help-0

[gritty bramble]

they're completely different

the solution for both is different

thats why you dont split -> do some operation -> add them back

you completely changed the function

[trail imp]

so, if I change the function, I cant recombine them back

[gritty bramble]

you cant go back and forth the two functions here, that means the simplification was all wrong

[gritty bramble]

yes

[trail imp]

alright, it's still a bit confusing but I shall practice more on this. thankyou for the help

[modern kernel]

But he made absolutely

No changes

Why is this not applicable?

[trail imp]

i wonder eh, ping me if you do find out the reason.

[modern kernel]

There probably is a different mistake, not this

I'm trying to find out

Hey hear me out

Wasn't the formula
1 - cos x/x = 0

[trail imp]

i am pretty sure sin x/x exists

[modern kernel]

Yes absolutely

But I don't think 1 - cos x/x² exists

It's 1 - cos x/x = 0

I believe

Correct me if I'm wrong

[trail imp]

yes, it does, try using l hopital on it.

you would get 1/2

[modern kernel]

You sure?

Won't the form be disrupted

If we do lhopital

[trail imp]

it's 0/0 form right, so differentiating it will give sinx/2x which is also 0/0 form

here, either you can l hopital again or use sinx/x=1

[modern kernel]

Hmm I see

The only possible conclusion I can draw is we can't use those formulas after lhopital

It'll change it

Either it's those formulas or only lhopital

[trail imp]

<@&268886789983436800>

[modern kernel]

I'm probably wrong however that's all I can think of

[trail imp]

same, but I still dont get why

[modern kernel]

Lhopital is a really awkward way

I don't know either

I'll research

[trail imp]

thankyou for your time. lemme know if you find something. I dont really have a teacher I could ask my doubts from.

[modern kernel]

don't thank me I barely contributed

I'll let you know my findings

[trail imp]

.close

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[gritty bramble]

well

then maybe you cant do allat if u apply l'hopital

but you also dont combine limits after you've split them, atleast thats what i read

[lone heart]

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[gray timber]

I wanted to derive cayley's formula using Matrix Tree Theorem. But I am not able to proceed as I am not able to solve this determinant of size (n-1)*(n-1). Can someone help please.

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@gray timber Has your question been resolved?

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@gray timber Has your question been resolved?

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[gray timber]

.reopen

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✅

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[velvet light]

quick qs but what does the circle at the top mean

[alpine sable]

degrees probably

a unit of angle

[velvet light]

this is the question it's clearly d or b

.close

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[glacial gyro]

i think d because the x axis on graph shows radians

[velvet light]

yeah i got it

[velvet light]

also how would u know when the ambigious case applies for qs like these

[glacial gyro]

im not quite sure what you mena

if it doesnt give you the triangle i would say draw it out most of the time

helps a lot

[severe portal]

.reopen

[lone heart]

✅

[glacial gyro]

also im new to the server could someone explain the function of the .close and .open commands?

[velvet light]

like u use sine rule to solve for thetaz

but then for some reason using ambigious case u have to do 180 - theta

[glacial gyro]

im not sure if there is a way to easily know every time

but drawing it out helps for me because you can see how even if the length of the dotted lines are the same it could form a triangle with each way

[velvet light]

so this would be two different lengths

what does that mean

because you can do that for any triangle can't u regardless of ambioguity

.close

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[ionic maple]

prove that if $5x^3 + 2x^2 y = 2y^3 + 5xy^2$ then $\frac{3x+4y}{4x+3y} \in Z$

[ocean seal]

24tudor24

[ionic maple]

I asked this question before and @ snorlaxboi told me to factorise 5x^3+2x^2y - 2y^3+5xy = 0 and that the solution would be clear.
(I dont know how to proceed)

[tiny sky]

the first step is as u've said to factorise

ig think of x as ur 'variable' and y as ur 'constant' - how would u factorise a cubic in general?

[ionic maple]

well yeah and that gives me
2(x^2 y - y^3) = 5(xy - x^3)

wait no

Ive got absolutely no fucking ideea

[tiny sky]

tbh there'll be many ways of going about trying to factorise this

but i think the easiest would be ||just to spot that x=y is a root of our polynomial so by factor theorem we can factor out (x-y)||

[ionic maple]

erm...what?

I need to prove for any x,y in R*

[tiny sky]

yeah so as in we have 5x^3 + 2x^2y - 2y^3 - 5xy^2 = 0

spot that x=y is a root of this equation so we can factor it out as

(x-y)(5x^2+7xy+2y^2) = 0

then you have a quadratic in x that you can solve

[tiny sky]

this works too but is essentially the same thing

[ionic maple]

no?
can you factor 0????

[tiny sky]

?

i'm confused?

like if i gave you a cubic in general

how would you factorise it?

[ionic maple]

this doesnt make sense.
if x=y then how can you do (x-y)(something)

[tiny sky]

if ur not happy factorising with x and y in the expression at the same time

[tiny sky]

you can do what they've done here and divide by x first to get a cubic in y/x

[ionic maple]

ohhh this makes sense

[left bridge]

I hope so

[raven onyx]

Yes, I think it's more intuitive

[ionic maple]

only thing is I dont get the last part

for 5x = -2y

like you replaced those in the ecuation and..?

[left bridge]

In the third case I use x=-2/5y and use it in the "then" fraction

[ionic maple]

yeah no I got that

but the fraction doesnt

[left bridge]

So I get a fraction with "y" only and simplify

[ionic maple]

OHHHHHHHHHHH

thanks

[left bridge]

Got it?

[ionic maple]

yep

thanks a lot

[left bridge]

Also in the second case I lost a minus in the answer

Answer should be -1, I just noticed that

But it doesnt matter that much overall, just be careful when writing

[raven onyx]

[ionic maple]

👍

[raven onyx]

[left bridge]

I guess u can close the question now if u don't need any more help with this one

Write .close

[ionic maple]

.close

[lone heart]

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[lone heart]

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[ornate moon]

can anyone help i have no idea how the denominator just dissapeared in the answers

[lavish cave]

they used the double angle identity for tan

[void nymph]

im not sure what you mean

oh like that

yes

[ornate moon]

but like 1-tan^215 doesnt equal 1

so where did it go lol

[lavish cave]

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

you can derive it from $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

[ocean seal]

south

[ornate moon]

oh so did they like not use the t formula or something

[lavish cave]

what do you mean by the t formula

[ornate moon]

im so done i have an exam in 9 hours

[lavish cave]

they used a formula correct

[ornate moon]

this thing i was meant to use apparently

[lavish cave]

yes that's correct

[void nymph]

yep

[wide cairn]

[lavish cave]

Weierstrass / half-angle sub

[analog pagoda]

the whole thing comes under multiple and sub multiple angles

[wide cairn]

when u get to calculus, they call it weierstrass but when you're in trig they call it half-angle

elitism

[ornate moon]

i hope i never get to calculus

[analog pagoda]

makes sense

[ornate moon]

wait but can anyone explain how they got from step 1 to 2 in the answers

[analog pagoda]

this one?

[ornate moon]

like i get that tan(2x15) = tan 30 but why did the rest just disapper

yep

[analog pagoda]

Its the formula for tan2x

x is 15

they just substituted it

[ornate moon]

which formula 😭😭

[lone heart]

@ornate moon Has your question been resolved?

[lone heart]

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[alpine sable]

if anyone can help me with these ixls that would be great plz

[small lance]

Post the questions

[alpine sable]

okay sec

heres the first

i need to get 30% on this ixl and thats likeee 5 questions right i think

[small lance]

yeah so, do you have any idea?

[alpine sable]

not really

i hate this part of math its what im not good at

[small lance]

Do you know that angle C is equivalent to angle D

[alpine sable]

nope

how would i find the angle

[small lance]

It’s shown on the graph

[alpine sable]

oh i see

the red line is the angle or width or something

[small lance]

It marks that the two angles are equal to each other

[alpine sable]

okay

what do i do now that i know that

[small lance]

Draw an isosceles triangle

[alpine sable]

with wat information tho

[small lance]

No, just draw one

Put the question aside

[alpine sable]

ok

ok now wat

@small lance

[small lance]

What are the status of the marked angle?

[alpine sable]

what does that mean

[small lance]

Are they equal or not?

[alpine sable]

yeah

i think so

[small lance]

How about the marked sides?

[alpine sable]

yes

[small lance]

e or n e

That’s it, move on to your question

[alpine sable]

ok im on it

so z is 11

?

[small lance]

Yes

[alpine sable]

oh nic

nice

here is the next queston

[small lance]

That’s the fundamental identity of isosceles triangle

[alpine sable]

is it just the same thing as long as its equal on both sides

[small lance]

Yes

[alpine sable]

what about this one cause this is def not equal

[small lance]

ig you can try this out on your own

[small lance]

How come?

[alpine sable]

one line is longer than the other

[small lance]

It has already told you the sides are qual

Equal

[small lance]

No, you should recognize the marks instead of looking with your bare eyes

[alpine sable]

what are the marks

sorry

what are they pointing out

[alpine sable]

i see

what are those pointing out

[alpine sable]

oh

so y = 46?

it was not 46

it was 67

sorry

last question

[small lance]

Why do you think it’s 46?

[alpine sable]

cause

and i thought it was the same on the other side

[small lance]

But the two sides are GH and HI

[alpine sable]

o

[small lance]

Look at this again

[alpine sable]

yes

[small lance]

So G is supposed to be equal to I

Not H

[alpine sable]

okay

[small lance]

What’s the sum of angles of a triangle?

[alpine sable]

uh

i dont know

[small lance]

…..

Alr, it’s 180°

[alpine sable]

the answer?

wait

im so confused

[small lance]

No

The sum of angles of a triangle is 180°

It’s one of the prior knowledge you have to pick up before these questions

[alpine sable]

i did not pay attention

in my classes

[small lance]

Damn

Do you mind watching some guidance on YouTube?

[alpine sable]

i could but i really wanna get all of this done

[small lance]

Alr

[small lance]

Just remember this

[alpine sable]

alright

[small lance]

And we can move on

[alpine sable]

im on this question

[small lance]

If the sum of angles is 180, what’s the magnitude of B+D

[alpine sable]

uh

what does magnitude mean

[small lance]

How big they are

or "what’s the value"

[alpine sable]

is it 180 - 88

cuz c = 88

[small lance]

yes it is

[alpine sable]

ok

92

[small lance]

We know that B=D, because BC=CD

So what’s the magnitude, which is known as the value, of B?

[alpine sable]

is it also 88

[small lance]

Why?

[alpine sable]

wait no

im confused again

[small lance]

B+D=92

B=D

[alpine sable]

um

is it gonna be like half of 92 or soemthing

[small lance]

Yes it is

[alpine sable]

46

so d is 46

[small lance]

Correct

That’s all, congrats

[alpine sable]

so the answer is a = 46

wait no thats d

[alpine sable]

oh

so it is 46

as the answer

yay thank you

.close

[lone heart]

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[supple crater]

heyyyy not really a math problem more like a sci cal problem

[supple crater]

idk how to type in X in my sci cal coz it doesnt show a button for it

[alpine sable]

So the square root of x^2 then, ig.

[jagged cobalt]

your calc doesnt seem like the type that can

but ill have a search

[supple crater]

tried it but doesnt work

[supple crater]

oh damn fr

[jagged cobalt]

yeah seems you cant really do that outside the statistics function

https://education.ti.com/en/customer-support/knowledge-base/scientific-elem-calculators/product-usage/12324

though you can have a look at this

does seem like more trouble than its worth though

[supple crater]

ight imma check it out

you can close the ticket now btw ion need help anymoe

[jagged cobalt]

.close

.close

[lone heart]

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[jagged cobalt]

oops

[lone heart]

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[mystic swallow]

okay first up

what's on the test

second

what exactly do you need help with

[low crescent]

everything

[jagged cobalt]

thats not very helpful

[small lance]

Direct and straightforward

[alpine sable]

I think you need to fresh up your mind and quit discord and social media.

I am also in 9th and I was earlier above average; now I am quite OP.

Concentration levels have been deteriorated by the excessive use of Chatgpts.

Oh you're in 10th

Be confident bro

[fading canopy]

can sm1 explain me how to do irrational and rational numbers ?

[alpine sable]

What do you mean 'do' irrational or rational numbers?

[fading canopy]

like solve them

[alpine sable]

Firstly, open yourself a new channel.
This is occupied.
Secondly give more context in the new channel you open.
#❓how-to-get-help

[fading canopy]

tq

[low crescent]

man i hate this life everything is complicated

[lone heart]

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[low crescent]

i thought i could get some help but it seem it's useless

[alpine sable]

hi

ask your question right away

[mystic swallow]

ok dude first of all

ur not being very helpful tbf

we can't exactly help if we don't know what you need help with

what topic(s) are you struggling with

[lone heart]

@low crescent Has your question been resolved?

[lone heart]

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[alpine sable]

simplify the expression

[winter lichen]

So i keep getting this wrong

-a^8+2.-a^8

this is for the top part

Wich i get -a^24

The bottom i find is -a^6

Wich leads me to an answer -a^18

So im wrong

İts not in the answers

[alpine sable]

sorry to brake your dream

but

its a^6

[winter lichen]

Aaa

Aight

So i take the bottom as a^6

How do i continiue

İs the top part right?

[alpine sable]

(-) * (-) = (+)

so based on this

what is

[winter lichen]

Ahh ok

My bad

[alpine sable]

$(-a^{5}) * (-a)^{3}$

[ocean seal]

daniel

[winter lichen]

a^8

[alpine sable]

correct

what is -a^0

[winter lichen]

So top part is a^24?

1

[alpine sable]

no top part is a^8 + 2*a^8

[winter lichen]

Okey

And the bottom is a^6

So top is a^8(1+2)

Or am i wrong?

[winter lichen]

So top part makes 3xa^8

While bottom is a^6

What now?

<@&286206848099549185>

[sterile trench]

its -a^0 which is -1

now do it

[winter lichen]

Ahh

Okey

Never thought about that

Ahhh

İs it -a^2?

-a^8/a^6

[winter lichen]

Okeyy

İ got it

Thanks y'all for helping

.close

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[brittle lake]

For 51, is it asking me to find f(1)*g(1)? Then find the derivative of that together?

Do I just find

Actually

Let me write it out

[gritty bramble]

first differentiate

then put x=1

[brittle lake]

Yeah

Let me see if I can write it out then you check it 🙏

[brittle lake]

Is this how they want me to do it?

Oh

I mixed them up

Bruh

Hold on

.close

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[brittle lake]

.reopen

[lone heart]

✅

[brittle lake]

51.a is this how I’m suppose to do this

[gray isle]

no, you seem to be still mixing up your values

[brittle lake]

I see

[gray isle]

separately, clearly state the values of
f(1)
g(1)
f'(1)
g'(1)

[brittle lake]

1
2
1
1/3

[gray isle]

they're all wrong

[brittle lake]

How g is the red graph

And f is the purple

Oh

Bruh

[gray isle]

yes,
can you tell me what exactly you're looking at / what you're doing to get those values

[brittle lake]

ok let me try again

2
3
1
1/3

[gray isle]

last two are wrong

you've mixed them up

[brittle lake]

1/3 then 1

Yea idk why I’m stupid I’m sorry

Ok thank you

I’ll pay you back one day

Probably not

.close

[lone heart]

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[gray isle]

don't self deprecate, just take a little more time / do things carefully

[lone heart]

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[narrow pier]

Is there a pdf , website , paper copy’s I can print that has like formulas , conversions, weight geometry that I can look at for reference

[narrow pier]

Something like this ?

[lone heart]

@narrow pier Has your question been resolved?

[lone heart]

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[muted pewter]

I need help solving this improper integral

[muted pewter]

I tried substituting x = sec u but ended up with integral sec u/ (1 + sec u) du

[tacit arch]

try subbing again with v = tan(u/2)

see some examples here: https://math24.net/weierstrass-substitution.html

[lone heart]

@muted pewter Has your question been resolved?

[muted pewter]

I tried using but the steps get too long

still do not come at an answer where I can subtitute limits

these are the other two questions I need help with:

[alpine sable]

test message

OHH COOL I CAN MESSAGE OVER HERE?

[muted pewter]

umm, you can make your own help room from here: https://discord.com/channels/268882317391429632/488120190538743810

[muted pewter]

Chatgpt showed me how can I get further ahead but I am not sure how tan became tanh at the end

[tacit arch]

don't use chatgpt for math

[muted pewter]

but are the steps except for the tanh part correct?

[lone heart]

@muted pewter Has your question been resolved?

[stark cosmos]

maybe for the first problem you can use that sin(x)cos(y) = 1/2(sin(x+y) + sin(x-y))

so you factor out cos^n(x)sin(nx) = cos^(n-1)(x) * cos(x)*sin(nx)

and then you exhaust all the n:s and hope you find some pattern

[lone heart]

@muted pewter Has your question been resolved?

[lone heart]

@muted pewter Has your question been resolved?

[lone heart]

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[stoic turtle]

so im doing end behaviour

[stoic turtle]

like as x approaches infinity, y approcahes infinity
as x approaches -infinity, y approaches -infinity

linear function when m>0

What would the end behaviour be for a linear function when m=0?

[wary root]

what does a liner function look like when m = 0

[stoic turtle]

parallel to x axis

[wary root]

yes

[stoic turtle]

how would you word that in end behaviour?

[wary root]

well which way does the line go

[stoic turtle]

both

[fallen verge]

the end behavior is the value at infinity

[wary root]

so what is happening when it reaches positive infinity on the right side

[fallen verge]

so if y=1, the end behavior is y=1

[stoic turtle]

i need like this format

[stoic turtle]

y value stays the same

[wary root]

oh i was thinking about this incorrectly

[stoic turtle]

do yk?

[wary root]

i think, give me a second

wait

so if say y = 1

[stoic turtle]

kk

[wary root]

then as x reaches positive infinity

then y reaches negative infinity?

since a line goes on forever?

wait no

[stoic turtle]

nah cuz then the m value would ahve to be negative

[wary root]

y never moves

[stoic turtle]

ye

[wary root]

only x

so as x reaches positive and negative infinty, y = 1?

[stoic turtle]

idk

i was just looking for the notation

cuz i dont know

and cant find it online

[wary root]

i guess
as x -> -inf, x -> +inf, y = 1

[stoic turtle]

yea ig

thx for the help

ill just write

y = wtv

.close

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[uneven isle]

it is a pretty long question but i will just cut to the chase

[uneven isle]

second half to the original question

an extra question was attached

asking if any geometric relationship can be drawn out from this

polynomial

scope is highschool math if that matters

i fail to realise what the geometric relationship is meant to be even after getting down to ix.

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[vital prawn]

Is there anyway to make this statement true

[lone heart]

Please don't occupy multiple help channels.

[acoustic sundial]

Don’t think so

Because (2k+1)!(2k+3) will always be bigger than (2k+2)!

[vital prawn]

ok thanks

.close

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[pale silo]

I don't know if it is possible to solve this or not

[tacit arch]

Did you try law of sines or cosines

[pale silo]

I tried sines but I would get an error when I did sin^-1

But I have not tried cosines

I don't think there was any errors in the work

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@pale silo Has your question been resolved?

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@pale silo Has your question been resolved?

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[upper rock]

1e please

I got help and I know that the power function is e^x

But idk the values and stuff

MOA

[cursive badger]

Do you know how to solve part d?

[upper rock]

ur here.

[cursive badger]

yeah 🦾 I made it

[upper rock]

oh shoot

i forgot about that

no i dont

[cursive badger]

Ok... So, we need to find f-1(1) right?

[upper rock]

yeah

[cursive badger]

let's say it's x.

i.e

f-1(1)=x

[upper rock]

yeah

[cursive badger]

Now what can we do?

let's take the function again on both sides...

i.e

f(f-1(1))=f(x)

[upper rock]

uh

oh i see

okay

so what does that do

[cursive badger]

So, what is f(f-1(1))?

[upper rock]

fx

[cursive badger]

Isn't it just 1?

f( f-1(1) )

[upper rock]

wait so f-1 and f cancel out?

so its just 1

[cursive badger]

Yeah...

[upper rock]

ohh ok

wait so

fx = 1

[cursive badger]

So, we have that:
1=f(x)

Can you now find x?

[upper rock]

you inverse f on both sides?

[cursive badger]

Oh no lol

Just find it throught he graph...

The value of x.

[upper rock]

oh

f1 is 2.8

or 2.7

[cursive badger]

Wait

are you telling the answer to f(x)=1?

[upper rock]

yeah

well

f(1)

[cursive badger]

Yeah...

[upper rock]

is that the same?

[cursive badger]

But we don't want f(1)

We want some x such that f(x)=1

[upper rock]

ohh what fx value makes 1

[cursive badger]

Yep

[upper rock]

id say

-2.2

[cursive badger]

yeah

around -2

[upper rock]

oh so the answer is 2

-2*

[cursive badger]

yep

[upper rock]

ah i see

so you fx both sides

and it gets rid of the inverse

so it becomes fx = 1

and then u just find x on the graph

[cursive badger]

Yeahh

remember this:

f( f-1 (x)) = x

for every function

Always use this to cancel f-1... It makes it easier...

[upper rock]

i see

got it

okay can we do e real quick

i have like 2 mins

[cursive badger]

Yeah sure

So, we know that the function is exponential...

right?

[upper rock]

yeah

[cursive badger]

so, let's suppose that the function is of this form:
$y = a^x$

[ocean seal]

moaforlife

[cursive badger]

for some a

[upper rock]

yeah

[cursive badger]

We need to find a...

[upper rock]

so log?

[cursive badger]

?

[upper rock]

log both sides

[cursive badger]

Oh no..

[upper rock]

so it becomes log y = x log a

oh

[cursive badger]

It will make things complicated...

[upper rock]

oh alright

[cursive badger]

You are given the graph right?

[upper rock]

yeah

its the same one

[cursive badger]

How about pluginging in the value of y and x, when x=1

what's the value of y when x=1?

[upper rock]

when x is 1

y is 2.8

[cursive badger]

yeah...

So, we have this:
$a^1=2.8$

[ocean seal]

moaforlife

[cursive badger]

Right?

[upper rock]

yeah

[cursive badger]

So, what's the value of a?

[upper rock]

just 2.8

right

[cursive badger]

Yep...

[upper rock]

wait so thats it?

a = 2.8

[cursive badger]

Yeah I guess...

Or maybe not...

Let's check for some values

[upper rock]

ok

[cursive badger]

check it for x=0

[upper rock]

when x is 0 y is 2

[cursive badger]

Our function is : y = 2.8^x

[cursive badger]

And what about our function?

[upper rock]

2 = 2.8^0

which is 2 = 1

so its not equal

[cursive badger]

Oh

we are doomed

So...

[upper rock]

😭

[cursive badger]

Our function is not of this form:
y = a^x

[hollow drum]

you are intercepting at y = 2

[cursive badger]

Maybe its of this form:
$y = a^x+k$

[ocean seal]

moaforlife

[cursive badger]

For some k

right?

[upper rock]

yeah

[cursive badger]

lets again plug in the value x=0

[upper rock]

cause its growing not linearly

[cursive badger]

yep

If we plugin x=0

[upper rock]

y = 2.8 ^ 0 + k

[cursive badger]

WHat does it tell us about k?

[upper rock]

2 = 1 + k

[cursive badger]

OH noo...

Forget about a=2.8

[upper rock]

oh

[cursive badger]

It was wrong...

Now a is something else...

What is the value of k?

plugin x=0

[upper rock]

so 2 = a^0 + k

[cursive badger]

Yeah..

So?

k=?

[upper rock]

uhh

k is 1?

cause a^0 is 1

[cursive badger]

Yep

Yep

So, out function became this:
$y = a^x + 1$

[ocean seal]

moaforlife

[cursive badger]

right?

[upper rock]

yeah\

[cursive badger]

Now plugin x=1

to find a

[upper rock]

2

ok so

[cursive badger]

Wait how 2?

[upper rock]

2 .8= a^1

nvm

i got it wrong

[cursive badger]

no no

2.8 = a^1 + 1

[upper rock]

oh

yeah

i forgot the 1

[cursive badger]

great

what is a?

[upper rock]

1.8

[cursive badger]

Yep

What about out final function?

[upper rock]

so itll be

y = 1.8^x + 1

[cursive badger]

Yep

You can check that this works...

Lets check for x=2

[hollow drum]

ye that wouldnt work

[upper rock]

x = 2 y = 3.8

3.8 = 3.24 + 1

oh.

[cursive badger]

Wait what?

NO

[hollow drum]

y = 1.8^x + 1 is asymptote to y=1

in the given graph the curve is asymptote to y=0

you are looking for a more general form that is y=ab^x

[cursive badger]

But how do we do it then?

[cursive badger]

Oh right!

[upper rock]

so we have a problem

[cursive badger]

Hey @upper rock

[upper rock]

i gotta go.

[cursive badger]

Oh

Ok wait

let me tell you...

assume the function of this form :

[hollow drum]

aight, if anything the equation is y=2(1.4)^x

[cursive badger]

$y = a b^x$

[ocean seal]

moaforlife

[cursive badger]

Where a and b are variables

And solve for a and b in the similar way...

This way you can get the right answer...

[upper rock]

so plug in

[cursive badger]

Yeah probably... I forgot about the general form 🤦

[upper rock]

0 for x

like before

[hollow drum]

for your own convenience @upper rock

[cursive badger]

Yeahhh

[upper rock]

ah ok

[cursive badger]

What does it tell??

b^0=1

So, a = 2

right?

[upper rock]

yeah

[cursive badger]

Our function becomes:
$y = 2b^x$

[ocean seal]

moaforlife

[cursive badger]

Now plugin x=1

[upper rock]

y is 2.8

2.8 = 2b

2.8/2 is

uhh

idk

[cursive badger]

1.4

[upper rock]

ok so

y = 2(1.4)^x

[cursive badger]

Yep

This works

[upper rock]

okay thank you guys

[cursive badger]

*FINALLY

[upper rock]

@hollow drum @cursive badger

appreciatei t

[cursive badger]

pleasure

cya!

[upper rock]

talk to you soon

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[stark reef]

Question about a convolution of piecewise functions.
Can you swipe from right to left , or can you only swipe left to right?

[stark reef]

if so does this yield different answers?

Nevermind, I didn't realise this switches the role of x(t) and h(t) due to the commutative attribute. woops

.close

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[snow lodge]

20log(x) = 20 how do u solve this

[snow lodge]

I just keep ending up with answers either x = 1 or x ~= 4.253

[tardy stag]

dividing both sides by 20 would be a good start

[snow lodge]

With that I end up with x = 1

Oh wait

[tardy stag]

sounds pretty good to me

[snow lodge]

OK I get it now

I got x = 1 cuz I raised 1 to 10

Instead of raising it to 1

Thanks lol

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[coarse ermine]

could i get some help picking which is u and which is dv

[coarse ermine]

(assuming u have to do integration by parts for this one)

wait

i think you have to do u sub first(?)

[fresh field]

You sure about starting with parts?

[coarse ermine]

no

[fresh field]

Isn't there something better to do ?

[coarse ermine]

I think you have to start with u sub

[fresh field]

I'd do so

[coarse ermine]

Alr alr

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[muted frost]

lock in UdayS this is NOT helping

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[rose horizon]

hello can anyone tell me why this domain is wrong?

[raw jetty]

how did tou get that answer?

[rose horizon]

haha actually i solved it while i was waiting for a solution. its actually -infinity,-3 U 3,4

thanks tho

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[raw jetty]

yda

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[old widget]

What is area of line AD when there's 2 triangles ABC and DBC

[cursive badger]

Mind posting the question?

[cursive badger]

Oh thanks

[cursive badger]

what do you mean by area of line? Length of line?

[old widget]

It means how big the line AD is

[cursive badger]

oh right!

[high marsh]

you can find the length of the green part first, because it is a right angled triangle

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[cerulean grove]

Let the function $f(x)$ be differentiable, and $f^{\prime }( x) > 0$. Define $g( x) = \int _0^xf( t)$d$t$. If $g(1)=1$ and $g(3)=7$, then the possible values of $g(2)$ are 2 or 3 or 4 or 5? I'm guessing 2 or 3 but I dont know exactly what thm it used

[ocean seal]

Laya

[pallid scarab]

well g is strictly convex

(FTC)

[lone heart]

@cerulean grove Has your question been resolved?

[cerulean grove]

could you elaborate a bit

[ocean seal]

Laya

[mortal trellis]

f' is the second derivative of g. so it tells you whether your function is convex or concave

[balmy anchor]

Hello guys I am 1st year engineering student! Can you guys explain this please

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[alpine sable]

g' = f

and g'' is always positive

so its concave

[halcyon mason]

$\int_0^3f(t)$d$t$ can also be written as $\int_0^1f(t)$d$t$ + $\int_1^3f(t)$d$t$ We know the first one is equal to 1 aka g(1) and we can transform the second one as u = t-1 and get $\int_0^2f(u+1)$d$u$
I don't know if this could be of any help but there's an idea

Oh yikes wajg

[mortal trellis]

if you draw it then its obvious that some of those points wont give a convex function. tho I am not quite sure how to prove it right now

uhh wait yeah thats easy. just def of convex

[ocean seal]

Jana

[halcyon mason]

Yeah that was my idea, but I am unsure if it's of any use

[cerulean grove]

Is this question too stupid to help

[mortal trellis]

you know the function is convex

convex means that the line segment connecting two points on the graph is always "above" the graph

in particular it needs to be "above" g(2)

and thats it

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[stuck cradle]

woops

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[stuck cradle]

can anyone evaluate this?

i just wanna make sure if gpt's solutions are correct according to these formulas given :

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[final shell]

can someone help me with this gravity dam problem. I know the general overview of Resultant force and Ry and Rx, I'm still confused how exactly do you get the weights and hydrostatic uplift in Ry

[sick slate]

I'm not a helper here and I have no idea how to do this type of thing, but I just wanna say that you should feel cool for learning this and to keep at it! You are cool!

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@final shell Has your question been resolved?

[final shell]

Thank youu!! I really needed that today ngl😆

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[heavy delta]

question about parametric equation of conic section
anyone know how to do it?

[heavy delta]

<@&286206848099549185>

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@heavy delta Has your question been resolved?

[heavy delta]

.close

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[opaque swift]

anybody know how to do this?

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Please don't occupy multiple help channels.

[solemn juniper]

!1c

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Please stick to your channel.

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@opaque swift Has your question been resolved?

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@opaque swift Has your question been resolved?

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Please stick to your channel.

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[tired raft]

What is the maximum number of rooks we can place on a 8x8 board such that every rook is attacked by exactly 2 other rooks

[tired raft]

I know the answer is 16 but how to prove 16 is the max

[lone heart]

@tired raft Has your question been resolved?

[tired raft]

<@&286206848099549185>

[lone heart]

@tired raft Has your question been resolved?

[tired raft]

<@&286206848099549185>

[alpine sable]

Hmm... Can we put the rooks anywhere on the board?

[lone heart]

@tired raft Has your question been resolved?

[tired raft]

Yes

[lone heart]

@tired raft Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

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[lone heart]

Available help channel!

Send your question here to claim the channel.

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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[tacit jewel]

is $α ∈ \mathbb R^+$ inclusive for 0

[ocean seal]

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

[livid sage]

that notation is ambiguous

[tacit jewel]

i c

how can someone make it less ambiguous

$\mathbb R^{x}_{0}$?

[ocean seal]

pyrate

[tacit jewel]

whats typical

[livid sage]

could use this

[tacit jewel]

i like that better

thanks

.close

[lone heart]

Channel closed

Closed by @tacit jewel

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[lone heart]

Available help channel!

Send your question here to claim the channel.

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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ripe berry]

i need help with a few questiond

[ripe berry]

is that correct

[rose sigil]

i think so

[ripe berry]

is that correct

[lone heart]

@ripe berry Has your question been resolved?

[hollow bear]

Seems right

[ripe berry]

hi

can u help i am taking a test and need my answers checked

[hollow bear]

Hmm sure

Only few problems I can do

[ripe berry]

okay

[hollow bear]

Seems right

I remember doing this

[ripe berry]

yeah its review

[hollow bear]

👍

[ripe berry]

would they both be negitive infini positive infini

[hollow bear]

Yes

[ripe berry]

okay thank u

[hollow bear]

Seems fine to me

[ripe berry]

would this have a vertical stretcxg

[hollow bear]

Nope

[ripe berry]

would this one be 6 or 10