help-0

[mellow reef]

wait

no

12/25?

wait no

still wrong

[rapid nova]

how many of 17 students that study physics also study chemistry?

[mellow reef]

13/25 * 17/25

[rapid nova]

that's c) btw lol

[rapid nova]

i'm not asking about the probabilities

let's focus on the number of such students

[mellow reef]

i dont know how much

[rapid nova]

we already know how many study both, don't we?

.

[mellow reef]

yes

[rapid nova]

so?

if there are 11 students that study both physics and chemistry

how many of 17 students that study physics also study chemistry?

[mellow reef]

11

[rapid nova]

yes

here you get c)

it's 11/25

now

how many of 17 students that study physics DO NOT study chemistry?

[mellow reef]

17/25 - 11/25?

[rapid nova]

just 17 - 11

[mellow reef]

so 6

[rapid nova]

and the probability then is 6/25

yes

[mellow reef]

oke

what about b

[rapid nova]

as for b), we also know the answer already

it's 19 students that study something

so 19/25

[mellow reef]

ohhhh

thank you

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[alpine sable]

I'm unsure if b needs to be updated or if it's still -1, 1

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@alpine sable Has your question been resolved?

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[median oar]

hmm, i just did a test, and one of the questions is peculiar (definition wise)

in class we said that we will call something a vector space if it is "closed under vector addition and scalar multiplication, and contains the 0 vector" and for something to be a subspace, it needs to be a vector space with the above definition within (or be itself) another vector space

[median oar]

one of the questions was asking if x = {(x_1, x_2, x_3) | x_1 ≥ 0, x_1, x_2, x_3 ∈ ℝ }

was a subspace

now according to what we define a subspace as in class it would seem like that is a subspace

but the axioms of what a vector space is includes having an additive inverse

which, this obviously does not

is there a way to glean this "additive inverse" property from just the definition provided?

[mortal trellis]

-1*v

also your definition forgets about all the commutativity, associativity etc stuff

[median oar]

those are definitely 4 characters to answer my question

and yeah we kinda just assume it's true?

it's just a first year first sem course so i suppose they aren't too rigorous with the definitions

and thank you

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[mortal trellis]

for subspace these three things are enough

because you know that all those other stuff hold in some bigger space

[median oar]

yeah we just defined vector space a bit weird

[mortal trellis]

R^n or space of all functions or something like that

[median oar]

but i guess if you have something that's a subspace in a vector space, and if those 3 conditions hold

then everything else is inherited from the parent vector space

[mortal trellis]

yes

[median oar]

👍

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@alpine sable Has your question been resolved?

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[fresh perch]

guys how do you get better at reading "tricky" problems?

[fresh perch]

I feel like half my issue is I don't know exactly what the restrictions of a problem are

https://i.imgur.com/N0Va6kR.png

for example here, I don't know what exactly left to right means

in the picture C is slightly to the left of B so is ACBD valid?

if its alphabetical is ABCD valid?

and if neither of the above are valid aren't the only ways you can go just ABD and ACD themselves?

[whole shell]

@fresh perch i think

you can only use

those paths youve drawn

[fresh perch]

that's not my drawing

[whole shell]

not you

as in

the question

so youre only expected to go ABD and ACD, but there are multiple ways to go ABD

and same with ACD

[fresh perch]

oh

each arrow

is a path

[whole shell]

yh

[fresh perch]

...

now it seems obvious

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@ivory pendant Has your question been resolved?

[alpine sable]

it wants you to simplify this or?

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[vale tiger]

Sarah wants to increase her mean average to 75%. What grade would she need to have in her extra (10th) class to achieve this?

Current Mean: 74.333 continued

[subtle birch]

Calculate

Sum

And then number of values

Clearly there are 9 values

We need 8

But to average 75 at 8 value the sum need to be 75×8

So the value you need to discard is just sum of those values-75×8

[vale tiger]

this is what I got

To find the grade Sarah would need in her extra (10th) class to achieve this, we can use the formula:

(sum of all grades + x) / 10 = 0.75

where x is the grade she needs to get in her 10th class to achieve a mean average of 75.

We know the sum of her current 9 grades is:

sum of all grades = 98 + 100 + 65 + 78 + 98 + 35 + 100 + 45 + 50 = 669

To achieve a sum of 600 with 8 grades, we need to discard one of the grades. Let's discard the lowest grade, which is 35.

So the sum of the remaining 8 grades is:

sum of all grades = 98 + 100 + 65 + 78 + 98 + 100 + 45 + 50 = 634

Now we can plug this into the formula and solve for x:

(634 + x) / 9 = 75

Multiplying both sides by 9 gives:

634 + x = 675

Subtracting 634 from both sides gives:

x = 41

Therefore, Sarah would need to get a grade of 41 in her 10th class to achieve a mean average of 75% with 10 grades.

,close

.cloe

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[white junco]

i need help with this question

[white junco]

so what im doing is

counting the number of numbers divisible by 3, which comes to be 300, the number of nubmers divisible by 4 which comes to be 225, then removing 2 times the number of numbers divisible by 12 since we need by either 3 or 4 , 2 times because we counted each of them twice and we don't need even once, now that we have removed nubmers divisible by 12 ( which means both 3 and 4 ) completely, we don't even need to remove those divisible by 48 as they are already removed!. so the answer according to me should be 325

but the solutions available on the internet are quite weird

and my answer ain't in the options even

[median oar]

why 2 times the number divisible by 12

[subtle birch]

Why would you remove it twice

You need to remove it only once

[lime elbow]

Isn't the answer 507?

[white junco]

no

[subtle birch]

Wait lemme try

[median oar]

by removing it twice you've removed 12

well

12 isn't a 3 digit number

[white junco]

see, we need numbers divisible by 3 or 4, so like we need to do A union B - A intersection B, whichi clearly is A + B - 2 times A intersection B

[median oar]

let's say 132

nope

A + B - 2 (A and B) is A xor B

A or B but not both

[white junco]

isn't that what we need?

[median oar]

nope

[white junco]

divisble by 3 or 4 but not both

[median oar]

we want to count numbers like 132

[white junco]

why would you do that

[median oar]

that is divisible by both 3 and 4

so we should count it

[white junco]

the question says either 3 or 4

clearly

[median oar]

either 3 or 4 means or

not xor

[median oar]

that is what it says

either 3 or 4

[white junco]

so we can include those divisible by 3 and 4

oh god thanks

[median oar]

that means divisible by 3 OR divisible by 4 OR divisible by 3 and 4

[white junco]

this is not very intuitive tho

[median oar]

it's implied

[white junco]

cuz when you say either 3 or 4, it feels like you don't want by both

[median oar]

it would say "that are divisible by either 3 or 4 but not both"

[median oar]

no if u want to say not both, you use xor or exclusive or

[white junco]

yea i gotchu

thank you!

[median oar]

or ⊕

[subtle birch]

Is it 432

[white junco]

yes

now it makes sense

[subtle birch]

Voohoo

What i did was find numbers divisible by 3 and 4 and the subtract divisible by 12 to get 450 then i just made a guess to nearest option

[white junco]

lmao

that's funny 😂

[lethal remnant]

You ignored the neither-this-nor-this case in that process.

[white junco]

what?

we didn't count neither this nor that numbers in the first place

[lethal remnant]

I think i misinterpreted it.

Seems like beard's answer should have come 0 if he did that

That's really 🤣

Funny

I'm solving this question. I think I can solve

[white junco]

yes it is easy

if u know eitehr a or b is or and not xor

[lethal remnant]

Well, that's not my process

I just don't get why I'm getting the answer 375

And it truly doesn't seem like I have done something wrong

Now, I need help with this question 🥲

[white junco]

what are you doing

[lethal remnant]

Vro I think the options are wrong. But I'm about to say a question of JEE-MAINS wrong. God save me🤧

Actually I'm using the LCM method.

[white junco]

no, optiosn are right

bros writing 4 an hr now

🙂

[lethal remnant]

From 1 to 12 there are only 5 numbers which are divisible by either 3 or 4.
The extreme 3 digit multiples of 12 are 108 and 996 (12×83).

Between, 96 and 108, there are 4 numbers (100, 102, 104, 105) that follow the condition and in between 996 and 1008 there's only one number (999) . So beyond these extremities, only 5 numbers are there. I'll add 5 later on.

I'm doing 83-9=74, (83 because 12 times 83 is the largest 3 digit multiple of LCM (12)) and then 74×5=370 because it'll count all the numbers that follow our condition. In this process, we don't need to subtract the number of numbers that are not divisible by 48 because we're not counting the multiples of LCM (and multiples of LCM (12) will entail the multiples of 48)

So, 370+5=375

Tell me, where I'm wrong?

Answer quickly, 'cause I wanna sleep 😴

And this question won't let me sleep

[white junco]

idk why ur complicating this much

[lethal remnant]

It's not complicated.

It's complex

In my imagination, I did it in a minute. It seems complicated in a para

[white junco]

just find number of numbers divisible by 3 (= 300), by 4 (=225) and then subtract those by 12 ( = 75) and then those by 48 (=18)

[lethal remnant]

No

[white junco]

yes

[lethal remnant]

I wanna know where I'm wrong

[white junco]

i gotta finish off goc before 12

urs is very unclear

[lethal remnant]

Goc?

I don't think so

[white junco]

why r u even considering nubmers like 96

we have to find for 3 digit numbers

[lethal remnant]

I'm considering but not counting it

96 is the multiple of 12 and numbers between the multiples of LCM follow the same pattern

[white junco]

what conditions do 100 to 105 follow

what r u talking about there

i don't know wha tu r doing

nubmers divisible by 12 should be counted

but those by 48 shouldnt

i gotta go now

[lethal remnant]

100 (4)
102 (3)
104 (4)
105 (3)

Bracketed numbers are the factors

[lethal remnant]

It won't be counted because it's either divisible by 3 or 4 and 12 is divisible by 3 and 4 both

[white junco]

why 74 * 5?

[white junco]

that's what's the key

we have to count by 3 and 4 both

cuz it's or which means union

[lethal remnant]

74 × 5 because there are seventy four 3 digit multiples of 12 and 5 is the numbers between any two multiples of 12 that follow our condition. You can think this through tomorrow. I'll also

[lethal remnant]

'Either' doesn't include 'and' afaik

[white junco]

in the math notation it does

a or b is clearly a union b i repeat

the either is what confused me and this help channel was started

now it's clear

[lethal remnant]

Well, you are not wrong. OR does mean union

But this question is of number system

And not sets

[chrome salmon]

I thought operational reasearch

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[lethal remnant]

So let me proceed further, if I add multiples of 12 in my counting, it'll become 375+74=449 and subtracting the number of multiples of 48 (which is 17) will give me 449-17=432, which is exactly what your supposed answer is. So, technically speaking I did it right

[lone heart]

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[lethal remnant]

For me, this topic isn't closed yet. Ping me if someone wants to discuss.

[lethal remnant]

And now having been seen this, I outright disagree that the answer is not 375. Sorry guys 🥲🤌

[subtle birch]

sure

[lethal remnant]

So, say

[subtle birch]

where are you stuck?

what is the thing you can not understand

I just used AP and though it is lengthy the process is too easy

[lethal remnant]

First tell me, this question is from which topic?

[subtle birch]

does that even matter?

[lethal remnant]

I believe it's from number system.

[subtle birch]

Number Theory

[lethal remnant]

I think it does

[subtle birch]

pretty much not

anyways proceed further

[lethal remnant]

I'm just going with the English. Either 3 or 4 means, we'll exclude numbers like 12 and count numbers like 3,4,6,8 and 9. I know or means union in maths but isn't that capital OR? (May be this is a stretch from my side).

[subtle birch]

No

You are absolutely wrong

either 3 or 4 means it is divisible by 3 or 4 or both

12 is counted as it is divisible either by 3 or 4

[lethal remnant]

In maths

[subtle birch]

we do not go further to assume that it shouldn't be both

[lethal remnant]

But 12 means 3 and 4 both

[subtle birch]

the numbers go like 3,4,6,8,9,12,15,16,18,21,24 and so on

[subtle birch]

yeah so what

it is divisibe by either of 3 or 4

we don't just exclude it

[lethal remnant]

It is divisible by both 3 and 4

[subtle birch]

yeah so?

[lethal remnant]

In inequalities, either or is used only when one of them is possible.

[subtle birch]

bru

[lethal remnant]

But not both

But we're somehow talking about sets

[subtle birch]

Here just common sense is to be used and not that inequalities stuff

we are not dealing with those

[lethal remnant]

I'm just saying

Because in maths we use either or differently in inequalities and differently in sets

[subtle birch]

there are some rules for inequalities doesn't mean we can use them anywhere

You're just being dumb

[lethal remnant]

I think you're creating a strawman

[tacit arch]

what math help do you even need

[tawny crown]

They don't seem to be needing help atm

[tacit arch]

great

@lethal remnant do .close

[lethal remnant]

.close

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[lethal remnant]

I know. Don't stress because of me.. I'm kinda stubborn on my beliefs

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[alpine sable]

?

this is =

[ivory pivot]

$=\sqrt{\leftroot{1/12}}$

[subtle birch]

yes

[lethal remnant]

Yes

[alpine sable]

we need to multiply 3 by 4 when we have this ?

[lethal remnant]

Yes

[alpine sable]

ok ty

[quasi scarab]

u can rewrite them to exponents

1/3 and 1/4

and a^b^c=a^(b* c)

[ocean seal]

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

[alpine sable]

.close

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[junior condor]

Prof decided that topology midterm would be an introduction to order theory question: if > is a partial order on X x X, where X is a topological space, we call X the carrier of the partial order. Does X need to be hausdorff?

[junior condor]

Plz say yes otherwise I lose 10%

I BEG MY LORDS

[rapid nova]

what 10%

is it a test?

[junior condor]

the exam I just handed in

[rapid nova]

we don't help with exams here

[junior condor]

It’s doomed

[limpid turret]

If it was already handed in, I believe it's fine

[junior condor]

? It’s now a past exam

[rapid nova]

uh, missed that part, sorry

[limpid turret]

You may have better luck if you ask in #point-set-topology. Also, try to sound less crestfallen. ❤️

[junior condor]

Bruh I crank through problems and then find out it’s no use forgot prof does research in order theory

Normal topology problems*

[worn fox]

From a Google the answer is yes

[crystal tiger]

Thank god

Bruh

Are you in my class

Dm me

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[wintry valve]

just need some help and pointers not to sure what I'm doing

[granite wing]

just generally, please add arrows on the axis in coordinate systems. additionally i think the red graph is a square too late at full speed.

for your question, we need the point where the graphs intersect if im interpreting the question correctly. in class, do you solve problems like this by calculating or just by looking at the graph ?

[wintry valve]

I think we did both ways

could you run me through both methods just the steps would be very helpful

[granite wing]

well graphically, just draw as precise as you can and give your best guess where the graphs intercept.

you know that you calculate the intercept by setting the functions equal right ?

so e.g. f(x)=g(x)

[wintry valve]

yup

[granite wing]

then we can notice that train a has a constant graph at the intercept, since it stays at the same height

[wintry valve]

aah I think I've got it now :)

thank you so much

[granite wing]

now you just have to get an expression for train b and you can calculate the intercept

[wintry valve]

tysm .close

.close

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[rapid kettle]

can a hyperbola intersect with its conjugate hyperbola ?

[limpid turret]

what are the asymptotes of both hyperbolas?

[rapid kettle]

no i am asking in general can they intersect or not ?

like is it possible ?

[limpid turret]

Consider the answer to this then

[rapid kettle]

actually i have not done asymptotes very thoroughly

[limpid turret]

It may be worth looking into then

[rapid kettle]

okay ill do so

thankyou

.close

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[lone heart]

@dusky shuttle Has your question been resolved?

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@dusky shuttle Has your question been resolved?

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@dusky shuttle Has your question been resolved?

[gusty gorge]

there's no interpolation

you're literally just adding width of interval * f(x_1) + width of interval * f(x_2) + ...

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[kindred sleet]

A non-linear second order differential equation with a constant e that is not the Euler’s number. Solve for discrete values of e such that psi, a function of u, does not diverge near infinity

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@kindred sleet Has your question been resolved?

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[south cove]

how would i find the distance between the origin and a point tangent to the circle connecting the origin?

[south cove]

i have no idea where to start with this

[slate monolith]

system of equations

with tangent line

solve for x and y

then pythagoras

[south cove]

How would such a system look like?

[slate monolith]

circle = tangent line

actually

yeah

i think that should work

mx = circle

[south cove]

well theres too many unknowns

mx = (y-8)^2 + x^2

[slate monolith]

do you know the x

[south cove]

nope

[jagged cobalt]

could construct a triangle with vertices being the center of the circle, the origin and the point on the tangent line. The side connecting the tangent line and the center has the length of the radius. and is perpendicular to the tangent line, thus creating a right angle. then the side connecting the center and the origin is obvious
and then use pythagoras

[south cove]

Yep it worked

i need this sort of mathematical thinking ability

thx

[lone heart]

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[rose forge]

im habinh a lot of trouble with this question

[slate monolith]

this is just memorization

[last ether]

Inscribed angle theorem

The inscribed angle is half the intercepted arc measure

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✅

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[lapis prawn]

Hello need help with volume and surface area of a very odd 3d shape

[lapis prawn]

I don’t know if there is a name for it but I will describe it when someone is here

[fierce prairie]

maybe send a pic

or describe it ig

[lapis prawn]

I can’t find any lol

[fallen verge]

is it described using equations?

[lapis prawn]

No visually sorry this is not very advanced math

Basically

Half cylinder wrapped around a sphere

[fallen verge]

do you have an image of the question?

[lapis prawn]

It is my personal idea I want to incorporate into my project

Imagine the server picture

But a sphere in the middle that overlaps to a certain amount

[fallen verge]

ok

why do you want to find the volume? perhaps it will help us understand a little better what you mean

[lapis prawn]

Creating a chess piece. I have a sphere as one component of the piece. I want to wrap a half cylinder around it as if it were a headband

If that makes sense

[fallen verge]

ok

[lapis prawn]

But the half cylinder will not fit snug against the sphere without subtracting some of its volume

I just don’t know what that shape is called

[fallen verge]

i dont think it has a name

[lapis prawn]

Perhaps if I find the volume of the cross section of the sphere that overlaps into the cylinder and subtract it from the cylinders equation?

[fallen verge]

so if im understanding this right, you want something like this

[lapis prawn]

Yes 🙂

[fallen verge]

im assuming you can use some calculus here

[lapis prawn]

So pretty much the shape in the server picture actually

What is that one called

[fallen verge]

the "donut" shape is formally known as a torus

[lapis prawn]

Ok ok yes

[fallen verge]

this i guess you can call it a "lune torus" but i looked it up and no such thing exists

[lapis prawn]

Torus with a sphere overlap in its center

[fallen verge]

so you can create a solid of revolution to find the volume of the solid

[lapis prawn]

Would you be able to walk me through this?

[fallen verge]

sure

for the sake of making numbers easy, im going to making the sphere have a radius of 17, and the torus part have a thickness of 16

[lapis prawn]

Ok

Would thickness be diameter of the torus?

[fallen verge]

so i can graph this shape like this

do you understand where i got that?

[lapis prawn]

I understand but I need to write it down one moment

Sphere radius extends to the torus center. ✅

Exactly what I’m going for yes.

[fallen verge]

when you mean "torus center" do you mean that if you set the torus on a table, the circle that touches the table is the center?

[lapis prawn]

I mean the center of a torus cross section sorry

I assume that’s incorrect

[fallen verge]

no i think we're on the same page

[lapis prawn]

Ok cool

I understand the current values

[fallen verge]

so now we need to write equations for the two circles

[lapis prawn]

Pi17sq and pi8sq?

[fallen verge]

they would look like this

i extended the circle since we only do the integral on the part from -8 to 8

[lapis prawn]

Ohhhh I now understand the graph

Perfect

[fallen verge]

do you know how to do the washer method for a solid of revolution?

[lapis prawn]

No

[fallen verge]

do you know how to find the volume of a solid of revolution?

[lapis prawn]

Ahh is this not practical to do without knowledge on calculus

😦

[fallen verge]

yeah sorry

if you dont mind, i can finish explaining this and you can come back to this when you are ready

[lapis prawn]

That is so sad I got too ambitious with my project

Yes go ahead actualu

[fallen verge]

i can explain a little bit about how to get a solid of revolution, but i suggest you learn a bit of calculus to understand the notation

[lapis prawn]

Yes due to circumstances I wasn’t able to continue my courses in advanced maths so I’m no longer able to do programs involving this sort of thing

I’m just trying to do as advanced as I can

I’m extending myself far beyond what I’m expected

[fallen verge]

heres what it would look like

[lapis prawn]

But I will still take note of this for later

[fallen verge]

essentially you have a larger circle, then you cut out a smaller circle

creating a "washer" shape

then you do that for the bounds which are in between -8 to 8

those are the intersections of the two shapes, the sphere and torus

[lapis prawn]

So what are the names of the processes used in this equation ?

[fallen verge]

solids of revolution

but i suggest learn about how to do derivatives and integrals before this

at this point, i dont think its possible to calculate by hand so im going to toss it to wolfram alpha

[lapis prawn]

Yes I will extend myself down the road but this is good to realise for now

At the end of the day I essentially wanted to know if it was within my ability and to know what processes are required to solve something like this

[fallen verge]

,w pi*fnInt((sqrt(64-x^2)+15)^2-(289-x^2),x,-8,8)

[lapis prawn]

I couldn’t find anything on what I was looking to do

[fallen verge]

oh boy wolfram is taking a while

[lapis prawn]

If I were to describe the task properly what would it be described as?

What exactly am I finding

[fallen verge]

finding the volume inside a torus, but outside a sphere

[lapis prawn]

Ok this is very helpful thank you

[ocean seal]

Results provided by WolframAlpha

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

Requested by GarlicBredFries#2110

[fallen verge]

oh finally lol

good luck with your studies!

[lapis prawn]

Is that result meant to be the total volume of the remaining torus?

[fallen verge]

yes that is the desired volume

fun fact: its exact value is 960(pi^2)

[lapis prawn]

Very interesting stuff is this advanced or around basic calculus

For a grade 11 advanced student… ?

[fallen verge]

this is still basic calculus

[lapis prawn]

Ok good to know it’s not far out of reach ahahaha

[fallen verge]

you can understand how to do something like this with knowledge of AP calculus and some creativity

if youre not in the US, its like usually covered in a Calculus 1 class

[lapis prawn]

Including solids of revolution?

[fallen verge]

solids of revolution are usually a calc 1 topic

[lapis prawn]

Nice to know thank u so much for this information the wolfram equation the entire equation for the answer correct?

[fallen verge]

yes

it should be

[lapis prawn]

Is there a way to save help transcript or should I just screenshot?

[fallen verge]

#help-0 message <- this is a link to the beginning of this conversation

you can save this somewhere and go back to it whenever you want

[lapis prawn]

How can I save it

Ohhhh

The link

Got it

Wont these chats be deleted?

[fallen verge]

they will not

the history of this chat should exist until forever

[lapis prawn]

Alright thanks heaps for that

[fallen verge]

no problem

[lapis prawn]

Keen to calculate the remaining surface area of the torus orrrr?

[fallen verge]

i forgot how to do surface area lol

there are formulas though

[lapis prawn]

Ok ahahahah all good, this is good stuff and I appreciate it I’ll close it now.

[fallen verge]

ill leave it up to you to figure out the surface area at some point

good luck!

[lapis prawn]

Wait one second,

[fallen verge]

oh

[lapis prawn]

If you were to wrap a half cylinder in a similar way around another cylinder, would there just be simple addition and volume calculations since there’s no curvature to subtract

Correct

[fallen verge]

you still have to subtract the inner volume

but its not curvy so the equation is easier

[lapis prawn]

Oh yeah I mean like it’s already a calculated half cylinder

So 1/2 cylinder with length of the circumference + regular cylinder

[fallen verge]

i dont think that will work

[lapis prawn]

Why’s this

[fallen verge]

think about a really thin cylinder

the outside would go around a greater distance than the inside

[lapis prawn]

There will be a gap where the cylinder is meant to meet won’t there

[fallen verge]

not a gap

it would overcount

[lapis prawn]

Is it another calculus question?

[fallen verge]

yeah you would probably have to use calculus

[lapis prawn]

Damnnnmmm I was keen to do that

[fallen verge]

fun fact: these shapes are called "napkin rings"

here is a derivation of the volume

[lapis prawn]

So u can’t (1/2 2pi r x circumference)

[fallen verge]

(credit: MAA)

apprently, you get a formula without needing calculus, but you need calculus to understand why it works

vsauce has a video about this:

oops didnt mean to send that again

https://www.youtube.com/watch?v=J51ncHP_BrY

[lapis prawn]

I don’t understand there is a rectangle inside the circle

Why

There shouldn’t be spherical caps

[fallen verge]

verical cross section

if you spin the shape around the y axis, the two top shapes make spherical caps, the left and right connect to make the napkin ring

[lapis prawn]

Could you visualise this?

[fallen verge]

i took this from the video

[lapis prawn]

No I understand

That makes sense now

The rectangle is the cylinder you can see through the sphere

[fallen verge]

yeah

[lapis prawn]

Wow this is really cool

It’s a shame I wasn’t able to continue my advanced courses

Basically removing the caps left after the cylinder leaves the caps that are revolutionary

Ok this last thing will clear up my last confusion

Ohhhh hold on

Once again I find my original question involved the volume of a torus but outside a cylinder

But I like the napkin ring

Because it has a non calculus formula

[elfin geyser]

🚶

[lone heart]

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[merry herald]

need help with this one

[remote heron]

write f(g) and g(f)

[merry herald]

wait i think i got it 🙂

.close

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[stoic cloud]

What is being utilized to find c?

[lone heart]

@stoic cloud Has your question been resolved?

[vale wigeon]

.close (discussion cont'd in another channel)

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[limpid dome]

rationalise the denominator (with steps)

[late agate]

time both numerator and denominator by root 5

@limpid dome

this is a different question

but apply the same idea with your one

[vale wigeon]

!nosols

[lone heart]

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.

[south trench]

i see

[limpid dome]

thats what i got but the answer in the text book is different

this is the answer in the textbook

[late agate]

they probably simplified the surds

show us what you did

[limpid dome]

how did they get this if they simplified?

[fallen verge]

looks like the denominator was mistyped

should be a 2

thats the only way the answer key works, otherwise you are correct

[late agate]

no that makes no sense, how would he have gotten the 2sqrt3

wait no

I'm dumb

[south trench]

yes it makes sense he is right

[late agate]

sqrt 12

bla bla bla

[south trench]

:3

[limpid dome]

thx

[fallen verge]

i suggest talking to your teacher

[vale wigeon]

is this pink writing yours or somebody else's?

[lone heart]

@limpid dome Has your question been resolved?

[lone heart]

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[tender dagger]

Is this just 0 :(

the only singularity is at z = 0 and the residue should be 0 there

[lone heart]

@tender dagger Has your question been resolved?

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@tender dagger Has your question been resolved?

[raw cairn]

there is a hint in the problem to use the Laurent series, have you tried to expand it?

or can you tell me the order of the pole at z = 0?

[lone heart]

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[desert oracle]

Can someone derive it, please?

[solar crown]

what r u confused ab?

[desert oracle]

I just don't know how to derive it.
I guess I don't know enough strategy for it.

Sorry, my native language is German, not English.

If you post an answer, please ping me.

[solar crown]

i don't think giving you the answer is helpful here :)

try using chain rule?

[solar crown]

you could rewrite it like

and then go from there, apply the chain rule

[desert oracle]

So? @solar crown

[lone heart]

@desert oracle Has your question been resolved?

[lone heart]

@desert oracle Has your question been resolved?

[solar crown]

yeah that's on track

[desert oracle]

Do I still have a mistake?

[solar crown]

i dont think so

[lone heart]

@desert oracle Has your question been resolved?

[lone heart]

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[dawn topaz]

I wanna practice radical equation
Gib questions

[ocean hawk]

These threads are for you to ask a question

[gray isle]

https://discordapp.com/channels/268882317391429632/908075848752590918/1088452662695051405

[dawn topaz]

...

[gray isle]

radical equation
is also vague

[dawn topaz]

):<

.close

[lone heart]

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[proper sky]

im not sure how to go about doing these

[copper rock]

the first one is actually a quadratic in disguise

use the fact that 9 = 3^2

[proper sky]

okay

let me try

[copper rock]

yes that’s good

can you find a way to “bring out” the square

and make it a quadratic

[proper sky]

is it a law of indices that im supposed to use?

[copper rock]

exponent laws

idk it might be called law of indices actually

oh yes

[proper sky]

Oh so wait

i gotta start memorizing those laws

[copper rock]

you’re going in the right direction

[proper sky]

Now i substitute right

[copper rock]

yea kinda

just make it a (3^x)^2 first

btw remember that

https://media.discordapp.net/attachments/1066433340921626654/1088458961478492205/348514594064695306.png

[proper sky]

so ill set y = 3^x

[copper rock]

yep

[proper sky]

im kinda stuck

[copper rock]

you can solve for y now

[proper sky]

now do i sqrt

[copper rock]

won’t work

it’s a quadratic so you need to use the quadratic formula

[proper sky]

Ohhhhh

I completely forgot

i think thats all

[copper rock]

well that’s what y is equal to

but you need to solve for x

[proper sky]

im stuck

[copper rock]

y = 5 or -2

but y is 3^x right

so 3^x = 5 or -2

[proper sky]

Do i have to use logs

[copper rock]

yup

[proper sky]

and i have to do the same for -2 right

no wait

you can't log a negative number

?

[copper rock]

yep

so theres only 1 solution

[proper sky]

Okay cool

Are you only able to do one question or do you wanna do the second one

[copper rock]

we can do the second one

have u tried it

[proper sky]

yeah on my test but i compelety had a brain melt

[copper rock]

ah rip

well maybe retry it now

[proper sky]

b)

that was my test

[copper rock]

ah i see

okay

how can you get rid of the 5^x on the right hand side

[proper sky]

Uh what about

can you multiply 125 x 5 then add the powers

would the bottom numbers have to both be the same

[copper rock]

the bases yeah

try it

[proper sky]

try what

[copper rock]

try turning 125 into 5 ^ something

then adding exponents

[proper sky]

Ohhhh

5^3

So 5^x+3

[copper rock]

yes

[proper sky]

jheez

[copper rock]

other way u couldve done

is divide both sides by 5^x

[proper sky]

yeah i was just about to say that

[copper rock]

here u can log base 5 both sides

[copper rock]

the left hand side here should be 5^4x * 5^-3

[proper sky]

Why

[copper rock]

cause 5^a * 5^b = 5^(a+b)

so 5^4x * 5^-3 = 5^(4x-3)

if it were a division itd be
5^4x / 5^3 = 5^(4x-3)

[proper sky]

What if i did

Yeah that

Can that work as well?

[copper rock]

yea

but it’s probably easier to do it by applying a log base 5 on both sides here

https://media.discordapp.net/attachments/490557019623915520/1088470445537366056/rn_image_picker_lib_temp_6d030cac-ab6e-4df1-a78d-80b2112993b7.jpg

[proper sky]

Yeah im doing this way instead rn

[copper rock]

alright

[proper sky]

i love the profile pic 😂

[copper rock]

perfect

[copper rock]

thanks lol

[proper sky]

What grade math do you think this is

[copper rock]

idk how grades work in ur country

here we learn this when were around 17

[proper sky]

i mean thats fair

im in grade 11

[copper rock]

that’s 16-17 year olds right

[proper sky]

Yeah i just turned 16

[copper rock]

ah and you’re in the us i assume

[proper sky]

Im Canadian but im live in the Caribbean

Grenada

I have an exam in about an hour that i not confident about at all

[copper rock]

that sounds cool

[copper rock]

oof

[proper sky]

Yeah its nice

[copper rock]

it explains the profile pic

[proper sky]

Oh yeah lol the island is pretty beautiful

Do you get paid for this?

[copper rock]

nope

doing it cause i like helping with math

and i’m procrastinating on other things lol

[proper sky]

Oh well in thank case thank you for your time

[copper rock]

you’re welcome

make sure you’re very familiar with the exponent laws

https://www.mathsisfun.com/algebra/exponent-laws.html

[lone heart]

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[alpine sable]

Need help moving forward with my leaning ladder problem. Specifically about the net torque = 0 equation.

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Anyone?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[gusty gorge]

I'm very lazy to actually read the problem

but you can set your hinge point anywhere if the ladder is stuck

and that means you can probably get rid of some frictional terms

and then it's just a matter of adding it all up

[alpine sable]

yea i did that

im stuck at whether or not i use sine or cosine

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

[gusty gorge]

huh? use the formula for torque

[alpine sable]

well yes but in one of my problems net torque had stuff and then rFcos(theta) instead of rFsin(theta)

so im wondering like uhhh

when does that occur

because the problem was pretty much identical to one where we used sine...

[gusty gorge]

torque is always computed as r x F

i.e. the magnitude is ||r|| ||F|| sin (theta), where theta is the angle between the force and r

maybe instead of memorizing the calculations, you should review the definitions

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

.close

[lone heart]

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[subtle ruin]

anyone help

<@&286206848099549185>

[vapid shuttle]

!15mins

[lone heart]

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

[subtle ruin]

lol sorry

[vapid shuttle]

To find the area of the square you need to know the length of one of its sides yes?

[subtle ruin]

yes

[vapid shuttle]

To find the length of one of its sides you need to know where its vertices are yes?

[subtle ruin]

ye

lemme add my friend too

@sterile palm hey join

[ocean seal]

AustinU

[subtle ruin]

it's written exam

[subtle ruin]

yeah

ik

[vapid shuttle]

Okay, then I'd just say, are there any clear points (x, y) that satisfy that equation?

[subtle ruin]

1,1

[vapid shuttle]

and

[subtle ruin]

2,1/2

3,1/3.....

[vapid shuttle]

Okay

here's a helpful note

if the point (2, 1/2) was a vertice of the square

then it would be shifted away from the origin

or else it would be rectangular

clearly the point (2, 1/2) can't be a vertice of the square, because the other vertices could then not possibly satisfy x^2y^2=1

[subtle ruin]

yes

[vapid shuttle]

so would you agree the point has to be of the form (x, y) where x=y or x=-y

[subtle ruin]

yes

true yeah

[vapid shuttle]

or else the square would be shifted off of the origin ( and then not satisfy the equation for all vertices)

[subtle ruin]

ye

[vapid shuttle]

so based on that, this could be one of the vertices

what would the other 3 be

[subtle ruin]

1,-1

-1,1

-1,-1

[vapid shuttle]

yea

and if those are the vertices

then you know the side length

then you know the area

[subtle ruin]

oh ok

noob me

[vapid shuttle]

tricky question

[subtle ruin]

what's the answer then

[vapid shuttle]

you tell me

[subtle ruin]

supposed to be 4 ryt

[vapid shuttle]

if these are the vertices, what is the length of each side?

yes 4

[subtle ruin]

no

it's 80

[vapid shuttle]

?

where did you get 80

[subtle ruin]

answer is 80 for that question

[vapid shuttle]

oh, we didn't read the second requirement: "the midpoints of its sides also lie on the same curve"

[subtle ruin]

yeah

that's were it gets tricky

[worn fox]

What did you mean by this

[subtle ruin]

no digital calculators

[vapid shuttle]

is this practice for the exam, or the actual exam

[subtle ruin]

model question

in mock test

today's

in school

[vapid shuttle]

Okay, well since you already know the answer, I can send you this solution I was able to find online

https://www.sarthaks.com/1058234/square-abcd-vertices-the-curve-the-midpoints-sides-also-the-same-curve-then-square-area-abcd

[subtle ruin]

💀

[vapid shuttle]

and maybe that will help you out

https://learn.careers360.com/engineering/question-a-square-abcd-has-all-its-vertices-on-the-curve-the-midpoints-of-its-sides-also-lie-on-the-same-curve-then-the-square-of-area-of-abcd-is-option-1-80option-2-60option-3-4option-4-25-120730/

here's another version of the same thing

[subtle ruin]

@vapid shuttle yeah I understood

it was easy

I didnt read question properly

[vapid shuttle]

so your question is resolved then?

[subtle ruin]

yes

.close()

oops

.close

[lone heart]

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[alpine sable]

Question 3
My working:

I applied the complement count. In doing so, I get three cases:

1. In three chosen numbers, one of them is 3 (1 way). In doing so, I have to choose either 1 or 2 (2 ways of choosing) so that minimum is not three, and exclude choosing 7 (remaining numbers in total are 7), since the maximum would then be 7. Hence, 1 × 2 × 7C1 is the desired number of ways for this case.
2. In three chosen numbers, one of them is 7 (1 way). In doing so, I have to choose either 8, 9 or 10 (3 ways of choosing) so that maximum is not seven, and exclude choosing 3 (remaining numbers in total are 7), since the minimum would then be 3. Hence, 1 × 3 × 7C1 is the desired number of ways for this case.
3. In three chosen numbers, neither of them is 3 and 7. Hence remaining numbers are 8. So 8C3 ways.

So in total, 5×(7C1) + 8C3 ways

By applying complement count, favourable ways are 10C3 - (5×(7C1) + 8C3) = 29
So probability of the event is 29/(10C3) = 29/120.

But, the answer is 11/40

Where am I going wrong in my logic here?

(Side note: It is not possible to choose three numbers from the set randomly that has 3 and 7, because one of the conditions would not get satisfied (after applying the complement count))

[merry depot]

what do you mean you can't choose 3 and 7 at the same time?

[alpine sable]

After I apply complement count, I have to form cases where the minimum is not 3 or maximum is not 7.

[merry depot]

your case one is "choose 3 and not 7"
case two is "choose 7 and not 3"
case three is "choose neither 7 nor 3"
where is "choose both 3 and 7"?

[alpine sable]

The side note is the reason why I am not forming the last case

If I do so, either minimum will be 3 or maximum will be 7

Oh wait

[merry depot]

oh, you're doing a compliment count?

[alpine sable]

Because it is "or", is it fine to form the last case?

[alpine sable]

Yes I'm taking the complement.

Oh god I'm sorry I mixed up inclusion exclusion and complement

Please ping me if you get back

[merry depot]

i think you're overcounting your cases when you choose 7 and when you choose 3.

Say you pick 7.
Then you pick a number larger than 7, say 10
then you pick any other number, say 8

That's the same as if you pick 7, 8, 10
But you counted them twice.

[alpine sable]

Oh I see

And in the other one, I counted the same case 3 times

[merry depot]

If you want to do this with inclusion/exclusion
You should do P(7 is max) + P(3 is min) - P(7 is max and 3 is min)

[alpine sable]

I'll try this out then

Thanks it worked out well

I got 33/120 = 11/40

👍

.close

[lone heart]

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[shut musk]

can someone explain this to me

ik you can just expand it all with binomial theorem but i think they used a shortcut

why did they equal it to x^0

i dont really get it

[alpine sable]

Yes. Instead of using the binomial theorem to find all coefficients, they used it to find just one

[alpine sable]

x^4*(1/x)^4=x^0=1

Okay I think I misunderstood the question

[shut musk]

they’re looking for a term in the expansion that isnt a coefficient of x

[alpine sable]

So first do you understand why the r-th term is (8 choose r)x^(8-r)(1/x)^r?

[shut musk]

well i know they looking for an unknown term so they sub in r

but i dont understand why its x^0

[alpine sable]

They're looking for a term independent of x

You have an x^2 term, and x^1 term, and similarily an x^(-1) term, an x^(-2) term, etc...

But all of these are dependent on x

The only power of x independent of x is x^0

So the only way for (8 choose r)x^(8-r)(1/x)^r to be independent of x is if x^(8-r)(1/x)^r=a constant

But you'll notice this simplifies to x^(8-2r)

And the only way for x^(8-2r) to be a constant (by that I mean to not be dependent on x) is if 8-2r=0

Which finally gives us an equation for r

[shut musk]

ahh ok i sorta understand what they’re doing now

Yeah i understand it now

thanks @alpine sable

.close

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[ripe basin]

how to do 5 power -2?

[echo socket]

This is how negative exponents are generally defined: $a^{-b} = \frac1{a^b}$

[ocean seal]

A Lonely Bean

[echo socket]

So, in this case, you have 5^(-2) = 1/5^2

[lone heart]

@ripe basin Has your question been resolved?

[echo socket]

What makes my explanation insufficient?

[lone heart]

@ripe basin Has your question been resolved?

[lone heart]

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[finite acorn]

Hey, I just need help to find angle ABC. I couldnt find the exact rule

[lone heart]

@finite acorn Has your question been resolved?

[finite acorn]

<@&286206848099549185> (15 minutes passed sorz just says to ping after 15 mins)

[alpine sable]

Yo

How can I help?

[finite acorn]

I just wanna find angle ABC

[alpine sable]

Okay, so what do you know about the equation so far?

.close

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[vapid shuttle]

for finding the laplace transform of tcos(bt) does this look so far so good?

or would there be an easier way

[surreal meadow]

not sure if that works, but what does is doing integration by parts twice

[vapid shuttle]

you can do integration this way

[surreal meadow]

then what you did is probably on the easier sides of things

as in there’s probably not many better ways

[vapid shuttle]

okay I wasn't sure cause that makes a mess to evaluate kind of

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@vapid shuttle Has your question been resolved?

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[plucky surge]

I'm learning Parallel vs Skew lines right now, and just wondering if the lines AB and GH are parallel, even though technically they're not on the same plane?

[vapid shuttle]

yes they are parallel