes, but that comes from the correct form

a^(log_a[x]) = x

[timid silo]

Btw why do we take log?

[normal oracle]

when?

[timid silo]

In cases like this

[normal oracle]

do you know the exponential function?

[timid silo]

No not yet

[normal oracle]

e^x, but in general

when you have base raised to x or similar

logarithm is the inverse function

so the way to lead with exponential we use logarithms

the same way to lead with x+1 = 2

we use subtraction

[timid silo]

So log is opposite of exponent

[normal oracle]

and to lead with multiplication we use division

not the opposite

the inverse function

[timid silo]

Log_a(a^x) = x

[timid silo]

Yes so it basically undoes it right?

[normal oracle]

Log_a(a^x) = x(log_a(a))

but log_a(a) = 1

[timid silo]

Ahh I understand

[normal oracle]

not undo

because nothing was done in the begining

the equation came like that

[timid silo]

I understand

Thank you so so so much

[normal oracle]

np

[timid silo]

I am really grateful for you time

Have a good day

[normal oracle]

god bless u

[timid silo]

God bless you too

.close

[obtuse pebble]

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[stiff bough]

What would the answer to this be? I was thinking it would be 4*-3 = -12 or maybe I would need to graph it so when t<-3, g(t) = 0 and if t >= 4, g(t) = 1?

[pseudo swift]

u is the step function right ?

[stiff bough]

yes

[pseudo swift]

alright

yeah graphing it is a good idea

it's not completely needed tho

when should a product of two step functions like these equal 1?

@stiff bough

[stiff bough]

when t is greater than 4?

[pseudo swift]

no you're skipping steps here

(and it's incorrect)

if a*b = 1, and a and b can only equal 0 or 1, what do a and b have to be ?

[stiff bough]

equals 1 when they are both 1?

[pseudo swift]

exactly

so what does this mean for the arguments of our step functions ?

[stiff bough]

i dont get it, wouldnt they just be 1 when t is greater than 4? and 0 else where?

[pseudo swift]

nope

[stiff bough]

oh would it just be u(-t+4)?

[pseudo swift]

second one ends up being 0 when you think about it, when t>=4

[pseudo swift]

what I'm getting at, is that, if you want the product to equal 1, you need both the arguments of your step functions to be >=0 at the same time

[stiff bough]

so t=4?

because they would both be 1 when t = 4?

[pseudo swift]

wdym t=4 ?

there's 2 inequalities to be satisfied here

if you want u(...)u(...) to be = 1

[stiff bough]

so u(...)u(...) = 1 when t >= 4, and u(...)u(...) = 0, when t < 4?

[pseudo swift]

how do you get that ?

[stiff bough]

Would the graph look like this?

[pseudo swift]

no

it's u(**-**t+4)

not u(t-4)

[stiff bough]

oh

so would it only be equal to 1 on the interval [-3 to 4]?

[pseudo swift]

yea

[stiff bough]

so the answer would look something like { 1, [-3, 4] and 0, elsewhere ?

[pseudo swift]

well that's your function u(t+3)u(-t+4)

what do you get if you integrate it ?

[stiff bough]

would this be the right way to write it?

[pseudo swift]

yeah the answer is 7

gg

[stiff bough]

thank you

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[obtuse pebble]

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[dark tundra]

Why does L(t) only add point P to the i+2j-8k vector

Ik it wouldn't be equal if we used Point Q

like t+2 , 2t+5, -8t-6 ?

[brisk matrix]

yes

it would be equal

[obtuse pebble]

@dark tundra Has your question been resolved?

[obtuse pebble]

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[eager siren]

this is what I have so far

I understand you need the cross product first

This is because a cross product will always give you the vector perpendicular to the vectors you multiplied?

Let me know if I have that logic right

I am just struggling on what to do from here

[timid silo]

so you already have ONE vector that is perpendicular to both a and b

[brisk matrix]

any vector parallel to that one will be orthogonal to the other 2 you started with

and when i say parallel i mean scalar multiple

can you think of multiplying that vector by some number so that the first coordinate is positive?

[eager siren]

I know I am supposed to do something with a unit vector...

is that what I need to multiply it by

[eager siren]

yeah but the first number needs to be a positive

[brisk matrix]

if you have a vector

then a parallel vector is of the form t * v

where t is a nonzero number

so like 2v is parallel to v

10v is parallel to v

do you know how to multiply a scalar and a vector together?

[eager siren]

yes

u like

distribute it

cuz scalar and vector gives scalar

iirc

[brisk matrix]

scalar and vector gives a vector

[eager siren]

oh typo

yes vector

so u distribute it in

[brisk matrix]

u x v gives you a vector orthogonal to u and v

let's call that vector w

right now you have w

[eager siren]

ok

yes

[brisk matrix]

any vector parallel to w is also orthogonal to u and v

[eager siren]

ah

ok makes sense

[brisk matrix]

so multiply by some scalar

[eager siren]

so we just find the parallel one

[brisk matrix]

so the first coordinate is positive

[eager siren]

Ok I understand

what scalar though

[brisk matrix]

any

[eager siren]

I remember there was something

really?

[brisk matrix]

as long as it makes the first coordinate positive

oh

the question wants unit as well

i will say you can still multiply by any

then you just normalize it after

[eager siren]

so if i multiply by -1

to make -5i positive

so it would be 5i-5k

[brisk matrix]

yes

and you want the unit vector in that direction

[eager siren]

I forgot unit vector stuff cuz I took break after mid term

I remember it was some formula

[brisk matrix]

you take the vector and divide it by it's length

that gives you a vector of length 1

[eager siren]

ok

and to find vector length

its

this?

[timid silo]

yes

[obtuse pebble]

@eager siren Has your question been resolved?

[eager siren]

ok so i got 5/sqrt(2)

now what

[timid silo]

this

[eager siren]

ah

ok

i got answer

i checked it in book and its right

ok thanks

For future reference right

when I get a problem like this asking me to find a vector orthogonal to whatever

in this case it was a and b

i would find the cross product

since it gives me a perpendicular vector

and if i need it to be positive or negative

i can multiply any scalar i want

and if it requires a unit vector

i just do what i did at the end?

[timid silo]

yes

[timid silo]

sometimes you only have one vector

let's say, to find a vector perpendicular to vector c

then you wanna consider this:

for vectors c and d to be perpendicular to each other, the dot product of c and d equals 0

you probably need additional information to find the actual vector d

[eager siren]

ok i will see if i can find a problem like that in the book to practice once i finish this homework

thank you though i will save this

[timid silo]

hope it helps

[eager siren]

.close

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[long knot]

Will 12x^2 be positive or negative here?

[brisk matrix]

do it step by step

-6x * 2x = -12x^2 like you said earlier

and now you have (x^2 + 1) * (-6) - (-12x^2)

[long knot]

Oh, I was just wondering if I should take into account the negative in the middle.

[brisk matrix]

i realize that

[brisk matrix]

i was hoping you'd be able to see here what happens to the - (-12x^2)

i wrote -12x up top my bad

it was meant to be -12x^2

[long knot]

Is this correct?

[brisk matrix]

yeah looks good

[long knot]

Ok, got the question right! Thanks!

.close

[obtuse pebble]

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[shrewd crow]

.open

.ask what is 2 + 3?

•ask what is 2 + 3?

[brisk matrix]

shitty jokes go in #chill

.close

[obtuse pebble]

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[shrewd crow]

.reopen

[obtuse pebble]

✅

[shrewd crow]

Tell me

[brisk matrix]

,w 2 + 3

[warm shale]

Results provided by WolframAlpha

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

Requested by maximofs#0

[brisk matrix]

.close

[obtuse pebble]

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[shrewd crow]

.reopen

[obtuse pebble]

✅

[shrewd crow]

Explain

[brisk matrix]

.close

[obtuse pebble]

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[hot hazel]

grow up and dont abuse help channels

[teal turret]

(explanation acquired)

[hidden garnet]

......... what the heck

[civic socket]

💀

[tardy epoch]

Haha muted

[obtuse pebble]

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[unkempt bramble]

Is A the bus

[tardy epoch]

Assume it is.

[unkempt bramble]

Ok

.close

[obtuse pebble]

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[violet ingot]

Having trouble figuring out what I'm doing wrong here

[violet ingot]

I'm trying to find the definition of the derivative

[rocky goblet]

how did you get 6.25?

also i think you accidentally turned this t^2 into a t

[rich silo]

and -6 next to that

i mean here

[obtuse pebble]

@violet ingot Has your question been resolved?

[violet ingot]

I squared it

Also the I foiled the -

[tough bolt]

Here's where you got wrong: the limit should be $$\lim_{h\to0}\frac{2.5(t+h)^2+6-(2.5t^2+6)}{h},$$ not $$\lim_{h\to0}\frac{2.5(t+h)^2+6-2.5t^2+6}{h}$$

[warm shale]

math_is_fun

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

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[wanton kraken]

I don't want to act too fast and choose an answer right away,
"In triangle ABC, the distance between the centroid and one of its sides equals to 6 cm. The same side is 15 cm. Find the surface of the triangle"

[wanton kraken]

one would just use S=1/2ab and double it

but I don't want to act too fast

[obtuse pebble]

@wanton kraken Has your question been resolved?

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

<@&286206848099549185>

I think I can use the signs for equal triangles

nvm

[patent tendon]

<@&286206848099549185> whats 1+0

[manic jewel]

1 👍

[patent tendon]

ty

[manic jewel]

I gotchu

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

<@&286206848099549185>

[patent tendon]

<@&286206848099549185> whats 4-4

[wanton kraken]

<@&268886789983436800> @patent tendon is spamming

intentionally

[patent tendon]

?

[wanton kraken]

im here still

[final tartan]

you're halfway there, just remember what is a centroid

[wanton kraken]

the point where all medians intersect

[final tartan]

correct, and what else do we know about it?

[wanton kraken]

it has ratios

1:2

from the heel to the apex

[final tartan]

I'm sorry English isn't my native language could you use a picture to describe what are heel and apex?

I found out what an apex is now what is a heel ?

when you're back you can ping me

[wanton kraken]

where the segment ends

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

this has been going on for HOURS

<@&286206848099549185>

[final tartan]

Sry abt that was busy, anyways the centroid of a triangle has a proprty that it cuts the medians into two part one of them is twice as long as the other

The short part is the one going from the centroid to the vertex while the long one is the one from the centroid to the side

[final tartan]

In this question we have the distance from one side to the centroid meaning it is the long side, meaning we have 2/3 of the hight.

[wanton kraken]

how do we have the height?

the medians intersect at the centroid

not the heights, no?

[wanton kraken]

I have this shit open for a fucking 10 hours

.close

[obtuse pebble]

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[oak aurora]

WTS: $#(A) + #(B) = #(A \cup B) + #(A \cap B)$

Lemma 0: If $n \ge c$, then there exists number whose $c$-th successor is $n$, we call this number $n-c$. Can be proven using induction.

Lemma 1: If $#(A)=n$ and $A' \subseteq A$ st $#(A')=c$, then $#(A\backslash A')=n-c$ . Can be proven using induction.

[warm shale]

Iusgnol

Iusgnol

[royal basin]

think there's a more elegant proof but it will require knowing about Cartesian products

[oak aurora]

Ok so, its kind of messy i apologise for that, have not done the very last part of it but i think it shouldn't be that hard to show injetivity and surjectivity. My question is whether theres a better way to do this since I haven't unlocked subtraction yet, i tried to save this by using Lemma 0, but i realise that in my proof, i am using some properties of subtraction that require much more than lemma 0.

Yes i do know about cartesian products

actually let me think about cartesian products for abit first, will let you know if i need help again

[royal basin]

here i intend to use them as a tool to construct disjoint copies of sets

@oak aurora i can put the solution i have in mind under a spoiler if thats ok w you?

[oak aurora]

yes sure

[royal basin]

to be clear you DO know cardinalities of disjoint sets add, yes?

[oak aurora]

yes

[royal basin]

||we construct two sets, one of which has cardinality #A + #B and the other has cardinality #(A ∪ B) + #(A ∩ B), then give an explicit bijection between them.

LEMMA 1: Sets A × {1}, B × {2}, (A ∪ B) × {3} and (A ∩ B) × {4} are pairwise disjoint.
PROOF: any ordered pair that belongs to two of these sets at once would have to have its second component equal to two different natural numbers, which is impossible.

let S = (A × {1}) ∪ (B × {2}) and let T = [(A ∪ B) × {3}] ∪ [(A ∩ B) × {4}].

construct a function f: S -> T as follows:
for all a ∈ A, f(a,1) = (a,3);
for all b ∈ B, f(b,2) = (b,4) if b ∈ A [equivalent to b ∈ A ∩ B], else (b,3).||

[obtuse pebble]

@oak aurora Has your question been resolved?

[oak aurora]

ah okay I gave up couldn't think of a solution involving cartesian products, I did manage to simplfiy my original proof instead of constructing a messy function based on 3 other bijections using the fact that cardinalities of disjoint sets add but it still required me to use a couple properties of subtraction.

[oak aurora]

This is definitely much more elegant and makes alot of sense, don't think I would never have thought of this, thanks alot for the help

.close

[obtuse pebble]

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[proven zealot]

Write 4 + 4i and 2√
3 + 2i in (r, φ) notation to find the quotient

[proven zealot]

this is how I did it

I was just not too sure about should I do 1/12 pi + 2kpi or not

I feel like I should add 2*k * pi

or I think the way the question is formulated I shouldn't add the 2* pi * ki at all

[obtuse pebble]

@proven zealot Has your question been resolved?

[tardy epoch]

you don't need the 2 pi k at all in either numerator or denominator in this problem

[obtuse pebble]

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[proven zealot]

ahh okay thank you

[proven zealot]

it's because how the question is formulated therefore I didn't need it right

[obtuse pebble]

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[glossy nest]

This is my work so far

I found the correct maximum value but not the correct minimum value

not sure where I went wrong

[obtuse pebble]

@glossy nest Has your question been resolved?

[glossy nest]

<@&286206848099549185>

[sleek moth]

I am sorry but i can't see the calculation

[glossy nest]

ill take a better pic

[glacial obsidian]

You did not consider the cases x=0 and y=0

Also near the end you wrote literally x^2 = -3

That can't be a solution by any means

[glossy nest]

True

[glossy nest]

In the case of x = 0 and y = 0, f(0,0) would be 0 then

[glacial obsidian]

But that point is not in x^4 + y^4 = 18, you should instead consider that only one of them is zero

[glossy nest]

When u say only one of them, you mean only either the x or y coordinate evaluated at zero

right?

[glacial obsidian]

Yes

[glossy nest]

.close

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[tight rapids]

how does the limit of the ratio of the two series show us that the two series are p much the same?

[tight rapids]

I get that if the limit is infinity then what that means that there not similar at all? but if its a finite number then they are similar?

[tardy epoch]

https://tutorial.math.lamar.edu/classes/calcII/seriescomptest.aspx

[tardy epoch]

proof is here

[tight rapids]

I kinda need it explained like im a 5 yr old not a proof, otherwise I can just accept the method and move on

basically if the comparison test doesn't work we j go to the limit comparison test right?

[obtuse pebble]

@tight rapids Has your question been resolved?

[obtuse pebble]

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[timid silo]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[timid silo]

hey try to consideer that x + 2 = (x + 1) + 1

every time you see a rational function like this, you wanna try to make it simpler

[timid silo]

for example by considering this

[autumn estuary]

@timid silo x+1 = t?

[timid silo]

yes

another way to think about this is by substitution

[autumn estuary]

with u and v?

what would be the substitution?

[timid silo]

you can sub t = x + 1

do u know how to do the integration by substitution from here?

[autumn estuary]

I am trying

not really

@timid silo I got x+1+ln(x+1) + c

can you check that?

[timid silo]

yes this is a correct ans

rmb that 1 is a constant

so u can just omit the 1

[timid silo]

well one thing

ln|x+1|, remember the absolute sign

otherwise, perfectly correct solution

[autumn estuary]

.close

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[polar idol]

needa lil push to get moving

never understood this perpendicular nonsense

[tardy epoch]

what definition of perpendicular were you given

[polar idol]

no clue dont remember

its 90 degrees

[tardy epoch]

look for it in your book/notes

that's for 2d vectors

you have 3d vectors

[polar idol]

oh...

its equal to 1?

[tardy epoch]

no

[tardy epoch]

.

[polar idol]

ohhh

cross product?

[polar idol]

coulda at least told me i was right

.close

[obtuse pebble]

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[polar idol]

needa lil push to get moving

[scarlet mason]

perpendicular means you want the dot product of the first lines direction with the second lines direction equal to zero

so if the second line is (x,y,z) = r0 + s(a,b,c), you want (2,-3,7) * (a,b,c) = 0

if two lines intersect at some point, then they lie in the same plane

[obtuse pebble]

@polar idol Has your question been resolved?

[obtuse pebble]

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[plain grove]

Hi guys I’m in year 8 math class, can you please check if I am on track?

[plain grove]

Is anything wrong?

I got an answer 2.8

Is that correct?

<@&286206848099549185>

[candid yarrow]

Yes, because you substituted the interest in instead of the principal

[plain grove]

So principal=Lin?

I not L**

[pearl echo]

i don´t understand much the numbers you are putting but the important thing is that the money earned is the 2% of the total invested. then you should figure how much was the initial investment using the percentage formula

[plain grove]

Do you know the answer

Please, anyone

?

[normal remnant]

Initial x 1.02 = Initial + 140

[plain grove]

So 140/0.02?

No no

That’s wrong

What’s the initial

[candid yarrow]

Why

[plain grove]

Oopssss. The initial, so it has to be more than the 140

So 7000 was the initial

And so 7000 x 140 x 0.02 = 19,600 (P)

Is that correct?

[normal remnant]

yeah

no

[candid yarrow]

Why did you use that formula

[plain grove]

P=IIN

[pearl echo]

part/total = percent/100
you want the total, so thats x, then the part is 140 and the percent is 2
substitute in the formula and clear x which is the total

[pearl echo]

that is correct

[plain grove]

So x =7000?

I’ve got another question

[blissful marlin]

@plain grove What have you tried so far?

Denote how it says monthly

[obtuse pebble]

@plain grove Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[placid smelt]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tranquil sonnet]

Where’s the work?

[placid smelt]

Do u even know how to do these?

[tranquil sonnet]

Yes lmao

[placid smelt]

nah u lying

u got the pre uni tag

[tranquil sonnet]

-1<2/x<1

For the first 1

-1<4x+20<1

Second

[placid smelt]

well u need to solve for x

but yeah

do u know how to solve the sum?

[tranquil sonnet]

yes a/1-r

[placid smelt]

and a will just be 1 in both

and r will just be 2/x in the first and -4(x-5) in the second

[tranquil sonnet]

Yes

[placid smelt]

nice work

ty

[tranquil sonnet]

I feel like I got quizzed but yw

[placid smelt]

lmao

I had to

cuz all the pre uni people who have come to ever help me

r complete trolls

and just waste time

u know ur stuff tho

[tranquil sonnet]

thanks lol

[placid smelt]

.close

[obtuse pebble]

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[whole dock]

Need help with 16

$(x+4)^2 = 12(1-y)$

[warm shale]

coldtee

[whole dock]

$x = X - 4 \
y = Y + 1$

[warm shale]

coldtee

[whole dock]

V(-4,1)

But

How

[bright quail]

i think the answer key might be wrong

graphing calculator concurs with ur answer

[whole dock]

I see

I hate this book

[bright quail]

what book is it?

[glossy storm]

is it ncert?

[whole dock]

Rd sharma

[glossy storm]

oh

[whole dock]

This isnt the first time

This makes it like 4 times

[bright quail]

hmm

[glossy storm]

ig you're using older editions?

[bright quail]

maybe it's an updated version with not updated answer?

yeah

[whole dock]

Yeah

I guess

Its 2021

[glossy storm]

bruh , it should've been updated then

so you're in class 11

jee?

[whole dock]

Mhm

[bright quail]

good luck!

when is jee

[obtuse pebble]

@whole dock Has your question been resolved?

[whole dock]

Thanks

[whole dock]

April next year perhaps

[bright quail]

ah ic

i believe u will do very well

[obtuse pebble]

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[bright quail]

lots of time to prepare

[obtuse pebble]

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[silver harness]

what is the difference between implicit and explicit

why did we add exponential to both side

in explicit form

[kind hawk]

explicit form is y=(formula only involving x)

implicit form is essentially (formula involving y and x)=0

it defines a function y(x) but you cant really read that function off

[silver harness]

so I have to add expontential to both sides?

[kind hawk]

in this case, yes. to solve for y

(which they havent completely done here but the last step is obvious so they omitted it)

[obtuse pebble]

@silver harness Has your question been resolved?

[silver harness]

may I know why -1/12

the integral of e^u

is e^u + c

[obtuse pebble]

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[wanton dagger]

how do I show this is true?

[wanton dagger]

so purely imaginary means Re(z)=0

do I set this expression equal to 0 and do something with it ig?

[gilded needle]

have you tried multiplying everything out?

[wanton dagger]

yes

is this fine?

[gilded needle]

no

[wanton dagger]

where did I go wrong?

[gilded needle]

what is $i\sqrt{3}$ times $i$?

[wanton dagger]

3i

[warm shale]

Bungo

[gilded needle]

sorry, typo, see updated version

[wanton dagger]

ok

i^2 sqrt 3

so let me redo that

ok so that

[gilded needle]

right

now can you simplify $i^2 \sqrt{3}$?

[warm shale]

Bungo

[wanton dagger]

so i^2 is -1

so -sqrt 3

[gilded needle]

right

so now look at the 1st and 4th terms

sqrt(3) and -sqrt(3)

[wanton dagger]

is that fine?