help-26

[deft holly]

ok

[deft holly]

this us basically what i mean by the 10C1

when you do 0.01x0.99^9 , youre fixing a certain disk to be defective

it can be any of the disks

so you do 10C1

out of the 10, one is chosen and is defective

[topaz sinew]

@opaque yoke Has your question been resolved?

[deft holly]

yo @opaque yoke is there a part of my explanation that doesnt make sense?

[opaque yoke]

im not sure y the binomial coefficient is needed. i just found that we can find it using n!/k!(n-k)! but im not sure y we do that

[deft holly]

you know Choose right?

like what its about

[opaque yoke]

nope

[deft holly]

ok

so the (nx) thing you see

is choose

its basically saying that if i have n objects, and i take x of them, i have nCx ways of picking them

[deft holly]

the formula they made for that is this

what youve found

n!/k!(n-k)!

reason why its important

its like saying i have 3 apples, and i know that one of them is rotten

nah this is a bad example

uh reason why its important is like

if you just do 0.01x0.99^9

its like sayig

ive already decided out of the ten disks, this is the defective one

feel free to butt in when im not making sense btw

this works better if you show where youre confused

[opaque yoke]

does it mean (nx) is the number of ways more than 1 defective discs can be chosen?

thats why we times the equation by this coefficient?

[deft holly]

so, cause (nx) isnt really the right way to write it, lets use nCx instead. its the same thing just different way of saying it

[deft holly]

nCx is the number of ways 1 defective disk can be chosen

do you have a casio calculator with you rn?

[opaque yoke]

yeah

[deft holly]

ok

press shift

then the divide button

does it come up with a C?

[opaque yoke]

ye

[deft holly]

ok

thats choose

type 10C1

tell me what the calc says

[opaque yoke]

10?

[deft holly]

yep

this is saying

from the 10 objects

there is 10 ways to choose one of them

makes sense?

[opaque yoke]

ye

[deft holly]

the C on the calculator is basically the n!/((k!)(n-k)!) formula built into it

[opaque yoke]

so back to my attempt, P(0)'s binomial coefficient is just 1 because theres only 1 way no defective discs are chosen?

[deft holly]

yep

[opaque yoke]

icic

icic that makes a lot more sense then

[deft holly]

nice

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[woven sigil]

Help

[earnest pine]

Can you decompose your matrix into symmetric matrices?

Also, do you know any info about T, as in is it linear transformation/map

[woven sigil]

I think

T is a linear transformation

[earnest pine]

Okay so then you know that T(u+v)=T(u) +T(v)

[woven sigil]

Yea

[earnest pine]

You also know that kT(u) = T(ku), but that’s not strictly necessary

[earnest pine]

Okay, so can you see what I’m hinting at?

[woven sigil]

Yes, but i cant envision two symmetric matrices that the sum of them would be the matrix i want

[earnest pine]

Okay, maybe you need more than 2

[earnest pine]

Sorry, this is misleading

You’re given a property, perhaps you can rip that matrix out, convert it and then add it back.

Then maybe you get a symmetric matrix

[woven sigil]

Oh thanks i think i got it

,rotate

[woven sigil]

Thanks

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[deft holly]

Part b

[deft holly]

this is how i have done it

i have two problems with this tho

1. whats the reasoning behind the similar triangles? Ik the right angle is a part of it, but idk what else there is.
2. is this the best way to do this or is there a more succinct/better method?

[cursive patrol]

about point 2, that is already an extremely succinct method

[deft holly]

oh

theyre

parallel

thats my bad lmao

[deft holly]

thanks

[cursive patrol]

oh ye that would be nice to note

yw

[deft holly]

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[deft holly]

am doing q9

my thought process was to find a, r, n to make sum

a and r are ok

finding n tho, i tried this, but after testing it, it seems is -2+3k

why is it what i thought + 1 term?

[topaz sinew]

@deft holly Has your question been resolved?

[deft holly]

<@&286206848099549185>

[rigid cloak]

Yo

Fr

[deft holly]

yoyoyo

[rigid cloak]

My guy

Look

Its a decreasing gp

Fr

Cz common ratio is less than one

[deft holly]

yeye

i just need help with the n part i explained

why is it +1 to what im thinking

[rigid cloak]

Aaj

Aah*

I did it

n came as 3k-2 💀☠️

[deft holly]

yea

but y

[rigid cloak]

But how will you find sum even then?

[deft holly]

i thought itd be 3k-3

[rigid cloak]

How ?

Lemme show

Imma explain

[deft holly]

a(r^n-1)/(r-1)

i got that mate

[rigid cloak]

Bro divide 2nd term by first term to get r

Do

Pls

[deft holly]

mate

[rigid cloak]

Yes

[deft holly]

i need help with n

not r

[rigid cloak]

n came as 3k-2

[deft holly]

yea but y

i thought itd be 3k-3

[rigid cloak]

By applying the condition that nth term of gp=a * (r)^(n-1)

I have taken 2^(11-6k)x as nth term of gp

I took first term as a

[deft holly]

ahhhhh thats where the 1 comes from

[rigid cloak]

And dividing 2nd by first term I got

R

Ok

[deft holly]

wait holup

lemme think real quick

[rigid cloak]

Ok

Fr

[deft holly]

like this

solve for n

5x is power of 1st term

Term n has a power of 5x + 2(n-1)

[rigid cloak]

Fr

[deft holly]

aight

makes sense

thanks for that mate

.close

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[rigid cloak]

2x(n-1)

But bro

Wait

@deft holly

Evaluate it

Don't you think r is -2x

?

[deft holly]

ah shit yea that

[rigid cloak]

2^(3x)/2^5x

Good ok

You good now

Have fun

[deft holly]

👍

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[neon iron]

Hello! Can someone help me solve this question?

[neon iron]

Having a hard time understanding domain restrictions

[pearl fog]

could you point out the middle piece of the piecewise function :3

[neon iron]

Haven’t gone to that part yet

We’re still learning about restrictions

I would be grateful if you can explain to me step by step on how to solve this problem. The education here kinda sucks especially for subjects like mathematics.

[pearl fog]

im pretty sure the middle piece of a piecewise function is the middle part of the graph, so in this case it would be this

could you figure out the domain of that

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@neon iron Has your question been resolved?

[neon iron]

sure!

So if I figure out the domain, then I will have my answer?

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@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

[neon iron]

ahhh 1 <= x < 3?

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[opaque crow]

how we get last transformating

[whole geode]

You have a common factor of 2t dt in the numerator and the denominator.

\begin{align*}
\frac{(4t^3 + 2t) \dd{t}}{(4t^3 - 2t) \dd{t}} &= \frac{2t (2t^2 + 1) \dd{t}}{2t (2t^2 - 1) \dd{t}} \
&= \frac{2t \dd{t} (2t^2 + 1)}{2t \dd{t} (2t^2 - 1) } \
&= \frac{\cancel{2t \dd{t}} (2t^2 + 1)}{\cancel{2t \dd{t}} (2t^2 - 1) } \
&= \frac{2t^2 + 1}{2t^2 - 1}
\end{align*}

[thorny flame]

OmnipotentEntity

[whole geode]

@opaque crow

[opaque crow]

ye im here

[whole geode]

Just wondering if this makes sense

[opaque crow]

aaaaa

okkkk thanks!

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[fluid sentinel]

I'm not sure how to approach solving this problem. Any ideas?

[sweet shard]

The +17 is extra

Or maybe incomplete

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[arctic sundial]

Don’t mind that 4 I fixed it

[loud oasis]

your formula doesn't match the amount of Mg you have in the diagram

[arctic sundial]

Wdym it’s 6

2+2+2

[merry hill]

6 is the total charge

[loud oasis]

there are 3 Mg. the charges are irrelevant for that

[arctic sundial]

So it should just be 2

[merry hill]

thats charge of 1 Mg

you want the number of Mg ions u used

and that is 3

[loud oasis]

the formula just counts how many there are, it doesn't care what the charges are

[arctic sundial]

.close

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[cinder rapids]

This is a bit of a weird one

[cinder rapids]

the little flair "using cas 4" refers to an example question which is this

so, because I've got no other data but this, i tried to copy what they did with the defining f(x) as the distance from the point to the values of x and y

but as it is the origin, the points are (0,0)

$f(x) = \sqrt{(0-x)^2+(0-(\frac{3}{x-1}+2))^2}$

[thorny flame]

suds

[cinder rapids]

but that wasn't it yk

[terse abyss]

That looks pretty good, I’m not sure how hyperbola with rule … works, but I suspect that’s where your problem is

[cinder rapids]

hm?

i dont understand

the calc is basically doing the derivative of the hyperbola = 0 to solve for x

[golden blade]

Since this is always non-negative, try to find any roots

[cinder rapids]

but i must have done something wrong because im not getting the right answer

[golden blade]

that should be your minimum

[cinder rapids]

but shouldn't i be following the example seeing as thats what its referring to - how am i supposed to solve it

[golden blade]

which example?

also are you supposed to compute it yourself, or with a calculator?

[cinder rapids]

with a calculator

[woeful drift]

i assume this means they're to use a calculator

[cinder rapids]

cas 4 is the example

CAS is the classpad calculator

sorry - i forgot its not a common calc/term

[pseudo jetty]

so this is the distance from the origin to a point on the hyperbola

[cinder rapids]

because f(x) is being used to define the distance from our points - the origin to the hyperbola, so its pythagoras of the distance between these

[pseudo jetty]

your goal is to minimise that function

[cinder rapids]

supposedly

[pseudo jetty]

that is, to find the derivative and set it equal to 0

[cinder rapids]

yes, which the calculator does for my when i do dy/dx f(x)

[pseudo jetty]

have you done that

[cinder rapids]

yes, which the function "solve(" does, but its not giving the right answer

[golden blade]

why not just find the roots, it's a square root function

[pseudo jetty]

because it won't be equal to 0 if the minimum distance is larger than 0

[cinder rapids]

i could, but this would be much faster and the textbook is wanting us to solve it in this way

[pseudo jetty]

what do you mean it's not giving the right answer

the question literally says that's the hyperbola

[golden blade]

well to spoiler it there is no minimum

[terse abyss]

Okay I’m going to sleep, go

[pseudo jetty]

can you at least show your working

[pseudo jetty]

[cinder rapids]

this is literally all my working lol, i define f(x) to be the function i sent above

[pseudo jetty]

okay and then follow the worked example

[cinder rapids]

$solve(\frac{dy}{dx} f(x) = 0$

[thorny flame]

suds

[pseudo jetty]

follow what theyve done here

[cinder rapids]

yes, thats what i've done but it gave x to be -0.35 and 2.92, which give a distance of 0.42 and 4.61 respectively

[cinder rapids]

i did

but the answers say that the distance is 0.10 units

[pseudo jetty]

okay lemme check

[cinder rapids]

so either i'm wrong or the answers are wrong

[pseudo jetty]

are you checking the right answers?

because im getting about 0.4145 as the min distance

[cinder rapids]

yes, i have checked that these are the right answers

it is possible that the answers are wrong

[pseudo jetty]

the answers are wrong then

the min distance occurs at x=-0.35ish

[thorny flame]

Result:

0.46657508258833i

[pseudo jetty]

and the min distance is about 0.4145

[cinder rapids]

yeah

[pseudo jetty]

you've solved it correctly

[cinder rapids]

maybe i should input the min distance of 0.1 and see what x value they wouldve got

[pseudo jetty]

there isnt a distance of 0.1

the min is 0.4blah

0.1 is not attainable

[cinder rapids]

unless by "origin" they mean the centre of the hyperbola instead of 0,0

[pseudo jetty]

that doesnt really make sense

[golden blade]

can you show it

if you dont mind

[pseudo jetty]

,w minimise sqrt(x^2 + (2 + 3/(x-1))^2)

[thorny flame]

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[pseudo jetty]

and that would just make the min distance larger not smaller

[cinder rapids]

idk its just this textbook is rarely wrong

but theres no other answer so ig thats it

[pseudo jetty]

well the answer is displayed plainly before you

so the textbook must be wrong if youre certain you're reading the right answers

[cinder rapids]

im certain lol

[pseudo jetty]

aight

[golden blade]

,w 1/(2sqrt(x^2 + (2 + 3/(x-1))^2)) * (2x + 2(2+3(x-1)) * (3/(x-1)^2))) = 0

[pseudo jetty]

then carry on with your day

just a textbook error

,w derivative of x^2 + (2 + 3/(x - 1))^2 = 0

[thorny flame]

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[golden blade]

I realized my function was wrong

and I am thinking how is snow making the impossible possible

[cinder rapids]

lol

[golden blade]

sorry for the confusion

[cinder rapids]

its all good

[pseudo jetty]

idk what you did but at least you've found your mistake

[golden blade]

x^2 + (2 + 3/(x - 1))^2

i had

x^2 - (2 + 3/(x - 1))^2

[pseudo jetty]

[golden blade]

🥹

[cursive patrol]

fun.

[sweet shard]

Change your username to signerror

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[pearl fog]

so basically there are 8 ways for O to win, which is by vertical left, vertical right, vertical middle, horizontal up, horizontal middle, horisontal down, diagonal upright, diagonal upleft

and out kf the 6 remaining empty tiles, x chooses 3, so 6c3×8

but we subtract the cases where x wins, so each of non diagonal has 2 possibilities where x wins, so subtract 12

wait hold on

.close holy shit i fucked up calculating the choose function 😭

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[pearl fog]

sorry for my silly

[long stirrup]

X can't win, there would be 2 X on the board, there's nothing to subtract

[pearl fog]

x moves first

[long stirrup]

oh yeah

sorry

[pearl fog]

all good, thanks for your time :>

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[jade knot]

can someone tell me the starting steps for this?

Ive tried:

substitution for u=e^2x +9 giving me a final result of 1/2 * integral of 1/(e^x * sqrt(u))

I've also tried u=e^x

but that just gives me u/(sqrt(u²+9))

and then Idk how to go from there

wait as I was writing this I think i figured something out, can I do integration by parts with after u=e^x?

nvm that doesn't work

[fair thorn]

that's wrong

the u-sub was correct

[jade knot]

the first one?

[fair thorn]

but your integral is not what it's supposed to be

mhm

[jade knot]

ight hold on

wait let me crop that

[vernal matrix]

(Your integral was alright, barring the lack of du)

[jade knot]

but then what can I do?

[vernal matrix]

Not what you implied here tho

[jade knot]

oh mb

but still tho

I thought it looked similar to arcsin integral

but that would require a '-' and not a '+'

arctan?

but isn't that 1/x²+a²

without sqrt

[vernal matrix]

Make another substitution

[jade knot]

oh

v=u²+9

and then you just have 1/sqrtv

and that's easy to integrat

no????

[vernal matrix]

I wouldn’t suggest that, no cause dv = 2u * du

[jade knot]

oh yeah fair enough

[gentle night]

Let u = e^x I think

[jade knot]

that was my first step

[gentle night]

So what ?

[vernal matrix]

The suggestion was that if you set u = 3tan(v)

[jade knot]

whaaat 😭 where does the 3tan(v) come from

[vernal matrix]

Try it

[jade knot]

but what is v

[gentle night]

$$\int\frac{u \cdot du}{\sqrt{u^2+9}} = \sqrt{u^2+9}$$
no?

[vernal matrix]

(Remember to change variables correctly)

[vernal matrix]

No, you didn’t change variables correctly either

[thorny flame]

Sherif Player

[jade knot]

wait so is u not e^x anymore?

[fair thorn]

what 😭

[jade knot]

i'm so confused

[vernal matrix]

[gentle night]

Ah yes

[vernal matrix]

Well, you have one integral, you got to the integral of 1/sqrt{u^2 + 9} du

[gentle night]

See it

[jade knot]

yes

[vernal matrix]

Which is somewhat annoying, but if you remember your trig identities, there should be one screaming at you there

That said, do you have a list of integrals that you know already, by any chance?

[jade knot]

yes

let me take a pic

it looks liek arctan without sqrt but that's the closest thing

[vernal matrix]

Damn, the one I wanted isn’t there

[jade knot]

😭

[gentle night]

u = 3 tan(theta)
substitution
Is that the one you are looking for ?

[jade knot]

huh

[gentle night]

Trig substitution

[vernal matrix]

What I suggested and has the problem that what you get isn’t listed

[gentle night]

Or you can use this if you got to the hyperbolic functions

[jade knot]

ok yeah

I got sec²v / (3sqrt(tan²v + 1))

[fallow heart]

Now you should remember what 1 + tan² simplifies down to

[jade knot]

which is just sec²v / (3sqrt(sec²v)

right?

and then that's just secv ,

[gentle night]

It should be sec(v) only

[jade knot]

yeah mb

and then just do integral of secv

[gentle night]

Yeah

[jade knot]

which is just 1/cosv

alright cool ty

[gentle night]

No

[jade knot]

the integral isn't 1/cov

but secv = 1/cosv

is it not

[jade knot]

oh I didn't know that existed, ty

[gentle night]

This is a common integeral

[jade knot]

what's the symbol next to the tan?

the i looking thing?

[vernal matrix]

Absolute values

[jade knot]

no not |

= ln|tan( ???

in the first row

[gentle night]

Ln

[vernal matrix]

Poor cropping, it’s the bottom line

[jade knot]

oh that's part of u

alr thx

.close

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[late canyon]

Hello guys, does anyone know the name of this property?

[late canyon]

What is this property called and how can I prove it?

[acoustic ocean]

Use e^alog(b) = b^a

Then, apply a first order approximation, of log(1+x), (convince urself that this is sufficient)

[late canyon]

Have you heard of this property though?

My professor just wrote it on the whiteboard and it's such an obscure knowledge that I couldn't even find it on Google

[acoustic ocean]

Like the name?

Sorry I haven’t

[late canyon]

Hmm

[acoustic ocean]

You may find it on a table of limit identities or smthn, but this is the first I’ve seen it

[late canyon]

Can you try it? It does seem like it works

I swear my professor brought up such an obscure equation and didn't even explain how it's proven

Anyone else can chime in?

[acoustic ocean]

Did some research on this limit, it seems you need to provide some information abt the limits of f and g for this to hold

It holds if f goes to 0, and g goes to infinity (so the limit is actually indeterminate)

[late canyon]

I just went to Desmos and it seems that if I put y = ln (1 + f(x)), then y approaches ln f(x) in higher x values

Hmm didn't seem to help at all

[acoustic ocean]

The limit ur taking is when x goes to a point, so considering large x won’t help

[late canyon]

But the property just says x approaches a even if it's small

[acoustic ocean]

Yh, and I’m saying it doesn’t approach infinity, so studying the behaviour as x grows won’t be too useful

Anyways, the limit isn’t generally true

Let f and h be both identity functions, and clearly the limits are unequal

[acoustic ocean]

This condition is sufficient to make it true, so perhaps ur prof meant to add this

[acoustic ocean]

Along with this identity

[late canyon]

Take this equation as an example

I used my professor's property and it did give the correct answer

For x approaching 0

Which is e^-2/7

[late canyon]

Wdym?

[acoustic ocean]

Yh, but does this function fit the general form of the limit?

[acoustic ocean]

.

[late canyon]

Uh

What does identity function mean again?

Oh y = x

[acoustic ocean]

Yh

[late canyon]

Wait what the heck

Even if f and h are identity functions it still applies

Becaust that would be (1+x)^x

So e^x^2, according to this rule

And if x = 0 then both of them would equal 1

[late canyon]

What does general form of limit mean?

[acoustic ocean]

Evaluated at a

(1+a)^a is not generally equal to e^a^2

And since it’s not indeterminate, me just substituting it in is fine, and removing the lims

If I take a = 1, you get 2=e, which is Ofc not true

[acoustic ocean]

(cosx)^4/xsin7x doesn’t indicate much if it’s not (1+f)^g or e^fg, and you wld need two limits to compare/verify

[late canyon]

Hmm

[acoustic ocean]

For some functions f and g

[late canyon]

Hmm you have a point, lemme desmos both of these

So if let's say x = 1

That implies e = 2

@acoustic ocean Is my professor wrong?!?!?!?!

[acoustic ocean]

No, the identity holds with further restrictions

[late canyon]

What are those restrictions?

[acoustic ocean]

.

[late canyon]

Right

[late canyon]

Wait where did you research about this

I wanna see the website/book from where you got these restrictions

[acoustic ocean]

Wrote out the Taylor expansion of ln(1+f), and considered when the part of it went to 0

[late canyon]

Dang I forgot about that lol

I've been out of education for 2 years lol

[acoustic ocean]

Nah u good

Anyways, if u take the Taylor expansion, you’ll see why the limit of f needs to go to 0

Then the limit identity works for any g, although it needs to go to infinity for an indeterminate form, (one where the limit identity isn’t trivial)

[late canyon]

indeterminate meaning?

Oh

just googled it

[acoustic ocean]

Somewhere where just substituting a doesn’t work

In which case the identity holds without the limit symbols

[late canyon]

Where did you thought of writing out the Taylor expansion though

You seem good at maths

Any books you recommend?

[acoustic ocean]

Because I know the first order approximation ln(1+x) ~ x

Use (1+f)^g = e^gln(1+f)

Then u see in the limit, ln(1+f) ~ f

So consider the Taylor expansion and see why the other terms tend to 0

[late canyon]

You mean ln(1+f) is approximately equal to f?

[acoustic ocean]

Also for books, I’d go for spivak, holy grail of calculus imo (although more of an analysis book)

[acoustic ocean]

Yes

[late canyon]

Hmm

[late canyon]

Wait but doesn't f need to approach 0?

That just gives e^0

[acoustic ocean]

Why g needs to go to infinity

[late canyon]

Oh so that's why g needs to be infinite

Right

Wait but how is ln(1+f) ~ f if f approaches 0?

Oh

Both approaches 0

[acoustic ocean]

Yh, approximation becomes stronger close to 0

[late canyon]

I never thought of using identity functions lol

When we assume that we approximate the result of the entire f(x) rather than using x one by one

Yeah it does seem like it if x = f

Wow

Thinking out of the box

Dude I gtg but thanks so much though for managing to explain such an obscure limit property that is barely anywhere on the internet @acoustic ocean

I actually really really appreciate it

You saved my life

[acoustic ocean]

Np m8

[late canyon]

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[dark skiff]

Hey Im uncertain if my solution is ‘legit’ . A is a vector

[dark skiff]

This is not hw or anything, Im just revising some things I studied ages ago.

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@dark skiff Has your question been resolved?

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[raw zenith]

What is this statement trying to say, the one which I have underlined

[raw zenith]

also what do they mean by complement?

complement are the elements that do not belong in the set but belong in the universal set right?

[proven tendon]

idk

[raw zenith]

So what does it mean here?

[proven tendon]

google.com

use it

[raw zenith]

Good idea

[proven tendon]

:))

[raw zenith]

Gemini says that M / N means set difference

wait I entered the wrong slash

it was supposed to be back slash

💀

same thing nvm

[glass maple]

sry a bit late

set difference is a name used

i think

im only aware of it being called relative complement

easiest way to understand this is to think about a venn diagram

of set A and B where they overlap

in the middle

or rather here, M and N

M \ N , or in english is relative complement of N in M is the part in M that is not in N

[topaz sinew]

@raw zenith Has your question been resolved?

[raw zenith]

umm, is that not set intersection?

[glass maple]

if N is a subset of M

then what part of M

is not in N

they used negated sign here since it is just anything that is not n

[raw zenith]

oh sorry, you said that the way to understand is by looking at that diagram, i thought you meant that overlap is what represented the diagram

[glass maple]

oh my bad

[raw zenith]

Nah I misread

[glass maple]

i should have explained it as a way of understanding

[raw zenith]

np lol

[glass maple]

so is it a bit clearer now?

[raw zenith]

thank you, I think get it

[glass maple]

the text here is just trying to mention a specific case of relative complement

[raw zenith]

Yeah totally, I knew what set difference was, I just did not know this notation

[glass maple]

oh ok

[raw zenith]

👋

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[gloomy sage]

Can anyone help explain this equation in a simpler way?

[neon iron]

you understood the question ?
cause it kind of doesn't make any sense

[gloomy sage]

Well i don't understand the part -2(x) = 6 over 4 then further

Though it's an example from a learning material

[neon iron]

say we have the quadratic equation
ax^2+bx+c
then we can say that
sum of roots is -b/a and
product of roots is c/a

inorder to prove this

[gloomy sage]

Yeah that's the case, i have solved how to find the sum and products of roots but when it comes to find the missing value I'm having trouble

[neon iron]

say the roots of quadratic equation ax^2+bx+c are x1 and x2
then we have
(x-x1)(x-x2)=ax^2+bx+c

equate of coeeficents of x, and constant on both sides

and you will have it

[neon iron]

could you be more specific,

[gloomy sage]

Ohh I meant, I'm struggling to find the missing values just like the ones in the equation especially when 6 over 4 showed up

[neon iron]

sorry,not sure how to help with that trouble

[gloomy sage]

Yeahh no worries

Thanks for the thought tho

[merry hill]

$-2x = \frac{6}{4}$ this part?

[thorny flame]

Astar777

[gloomy sage]

Yep

[merry hill]

so you are confused about how to find x here?

[gloomy sage]

Yeah and I don't understand how they found the value of "k"

[merry hill]

ok so you know what sum of roots and product of roots is equal to, right?

[gloomy sage]

Yeah well, but if there's only 1 root how do i solve it?

[merry hill]

you are given one root

and the quadratic

now use this info: sum is -b/a and product is c/a

if you write the product of roots, its -2x = 6/4

so x= -3/4

now you have both of the roots

now if you look at the quadratic

k is in place where b goes, right?

[gloomy sage]

Yep

[merry hill]

and what is sum of roots equal to?

-b/a, right?

so you can write -2 + (-3/4) = -k/1

[gloomy sage]

I see.. I'm quite a slow learner though that makes it simpler

[merry hill]

there's a typo here

it should be 4x^2

[gloomy sage]

Ohhh now it get it

[merry hill]

so -k/4

because a is 4

[gloomy sage]

I was wondering where they got the 4

So yeah

[merry hill]

so u understand how we used this info to find k?

[gloomy sage]

Mhm thanks alot

[merry hill]

[gloomy sage]

Have a good dayyy

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[tribal lance]

Anyone know if what I have done here is correct?

[wary tulip]

i love your handwriting

[tribal lance]

Thanks 😅

[wary tulip]

yes i think it's ok

[tribal lance]

Gotcha, thanks!

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[midnight sinew]

I need to show that $(n,m)=(n+m,[n,m])$ where $(n,m)$ and $[n,m]$ denotes gcd and LCM respectively

[thorny flame]

somethingwrong

[midnight sinew]

So i was able to show that any divisor that divides lhs will divide rhs

But im not sure how to show that any divisor that divides rhs will divide lhs

so to rephrase the question, how can i show that $d|n+m$ and $d|[m,n]$ implies $d|n$

[thorny flame]

somethingwrong

[hearty turret]

i guess you can write n=dx, m=dy where d=gcd(n,m)

then it suffices to prove that for coprime x,y

gcd(x+y, xy) = 1

[midnight sinew]

ah okay i got it, thanks alot

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[thorny remnant]

*hint left multiply bot sides by P

[strong sable]

nvm I got it now

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[lime delta]

I'm mad geeked 😭

So like domain

Do I put two different things cus 2 lines

So like -inf < x < 1

Then 1 < x < inf

[merry hill]

how is the first graph discrete

[lime delta]

Cus not connect

[merry hill]

nah

[vernal matrix]

1 is a valid element of the domain, you have f(1) being about 1

[merry hill]

the domain written in ur image is correct

the graph isnt continuous, sure, but its still defined for f(1) so 1 is in domain

[lime delta]

Oh I'm geeked it's neither discreet or continuous

[merry hill]

yeah

[lime delta]

So how do I do the domain and range

If it's neither

[vernal matrix]

how are they meaning to use the word "continuous" there

[merry hill]

whole graph continous ig

[lime delta]

Is -inf < x < 1 correct

[merry hill]

no

-inf < x < inf

[lime delta]

What about the 1

[merry hill]

the function is defined at 1

so its in domain

its just that the graph isnt continous at 1

but that doesnt matter for domain as long as its defined there

[lime delta]

😭

[vernal matrix]

I feel they're using it not in the sense of "continuous function, no jumps" continuous, but "not discrete, can take any number in an interval" continuous

Which is a very poor choice of wording

(also do you know the difference between hollow and solid dots on graphs?)

[lime delta]

For range is it also -inf < x < inf

[merry hill]

range is what output u get

which is on y axis

[vernal matrix]

I really don't like the way they've graphed this one bit at all

[merry hill]

same

[lime delta]

Me no no get it 😔

[merry hill]

do you know what domain and range is?

[lime delta]

Domain is like the x and y is range

[merry hill]

yeah so check the graph

[lime delta]

Will there be two seperate things for y?

[merry hill]

nah

[lime delta]

Okay I gotta go

Sorry

[merry hill]

okay

[lime delta]

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[topaz sinew]

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

[neon iron]

1

[brave coral]

Start with the basics: Bring the like terms together

(or rather, bring the terms containing x to one side)

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@neon iron Has your question been resolved?

[neon iron]

i finished ty tho

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[finite pond]

idk if i've done something wrong

[finite pond]

what have i done wrong

i got the answer 8.2 (rounded to 1dp)

[shut garden]

What is x?

[finite pond]

469/180 pi

[shut garden]

≈ 42.6 degrees

[finite pond]

howd u get that

[snow tiger]

Where does the right side of your equation even come from

[finite pond]

i did 14pi / 360

then ans x 67

[finite pond]

howd u get that

[shut garden]

What is ?

[finite pond]

it says they both have the same arc length

so the arc length of sector 2 is 469/180 pi

so i just did inverse thing

[snow tiger]

the left side is correct but i dont understand where you get the right side from

22pi/360 is just the arc length of sector 2 with degree 1

[finite pond]

nvm i figured out where i went wrong

it shouldve been 22/360 pi

not 22pi / 360

[finite pond]

the 22 pi comes from

the circumference

so 11 is the radius

[snow tiger]

yes

[finite pond]

the circumference would be 22 pi

[snow tiger]

why are you dividing it by 360

[finite pond]

so i can multiply that by the angle

and then that would give me the vector arc

[snow tiger]

but you already have the arc?

[finite pond]

i know but i was just writing what i would have to do

in order to get the arc

the one piece that was missing was the angle

the ? is the angle

22/360 pi multiplied by the angle would be the length of the vector arc

[snow tiger]

oh

i see now

but why is x on that side of the equation now

[finite pond]

wdym

[snow tiger]

you have 22pi/360 * x * angle

[finite pond]

x is just a multiply sign

[snow tiger]

oh ok lol

[finite pond]

wait

yeah

[snow tiger]

but then your original equation is correct

not 22/360pi

[finite pond]

it wasnt that

i thought it was

but i messed it up

cuz that gives

8.1.....

so yk

the other one gives 41 or smth

[snow tiger]

um i got 42.6 with that equation

go over it again it should be right

[finite pond]

with this?

or 22pi/360

[snow tiger]

22pi/360

the first thing you had

[finite pond]

maybe we got different calculators

ill send u what i got

[snow tiger]

send your work

[snow tiger]

yes

and then

[finite pond]

do that divided by 469/180 pi

and u get sm random decimal

[snow tiger]

you do 469/180pi

and then divide that result by 0.191

look at your equation and solve for the angle first and then put everything in the calculator it is easier

[finite pond]

idk but i got the correct answer by doing

469/180 pi / 22/360 pi

when i did 469/180 pi / 22pi/360 it did sm random answer

[finite pond]

omg

im such a dumbass

i didnt it other way around

did it*

i did (22pi / 360) / (469/10 pi)

🤦‍♂️🤦‍♂️

[snow tiger]

Well there you go

[finite pond]

oh well

we learn from our mistakes 🤷‍♂️

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[heady mulch]

hello

[heady mulch]

Exercise 1 (P). Let m, n ∈ Z. We say that m divides n (or n is divisible by m) if and only if there exists a k ∈ Z such that n = km. In mathematical symbols, we write for m, n ∈ Z
m | n ⇔ ∃k ∈ Z : n = km.
An integer is said to be even if it is divisible by 2, and odd otherwise.

(i) Show that if an integer is even, then its square is divisible by 4.

i need help for the first

[compact moss]

use the definition provided first

[heady mulch]

n = km ?

[compact moss]

yes but if its even its divisible by 2

so 2|n

[heady mulch]

n=2k ?

[compact moss]

yes and now square it

[heady mulch]

n²=4k²

[compact moss]

and remember definitions are iff so <->

[heady mulch]

4 | n²

[compact moss]

yes you can let k^2=k' just so it fits better with the definition

n^2=4k'

[heady mulch]

but its for n² not n

[compact moss]

yes thats what the question was asking for

oh you mean when using the definition in reverse?

you can also let n^2=n'

so n'=4k'

[heady mulch]

Ok

ty

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@heady mulch Has your question been resolved?

[heady mulch]

@compact moss for "(ii) Show that if an integer is odd, then its square is also odd."

do i need to do n = 2k+1 ?

[compact moss]

yes since that means 2 not divides n which was the definition they gave you for an odd number

@heady mulch

[heady mulch]

ok i have a last question

"Show that if the square of an integer is even, then that number is even too." i did not understand that question

i need to start by n²=2k ?

[compact moss]

you technically already proved it by (i) and (ii)

the integer is either even or odd (divisible by 2 or not). you showed that if the integer is even it has even square and if its odd it has odd square. so the integer must have been even for the square to be even.

@heady mulch

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@heady mulch Has your question been resolved?

[sweet shard]

<@&268886789983436800> spam

[heady mulch]

ok thank you

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[flint widget]

need help with boolean expressions
Using the Identities and Laws of Boolean Algebra, simplify the following expressions. List
the specific law used for every simplification.
A∗B∗~C+A∗B+A∗C+A

[flint widget]

<@&286206848099549185>

[rigid ivy]

!15m

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[vivid pivot]

Did i do these right?

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[empty locust]

i believe yes

[lucid tiger]

.close

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Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly snow]

HELP ME PLEASE

[chilly snow]

i just need to do the last quadrant

im begging

<@&286206848099549185> its been 15 mins

[grave wolf]

u got it bro just rotate the shape

and make sure the x, y values align

i believe in u

[stark storm]

Ok ok ok

IIm here now

Daddys here now

ROTATE THE SHAPE

[chilly snow]

thank you vro 🙏

this???

[grave wolf]

that should be it

[chilly snow]

it was wrong

😔

[grave wolf]

what

[stark storm]

Wait i am learnin this in school rn too...omg twintwinsss

Ok then reflection

Reflect it

[chilly snow]

yayyayya

[chilly snow]

mk

[grave wolf]

no.. it clearly says rotate 😭

[stark storm]

No bro

You dont know

I know

[chilly snow]

IT IS

every one of my classmates

[grave wolf]

i do know i read the problem

[chilly snow]

have been struggling on it

[grave wolf]

erm excuse me i take calculus 🤓🤓🤓

[stark storm]

Brielle lets be smart here

Ok reflect

[grave wolf]

did u guys ever do ixl

[stark storm]

Ts aint calculus

[grave wolf]

ik man

[stark storm]

Yes

In fact

[grave wolf]

fuck ixl

[chilly snow]

ixl sucks

[stark storm]

What is ixl!!

[chilly snow]

its a math program

it sucks so bad

it takes you down by 10 points if you get smt wrong

[grave wolf]

prodigy

[grave wolf]

LITERALLY

[chilly snow]

prodigy was fun

[stark storm]

StaStay with me now guys...so when we take the uh pentagon

Prodigy was my shit bro

[grave wolf]

that is NOT a pentagon that’s a trapezoid 😭

[stark storm]

HeLP

Mb mb

I knwo what im doing bro

[grave wolf]

ur right ur right

[stark storm]

Oh yeah i accidenlu ok but stikk

[grave wolf]

let mitch do his thing

mitch u take the reigns

[stark storm]

So basically guys

[grave wolf]

all of our eggs are in YOUR basket

[chilly snow]

AW

[stark storm]

So we gonna have to

[grave wolf]

cuteee

[stark storm]

Acriss the x axis

We fonna have ro

[chilly snow]

im all ears

[stark storm]

Stay with me now

So

OK SO

Its a rotate bri idk

[chilly snow]

i believe in you mitch

[stark storm]

We can

So we gotta rotate this

Yeah

ItIts ararotate

Jut put it

[chilly snow]

i got it wrong

[stark storm]

Its lying to you

[chilly snow]

real

[stark storm]

Oopsie

No your correct the program is js incorrect bro

YeAH

This shit laced

Skip to nect probalem

Email techer

[chilly snow]

i dont have the guts to do that

the teacher said we had to ask for help

and then like

[stark storm]

I fear i might have to google this one

BRO USE PHTOMATPHTOMATH