help-39

[golden bramble]

yes

[buoyant garnet]

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[golden bramble]

du = dy in this case

Your variable is y

[buoyant garnet]

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[golden bramble]

yes

and then your bounds become…

[buoyant garnet]

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[golden bramble]

Plug in

y=0, u is…

[buoyant garnet]

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[golden bramble]

No, lower is 1

[buoyant garnet]

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[golden bramble]

upper, if y=3, u=?

[buoyant garnet]

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[golden bramble]

Yes

[buoyant garnet]

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[golden bramble]

why u^2/3?

[buoyant garnet]

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[golden bramble]

ok

But it’s a definite integral

[buoyant garnet]

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[golden bramble]

So how do you evaluate the definite integral?

[buoyant garnet]

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[golden bramble]

Yes

[buoyant garnet]

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[buoyant garnet]

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[golden bramble]

where the du when?

[buoyant garnet]

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[golden bramble]

So du is some expression in terms of x, yeah?

[buoyant garnet]

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[golden bramble]

No, your bounds are already in terms of u

So just plug it in directly

[buoyant garnet]

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[buoyant garnet]

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[golden bramble]

the idea is to make that expression appear in the original integral

[buoyant garnet]

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[golden bramble]

so that you can get rid of everything not equal to u

[buoyant garnet]

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[golden bramble]

$du=6x^2dx\Rightarrow\frac{1}{6}du=x^2dx$

[jolly parrot]

CST (please ping when replying)

[buoyant garnet]

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[golden bramble]

Hence why a 1/6 suddenly appears

[buoyant garnet]

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[golden bramble]

And the x^2 dx is replaced by 1/6 du

[buoyant garnet]

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[golden bramble]

it’s substituted in

[buoyant garnet]

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[pearl pond]

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[pearl pond]

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[reef steppe]

I understand this concept

[reef steppe]

Wesaythattwonon-zero vectorsv and w are parallel if they have same or opposite directions.
That is, v= 0 and w= 0 are parallel if either v = w or v = w. Show that this means
v =kw for some non-zero scalar k and that k > 0 if the vectors have the same direction and
k <0 if they point in opposite directions

oops

but I don't understand how I would write the proof

I get that if the unit vectors of the two vectors are equivalent, the vectors are parallel

If anyone could tell me how to prove it that would be great

nvm

.close

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[radiant epoch]

i need help

[radiant epoch]

i got part 1 by just subbing 1

i think theyre tryna show that everything else cancels

idk how to apply this for part b

[cobalt hinge]

You have several methods here. One of them is Vieta’s formulae, though it might not be the cleanest.

[radiant epoch]

i think im suppoed to use vietes cause its what we learned in class

tdy

[cobalt hinge]

Okay

[radiant epoch]

what are some of the others?

[cobalt hinge]

Quadratic formula, completing the square. etc. But they’re more tedious.

[cobalt hinge]

Yeah, those are Vieta’s formulae.

[radiant epoch]

how do i use viete here

im getting the second root

as (a-c)/(b-c) - 1

broooo 😭

.close

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[radiant epoch]

.close

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[mental surge]

Hello, is it possible to use squeeze theorem on this?

[verbal whale]

Nope, because you're around 0 as limit point

[pearl pond]

@mental surge Has your question been resolved?

[compact ridge]

https://www.desmos.com/calculator/69t73uiwwq nope

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[still dirge]

Can someone explain what happened here in the rounded area?

[still dirge]

Just the denominator

[compact ridge]

ah, so you have the constant * 1/0 form, which is either infinity or -infinity

but look at the sign of what you have

you have a negative number divided by something positive

so this will be negative

and hence you must get -infinity

[still dirge]

Can you elaborate how to multiply (x-1)(x+1) if for example we have x->1^-

[compact ridge]

that would be $0^- \cdot 2$

[jolly parrot]

south

[compact ridge]

so now the denominator would be negative

so $\frac{-3}{0^- \cdot 2} = +\infty$, negative/negative = positive

[jolly parrot]

south

[still dirge]

But how to put 1^- instead of X and multiply everything?

Can you start from there?

[compact ridge]

$1^- - 1 = 0^-$ and $1^- + 1 = 2^-$

but $2^- = 2$

[jolly parrot]

south

[still dirge]

is -1^+ something like 0.9999999999...

[compact ridge]

it's something like -0.9999999.....

[still dirge]

ah, that's what I meant, just forgot the minus

[compact ridge]

-1 from a bit to the right, yes

[still dirge]

but 1^+ is 1.00000000....

[compact ridge]

yeah the 1 never appears, well it appears after infinitely many digits

the point is, we only care about what happens near 0

because the thing about 0 is that you can approach it from two directions, resulting in two different signs of infinity

if the denominator is not 0 then we can just divide normally and get a normal number

[still dirge]

so to solve the part in pink area, do I put in there, instead of X something like 1^+ ?

[compact ridge]

you substitute in $x = 1^+$

[jolly parrot]

south

[still dirge]

in this task we are suppose to find "asimptote"

vertical asimptote and horizontal

was it possible that we don't find them here?

[compact ridge]

you don't need to know those to find the limit

those are two different tasks

well, I think your teacher or whoever is writing this is trying to justify that x = -1, 1 are asymptotes

by showing how f(x) goes to plusminus infinity around x = 1 and x = -1

[still dirge]

yeah, but is it possible that she can't justify it?

[compact ridge]

wdym?

[still dirge]

what would the solution have to be so that there aren't vertical asymptotes?

[compact ridge]

that's impossible

[still dirge]

ah okay

[compact ridge]

the denominator would need to be never equal to 0 to have no vertical asymptotes

[still dirge]

I mean, we are proving something that's always truth, haha, just funny

thanks for help @compact ridge

.solved

[pearl pond]

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[compact ridge]

I agree that this is totally unnecessary, yeah

x^2 - 1 = 0 gives x = -1, 1

no worries!

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[cobalt hinge]

Find the area inside $r=2$ and outside $r=1+2\sin(\theta)$. Not quite sure what to do here, I graphed it and got confused.

[jolly parrot]

;(

;(

[pearl pond]

@cobalt hinge Has your question been resolved?

[meager frost]

Won't it be easier to convert it into Cartesian coordinate system first?

[meager frost]

I prefer that because it's easier to visualise.

[cobalt hinge]

It's required to do it in polar.

And besides, its an odd looking shape in Cartesian anyways.

[pearl pond]

@cobalt hinge Has your question been resolved?

[cobalt hinge]

.close

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[little comet]

heya, not too familiar with infinimums so i thought i'd just check;
i have the minimum eigenvalue of $A^TA$ as $\inf_{x \in R^m \setminus {0}} \dfrac{||Ax||^2_2}{||x||^2_2}$.
I need to take the square root of this to get an equation for singular values, can i just strip away the squares from the numerator/denominator or do i have to do something weird to the infinimum?

[jolly parrot]

ChiliLion

[coarse harbor]

Square root is continuous, so you can interchange it with inf just fine

[little comet]

i guess that makes sense

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[little comet]

.reopon

.reopen

[pearl pond]

✅

[little comet]

.close

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[errant fable]

AE=DE, m(EAB)=90°, m(ADB)=2.m(BAD)

show AF÷DF=2

[errant fable]

tried many stuff but couldnt get to anywhere

[pearl pond]

@errant fable Has your question been resolved?

[errant fable]

<@&286206848099549185>

[pearl pond]

@errant fable Has your question been resolved?

[cinder thistle]

hm

assume we start with D and A

what is the locus of B?

[errant fable]

let me learn what locus is first

1 sec

i think learning by myself will take more than 1 sec

[errant fable]

can you explain?

[cinder thistle]

i was just thinking

[errant fable]

my work so far

M is midpoint of AF

if i can show AL, DL’, and EK intersect at 1 point, its sufficent

or this is the other way around

[cinder thistle]

how do you define L'?

[errant fable]

let AM=FD and try to show FM is also equal

[errant fable]

top image: extension of EM intersects

bottom image: reflection of L with respect to EK

[cinder thistle]

i am assuming you dont want a bash

[errant fable]

no complex pls

[cinder thistle]

yes

i wont do that

let us assume we have A E and B, can we find D?

hm

A,B,F seems nice

that gives us to much info

i think we need to do ceva

[errant fable]

i tried this:

Construct a parallel to ED on point A

K becomes the center of a circle

[cinder thistle]

i think you need to do trig ceva or somtehing along those lines (consider cevians of traingle ADE)

consider EA intersect BD and ED intrsect AB

[errant fable]

im constantly looking for equal cevas

[pearl pond]

@errant fable Has your question been resolved?

[pearl pond]

@errant fable Has your question been resolved?

[errant fable]

yeag

(E,K;F,B) harmonic, G(E,K;F,B)=G(H,D;F,A)=1

so (A,F;H,D) also harmonic

3-1-2 ratio

still gonna try to solve w/o that

[glass salmon]

yeah so that just solves it

[pearl pond]

@errant fable Has your question been resolved?

[pearl pond]

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[latent elbow]

can people help with engineering mechanics here?

[latent elbow]

its a pretty simple moment question imo but im not understanding the method for solving well

[pearl pond]

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[scarlet glade]

idk what the last 2 are, i get 1+(y^2/(y^2+y^6)) for B

[scarlet glade]

is B 1???

idk

[toxic lichen]

so you got that $h(y,y) = 1 + \frac{y^2}{y^2+y^6}$?

[jolly parrot]

ann.in.a.teacup

[toxic lichen]

@scarlet glade

[scarlet glade]

uhh yeah??

i think

[toxic lichen]

ok, and if y approaches 0, what does this go to?

[scarlet glade]

1+0/0

oh

[toxic lichen]

0/0 is not an answer. it means you have to do more work.

[scarlet glade]

1

just 1

[toxic lichen]

are you sure?

so you are saying that $\lim_{y \to 0} \frac{y^2}{y^2+y^6} = 0$?

[jolly parrot]

ann.in.a.teacup

[scarlet glade]

doesn't the base get super big?

ohhhhh

[toxic lichen]

does it?

y goes to zero remember.

not to infinity.

[scarlet glade]

no its small

so 1/2

[toxic lichen]

why?

[scarlet glade]

no

uh

just 0

[toxic lichen]

it feels as if you are trying to throw guesses at a wall until one sticks. don't do that.

you can tinker with the fraction a little bit to work it out properly.

[odd surge]

try to see if you can cancel anything

[scarlet glade]

i was thinking cuz 1^6 + 1^2 = 2

1/2

but thats wrongh

[toxic lichen]

y goes to 0 not to 1.

[scarlet glade]

yeah

[odd surge]

how can you simplify $\frac{y^2}{y^2+y^6}$?

[jolly parrot]

DiamondPanda16

[toxic lichen]

would you like to try and see for yourself how you can simplify y^2/(y^2+y^6), or would you like a hint?

[scarlet glade]

so 1/(1+y^4)

[toxic lichen]

ok, so you simplified it already, good

so what's the limit of this fraction then

[scarlet glade]

so if y is zero thats 1

[toxic lichen]

yes

so then, not forgetting the extra +1 we had orirginally, what will be the answer to part b?

[scarlet glade]

and then i do the same for c i assume

so the answer is 2

[toxic lichen]

yes

and yes the process is the same for part c

[scarlet glade]

c is also 2

[toxic lichen]

is that so? why?

[scarlet glade]

1+(1/(1+(0)^4)

[toxic lichen]

hold on.

what is $h(y^2, y)$?

[jolly parrot]

ann.in.a.teacup

[scarlet glade]

(y^4+y^6)/(y^4+y^6)+y^4/(y^4+y^6)

sry, idk latex

[odd surge]

$\frac{y^4+y^6}{y^4+y^6}+\frac{y^4}{y^4+y^6}$

[jolly parrot]

DiamondPanda16

[odd surge]

well this is what you typed

is this what you meant

[scarlet glade]

yeah

[toxic lichen]

ok, but this does not simplify to 1 + 1/(1 + y^4) again

[scarlet glade]

oh, 1+1/(1+y^2)

oh ok, ty

[pearl pond]

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[supple roost]

23 sin^2x+4sinxcosx=20

[rigid mist]

what about it?

[supple roost]

General solution

What?

23-23cos^2x+4sinxcosx=20

[plush moss]

Use double angle formula for sine, 4sinxcosx=2sin(2x)

Now we have $23\sin^2(x)+2\sin(2x)=20$

[jolly parrot]

denzio321

[plush moss]

Let's move the 23 sin^2(x) to the right side

Now we have $2\sin(2x)=20-23\sin^2(x)$

[jolly parrot]

denzio321

[ornate parrot]

I have I guess something to say

I got curious about the theories of divisions of primary numbers

[plush moss]

Recall that cos(2x)=1-2sin^2 x

[toxic lichen]

!occupied

[pearl pond]

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).

[toxic lichen]

you need to take your own help channel for your question

[ornate parrot]

how I am still new and I don't know how

[toxic lichen]

#❓how-to-get-help has instructions

[plush moss]

Read #❓how-to-get-help

[ornate parrot]

thx

[plush moss]

We can then express
$$20-23\sin^2(x)$$
As
$$\frac{23}{2}(1-2\sin^2(x))+8.5$$
Now we have $2\sin(2x)=\frac{23}{2}(1-2\sin^2(x))+8.5 $

[jolly parrot]

denzio321

[plush moss]

With the double angle formula for cosine, we get this down to
$2\sin(2x)=\frac{23}{2}\cos(2x)+8.5$

[jolly parrot]

denzio321

[plush moss]

We then rearrange to get
$2\sin(2x)-\frac{23}{2}\cos(2x)=8.5$

[jolly parrot]

denzio321

[plush moss]

Now we will utilise the R formula

[toxic lichen]

btw just as a heads-up:

!nosols

[pearl pond]

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.

[plush moss]

Welp I was getting too caught up mb

I guess this is a good enough place to stop

@supple roost

Are you able to continue from here?

[supple roost]

Let me read all the discussion

[plush moss]

@supple roost any troubles?

[supple roost]

@plush moss

[supple roost]

I did not understand it

[plush moss]

That's ok too if you want

[plush moss]

Ok uh

[supple roost]

how I change sin2x into cos2x

[plush moss]

Hold on ill explain

I guess the term R formula isnt too common so

For a function
$$a\sin(\theta)\pm b\cos(\theta)$$
We can express it in the form
$$R\sin(\theta \pm \alpha)$$

[jolly parrot]

denzio321

[plush moss]

Have you seen something like this in class before @supple roost

[supple roost]

ohh i got it

Yes maximum minimum and multiply and divide

And then one formula

Wait let me finish it

But for what values we have 4 and 23

I meant sinb=4

Cosb=4

@plush moss

[supple roost]

But i don't think they gave the whole solution many steps are still left

[toxic lichen]

because i stopped him from dumping the entire solution

[plush moss]

Mb bruh

[plush moss]

This should be in your textbook I figure

Just reference that

I can't really explain this that well

[supple roost]

sin(a+b)=sinacosb+cosasinb

[plush moss]

Well it's a bit more complicated than this

[supple roost]

Ohh wait

[plush moss]

Yea that

[supple roost]

√545

Damn

√545(4/√545sin(2x)-23/√545cos2x)=17

[supple roost]

Is this correct?

[plush moss]

Yea not fully correct

A few more steps

Bear with me

[supple roost]

( \sqrt{545} \left( 4/\sqrt{545} \sin(2x) - 23/\sqrt{545} \cos(2x) \right) = 17 )

[jolly parrot]

Abdul sharma

[supple roost]

Sure

Give me directions

[plush moss]

So we know that 4/sqrt(545) can be expressed as cos(alpha)

Where alpha is some acute angle

since cos(alpha) = 4/sqrt(545)

How would you find alpha @supple roost

Use a calculator

[supple roost]

,w arccos(4/√545)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by chrome08014

[supple roost]

Lmao

The question makers should have given a good angle

Matiyabut thanks a lot because I have done some new methods

[golden bramble]

this might be the craziest solution I have ever seen lmao

[supple roost]

Yeah same thanks mate

.close

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[golden bramble]

There’s probably an easier way but whatever

[supple roost]

I guess there is mistake in the question

[pearl pond]

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[plush moss]

No it's fine give it to 5s.f

[plush moss]

Welp

@supple roost can you reclaim your channel

.close

[pearl pond]

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[supple roost]

.reopen

@plush moss

[supple roost]

5 sf?

[plush moss]

So 1.3986

Round it to 5 significant figures or whatever your school wants you to

[plush moss]

So yes now we can express this whole equation as
$$\sqrt{545}\sin(2x-1.3986)=17$$

[jolly parrot]

denzio321

[plush moss]

Now solving is trivial-ish

With a calculator

@supple roost

See that wasn't that bad

A bit long

But we didn't do anything too extreme

[plush moss]

please share

this is a bit convoluted

[golden bramble]

From the beginning, use sin^2+cos^2=1 on the 20

@supple roost

[pearl pond]

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[ripe pagoda]

The median AD of the triangle ABC is bisected at E and BE meets AC at F. Find AF:FC

I have to solve it using vectors
I'm assuming B to be the reference point here
Also assuming that AF:FC = K:1
ill use lowercase letters for vector notation since idk how to type it out
i know:
f - b = (K(c - b) + (a - b))/(k + 1)
(c - b)/2 = d - b
(e - b) = ((a - b) + (d - b))/2
i tried solving for these but im getting random results and not getting a value for K

[pearl pond]

@ripe pagoda Has your question been resolved?

[ripe pagoda]

<@&286206848099549185>

[pearl pond]

@ripe pagoda Has your question been resolved?

[ripe pagoda]

.close

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[orchid token]

can i get some hlep with this

[orchid token]

i got sin(o)/ sinx

which is 0/0 if the lim is as x approaches 0

[rough forge]

sin(x) ~ x for small x are asymptotically equivalent

so consider tan(x)/sin(x)

[orchid token]

idk what this mean

[rough forge]

it basically means that sin(x) and x behave pretty much the same as x -> 0

[open rivet]

lhopital?

[light helm]

divide numerator and denom by cos(x)
then apply limit identity

[orchid token]

oh i see

i realized i wasn't supposed to do this question actually

so thank you for the help anyway

.close

[pearl pond]

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[pearl pond]

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[toxic lichen]

<@&268886789983436800> more of these scammers

[bitter lodge]

@midnight haven (sry for ping (not rly cause its your fault))

[pearl pond]

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[bitter lodge]

<@&268886789983436800>

[open blaze]

Barely sniped

[bitter lodge]

Wdym barely sniped

[radiant terrace]

Jeez more?

[bitter lodge]

Were you about to ping before i did

[open blaze]

No someone else banned before me

[pearl pond]

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[earnest flint]

in evaluating limits and approaching from the right in constructing table values

its this order right? .1, .01, .001

Thats what i see in yt

but my teacher does it like .001, .01, .1

and idk who to follow

[open rivet]

it doesnt matter

you can also do .01, .1, .001

as long as you take the right conclusion

like, if you have f(x)=1/x

then f(0.1)=10
f(0.01)=100
f(0.001)=1000

but you can also write it as
f(0.001)=1000
f(0.01)=100
f(0.1)=10

in either case you see that as x gets smaller, f(x) gets bigger

[pearl pond]

@earnest flint Has your question been resolved?

[pearl pond]

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[vivid pilot]

why is the correct answer true?

[open rivet]

gcd(a,b)=gcd(a,b-n*a) with n in Z

[rough forge]

literally applying bezout

[cunning comet]

if a number t would divide a and t would also divide b then what would that mean for t and ax+by?

[vivid pilot]

so any of a,x,b,y can be numbers in gcd(_, _)=1?

[rough forge]

yes

well any a,b,x,y where ax+by=1 holds of course

because then you find for every pair two integers where this equation holds

[vivid pilot]

right

thank you for your help!

.close

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[wet latch]

dont know how to start

[wet latch]

i think its some sort of trig sub?

i just dont know how to deal with the 26x

[unkempt yacht]

complete the square

[agile valley]

then trig sub

[pearl pond]

@wet latch Has your question been resolved?

[wet latch]

do i have to u sub too

i got 169-(x-13)^2 under the root

i dont know how to deal with the x-13

should i replace it with u?

im so lost

[agile valley]

you could if you want

it's not necessary

[wet latch]

first half of page is me trying it

without u sub

i got lost though

did i do something wrong?

im trying it with u sub now

[plucky yoke]

Wouldn’t be sin

If it was x^2-a^2 that’s secθ

[wet latch]

oh

[plucky yoke]

sec^2x-1 = tan^2x

[wet latch]

isnt 169-(x-13)^2 a^2-x^2 though?

[agile valley]

yes

[pearl pond]

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[plucky yoke]

Ohh I see how you mean

Sorry bout that

[pearl pond]

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[noble sleet]

I've got multiple of these questions wrong. Why is B correct, not C?

[noble sleet]

Like this too?

[compact ridge]

let Priya earn x before her pay rise

do you agree that x * 1.02 = 357?

[noble sleet]

yes

[compact ridge]

yeah, then divide by 1.02 on both sides

and you get x = 357/1.02

[noble sleet]

Oh so reverse percentage it has to be division

[compact ridge]

yep!

[noble sleet]

Alright tysm (:

[compact ridge]

so like we don't know what she earned before, but after she gets a bonus (so that's multiplication), she earns 357

if you undo multiplication you get division

it's a reverse problem, ok nw

[noble sleet]

ty one more thing

[compact ridge]

ah

[noble sleet]

Why is this C, not A? Is it bcos 1.75 is 75% if im right?

[compact ridge]

this wording is very very sneaky

before you do anything you already have 100% of the original

so if you want it to increase by 175%, it's just 100% + 175%

so a 100% increase is 100% + 100% = 200% similarly

[noble sleet]

https://tenor.com/view/sponge-bob-smooth-see-ya-you-got-it-hell-yeah-gif-10292846113768095909

lol

ty i understand

[compact ridge]

if you want 1.75 times the original that's only a 75% increase also

np!

[noble sleet]

Thnks man you answered my questions hve a good one (:

.close

[compact ridge]

cheers, have a good one mate

[pearl pond]

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[supple roost]

,w π/2 to π int (floor(cos2x))

[jolly parrot]

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[supple roost]

My doubt is in this question at π we will have value =1

We will not include it in integration?

[bitter lodge]

But for how long (in terms of delta x) is that value equal to 1

[supple roost]

What??

[supple roost]

Exactly at that point

no long

[bitter lodge]

Hence, its too small to have a meaningful effect on the integral

[supple roost]

So a point has no area?

,w area of a point

[jolly parrot]

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[bitter lodge]

0

[supple roost]

Thank you sir

[compact ridge]

fundamental concept of integration, btw

[supple roost]

Thanks

,w 0 to 10π intfloor(arctanx)

[jolly parrot]

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[pearl pond]

@supple roost Has your question been resolved?

[supple roost]

.close

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[sharp smelt]

Trying to find the arc length here

[sharp smelt]

So $\int_{0}^{\pi /2} \sqrt{ ((3\cos^2(t)\sin(t))^2+ (3 \sin^2(t) \cos(t))^2}dt$

which would give me $\int_{0}^{\frac{\pi}{2}} 3(sin(t)+cos(t))dt$

[jolly parrot]

What a wonderful world it is !

[toxic lichen]

\cos and \sin

[toxic lichen]

i am not sure about this rewrite

[jolly parrot]

What a wonderful world it is !

[sharp smelt]

I then have

$

$int_{0}^{\pi /2} \sqrt{9 \cos^4(t) \sin^2(t) + 9\sin^4(t) \cos^2}(t)dt $

$\int_{0}^{\pi /2} \sqrt{9 \cos^4(t) \sin^2(t) + 9\sin^4(t) \cos^2}(t)dt$

oops

[jolly parrot]

What a wonderful world it is !

[sharp smelt]

It should be $\int_{0}^{\pi/2} 3 \sin(t) \cos(t) dt$

[jolly parrot]

What a wonderful world it is !

[sharp smelt]

should it not

[toxic lichen]

why did the last (t) end up outside the root lol

but ok that seems more like it now

[sharp smelt]

} in wrong place

now this is easy enough to integrate

3/2

[pearl pond]

@sharp smelt Has your question been resolved?

[pearl pond]

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[dapper kraken]

ABC is a triangle such that ABC=ACB=70, if P is in ABC such that PCA=40 and AP=BC then find PAB

[dapper kraken]

im not really sure how to do this, any hintd/ideas?

extending CP so AB and CP intersect at D seems useful, but seems and i couldnt really get anything

is this some sort of trig bash? (note i need to do this by hand)

[pure rapids]

Did you try moving APC such that AP overlays BC

Idk if that works

[dapper kraken]

what?

[cyan bronze]

BC/sin40 = AC/sin70

from the big triangle

then from APC, you get AP/sin40 = AC/sin(100+x)

where x is angle PAB

and AP=BC

then youre done

[dapper kraken]

how did you get APC=100+x

oh nvn

[cyan bronze]

if PAB is x, then PAC is 40-x, since BAC is 180-(70+70)

[dapper kraken]

so sin(70)=sin(100+x) so x=10?

[cyan bronze]

yes

[dapper kraken]

ok ty

.close

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[rancid steppe]

If x-y = 10 and x/y = 5/3 find x and y

[rancid steppe]

we have to solve this whout using systems of equations btw

[cyan bronze]

What does this even mean

[rancid steppe]

poportions

Plug in a K i think but idk how to do it exactly

[cyan bronze]

Ah so like x=5k and y=3k

[odd surge]

just find x in terms of y and then substitute it into the first equation

[rancid steppe]

Yea but i do not understand what does this K even mean

[cyan bronze]

K is a constant of proportionality

If x/y is 5/3

It means x is 5 times somethjng and y is 3 times of that same thing

were letting k = Something

[rancid steppe]

right

$\frac{x}{y} = \frac{5}{3} = K$

[jolly parrot]

Simon

[cyan bronze]

No

[rancid steppe]

?

[cyan bronze]

Thats not what were saying here

[cyan bronze]

.

[rancid steppe]

we did it like this in other examples

[cyan bronze]

There is a difference

In what you wrote

And this

[rancid steppe]

what difference

[cyan bronze]

Its not a/b

Its a/4

So that should be x/5

[rancid steppe]

but the pic i showed rn is from a other problem similar

[cyan bronze]

Since you're equating it with k

[cyan bronze]

Yes i know

But you arent following what you did in that problem

[cyan bronze]

Just replace a and b with x and y and 4 and 3 with 5 and 3

Youll get what i mean

You modified the left equality, which holds only independently now

You cant equate that to k

Aftrr modifying

[rancid steppe]

i am very sorry i just do not get it

[cyan bronze]

Look at the original problem here

What does it say

It should be something like a/b=4/3

[rancid steppe]

x/y = 5/3 yea

[cyan bronze]

Yes so for the a/b=4/3

We chose k as a/4=b/3=k

Now do the same thing with x/y=5/3

x/5=y/3=k

So x=5k and y=3k as we expected

[rancid steppe]

$\frac{a}{4} = \frac{b}{3} = k$

[jolly parrot]

Simon

[rancid steppe]

this is what you said?

[cyan bronze]

Yes

[rancid steppe]

but how did we get from a/b = 4/3 to this

[cyan bronze]

One way to think is multiplying both sides by b

And then dividing both sides by 4

To get that a/4=b/3

And we call this thing k

This thing is a/4 or b/3

But you should just also be able to see that to satisfy a/b=4/3. I can pick an arbitrary number k

And write a=4k and b=3k

Because 4k/3k is just 4/3

The choice of k is decided by the first equation

x-y=10 or whatever

[rancid steppe]

5K - 3k = 10
2k = 10
k = 5

is this ok?

[pearl pond]

@rancid steppe Has your question been resolved?

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[stoic imp]

The polynomial $P \in \mathbb{R}[x]$ of minimal degree that has as roots all $z \in \mathbb{C}$ such that $z^2 |z| = i z^5 \overline{z}$ and such that $P(i) = -6i$ is

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[toxic lichen]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stoic imp]

1

[toxic lichen]

start by solving the equation $z^2|z|=iz^5\overline z$

[jolly parrot]

ann.in.a.teacup

[toxic lichen]

to see what roots P has to have

[stoic imp]

that's where I got stuck

[toxic lichen]

ok right. first off z=0 clearly works, yes?

[stoic imp]

yea

[toxic lichen]

so let's look at nonzero z for convenience

[stoic imp]

them divide by z?

[toxic lichen]

divide by z^2 actually

might as well

and also write $z\overline z$ as $|z|^2$ on the right, and then divide by $|z|$

[jolly parrot]

ann.in.a.teacup

[toxic lichen]

just to remove as much crap as we can from the equation at this point

iz^2 |z| = 1 is what you should be left with, i think.

[toxic lichen]

you have |z|^2 on the right not just |z|

everything correct except for that

(but the next step is figuring out |z|=1 anyway...)

[stoic imp]

wait, yeah let me correct it

[toxic lichen]

youll get two values of z

[cursive wraith]

so now, how can we get that |z| = 1?

[toxic lichen]

they and their conjugates will be the roots of P

[stoic imp]

I don't follow

[stoic imp]

idk

demoivre?

[toxic lichen]

no need for demoivre

just take modulus on both sides

it will fall out

[stoic imp]

how?

|z| is already the modulus

how do I double modulus that

[cursive wraith]

a = b -> |a| = |b|

[toxic lichen]

$\absv{iz^2|z|} = \absv{-i}$

[jolly parrot]

ann.in.a.teacup

[toxic lichen]

this is what i meant

[cursive wraith]

|1| on the RHS

[toxic lichen]

if you are asking how to simplify | |z| |: the modulus of a positive real is just that number itself

[toxic lichen]

sure yeah i was just demonstrating the procedure

oh

yes my bad sorry |1|

wasn't TECHNICALLY wrong

[cursive wraith]

but def confusing xxddd

[stoic imp]

I don't know the modulus of |z|

[toxic lichen]

if you are asking how to simplify | |z| |: the modulus of a positive real is just that number itself

[stoic imp]

1 = |z^2 |z||

[stoic imp]

ok

so ||z|| = |z|

[toxic lichen]

yes

[stoic imp]

1 = |z^2| . |z|

1 = |z|^3

[toxic lichen]

yes

[stoic imp]

|z| = 1

[toxic lichen]

therefore |z| = ?

yes

[stoic imp]

niice

[toxic lichen]

now back to the original equation

you get iz^2=1

find z

[stoic imp]

demoivre

?

[toxic lichen]

if you want

[stoic imp]

idk what am i doing

[toxic lichen]

you went down an extremely laborious route

only to arrive at z = ±sqrt(-i) lol

[stoic imp]

wdym?

the quadratic formula was easiest and j should have tried it first?

,w solve iz^2 -1 = 0

[jolly parrot]

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[toxic lichen]

i mean your equation is just z^2=-i to begin with

you don't even need QF

just how to take roots of complex numbers

[stoic imp]

z = +- sqrt(-i)

[stoic imp]

what the hell

[toxic lichen]

give me a moment

actually im gonna draw some more stuff out

[stoic imp]

in the argand plane?

[toxic lichen]

armando?

... yeah that

[stoic imp]

im on my phone, sry

[toxic lichen]

red are the points satisfying the original equation

these have to be roots of P

blue are their conjugates which also have to be roots of P

[stoic imp]

but why 45 deg, 135deg, 225deg,315 deg

[fringe robin]

Renato are you Chinese by any chance

or speak Chinese

[stoic imp]

what? fuck no im from south America

[fringe robin]

where does your name come from?

[stoic imp]

born and raised

[fringe robin]

i see

[stoic imp]

is random chars

[fringe robin]

Renato

[stoic imp]

ohh is what my mother decided was OK ish name

[fringe robin]

ah ok

gotcha

[stoic imp]

as of now, i don't

[stoic imp]

you put it in polar form and this is from unit circle cos(x) = 0 and sin(x)=1

[toxic lichen]

z^2 = -i

-i has argument -90°

half that to get -45° a.k.a. 315° -- that's the southeastern point

opposite that is -45° + 180°, or 135°

[stoic imp]

why halfed

🤔

[toxic lichen]

because you take the square root

[stoic imp]

yea in the argand plane -i clearly has arg -90 i agree

[toxic lichen]

when you take the square root, the argument is halved

it is like with other exponents e.g. arg(z^2) = 2 arg(z) [with adjustments by ±360°]

[stoic imp]

ahh

yeah is an argument property

z = +- sqrt(-i)

arg(z) = 1/2 arg(-i) + 2kpi

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

I will just continue this later

because nobody is helping and seems like tedious but not hard

. close

I will do alcumus practice and then continue with this on Monday

@odd surge

[odd surge]

yeah?

[stoic imp]

nothing

[stoic imp]

answer for the other guy is 2^24?

[odd surge]

yeah

[stoic imp]

okay thanks

. close

.close

[pearl pond]

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[stoic imp]

A circle is circumscribed about an equilateral triangle with side lengths of $9$ units each. What is the area of the circle, in square units? Express your answer in terms of $\pi$.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

81pi?

[autumn fossil]

Why?

is the radius 9?

[stoic imp]

radius is 9/2

[odd surge]

why?

[autumn fossil]

well, this is also wrong

but even if it was, the area stil lwouldnt be 81pi

[stoic imp]

xd

let me do a drawing

[autumn fossil]

you're confusing circumscribed and inscribed

[stoic imp]

ahh

what now?

[autumn fossil]

now you should probably calculate the radius

you know pretty much all the angles

there are several isosceles triangles

try splitting one of them into 2 right triangels

and then do some trig

[stoic imp]

what trig?

[stoic imp]

why isoceles

[odd surge]

that looked so cursed

whatever it was

[stoic imp]

this is what the other guy refers to

[odd surge]

this is the general case but your triangle is equilateral so there's even more equal lengths

[stoic imp]

help me panda im stuck

[odd surge]

draw an equilateral triangle and lines from the centre to the corners

you see the triangles?

[stoic imp]

yeah

[odd surge]

and then just find the length you need using trig in one of the triangles

[stoic imp]

how

[odd surge]

see the small right angle triangle

[odd surge]

what's the angles in it

[stoic imp]

90

[odd surge]

yeah but the other angles

what are the angles in an equilateral triangle

[stoic imp]

45 ig

[odd surge]

no...

[stoic imp]

60 60 60

[odd surge]

yeah

so in the small triangle

one of the angles is

[stoic imp]

30 ig

[odd surge]

yes

and what's the side length of the base

[stoic imp]

90 + 30 = 120

[stoic imp]

4.5

[odd surge]

what

yes

so now can you find the hypotenuse

if the angle is 30 and the adjacent side is 4.5

[stoic imp]

hyp = (4.5)/cos(30)

hyp = (9/2)/cos(pi/6)

,w (9/2)/cos(pi/6)

[jolly parrot]

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[stoic imp]

,w pi * (3sqrt(3))^2

[jolly parrot]

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[stoic imp]

bro it was hard

[odd surge]

they expect you to just know AM = 2/3 AO

you clearly didn't know that which is why we were showing you how to prove it

[stoic imp]

why is that

that's crazy

[odd surge]

idk it's a useful fact

[stoic imp]

it was hard for me though

but we managed lmao

thanks

.solved

[pearl pond]

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[pearl pond]

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[worn goblet]

@toxic lichen

[toxic lichen]

half of your x's look like u's

but let me look this over

ok i see the mistake

cos(4x) integrates to sin(4x)/4

[worn goblet]

Oh they actually do

[toxic lichen]

you forgor that 1/4 factor and that is why you've got 1/8 instead of 1/32 coefficient in the first line of working

[worn goblet]

Ohh

I'm mentally disabled

Sorry

.close

[pearl pond]

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[timber tartan]

nah ur good

i make sillier mistakes

differentiates when asked to integrate

[worn goblet]

I have 100marks integration paper tomorrow

😭😭😭

[toxic lichen]

.reopen

[pearl pond]

✅

[toxic lichen]

im not sure using this as a self-putdown is a good idea

[worn goblet]

Lmao

[toxic lichen]

in fact i would even say it is a bad idea

[timber tartan]

negative self talk is harmful

[worn goblet]

But it's always better than to talk yourself up too much and have nothing to show for it

[toxic lichen]

brainfarts happen to all of us, but just beating yourself up like this -- regardless of if you are actually disabled -- does you no good at all

[toxic lichen]

there's an even better alternative: you could extend yourself some grace

EVERYBODY makes mistakes

even terry fucking tao makes mistakes

alright?

that sort of thing happens all the time and the mere presence of a mistake is NEVER a value judgment on your own character or a singlehanded indication of any kind of disability mental or otherwise

like yeah also disability should not be stigmatized but thats a separate convo

[worn goblet]

What I think is that if u talk shit about yourself and worst case scenario happens atleast you said it this is better than talking too much and having shit to show for it according to me but what you're saying makes more sense now lol

[toxic lichen]

you should just not talk shit about yourself

i would say "simple as" but i unfortunately know firsthand that it is really not simple as

and it takes practice

to unlearn negative self-talk

its a bad habit

[worn goblet]

But it's so fun to talk shit about yourself and infact sometimes do stuff which is not total shit

Thanks again

[worn goblet]

Btw who's terey tao

[fierce totem]

uh-oh

what happened here

[worn goblet]

hey i know this

your pfp

its the character from hollow knight right

hornet

[fierce totem]

its the character from a game that is almost similar to hollow knight

[worn goblet]

metroidvania too?

[fierce totem]

ye

Rain World, a pretty good game

[worn goblet]

only metroidvania i enjoyed is blasphemous

[worn goblet]

will check it out thanks

[soft leaf]

sometimes i kinda degrade myself when i make mistakes as a way to cope w it but yea, sometimes doing something too much is badd

[worn goblet]

makes sense thanks!

[toxic lichen]

Terence Tao. famous mathematician. look him up

[worn goblet]

ohh is he a proff at your uni?

[toxic lichen]

no

[worn goblet]

ohhh then he must be real famous

[toxic lichen]

that he is

[worn goblet]

what for?

[toxic lichen]

https://en.wikipedia.org/wiki/Terence_Tao?wprov=sfla1

[worn goblet]

btw is the 4/rootn officially accepted formula for vorinoi diagram perimeter of each cell thing

[worn goblet]

daaammmnn

[pearl pond]

@worn goblet Has your question been resolved?

[pearl pond]

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[silver nest]

How many 4, 5, 6 digits long numbers I can create with the digits {0, 1, 2, 3, 4} if those digits can repeat the any amount of times
for example 00012 is valid, 11111 is valid, 432100 also valid, 4444 also valid....

[silver nest]

is it correct to sum the cases of 4 digits + cases of 5 didgits, + cases of 6 digits?

something like

[vestal tapir]

5^4+5^5+5^6

[silver nest]

yeah

btw, what keys do you press to make the exponent symbol?

[vestal tapir]

shift-6

[silver nest]

thank you

.close

[pearl pond]

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[pearl pond]

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[open rivet]

<@&268886789983436800>

[pearl pond]

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[mild chasm]

hello everybody