help-39

[brave sluice]

look at the angle supplementary to the 90

then, there's a triangle so you can find the third angle of the triangle

[stoic imp]

?

[stoic imp]

Yeah

[open blaze]

Did someone delete a ping?

[radiant terrace]

K, L and X make a 30, 60, 90 triangle ig

No derivada responded to it

[brave sluice]

the right angle closer to the 30 and on top of l

[radiant terrace]

Okay yah I see the trick

[brave sluice]

it's a right angle

[stoic imp]

Can you make a drowing

[radiant terrace]

Here is a bad one

[radiant terrace]

When you have two intersecting lines like this the two x angles are the same

[stoic imp]

X=60

[radiant terrace]

Three of your lines make a 30 60 90 triangle where you can use the fact I mentioned to find x

Yep x=60

[stoic imp]

Ohhh

[radiant terrace]

The red mark here is also x just like my drawing.

[stoic imp]

.solved

[pearl pond]

Channel closed

Closed by @stoic imp

Use .reopen if this was a mistake.

[pearl pond]

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[orchid path]

how would one go about proving cos is bounded between 1 and -1. I'm thinking of hypotenuse longer than opposite (from pythagoras) or cauchy schwartz (u.v=|u||v|cos (theta))

maybe my question is what other ways can one prove the above proposition

[orchid path]

js a random question haha

[tropic saddle]

well the first question is how you even define what cos is

[orchid path]

that's what I thought

I forgot how it was even defined

[grand pilot]

The x-coordinate of a certain point on a unit circle cannot exceed 1 or be smaller than -1

[bitter herald]

you have quite a few options. Solution to y'' + y' = 0, power series, unit circle, ...

[orchid path]

good point

yea could do its maclaurin series

such a pointless question that was from me

thanks

.close

[pearl pond]

Channel closed

Closed by @orchid path

Use .reopen if this was a mistake.

[tropic saddle]

with a series its probably painful

[orchid path]

exactly

[fringe raft]

grr

[warm current]

Not really. That's how mathematicians ended up with all these equivalent ways to define it

[orchid path]

true

which definition would occur first though

[fringe raft]

taking an analysis course, the power seires definition probably

[orchid path]

it would be geometry for euclid

[fringe raft]

it's not very natural, but easiest to start off with in analysis

[orchid path]

yes I agree

[fringe raft]

from there you can easily arrive at eulers formula exp(ix) = cos(x) + isin(x), then multiply by its complex conjugate to get cos^2 + sin^2 = 1

and then it's clear that both cos^2 and sin^2 can't have a value greater than 1

[warm current]

That's the beauty of all of them being equivalent. Any could be first and you can always get the others

[orchid path]

I was thinking some areas of maths must first be define before others but yes

[summer gorge]

i don't think so..

[warm current]

If you want a challenge, prove that cos is strictly decreasing from 0 to pi

[fringe raft]

which part

[spare lark]

I think the easier is geometric pythagora imo

[warm current]

Should we reopen this?

[orchid path]

.reopen

[pearl pond]

✅

[spare lark]

You the chief

[orchid path]

sure

[orchid path]

from which definition of cos

[summer gorge]

the easy part analysis is a great method dunno about it being easy if you are only given the basic tools

[orchid path]

geometry is more difficult to axiomatize but sure

[warm current]

Definitely the easiest, but as violeta question showed (and i had these same questions myself once), you eventually want a more rigorous way to prove the trig identities

[orchid path]

better for an intuitive understanding

[warm current]

Any, really

[spare lark]

Ah i meant from this thing like the first one to come

[sage bloom]

how to prove?

i m curious

[warm current]

Yeah, "easy" really falls on the reader. For some, even geometry is really hard

[summer gorge]

i agree

[warm current]

I did it using the Taylor series of sin and then Taylor's theorem

The exact proof, while not overly difficult, is too much to say in just a few short sentences

[sage bloom]

o

[orchid path]

I haven't done analysis yet can't deal with the sign alternating series to prove its strictly devreasing

[fringe raft]

yeah was just saying, if you are purely only considering analysis, then from that standpoint comparing it to other possible definitions you can do using analysis, this just seems the simplest, only requiring you to know about limits, no derivatives/integrals, but outside of analysis it would be not as easy yes

[warm current]

Are you familiar with Taylor series, or basic differential equations?

[orchid path]

I think so

need a quick recap for Taylor series prob

[warm current]

You could start there to prove a lot of this

[warm current]

Yeah they are very easy to forget

[summer gorge]

it is not when you have to approximate everyday

[warm current]

As rbit said earlier though, if you want to prove just that cos is bounded, all you need to show is cos^2+sin^2=1. From there, you can easily show the bounding

[orchid path]

how do I proceed?

[fringe raft]

you want to show euler's formula

what was the series for exp(x)?

[orchid path]

I'm trying to show uh cos x strictly decreasing between 0 and \pi

just to clarify

[fringe raft]

oh that, one sec

the first goal is to show it is decreasing between 0 and 2, from there we can extend it

[orchid path]

give small hints if you can

okay

is 2 completely arbitary

[warm current]

Consider how $\frac{d}{dx}[\cos x]=-\sin x$. Show that $\sin x>0$ for $x$ between 0 and 2

[jolly parrot]

SWR

[fringe raft]

can even be done without derivatives though

[warm current]

2 is an integer between pi/2 and pi. It's easier to work with than some irrational number

[warm current]

(If you have already proven that cos(pi)=-1)

[fringe raft]

right, that depends on the definition of pi

[orchid path]

can it be done without a course in analysis?

[warm current]

Eh i could be wrong there actually

[warm current]

It gets a little fuzzy. Hard to say yes or no

Really, imo, all you need is decent calculus knowledge

[orchid path]

well at least I'm kind of stuck in showing the series is decreasing

[fringe raft]

at least I remember it being done using the cos x - cos y identity

[warm current]

(But calculus is just analysis)

[warm current]

Yeah there may be several ways to do it. That's why i can't say definitively one way or another what all you need to do this

[fringe raft]

hmm, assume x, y in [0, 2], then essentially we want to show that x > y -> cos(x) < cos(y)

[warm current]

That was my first attempt way back when i did this, but i ended up struggling, and went with the sin approach. But maybe rbit has a better idea than myself

[fringe raft]

or in other words, assume that x - y > 0, then show that cos(x) - cos(y) < 0

[orchid path]

okay so I let x = y +k for positive k

then I guess I compsre each term in the series?

[fringe raft]

here the sum to product identity for cos arrives

(you would need to prove that identity first, using eulers formula, but whatever)

[orchid path]

-2sin[(x+y)/2]sin[(x-y)/2]

[warm current]

(This gets ugly because now we need to consider complex analysis, not that that is necessarily a bad thing)

[orchid path]

I genuine forgot

[fringe raft]

(minus)

$\cos x - \cos y = -2 \sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right)$

[jolly parrot]

rbit

[orchid path]

well both quantity in the sine is posotive

positive

[fringe raft]

good

and we want to show this whole thing is negative yeah?

[orchid path]

and less than 2

yea

[fringe raft]

or $\sin \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right) > 0$

[jolly parrot]

rbit

[orchid path]

exactly

this means proving both sindhave same sign

[fringe raft]

right

which sign is it?

[orchid path]

positive since 2 < pi

but then this requires that sin x is positive between 0 and pi

ig that's trivially enough to prove

[fringe raft]

so instead of showing cos is decreasing on [0, 2], we could simplify this problem to showing that sin is positive on [0, 2] (which is much better at least)

[orchid path]

yes

let me try

[fringe raft]

showing that is very doable I think

[warm current]

Funny that you are doing it without derivatives but are now proving the exact thing i had to prove

[orchid path]

am I using the series for sine

[fringe raft]

yeah

[fringe raft]

oh I can already see

the derivative just skips having to use the identity

[orchid path]

ah yes

that would have been easier

[orchid path]

my brain is not working 😭

[fringe raft]

$\sin x = \frac{x^1}{1!} - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$

[jolly parrot]

rbit

[fringe raft]

hmm I kinda don't want to spoil it

[orchid path]

yea don't

do I write sin x using exp and imaginary units

the series is kinda troublesome

or do I show frac (x^i)(i!) - frac(x^(i+2))((i+2)!) positive for x in [0, 2]

for natural number I (I assume it works for even numbers as well)

my phone is running out of battery sadly and I'm not at home can't change it I'll probably come back couple of hours later 😢

[fringe raft]

yeah very good

[orchid path]

I'll have a think

induction?

[orchid path]

how can one justify that it implies the whole infinite sum is positive though

not sure how one deals with infinite series

[fringe raft]

it shows that every partial sum is positive

there's a theorem that if a_n >= b, then lim a_n >= b

[orchid path]

frac((i+1)(i+2)x^i - x^(i+2)) ((i+2)!)

x^i[(i+1)(i+2)-x^2]

numerator

yea this is obv now

[fringe raft]

maybe you write it down on paper if you have the time

[orchid path]

i+1 times i+2 is at least 6

greater than 4 = 2^2

this prob isn't rigorous but I think it's good enough

1% battery

I think I've proved it

thanks good exercise

[warm current]

gg

[orchid path]

gg

.clowed

.close

[pearl pond]

Channel closed

Closed by @orchid path

Use .reopen if this was a mistake.

[orchid path]

ty guys

time we'll spent

wait

.reopen

[pearl pond]

✅

[orchid path]

pi^2 > 6 I'm in trouble

I'll try again later when I'm back home

[fringe raft]

we only showed it up to 2

you need a whole different method for the rest

[warm current]

2²<6

[orchid path]

do I do the same for each adjacent terms excluding the first

tell me before phone dies

pls

[fringe raft]

you need to use symmetries of the cos function

[orchid path]

there isnt symmetry

well cosx equals cos-x

[fringe raft]

cos is full of little symmetries

[orchid path]

bur aren't we working with sin

[fringe raft]

we are done with sin

we now know cos is decreasing on [0, pi/2]

use symmetry to apply that to [pi/2, pi]

[orchid path]

cos( pi/2 + x )= -cos( pi/2 - x)?

yea no I dont have enough time

[fringe raft]

yeah I think some of these formulas

[orchid path]

phone is dying

need to prove that though

[warm current]

If it's not urgent, you can just ask when you get back

[orchid path]

it u guys are still here yh

;thumbsy

[warm current]

I'll be around. You can ping me if you want for this question, but this is @fringe raft's approach, so I don't know how much I could help you there

[fringe raft]

what would have been yours?

[warm current]

Mine was more abstract because I didn't even assume cos(pi) was -1 yet. My approach is the same, showing that every two sequential terms are positive, I proved it just for [0, 2]. I did get to the same x²<6 requirement though, so maybe our approaches are identical

[fringe raft]

like defining pi/2 as the first positive root of cos allows us to make such assumptions, but yeah there's other definitions that would make it more tricky

[pearl pond]

@orchid path Has your question been resolved?

[warm current]

This is tricky as you must prove first that cos has a root. And then you need to consider your definitions further. Really, the trig functions are quite a thing to think about sometimes

[fringe raft]

I just need to find a positive and a negative values and it's continuous

[warm current]

Yeah. The tricky part here though is that finding the root won't guarantee that the cos is strictly decreasing from 0 to 2.

Here's what I had to do:

• Show sin x > 0 for 0<x<2
• Corollary is that cos x is strictly decreasing in the interval (0, 2)
• Show that cos(2) < 0 using Taylor's theorem
• Deduce that a root cos(x)=0 must exist for 0<x<2
• Because cos x is strictly decreasing for 0<x<2, this root must be unique.

The uniqueness is vital. Without it, you couldn't prove that cosine is strictly decreasing using just the fact that a root exists (at least for 0<x<2)

[fringe raft]

oh right

[pearl pond]

Channel closed

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[quasi jay]

What’s the formula for example I wanna count how much is the sum of numbers from 1 to 100

[warm current]

$\sum_{k=1}^{100} k$

[jolly parrot]

SWR

[quasi jay]

Idk what this is

I’m a freshman

[warm current]

Summation symbol, which is just the Greek letter Capital Sigma

[orchid path]

.reopen

[feral sedge]

damn i was off by 2 years

[orchid path]

ooos

oops

[warm current]

create new channel, I'm still around

[quasi jay]

Help me first?

Can u maybe explain?

I still haven’t learned this

Can u help me?

[warm current]

tbf, I was helping violet a while earlier, but I'm helping you both at the same time

[quasi jay]

Can u help?

Or not

[warm current]

Okay, I need a little more context, what are you trying to do and why?

[quasi jay]

I’m m losing patience

[quasi jay]

I want like

[warm current]

!volunteer

[pearl pond]

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

[quasi jay]

1+2+3+4+5+6 etc to 100

How to calculate that

Cause I ain’t spending hours clicking on calculator

[feral sedge]

wait that was someone else nvm

[quasi jay]

Isn’t it

[warm current]

$1+2+3+...+n=\frac{n(n+1)}2$

[jolly parrot]

SWR

[quasi jay]

,w 100(100+1)/2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by .arianag

[quasi jay]

Why divided by 2?

,calc 1001001

[warm current]

think of adding just the first and last numbers over and over

[jolly parrot]

Result:

10000

[warm current]

1+100=101
2+99=101
2+98=101
...
But where do you stop?
50+51=101

You stop at 50, halfway to 100

[quasi jay]

ok

[warm current]

you've added 101 a total of half of 100 times

[quasi jay]

Ale

Ty

Alr

,close

[south inlet]

.close

[pearl pond]

Channel closed

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

what does it mean that a subfield of C is isomorphic to R?

[orchid path]

It means that there is a one-to=one corresponding between the two that preserves the operations defined, i.e. + and .

[plucky python]

there's a subset of C that is a field under the same + and x

and is isomorphic to R

(i.e. like the obvious copy of R in C)

[orchid path]

{one, two, three, four, …} is isomorphic to N for example (the naturals)

Provided we define addition and multiplication the proper way

But of course this isn’t the best example since N is itself not a field

[pearl pond]

@desert solar Has your question been resolved?

[desert solar]

Thanks very much

[pearl pond]

Channel closed

Closed by @desert solar

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[pearl pond]

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

Please don't occupy multiple help channels.

[quasi jay]

1/3 * 3/1

3:3 = 1

30:1=30

How is it not 30?

[grizzled dust]

what is the proper order of operations?

[quasi jay]

Division and multiplication are on the same place, smartie.

[vast escarp]

270?

[grizzled dust]

so why are you doing the multiplication first?

[quasi jay]

Doesn’t matter

What I do first

Since

It’s in the same place

[grizzled dust]

are you sure it doesn't?

try it the other way

30 / (1/3) = ?

[vast escarp]

90

[grizzled dust]

!noans

[pearl pond]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[quasi jay]

90/1*30=270

Ik how do to this

[vast escarp]

Is this it?

[quasi jay]

This is like 5th grade

But like

Why said division is first

Than multiplication

[vast escarp]

Left to right

[grizzled dust]

stop please, this isnt your channel, if you are going to help the asker learn then great, but it's not your place to blurt out answers, the goal is to help them arrive at the answer themselves

[vast escarp]

Apologies

[quasi jay]

@spiral pivot @vast escarp @glacial sequoia @vestal tapir @main oxide SEE, here they used “:” as division.

[quasi jay]

It’s fine

[vestal tapir]

multiplication before division is a troll

[quasi jay]

How?

I did that first

Cuz it’s easy to do that way

[vestal tapir]

nobody does that

[spare lark]

Avengers assemble !

[vast escarp]

Yeah, sometimes also called a ratio

[grizzled dust]

$(a / b) * c \neq a / (b *c )$

[jolly parrot]

from math import sqrt

[grizzled dust]

in general

[quasi jay]

And when finding ratio, you divide?

[vestal tapir]

there are acronyms like pemdas, they are troll

[spare lark]

Bodmas

[vast escarp]

Yeah

[regal herald]

Bidmas

[spare lark]

Xmas

[vestal tapir]

division before multiplication does give the same answer on the other hand

[quasi jay]

Pemdas, multiplication before division

[grizzled dust]

$\f{ac}{b} = (a / b) * c \neq a / (b *c ) = \f{a}{bc}$

[jolly parrot]

from math import sqrt

[vestal tapir]

so pedmas is less troll

[sharp vigil]

pemdas should really be pe(md)(as)

[grizzled dust]

think of it this way, if you just do the multiplication first you are changing the problem from what is on the left to what is on the right

[vestal tapir]

it should be pedmsa ideally

[regal herald]

md and as are generally of equal priority, you can switch them

[vestal tapir]

then it works even if you get trolled

[sharp vigil]

but equal priority operations should be done from left to right

[vast escarp]

Just like a calculator

[vestal tapir]

you can't

[quasi jay]

Okay but:

you use “:” as division
“*” or “•” as multiplication
tg as tangens
“,” as decimal point
“.” or spaces As thousand separator

[vast escarp]

"*" is commonly used in coding and programming

I math mostly "×" and "·"

[quasi jay]

x is complicated to do

• is easier

[vast escarp]

Agreed

[grizzled dust]

you should get rid of the habit of using x for multiplication it just gets confused with the variable x

[quasi jay]

Also x can be mistaken as algebraic “x”

[quasi jay]

Fr

[grizzled dust]

* is next to the /-+ symbols on every standard keyboard, it's kind of self explanatory what it is

[quasi jay]

Also no one uses

•

•

As division

[vast escarp]

What if we are doing vectors?

[quasi jay]

I hate when people do /….

As division

It gets confused as fraction

So damn annoying

[main oxide]

fractions are division

[quasi jay]

Ik…

But like

Not always u want to fractionize it

But just to divide

[grizzled dust]

then x is usually used for cross product, vector variables are often denoted with stuff like a bar or arrow symbol above them so it's clearer in that context that x isn't a variable, and the typescript for cross product is usually quite different from the letter x

[glacial sequoia]

okay?

[vast escarp]

Makes sense

[quasi jay]

Js proving my point 🥰

[glacial sequoia]

i don't think i ever disputed you lol but sure

[grizzled dust]

plz don't mass ping people like that randomly

btw

[glacial sequoia]

power rangers ahh

[quasi jay]

It’s js pinging

U can always mute the server

Or just phone

[sharp vigil]

so to conclude we have
30 ÷ (1/3) * 3
30 ÷ (1/3) = 90
90 * 3 = 270

where i used ÷ instead of : but they mean the same

[glacial sequoia]

[quasi jay]

We have only two variables, Cloud.

We’re talking about doing 1/3 * 3 first or 30:1/3

[sharp vigil]

like i said, it goes left to right

[quasi jay]

.close

[pearl pond]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[mental wigeon]

Do you guys accept sqrt(x^2)=|x| as a definition?

[midnight haven]

For real numbers, I believe so

[mental wigeon]

isnt it a consecuence of the previous definition of absolute value and square root?

[vast escarp]

Yes

[midnight haven]

Yup

[vast escarp]

√x^2 is the distance of a real number x from 0 and distance cannot be negative

[mental wigeon]

how is that, a consecuence cannot be a definition

[vast escarp]

It is not a consequence per say

[mental wigeon]

i see directly as a consecuence of those two things i said before

i used to use that as a definition

but now i am not sure is correct to call it like that

[gleaming moss]

That doesn’t seem right

[vast escarp]

Where do you disagree?

[gleaming moss]

x = 2 = 0 +2. sqrt(x) = sqrt(2) = 1.41… = 0 + 1.41…

[vast escarp]

I mean sqrt(x^2)

=abs(x)

[mental wigeon]

in any case, i see it as a consecuence of the real definition of absolute value + the definition of square root

[gleaming moss]

Ah right, I misread mb

[valid peak]

as a definition of |x| for a real number x, yes

[mental wigeon]

what do you think about what i say about being the consecuence?

[vast escarp]

How would you define absolute value?

[valid peak]

any other definition of |x| will be equivalent, that's kind of the point

[mental wigeon]

like everbody

[valid peak]

so something like this is okay

particularly I would just define |x| as x if x>0, -x otherwise

[mental wigeon]

[
|x| =
\begin{cases}
x, & \text{if } x \geq 0, \
-x, & \text{if } x < 0.
\end{cases}
]

[jolly parrot]

Mary

[valid peak]

since that doesn't require you to define anything else like sqrt(x^2) would

[mental wigeon]

yeah i know all this i was just worried about using the word definition

[vast escarp]

Consider the function f(x) = sqrt(x) for x in R+ U {0} f(x)≥0

[mental wigeon]

that has nothing to do with the question anyways

[vast escarp]

By this I'm trying to say that output of any sqrt() is always not negative

[mental wigeon]

I know that, is not a math problem is a definition problem

because definitions dont need to be proved

but i can prove sqrt(x^2)=|x|

[mental wigeon]

but i cant prove this

[vast escarp]

I would like to see the proof

[pearl pond]

@mental wigeon Has your question been resolved?

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[livid bolt]

is this a shape with 5 sides?

[surreal relic]

If non of them are colinear then yes

But not a close shape

So vector sum may or may not be 0

[livid bolt]

wtf

i hate matehmatics

[surreal relic]

.-.

They are telling you to illustrate it

Not find a finite value

[livid bolt]

so the answers can vary for part 4?

[surreal relic]

Just connect E and O

Make a closed shape

Then vector sum laws

[livid bolt]

wait..

did they draw the sahpe for me alr..

[surreal relic]

That's for question 4 right

Question 3 is independent of it

[livid bolt]

oh

):

ok thank u

i just woke up my tutor had to knock on my door

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[frank shale]

am I missing something, why aren’t these adding up? I’m mixing up the signs somewhere

[dense jasper]

?

[frank shale]

yeah nvm, I got it. I wrote down the wrong angle

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[bitter herald]

Not sure if this is totally appropriate to ask here, but may I get some help with this? Would love to provide what I have done so far, but I haven't been able to take the first step with this yet

[bitter herald]

A nudge might be helpful

[pearl pond]

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[green dew]

cant u use the base change formula

for logs

[midnight flicker]

what formula

[supple silo]

Man in the Logarithm Maths question u sent...in that 2 cases will be made
Condition 1 : -
We consider the base of logarith to be greater than 1 and solv the logarith nomally with with the same inequality sign in the question
Condition 2 :-
in this we had to consider the base of Logarith to be greater than 1 and we will solve it normally...just a twist that the inwquality sign given in question in while opening the log will be revresed and solve for x

Dont forogt to take the domain for X ....take (x/2)-1 > 0 and 6-x > 0 ...and take intersection as u will get domain

Finnaly the answer must get into domain
if not eliminate and u will get the finnals answer

THEN FINNALY :-
CHECK FOR THE VALUE OF X
put the value of x u got and in 1 conditon the base will become -ve or will not define log fxn so eliminate that value

Hope this works 4 u

[craggy hull]

guys where can i ask for help

calculating a problem

[urban galleon]

See #❓how-to-get-help

[green dew]

but really, the base must be strictly positive

same for what ur applying the logarithm on

so ur forcing a condition: 2 < x < 6

[pearl pond]

@midnight flicker Has your question been resolved?

[midnight flicker]

ty

🙂

[pearl pond]

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[prisma vortex]

would n be the midpoint and why? and if not how would we do this question

[marble sonnet]

How is P defined?

[strong relic]

its shown that p is the midpoint

[strong relic]

its said in the question

[prisma vortex]

im a bit confused on how to approach this one without knowing n is middle height

[strong relic]

?

n is the midpoint of ac

[marble sonnet]

N is infact the middle "hieght" aswell

[prisma vortex]

why is that?

[strong relic]

what is middle height

[prisma vortex]

like the middle of the height of abc

[marble sonnet]

You can see that by drawing 2 similar right triangles 1 with hypotenuse CN and the other with hypotenuse CA

[prisma vortex]

like do they line up

[strong relic]

oh

yes they do

can be shown with similar triangles

[prisma vortex]

how so?

[marble sonnet]

Drop a perpendicular down from both A and N

[prisma vortex]

ye

[marble sonnet]

Lets call the foot of this perpendicular x and y respectively

ACX is similar to NCY

Because they share 2 angles (the 90° and the common angle at C)

[prisma vortex]

and nc is half ac so ny is half ax?

[marble sonnet]

Yes

[prisma vortex]

js wondering tho is there a way to do it without drawing that extra line? cause i just saw that ncm is similar to abc

so im wondering if you can prove it using that

[marble sonnet]

What happens to the area of a figure when you half one of its lengths but keep the shape similar

[prisma vortex]

the other lengths half?

[marble sonnet]

Yes
And the hight is also a length and scales as such

Since you halfed the base you halfed the hight

This argument is basically intuitively the same as what i said earlier but without imagining that extra line

[prisma vortex]

srry my dad called me

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@prisma vortex Has your question been resolved?

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[dapper nova]

ax^2+128x+c
In the given expression, a and c are positive constants. If px+q is a factor of the expression, where p and q are positive constants, what is the greatest possible value of ac?

[dapper nova]

I'm not sure how to even start

My understanding is that the implication of there being a root px+q is that the discriminate must be positive/zero

But I do not get why p and q must be positive (or what that would imply)

[dense jasper]

px+q is a factor -> the value of x that makes px+q equal to zero is a root

So what does this tell you?

As a hint for the question at large, consider ||the discriminant||.

[dapper nova]

Yes, I already considered that

My question is about why p and q have to be positive

Is it just extra information to confuse me

[dense jasper]

px+q is to impose some condition about the roots, the fact that p and q are positive is so that the maximum is actually attainable (you should see what I mean once you do the question)

[dapper nova]

bc you would get
ac < 64^2

[dense jasper]

yeah

now read the second part of what I said

Ex. If p and q had one of them being negative and the other being positive, then that wouldn't be attainable

also, it should be ac < 64^2 (which is 4096 iirc?)

[dapper nova]

yes

Why would it not be attainable

Sorry I don't understand 😭

[dense jasper]

Go solve for x in the case that ac=64^2

[dapper nova]

-64/a

[dense jasper]

see what I mean now?

[dapper nova]

I'm sorry I still don't understand

[dense jasper]

a > 0 -> -64/a < 0

[dapper nova]

px+q with p,q >0 implies a negative root

[dense jasper]

but if you had something like this

[dapper nova]

-64/a is a negative root

[dense jasper]

then -64/a being a root wouldn't be possible

since px+q being a factor would give you a positive root

[dapper nova]

Ohh

[dense jasper]

(technically, you could have p and q both be negative, but that's weird since you're just throwing in a factor of -1.)

[dapper nova]

from the quadratic formula both roots are negative, ah I see

[dense jasper]

quadratic formula is too much effort imo

$ax^2+128x+\frac{64^2}{a}=a \left(x^2+\frac{128}{a} x+\frac{64^2}{a^2} \right)=a \left(x^2 +2 \cdot \frac{64}{a} x+\left(\frac{64}{a} \right)^2 \right)=a \left(x+\frac{64}{a} \right)^2$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

looks long when you write it out

but you can jump from the start to the end if you're decently familiar with the form of the expansion of (a+b)^2

I just included the extra detail cuz why not

[dapper nova]

So, for the question,
Since there is a px+q factor, it implies that the expression is factorable, this means the discriminate is positive or zero. The implication of p, q > 0 is that it is a negative root and this is important bc both roots according to the quadratic formula must be negative

[dapper nova]

Yeah but I mean I just want to understand it 😭

For the SATs

[dense jasper]

Since there is a px+q factor, it implies that the expression is factorable over the rational numbers, this means the discriminate is positive or zero since the coefficients are all real, meaning that both roots must also be real. The implication of p, q > 0 is that it is a negative root and this is important bc both roots according to the quadratic formula must be negative

[dense jasper]

ah SATs

Yeah I just went in and rawdogged it lol

anyway

[pearl pond]

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

[dense jasper]

if you don't have anything else ^

[dapper nova]

Just one more thing

[dense jasper]

?

[dapper nova]

Similar principle for this right

U sub u=z^9

[dense jasper]

yeah

[dapper nova]

Ty!!

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[stoic imp]

George is about to get a certain amount of change less than one dollar from the cash register. If he gets the most quarters possible and the rest in pennies, he would need to receive $3$ pennies to meet the amount. If he gets the most dimes possible and the rest in pennies, he would need to receive $8$ pennies to meet the amount. What is the sum, in cents, of the possible amounts of change that he is trying to get?

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[last summit]

george 😂

[wild fable]

any georges here

[last summit]

@tough seal

[wild fable]

no way

bannable offense

[pearl pond]

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

[dense jasper]

Hint: ||Chinese Remainder Theorem||

[stoic imp]

I am bad with crt, can I get any help, I lack the mathematical maturity

[dense jasper]

What have you done so far? There's an implicit hint before you can use crt.

[stoic imp]

I am going to be honest, I wanted first some help understsnding the problem statement

idk if you are up to dat

george is about to get a certain amount of change less than one dollar from the cash register

[dense jasper]

George is shopping. He is getting change. Amount of change is less than 1 dollar.

[stoic imp]

ok good

if he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount

which amount?

[dense jasper]

george is about to get a certain amount of change less than one dollar from the cash register

[stoic imp]

quarter means 0.25

penny means 0.01?

[dense jasper]

yeah

american money:
quarter = 25 cents
dime - 10 cents
nickel = 5 cents
penny = 1 cent

[stoic imp]

like if they only give him in quarters the residue is 3 pennies

[dense jasper]

the word "remainder" feels more natural than "residue", but yeah

basically the change is 3 pennies + some number of quarters

r u there?

cause if not

imma dip

alright cya @stoic imp

[stoic imp]

[stoic imp]

is okay, is better if I give this one more thought

.close

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[midnight haven]

Could someone help me with the backward and forward phase in guass Jordan elimination

[sharp smelt]

How are you already doing guass jordan

[midnight haven]

Lol what

[sharp smelt]

Weren't you just doing Gaussian

[midnight haven]

I think they are explaining both at the same time

idk what the heck is going on

[sharp smelt]

I'm just suprised as I'll only be doing this in my 2nd or third year

[midnight haven]

All of this was already done in ha

Hs

[midnight haven]

I thought you completed LA

[sharp smelt]

I focused on more abstract LA

linear transformations

vector and inner product spaces

[midnight haven]

Oh taking matrices as vectors is called abstract LA?

[sharp smelt]

Well, it depends

I just mean to say I hardly computed anything

[midnight haven]

I have no idea what computed means

[sharp smelt]

I mean I only row reduced stuff at te beginning of my course

But I digress

sorry

[midnight haven]

Lol I figured it out

.solved

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[midnight haven]

whats the question

Guass Jordan directly gives us the free variables where as in case of guass we have to do some basic algebra

But in the end they both are the same right ?

Btuh

Nvm

Sorry

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[midnight haven]

.reopen

[pearl pond]

✅

[midnight haven]

,rccw

[jolly parrot]

Couldn't find an attached image in the last 10 messages.

Dec 22, 2024, 4:54 AM

[midnight haven]

,rccw

[midnight haven]

They didn't discuss "dimension theorem"

What is that?

[sharp vigil]

did they mention the "free variable theorem for homogenous systems"? those are most likely the same thing

[midnight haven]

Ph

Oh

They did

[sharp vigil]

both of them are really different statements of the rank-nullity theorem, but it won't be stated in that language until later

[midnight haven]

Okay!

Thanks

.close

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[sterile turtle]

for part b

[sterile turtle]

it wants me to express in terms of w and theta, but if i differentiate it from there, arent they 2 variables?

[rare burrow]

what's wrong with expressing it in terms of two variables?

[sterile turtle]

then i cant differentiate in respect to 1 variable without differentiating the other no?

[merry carbon]

w is fixed from what I read, so you can treat it as constant and then you get a function only of one variable, theta

[sterile turtle]

could i get an explanation of that

i thought w was dependent on theta

[merry carbon]

“…bending a long rectangular metal strip of width w…” so w is given to you to begin with

That forces theta and r to be dependent on each other, but because you’re given a fixed width to begin with, w doesn't change

[sterile turtle]

ok

[merry carbon]

It’s like those other problems where you’re given e.g. a fixed perimeter you can work with and wanna maximise the area of some shape, if you’ve dealt with those before(?)

[sterile turtle]

yep

i was treating r as a constant so thats what confused me

im still getting it wrong tho lmao

why am i off by a factor of 1/4

OHHH

WAIT A SEC

nvm i got it lmao

thanks for explanations

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[bitter herald]

is there a nice way to take the inverse fourier transform of this without resorting to the definition

[bitter herald]

not quite well-versed in the properties yet but i am sure something would work out

ok so since H(jw) is odd and purely imaginary, then h(t) is real and odd too

what extra information can I retrieve

[bitter herald]

Actually, H(jw) being odd and purely imaginary is a necessary condition for h(t) being real and odd, but I am not sure if it is also sufficient

Actually it's probably not sufficient...The system is specified as causal, so h(t) = 0, t < 0

So this doesn't help

[pearl pond]

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[pearl pond]

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[sharp smelt]

Prove that if $lim(a_n)=a; lim(b_n)=b$ , then $lim(a_n/b_n) = a/b$
\
Let $\varepsilon >0$. we thus have
\
$\abs{\frac{b_n}{a_n}- \frac{b}{a}}$
\
=$\abs{\frac{b_n}{a_n} - \frac{b}{a_n}+ \frac{b}{a_n}- \frac{b}{a}$
\
$\abs{\frac{1}{a_n} ( b_n - b) + b(\frac{1}{a_n} - \frac{1}{a})< $
\

[sharp smelt]

Abbott has a simpler method, just wanted to know if this is right too

[inland ivy]

What is this?

You just wrote two expressions

[sharp smelt]

They are equal to one another

[jolly parrot]

ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp smelt]

Prove that if $\lim(a_n) = a$ and $\lim(b_n) = b$, then $\lim\left(\frac{a_n}{b_n}\right) = \frac{a}{b}$.

Let $\varepsilon > 0$. We thus have

[
\left| \frac{a_n}{b_n} - \frac{a}{b} \right|
]

[
= \left| \frac{a_n}{b_n} - \frac{a}{b_n} + \frac{a}{b_n} - \frac{a}{b} \right|
]

[
= \left| \frac{1}{b_n} (a_n - a) + a \left( \frac{1}{b_n} - \frac{1}{b} \right) \right|
]

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[sharp smelt]

hmm

probably a bad idea

yeah , def a bad idea

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[sharp smelt]

sorry

[pearl pond]

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[buoyant relic]

(images are the wrong way round) Am i allowed to conclude that, since $AC$=$O$ in $\Delta ABC$, and $sin\theta=\frac{O}{H}$, since $\frac{4}{5}$ is in its simplest form, $4=O=AC$? or is this only correct by chance

[jolly parrot]

Michele

[compact ridge]

only by chance, so you're just guessing

who's to say that O = 2 and H = 2.5 isn't possible

[buoyant relic]

Thought so

[compact ridge]

or O = 0.4 and H = 0.5

infinitely many possibilities

[compact ridge]

actually sine is the wrong trig function to use

adjacent = 3 and hypotenuse = BC, so you can use cos to find an expression for BC

similarly, opposite = AC and adjacent = 3, so you need tan here

[buoyant relic]

yeah I think I figured it out, $sin/theta=4/5$, then used pythag identity to get $cos/theta=3/5$, then subbed into trig to get $4/3$, then $3tan/theta$, giving $4$

[jolly parrot]

Michele

[buoyant relic]

Mb for formatting I’m on my phone

[compact ridge]

don't use 4 and 5 for the distances

you can work it out from AB = 3 and angle ABC = theta alone

4 and 5 are the speeds, so you divide the distance BC by 4

and the distance CD by 5

to get the times for BC and CD

[buoyant relic]

Ahhhh I can see that

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[ripe pagoda]

Eight players 𝑆1, 𝑆2, 𝑆3, . . . , 𝑆8 play in a tournament. They are divided into four pairs at random. From each
pair a winner is decided. It is given that the chance of winning of any player in a given match is same as that
of his opponent, and the probability of any game ending up as a draw is 0.5.
1)Find probability that S1 and S2 win their match
2)Find chance that atleast one of S1 or S2 is among the winners

[ripe pagoda]

for the first question
is it correct for me to assume that the chance of winning is also 0.5?

if that is so
im approaching it like
prob that S1 wins = 0.5 x 0.5 = 0.25
one 0.5 for his winning chance and another for the draw chance
and then prob that both of them win is
0.25 x 0.25

which is 1/16

and for the second, i have no clue

tried to find the chance that none of them win

getting stuck

[sick adder]

What'd you have for finding the chance that neither of them win their matches?

[edgy stone]

how many winners are there

[sick adder]

I'm assuming this assumes s1 and s2 are not against each other

[edgy stone]

or, let's go like this

how many ways are there to pick the first winner out of all 8 people if they can't be S1 or S2

then how many ways are there to pick the second winner after that

etc

[ripe pagoda]

im sorry, i dont get this one?

[edgy stone]

S1, S2, S3, S4, S5, S6, S7, S8. pick one of these, but you can't choose S1 or S2.

choose

[ripe pagoda]

6 ways right

[edgy stone]

sure

which one did you pick?

[ripe pagoda]

S4

[edgy stone]

sure

now pick the second winner out of S1, S2, S3, S5, S6, S7, S8

how many choices

[ripe pagoda]

i can pick S1 and S2 in this case?

[edgy stone]

no

[ripe pagoda]

5 then

[edgy stone]

yes

[ripe pagoda]

then 4, 3, 2

1

[edgy stone]

right

but you only need four winners

so it's 6, 5, 4, 3

[ripe pagoda]

oh right yeah mb

[edgy stone]

so in total 6*5*4*3 ways to pick 4 winners if you can't choose S1 or S2

[ripe pagoda]

uh huh

[edgy stone]

now, how many ways are there to pick 4 winners if you can choose S1 and S2

and then you just divide one by the other and you're done

[ripe pagoda]

and 8x7x6x5

and then

yeah i got it

thank you

my options arent matching tho

[ripe pagoda]

tried that their losing probability is 1/4
and it can be a draw too so its
1/4+1/2 = 3/4
and then both of them dont win is
3/4 * 3/4
so 9/16
then 1 - 9/16
but still dont match up

[pearl pond]

@ripe pagoda Has your question been resolved?

[ripe pagoda]

.close

[pearl pond]

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[pearl pond]

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[fickle patio]

Hi, I wanna ask is my blue working wrong, cause I think the L should be half period which is suppose to be 1, but the integral should be -1 till 1 which can I write like the red one where I change that to 0 till 2?

Or is my blue working is correct? Because I don’t have the answer, my teacher didn’t discuss that

<@&286206848099549185> I think the red one is correct?

[plush bramble]

Where do you use that f(2k)=2 for all integers k

[fickle patio]

Nvm, I figure it out, the L should be 1, so the red should be correct since I got the answer already

Thanks

.close

[pearl pond]

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[pearl pond]

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[analog forge]

how do you do this? not sure how to combine the different parts of it

[fallen forge]

still need?

I can show

[analog forge]

yes pls

[tranquil estuary]

9c2 * 8c2 * 13c1 = 36 x 28 x 13 = 13104

[quiet tendon]

yeah uh just telling someone an answer doesn't help them

[fallen forge]

we have one restriction - 2 members from each gender. It means that we may have 2W + 3M, 3W + 2M. As you can see, summation gives 5 memebrs in total

[fallen forge]

yep

[tranquil estuary]

Oh wait lol, thought he needed answer

[analog forge]

hmm okay

[fallen forge]

@analog forge clear?

[analog forge]

yes

[tranquil estuary]

2 members of each gender, so choose 2 from male, 2 from female, 1 from the remaining i.e. 13

[fallen forge]

now, let's look at first instance. 2W and 3M. we need to choose 2W from 9W without any restrction of order( 9C2 ). 3M from 8M -> (8C3)

clear?

let's move on to the next instance

3W and 2M. 3women from 9 give us 9C3. 2 men from 8 give us 8C2.

clear?

[analog forge]

yep

so then how would you combine them

[fallen forge]

now, for first instance: 9C2 multiplied by 8C3
for second instance: 9C3 multiplied by 8C2

if you do not know why we need to multiply, i can explain

[real ridge]

jee aspirant?? anyone

[fallen forge]

@analog forge clear?

[analog forge]

i kind of get it

[fallen forge]

you know the difference between dependent and independent event?

[analog forge]

like the combination of every group of men with every group of women?

[analog forge]

no

[fallen forge]

look

in our case, the events are independent to each other. It means that HOW we select the n-th women/men does not affect the next n+1th

[analog forge]

okay

[fallen forge]

the choice of women does not affect the choice of men, and vice versa.

[analog forge]

yeah that makes sense

[fallen forge]

on the other hand, we have dependent events

imagine that you have a box with colorful balls

[analog forge]

i think i kind of know where you're going with this

like you take one out but that changes the total amount left

[fallen forge]

not that

lemme finish then

and imagine that you have a task to take red ball first and then black one.

3 red and 2 black

the number of possible ways to get red as first is just 3

that was FIRST EVENT

and now, the issue comes. we have to check 2 cases.

The number of ways to draw a black ball depends on whether the first ball was red or not. IF it was red, then we have 2 possible ways. but if it was black first, then our task in such case cannot be carried out

but

if it possible to you, look at youtube for dependent and independent events, because I am feeling that my example kinda confusing

[analog forge]

okay so what you're saying that for dependent it specifically depends on the outcome of the previous draw, not that a previous draw happened

right?

[fallen forge]

yep

like, A and B are dependent if the occurence of one will affect the occurence of other

[fallen forge]

keyword here is OUTCOME

by the way, may I continue with genders? or you got it?

[analog forge]

yes that makes sense

[analog forge]

we had two cases

9C2 * 8C3 and 9C3 * 8C2

[fallen forge]

yep

[analog forge]

so would these then be added together?

after you multiply

[fallen forge]

yeah. we mulptiply them in order find the total number of combination in that particular scenario

and then, we just sum the number of combinations from both scenatios

[analog forge]

yeah that makes sense thanks

was the other person's method that they were saying before correct?

because they get a different answer

[fallen forge]

here I know only you

i did not read anything else except your messages

[analog forge]

this one

[analog forge]

thanks for the help btw

[fallen forge]

13C1. what?

[fallen forge]

thanks to you i revised combinatorics

[analog forge]

haha

[pearl pond]

@analog forge Has your question been resolved?

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[cinder zodiac]

how can I solve this

[urban galleon]

Inscribed angles subtended by the same arc are congruent

[cinder zodiac]

I dont really understand

english is not my first language

like out of these 4 which one is it

[autumn trellis]

use this diagram

[autumn fossil]

The point is that no matter where you draw the angles, they are gonna be the same

as long as they "start" at the same points on the circles

all the blue angles and the black angles are the same

[cinder zodiac]

like that means its 58?

or no its 65

[urban galleon]

Angle V is 65 degrees yes

[cinder zodiac]

thanks for the help, appercite it

also I have this question,
A candy bag is labeled 125 g. The weight of the contents of a large number of bags is normally distributed with a mean of 129 g and a standard deviation of 2 g.

What percentage of the bags weigh

a) between 127 g and 129 g b) more than 131 g c) less than 131 g

My answer that I get is a) 34.1% b) 16% c) 84%

Am I correct? and sorry for being too annyoing im stressed af

[plush bramble]

Show your work

[cinder zodiac]

sorry for my handwriting hope u can understand

[plush bramble]

Why did you round to 34.1% in part a) but not the rest

Just round to the tenths place in all 3 numbers and you're fine.

[cinder zodiac]

so my answer was right but calc was off?

a) 34.1%
b) 15.9%
c) 84.1%
or I guess that will be like this?

[pearl pond]

@cinder zodiac Has your question been resolved?

[pearl pond]

@cinder zodiac Has your question been resolved?

[pearl pond]

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[dense iris]

11b has me at a loss

[brave sluice]

what did you get for dy/dx?

maybe we can use that

[dense iris]

omg

point p is a turning point innit

thats why the mark scheme sets dy/dx = to 0

am i cooking?

is that right?

[brave sluice]

does it set dy/dx to 0?

or just set the denominator to 0

[dense iris]

just the denominator i think actually

yeah wait

I dont get it

[brave sluice]

the slope explodes to infinity when the denominator is 0

line l is vertical

which is "infinite slope"

so set dy/dx = infinity which happens when the denominator is 0

^ this is naughty using infinity but u get what i mean

[dense iris]

ahh yeah i get u

thanks

.close

[pearl pond]

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[thick trench]

Does anyone recognise that notation with the circle and minus sign?

q is some power of an even prime here, not sure if that is useful tho

.close

[pearl pond]

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[nocturne plover]

some power of 2

you mean

[solid pier]

i bet it’s a symbol indicating an inverse but i didn’t check (the only thing to check would be compatibility with oplus)

i know you closed it already but i figured i’d shout an idea out

[worthy lance]

It is the custom operations described

[pearl pond]

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[sour blaze]

What are the solutions of this differential equation : $$y'=yf$$, $f$ a function

[jolly parrot]

Aquamy

[sharp smelt]

Have you tried solving the ode

[sour blaze]

It's the first time I meet this differential equation

This form

I tried

[inland ivy]

Separate the variables

[sour blaze]

I thought it's Ce^f×x, C a real number

[inland ivy]

Not exactly

!show

[pearl pond]

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

[sour blaze]

How

Do I resolve y' = y ?

And I resolve the first differential equation after ?

[spiral pivot]

Divide by y

You get $\frac{y'}{y} = f$

[jolly parrot]

OmnipotentEntity

[spiral pivot]

The left is the derivative of ln

[sour blaze]

Ok thanks

I just want an indication

[spiral pivot]

So I am assuming that f = f(x) and y = y(x)

You'll get $\int \frac{y'}{y} , \dd{x} = \int f(x) , \dd{x}$

[jolly parrot]

OmnipotentEntity

[spiral pivot]

So $y = \mathrm{e}^{\int f(x) , \dd{x}}$

[jolly parrot]

OmnipotentEntity

[spiral pivot]

If we let F(x) be the anti derivative of f(x) then we have $y = \mathrm{e}^{F(x) + C} = k \mathrm{e}^{F(x)}$

[jolly parrot]

OmnipotentEntity

[spiral pivot]

@sour blaze lmk if you have any questions about me this

[pearl pond]

@sour blaze Has your question been resolved?

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[glacial tangle]

Hi, why is median better than average?

[pearl pond]

Please don't occupy multiple help channels.

[prime bramble]

what do you mean?

the median is a certain kind of average

[regal heath]

in what context

[prime bramble]

by average, do you mean the mean?

[urban galleon]

I wouldn't generalize one thing as being "better" than another

[mint spruce]

Median is just the middle number (or the average of the two middle numbers if you have an even amount)

So if you’ve got a couple of big numbers at the end then that’ll impact the average but not the median

[pearl pond]

@glacial tangle Has your question been resolved?

[vestal tapir]

i also don't get it