help-43

[floral kelp]

in the mark scheme, they did 1-(13/27) for P(1<=X<=2 and X>2)

[kind viper]

P(1<=X<=2 and X>2)
and?

are you absolutely sure you mean "and" there

[floral kelp]

not really

the intersection

[kind viper]

the intersection?

[floral kelp]

😭

[kind viper]

again are you sure you mean the intersection?

cause that's empty.

[floral kelp]

∩

[kind viper]

no value of X can possibly satisfy both of these at once:

• 1 ≤ X ≤ 2
• X > 2

it is impossible to be both ≤2 and >2 at the same time, capisce?

are you really sure you mean ∩ and not ∪ ?

[floral kelp]

wait i thought if the probability question uses "given that", it should be using ∩

[floral kelp]

oh yes thats true

[kind viper]

oh, so you got confused at the wording

you thought they were setting up some kind of conditional probability rather than giving you some info for free.

they were doing the latter.

if they had named an event after the "given that", e.g. if they had written

Given that 1 ≤ X ≤ 2, ...
then you could read it as a conditional probability.

but right now, no, they just pre-calculate a probability value for you.

[floral kelp]

ohhh

is there an exact formula for that

i wanted to make sure, they did [1-(13/27)]/2 bcs 0<=1 is the same probability as 2<=3 right?

[kind viper]

for what

[floral kelp]

for questions like 1c

im almost balding

[kind viper]

ideally you should not be engaging in whats-the-formula-ism like that

but the idea behind question 1c is to exploit the fact that your distribution is symmetric and centered about 1.5

bcs 0<=1 is the same probability as 2<=3 right?
this is poorly notated, but correct.

[floral kelp]

how do we know that the distribution is symmetric

[floral kelp]

okaayyy thankuu

[kind viper]

its curve is a parabola

well ig more formally we could say the density function satisfies f(2*1.5 - x) = f(x) for all x.

[floral kelp]

how do we know

[floral kelp]

okieokie thankyouthankyou!

i think i get itt

will close the channel

thanks once again

.close

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[floral kelp]

guys, for 2b, how do we determine that p is the lower limit and not the upper limit?

[floral kelp]

i was speculating that they know it from this

"25% contain more", so if shaded in the distribution curve, right side will be shaded

like this, so p is the lower limit and the upper limit is 10 (as stated in the question)

OKAYU THANKUU

thankyou chartbit for the reaction

i get it now

[wheat pasture]

Awwww (also hiiiii )

[floral kelp]

heeyyyy !!

ok so i have to get back studying

thanku sm once again

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[regal osprey]

Hi im stuck w this qn, the graph qn, the domain that i got was y = -x and i drew it w open and close points alr its still wrong, is there something wrong w the domain that i got or?

[cursive harbor]

The domain shouldn't be an equality

[regal osprey]

the domain is x+y >0?

[sturdy star]

yes

[regal osprey]

so its not like the graph i plotted?

[sturdy star]

with the dashed line should be right

[vale star]

It looks like the line is passing through (0,0)
And summing up them doesnt give u a value greater than 0
Maybe that's the issue? I m not sure

[regal osprey]

ahh i also not v sure

[regal osprey]

u meant the blue line being a dash?

[vale star]

But ur way should be right,idk if we r missing something

[sturdy star]

ru able to put closed circles

try put a closed circle at origin

[regal osprey]

3 points? so (-11,11) (11,-11) and (0,0)? but the first 2 is it open circle? cause first tries i used close and its wrong and now i used open also wrong

like this?

[sturdy star]

like a dashed line with a closed circle at origin

line is dashed tho

[regal osprey]

i couldn't get the dash line :/

only if its a curve, i can make it dash

omg omg i solve it thks guys!!

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[lavish gull]

og question statement please.

[icy topaz]

The message deleted automatically twice.

wait

[kind viper]

were you trying to say angle F A G

[icy topaz]

yes

[kind viper]

say it as GAF so it doesn't get caught in the filter

[icy topaz]

okay

thanks

[lavish gull]

actually, you should receive an error saying exactly what is filtered

we've heard reports that some people were not seeing them

[icy topaz]

Let a circle be.
Let DE and BC be two parallel chords with the lengths 46 nanometres and 18 nanometres respectively.
If angle DAF = angle GAF = angle GAE, find the value of the radius of the circle in yottametres.

I found DF and EG to be equal and to be equal to 18, and FG to 10.

[kind viper]

so AC and AB trisect angle DAE, is that what you're saying

[icy topaz]

Yes.

[kind viper]

let angle CAB = 2t and let the radius of the circle be r, then CB = 2r sin(t) and DE = 2r sin(3t)

[icy topaz]

46 = 2r sin(3t) and 18 = 2r sin(t)
Hence sin(3t)/sin(t) = 23/9

[kind viper]

yeah keep going

[icy topaz]

sin 3t = 3sint - 4 sin^3 t

Hence,
3 - 4 sin^2 t = 23/9
So sint = 1/3

[icy topaz]

Now I shall just put this in the second equation.

Thanks.

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[chilly basalt]

,rotate

[kind viper]

!status

[compact pewter]

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

[pseudo wasp]

Smallest number I could find

Hoe will I find 5n + 2013

[kind viper]

why not move the six to the very front

that'll make your number even smaller

[pseudo wasp]

How will that make a difference

[kind viper]

reread my last msg

[pseudo wasp]

Well ig what you say is true

[kind viper]

you need N to be the smallest integer with digit sum = 2013

[pseudo wasp]

But tbh that doesn't solve the problem of how to get 5n

[kind viper]

N = 6999.....999
     ^~~~~~~~~~~
      223 nines

[pseudo wasp]

Okay....... So s(5n+2013 )how to get

[kind viper]

patience

[pseudo wasp]

Okie

mam

[kind viper]

this number has a lot of nines at the end

what do you know about numbers that end with a lot of nines

[pseudo wasp]

ya

Ehh conver to ten and subtract one☠️

[kind viper]

...

[pseudo wasp]

[kind viper]

okay that made no sense to me but maybe you have some sort of right idea

tell me: what will N+1 look like?

[pseudo wasp]

Like convert it 7 with 223 zeroes - 1

[pseudo wasp]

something like 7 with 222 zeroes and one 1

in thar order

[kind viper]

you add 1 to a number ending in 9, and the result ends in a one?

[pseudo wasp]

Well I guess add and subtract

[kind viper]

...

[pseudo wasp]

wait I am giving definite answer

[pseudo wasp]

Didn't I already answer tho

[kind viper]

you should be answering the questions exactly as i asked you instead of trying to play 5D chess with me and make wild guesses as to what i might want

[pseudo wasp]

N+1 should be you should be 7 with 223 zeroes

[kind viper]

if you want to yap that's fine but first answer the question as i ask you (exactly, no more, no less) and then make your addition

yes so N+1 = 7 × 10^223

[pseudo wasp]

Ya

[kind viper]

N = 7 × 10^223 - 1, so 5N+2013 is?

[pseudo wasp]

2013?

[pseudo wasp]

ehhhh 350000000......... 2008

[kind viper]

why does it end with 2012?

why with 2007

[pseudo wasp]

2013 added and 5 subtracted right?

[kind viper]

2013 - 5 = 2007 ?

[pseudo wasp]

oh shit

2008 I mean

[kind viper]

ok

i hope you now see how to find the digit sum of this

[pseudo wasp]

18 lmao?

.close

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[rapid pike]

Hii, I need help factoring 5 and 6

[stuck haven]

You have to factorise LHS or including rhs

?

[rapid pike]

you can do that too

but recommended to rhs

[whole flax]

doesn't look possible

,w factor y^2 -2xy - 2x -6

[boreal girder]

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[rapid pike]

ermahgerd

[kind viper]

can you post the full problem statement

looks very sus rn

[rapid pike]

its a diophantine equations

[twin idol]

y^2 - 2x(y-1) = 6

[whole flax]

oh okay

then one side works

[rapid pike]

should i upload an example

[kind viper]

that sounds like important info, why didnt you say that first

[rapid pike]

i thought no one would know

[whole flax]

Hint: ||add -1 to both sides and see if you can figure out what to do||

[kind viper]

bruh what.

you should say upfront it's a diophantine eq.

[rapid pike]

[whole flax]

continue from this

[kind viper]

that makes the approach like way different.

don't sotrue me.

[rapid pike]

alright boss

[kind viper]

that was just obstructive behavior from you, hiding the diophantine part of the thing from us

[rapid pike]

hmm ok ok

[kind viper]

anyway add & subtract x^2 to get

y^2 - 2xy + x^2 - x^2 - 2x = 6

[stuck haven]

Good catch

[rapid pike]

i wasnt so sure if it was called that

[kind viper]

what is it called in your language?

[stuck haven]

She's gatekeeping 😭

[kind viper]

she*

[rapid pike]

диофантын тэгшитгэл

[stuck haven]

Shit

[kind viper]

she*

[stuck haven]

My bad

[rapid pike]

sorry 3nglish is my 3rd lang

[kind viper]

well the name Diophantus is in there still

yes it is a diophantine equation

anyway your thing simplifies down to, i believe,
(y-x)^2 - (x+1)^2 = 5

[rapid pike]

that makes sense

[kind viper]

so it was not a factorization that you needed

well, we're still going to get a factorization bc this is a difference of squares

[rapid pike]

shoyld i send an example

[kind viper]

and from there, there's only a handful of ways to break 5 down into two factors...

[rapid pike]

yeh

ill work it out

[twin idol]

[rapid pike]

any hints on 6?

[whole flax]

actually for the 6th one you again add -1 to both sides

[pine thicket]

@kind viper you can read russian right?

[rapid pike]

ohh yeahh

[rapid pike]

its mongolian lol

still very similar

[pine thicket]

this looks like russain

oh mb

[kind viper]

it's mongolian written in cyrillic script.

in russian it would be диофантово уравнение

[rapid pike]

yes yes

[pine thicket]

those look the same to me

[rapid pike]

thanks for helping guys, i think i got it

[kind viper]

ok and two very different devanagari words would probably look the same to me since i cannot read that fluently.

so i think thats just a symptom of you being unaccustomed to cyrillic.

[pine thicket]

i was thinking of the same thing

like you wouldnt be able to diff bw hindi and sanskrit

[rapid pike]

avg mongolian textbook for 7th graders

[kind viper]

you can factorize the LHS as (x + ___)(x + ___) once you add a suitable constant to both sides

[kind viper]

this is for grade 7?

[whole flax]

out of curiosity can you read the vertical native script

[kind viper]

damn mongolian schools are cracked

[rapid pike]

i can

i can try writing ur name

in vertical

[rapid pike]

yeh

its amazing that a lot of ppl

are familiar to mongolian lang

so cool

[rapid pike]

here

[whole flax]

so cool

do they teach it you in school

[rapid pike]

yeh from grade 6

[whole flax]

amazing

[rapid pike]

thanks yall im dun with ts chapter

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[covert crater]

,rotate

[pseudo wasp]

Idk got stuck after wards

[kind crane]

,rotate

[kind viper]

try first restricting yourself to just odd k

[pseudo wasp]

And then go for even k?

Can ya explain thought process?

[kind viper]

do you mean "why do we bother breaking down by even vs. odd k" or do you mean "how do you do it for odd k" or maybe "how do you do it for even k"

[pseudo wasp]

No , why ut occurs to take for odd k

Like why that occurs to you

[kind viper]

if k = 2m+1 then you can let n be the middle integer in the sum and write the sum itself as (n-m) + (n-m+1) + ... + (n-1) + n + (n+1) + (n+2) + ... + (n+m-1) + (n+m)

and then that sum works out to simply kn

which is quite nice and makes subsequent reasoning cleaner

[pseudo wasp]

Well I tried can ta work smty put

[kind viper]

but again this only covers odd k so you still have to look at sums of even length

Well I tried can ta work smty put
i have no idea what "can ta work smty put" means.

[pseudo wasp]

I tried ta work smth out I meant

[kind crane]

does ta = to

[pseudo wasp]

Beuh....

[kind crane]

does beuh mean yes?

[pseudo wasp]

Bruh

Lmao where are ya from ?

Your eng is not corrupted at all

[kind crane]

[pseudo wasp]

Eh????

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[icy topaz]

If sin^8 (x) + cos^8 (x) = k, find the value of sin^9 (x) + cos^9 (x) in terms of k

[compact pewter]

Please don't occupy multiple help channels.

[icy topaz]

I will not. This is the first and last time.

[potent berry]

[whole flax]

Rewrite the second expression with the a^3 + b^3 formula I suppose

[thorny walrus]

try writing in euler form

may help

[compact pewter]

@icy topaz Has your question been resolved?

[icy topaz]

I have not learnt calculus.

[kind crane]

can you close your other channel

[kind crane]

the a^3 + b^3 formula is not calculus

[icy topaz]

This.

The person said to write it in "Euler's form", which implies it must be connected to e, which is taught in calculus.

[kind crane]

it varies by grade

euler's formula is taught in high school before calculus in america https://www.onlinemathlearning.com/demoivre-euler.html

[forest token]

maybe write it as a perfect square

(sin^4x + cos^4x)^2-2 sin^4x cos^4x = k

then do the same thing with sin^4x + cos^4x

and convert everything to sin 2 theta?

it may work im not sure

[compact pewter]

@icy topaz Has your question been resolved?

[compact pewter]

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[pseudo willow]

One message removed from a suspended account.

[soft stone]

That's lack of study sadly 🙁

It's like teach me the whole point of trigonometry bc I didn't do myself

In any case I'd help. First I would must assume X is acute bc otherwise the ratios wouldn't be unique

[pseudo willow]

One message removed from a suspended account.

[soft stone]

Now for example to calculate the cosine you can use the fundamental formula

$$\sin^2(X)+\cos^2(X)=1$$

[boreal girder]

Categorist

[soft stone]

Since you know sine you can isolate cosine

[ripe saffron]

Draw a triangle.

[pseudo willow]

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[pseudo willow]

One message removed from a suspended account.

[soft stone]

Now dividing by cos^2 you have that

$$1+\cos^2(X)=\tan^2(X)$$

[boreal girder]

Categorist

[soft stone]

Since now you know cos, you can calculate tan

[dawn wadi]

Then use SOH CAH TOA acronym

[soft stone]

And with sin, tan and cos you have it all pretty much straightforwardly

[pseudo willow]

One message removed from a suspended account.

[dawn wadi]

Nope

[ripe saffron]

No, you just need to study the basics.

[dawn wadi]

You're not

[pseudo willow]

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One message removed from a suspended account.

[twin idol]

They're basically reciprocals of the 3 you learned

[ripe saffron]

So you have opp and hyp, find adj and get all 5 remaining.

[twin idol]

sin(x) = 1/csc(x) and so on

[pseudo willow]

One message removed from a suspended account.

One message removed from a suspended account.

[twin idol]

Hmm

[pseudo willow]

One message removed from a suspended account.

[twin idol]

Did you learn about trigonometry through triangles or through the unit circle?

[pseudo willow]

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[twin idol]

That's a good start

[soft stone]

Sorry it would be

$$\tan^2(X)+1 =\frac{1}{\cos^2(X)}$$

[boreal girder]

Categorist

[twin idol]

Trigonometry is easier when you understand how it's linked to the unit circle

Anyways

Try drawing a right angle triangle

[pseudo willow]

One message removed from a suspended account.

[twin idol]

And label one of the smaller angles as "x"

[pseudo willow]

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[twin idol]

Hmm

You know pythagoras right?

[pseudo willow]

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[twin idol]

You'll need to use that to write down the ratio for cosine

[pseudo willow]

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[soft stone]

I insist with

sin²(X)+cos²(X)=1 to get cos
tan(X)=sin(X)/cos(X) to get tan

sec(X)=1/cos(X) to get sec
cosec(X)=1/sin(X) to get cosec
cotan(X)=1/tan(X) to get cotan

[twin idol]

Those numbers are for sine

Yeah 3 is opp

[pseudo willow]

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One message removed from a suspended account.

[twin idol]

Yep

[pseudo willow]

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One message removed from a suspended account.

[twin idol]

Ye pythagoras

Don't forget to erase the 8/3 thingy for cos x

[pseudo willow]

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[dawn wadi]

It would be a good idea to mark out angle x

[twin idol]

,w sqrt(64-9)

[pseudo willow]

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[soft stone]

😑

[dawn wadi]

X² remember

[twin idol]

You forgot to square the missing side

[pseudo willow]

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[twin idol]

,w (64-9)^1/2

[pseudo willow]

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One message removed from a suspended account.

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[twin idol]

,w calculate sqrt(64-9)

[pseudo willow]

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[boreal girder]

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Requested by 1vul

[pseudo willow]

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[twin idol]

This right here

[pseudo willow]

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[twin idol]

The unit circle explanation will be intuitive when you see animation about it

[twin idol]

Yeah

You have the hyp = 8, and the opposite = 3 and the adjacent = root(55)

[pseudo willow]

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[twin idol]

You can go ahead and find cos(x) and tan(x)

[pseudo willow]

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[twin idol]

cos(x) is incorrect but the rest is correct

[pseudo willow]

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One message removed from a suspended account.

[twin idol]

The triangle is how you'll work with trig functions but the unit circle is why the trig functions behave the way they do

[pseudo willow]

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[twin idol]

https://youtube.com/shorts/gGYmDhpvNv4?feature=shared

[pseudo willow]

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[pseudo willow]

One message removed from a suspended account.

[twin idol]

This one too. The angle changes here https://youtube.com/shorts/DHShSFnQ3rY?feature=shared

[pseudo willow]

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[pseudo willow]

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[twin idol]

Yeah

This shows the 6 functions together with their visualization on the unit circle

https://youtu.be/iBrpZ7vn_Mw?feature=shared

[pseudo willow]

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[twin idol]

The triangle is a consequence of our coordinate system

We want to describe the points on the circle using (x,y) coordinates

[pseudo willow]

One message removed from a suspended account.

[twin idol]

Turns out the best way to do that is by using sin(x) and cos(x)

[pseudo willow]

One message removed from a suspended account.

[twin idol]

That's why they were made

[soft stone]

This is simpler for this question

[pseudo willow]

One message removed from a suspended account.

[twin idol]

Yeah

If you have a bigger or smaller circle, you'll multiply the sin and cos by the radius of said circle

To get the coordinates

[pseudo willow]

One message removed from a suspended account.

[twin idol]

Yeah that will help you immensly

And ask yourself questions about it

And try to answer them, and if you can't understand, we'll be glad to help out when you ask us

[pseudo willow]

One message removed from a suspended account.

[twin idol]

The help channels will always be open and there will be helpers always available :)

[pseudo willow]

One message removed from a suspended account.

[compact pewter]

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[glacial prairie]

Can someone please help me solve

$838500/1.150.84000(cotx)=0.3*(1-20/250)3000.84000(sin2x)$?

My problem is that I need real roots for this equation

[boreal girder]

MichaelRafto

[glacial prairie]

I need x values ranging from 0 to 45

[weak cobalt]

Why the extra coefficients?

And why would you expect non real roots?

[glacial prairie]

It's a question about concrete walls, the tutor assumes two equations as equal and this i what i got

Usually it's a lot simpler but ye

I think he just assumed a value and went along with it

[weak cobalt]

So do you just want an approximate value?

[glacial prairie]

I want a value from 0 to 45 if possible

[weak cobalt]

In degrees, I assume?

[glacial prairie]

Yes

[weak cobalt]

Let me rephrase my question. Do you want to simplify this manually to get a short and simple equation (and then maybe plug it in a calculator)? Or do you just want to use a calculator to get a solution and the process doesn't interest you?

[glacial prairie]

Sure imma simplify it as much as I can

$113,8586956522cotx=264.960sin2x$

[boreal girder]

MichaelRafto

[glacial prairie]

$4.400,3360638523=\frac{sin2x}{cotx}$

[weak cobalt]

Uh... you sure you did that right?

[boreal girder]

MichaelRafto

[weak cobalt]

Please don't mix commas and dots

[dim whale]

They're not -- many countries use the comma as a decimal point and the point to separate large numbers

[glacial prairie]

That would include our country

4,400.3360638523

[weak cobalt]

I don't really care, you don't need a thousands separator

It just makes it confusing for everyone else

[glacial prairie]

$4,400.3360638523=\frac{sin2x}{cotx}$

[boreal girder]

MichaelRafto

[glacial prairie]

Both written to avoid confusion

[weak cobalt]

?? That's even more confusing

Anyway whatever your big number is, call it N, I'm not going to double check that

N = sin(2x)/cot(x)

Do you know how to rewrite sin(2x) and cot(x)?

[glacial prairie]

$sin2x=2sinxcosx$, $cotx=cosx/sinx$

[boreal girder]

MichaelRafto

[weak cobalt]

Great

[glacial prairie]

$4400,3360638523=2sin^{2}x$

[weak cobalt]

sin, not cos

And the square is on the function, not on x

[glacial prairie]

Right, sorry

[boreal girder]

MichaelRafto

[weak cobalt]

Right so it should be easy to solve now

[glacial prairie]

$2200,1680319262=sin^{2}x$

[boreal girder]

MichaelRafto

[glacial prairie]

Yup it's a lot bigger than 1

Ig I'm just going to ask my tutor when I get the chance but thank you for your help man

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[weak cobalt]

Sorry I went AFK. I didn't realize what your actual problem was because of all the coefficient stuff thrown on top.

[weak cobalt]

Point is, when you divided by cot(x), you should have put aside a case for cot(x) = 0

That's basically the only solution to your equation

It still doesn't satisfy 0 <= x <= 45º though

[glacial prairie]

I see

Still, thank you

[weak cobalt]

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[weak gazelle]

Hello guys, i need some help

[dim whale]

Just go ahead and put your question in

[weak gazelle]

So i am studying rn about probabilitys and there are some symbolsa that i really dont understand

these

[dim whale]

Okay -- U represents the union

[weak gazelle]

and that means?

[bleak dock]

this is the intersection

all of the coloured regions is the union

[dim whale]

Example:
let P(A) = P(even number on a 6-sided die) P(2,4, or 6)
let P(B) P(multiple of 3 on a six-sided die) = P(3, or 6)
so P(A U B) = P*(2,3,4,6)

P(A) = 3, P(B) = 2, but P(A U B) = 4

[weak gazelle]

so its basically the information that is in both a and b

[bleak dock]

you mean P(A) = 3/6 and so on

or you mean n(A) = 3

[dim whale]

yeah -- thanks south

[bleak dock]

be careful when you say that

the union is the info in a or in b

[bleak dock]

this would be the intersection

[dim whale]

P(A) = 3/6, P(B) = 2/6 and P(A U B) = 4/6

P(A | B) is the probability A is true if we already know that B is true

[weak gazelle]

could you maybe explain it a bit further then?

[bleak dock]

do you have a textbook?

[weak gazelle]

yes i do

[bleak dock]

yeah then follow that

or you could also watch and practice on Khan Academy

https://www.khanacademy.org/math/ka-math-class-11/x0419e5b3b578592a:sets-ncert-new/x0419e5b3b578592a:sets-and-their-representations/v/what-are-sets

[weak gazelle]

okay ill try that then thank you

[bleak dock]

no worries!

[dim whale]

feel free to come back if you're not able to make sense of your text or videos still aren't making it clicik

[weak gazelle]

i will, thank you again

[bleak dock]

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[dusky quartz]

is there a convenient identity here?

[dusky quartz]

Correction to the problem. Swapped the plus signs to minus signs here.
Given $x+y+z+\sqrt{x+y+z}=182$, and
$x-y-z-\sqrt{x-y-z}=156$.
\

Find $\sqrt{\sqrt{x^2-y^2-z^2-2yz}}$

[boreal girder]

laestia

[short lantern]

the trick is to guess that (x,y,z)=(13,0,0)

in all seriousness, id have no clue

[dusky quartz]

aint it x=169

[short lantern]

whoops yeah

[dusky quartz]

thats sqrtx

[short lantern]

it could help to let y+z=b, then the problem reduces and y,z pairs to b

even the sqrt would have b^×

b^2

[dusky quartz]

im slightly sure thats the trick considering how the bottom expression above is conveniently (y-z)^2

[short lantern]

its -(y+z)^2

[dusky quartz]

oh shit u right

,, x+k+\sqrt{x+k}=182, x-k-\sqrt{x-k}=156

[boreal girder]

laestia

[dusky quartz]

goal: find x²-k²

2x=169×2
x=169

k=0

√√x²=13

you were right on the money

@short lantern congrats 726, you cooked

what a convenient guess

[short lantern]

i always do

[short lantern]

but yeah how did you make this jump

[dusky quartz]

oh wait

it's

2x+sqrt(x+k)-sqrt(x-k)

im blind

[calm trout]

i mean the original equations are just a^2 + a = 182, b^2 - b = 156

just solve the quadratics and youre done

[short lantern]

you should use a different var other than b, but yeah

[dusky quartz]

a+b=2x
a=(-1±27)/2
b=(-1±25)/2

[calm trout]

what

[dusky quartz]

the two results for a and b conveniently add up to 26

[calm trout]

[dusky quartz]

uuuh

a²+b²

[short lantern]

we used b for y+z

[calm trout]

clearly not as i see a k

[dusky quartz]

uf

[short lantern]

ok im just sleepless

[calm trout]

a = sqrt(x + y + z)
b = sqrt(x - y - z)
the desired quantity is sqrt(ab)

[short lantern]

goodnight

[dusky quartz]

blaah

the answer is 13 but i still got lucky with the assumptions

[calm trout]

well

,w a^2 + a = 182

[boreal girder]

Results provided by WolframAlpha

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[calm trout]

,w b^2 - b = 156

[boreal girder]

Results provided by WolframAlpha

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Requested by tearfox

[calm trout]

you can see that a = b = 13

[dusky quartz]

bah

yeah

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[molten badger]

In space,given 2 point $A(0,0,10)$ and $B(3,4,6)$. For point $M$ satisfies $\triangle OMA$ is an acute triangle and $[OAM] = 15$. Determine the minimum values of $MB$

[boreal girder]

Alexis_Fx

[kind viper]

minimum values?

[molten badger]

value*

Sorry

[kind viper]

ok

progress thus far?

[molten badger]

So the locus of all points M for [OAM]=15 is a cylinder and I also found out that for angle MOA and MAO to be acute the z value of M has to be around 0 to 10

[kind viper]

it's a cylinder with what radius and what line as its axis?

[molten badger]

For OMA I used cosine law cos(OMA)>=0 so MA^2 + MO^2 >= OA^2 and I'm stuck here

[molten badger]

It's around the z-axis and with radius 3 (idk if I word this right)

[kind viper]

why radius 3

[molten badger]

Segment OA is in z-axis so d(M,Oz)•OA•1/2=15 => d(M,Oz)=3

[kind viper]

oh you did it based on area right ok

thing is, M cannot go on or inside the sphere whose diameter is OA

otherwise angle OMA will be obtuse

[lofty stirrup]

how do you guys understand this...

Im still stuck on ratios💀

[molten badger]

I haven't thought of that

Give me a min

[kind viper]

i mean, there's people here from a lot of different levels of education

[lofty stirrup]

Ig

[kind viper]

but also if you wanna complain about feeling like you suck at math that's better done in #chill

[kind viper]

i have a feeling the question is not quite well-posed btw because we won't be able to minimize MB (or rather the point where the min would appear will also be one where the angle at M is actually right and not acute)

[molten badger]

Hold on, that's my fault

[molten badger]

Okay so, that's not hard to prove, and I guess minimum occurs at one of the intersection points

[molten badger]

I translated it wrong in that part sorry

So minimum occurs at z=9 or z=1

[thorny walrus]

Usually the long but always working way is to make a single parameter fxn and then differentiate and check at end points of that single parameter fxn for minima and maxima

How does the bot have a gender hmmm

[molten badger]

I'm kinda confused rn. I can find MB in both cases using geometry. But if I use a vector I get $(x-3)^2 + (y-4)^2 + (z-6)^2$

[boreal girder]

Alexis_Fx

[molten badger]

expanding this gives -6x-8y which is unknown or am I missing something

Yeah I figured it out

So we have to find the maximum for 6x+8y

x and y are points in circle so I guess we can sub x=cos(t) and y=sin(t)

-2(3cos(t)+4sin(t))

Its minimum is -2•5 but i can't get the same value for MB

Can someone check please, i got MB= sqrt(13) and this's also the correct answer according to the answer key

But I can't get the answer using vector somehow

[compact pewter]

@molten badger Has your question been resolved?

[molten badger]

Oh yeah I see, it supposed to be x=3cos(t) and y=3cos(t) since the radius is 3

thanks Ann

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[stark schooner]

can someone help me

[stark schooner]

i couldnt post the working out due to camera difficulties

but my answe is

1120.5mm^2

but it is incorrect (770mm^2)

how is it 770mm^2?

i followed this guide through my maths class today for this question

shouldnt it be correct?

[stark schooner]

i figured since the example question here, has the same amount of given information as the other question, it would be the same basically?

[compact pewter]

@stark schooner Has your question been resolved?

[stark schooner]

dawg u srs rn?
\

i give up

[compact pewter]

@stark schooner Has your question been resolved?

[glacial prairie]

Wait man, you there?

[compact pewter]

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[sullen gorge]

$$\int_1^4 (x^2 - x) ,dx$$

[boreal girder]

Normie

[twin idol]

Alright is there a problem with evaluating it?

[glacial prairie]

!status

[compact pewter]

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

[sullen gorge]

I seem to have done it but not quite the answer

[twin idol]

Can you show us how you did it?

[sullen gorge]

I get 25/2 instead of 27/2

[lavish gull]

show work.

[glacial prairie]

what are the antiderivatives that you took?

[sullen gorge]

Let see

[twin idol]

Ok the calculus part looks good

[sullen gorge]

Which part?

[twin idol]

The antiderivatives you took

And the evaluation

[upper bane]

which part do you want us to look at though

[twin idol]

It may be an algebra problem

[upper bane]

this is incorrect btw

[sullen gorge]

Lmao

I see

I thought it's Upper limit - Lower limit

[weak cobalt]

(126-45)/6 is correct

Just compute it

[twin idol]

It is antiderivative at upper limit - antiderivative at lower limit

[upper bane]

you opened the brackets wrongly

they're supposed to be like this

[glacial prairie]

the left part that you erased is correct at least until the last =>

[upper bane]

so the final - should have been a +

[sullen gorge]

Oh right

This gives me the correct answer

The answer was right in front me bruh

Thank you though 🙂

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[sullen gorge]

.reopen

[compact pewter]

✅

[sullen gorge]

Okay so this question is a bit tougher

[twin idol]

We'll help out :)

[sullen gorge]

I am at the 5th question

Basically I got this wrong for some reason

[winged lion]

Whats the solution?

[sullen gorge]

e - 1/(e)

[winged lion]

You serious

That just e¹-1/e¹

That what you had already

[sullen gorge]

Oh...

Damn this

I am not trolling sorry

This is literal 🤪

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[next palm]

I dont understand what is wrong why dose it give 2 diffrent results?

[kind viper]

cos(phi)/sin(phi) = cot(phi) and not tan(phi)

so consider that (and also neither cos(phi) nor sin(phi) are known to be sign-constant so you cannot do division like this...)

[next palm]

Ok so how is it supposed to be done?

[next palm]

Also i did put cot we were js taught to write it ctg

[kind viper]

what exactly are you trying to accomplish here anyway

rewrite the inequality x ≤ y in polar coordinates?

make a sketch of it and note what angles phi actually goes over. and free yourself of this entire headache with case work.

[next palm]

Yes i think

[kind viper]

you think?

you are the one with the problem.

you cannot be saying "i think". if you are even remotely unsure then you should post the original problem.

[next palm]

.

[kind viper]

you need to find the double integral of x dx dy over some region...?

except your diagram is too cluttered and microscopic and i can't really tell what region is meant...

[next palm]

Yes need to find the area that meets those conditions

@kind viper

[kind viper]

so... $0 \leq \rho \leq 1; \pi/4 \leq \varphi \leq \pi/2$ it looks like.

[boreal girder]

Ann

[next palm]

No sry did it worng

[kind viper]

ugh

do you have a picture or a written description of the damn thing?

[next palm]

Its js the x region excluding the region set before

[kind viper]

ok and that big circle is... what, rho = 4 sin(phi)?

[kind viper]

then it's 1 ≤ rho ≤ 4 sin(phi) and still pi/4 ≤ phi ≤ pi/2.

[next palm]

Yeah i know that i was asking why this was like this

[kind viper]

btw in the future: http://xyproblem.info/ so that neither of us has to get frustrated as we try to solve a completely different problem than intended

i dont think you should be trying to write these algebraically at all unless you want to make a million mistakes every time

[next palm]

@kind viper

[kind viper]

do it graphically with your diagram

[next palm]

Cant have to prove

[kind viper]

note the polar coordinates of endpoints and vertices of your region

ok well if you "have to prove" then talk to your teacher about it and how they want it done. idk.

[next palm]

I did it how they taught but u pointed to it being flawed

[next palm]

.

[ripe ether]

plug in l=3/4pi to the left deductions

dividing by negative number flips the sign and you dunno if cos(l) is negative or not

and your deduction pretended it is positive

[next palm]

But sin(phi) cant be negative?

Ok ok wdym by plug in

[ripe ether]

if you can prove that then sure. But do you hhave that cos(phi) cant be negative?

for the left deduction to make sense

[ripe ether]

also in that case you should say it in the solution so its easier to understand (because sin(phi)>=0, we have...)

[next palm]

Ok but wdym by plug in

Is there an outlet im missing?

[ripe ether]

look at the left hand deduction, plug in l=3/4pi, see where the deduction went wrong

[next palm]

How do i prove i have to find this region

X>0 and x<=y

Phi E[?,?]

[ripe ether]

why did you use cos and sin?

[next palm]

X = rcos(phi)
Y = rsin(phi)

Y^2 + X^2 = r^2

[ripe ether]

ok cool so you are trying to find whenn rcos(phi)<=rsin(phi)

given that both rsin(phi) and rcos(phi) are positive

[next palm]

...

[ripe ether]

if both are positive then both results are correct

maybe not the last step id need to check

[next palm]

but

The results are diffrent

How can they be correct if one if one of the is worng

One is saying phi >= pi/4 and the other is saying the opposite

That why i came asking why

[ripe ether]

cot(l)<=1 gives you l>=pi/4.

[next palm]

How?

Is that js how cot works?

[next palm]

Is that trigonometry rule were if u use cos or cot the sign changes

[ripe ether]

cot is decreasing on the interval [0,pi]

[next palm]

Yes or no,y or n,1 or 0.

[ripe ether]

idk what u mean

idk how cot decreasing has things to do with sign change

[next palm]

IDK WHAT COT DECREASING AS ANYTHINV TO DO WITH THIZ

[ripe ether]

since cot(pi/4)=1, using the fact that cot is decreasing, we know that for all l>=pi/4 inside of [0,pi], cot(l)<=1

in particular, l>=1/4pi should be part of the solution to cos(l)<=1

and your line "cot(l)<=1, so l<=pi/4" would be wrong

to summarize: just because cot(pi/4)=1, doesn't mean the solution to cot(l)<=1 is l<=pi/4

[next palm]

Ok

.close

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[frail anchor]

60

[soft stone]

So 60 for fixed y is an exponential ae^x, innit?

And for fixed x it is a cosine bcos(y)

So does that not help you?

[frail anchor]

hmm true

yeah I got the answer by observing the graphs they gave but how can I think about it's contour map

when z = constant

nvm i can just get that aswell from the graphs they gave lol

[soft stone]

Yes

Visually sectioning them

[frail anchor]

precision does not matter when we draw contour maps right

[soft stone]

The precisioner the better

[frail anchor]

hmm

I don't think I can get precision on this one as it's not any recognizable curve

$$ k =e^{x}\cos{y} $$

[soft stone]

Aren't you supposed to just choose the figure I to VI given?

[frail anchor]

no LaTeX

[soft stone]

That is what the question text says

[frail anchor]

yes but this is not some homework I am just trying to speed run multivariable calculus for physiks

so only understanding is my focus

[soft stone]

I don't now much about physics but I don't think you'd ever need to sketch countour maps by hand

[frail anchor]

ah yea true but it's always nice to learn something completely , anyway thanks ❤️

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[soft stone]

Oh I remember doing that in ODE but it was with linear system of ordinary differential equations

We sketched contour maps by hand reducing to jordan form etc haha lmao

But don't worry

[compact pewter]

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[frail anchor]

i am planning to read simmons DE after I finish this lol so it might helpful

[frail anchor]

.close

[compact pewter]

Channel closed

Closed by @frail anchor

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[reef nacelle]

I keep making mistakes in my irrational equation problems, but I can't tell where. Everytime its eathier the discriminant is negative or the check is wrong

[trail cave]

uh what are your answers for the problems

[devout latch]

For the first one

there is a little sign error messing up the quadratic

[devout latch]

.

<@&268886789983436800>

[reef nacelle]

oh my god 🥀 but everything else is fine right?

in this first one

also i went to wash my face thats why i wasnt answering

[devout latch]

well you have to adjust calculations

[reef nacelle]

i feel much better to tackle my shit now

[reef nacelle]

well yeah but was this the only mistake in the equation?

like that the rest of the steps were fine besides this error you pointed out

[devout latch]

up to solving the quadratic equation it's the only mistake. I've not parsed further

[reef nacelle]

alr, please let me know with the other ones (take your time), im gonna fix the mistake in the meantime

[devout latch]

again small sign error

[reef nacelle]

im going to fucking loose it

why is it always such stupid mistakes with me 🥀

[devout latch]

pro tip: take your time and review each steps before going to the next

[reef nacelle]

yeah tbh i was having a bit of a crashout just mass doing these because i was pannicking that if i just didnt forget how to do irrational equations entirely

kinda relieved its so far just me forgetting minuses and pluses

[devout latch]

there is some trickery incoming, you have to test the solution you get out of the quadratic because some of them won’t work all the time

but the main part is the quadratic

[reef nacelle]

the test you mean the P(no) L(no) test?

[devout latch]

yeah that seems to be that verification

[reef nacelle]

yea ive been doing it in some of the problems when the discriminant i got wasnt fucking negative

[devout latch]

finally adding instead of subtracting

[devout latch]

that should be it for now

another pro tip: send one question at a time, because it can be tedious for us to verify or deal with multiple problems at once in one channel

[compact pewter]

@reef nacelle Has your question been resolved?

[reef nacelle]

alr, sorry

[reef nacelle]

why does the minus go away with the square?

[devout latch]

is ok, but just so you know

[reef nacelle]

yea so it all just boils down to me to not paying attention to negative or positive values in numbers and counting them wrong

[devout latch]

$(-\sqrt{5+x})^2=((-1)\sqrt{5+x})^2$

[boreal girder]

pola_touche

[reef nacelle]

so the minus is actually assigned to a 1 that isn't visible?

[devout latch]

then if you square this you have (-1)^2(5+x)

[devout latch]

i think the confusion came when you put the - beside the square root here without clearly indicating on what the square was

when you distribute (a+b)^2=a^2+2ab+b^2. here a=3, b=-sqrt{x+5}

[reef nacelle]

oh i didnt know that, this is kinda my first time coming across an equation where i had to expand a part of it with the squared being negative

or having a negative in this case

[devout latch]

this is another way to see why you need these brackets

if you « foil » the expression

[reef nacelle]

i see

[reef nacelle]

alr, im gonna fix/do more, so in the meantime imma close this channel for others. Thanks for the help

.close

[compact pewter]

Channel closed

Closed by @reef nacelle

Use .reopen if this was a mistake.

[compact pewter]

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.

[glacial tusk]

i need help on a question |-2k|/7=4

[glacial tusk]

i did the first step i think but im not sure if im on the right path

[leaden fulcrum]

what did you do?

[glacial tusk]

that

[upper bane]

multiply both sides by 7 first

[leaden fulcrum]

yea

[upper bane]

then do the +/-

[leaden fulcrum]

beat me to it

[glacial tusk]

so i’ll multiply 4 and 7

by 7

[leaden fulcrum]

yea

then u can take the absolute value

manipulate what is outside the absolute value first

[glacial tusk]

do i make the 2 equations now?

[leaden fulcrum]

yep

[glacial tusk]

so -2k/49=28

right??!!

[upper bane]

uh nononono

[leaden fulcrum]

wat

u multiply both sides by 7

[upper bane]

this is how you multiply both sides by something

don't multiply the denominator

[glacial tusk]

mmmm i seee

[leaden fulcrum]

what u did was divide the left hand side

[glacial tusk]

if you wanted to cancel out the 7 of the denominator wouldn’t you subtract?

or no

[leaden fulcrum]

no

u would multiply 7 in the numerator

[glacial tusk]

that’s only if i didn’t have fractions right

[leaden fulcrum]

7/7

[glacial tusk]

49

[leaden fulcrum]

what youre supposed to do is muliply 1/7 by 7

and youll get 7/7

[glacial tusk]

how would i write that on paper

[leaden fulcrum]

like this

[glacial tusk]

i see

then would i do the two equations

[leaden fulcrum]

yea

what would the 2 equations be?

[glacial tusk]

|-2k| =28 and |-2k|=-28

right

[leaden fulcrum]

u can remove the absolute value when u get your 2 equations

[glacial tusk]

so both of them would have 2k

not the negative?

[leaden fulcrum]

no

[glacial tusk]

mm i see

[leaden fulcrum]

this is right u just need to remove the absolute value for both equations

[glacial tusk]

sooo

i remove the brackets

[leaden fulcrum]

yes

and now u can solve for k

[glacial tusk]

then i divide them by 2

negative 2

[leaden fulcrum]

yes

[glacial tusk]

so the answer is 24 and -24

[leaden fulcrum]

what happens when u divide 28 by 2

what did you do to get 24?

[glacial tusk]

whoops i wrote 48 instead of 28

it’s 14

right

[leaden fulcrum]

correct

[glacial tusk]

oaky i have one more

if that’s okay with you

[leaden fulcrum]

have u tried solving it yourself first?

but sure

[glacial tusk]

your right

i’ll try

[leaden fulcrum]

if its the same type of questions what ive told you would apply

[glacial tusk]

i got this

[leaden fulcrum]

ok well the answers are right

but the steps are wrong

[glacial tusk]

i see

what did i get working

wrong

[leaden fulcrum]

take the absolute value first

since the 1/5 is inside the absolute value

then u can multiply by 5

[glacial tusk]

so 25 or 5

[leaden fulcrum]

whats 2*5

your answers are correct

but the order of your steps are wrong

[glacial tusk]

ohhh

[leaden fulcrum]

can you explain to me what is wrong with the steps?

[glacial tusk]

uhh

ermm

i need to take the absolute value out

to be able to multiply

i think

[leaden fulcrum]

yes

[glacial tusk]

so would i multiply 5x5

or 2 time 5

[leaden fulcrum]

?

on the left side

and right side

but u would have k=-2(5) k=2(5)

[glacial tusk]

how did you get that

[upper bane]

the same way as you did with the other questions

in each equation, multiply both sides by 5

[glacial tusk]

ohhh i seee

then i would mutiply 5

by both sides

right

[upper bane]

mhm