help-0

[alpine sable]

but from what i am seeing online

it boils down to convention or whatever

[waxen osprey]

yh so from my understanding

let us say we have a function h(x) and we want to find fourier transform of this

[alpine sable]

Yeah

[waxen osprey]

well what I actually mean is we want to approximate the FT for N frequencies with Nyquist frequency

you know what Nyquist frequency is right?

[alpine sable]

yes

[waxen osprey]

so let us space the time lets say t_n where $t_n=n\delta$ so equally spaced

[ocean seal]

3.141593
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[waxen osprey]

n=0,1,2,3,4,....N-1 is the usual convention (we start from 0)

so in other words we are sampling h(t_n) for n=0,1,2,... N-1

okay so we need frequency to be within nyquist frequency so $f_n=n/(N\delta)$

[ocean seal]

3.141593

[waxen osprey]

so the discrete fourier transform plugging it into integral $h(t)e^{-2pif_nt}$ and then using riemann sums to approximate the integral

[ocean seal]

3.141593

[alpine sable]

yeah im with u so far

[waxen osprey]

actually Im not too good with latex I have some very good notes on this from my lecturer

Ill send you it

coz I cant write integrals and Sigma sum signs using Latex lol

but it is very clear so you will be okay

[alpine sable]

its okay

[boreal verge]

\int and \sum btw

and just put _ and ^ for lower and upper bounds

[waxen osprey]

https://www.damtp.cam.ac.uk/user/examples/B8Lc.pdf

Page 32 onwards

[waxen osprey]

thanks!

[alpine sable]

alright thanks lots!

will look into it i think i figured out my question

will give ur notes a shot too

thanks again

[waxen osprey]

actually its page 30 onwards

[alpine sable]

have a nice one :))

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[alpine sable]

got it!

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[rustic junco]

Can someone explain what splitting the middle term is

[ruby current]

writing x^2 - (a+b)x + ab as x^2 - ax - bx + ab

[rustic junco]

Thanks

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[surreal oracle]

why is x > 2 ?

not x > 1

i dont get it

[keen mason]

it should be x != 2

not just below 2

think the question a bit wrong

if x = 2

the denom would be 0, and answer would be undefined

[surreal oracle]

would that be the range as well?

[plush wagon]

it is greater than 2 because anything below 2 is put in f^-1(x) than the function will be negative and in question it is clearly stated that the function f(x) that is always positive

[surreal oracle]

x > 2

[surreal oracle]

right

ok thanks

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[glad osprey]

is this working out right so far

this is the question

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[ancient ivy]

With the derivative of the fonction f(x) (with an optional y ) show that :

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[formal obsidian]

Hi

Can someone help me

Please

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[charred cobalt]

I got a basic equation but can't find someone who solves it without log its 2^x+2 = 12 how can i solve for x (x+2 is the power)

[royal plank]

use exponent rules like $a^{b}\cdot a^c=a^{b+c}$

[ocean seal]

Jigglyproff

[royal plank]

this makes it very easy, but 'in your head' you will still effectively use log

[charred cobalt]

ok i'll try

[mortal trellis]

use () around the exponent. 2^(x+2)

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[weak widget]

would love some help on this

[weak widget]

out of these three examples why does my formula not hold true on equation 1?

[outer lark]

[weak widget]

sorry one moment

ill repost

[outer lark]

Why was it marked incorrect?

[weak widget]

logically thinking if you were purchasing something for 15 dollars with 72% tax it wouldnt cost 53 dollars

should be closer to a fee of 10.80 so a total of 25.80

my forumla allows this to be true for the other 2 equations but not when the % is higher

[tacit arch]

show the question to these calculations

[weak widget]

Ill send the example they gave

This is the equation exactly as they wrote I was basing my math off of

its the calculation of a fee on top of the purchase price in % form

then translated back to a dollar amount

question states "what is the offering price that you would pay per unit"

[outer lark]

I dont think this is the solution to your question as I don't understand but you can play around with the slider here if it helps visualize stuff https://www.desmos.com/calculator/jxppgqkwyr ,

it holds between a VERY

low percentage

like above 6% it breaks

[weak widget]

like the formula is only true up to 6% ish?

[outer lark]

yes

[weak widget]

thats so dumb

[outer lark]

go on that link and slide the a bar

[weak widget]

its extremely unlikely but technically this product could have any % of front end load fee

[outer lark]

sounds like a question for google

[weak widget]

I've tried that haha

thanks for your help though

i appreciate it

[outer lark]

im reading it cant go above 8.5%

[weak widget]

perfect good to know

[outer lark]

from 'FINRA' rules

👍

[weak widget]

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[alpine sable]

Why is it 36^6-26^6

I understand that 36 is 26 letter + 10 digits

But why did we have to subtract 26^6

Nvm

I got it 😭

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[carmine kelp]

** I can't, partly, understand the logic behind this-**,Correct me if i'm wrong, in this case, we are dividing a whole number by a fraction. The method this girl uses explained (correct me if i'm wrong) Is to get 1 / fraction. Which, logically, because it is how many times the fraction can be inside 1 - (atleast rethorically, because its likely not going to be a whole number) -, when its multiplied by 8 in this case, it's going to give us the answer for the division. However, later in this video, she explains that we can simplify this method by only multiplying the number by the fraction reciprocal. However, this would mean that 1 / (any fraction), is always a reciprocal of the original. I can not understand why this is. And that is what I would like to be explained

[marsh rapids]

So you're basically asking why 1/(a/b) = b/a ?

In this example with a = 3, b = 4

And then multiplied by 8

[manic jasper]

well b/a = 1.3 recurring

wait im confused it makes sense

[carmine kelp]

Exactly

It should be true in all cases

[marsh rapids]

It is

Assume that 1/(a/b) can be represented as a fraction c/d
Then multiplying by a/b we get 1 = c/d * a/b = ac/bd
Therefore c = b, d = a is a solution as ab/ab = 1. So are a bunch of others like 3b and 3d, but they all make the same fraction.

Therefore 1/(a/b) = b/a

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[next pulsar]

how is the inequality and the PSD matrix equivalent statements? I don't see it

[next pulsar]

I doubt this is needed for my specific question since it's probably something obvious i am missing, but this is from this wiki page https://en.wikipedia.org/wiki/Nevanlinna–Pick_interpolation

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[steady basin]

confused on how they got this line

[alpine sable]

e is the identity

so if you multiply one identity with the other, it fixes them

[steady basin]

wdym fixes them

[alpine sable]

1.9a and 1.9b

ae=a

then do appropriate substitutions for a as indicated

Oh, it's just proof that u can't have two identity elements in a group

[steady basin]

sry im a bit confused

[alpine sable]

e_1 * e_2 = e_1 because e_2 is an identity

this comes from 1.9b

[steady basin]

yh

[alpine sable]

similarly, e_1*e_2=e_2

[steady basin]

yh

[alpine sable]

so e_1=e_2

[steady basin]

ight ok

think im overthinking it

tbh

[alpine sable]

well basic

a•e = a right ?

[steady basin]

yh

[alpine sable]

So, let's say in ur group, u have two identity element e_1 and e_2

[steady basin]

yh

[alpine sable]

since we just said that a•e is a
We can say that
a•e_1 = a and also
a•e_2 = a
Right ?

[steady basin]

ye

agree

[alpine sable]

So, in the end, we have a•e_1 = a•e_2
Okay ?

[steady basin]

ye

then the a's canel

so e1 = e2?

[alpine sable]

Yep

This basically proves that

[steady basin]

yh kl

easy

lol

[alpine sable]

"in a group A closed for operation * it holds that the group only contains one and only one identity element"
When u try to force two identity elements like e_1 and e_2 u just end up with them being equal

[steady basin]

also what does it mean for an element to be unique?

[alpine sable]

distinct, basically separate from each other
I can give few examples of it but do u know multiplicative group of real numbers for it then ?

[steady basin]

i think so

just give the examples

[alpine sable]

Well, distinct means unique, for example we have axiom states that every element has a "unique" inverse
If I pick any number, let's say 3 from muktiplicative group of real numbers, it's unique number will be 1/3 since multiplication of them will give me neutral element
Similarly 4 will have inverse 1/4 note that when u change the numbers, the inverse also changes, that's what it means for inverses to be distinct

Technically, I could also say that inverse of all elements is 0
3×0 = 0
4×0 = 0
......
But it's not "distinct" it's just 0s

[steady basin]

i see

[pseudo ice]

https://tenor.com/view/what-do-you-mean-by-that-druski-gif-20560325

[limpid spade]

dj khaled anoda one

[alpine sable]

What is the identity element of multiplicative group of real numbers ?

[pseudo ice]

1

[alpine sable]

The question I want to ask u is why ?
Why is it 1 ?

It can also be 0

[steady basin]

everything times 1 gives the same result back

everything times 0 gives 0

[alpine sable]

Same with 0, multiply anything by 0 u get same result
3×0 = 0
4×0 = 0
5×0= 0

[steady basin]

trivial

[pseudo ice]

But it doesn't satisfy the a*e = a

If you took e = 0

[alpine sable]

Yep, but there's also a fundamental identity it breaks

Apart from a•e

And the identity it breaks is also the reason why 0 is not e

And also, that identity will explain what "unique" means

Another way to say would just be set of all a is bijective to set of a^-1

That's one way to say what unique invserse means

[pseudo ice]

Hmmm fair fair I guess, maybe I'm just tripping on reading that, knowing how it should look

[steady basin]

Is this always true for addition and multiplicative groups?

[alpine sable]

Yeah

[pseudo ice]

That's generally how you'd write the identities and inverses

[steady basin]

ok

[alpine sable]

@steady basin are u in university ?

[steady basin]

ye

1st year

semester 1

[alpine sable]

That's wonderful ^°^

[steady basin]

hbu

[alpine sable]

I wanna go to uni too ❤️

[pseudo ice]

Of course, there's e.g. S_n, the set of permutations of a set n, but I haven't seen it written like the multiplication way w/ "e = 1"

[alpine sable]

Senior Year Highschool in India ❤️

[steady basin]

danggg

gl

[alpine sable]

Thank you ^°^

[pseudo ice]

Groups

[lone heart]

@steady basin Has your question been resolved?

[steady basin]

im trying to do this question

so first off imma show if its abelian

which means that the operation X is commutative correct?

[pseudo ice]

What a complex question

[steady basin]

thats not the same as the question is it?

since ad - bc is not the same as bc - ad

ie doesnt that mean that it isnt Abelian?

or have i made a mistake

?

[pseudo ice]

Hmmmm

[steady basin]

this is even more funny now

lol

[steady basin]

u get my point tho right?

[pseudo ice]

Yeah I see what you mean, it shouldn’t be that way though

Unless I’m being slow haha

[steady basin]

imma prove its a group at least

then ill show the mark scheme

[pseudo ice]

,w (a+bi)*(c+di)

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by chartbit#3171

[pseudo ice]

Damn you Wolfie 😒

Didn’t need you anyway, can work on paper, how bout dah

[minor needle]

for it c is speed of light

xd

[pseudo ice]

Anyways think they should have that def as (ac-bd, ad + bc)

[pseudo ice]

My saviour

[pseudo ice]

Yeah voting on their definition

[steady basin]

r u taking the piss

not u chartbit

look at this

and i just solved the associativity thing

...

so i gotta do it again now

[pseudo ice]

Hahah, see how all of us spotted that they were wrong 😂

They’re really violating

[steady basin]

first the calculus yesterday

now this

...

[pseudo ice]

You know what, I can never say I’m bad again after seeing that

Even I’m not that bad 😒

Like how long does it take to check the product of complex numbers ffs

[hollow sparrow]

lol , SA is very anger , relax man 😆
it's happened

You have to hope that they don't make any mistakes on the exam paper

[steady basin]

true

lol

[pseudo ice]

[steady basin]

so im trying to prove G1 associativity

is this correct?

but anyway is there an easier method

seems like ms did an easier method

[pseudo ice]

Hmmm, imo not really, unless you spot things? What did they do, skip steps?

[pseudo ice]

Skip some steps it is then

[steady basin]

how do i make this thing

look like

[pseudo ice]

imo just expand those you have there and they should be "obviously" the same

[steady basin]

or vice versa

ok

[pseudo ice]

If you show they get to the same thing then you're done

[steady basin]

so for G2 which is identity element

do i do (a,b) x (1,0)?

and (1,0) x (c,d)

?

[pseudo ice]

to both

[lone heart]

Please read #❓how-to-get-help

[steady basin]

ok ive done identity element

that was easy

now G3

which is inverse

hmmm

[pseudo ice]

Yeah, hmmm

[steady basin]

so (a,b) times something has to equal (1,0)

in terms of the group definition of X

hmmmm

cuz (a,b) x inverse = (1,0)

[pseudo ice]

Yep, that already looks like pain to try and find that way tbh

[steady basin]

hmmmm

[pseudo ice]

This chat makes interesting reading tho hahah

cough cough

[steady basin]

lmao

any idea then?

or u wanna see MS

?

[pseudo ice]

I wonder

[steady basin]

MS?

ok

[pseudo ice]

Think about it for a bit

Remember the whole mention of complex numbers and everything

What's the inverse of a + bi as a complex number?

[steady basin]

1/a+bi?

[pseudo ice]

make the denom real?

[steady basin]

wait sry gimme sec chartbit

might have a method

had a lightbulb

moment

@pseudo ice

how do i solve this sim eqs lol

[pseudo ice]

That's pretty peak imo, wouldn't do that tbh

[keen plinth]

it is solvable

[steady basin]

ok chartbit

lets go back to ur method

lol

[pseudo ice]

My guess is that you're to notice the above, but...

[pseudo ice]

...I'm interested now

[pseudo ice]

That said, via mine, you're matching (a,b) to a + bi and so on

[steady basin]

might cba in like 5mins

spent way too long on this and need to move on lol

[keen plinth]

,w rref {{a, -b, 1}, {b, a, 0}}

[ocean seal]

Results provided by WolframAlpha

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

Requested by 夢雪#2250

[keen plinth]

it's right

[pseudo ice]

Oh damn fair enough

[steady basin]

imma show MS

[keen plinth]

[pseudo ice]

Yep that gets you there, whatever way you get there tbh

[keen plinth]

they did difference of squares

[steady basin]

note how its exactly the same as how they make you "discover" complex inverses (ie. multiply by conjugate, notice that its real, and scale down by appropriate factor)

[keen plinth]

no normal people just rationalise the denominator

[magic haven]

copypasting my messages into here 😭😭

[steady basin]

eh?

[magic haven]

you know what youve done.

[steady basin]

if i speak i am in big trouble

[keen moss]

?

[steady basin]

?

[keen moss]

?

[steady basin]

?

[keen moss]

??

[magic haven]

he asked me the same question in a diff server and then cross-posted something i said

[steady basin]

is it now

[magic haven]

im only pretending to be mad about it i dont really care but it is very very funny to me

[hoary heron]

lmaooooooo

[steady basin]

@magic haven if u multiply this inverse by (c,d) tho, it wont give (1,0) will it?

[magic haven]

What

no, bc its the inverse of (a,b)

[steady basin]

ok

.close

[lone heart]

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[warm socket]

I don't understand why we lost a sec^2(x)

I understand the trig substitution and the U-sub, but why did that second sec^2(x) just vanish from the equation?

[manic jasper]

well what does sec^2(x) =

[limpid spade]

What's derivative of tanx?

[proper dust]

its a trig identity

sec2 x = tan2x + 1

[warm socket]

wouldn't we have two of them then?

like (tan^2x +1)^2

[manic jasper]

no bc

[proper dust]

ooh yeah

[manic jasper]

sec^2(x) = tan^2x + 1

[warm socket]

I see, so it would be (tan^2x +1) tan^6x(tan^2x+1)?

[manic jasper]

yeah but

most likely be writting as

(tan^2(x) + 1)^2 tan^6(x)

[warm socket]

Good good, so the website that is showing the steps is incorrect?

[limpid spade]

Why would it be incorrect

[limpid spade]

What can u do with this if u wrote it like this

[outer lark]

dx -> du ...

[warm socket]

since it doesn't square the tan^2x+1 to accomodate having two sec^x

[manic jasper]

wait

[outer lark]

no

it does

[manic jasper]

where does the other sec go

[outer lark]

its changing from dx to du

the integrand is with respect to x

[manic jasper]

yes bc they are subbing in u for tan

[proper dust]

OH we forgot lmao

[manic jasper]

so u = tan

[limpid spade]

Yeah

[outer lark]

therefor du =...

[manic jasper]

OOOO

YEAH

[proper dust]

LMAOOO

[manic jasper]

no you right

when i saw du i thought u = x and idk why

[warm socket]

Ohh i see

that helped

.close

[lone heart]

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[vestal adder]

Hi, i need help with a very large matrix equation that I am not sure can be solved, Im using google sheets to do it in: its a matrix equation of a 1x23 * 23x5 = 1x5 matrix and i need to find the 1x23 matrix(for my purposes its amount * mods = values and i need to find amount)

[vestal adder]

Im not sure where to start, as i have a very basic understanding of linear algebra

and i cant do non square matrices

[plain flame]

there is no unique solution

[vestal adder]

yes i know

is there a way to find all solutions?

[plain flame]

oh absolutely

but i personally dont know of any software that can handle this size

im sure it is out there

try #computing-software

[vestal adder]

ok thanks

.close

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[vestal adder]

.reopen

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[vestal adder]

@plain flame is there an algorithm i could do in google sheets?

[plain flame]

unfortunately i have no experience with that

[vestal adder]

shoot ok

ill figure something out thank you

.close

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[heady berry]

Please help with part B

thanks

[lone heart]

@heady berry Has your question been resolved?

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@heady berry Has your question been resolved?

[heady berry]

please help 🥺

[manic jasper]

hey

good luck

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[signal mountain]

please help

[last ether]

lim (A + B + C) = lim A + lim B + lim C

But note something about each fraction

What's true about the denominator versus the numerator

[signal mountain]

riemann series

?

[last ether]

You remember what you can do if you have limits to infinity?

You can erase low order terms

Each fraction essentially becomes n/n^2 = 1/n

[signal mountain]

kk

[lone heart]

@signal mountain Has your question been resolved?

[naive valley]

question seems a bit dodgy, how many terms are represented by the ... ?

and how does 1/n fit the pattern ?

[tacit arch]

$\frac{n+k}{n^2+k^2} = \frac{1 +k/n}{1+(k/n)^2}\cdot \frac{1}{n}$

[ocean seal]

riemann

[tacit arch]

should be able to turn that into a riemann sum now

[modern topaz]

@tacit arch sum

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[signal mountain]

oh

[lone heart]

Please don't occupy multiple help channels.

@signal mountain Has your question been resolved?

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[velvet kernel]

if I were to pay for a service that costs 170.823 USD / hour, how much would it use that service for 45.123 minutes?

[lone heart]

@velvet kernel Has your question been resolved?

[sonic ridge]

well an hour is just 60 mins

u can express it the cost per hour as a fraction/ratio

and that would be equal to the ratio of x dollars per 45.123 mins

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[desert iron]

~128.5 dollars

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[serene lava]

Hello, I have some trouble computing the integral $\int_{\gamma} \overline{z}\ dz$ where $\gamma$ is the line $\gamma(t)=t+it$. The result I got from this is 1.

[serene lava]

Doing the same integral but for $\gamma_{1}=t$ and $\gamma_{2}=1+it$ I dont get the same results

[ocean seal]

econo

[ocean seal]

econo

[serene lava]

shouldve been \overline, my bad

my question is: why aren't they the same?

im thinking theyre analytic and should be the same according to the independence of path

[worn fox]

what do you think is analytic?

[serene lava]

$f(z)=\overline{z}$

[ocean seal]

econo

[worn fox]

its not analytic

[serene lava]

oo

from C-R eqs, right...

thanks !

[worn fox]

yep, not differentiable

[serene lava]

saved my sanity !

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[alpine sable]

How do I solve q 12?

[alpine sable]

Why are we omitting R0 here btw?

I got all the answers right by omitting R0

[worn fox]

wdym omit R0

[alpine sable]

Sorry I just realised my mistake

Thx

.close

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[lone heart]

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[spare fern]

Could someone help me out with this question

[spare fern]

Here is my working

I'm just confused on how to represent it as an inequality

[lyric stream]

Looks correct

[spare fern]

so how would I represent it as an inequality

[lyric stream]

As a singular inequality?

[spare fern]

like the selectable visible answers above

alright

[vale wigeon]

bad!

[lyric stream]

oh ic

sorry

[spare fern]

-8 >= x >= 2

[vale wigeon]

this is bad. don't ever write self-contradictory inequality chains like this one.

when you chain several inequalities together, they are joined by a logical AND, not by a logical OR.

[spare fern]

?

I just followed what the instructional video did on past example questions

[vale wigeon]

can you show the instruction video and timestamp

[spare fern]

ok

[vale wigeon]

i think there might be an important difference between your current problem and whichever ones are shown there

[spare fern]

so is this correct?

[alpine sable]

You are saying that there is some value x that is lesser than -8 but also greater than 2. Does that make sense to you?

[tacit arch]

No number satisfies that double inequality

[alpine sable]

One of the thing that it implies is that -8 > 2

[fallen verge]

oh yooo riemann you got honorable?

v cool

[vale wigeon]

ah see

the thing is

inequalities that look like |x| ≤ a and those that look like |x| ≥ a are actually quite different in the nature of their solution set

[spare fern]

oh ok

[vale wigeon]

the solution set of |x| ≤ a is an interval, while that of |x| ≥ a is a union of two rays.

but the shot from the instruction video does not help illustrate that difference.

although, i suppose that solving it the way they've shown, where you first find the points at which equality happens and then draw it on a number line, is valid...

it feels a bit too mechanical

[spare fern]

oh I see

although according to my working x > = -8 and x >= 2

[alpine sable]

So x can be in all numbers from 2 to infinity ?

[spare fern]

but it does not appear on the selectable answers

[alpine sable]

It does

There is indeed an option in the ones u r given where x is between -8 and 2

[vale wigeon]

brb

[alpine sable]

Is it too bad if I am watching Lex Podcast rn too ?

[spare fern]

hahahah that's all good just try not to fall asleep

[worn fox]

you should show how you came to this conclusion, particularly how you decided if its <= or >=

[spare fern]

ok so as per my working I didn't divide by a negative number hence the inequality sign shouldn't change right

[alpine sable]

Bigger number gets the bigger side
Smaller number gets the smaller side, don't forget

[worn fox]

your working is you solving an equation which is fine

but you haven't shown how you've concluded your answer for the inequality

[alpine sable]

I think ur answer is correct in terms -8, 2 but what ur problem is how u use ≤ and ≥

[spare fern]

yeah I'm pretty confused

[worn fox]

how did you arrive at x >= -8?

[spare fern]

hold on I'll send more working

[worn fox]

your right hand side should read $-(3x+9) \geq 15$

[ocean seal]

ΣAC

[worn fox]

and then if wanted to move the negative to the other side your inequality sign would flip

[spare fern]

Oh true I forgot about that

But this question for instance

Just has the product as negative on the second equation (right)

How would I know whether to make the whole equation negative vs just make the produce negative?

[vale wigeon]

i want to say something but it feels like theres a lot to address here.

[worn fox]

go ahead

i feel like the issue will be with them not really understanding what abs is doing

[vale wigeon]

your school, lex, would apparently have you believe that inequalities are just broken equations and can be treated as equations with an extra complication regarding the sign in the middle.

[spare fern]

More or less yes, although I do know that certain algebraic operations may alter the inequality.

[vale wigeon]

well for shame

this kind of thinking betrays you when you try to apply techniques from absolute value equations blindly to absolute value inequalities

are you familiar with the geometric meaning of absolute value?

[spare fern]

The distance a number is away from zero on a number line?

[vale wigeon]

yes

or, more generally, |a-b| is the distance between a and b.

[spare fern]

Ok, so in the context of this question what am I missing?

[vale wigeon]

|3x+9| ≥ 15

it's this one that you're doing, yes?

[spare fern]

Yes

[vale wigeon]

well, you're missing the fact that the inequality |3x+9| ≥ 15 has precious little to do with the inequalities 3x+9 ≥ 15 and 3x+9 ≥ -15.

you could commit yourself to casework, but it requires careful manipulation and thorough understanding of what you're doing. not mere treating-inequalities-as-broken-equations.

this is the kind of picture that i think would be helpful for you to have in mind.

of course, your inequality will require some adjustment to be put into one of these forms (and even then there will be a tiny difference)

from |3x + 9| ≥ 15, divide both sides by 3 to get |x + 3| ≥ 5.

and then realize that x + 3 is the same thing as x - (-3), thus the inequality can be read as "x is at least 5 units away from -3"

does this make sense to you?

[spare fern]

Somewhat

[vale wigeon]

can you point to a part that does not make sense?

[spare fern]

How would I find the exact inequality of x with the information “x is at least 5 units away from -3”?

[vale wigeon]

so the part that doesn't make sense is "but how do i proceed further"...?

[spare fern]

Yes

[vale wigeon]

okay, but is there anything in what i've written that requires elaboration?

[spare fern]

No

[vale wigeon]

right

well, i'd like for you to draw a number line

mark on it the point -3

and then mark off points that are 5 units away from it in both directions

do this and show me the result

[spare fern]

Okay

[vale wigeon]

you forgot the absolute value bars in that inequality

it's |x+3| ≥ 5, not x+3 ≥ 5

anyway

[spare fern]

Right

[vale wigeon]

can you now take your number line and illustrate on it the set of all points which are 5 units or further away from -3?

[spare fern]

But I just did?

[vale wigeon]

no, you did not. you only marked the points which are exactly 5 units away.

i want you to shade in the points that are 5 units or further away from -3.

[spare fern]

Including -3?

[vale wigeon]

what do you mean, "including -3"?

do you think the point -3 fits the description of "a point that is 5 units or further away from -3"?

[spare fern]

No I mean I’m just used to illustrating open and closed circles when representing inequalities, hence it’s easier for me to use that notation.

[vale wigeon]

ok sure but

(a) you have not yet shown any intervals on the number line

[lavish sky]

Anyone problem from this?

[vale wigeon]

(b) the question of whether to use open or filled circles at the endpoints really is answered by the statement of the task i gave you

but

(c) -3 is not an endpoint

so asking "do i include -3?" is a bit nonsensical

the point -3 does NOT belong to the set of all points that are at least five units away from it.

[spare fern]

So do I just add to arrows on each opposing side of the number line

[vale wigeon]

no you do not "just add arrows"...

[vale wigeon]

look at this

[spare fern]

Then?

[vale wigeon]

you're supposed to highlight a subset of the number line

do it in a way that makes it obvious what is highlighted and what is not...

[spare fern]

Oh I see

That seems a lot more intuitive

[vale wigeon]

i suppose that could work. but then you risk agonizing over which one you're supposed to flip and which one not.

[lavish sky]

Look, in the absolute value, the coefficient is just an obstacle, you should simplify the parentheses as much as you can. But don't be blind to addition and subtraction.

On the other hand, when the sign is towards the absolute value, we consider the outer 2 roots....

[spare fern]

Why did the signs flip on the last line?

[lavish sky]

You remember this?

If the sign is on the side of the number, the number goes between the absolute value. If the sign of the side is absolute value. The smaller number is equal to the absolute value along with the negative and the larger number is equal to the absolute value

It's a kind of formula. Try it and you will see that it will not go wrong

[lone heart]

@spare fern Has your question been resolved?

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[alpine sable]

I'm doing the advanced euclidean algorithm
but for some reason it just won't work

https://docs.google.com/document/d/1tFqTZvXmbf-Ib0IVbuPV3FJWlpdCZt063xrQid-Yaxc/edit?usp=sharing
Google Docs
advanced euclidean algorithm example
a=11 b=7 q=1 r=4 i a b q r x y 1 11 7 1 4 2 7 4 1 3 -1 3 4 3 1 1 1 -1 4 3 1 3 0 0 1 ggT(11,7)=1 Xi+1 - (qi * yi+1) = a * x + b * y = ggT(a,b)

here's the table
the formula ax*by=ggT(a,b) is valid all throughout until I get to the second last line and I don't get why
can someone explain what I'm doing wrong?

[lavish sky]

Hi

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

.reopen

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✅

[alpine sable]

yes hi

do you know what I did wrong

<@&286206848099549185> someone help out please

[manic jasper]

what are you struggiling with

[rancid vault]

fractions

every kind of fractions

[manic jasper]

go to ur channel

[rancid vault]

addition subtraction multiplication divison

mixed

simplifying

[manic jasper]

go helpv9

[limpid turret]

@alpine sable if you still need help in 10 hours I'll check it out

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@alpine sable Has your question been resolved?

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[viral pine]

Balls

[viral pine]

could anyone give some pointers on how to go about doing q4i

[wanton pebble]

equate left and right side

[viral pine]

how do i get rid of the power

[wanton pebble]

why did you do this

[viral pine]

was I not supposed to econs

epanco*

expnd*

expand*

[wanton pebble]

10x-6= A(5x-2) +B

find A B

that was the question right

[viral pine]

wait

I wasn’t supposed to equate the

y = (10x+6)/(5x-2)^2 together?

[wanton pebble]

noooo

[viral pine]

Ok solved it

2 and -2

[wanton pebble]

see equate constant terms and terms with x seperately

[wanton pebble]

yp

[viral pine]

is part two correct

[alpine sable]

10 hours?

man that's a lot

[wanton pebble]

no

[viral pine]

why

how wher

[wanton pebble]

if are using product rule atleast find derivative of both terms

(uv)' = u'v+ v'u

or u can use quotient rule

and yo have to integrate y

not the result you got from differentiation

and for integration you have to simply y by using value of 10x-6 in first part

[viral pine]

@wanton pebble is it possible to do them using neither product / quotient rule

because we weren’t taught those

[wanton pebble]

i dont know

these are basic rule

so you were taught integration but not product rule?

[viral pine]

they aren’t included in the syllabus im being taught

[wanton pebble]

okie

[viral pine]

the calculus im being taught is actually really easy

[wanton pebble]

https://youtu.be/4JlYwRRmZws

[viral pine]

I should be able to solve the question

without using any of the two

rules

[wanton pebble]

yes

put 10x+6=2(5x+2) -2

the solve

then*

[viral pine]

what’d I do with x next

[wanton pebble]

see then it become 2/(5x+2) -2/(5x+2)²

then u can the u can diff. them separately

[lone heart]

@viral pine Has your question been resolved?

[lone heart]

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[lilac iron]

help pls

[manic jasper]

so

whats the angle rules

[lilac iron]

1 sec

this is as far as i got

sorry for my handwriting

[manic jasper]

dwdw

right lets start from the beginning and note down

what we know

so lets give AC and AB's length the variable y

We know m = 15 degrees

so what do we notice about

triangle ACE and triangle BDE

[lilac iron]

they're congruent

[manic jasper]

so what rule do we apply

[lilac iron]

uhhh

[manic jasper]

there are 4

[lilac iron]

you mean angle angle angle?

[manic jasper]

so for the SSS rule to apply

what do both triangles need to be

ill give you a website to refer

https://tutorme.com/blog/post/triangle-rules/

[manic jasper]

so you want to apply the AAA rule

[lilac iron]

this is the only option i see

[manic jasper]

well

we know both are right angle

and none of the triangles sides are equal

we have to use the AAA rule

[lilac iron]

yea

[manic jasper]

so we know both shapes have the angles 15,90 and 75 in them

Now we move to the triangle

ABE

what angle can we first find out

[lilac iron]

AEB

[manic jasper]

and what would it be

[lilac iron]

105

[manic jasper]

okay good write it

and we know the base of that triangle

what is it

[lilac iron]

y

[manic jasper]

good

[lilac iron]

as you said earlier. though i dont really understand how did you find out CAB was isosceles

[manic jasper]

well

lets apply pythagorus theorum then

so we know

the hypotunese

is 2y

so c^2 = 2y^2

[lilac iron]

wait... what triangle are we talking about now?

[manic jasper]

ACE

ill show you why

so we know c = 2y

and we know both angles

so what rule do we use to find out a and b

[lilac iron]

SAS?

[manic jasper]

no calculation rule this time

what equation

do we use

[lilac iron]

i mean ASA

[manic jasper]

So

what would the equation be

[lilac iron]

uhh

[manic jasper]

hint: sin tan cos

[lilac iron]

oh

[manic jasper]

do you know?

[lilac iron]

you mean soh cah toa?

oh sines theorem?

[manic jasper]

yes

so we know sinA/a = sinB/b = sinC/c

[lilac iron]

no im afraid i dont know the sines theorem

[manic jasper]

thats it here

[lilac iron]

ight

[manic jasper]

so what one can we fill out

wait

no im being stupid im

im being dumb

you were right

we use sohcahtoa

[lilac iron]

yipee

[manic jasper]

dwdw mbm

so we know the hypottunese

we know all the angles

what side we finding

[lilac iron]

AC?

[manic jasper]

yeah we can do adjacent

what angle would we use then

[lilac iron]

cos15?

[manic jasper]

yes

cos15 = 2y/the adjacent

[lilac iron]

but we're left w 2 variables now

[manic jasper]

well no lengths are given

no measurements are needed if we make y = something

bc no matter what we make y the equation will work

[lilac iron]

hm

i think there is a way to solve this without trig

it's from a 7th grade math book

[manic jasper]

there is

[lilac iron]

y know how?

[manic jasper]

yes

i can show you

@lilac iron

[lilac iron]

oh no thanks, i figured it out

i appreciate your help

.close

[lone heart]

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[prisma laurel]

bro all these math equations are too complicated for me i joined because i thought i was gonna get sum yr 7 math💀

[prisma laurel]

whyd it pin

[vague mirage]

Bro this is a help channel

Don't text in this channel

Read #rules

[gray isle]

the auto pin by the bot is intended to make it convenient for helpers to keep track of the original question

[vague mirage]

.close

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.reopen

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✅

[vague mirage]

@prisma laurel wait do you have any questions?

[prisma laurel]

yes whats pi?

[gray isle]

pi is the ratio of the circumference of a circle to its diameter

[prisma laurel]

uh i dont understand so im going to google how to explain pi for 6 year olds thank you though

[keen mason]

True but u can ask for help too

which is mainly why ppl join

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[polar breach]

How do I do 4

[wanton pebble]

calculate 2a+2b and 2a-2b

first

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[lavish sky]

Which one is

What it wants? Because u can give any N number

[ocean seal]

Intru[S]

Intru[S]

Intru[S]

Intru[S]

[wise comet]

What's wrong with $n=1$? Then the inequality is true for each fixed $\varepsilon>0$.

[ocean seal]

RiemannZeta

[wise comet]

To clarify, I understand the question to be:

Fix $\varepsilon>0$. Does there exist an $n_\varepsilon\in\Bbb N$ such that $\frac{3}{10}-\varepsilon<\frac{3}{10^{n_\varepsilon}}\leq\frac{3}{10}$?

[ocean seal]

RiemannZeta

Intru[S]

[wise comet]

I'm not sure your question makes much sense. Once $n=2$ or larger, you're looking at $3/10^2=0.03$. In order for this to make sense, you'd have to take $\varepsilon$ larger (as opposed to smaller). So $n=1$ is the only sensible choice.

[ocean seal]

RiemannZeta

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[acoustic compass]

How do you do 49 ^ .75 without a calculator?

[minor needle]

use properties of exponents

[alpine sable]

0.75=3/4

[boreal verge]

rewrite .75 into a fraction and properties of exponents

[acoustic compass]

so just make it a fraction?

jeez I hate fractions 😔 alright

thanks yall

[median osprey]

Express 49 as 7^2

You will have to calculate square root manually in the end

By the standard method

[gray isle]

supposedly the goal is just to simplify (and not give a decimal approximation)

[acoustic compass]

I'm not understanding, generally 49 * 49 would be 49 ^ 2 right so how is 49 ^ .75 not just 49 ^ (2*.75)

oh wait that gives me 343

thats correct

[median osprey]

(7²)^3/4= 7 ^3/2 = 343 ^1/2

[acoustic compass]

but nvm becuz I can't get that

without a calculator

[median osprey]

Now you will have to calc √343

Which can be easily done manually

[acoustic compass]

wait why would I have to square root I'm confused

[median osprey]

Because here we get power 1/2

To 343

[boreal verge]

$49^{\frac{3}{4}} = \sqrt[4]{49^3}=\sqrt[4]{7^6}=\sqrt{7^3}=7\sqrt{7}$

how did you do ordered roots again

[ocean seal]

Kel.plush

[boreal verge]

there

[median osprey]

Yea it's better to reduce √343 as 7√7

[acoustic compass]

wait so how do u get 7 * sqrt(7) without a calculator

[boreal verge]

its 7²*7

so 7 * sqrt 7

and sqrt 7 in decimal value you gotta use an algorithm

i did one last year that i completely forgot unfortunately

[gray isle]

note that you didn't have to explicitly cube 7

[boreal verge]

did as in learned

[median osprey]

√(7×7×7)= √(7×7)×√7=7√7

[gray isle]

considering exponent laws again,
7^(3/2) = 7 * 7^(1/2)

[acoustic compass]

wait wait let me explain the original problem was this 49 ^ 3/2, I don't like fractions so I turn them to decimals in my head automatically which is 49 ^ .75 it says answer is 343 which is correct

but the only problem is I can get that answer easilly

with a calculator

but this is just practice for the test, of which I can't use a calculator

[median osprey]

3/2=1.5

Not 0.75

[boreal verge]

yeah 3/2 is 1.5

[median osprey]

You made a mistake in converting it into decimal

[acoustic compass]

oh wait ye I'm thinking about 2/3

[boreal verge]

also i dont think theres a human way to compute decimal exponents

[boreal verge]

thats 0.66666....

[acoustic compass]

it's fine I got 1.5 the first time I jsut for some reason remembered and typed .75

[acoustic compass]

But okay so this is the problem

so I literally cannot do it that way

😔

Oh thats y i said .75

49 ^ (2*.75)

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[rugged pebble]

how do i find the range of p:x ->-2|x+4|+10

[outer lark]

notice how -2|x+4| is always negative

so you want an x value that makes this 'least negative'

and that will give you your maximum value.

[rugged pebble]

x is not always negative

that a modulus

[outer lark]

[rugged pebble]

there is a pciture of the graph which looks like a upward arrow, like -x^2 but pointy

[outer lark]

-2 * |x+4| ...

[rugged pebble]

?

[outer lark]

is always negative..

[rugged pebble]

so it would be 2|x+4| +10 and -2|x+4|+10

[outer lark]

huh

[rugged pebble]

as it happens when you calculate a function with a mod

[outer lark]

if given -2 |x+4| + 10

You can see from observation that IF THERE IS ANY VALUE INSIDE THE ||

then we will be subtracting from the 10

10 - 2| SOMETHING |

if something is 0

then we have a maximum of 10

this will then piss off to negative infinity on both ends

,w plot -2|x+4|+10

[ocean seal]

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[outer lark]

hence your range is (-inf,10] ig

[alpine sable]

If you pull the - out or even expand the function you have to remove the mod.

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@rugged pebble Has your question been resolved?

[rugged pebble]

yes i agree

the answer says p(x) <= 10

which i guess makes sense

thank you

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