help-10

[gilded jungle]

You're welcome-

[pale valley]

alright guys i’m cooked either way thanks for all the help my paper is liek soaked in tears rn 😭

[rare hollow]

oops i forgot u could divide both sides by 7

so it would be log_4 (v-9) = 0

[gilded jungle]

It's OK, we'll help you.

[pale valley]

oh that’s what the other guy did

[rare hollow]

you can rewrite it to 4^0 = v-9

so v = 10

[gilded jungle]

That's what I got as well-

Alright, next question.

[pale valley]

OH MY GOD i see how it’s 10

okay

yeah next of many

sigh

[gilded jungle]

log₂(9) + log₂(x) = 2.
Huh.

[pale valley]

oh it’s log_2(9)

it’s a lil blurry at the top

[gilded jungle]

Oh, sorry-

[pale valley]

no it’s ok! i took the pic fast

i think this is the product rule?

[gilded jungle]

OK, should be easier than the last question.

[gilded jungle]

Mhm.

[pale valley]

i pray

okay so will it be log_2(9)(x)?

[gilded jungle]

No, log₂(9x), maybe-

[pale valley]

right yeah i meant that oops

so rhsn 2^2 = 9x?

[gilded jungle]

Yup.

9x = 4, should be easy from here.

[pale valley]

yeah 4/9 is the answer right

?

she said not to simplify just keep fractions

*my teacher

[gilded jungle]

Yup.

[pale valley]

yay

[gilded jungle]

What is there to simplify-

[pale valley]

darn i meant decimals not simplify

whops

okey next one is quotient right

[gilded jungle]

Next, log₇(x) - log₇(5) = 1. Should be the quotient rule.

[pale valley]

okay so far i have log_7 x/5 = 1

[gilded jungle]

That means x/5 = 7¹.

[pale valley]

yes

i’m tryna work it out rn

x=35?

[gilded jungle]

Yup, because:
x/5 = 7
x = 7(5)
x = 35.

Next, log₇(4x) + log₇(4) = 2.

Once again, use the product rule.

[pale valley]

okey!

x=8?

[gilded jungle]

Uhh, I don't think so...

Lemme see.

log₇(4x) + log₇(x) = 2.
log₇(4x * x) = 2.
4x * x = 7².

Is this right?

[pale valley]

lemme see

oh crap i messed up the power

okay gimme a sec

49/16?

[gilded jungle]

Alright:
4x² = 7².
√(4x²) = √(7²).
2x = 7.
This should be right.

[pale valley]

wait how

i did 4x • 4 so 16x so log_7(16x)=2 and then i did 7^2=16x

[gilded jungle]

Oh crap, I messed up. UGHHHHH-

[pale valley]

it’s okay!!!

log is …… something

[gilded jungle]

log₇(4x) + log₇(4) = 2.
log₇(4x * 4) = 2.
4x * 4 = 7².
16x = 49.
x = 49/16. You were right.

[gilded jungle]

I've dealt with worse.

Next, log₆(4x) + log₆(4) = log₆(32).

[pale valley]

yikes….

okay this is the log=log stuff

x=2?

[gilded jungle]

log₆(4x) + log₆(4) = log₆(32).
log₆(4x * 4) = log₆(32).
log₆(16x) = log₆(32).
16x = 32.
x = 2.

[pale valley]

yay !

[gilded jungle]

Alright! Next, log₆(8) - log₆(x - 8) = 1.

[pale valley]

quotient i think

*rule

[gilded jungle]

Yup. Shouldn't be too bad for you.

[pale valley]

i’m not sure what to do about the x-8

oh wait

bruh

nevermind

am i on the right track….

[gilded jungle]

Mhm. Yup, that's right.

[pale valley]

do i multiply by 8 on both sides?

[gilded jungle]

No, that becomes 6(8) = 64/(x - 8).

Alright, how do you get rid of the (x - 8) term in the denominator?

[pale valley]

☹️ im not sure….

[gilded jungle]

||Hint: (x/y) * y = x.||

Hello?

[pale valley]

im probably dumb is it like x-8 multiply on both sides

i was contemplating

[gilded jungle]

It is.

[pale valley]

oh my gosh i’m einstein

yippee

ok i’ll solve like normal is this correct before i solve
6(x-8)=8

[gilded jungle]

Mhm.

[pale valley]

is the answer 56/6

or 28/3?

[gilded jungle]

Yup.

[pale valley]

yayyy

[gilded jungle]

Great job so far!

[pale valley]

thank you 😞 also to you too cuz this must be exhausting

[gilded jungle]

Last, log₉(4) - log₉(2x) = log₉(20).

[gilded jungle]

Not really-

[pale valley]

oh

[pale valley]

i think the picture is a little blurry it’s actually log_9 4 - log_9 blah blah blah

there’s not an equal sign

in the middle of the first two i mean

[gilded jungle]

Oop, typo.

You can solve this, right?-

[pale valley]

yeah i started out is this right so far?

(4)(2x)=20

[gilded jungle]

Mhm.

[pale valley]

so 20/8?

[gilded jungle]

AFK, gimme a bit.

Alright, I'm back.

[pale valley]

hai

[gilded jungle]

Yup.

[pale valley]

YAY

thank you sm for helping me you’re an angel i just saved like a trillion more tears

i’m not cooked anymore !!

[gilded jungle]

You're welcome! If you have any more questions, feel free to DM me.

[pale valley]

thank you!! have a goodnight

[gilded jungle]

You too!

Also,

!done

[obtuse pebble]

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

[pale valley]

.close

[obtuse pebble]

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[obtuse pebble]

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[warped zinc]

How would I find m> ±6√2

[warped zinc]

Question is to find what values of m make the quadratic have 2 real roots

Quadratic is 2x²+mx+9

So I used discriminant

[tranquil geyser]

In the quadratic formula
You have x=[-b±√(b²-4ac)]/2a

What happens if b²<4ac

[warped zinc]

If b²-4ac>0 there are two real roots

[warped zinc]

I simplified that formula to this

[tranquil geyser]

What happens if b² < 4ac

[warped zinc]

Not sure

Hmmm

[warped zinc]

No real roots?

[tranquil geyser]

Yes

So solve b² ≥ 4ac

[warped zinc]

m≥±6√2

[tranquil geyser]

Solve it and show your steps

[tranquil geyser]

,rccw

[tranquil geyser]

Do you know how to solve quadratic inequalities

This is wrong

[warped zinc]

Nope

Not really

How would I

[tranquil geyser]

You can use a table like this

[warped zinc]

I remember using a number line

[obtuse pebble]

@warped zinc Has your question been resolved?

[warped zinc]

<@&286206848099549185>

[topaz island]

hello, what is the question?

@warped zinc

[warped zinc]

Idk how to solve quadratic inequalities

[warped zinc]

I'm at this step

[topaz island]

which ones? is there a specific question?

or just in general?

[warped zinc]

Click the msg I replied to

I explain it there

[topaz island]

Question is to find what values of m make the quadratic have 2 real roots
Quadratic is 2x²+mx+9

this is the question i assume

ok so, when does a quadratic have 2 real roots, like what conditions need to be met for this to occur

[warped zinc]

The discriminate is greater than 0

[topaz island]

yes

[warped zinc]

B²-4ac

[topaz island]

yes

[topaz island]

thats your answer then

the ranges for the values are when m>6sqrt(2) and when m<-6sqrt(2)

[warped zinc]

But why does the inequality flip

I thought it is just

M>6√2

M>-6√2

[topaz island]

it means the same thing in this context

[warped zinc]

Is there a way to visualize it

Better

[topaz island]

it is everything that is less than the one on the left or greater than the one on the right, if that makes sense

[obtuse pebble]

@warped zinc Has your question been resolved?

[tranquil geyser]

Back

[tranquil geyser]

Have a look at

,w Plot y=x²-4

[warm shale]

Results provided by WolframAlpha

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

Requested by ishigamisenku_

[tranquil geyser]

At what value is the curve above 0?

[obtuse pebble]

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[rocky tundra]

this is probably me being dumb bot how do i solve this?

this is my attempted work

[rocky tundra]

wait nevermind i just forgot how to solve a system

[raven spire]

bro hold on

you're doing more Onion stuff than just forgetting how to solve a system

sin(pi/3) is not 1/2 and cos(pi/3) is not √3/2

[rocky tundra]

is it not

wait holy ;8,&0

imm kms

how the hell did i make it to DE 😍😍😍

[raven spire]

[rocky tundra]

i think i might be special

!close

.close

[obtuse pebble]

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[obtuse pebble]

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

Just need help explaining how we get from the second step to x^2. Thanks

[stable hill]

log laws

a*log(b)=log(b^a)

if youre talking about the next step, b^log base b (c) = c
ln(c)=log base e(c)

[empty dune]

Right, thanks

.close

[obtuse pebble]

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

Can someone check my work real quick?

[kind rapids]

I suppose this is correct since x will grow faster than ln(x)

[dark stirrup]

How did you get to $\frac{-\ln x}{1/x}$?

[warm shale]

SWR

[kind rapids]

oh fuck that is -xln(x)

[dark stirrup]

Hm?

[kind rapids]

shit i guess i need to do smth else

well 1/1/x is just x?

wait wtf did i do

how is 1/x lim x-> infinte even infinite

imma try again

💀

[restive gorge]

Maybe you meant x(-ln(x)) not x-ln(x)

[kind rapids]

ye

hmmm

[dark stirrup]

!original if possible

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[snow cradle]

[restive gorge]

Ok yea

Intuitively any ideas

[kind rapids]

maybe split them?

[restive gorge]

No

[kind rapids]

oh

[snow cradle]

rewrite x as ln(e^x)

[kind rapids]

then no idea tbf

aaaaaaa

why do i not see thissss

okay thank you

[vernal wigeon]

Try to factor by x and you will get something known

[dark stirrup]

No. Not aaaaa, it was @snow cradle

[kind rapids]

thank you mmmm

so i now have lim x-> infinite ln(e^x/x), then i want to do limit x-> infinite e^x/x and solve that and then plug that back in ?

so this is correct?

[restive gorge]

It's funny you are using the same argument for e^x/x you could have used for x-ln(x)

[kind rapids]

what

wait

was that allowed?

[restive gorge]

You are concluding e^x/x goes to infinity because "an exponential dominates a polynomial"

But so

[kind rapids]

does x over ln(x)

[restive gorge]

You could have done the same for x-ln(x), a polynomial dominates any log

[kind rapids]

was that enough too?

[restive gorge]

I am just saying

You could technically apply L'Hopital for e^x/x if you wanna make it really obv

But I was just trying to give that intuition

[kind rapids]

aaah

[kind rapids]

imma do this that is probably what they want

smart smart

[kind rapids]

is this one correct

?

[restive gorge]

yup

[kind rapids]

lets gooo

thank you

why am i allowed to put the lim on the power of e?

is it cuz e is bijective?

[restive gorge]

it's because e^x is a continuous function

[kind rapids]

Aaah okay so i am allowed to do it as long as it is a continious function?

[kind rapids]

Aaaah ye we did this proof

[obtuse pebble]

@kind rapids Has your question been resolved?

[raven spire]

!show

[obtuse pebble]

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

[kind rapids]

i feel like i did smth wrong just because it took so long

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[raven spire]

show work +_+

[kind rapids]

this is my work

[raven spire]

XD both are you, mb

gimme a sec

[kind rapids]

no worries its just cuz if i am not home i can quickly send stuff to my ipad and laptop

[raven spire]

Ayeeeeeee

T-T

alr I'll tell you mistake later, but you'll have to scratch that anyways so can you follow me for a bit?

[kind rapids]

i am all ears!

[raven spire]

$L = \lim_{x \to 0^+} (\tan 2x)^x$

[warm shale]

Arya

[raven spire]

$\ln L = \lim_{x \to 0^+} x \ln (\tan 2x)$

[warm shale]

Arya

[raven spire]

$\ln L = \lim_{x \to 0^+} \frac{\ln \tan 2x}{\frac{1}{x}}$

[warm shale]

Arya

[kind rapids]

oh wow

i thought of the arctan

instead of the tan

[raven spire]

$= \frac{\frac{2\sec^2 2x}{\tan 2x}}{-\frac{1}{x^2}}$

[warm shale]

Arya

[kind rapids]

wow

i havent had sec :-:

[raven spire]

XD

[kind rapids]

dy/dx tan(x) = 1/tan²x?

[raven spire]

no it's sec² x

[kind rapids]

huh

oh

1/cos²x

[raven spire]

d/dx (tan x) = sec² x

[kind rapids]

sec = 1/cos x?

[raven spire]

yes

[kind rapids]

that is so much easier to write

imma use that next time

[raven spire]

btw here sec² 0 = 1, so it's easier to just plug it 1

$= -\frac{2\sec^ 2x \cdot x^2}{\tan 2x}$

[kind rapids]

i dont think our prof allows us to we must use all limit rules and stuff first :/

[warm shale]

Arya

[raven spire]

= 0

do you get it?

[kind rapids]

yes thank you

wait

i think imma try to solve thisn ow

now

[raven spire]

$\implies \ln L = 0 \implies L = 1$

[warm shale]

Arya

[kind rapids]

i have to split tan(2x) into sin(2x)/cos(2x) right?

and split the limits?

[raven spire]

btw there's an alternative way that does not require you to use L'Hopital or any of this

[kind rapids]

oh can you show me that afterwards?

[raven spire]

sure. but for now, no you do not need to split tan 2x into sin 2x / cos 2x because the derivative of tan x is sec² x is known.

You can apply chain rule for tan 2x

[kind rapids]

wait u do l'hospital again?

[raven spire]

wait, what are you talking about?

[kind rapids]

to solve this

[raven spire]

me applied Lhopital here

[kind rapids]

how do i solve this?

[raven spire]

$= -\sec^2 x \cdot \frac{2x}{\tan 2x} \cdot x$

[warm shale]

Arya

[raven spire]

where (2x/tan 2x) approaches 1, sec² x approaches 1, and x approaches 0

so limit is 0

[kind rapids]

but 2x/tan(2x) is 0/0 ?

and it is not a special limit right?

[raven spire]

is it not a basic limit? I thought you guys would've learnt $\lim_{x \to 0} \frac{\tan x}{x} = 1$ as a standard limit

[warm shale]

Arya

[kind rapids]

let me check to be sure

[raven spire]

you can lhopital it ig

[kind rapids]

again l'hopital? wouldnt that make it so much more complicated with all product and quotient rules?

[raven spire]

just for (tan x)/x, and you can just state \lim (tan x) / x = 1

[kind rapids]

oh so i split them up the limits and only do it for that one?

that is smart

[raven spire]

✅

[kind rapids]

thank you!

nope just checked we arent allowed to use 2x/tan(2x) = 1

[raven spire]

🤦

Ig then you derive that \lim (tan x) / x = 1

using lhopital

[kind rapids]

how would i write that down mathematically

i now have lim sec² x * lim 2x/tan(2x) * lim x = ?

how can i like state that i only use LH for lim 2x/tan(2x)?

[raven spire]

yes

[raven spire]

Seperately =_+ Did you not seperate limits in your work already because they were seperately finite

you just need to show lim 2x/(tan 2x) is finite as are the rest and seperate them like so

[kind rapids]

aaaaah okay

u said there was a way faster way?

[raven spire]

not necessarily faster :c

[kind rapids]

well i am still curious if you dont mind showing me

[raven spire]

$2\ln L = \lim_{y \to 0^+} (\tan^{-1} y) (\ln y)$

[kind rapids]

wait what

[warm shale]

Arya

[kind rapids]

i dont get this step

[raven spire]

I took tan 2x = y

[kind rapids]

ooooh

okay

[raven spire]

$2\ln L = \lim_{y \to 0^+} \frac{\tan^{-1} y}{y} \cdot \lim_{y \to 0^+} (y\ln y)$

[warm shale]

Arya

[raven spire]

and the left limit is 1

right limit is 0 you'll have to show (is easy)

[kind rapids]

the left one is arctan right?

[raven spire]

Yesh

[kind rapids]

arctan 0 is 1?

[raven spire]

=_=

its 0/0 form, but the limit is 1

[kind rapids]

ooooh

aaah mb i thought it was arctan * y / y

so only arctan

[raven spire]

oiii =_=

[kind rapids]

ye this way is for the smart people tho

[raven spire]

:p you need to build basics strong to survive calculus

[kind rapids]

ye it is going so fast 💀

anyways thank you very much for the help

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[kind rapids]

.reopen

[obtuse pebble]

✅

[kind rapids]

why the 2 ln L btw?

[raven spire]

oh because I put y = tan (2x) and here x is sitting alone

[kind rapids]

aaah

got it

thank you

.close

[obtuse pebble]

Channel closed

Closed by @kind rapids

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[loud sand]

can someone tell me if i’m write

[loud sand]

everything else is saying i’m wrong

[crimson pike]

,rccw

[half plinth]

It’s not wrong… so far. It’s incomplete

[loud sand]

so just find where csc is 2 root theee over three right?

[loud sand]

?

[crimson pike]

just a command to rotate the image

[loud sand]

oh alright

[half plinth]

First, do you know another way to write csc?

[loud sand]

1 over sin

[half plinth]

I’d start by swapping that in

[loud sand]

from the beggining?

[half plinth]

Once you reach csc = …

[loud sand]

so three times 1 over since squared theta

oh

alright

[loud sand]

1 over sin theta = 2root three over three?

[half plinth]

Right, and what would make it easier to isolate theta

[loud sand]

so just take the the reciprocal of both sides?

[half plinth]

Yes

[loud sand]

alright thanks bro i don’t know why i had trouble with this

.close

[obtuse pebble]

Channel closed

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[loud sand]

.reopen

[obtuse pebble]

✅

[loud sand]

can someone explain how i got this wrong on my quiz? ive chucked it a million times and cant find my mistake

[polar fossil]

,rccw

[polar fossil]

dunno

those are the only solutions in [0, 2π)

[obtuse pebble]

@loud sand Has your question been resolved?

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[gloomy vector]

$\sum^{20}_{n=1}\frac{n}{n^4+n^2+1}$

[warm shale]

skissue.in.a.teacup

[gloomy vector]

not sure how to do this

maybe some sort of fraction decomp? but idk how to factor n^4+n^2+1

[ionic oar]

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

oh yeah and after fraction decomp the terms cancel

@gloomy vector

[gloomy vector]

oh

[polar fossil]

neat

[gloomy vector]

an^2+an+a-bn^2+bn-b
a=b=1/2

$\frac{1}{2}\sum^{20}_{n=1}\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}$

[warm shale]

skissue.in.a.teacup

[gloomy vector]

so (1/2)(1-1/421)

210/421

ok ty

.close

[obtuse pebble]

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[soft loom]

How should solve this

[velvet quail]

what have u tried

[soft loom]

I have made the drawing of the graph

Then angle b = 90

[velvet quail]

show diag

[soft loom]

It should look like this

[velvet quail]

,rccw

[soft loom]

Sorry for no labeling

[velvet quail]

name it

[soft loom]

You want me to take the picture again

[velvet quail]

it wud be better

[soft loom]

,rccw

[soft loom]

,rccw

[soft loom]

Ab=bc=cd

Angle ABC is 90 degree

Angle BCD is 150 degree

[velvet quail]

get 1 eqn by equating sum of int angles of a quad = 360

get another by taking triangle abd

get another by taking triangle acd

or u can also use triangle agd

and find out x

[soft loom]

A is 120 right

X+y = 60

Then I'm stuck there

[velvet quail]

get 1 eqn by equating sum of int angles of a quad = 360
get another by taking triangle abd
get another by taking triangle acd

these three dont work?

i didnt check

they dont work

hmm

[soft loom]

Yeah

They all lead to x+y = 60

[obtuse pebble]

@soft loom Has your question been resolved?

[velvet quail]

got any ideas?

[soft loom]

Nah

[velvet quail]

do u think this can be a cyclic quad?

[soft loom]

Uhh idk

I found youtube explanation

Ty

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[velvet quail]

send link

lemme also watch

@soft loom

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[umbral cobalt]

Guys, im trying to understand a topic on direct and inverse proportion that I just can't quite understand, so I have a question here and an answer.
y is directly proportional to the square of x.
find the percentage increase in y when x is increased by 15%

[umbral cobalt]

the answer is 32.25 if im nto mistaken but I am kind of lost on how to get there, and the usual method for proportion just isn't working

[gritty bolt]

y is directly proportional to the square of x.
So, y = k*x^2 ...(1)

Now I'd increment x by 15%
So new x will become x+(15% of x).
x'=115x/100
Now, let's find y when x is x'

y' = k(115x/100)^ ...(2)

We'll get relation between y' and y when we divide equation (1) and (2)

y' = (115/100)^2 y
y' = 1.3225y
y' = y + 0.3225y
So, y' is 32.25% more than y

[obtuse pebble]

@umbral cobalt Has your question been resolved?

[umbral cobalt]

yes thank you viviiix!

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[gleaming acorn]

I ended up with tanx = 1/5

Not sure where to go from there, is that incorect?

[brazen gorge]

,w 5t^2 + 2t - 3 = 0

[warm shale]

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[brazen gorge]

nope

[gleaming acorn]

Lemme ry again

[brazen gorge]

wait

everything up to this is definitely correct

how did you get 1/5 from that?

[gleaming acorn]

Not sure actually

Nvm i see

Ok thanks

[brazen gorge]

5x (x+1) - 3(x+1) = (5x-3)(x+1)
so roots are indeed x=3/5 and x=-1

[gleaming acorn]

Got the answer

thanks

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[zinc tree]

Hello everyone, I need help with this problem, I don't understand how I'm supposed to solve it

[zinc tree]

Sorry, here's the english one

[zinc tree]

<@&286206848099549185>

[obtuse pebble]

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[spring rampart]

pw?

[obtuse pebble]

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[steady totem]

What if I add and subtract equation 1 and 2 to every question would it be okay?

Just like I did to this question

[idle mesa]

You don't have to do this to all the questions

It will be time consuming and I don't think you will get the correct answer too

You can use this method only when the coeff of eq 1 is exchanged in eq 2

[steady totem]

Ohk i get it

Thanx buddy

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[idle mesa]

No problem

[royal basin]

adding and subtracting equations shouldn't be treated as a "do this every single time" thing, they should be tools in your toolbox

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[fast vale]

why is this wrong?

[azure wraith]

how did

u get just x^5

anyway

[fast vale]

factoring out x, i can see that the left nd right limits arent the same

but how do i know not to do what i did initially

[azure wraith]

direct substitution first

it gives

63/0

this is an asymptote form

not indeterminate form

[fast vale]

so if its asymptote form, i shud look at the left and rright limits next?

[azure wraith]

ye to see if it goes any infinity or not

or just dne

https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/cs1-strategy-in-finding-limits/a/limit-strategies-flow-chart

[fast vale]

thanks!

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[kind rapids]

i got this so far: $\frac{1}{\theta^{n}}e^{-\frac{1}{\theta}\sum(x-\lambda)}\mathbf{1}(\min X_i \geq \lambda)$

[warm shale]

Brett

[kind rapids]

idk how to continue :/

g(T1(X),T2(X),theta) * h(X)
is what I want to get

[obtuse pebble]

@kind rapids Has your question been resolved?

[kind rapids]

<@&286206848099549185>

[kind rapids]

Poor statistics

[kind rapids]

<@&286206848099549185>

[compact star]

I think h(X) can be the constant function 1 in this case

[kind rapids]

Hmm but how can I get the sum of xi seperated?

the indicator function is already seperated

[compact star]

you want to apply this i think

[kind rapids]

yep

[compact star]

properties of the exponential function imo

yeah you spotted the min the product of indicator functions

[kind rapids]

aaah

so I can use the expo family

for the other one?

by just using e^ln?

and then find my T(Xi) ?

[kind rapids]

yes but I have a sum now tho :-:

sum y-u is sum y - nu?

[compact star]

works for sums too

[kind rapids]

alright

okay got it

thank you :)

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[lost barn]

hey guys i was wondering if there’s an easy way to identify if something is direct or inverse? its going to come as a multiple choice question for me and its kinda confusing matching the formulas and the questions

[red ice]

right, so for y = 3/x, that's inverse variation

k = 3

[atomic marsh]

what is written in red on the right should be enough for you

[red ice]

if you choose another one, say y/11 = x/8

can you isolate y?

you can think of x/8 as 1/8 * x also

[lost barn]

since its division right?

[lost barn]

yea but thats what im struggling with i cant match them up

[red ice]

you literally wrote that inverse variation means y = k/x

also, k is a constant (number)

[lost barn]

those were my class notes

[red ice]

yes

[lost barn]

OH RIGHT since x is 1

[red ice]

no.....

[lost barn]

oh 💔

so we cant do 1/8?

[red ice]

it seems like you really really need to review basic algebra

say y/11 = 5

what can you do to both sides to solve for y?

[lost barn]

multiply right? so 5x11 is 55

y is 55

[red ice]

yep!

[lost barn]

omg yayyy but how would i know if its inverse or direct or neither

[red ice]

so same principle, $\frac{y}{11} = \frac{x}{8} \implies y = 11 \cdot \frac{x}{8}$

[warm shale]

south

[red ice]

I think you are on the right track with asking 'is it division?'

for direct variation, some number is being multiplied with x

for inverse variation, some number is being divided by x

[lost barn]

oh so this is neither correct? since its 11 times x/8 but i dont know how i should see it should i see it as division since its x/8 or should i see it as multiplication since its 11 x x/8

[red ice]

x/8 is x being divided by some number

so that's not relevant here

[lost barn]

omgg so its inverse

?

[red ice]

no, $y = \frac{11}{8} \cdot x$

[warm shale]

south

[red ice]

you should be able to figure it out from here

[lost barn]

division so its direct 11 divided by 8 times x sorry i switched up inverse and direct for a sec

[red ice]

yep! it's direct!

[lost barn]

YAYY okay thanks i get it now

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[red ice]

the answers for the other two are:

bottom left - ||inverse (k = 7/5) ||
top left - ||neither! y = 2/3 so that's constant variation||

try it out before you spoil it

no worries!

[lost barn]

i will tyyy!!!!

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[versed stratus]

so X=1/2; Y=1

[red ice]

should be geometric probability using the discriminant >= 0 condition

[versed stratus]

I need to find P(X^2≥Y)

I suppose I do this using a double integral now?

[red ice]

ah this is a nice integration question

[red ice]

nope, it's single-variable

[versed stratus]

hmm, surely we want to examine the joint PDF

[red ice]

the assumption is that X, Y are independent

otherwise yes we'd need to consider joint and marginal distributions etc

[versed stratus]

so f(x,y)=f(x)f(y)

[red ice]

ye

[versed stratus]

and unless I;m tripping $X^2= \int_{-1}^{1} 1/4 dx$

[warm shale]

wai

[red ice]

I was thinking of this

[versed stratus]

mhm

That does make sense

Thanks!

[red ice]

no worries!

[versed stratus]

How would this work if I wanted to use LOTUS

[red ice]

no idea

[versed stratus]

nvm, that's for expectation

tq

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[lament coral]

im getting the answer as k=39/2 , by equating the determinant as zero and solving for k, how is the answer 1/4

[royal basin]

can you show your work for the det

this sort of calculation is notoriously error-prone

[red ice]

,w 3a-2b+2c=3, a+39b/2-3c=0, 4a+b+2c=7

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[red ice]

,w 3a-2b+2c=3, a+b/4-3c=0, 4a+b+2c=7

[warm shale]

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[latent quiver]

What you did is basically find where the system is inconsistent by setting det(A) = 0

[red ice]

oh LMAO

[lament coral]

currently i got no clue how to solve this 🫠

[drifting coral]

Doesnt a consistent equation have det=0 tho?

[royal basin]

ok sorry i got a phone call and couldnt check it

[red ice]

surely it's just all real k except 39/2 then

[royal basin]

uhhh right ok so this is a non homog system tho.

this is mega weird cause i dont see any mistake in the det

but let's have WA do it anyway

[red ice]

I already did it

with WA

[royal basin]

,w det {{3,-2,2},{1,k,-3},{4,1,2}}

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[royal basin]

if det weren't zero the system would be guaranteed consistent

,w {{3,-2,2},{1,1/4,-3},{4,1,2}}^-1 * {{3},{0},{7}}

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[royal basin]

their 1/4 gives a sol

[royal basin]

diagnosis: question is borked

[lament coral]

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[remote atlas]

I've been taking seperable differential equations till I reached this step:

If $\frac{dy}{dx} = f(x) \cdot g(y)$ then $\int \frac{1}{g(y)} dy = \int f(x)dx$

[warm shale]

SeaSamak

[remote atlas]

But I couldn't come to a definite conclusion as to how we jumped from the first step to the second one

If $\frac{1}{g(y)} dy = f(x)dx$ then why is it that when we integrate both sides with respect to the differentials already present, the equation is true?

[warm shale]

SeaSamak

[remote atlas]

I'd really appreciate if someone could explain how that makes sense graphically perhaps through Desmos/Geogebra or lead me towards intuiting it

Perhaps through a particular example

[worn yoke]

``under the hood'' what we are really doing is integrating both sides of the equation
[ \frac{1}{g(y)} \odv yx = f(x) ]
with respect to $x$:
[ \int \frac 1{g(y(x))} \odv yx \dd x = \int f(x) \dd x ]
the first integral can be converted to $\dd y$ by substituting $u = y$, which leaves us with
[ \int \frac{1}{g(y(x))} \odv yx \dd x = \int \frac{1}{g(y)} \dd y ]

[remote atlas]

Should dx be here?

[warm shale]

κλαουντ ☁ (cloud)

[remote atlas]

I see, from that standpoint it makes sense!

[remote atlas]

Is there a graphical way to explain how we can directly jump from the first to the second step here?

Like through area with respect to y and area with respect to x

But that answers the question for now, thank you!

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[tame roost]

Hey mathers

[tame roost]

So

x' + x/t = e^t

homogenious solution is xh = C*1/t

But Im struggling at the particular one

using xp = a*e^t does not work

[trail musk]

My guess is try a log(t)

Just a guess

[tame roost]

doesnt work

this is the particular solution

That Im supposed to get

I tried doing xp = a(t) * e^t where a is not constant and a function of t

but then I end up with a' + a (1 + 1/t) = 1

this can not be solved imo

[royal basin]

integrating factor t

tx' + x = te^t

this integrates quite nicely

[tame roost]

Im sorry but in which way

homogenious will be same so particular solution

would be xp=a*e^t ?

[obtuse pebble]

@tame roost Has your question been resolved?

[tame roost]

whatever I found a way still thanks

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[karmic hedge]

I know that x^-1 turns into 1/x

But I dont really know what algebra to do here

Like does it become -1/(x)(x - 1)

Or -1/(x + x - 1)

Ah nvm I found out

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[karmic hedge]

I follow the algebra but I dont really understand whats happening here

What are we doing and why?

I think I understand that f(x) is between the 2 given functions in the given interval

And the limit is within that interval

But I dont really get anything else

[unreal musk]

You're "sandwiched" between two things, and both of them go to the same thing, so the middle has to go to the same thing too

[dusk widget]

the squeeze theorem says that if g(x) <= f(x) <= h(x) on an interval containing some point a and the limits of g and h at a are equal, then the limit of f at a has the same value

put more plainly, f gets squeezed between g and h at a

[karmic hedge]

So at f(0) the limits of all 3 are equal?

Its not really 'between' if its the same tho right?

[dusk widget]

at x = 0, f(0) is not necessarily defined :p

[karmic hedge]

Or am I focusing too much on the semantics

Okok haha

[tepid mountain]

The sandwich theorem, yummy.

[dusk widget]

there's one country that calls it the "two policemen and drunkard theorem" iirc

I always found that to be the most colourful name

[karmic hedge]

Hahaha I think this will be easy to remember

But ok I think I get it

Thank you everyone!

❤️

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[night plume]

katherin

[night plume]

e

still here?

still abit confused as to why my 3. is not transitive

i can see that its symmetric because we can have arrows that goes back and forth

[mortal blade]

Why is the first not transitive

3 is not transitive since you have (2,4) and (4,2) but no (2,2)

[obtuse pebble]

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[pulsar terrace]

0.999.... = 1

Is this a true statement, if not give reasons

[full trench]

It is indeed a true statement.

[pulsar terrace]

But 0.99999 is kept repeating but never reaches 1

[pastel flint]

yes

[full trench]

it does indeed reach one.

1/3 = 0.333⋯
3 • ¹⁄₃ = 3 • 0.333⋯
1 = 0.999⋯

this is just 1 of MANY proofs

[pastel flint]

it does reach 1 at infinite iterations of 9

[pulsar terrace]

1 ÷ 3 = 0.33333 is true
But 1÷ 1 is just 1

[pastel flint]

multiply both sides by 3

[full trench]

1/1 is just 1.

[trim portal]

and whats ur point?

[pulsar terrace]

Yes its not 0.999

[trim portal]

1 and 0.99999... is the same thing

so it's both 1 and 0.9999...

[full trench]

We’re not here to debate with you. Would you like proofs that 1 does = 0.999⋯?

[trim portal]

this is a solid debate for now imo

[full trench]

for now, yes, but i wouldn’t want this to degrade into a “yes no” thingy

[pulsar terrace]

I already know proofs of that 1 = 0.999...

[full trench]

so what exactly are you confused about?

are you confused about the proofs?

[pulsar terrace]

I was just confused since some people say 1 doesnt equal 0.999

So i just came to double checc it

[timid silo]

if you don't put "..." then they're not equal and you're correct

[trim portal]

"1" and "0.999..." are just different "names" for the same number 1

[timid silo]

but if you put the "..." then they're equal

[trim portal]

they're wrong (if they mean 0.999...). Some people say that earth is flat, but that doesnt mean they're right

[full trench]

Well stubbornness for starters lol.

[pulsar terrace]

I just forgot to put "..."

[pastel flint]

bc 0.999 and 0.999...... is not the same one terminates at 3rd 9 the other goes on forever

[full trench]

Some people just refuse to believe the proofs, continuing on from MathIsAlwaysRight

[deep mica]

hello guys sorry interrupting i still struggling understand exponential and arithmetic can anyone explaining to me like i was 5 years old?

[timid silo]

open a new channel

[pulsar terrace]

@deep mica #help-22

[full trench]

This channel’s occupied.

[pulsar terrace]

.close

[obtuse pebble]

Channel closed

Closed by @pulsar terrace

Use .reopen if this was a mistake.

[trim portal]

this one isnt currently available, u gotta choose one of these

[pulsar terrace]

Yes

#help-21｜아리스킨충1

[obtuse pebble]

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[regal prism]

can anyone give me clues about the D one?

[past sand]

What did you answer for a,b,c?

[regal prism]

90
900
900

[past sand]

Ok, now what would be the number of 4-digit palindromes with only two different non-zero digits?

[regal prism]

2?

[past sand]

Since there are so few, can you list them? Just take the digits 8 and 9

[regal prism]

89 or 98 (mirror)

[past sand]

These aren't 4-digit palindromes

[regal prism]

8998 or 9889

[past sand]

Ok, I'm guessing you included the additional constraint of having at least one of each?

[regal prism]

yes

[past sand]

Don't

How many 16-digit palindromes with only 8 and 9, not necessarily with at least one of each?

[regal prism]

2^8?

[past sand]

Yes, now how many are there with at least one of each digit?

Would you know how to easily find that?

[regal prism]

2^8-2

[past sand]

Exactly

[regal prism]

complementary counting

[past sand]

At what point does 2^k - 2 become greater than 2004?

[regal prism]

k = 11

[past sand]

Yes, and that corresponds to how many digits of the palindrome?

[regal prism]

22 or 21

[past sand]

Nice

[regal prism]

complementary isn't natural or thinkable for me, what should i do?

also i'm preparing for big exam and combinatorics ins't major topic, is it worth to invest too much on it?

[past sand]

Learn the principle of inclusion-exclusion and when to apply it (that would just come with practice I guess)

[past sand]

I can't answer that

[regal prism]

okay, huge thankss

.close

[obtuse pebble]

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[deep perch]

This is the question

[deep perch]

This is the answer

Why do they use -3 and not 0? Isnt the second bracket on the first line supposed to be (X-t)2 (X-0)??

How can they sub -3 into the F(x) This makes sense and the second bracket as well??? So confused..

[timid silo]

no, because we have one root (3,0)

so when x = 3, the graph is 0, so (x-3)

[deep perch]

But thats the y value of 0

isnt the bracket supposed to be the opposite sign of the X value??

[timid silo]

would be (x-0) if you have a root on x =0, but that is not the case

you have a root on x = 3

[deep perch]

AAa but isnt thwe root 0 ??

OH WAIT

[timid silo]

yeah the y-value of the root is 0

[deep perch]

I was looking at the y interceptttt

[timid silo]

but we need the x-value

[deep perch]

OHHHHH

I get it nowww TY TY

Goofy moment

[timid silo]

you're welcome

and also

we have a double root on a x = t

[deep perch]

Yes yes thats why its squared

[timid silo]

yeah

[deep perch]

.close

[obtuse pebble]

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[stoic palm]

https://www.damtp.cam.ac.uk/user/examples/A3a.pdf Q12

[stoic palm]

Me trying to use the chain rule

apparently the dp/dv on both sides don't cancel

but like idk what you're supposed to do for part i

<@&286206848099549185>

[obtuse pebble]

@stoic palm Has your question been resolved?

[novel elk]

Usually what you start doing in this types of exercises is writing the differentials of p instead of the derivatives, it would go like this: $p=p(V,S)\Longrightarrow dp=\frac{\partial p}{\partial V} \bigg|_S dV+\frac{\partial p}{\partial S} \bigg|_V dS$. Since $S=S(V,T)$ you can do a similar expansion for the entropy and substitute in the previous expression. Finally to get the relation you are looking for you just need to write $p=p(V,T)$ the same way and compare the terms with the same differentials. In the last step you need to write $p$ as a function of the $T$ and $V$, because when you write $p(V,S(V,T))$ it means you can write $p$ just as a function of $T$ and $V$ (by substituting $S$ by $S(V,T)$ in the expression of $p$)

[warm shale]

KonoEmllikDa

[stoic palm]

tysm

.close

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[timid silo]

Hi, can anyone clarify how you suppose to define transformation for the RHS of this equality. Cause for volume integral I just used Jacobian but for surface integral you cant do that cause you ended up with 3 by 2 matrix. So it turns out that scale factor is a magnitude of a cross product of tangents with respect to each angle.

[timid silo]

Here is what I have done

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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[ember sequoia]

Proof for formulas to find no of squares in a given grind shape

[obtuse pebble]

@ember sequoia Has your question been resolved?

[obtuse pebble]

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[cosmic flame]

differentiate the F(x) function

didnt amount to much\

[tardy epoch]

correct

show your F'(x)

[cosmic flame]

2f(x)f'(x)-(f'(x)f(x^2-x)+f(x)f'(x^2-x)(2x-1))

[tardy epoch]

almost. you have a sign error

oh fixed

yea just plug in 2 now

f(2) = 1, f'(2) = slope of a line you have to find, f(2^2 - 2) = f(4-2), etc.

[cosmic flame]

mustve done something wrong since i got -3

.close

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[eager hawk]

Hi?

[eager hawk]

10th question

[royal basin]

,rccw

[eager hawk]

Anyhelp?

[royal basin]

sorry, can't/don't want to help with this one

[eager hawk]

Okaay

<@&286206848099549185>

[blazing pine]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[blazing pine]

But i'll help

[eager hawk]

..

[blazing pine]

Have you tried graphing the functions

[blazing pine]

??

[eager hawk]

Yh

Here u go

Im not sure about d

[blazing pine]

This is f(x)