help-27

[woven radish]

Swas.py

[bold prism]

(-pi/2, pi/2)

[merry seal]

ah

tan was different right right

[ruby shell]

oh so its equal to range of tan-1

[bold prism]

so we have ascertained that domain of tan inverse is R

[ruby shell]

but how

[bold prism]

in other words no restriction on anything inside it

[ruby shell]

what about the log part

[bold prism]

we move to that next

for logx domain is x>0

so for your function, $5x^2 - 8x + 4 > 0$

[woven radish]

Swas.py

[ruby shell]

yes

[bold prism]

see if you can factor this

[ruby shell]

no middle splitting idk why

+4 is there

[bold prism]

yes

use quadratic formula

[ruby shell]

okay lemme try

[bold prism]

💀

[devout snow]

@ruby shell Has your question been resolved?

[devout snow]

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[restive river]

.reopen

Hi everyone, I have a question: A ship is flooding and will only last 10 minutes, it is drifting at a constant speed of 100 km/h and the harbor is 1200 km away. A rescue boat is sent from the harbor but it takes 6 min to get ready and once the rescue boat reaches the ship it takes 2 min to dock and fix the flooding, what is the minimum speed in km/h the rescue boat can travel and save the ship

[restive river]

<@&286206848099549185>

[signal veldt]

is it drifting towards the dock?

[restive river]

Yes

[signal veldt]

cool right so the boat needs to get there in 8 mins

to save it

in 8 minutes time how far will the ship be from the dock?

[restive river]

1186.6km away from the dock

[signal veldt]

exactly

the rescue boat takes 6 mins to get ready

so only 2 mins to travel

2 mins to travel 1186.66km

basically the ship is a gonna haha

[silent viper]

Maybe 4?

Because lifetime is 10 minutes

[signal veldt]

but they need 2 mins to save it

once they reach it

[silent viper]

Still remains 2 minutes

10-6-2

[signal veldt]

the ship will sink in 10, the ship will sink if the boat reaches it over 8 mins because they need 2 mins to save it

the rescue ship needs 6 mins to leave the dock

so 2 mins of travel

[silent viper]

Oh

True

[signal veldt]

all good i didn't notice that at first

[restive river]

Oh so is the answer 35580 km/h

[silent viper]

Poor boat

[signal veldt]

1186k.66km repeating in 2 minutes is about 35600 km/h

boat is screwed haha

2x rocket speed

[restive river]

ah ok thanks

rip to the boat lol

.close

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[deep quarry]

I don't find it quite clear as to what the question is asking can anyone help me?

[deep quarry]

<@&286206848099549185>

someone

please

[stray egret]

its asking

maximize P in the pink region

[deep quarry]

wdym

so like P > 2x +3y?

[lime ore]

you have a pink region which consists of a bunch of points with x and y coordinates right

[deep quarry]

yea

[lime ore]

so you can pick any point inside that region an put it into the function which is again also written in terms of x and y

[deep quarry]

yes

so then what would maximizing it be

[lime ore]

now what the question wants you to do is to find the point in the pink region that makes it so P is as big as possible when filling in the coordinates of that point

[deep quarry]

And P is the shaded region?

[lime ore]

no P is the function you got

[deep quarry]

o

[lime ore]

2x+3y

[deep quarry]

yea

so to make it as big as possible

[lime ore]

yes

[deep quarry]

would I go for the point they intersect?

im confused how I maximize a coordinate

[lime ore]

it would be usefull to calculate the intersection but it probably isnt the point you are serching for

you dont maximize the coordinates, you search the coordinates for which P would be the biggest it can be

[deep quarry]

The function

so

Is P including both lines

[lime ore]

P doesnt include or exclude anything, but those lines are included in the pink region

[deep quarry]

could you draw like a model or something because I don't really understand what you mean

[lime ore]

i dont really know how i'd draw a model for this problem but you basically have a bunch of points represented by the pink region right

[deep quarry]

yes

[lime ore]

choose one point in that region

[deep quarry]

yea

[lime ore]

put it in chat i mean

[deep quarry]

ill go with 6,5

[lime ore]

sure, now plug that into P

[deep quarry]

So P=12+15

Oh I see

[lime ore]

yeah

[deep quarry]

So which coordinate

But how would I figure that out

[lime ore]

so you want to make P as big as possible while staying in the pink zone

[deep quarry]

yes

[lime ore]

first of all look at what P is

[deep quarry]

yes

[lime ore]

what is it?

it's really simple no need to overthink

[deep quarry]

a function of 2 lines that form a shaded region including all points that are valid in the equation P=2x+3y

[lime ore]

no P is a sum

meaning that if x and y are positive the sum will only get bigger if x and y get bigger

[deep quarry]

ye

[lime ore]

and can you tell anything about the pink region

[deep quarry]

yes

[lime ore]

or at least about the points in it

tell me

again pretty simple no need to overthink

[deep quarry]

They are all inncluded in the function?

[lime ore]

think about this

[lime ore]

no for every point in the pink region the x and y coordinates are greater or equal to 0

meaning every x and y you could plug in is positive

[deep quarry]

they are all ohh

I thought it kept going

[lime ore]

no it doesnt

but the fact theyre both positive means you have to search for x and y that are as big as possible

[deep quarry]

is there an equation I can use to figure that out

[lime ore]

meaning you'd have to look at the (non-zero) boundary of the region

[lime ore]

yes there is can you tell me what you think it can be

[deep quarry]

uhm

well

[lime ore]

no need to put in the actual function at the moment just say how you think you'd go about finding it

[deep quarry]

You would have to plug in all the points and calculate the one that would produce the largest sum

and you could rule out ones along the axis

[lime ore]

yes but can you rule out any other points

[deep quarry]

all the ones inside the shaded region

[lime ore]

exactly

[deep quarry]

it would be on the line itself right

[opal token]

isnt it where the points intersect?

[deep quarry]

i thought so idk

[lime ore]

not necessarily

[lime ore]

yes indeed

[opal token]

in this specific case though

[lime ore]

i wouldnt know i havent computed it

but even if it is so in this case, it might be different on for example an exam

[deep quarry]

Well the values decrease as you stray from the intersection

[opal token]

yea ofc im just saying with this specific example

[deep quarry]

sec brb

[lime ore]

im not sure wait a sec

[deep quarry]

back

[lime ore]

yeah so you know the point where P is maximized is on the outer line right

do you know a way we could plug every point on that outer line into P at once?

[deep quarry]

< > =

greater than

less than

[lime ore]

no

[deep quarry]

o

[opal token]

yea im pretty sure with this case its (6,5)

[deep quarry]

Why wouldnt it be less than or equal to P

[opal token]

basically where they intersect

[deep quarry]

that would include all ranging from least to greatest

[opal token]

no this is just where p is at a maximum within the shaded region

[lime ore]

you would have to plug in the functions depicting the boundary lines into P

[opal token]

which points when inputted into the function gives the largest possible value for p

[deep quarry]

yea

ive done this

[opal token]

ofc outside the shaded region you can get infinite values for p

[deep quarry]

and the point they share is the intersection

[opal token]

but this is asking for the shaded region

[deep quarry]

but solid lines indicate inclusion

[lime ore]

yeah but that doesnt really matter here

[deep quarry]

so would it be the intersection though

looking at the line

thats the highest value

[opal token]

in this case yea

i dont know if thats always the case but in this specific event it is

[lime ore]

you basically have to solve for max of P in the region the first function is its boundary and then also look at the same for the other function

[deep quarry]

hm

so it would be 27

@lime ore

heres a solution I found online

What they did is plot in corners to see which was larger and concluded the intersection was

this way they could get a good idea all over

[lime ore]

yeah its a valid method but not in all cases

[deep quarry]

so idk about there being an equation to solve it

[lime ore]

what grade are you in?

[deep quarry]

do you know if there is

8

[lime ore]

im sorry im europian and dont actually know what that means

how old are you

and have you already learned about derivatives?

[deep quarry]

14

uhm

no

I just started pre cal

im in the first unit

[lime ore]

dont worry

there is a solvable equation using derivatives to solve maximum problems

[deep quarry]

o

well the lesson I just took had to do with calculating intersections using matrices and linear equations

[lime ore]

yeah that's pretty usefull stuff too

especially the matrices in my opinion

[deep quarry]

yea

for what I'm doing matrices are obsolete to just substituting

unless I have a clac

calc*

calculator

took me 15 minutes to write out a whole equation for X

the A^-1 stuff

[lime ore]

yeah atm it might seem obsolete bcs the problems arent really that hard but later on you'll have to do more difficult stuff and having experiencing working with matrices will help

[deep quarry]

ye

quick question:

This would be (3,2) right?

i did x4 on top x 3 on bottom

[lime ore]

yeah that's pretty easy by hand but personally i'd instantly just use matrices

wait a sec

yup 3,2 is right

[deep quarry]

k

ty

[lime ore]

np 🙂

[deep quarry]

when this says y coord solution would that just be what y=

which would be 3.8

[winter patrol]

how are you getting 3.8

[deep quarry]

idk

whats it asking for

[supple knot]

Two equations and two unknowns.

[deep quarry]

yea so isnt the answer 3.8?

or did I do it wrong

[supple knot]

Do this.

I.e. show your work

[deep quarry]

oke

[supple knot]

An answer to a problem like this without work is not enough

[devout snow]

@deep quarry Has your question been resolved?

[deep quarry]

sorry i was afk

heres my work

2(5x+2y=21) -----> 10x+4y=42
5(-2x+6y=-34) --> -10x+30y=-170

Add them together

0x+34y=-128

34y=-128

y= -128/34

y= 3.8 (to the nearest tenth)

<@&286206848099549185>

But I am not sure if that is what the question is asking for

which is what I want to know

[deep quarry]

@supple knot

[supple knot]

That looks right.

[deep quarry]

ok thank you

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[digital dune]

Quick question guys. Say I am solving an inequality, where some of the critical points need to be approximated numerically. How (in terms of notation) would I write down the solution to the inequality based on the fact that I only have approximate values for critical points? Thanks

[wooden veldt]

Perhaps you want to use $\lesssim$ or $\lessapprox$

[woven radish]

ΣAC

[digital dune]

ah i didnt know that notation existed

what is the difference between each sign?

is the second one more precise?

[wooden veldt]

Actually have no idea lol

[digital dune]

is there a name for the symbol

[wooden veldt]

Probs like "approximately less than"

[digital dune]

ill have a look online

thanks

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[green wagon]

if i got three propositions A, B and C such that
A implies B
B implies C
C implies A
then A, B, C are all equivalent to each other right?

[wooden veldt]

Yes

[devout snow]

@green wagon Has your question been resolved?

[green wagon]

thank you!

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[prisma frost]

For this question would I still get a mark for saying m=10 and -10 since on the mark scheme the answer is 9<m<11 and -9<m<-11

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[restive river]

What is the rule of congruency?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[restive river]

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[karmic radish]

find cos^ 4 a minus sin^4 a in terms of Cos A

[winter patrol]

what have you tried

[karmic radish]

tried writing in in the form of a^2 - b^2

[fringe parrot]

won't that be cos(2*a)?

either that or negative

[karmic radish]

?

[fringe parrot]

?

[winter patrol]

what did you do after that

[karmic radish]

1(cos sqaure a - sin square a)

[fringe parrot]

yup

and that cos^2(a) - sin^2(a) = cos(2a)

see double angle formula

[karmic radish]

didnt get it

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[teal gust]

Hi. Could someone help explain how I can factor this perfect square trinomial? I'm still confused with these ones, unfortunately.

[royal current]

It's just (6x+1) ²

(6x) ² +2*6x+1²

[teal gust]

Am I just supposed to know that, or like?

[placid narwhal]

Look at the 36x^2

Its just (6x)^2

[teal gust]

Right.

[cosmic jacinth]

$a^2 + 2ab + b^2 = (a+b)^2$

[woven radish]

Muhammad Hussaini

[restive river]

this is the short cut method

[placid narwhal]

@teal gust try factoring this.
$$64x^2 + 16x + 1$$

[woven radish]

Pluton

[teal gust]

Ah, I see.

[restive river]

and the long cut method is middle term factor

[teal gust]

Is that (8x +1)^2?

[restive river]

ya

[teal gust]

Ah, I think I got it.

Thank you all. :)

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[restive river]

help with u sub
x * sqrt(3x+1)

[restive river]

i dont understand what i did wrong
u = 3x + 1
du = 3
so

1/9 * (u-1) * sqrt(u)

[royal current]

Try putting (3x+1)^(3/2) =t
(3x+1)^(1/2)dx=2dt/9
x=(t^(2/3)-1)/3

[restive river]

ill try hold on

im gonna review u sub again im missing something important

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[lucid peak]

Determine the volume of the rotating object if the area A is rotated to the rotation axis y = 0

[lucid peak]

area A is that area between the green/red curve

so essentially it'll make like a bowl

i genuinely have no idea and there are multiple formulas online that i dunno how to correlate

[devout snow]

@lucid peak Has your question been resolved?

[twin flame]

So what formula have you thought about using

Ok so to revolve a curve around the x axis, the integral we need is

[lucid peak]

this

[twin flame]

Yes and this is the integral for revolving around the y axis, which is what we need

[lucid peak]

but im gonna have to convert from y = x to x = y if u kno what that means

(that x^2 is basically the converted function)

[twin flame]

Yes, you will need to integrate with respect to y so that you find the area against the y axis. So you will have to find the inverse of both of those functions

[lucid peak]

there's also this too 🙃

oh wait

[twin flame]

This will revolve around the x axis

[lucid peak]

the rotation on x and y axis

is different

hmm yes

it appeard i have also used the wrong formula :)))

[twin flame]

Hehe, so you need to first find the inverse of both functions then integrate those

With respect to y

There is also a pi missing from the first formula

[lucid peak]

from this?

[twin flame]

Yes

Like in the second

[lucid peak]

my bad i just finished erasing all my answer lmao

[lucid peak]

the inverse?

[twin flame]

Yes, not $y=x^2$ but $x=\sqrt{y}$

[woven radish]

Social Capital Gainer

[lucid peak]

ah yeah that's the "conversion" lmao

[twin flame]

Mhm, I see

[lucid peak]

inversed, aight noted

wait so

i have to inverse both functions

[twin flame]

Yes, you need to find the inverse of both. This is because the integral is against the y axis, so you need to integrate with respect to y

[lucid peak]

so then this should be the second function's inversed state?

[lucid peak]

noted

[twin flame]

Not sure about this

[lucid peak]

from this

it's all just about moving the numbers to the y's side right?

[twin flame]

Ye, but I think you have made some small mistakes in doing that

Just one really

[lucid peak]

im assuming it's related to the /2?

[twin flame]

Yes, you have divided by 2 instead of multiplied. It is already a division

[lucid peak]

OH

OH RIGHT

i blundered so bad thanks a lot

[twin flame]

No, it was just one thing

[twin flame]

Yes

[lucid peak]

most of my mistakes always from just that "1 thing" :")

okay inversed completed

time to integrate?

oh wait no

[twin flame]

Don't worry about it, it is just practice

[lucid peak]

power them all to 2 aye?

[twin flame]

Yes, but we also need to know what the limits of integration are

[lucid peak]

true lule

[lucid peak]

what does that mean?

oh the

[twin flame]

When you do a definite integral, the upper and lower numbers

[lucid peak]

i think it's 2 and -2?

oh wait no this is a different case ummm

okay i have no idea if it keeps spinning im assuming it'll just keep spinning

[twin flame]

Yeah, we don't need to worry about the revolution itself, but from what lower value of y to what upper value of y are we integrating

[lucid peak]

so then it's still 2 and -2

[twin flame]

How did you get -2?

[lucid peak]

cus they're the X values of the intersection

[twin flame]

Yeah, those are the x coordinates, but remember that we are integrating with respect to y. So we are putting in y values this time

[lucid peak]

OHHHHH

hmm that does make sense yeah damn

oh right

[twin flame]

Hehe, nice

[lucid peak]

the problem did mention sumn about y = 0

Determine the volume of the rotating object if the area A is rotated to the rotation axis y = 0

[twin flame]

Yes, that is theblower limit we need the upper.

[lucid peak]

and from the object its limit is y = 4

so it's 4 and 0?

[twin flame]

I need tl sort something real quick will be back in a few mins

Yes thats right

[lucid peak]

sure no rush c:

ait imma try integrating to wait

wait oh right, what's the relation between the 2 function?

[lucid peak]

im assuming it's green - red

[twin flame]

Ok, so have you squared both first

[lucid peak]

,rotate

[lucid peak]

i integrated them independently ehe

[twin flame]

Yeah, that is the right thing to do in this case, but the reason is still something need to take account of

So imagine the block made by revolving just the red curve

To get the bowl, we need to make a cavity in the shape of what comes from revolving the green curve

[lucid peak]

excuse me just posting it here again to visualize haha

[twin flame]

Yea no problem

So we need to remove the green block. But to get the volume of the green block, the limits of integration are not the same as the red one

One minute again

[lucid peak]

also by revolving and making only the red bowl, should it look like this?

or this one

im guessing the former is the correct

[twin flame]

The second, the red curve is he outside of the bowl, the green is the inside

Sorry for that ambiguity

It's more of a cup than a bowl

With a curved bottom

[lucid peak]

Ouh, okay

Wait okay how do i know that btw sorry for not knowing

[twin flame]

Anyway, we need to subtract the green volume from the red volume

[lucid peak]

Ah yeah sure2

We’ll get to that l8r

[lucid peak]

so if the red one is 8, and green one is 0

then the volume's 8 pi?

8π

[twin flame]

The red volume integral is correct. The green volume integral is also correct but it shouldn't go from 0 to 4, but from the bottom of the green curve to 4

[lucid peak]

which is 2

gotcha brb

[twin flame]

Ye, so from 2 to 4

Oke

[lucid peak]

my bad lmfao

it's the y it's always the Y

it's 4!

[twin flame]

The volume?

[lucid peak]

and the integrated function of sqrt(2y-4)

ahah

4π it is

[twin flame]

Ye that looks right to me

[lucid peak]

lmao cool

okay going back to that question um, how'd u know which direction do the "infinite rectangles" as i call em go?

[twin flame]

Oke, so in a normal integral of a function of x, the the rectangles have a height given by the function and the infinitesimal side is along the x axis, hence the dx in the integral

So they 'stack' from left to right, if that makes sense

[lucid peak]

Yep ive seen the example

Of more exaggerated kind of look

Ah yeah but how about the up or down?

[twin flame]

So when we revolve these rectangles to get cylindrical discs, they again stack from left to right and they are centred on the x axis

If we want to get the rectangles with the infinitesimal length going up the y axis, then the length of the rectangle is no longer the y value of the curve but rather the horizontal distance from the y axis. This is in fact the height of the inverse function, so we find the inverse function and integrate this with respect to y to get the length to the y axis

In this case, the discs stack vertically, from the bottom to the top, and not from left to right

[lucid peak]

Ah wait holdon

So with the process of inversion weve made the rectangles lie on their long side…?

Horizontally*

Im a bit lost since u mentioned the cylindrical disks lmfao

No offense, im stupid dw

[twin flame]

No you are not stupid

I'll try and find a diagram

[lucid peak]

Ahahahsdd okay ill wait

[twin flame]

calculus - Why is the volume of revolution of a function f(x) the ...

Ignore the text above the image

I didn't write that

[lucid peak]

Oo they rotate on the x axis

[twin flame]

So the rectangles become discs like in this image when they are revolved

Ye

[lucid peak]

Fr i thought it’s the other way around

The Y acis lmao

[twin flame]

Yeah, this is for an x axis revolution

I will find one for a y axis

Revolution

[lucid peak]

Wait so the y axis is literally just that

[lucid peak]

But the rectangles go horizontally right?

[twin flame]

Yes, they go horizontally making horizontal discs when they are revolved

[lucid peak]

Aaand that’s what the inversion does

Unfortunately i gotta go now, thank you so much for an even more thorough explanation haha @twin flame

Aah yep got it now 👍🏻

[twin flame]

Oke, sorry for not finding the images in time, but it looks the same just vertically

Oke, nice

[lucid peak]

Lmfao it’s fine, still have it saved

[twin flame]

Nice ok

[lucid peak]

Iss nice, thank you so much again :)))

[twin flame]

No problem at all!

[lucid peak]

Ehe

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[devout snow]

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[rustic barn]

If I have P(c1 < T < c2), where T is a R.V. having some distribution then is it always true that this is equal to P(T<c2)- P(T<c1)? I know it's true if T is distributed normally.

For other cases, we have
P(c1 < T < c2)= P(T<c2)+P(T>c1)=P(T<c2)+1-P(T<c1)

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[high badge]

1. Your book club has 14 members. In how many ways can you select a president and a vice-president?
2. Your book club has 14 members. In how many ways can you choose two people to represent you at the national competition?

for the first one, it's 14*13, and for the second, it's 14 choose 2, right? but why is the second one 14 choose 2? does it mean the candidates are not different? because I thought the 2 people had to be different human beings anyways...

[low coyote]

You're confusing differentness and order

14 choose 2 means the order does not matter but they are 2 different people since they are two

14 choose 2 chooses subsets and sets have not repeated elements

There are 14·13 ordered couples and 14 choose 2 unordered couples

Both times are different people

inside couple

@high badge

[high badge]

Ohh I see, like the title president vs Vice President

[keen dust]

For the first one, a chosen set will may have meaning like <President, Vice> for each candidate. So a set <a ,b> and <b, a> have different means, which make them unique. However, in the second situation. A chosen set <a, b> and <b, a> have same meanings which make them repeat. That's my opinion.

[high badge]

Got it, thanks so much!!

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[craggy oxide]

Find the solutions of the form

[craggy oxide]

to the partial differential equation

I've done these types of problems some time ago but have forgotten, looked at the right answer and it's so confusing

[wicked rover]

plug the form into the PDE

[craggy oxide]

can you give some insight to what's going on there?

[wicked rover]

specifically what? theres a lot of ?s

[low coyote]

I think using r(x,y)=x^2+y^2 makes notation unnecessarily difficult

[craggy oxide]

why r^2 = x^2 + y^2 for example

why not just r =

[low coyote]

Ah

Anyway, i would simply use x and y

[craggy oxide]

and then where do these equalities come from?

rx, derivative of r with respect to x is 2x

rxx, derivative of r with respect to x two times is not 2r*rx

it's rxx = 2

[low coyote]

r²=x²+y² deriving both sides with respect to x we have 2r·r_x=2x

[craggy oxide]

rx^2 is (rx)^2 = (2x)^2 = 4x^2

ahh..

[low coyote]

you see?

[craggy oxide]

okay, sure but why r^2 = x^2+y^2?

why not just r

[low coyote]

To avoid square roots

[craggy oxide]

why would you have square roots though

[low coyote]

simply handy notation

Well, you're right, actually i don't know

Maybe it is for having 2r·r_x=2x so "2" cancells out

To avoid "2" and keep "r"

if r=x²+y² then r_x=2x

you don't get r

[craggy oxide]

why would you want r there?

[low coyote]

I dunno

But it's a possible reason

[craggy oxide]

thought the whole point was to just do the derivatives, then replace them with x and y

I just don't get the overall goal here

we solve for fxx and fyy

then put them into the equation

and now what?

usually solve for g, the function

which I think you first can do when you have isolated g'' from x and y and then take the integral until you get function g

but here she keeps the r

rip this test

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[restive river]

So basically I tried to subtract the area of quarter cricle from the area of semi circle, but its equal 0 and im going insane lmao. What should I do?

[zealous oyster]

How did you get 0?

Thats what i would’ve done

[restive river]

I Dont know really. I mean the area of semi circle is (π9^2)/2 and the area of quarter circle is (π(9 √2)^2)/4, right?

[zealous oyster]

Where did the sqrt of 2 come from

Its just divided by 4

[coral pecan]

Yes

The forumla for the area of quater circle is pi * r^2/4

[restive river]

Oh god

Okay thanks guys

[coral pecan]

👋

[restive river]

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[sweet minnow]

If you have a random variable defined by discrete values eg colours, is it acceptable to find expectation by assigning each colour a number?

[graceful cosmos]

The problem with that is that the magnitude of the number matters

If red = 1
And green = 2
Then green is somehow double of red?

[sweet minnow]

yeah thats what im confused by

[graceful cosmos]

Or, if we add that blue = 3
And then we decide 50/50 between red and blue, the expectation becomes green

Which is um

[sweet minnow]

can you just not take an expectation if the data isnt numerical?

[graceful cosmos]

There's other senses of "values we expect to get", like the median or mode.

But the expectation is not what you want here, I think

[sweet minnow]

cheers

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[restive river]

I’ve been trying to solve this specific integral problem using trig substitutions, I got to a solution but when I look at the answer on the textbook I got the problem from it shows my answer as wrong and I genuinely can’t find out where I went wrong with solving this

[restive river]

I can provide neater work if needed

[gentle estuary]

Is the integral $\int \frac{\dd x}{x\sqrt{a^2 + x^2}}$?

[woven radish]

castroploiin

[restive river]

Yes

Sorry the lighting kinda made it hard to see

I’m realizing I forgot to include the 2 in the natural log in the last few steps

It still doesn’t give me the right answer though

[gentle estuary]

I think u = √(a² + x²) is an easier substitution

du = x/√(a² + x²) dx ⟹ dx = √(a² + x²)/x du which makes the integral 1/(u² - a²)

[restive river]

Ohhh okay, then would I apply trig sub after that?

[gentle estuary]

It's a standard integral, but if you wanted to sub again, you can rewrite the above expression as -1/a² * 1/(1 - (u/a)²) and then substitute in t = iu/a

That gets you to another standard integral, but this time without an a in the integral.

[restive river]

I understand now and got to the solution I needed to, thank you!

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[restive river]

Assuming IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. If someone's IQ score is 85, what it their percentile?

I have no idea how to solve this so please ping me before explaining first :)

[supple knot]

use z score

[restive river]

I am quite new to it. Can you tell me what exactly is that?

And how does that relate to percentile?

[supple knot]

https://www.khanacademy.org/math/ap-statistics/density-curves-normal-distribution-ap/measuring-position/v/z-score-introduction

[restive river]

Thank you!

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[rare yoke]

How to prove by recurrence that 21ⁿ is congruent to 1+20n(mod 100)

[wanton star]

suppose 21^n = 1 + 20n mod 100, then prove 21^(n+1) = 1 + 20(n+1) mod 100

(but first check n=0 :D)

[devout snow]

@rare yoke Has your question been resolved?

[rare yoke]

Ik the method but i coudn't prove it

[wanton star]

what have you tried

[rare yoke]

21ⁿ×21= (1+20n)×21
=21 +20n×(20+1)
=21+ 20(n+1) + 20n

[wanton star]

how did you get to the last expression =21+ 20(n+1) + 20n

[waxen hazel]

I disagree. I’ve determined that the 20 shouldn’t be rounded to the nearest decimal, but it should be divided to the nearest ten because of the fact that its.

I think when we observe the patterns, it will display a lot of downsides of treating the theory of quantitatively is that it isn’t time efficient, as it’ll take up a lot of time just to find an inaccurate answer. I recommend we utilize the grens theory, as it shows a more accurate response.

[rare yoke]

it made me mad idk anymore
The question isn't hard but it is flying over my head

[wanton star]

well, the 20(n+1) should actually be 20n*20 = 400n = 0 mod 100

so you get 21 + 20n

[rare yoke]

Knew it was very simple lmao
Dk how i didn't find it

Thanks

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[wooden zodiac]

The volume enclosed by the surface ..... and the planes ....

[wooden zodiac]

how do u even do this

[devout snow]

@wooden zodiac Has your question been resolved?

[devout snow]

@wooden zodiac Has your question been resolved?

[devout snow]

@wooden zodiac Has your question been resolved?

[devout snow]

@wooden zodiac Has your question been resolved?

[wispy folio]

uhh

[wooden zodiac]

Yes

[devout snow]

@wooden zodiac Has your question been resolved?

[steel sage]

let x=rcos^3(u),y=rsin^3(u),z=v then dxdydz=3r^2cos^2(u)sin^2(u)drdudv

=3r^2cos^2(u)sin^2(u)(a-rcos^3(u)-rsin^3(u))dudr

=integral a^4cos^2(u)sin^2(u)du-(3a^4/4)(integral cos^5(u)sin^2(u)du+integral cos^2(u)sin^5(u)du)

There are existing results of integral cos^m(u)du and integral sin^n(u)du

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[sharp sinew]

for this question here

[sharp sinew]

why can't i resolve forces in the normal reaction force direction to find m

the highlighted stuff i guess

[steel sage]

The force given by the slope, I call it F
mgcos(30 degrees)+5gcos(30 degrees)=F, 5gsin(30 degrees)-mgsin(30 degrees)=0.5m
This is a system of two linear equations of m and F, surely you can solve for m

[sharp sinew]

is my answer for mass right?

@steel sage

[steel sage]

Okay I start calculating,wait a minute

Sorry I noticed that only the second equation was enough

[steel sage]

Correct

[sharp sinew]

are the highlighted equations right too?

because im confused as to why the highlighted equation didn't give the same mass value...

[limpid pilot]

Nice clean writing 😁

[steel sage]

No

[sharp sinew]

could you explain where i went wrong

[steel sage]

Should be the first equation in this context

You forgot the force given by slope

5gsin(60)-mgsin(60) is not zero, instead, 5gsin(60)+mgsin(60) equals the value of the force given by the slope. Which is irrelevant in this question though

[sharp sinew]

it says smooth slope

so surely there isn't any force given by the slope

[steel sage]

I am searching what it’s called in English

Not sure, support?

It’s smooth, doesn’t mean that it doesn’t have support

Okay found it: normal force

Not support

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[eager smelt]

.help

[devout snow]

Commands:
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Type .help <command name> for more info on a command.

[eager smelt]

Does someone know how to solve questions written at the top?

[eager smelt]

Question*

[devout snow]

@eager smelt Has your question been resolved?

[charred pebble]

its a quadratic equation so just use quadratic formula

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[tame sage]

help please

[tame sage]

i got 16/3 but its wrong

[pseudo basin]

how did you get 16/3?

show your work

[tame sage]

so i used the formula

BD=AD^2/DC

and using pyhtagorean theorm CD=3

so BD=16/3

[pseudo basin]

how did you get that CD = 3?

oh nevermind

was looking at the wrong thing

16/3 seems correct... is this answer getting rejected?

[tame sage]

yes

maybe answer key issuse can u help me with another one?

[pseudo basin]

you'll have to post the problem before i can tell if i can help you

[tame sage]

this one

\

[ruby zenith]

try the sine rule

[tame sage]

?

wuts a sine

[neon folio]

a trig function

[tame sage]

my teacher says trig is bad

[neon folio]

trig is something smart people use

who want to get to the solution quickly

[tame sage]

my teacher said trig is bad for the topic we ar elearning rn

[neon folio]

and more efficiently

[neon folio]

yeah

probably

)

[tame sage]

ok can u tell me what trig is exactly

[neon folio]

ok

its short for

trygonometric

[tame sage]

wuts that

[neon folio]

for ex

there is sine

cosine

tangent

cotangent

[tame sage]

this is already confusing

idk what does r

[neon folio]

youll learn

[tame sage]

ok

[neon folio]

u in class rn?

or what

[tame sage]

its hw

[neon folio]

then how did you talk to your teacher

about trig

[tame sage]

i didnt

i just wanted to learn it

before hand

since u said its easioer

[neon folio]

anyway

i dont really know how to solve it

[tame sage]

oh

[neon folio]

you can tag helpers

[tame sage]

ok

[neon folio]

since it ha been 30 min

[tame sage]

<@&286206848099549185>

[devout snow]

@tame sage Has your question been resolved?

[neon folio]

cmon cedric

[hollow night]

ffs im struggling with latex again lol

[neon folio]

haha

i know how you feel

what do you have in mind?

[hollow night]

wait nvm my method wont even work

[neon folio]

k

[devout snow]

@tame sage Has your question been resolved?

[trim flame]

ΔADC~ΔCDB~ΔACB, hence AD/DC=DC/BD
If we assume AD and DB are literally 4cm and 25cm, then we have 4/DC=DC/25 which gives DC=10
Then AC²=4²+10² and CB²=10²+25² and you can just divide AC by CB to get the answer

[tame sage]

i gto 4:25 its wrong

[trim flame]

4:25 is AC²:CB²?

[tame sage]

oh

so its

16

to

625

@trim flame ?

[devout snow]

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[trim flame]

AC²/CB²=116/725=16/100=4/25
AC/CB=2/5

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[tame sage]

oh ok

[tame sage]

.close

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[dusk dawn]

???

[jolly sandal]

there is no question ı just didnt understand what has to be done for the top to equal to the expression at the bottom

[gleaming socket]

Inverse function?

[jolly sandal]

off

the the question ask for the domain

wait ıll explain better

here is the full question

Im asking for the question b

I didnt understand how to express the domain properly

[dusk dawn]

is this a or b

the ms

[jolly sandal]

b

[gleaming socket]

For inverse functions you swap the y and x no?

[jolly sandal]

I have

that part is clear

[jolly sandal]

the answer is to b is this

[gleaming socket]

Oh then the range of the original function is the domain of the inverse function

[jolly sandal]

I just didnt understand the steps to lead the final answer

[jolly sandal]

is it posible to solve like that

[dusk dawn]

yes for sure

[jolly sandal]

ıll try

I couldnt do it

[dusk dawn]

ok let me try

[gleaming socket]

Well If you look at the inverse function, cos (π/4 - x^2) what is it's domain?

[jolly sandal]

I only struggled at that part

[dusk dawn]

okay what is the range of this

@jolly sandal normally

what is the range of the inverse cosine function

[jolly sandal]

is it 0<= x<= 1 / 2^1/2

is it clear ım sorry

[jolly sandal]

-1<x< 1

'nclued

[dusk dawn]

i think you mean pi/4

wiat

no im wrong

yeah youre right

[jolly sandal]

there is aarccos

[dusk dawn]

thats domain

but ok we're getting somehwere

[jolly sandal]

yee 😎

[dusk dawn]

huh thats strange

[dusk dawn]

shouldnt this be the ans

[jolly sandal]

whats that

ans

answer? for what

[dusk dawn]

nvm

im dumb

[dusk dawn]

how did you get the rhs part

[jolly sandal]

hmm I dont know how to write it from comp

so ill use paint wait

[dusk dawn]

i think youre overcomplicating it @jolly sandal

[gleaming socket]

So domain of the original function is (x/4) - arccosx < 0 right?

[dusk dawn]

nah its not that deep

[jolly sandal]

using paint or the question?

[dusk dawn]

okay

just in general

what is the smallest that can be inside the square root

[jolly sandal]

0

[dusk dawn]

great

and what is the largest that can be in the square root

[jolly sandal]

1

[dusk dawn]

that would be assuming arccos(x) = 0

[dusk dawn]

why?

recall that the range of arccos(x) is 0 to pi inclusive

so if you want to maximise whats in teh square root

you have to minimise arccos(x)

since its subrtracting

does that make sense?

[jolly sandal]

for it to be the least value it can be?

[dusk dawn]

no the maximum

you said minimum is 0

which is the lower bound of the range

[jolly sandal]

pi/4

[dusk dawn]

great

and then if you apply the square root

it would be sqrt(pi/4) right?

so you have your upper bound and lower bound of the range of f(x) function

which becomes your domain of the inverse function

done

[jolly sandal]

OHHHMYGOODD

I understand hs

thxx

[dusk dawn]

awesome :)

no worries!

glad to help

.close

[devout snow]

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[agile narwhal]

can someone help me to prove b) ?

[gritty parcel]

This channel is already claimed

#❓how-to-get-help

One of the available channels

[agile narwhal]

.close

[devout snow]

Channel closed

Closed by @agile narwhal

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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.

[mental spire]

I need help making an equation for radicals units using a graph

[mental spire]

would it be x^2

[gleaming socket]

Hmm there's only one x intercept right?

Do you know the asymptote here?

[mental spire]

No

I don’t

[gleaming socket]

What are the green lines in the graph?

[mental spire]

Ik that is but like the number

2?

[devout snow]

@mental spire Has your question been resolved?

[plush knot]

What are asymptotes?

@mental spire

[mental spire]

2

[plush knot]

Oh I was referring to the definition, but that's fine

There are 2 of them

[mental spire]

Yeaa

Green line

[plush knot]

Asymptotes are lines that our function never touches

[mental spire]

Yea

[plush knot]

So in our case, our asymptotes would be:
||x = -2||
||y = 2||

[mental spire]

Ok

How does it help with the equation

[plush knot]

They are really useful, do you remember the formulas for asymptotes?

[mental spire]

He never gave us any formulas

[plush knot]

Oh, ok

Let me try something else real quick then

[mental spire]

Ok thx

[plush knot]

Ok!

Are you familiar with translations?

@mental spire ?

[mental spire]

Umm kinda

[plush knot]

Great!

So, this is a homographic function

And, if it's center is at the origin (In our case it isn't!), it's equation looks like
y = k / x,
where k is a constant

[mental spire]

What is k?

On the graph

Or I have to find

[plush knot]

We'll find it later

follow me for now

Our center is the intersection between the asymptotes

In this case, the center is (-2; 2)

Do you agree?

[mental spire]

Yea

[plush knot]

Ok!

If y = k/x is used when our center is at O(0; 0), we can translate it to where our center currently is

So we have a function that represents our center and then we can look for k!

[mental spire]

So the purple lines would move

As well?

[plush knot]

Wait, i'll make a graph real quick

[mental spire]

I

K

[plush knot]

The purple function is the one we are looking for

The green one is a function in the form y = k/x, because it's center is at (0; 0)

So we can write our purple function as a translation of the green one

Do you agree?

[mental spire]

Don’t we need a point

To write it

On the purple or green

[plush knot]

yes! To translate the green function to get the purple one you need a point

The center is really good in this case

[mental spire]

It won’t be exactly

A value

On the y

[plush knot]

What do you mean?

[mental spire]

The green line

For a point

U said middle

[plush knot]

where's the green function's centre?

[mental spire]

-3,2.something

[plush knot]

No no

not it's vertex

it's center

The intersection of it's asymptotes

So that would be (0, 0)

[mental spire]

Oh#

So u move the

Black line ?

[plush knot]

Wait

you can't move asymptotes without moving all the function

The generic form of an homographic function with center in (0, 0) is y = k / x

We are looking for an homographic function with center (-2, 2)

So we can just translate it

[mental spire]

Ok

[plush knot]

Ok!

So, to translate a function:
x → x - (x of the point of translation)
y → y - (y of the point of translation)

So in our case, translating y = k/x by (-2, 2), how would it become?

[mental spire]

X-2 and y-2v

X+2

[plush knot]

Yeah!

So,
(y - 2) = k / (x + 2)

[mental spire]

So we find k?

[plush knot]

Yes! this will hold true for any point of our function

Pick a given point and plug it in, then solve for k

[mental spire]

1.5 and 0

[plush knot]

plug them in and find k

[mental spire]

How would I do

K/(1.5+2)

Added them and move ?

[plush knot]

Let's solve it together

[mental spire]

Ok

[plush knot]

$y - 2 = \frac{k}{x+2}$

[woven radish]

Nonna