help-49

[exotic basalt]

do u see in the first one the shape is a trapezium

and all the other faces are rectangles

[wicked beacon]

Yes

Like the base looks like a trapezoid?

[exotic basalt]

yes

so would u agree we would have to find the area of the trapezoid first

before finding the volume

i’ll assume yes

area of trapezoid is 0.5 x height x (top+bottom)

[wicked beacon]

Uhh

So I’d do

(13+21)

[exotic basalt]

not 21

[wicked beacon]

Would the height be 21?

[wicked beacon]

Sorry I mean 20

[exotic basalt]

ah yes then

[wicked beacon]

13+20

[exotic basalt]

and u would get something like 346.5

check if u got it correct

wait its asking about surface area MY BAD 😭😭😭

[wicked beacon]

I got that

[exotic basalt]

but at least we’re in the right direction

[wicked beacon]

Yayy

[exotic basalt]

if theres a trapezoid shape in the front

surely there must be one in the back

so multiply that area by 2

thats 2 faces down 4 more to go

[wicked beacon]

Yes

[exotic basalt]

sadly the other faces are all different

but we can start with the left first

[wicked beacon]

😞😞😞

[exotic basalt]

its a rectangle with length 28 and width 21

[wicked beacon]

It’s all rectangles

So I just find those

And add them right

[exotic basalt]

but they come in different shapes and sizes

[exotic basalt]

yes essentially

[wicked beacon]

I think I get it now

[exotic basalt]

ok for question 3

[wicked beacon]

I’m not worried abt the math part since it’s simple numbers

[exotic basalt]

oh that ones a doozy

[wicked beacon]

Oh

[exotic basalt]

lets check the cube first

5 out of 6 sides are exposed to the outside

only 1 is touching the semicylinder

so we find the area of 1 side then multiply by 5

[wicked beacon]

So 12 x12

[exotic basalt]

yes

[wicked beacon]

144 x 5

[exotic basalt]

correct

[wicked beacon]

Ok I got 720

[exotic basalt]

nice

now lets panic about the cylinder part

hmmm

do u know the general formula for s.a. of cylinder

[exotic basalt]

correct

[wicked beacon]

This so total surface area tho

[exotic basalt]

do u know which term represents which

[wicked beacon]

Yes

This is lateral

[exotic basalt]

great

so we usually find 2 circles and a lateral

but we have 2 semicircles here, and half a lateral

isnt that just half the total surface area

[wicked beacon]

I think?

[exotic basalt]

halve the circles, and halve the lateral, easy

so its cube s.a. + (1/2) cylinder s.a.

[wicked beacon]

So the radius is 6

Height is 12

[exotic basalt]

u should get 216pi if im not mistaken

[wicked beacon]

Ok I got

[wicked beacon]

Uhh

I got 339.292

[exotic basalt]

2(pi)(6)(12)+2(pi)(6^2)

144pi+72pi=216pi

thats what i did

[wicked beacon]

Oh

I still got the same answer in the end

[exotic basalt]

ye

[wicked beacon]

After I added 720

[exotic basalt]

oh u halved the whole thinng already

nice

so ur answer is equivalent to 108pi

[wicked beacon]

Yea

[exotic basalt]

so yeah ur correct

[wicked beacon]

Yass!

[exotic basalt]

so add the 720 to ur 339

then round it to what it tells u

[wicked beacon]

1059.3

[exotic basalt]

that should be it

[wicked beacon]

Ty!!

[exotic basalt]

np

!done

[midnight plank]

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[wicked beacon]

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[halcyon oar]

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[plush fossil]

In part a we proved that ∀n ∈ N 1/sqrt(n+1) > sqrt(n+1) - sqrt(n)

[plush fossil]

and im on the inductive step and im at a stop because I could use the previous proof to say the same for k + 1 but i dont know how to get to the point where it shows that pattern of addition is > sqrt(k+1)

[flint hare]

show your exact progress

where did you stop exactly

did you just write down 1/sqrt(1)+...+1/sqrt(k)+1/sqrt(k+1)?

[plush fossil]

hold on im rewriting everything i did because the pencil i used comes out all light and its hard to take pictures of

[plush fossil]

well actually its just that but with > sqrt(k+1) - sqrt(k)

but i dont think i can just add the sqrt(k) on both sides and call it

[flint hare]

-sqrt(k)?

[plush fossil]

this was part a of the problem

which i already did and the prof said to use it for part b

[flint hare]

$\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k+1}}$

[grand pond]

Erk Gah

$\frac{1}{\sqrt{1}}+...+\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k+1}}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 285089887798034436.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[flint hare]

$\frac{1}{ \sqrt{1}}+...+ \frac{1}{ \sqrt{k}+\frac{1}{ \sqrt{k+1}}$

[grand pond]

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

[flint hare]

huh

$\frac{1}{ \sqrt{1}}$

[grand pond]

Erk Gah

[flint hare]

$\frac{1}{ \sqrt{1}} + ... + \frac{1}{ \sqrt{k}} + \frac{1}{ \sqrt{k+1}}$

[grand pond]

Erk Gah

[flint hare]

our inductive hypothesis is that it's true for n=k

[flint hare]

so can you derive any sort of relation to this from that?

if 1/sqrt(1)+...+1/sqrt(k)>sqrt(k), then what does that mean for 1/sqrt(1)+...+1/sqrt(k)+1/sqrt(k+1)?

[plush fossil]

the first is > sqrt(k) so the k + 1 should be the same as well

[flint hare]

can you try writing that out?

we can set a inequality from our inductive hypothesis

[flint hare]

the hint is to make this greater than something, it should be pretty straightforward just from the inductive hypothesis

[plush fossil]

im trying to be careful whether i can say that the whole thing is greater than sqrt(k)

i mean i feel like thats a given but could I just say that is the question

[flint hare]

that is true, but not useful in this case

the way forward is something really similar to that

you don't have to set "the whole thing" to be greater than just sqrt(k)

there's a inequality that is more useful to us that we can set up

[plush fossil]

im not sure what you mean here

are you talking about the part a?

[flint hare]

no, not yet

$\underbrace{\frac{1}{ \sqrt{1}} + ... + \frac{1}{ \sqrt{k}}}_{> \sqrt{k}}+ \frac{1}{ \sqrt{k+1}}$

[grand pond]

Erk Gah

[flint hare]

this is true, correct?

it's just the inductive hypothesis

i'm being not very direct on purpose, because it's a one step thing

you'll just be able to use a) straight away after setting up the inequality

[plush fossil]

ok wow thats exactly what i wrote down on my paper but yeah its true

like counting the fact that i skipped the 1/sqrt2 + 1/sqrt3

sorry its just funny to me

[flint hare]

too much of a hassle to type all that lol

anyway, if just that part of the expression is already greater than sqrt(k), we can swap it for sqrt(k) to create a inequality, correct?

leaving the 1/sqrt(k+1) alone

just swapping the underbraced part

[plush fossil]

what would both sides of the inequality even look like

[flint hare]

$\underbrace{\frac{1}{ \sqrt{1}} + ... + \frac{1}{ \sqrt{k}}}_{> \sqrt{k}}+ \frac{1}{ \sqrt{k+1}}>\sqrt{k}+\frac{1}{ \sqrt{k+1}}$

[grand pond]

Erk Gah

[flint hare]

this is what i mean

[plush fossil]

ohhhh ok

i was about to say because for the way you said it sounded weird but ok thats kinda what I thought

[flint hare]

yes, we know that sum is greater than sqrt(k), so we can affirm that the expression is greater than sqrt(k)+1/sqrt(k+1)

so you should be able to use a) directly now

can you do it?

[plush fossil]

ok so please correct me if im wrong

but i can use part a for the 1/sqrt(k+1)

[flint hare]

yes

[plush fossil]

and then basically do the same thing as before

[flint hare]

correct

try writing it out

[plush fossil]

idk how to use the bot so i just wrote it in paint

but i see it now

[flint hare]

exactly

[plush fossil]

sqrt(k)'s cancel out on the right so the whole thing is > sqrt(k+1)

making the hypothesis valid for k+1

[flint hare]

yes!

[plush fossil]

okkk damn

i just though using the inequalities like that was kind of risky but i see that it works just fine

[flint hare]

of course we need to justify our steps, we don't do it just because

[flint hare]

this works because we know 1/sqrt(k+1) is always a positive number

actually even if it was negative it would work here i think

anyway, we can do that because we have assumed that it's true for n=k

$\underbrace{\frac{1}{ \sqrt{1}} + ... + \frac{1}{ \sqrt{k}}}_{> \sqrt{k}}+ \frac{1}{ \sqrt{k+1}}>\sqrt{k}$

[grand pond]

Erk Gah

[flint hare]

we could've also done this, as you first suggested

but that's not very useful here, it doesn't lead us anywhere

unfortunately in math true things doesn't always lead to something that'll help us solve a problem

so you have to think of a 'clever' way to use something that is true that will also help you solve your problem

[plush fossil]

alright

glad i can understand it because finals are coming up and i dont know what i wouldve done if i couldnt lol

oh wait are you able to help with another or do i have to close this one and open another

[flint hare]

sure you can send it here

[flint hare]

have you tried it yet?

[plush fossil]

ok so for this one i have the part before the whole (k+1) fraction replaced with the induction hypothesis

and then i made common denominators and turned it into one whole fraction

and i tried expanding it but I was getting lost

so i went back and have something like this

[flint hare]

it'd be easier if you could send a picture

[plush fossil]

this is where im at so far

[flint hare]

seems right so far

just missing the square here

[plush fossil]

oh yeah

wait no

the (k+1) is factored out its not squared

forgot an extra parenthesis

[flint hare]

ah i see

well try to keep going

[plush fossil]

with the second parenthesis group expanded i cant really see much that i can do

[flint hare]

what did you get?

[plush fossil]

like what i have is close to the hypothesis for k+1 but i need that (2k+1) factor out of the denominator

[plush fossil]

ill draw it

[flint hare]

wait

$\frac{\left(n+1\right) \left(2n^2+5n+2\right)}{2(2n+1)(2n+3)}$

did you get this?

[grand pond]

Erk Gah

[plush fossil]

yeah

[flint hare]

try factoring the quadratic function

[plush fossil]

ok ok i got it

sometimes i just dont do the simple things lmao

but yeah factored it out

2(n+1/2) = 2n+1 which cancels out with the factor in the denominator

and then thats the (k+1) hypothesis

alright thank you so much

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[small zenith]

hi does anyone mind solving this for me cuz my calculator program doesnt work

it shows the determinant to be 0??

and idk maybe im too sleepy but i cant solve it either

[sudden yacht]

The determinant of what?

[dawn dagger]

when you use calculator you dont learn silly

[small zenith]

this as a matrix

[small zenith]

ikik but like ive literally tried to resolve them

am i missing something

[sudden yacht]

But yes, the determinant of the coefficient matrix is 0

[small zenith]

well then doesnt that mean there r infinite solutions

im getting some numbers here

(theres a typo^)

[sudden yacht]

Not necessarily, there might be no solutions as well

[small zenith]

then how did the guy get the solutions i.e. x = -2/7 etc.?

[sudden yacht]

Mmh there's a typo there

[small zenith]

yeah it's his typo

[sudden yacht]

That's why he gets a unique solution

[small zenith]

here's the question

i still dont get how he got z = -1/7 tho

cuz i feel like this has infinite solutions even it's not supposed to

his equations r correct

i got these as well

[small zenith]

if i sub his numbers in the program it still doesnt work tho

it gives x = 5, y = -2/5, z = -1/5

why didnt he let z = t; but instead he got a set of numbers as the solution?

<@&286206848099549185>

[midnight plank]

@small zenith Has your question been resolved?

[small zenith]

yeah lol not a lot of ppl r free to help @frozen widget

[ornate hearth]

This is your question @small zenith

[small zenith]

erm it kinda evolved into

[small zenith]

this

[frozen widget]

Why did u mention me??

[small zenith]

cuz u reacted on the bot msg??

wait nvm my friend got me *cries bc ive been stuck on this for an hour *

.close

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[graceful drum]

Let $a_1 \leq a_2 \leq ... \leq a_n$ be natural numbers satisfying $a_1+a_2+...+a_n = 2n$
and $a_n \neq n + 1$. Prove that, if n is even, there exists a subset A of ${1, 2, ..., n}$ such that, $\sum_{i\in A}a_i = n$

[grand pond]

CherryMan

[graceful drum]

We're supposed to use the pigeon hole principle here but I have no idea how to proceed

Only thing I found is that $1 \leq a_i \leq n$

[grand pond]

CherryMan

[graceful drum]

<@&286206848099549185>

Any ideas would be helpful! I dont need a complete answer

<@&286206848099549185>

[karmic snow]

Does {1,2,.., n} use the same n as a1+a2+a3+...+an = 2n?

[graceful drum]

Yes.

<@&286206848099549185>

[exotic stratus]

Well take the case where $\forall i, j, a_i = a_j = 2$

[grand pond]

@exotic stratus

[graceful drum]

I feel that taking this extremely specific case (I know that if all a_i are equal they have to be 2) won't result in any meaningful result?

[exotic stratus]

Yea by itself yes

[grand pond]

@exotic stratus

[exotic stratus]

Use this

[graceful drum]

I dont get how this idea will be used

Also we gotta use the pigeon hole principle here

[exotic stratus]

Well yea

[graceful drum]

???

[exotic stratus]

Give me a sec, I am thinking of more hints I can give you without revealing the answer lol

[graceful drum]

kinda desperate, so would it be ok with u if u could send the solution in dms?

[midnight plank]

@graceful drum Has your question been resolved?

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@graceful drum Has your question been resolved?

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[willow lotus]

if you have an integral where we're integrating with respect to t, for example, and we have stuff like "x^2" in our integrand, would "x" be a constant? (even though it's not necessarily a number)

[dreamy lichen]

you can treat it as a constant

the question whether it actually is a constant is context dependent

but during the integration, you can certainly treat it as a constant

unless x depends on t

[willow lotus]

I see

another similar example, if you're trying to find the surface integral of a sphere, does your "r" have to be a constant?

[dreamy lichen]

depends on what "r" is

[willow lotus]

your "r" as in the radius

[dreamy lichen]

if it's just the radius of the sphere, then yeah, it's a constant

[willow lotus]

does it have to be?

or can it be a variable

[visual tiger]

you're integrating with respect to the angles of the sphere aren't you?

[willow lotus]

yes

[visual tiger]

with respect to those

r doesn't vary

so r is treated as a constant inside the integrals

[willow lotus]

i ask this because in physics (gauss's law), when finding the electric field at some variable distance "r", we put "r" as the radius of an imaginary gaussian sphere

and we integrate over that sphere

[visual tiger]

I'm lacking context in that case

if someone knows what this is about

[midnight plank]

@willow lotus Has your question been resolved?

[midnight plank]

@willow lotus Has your question been resolved?

[willow lotus]

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[vital sinew]

I have two statements of Faulhaber's formula, one given by Wolfram Mathworld and the other given by Wikipedia. In theory these should be equivalent, but Wikipedia's version says \sum_{i=1}^{n-1} i^0=n, which is false, and Wolfram gives n-1, which is correct. Is Wikipedia just wrong here?

[last slate]

I have no idea but why even consider what wikipedia has to say. Its clearly not a credible source

[vital sinew]

Wikipedia tends to be credible regarding such things. On the whole, their math articles are pretty good.

[midnight plank]

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[vital sinew]

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[proud abyss]

what exactly does enumeration mean here? like why is it written separately?

[lyric charm]

enumeration ≈ counting

[proud abyss]

so that's a separate topic? i thought it is the same thing

[lyric charm]

eh

idk, i wouldn't take it that seriously in the topic listing

[proud abyss]

okay, fair, thank you

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[safe onyx]

I said that x should be between -8 and 8

Then I plugged them in

[dusty portal]

Hmm

Does $\sum_{n\ge 0}(-1)^n$ converge?

[grand pond]

;(

[safe onyx]

no

lim n --> inf |(-1)^n| !== 0

[fallow scarab]

whata do you get when you plug x=8 into the series?

[dusty portal]

Therefore should 8 be considered?

[safe onyx]

ohh I did
(-1)^n/8^n instead of just 8

thank you

it should be
(-1)^n/8 which would not converge

thanks

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[fathom onyx]

@tidal turret

!don

!done

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[tidal turret]

the fuck?

[fathom onyx]

iirc you'd answered this question in another channel?

[tidal turret]

no bro

the fuck?

[fathom onyx]

I mistyped "done"

[tidal turret]

?

[fathom onyx]

Wait wHAT

[tidal turret]

help me please

the fuck?¿

[fathom onyx]

I could swear I did see this

[tidal turret]

no

[fathom onyx]

oh shit I was looking at #calculus message

[tidal turret]

yes but people didnt replied

the fk

Let \begin{align*} A &= {(x, y) \in \mathbb{R}^2 \mid -2 \leq x \leq 3; 1 \leq y \leq 4} \ B &= {(x, y) \in \mathbb{R}^2 \mid 0 \leq x + y \leq 5}\end{align*} be sets in (\mathbb{R}^2), and let (P = (1, -1)), (Q = (-2, 2)), and (R = (-1, 3)) be points. (\)

a) Determine which of the points (P), (Q), (R) belong to (A \cap B) and which belong to (A \cup B. \ )

b) Let (C = {(x, y) \in \mathbb{R}^2 \mid y = 2x}). Find (A \cap C) and determine whether (A \cap C \subset B).

[fathom onyx]

Yh I just found out you'd put out an answer but weren't sure and no one responded

dw I can read it

Verbatim:

A n B = {(x, y) € R^2 | 0 <= x + y <= 5; -2 <= x <= 3; 1 <= y <= 4}
P = (1,-1) = (p1,p2)
Q = (-2,2) = (q1,q2)
R = (-1,3) = (r1, r2) (edited)
[21:40]
-------------------------------
p1 + p2 = 0 € [0,5]  

q1 + q2 = 0 € [0,5]  

r1 + r2 = 2 € [0,5]  (edited)
[21:44]
--------------------------
p1 = 1 € [-2,3]  
p2 = -1 € [1,4]  

q1 = -2 € [-2,3]  
q2 = 2 € [1,4]  

r1 = -1 € [-2,3]  
r2 = 3 € [1,4]  (edited)
[21:44]
---------------------------------------------------------------------------------
P € A n B  
Q € AnB  
R € AnB  (edited)
[21:45]

Gimme a moment lemme see if I get to the same numbers

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[fathom onyx]

Sorry, still working on it

(pero gracias por la traducción)

[tidal turret]

Let \begin{align*} A &= {(x, y) \in \mathbb{R}^2 \mid -2 \leq x \leq 3; 1 \leq y \leq 4} \ B &= {(x, y) \in \mathbb{R}^2 \mid 0 \leq x + y \leq 5}\end{align*} be sets in (\mathbb{R}^2), and let (P = (1, -1)), (Q = (-2, 2)), and (R = (-1, 3)) be points. (\)

a) Determine which of the points (P), (Q), (R) belong to (A \cap B) and which belong to (A \cup B. \ )

b) Let (C = {(x, y) \in \mathbb{R}^2 \mid y = 2x}). Find (A \cap C) and determine whether (A \cap C \subset B).

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

A n B = {(x, y) € R^2 | 0 <= x + y <= 5; -2 <= x <= 3; 1 <= y <= 4}
P = (1,-1) = (p1,p2)
Q = (-2,2) = (q1,q2)
R = (-1,3) = (r1, r2)

p1 + p2 = 0 € [0,5] ✅

q1 + q2 = 0 € [0,5] ✅

r1 + r2 = 2 € [0,5] ✅

p1 = 1 € [-2,3] ✅
p2 = -1 € [1,4] ❌

q1 = -2 € [-2,3] ✅
q2 = 2 € [1,4] ✅

r1 = -1 € [-2,3] ✅
r2 = 3 € [1,4] ✅

P € A n B ❌
Q € AnB ✅
R € AnB ✅

[fathom onyx]

If it helps - you can also answer this graphically

But yh P isn't in the intersection

[tidal turret]

can u explain this graph?

I dont get it, why a rectangle

[fathom onyx]

Ah, I should have shown the preceding work

A is the blue rectangle, as it's defined by x being between some values and y being between some values

x and y are independent of each other here

Thus it's a rectangle (I've writen another way of writing that set)

@tidal turret Do you need assistance with b?

[tidal turret]

?

why are you cross multiplying

[-2,3] x [1,4]

@fathom onyx

[fathom onyx]

This isn't cross-multiplying, this is notation shorthand for what A is

Definitionally it's what A is as described in the original question

[tidal turret]

https://math.stackexchange.com/questions/229083/the-cross-product-of-two-sets

[fathom onyx]

Oh right yh that works

[fathom onyx]

But that is still what we've got here

[tidal turret]

bro its cross product between two sets

[fathom onyx]

yh ik I'm walking that back

But my point still stands; the set A is a cross-product

[tidal turret]

ok

where is A plotted?

[fathom onyx]

A is the blue rectangle

[tidal turret]

is A the light blue rectangle?

[fathom onyx]

ye

[tidal turret]

where is B

?

[fathom onyx]

the red region

I'd used arrows, but maybe I didn't make that clear enough

[tidal turret]

dont really get it

how did you plotted it ?

[fathom onyx]

Are you familiar with plotting inequalities on graphs in general

[fathom onyx]

(ignoring the sets for now)

It's two inequalities

[tidal turret]

a little bit

[fathom onyx]

one, 0 <= x + y

two, x + y <= 5

Both of those have to be true here, so we take the intersection of those graphs; try drawing both of them

[fathom onyx]

So the intersection is what B is

[tidal turret]

how do you know how to plot 0 <= x + y

[fathom onyx]

You know how to plot 0 = x + y, right?

[tidal turret]

I dont

well I kinda do

[fathom onyx]

If it helps, rearrange to make y the subject

[tidal turret]

just (-1,1); (-2,2); (-3,3); (1,-1); (2,-2); (3,-3) so in general (y,-y) or (-x,x)

[tidal turret]

is hard

[fathom onyx]

Yh, it's a straight line, so you only need two points

But yh, the x-coord is the negative of the y-coord

Now for the inequality

You only really need to test one point that's not on the boundary

Typically the origin's a good bet because it's simple to check

But since that's on the boundary (i.e. the line x+y=0), we'll just use a different point, say (1,1)

Q: Is (1, 1) in the region 0 <= x + y ?

[tidal turret]

how to plot 5 = x + y

[fathom onyx]

Same as before

You just need two points to find and connect

Easy options include setting x = 0 and y = 0

[tidal turret]

5 = 0 + 0?

[fathom onyx]

I mean, set x =0 and see what point you get

Similarly set y = 0 and see what point you get

Connect the two points together

[tidal turret]

when x = 0
5 = x + y
5 = 0 + y
y = 5
(x,y) = (0,5)

when y = 0
5 = x
(x,y) = (5,0)

[fathom onyx]

(without meaning to sound disrespectful, this should be familiar high-school work)

ye

[tidal turret]

this is why I need help, idk

is lowkey confusing ngl

specially plotting 0 <= x + y <= 5

because

0 = x + y is understandable

0 < x + y aswell

x + y = 5 is more challenging

[tidal turret]

how

[fathom onyx]

wdym how

(0,5) and (5,0) are two points on the line x + y = 5

Draw the line connecting them

[tidal turret]

ok

y = mx + b

P1 = (0,5)
P2 = (5,0)

5 = b

y = 5

[fathom onyx]

Bro

You do not need to convert anything

I'm assuming you have a sketch of a graph, yes?

[tidal turret]

how do I draw a line that passes through two points in R2

[fathom onyx]

...

The same way you draw a line passing through two points on a graph

R2 is the xy-plane

i.e. you have an x-axis and a y-axis

That's any 2D graph space

[tidal turret]

is like the line that passes through the red and the blue dot

[fathom onyx]

Draw this by hand, you'll be doing yourself a favour

[tidal turret]

but idk how to get the parametric equation of the line

[fathom onyx]

(and in any case, you already have the equation of the line if you're trying to graph it - it's x+y = 5)

[fathom onyx]

...Okay, but why are you trying to parametrise it?

[tidal turret]

I am trying to get the equation of the line knowing two points in R2

idk how to do it tbh

[fathom onyx]

...You have the equation of the line

[rigid tangle]

I have zero contributions except that a hair strand was on my screen at the perfect place to make me think it was the graph, funny moment

[tidal turret]

where?

[fathom onyx]

It's x + y = 5

That is literally a line

[tidal turret]

ok, that did the trick

how do you draw it manually by hand?

we know x + y = 5 passes through (5,0) and (0,5)

[fathom onyx]

By using this amazing invention called a ruler

[tidal turret]

ok

that is true

so we know

x + y <= 5 is this portion of the graph

[fathom onyx]

uh-huh

[tidal turret]

and 0 <= x + y is this part of the graph

how do I plot 0 <= x + y <= 5

[fathom onyx]

It's the region where both 0 <= x + y AND x + y <= 5

[fathom onyx]

i.e. the intersection

[tidal turret]

the intersection!

[fathom onyx]

The dark purple band there

[tidal turret]

yeah

ok

now how do I draw the retangle ([-2,3] x [1,4])

[fathom onyx]

A x B?

x?

Ah

Well it's just a rectangle

You're drawing a rectangle of points where x lies between -2 and 3

And where y lies between 1 and 4

[tidal turret]

how do I draw the rectangle

oof

[fathom onyx]

Hold on - do you mean how to draw the rectangle in Desmos?

Because honestly you're better off drawing it by hand

I just drew those graphs on a tablet for example

[tidal turret]

oh

[fathom onyx]

If you're wondering about it looking neat, that just comes with over a decade of maths practice and some years of art practice hehe

I mean that is one way of doing it 🤣

I mean at this point this is more a discussion on how to use Desmos than the question itself at hand

[fathom onyx]

But yh, you can now see the union and the intersection, yh?

[tidal turret]

why is A equal to the cross product of [-2,3] x [1,4]

[fathom onyx]

Because by definition, A is the set of values (x,y) in R2 such that x is in [-2, 3] and y is in [1,4]

[tidal turret]

because as I understand A = {(x,y) € R^2 | -2 <= x <= 3 ; 1 <= y <= 4 }

[fathom onyx]

Yes

[tidal turret]

why cross product

[fathom onyx]

-2 <= x <= 3 is equivalent to x lying in the interval [-2,3]

[tidal turret]

how does the cross product between sets work?

[fathom onyx]

R^2 is just a cross product of R and R

You'd sent a definition earlier, that $(x,y) \in A \times B$ precisely when $x \in A$ and $y \in B$

[grand pond]

Waes (Wires)

[tidal turret]

how ?

[fathom onyx]

wdym "how"

[fathom onyx]

This is literally a definition

[tidal turret]

can you help me understand this diagram btw

[fathom onyx]

You take every element in A, and pair it with every element in B

The result is a big list of pairs (a,b) where a is in A and b is in B

[tidal turret]

ok. got it

[tidal turret]

is jusst strange

[fathom onyx]

What is?

[tidal turret]

like how does it create a rectangle dont get it

also, x €[-2,3]; y € [1,4]

when you do [-2,3]x[1,4] you are doing the cartesian product of two sets

[fathom onyx]

Yes

[tidal turret]

but for example [-2,3] is referring to X
and [1,4] is referring to Y

[fathom onyx]

Yes

That is generally assumed in R2

[tidal turret]

is just weird, why [-2,3] x [1,4]

[tidal turret]

X € R
Y € R

X x Y € R^2

is just weird, why are you describing A as a cartesian product

[fathom onyx]

You're abusing notation here

x € R, y € R leads to (x,y) being an element of R^2

[fathom onyx]

A = [-2,3] x [1,4] is a subset (not an element) of R^2

[tidal turret]

X ⊆ R
Y ⊆ R

X x Y ⊆ R^2

[fathom onyx]

yes...?

[tidal turret]

how do i plot the cartesian product of two sets or intervals

[fathom onyx]

Maxim of clarity needed

Do you mean in general, or do you mean in Desmos

[tidal turret]

in general

[fathom onyx]

Well then you've got an example right here

[tidal turret]

forget about desmos for a second, I am having trouble understanding how to plot the cartesian product of two sets

[fathom onyx]

Are you good with drawing straight lines?

[tidal turret]

sort of

[fathom onyx]

i.e. given a line's equation, can you draw it?

[tidal turret]

I think so

in 2d yeah, in 3d maybe

[fathom onyx]

This is 2D, that's all that's needed here

[tidal turret]

ok

[fathom onyx]

You need (massive emphasis) to be strongly (massive emphasis again) familiar and confident in doing these before you can get further

[tidal turret]

test me if you want ,I think I can handle lines in 2d

[fathom onyx]

That being said - if you've got the product of intervals [a, b] x [c, d] (where we're assuming a < b and c < d), you sketch the lines x = a, x = b, y = c and y = d

The rectangular bounded region you make is the region you want

[tidal turret]

ok

A = . . .
i) -2 <= x <= 3
ii) 1 <= y <= 4

something like this I pressume

[fathom onyx]

yep

[tidal turret]

wait a second bro

look at this

why are you doing the lines and not plotting the inequalities?

if the rectangle is the figure of [-2,3] x [1,4]

,w CartesianProduct[{-2,3}, {1,4}]

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

ok, now we are cooking.

[fathom onyx]

No

It's interpreting those as the sets {-2, 3} and {1,4}

Conveniently, that does give you the coordinates of the vertices

A the region is the INSIDE and BOUNDARY of this rectangle

But to make any sense of that you need the rectangle in the first place

[tidal turret]

so its not the cartesian product between two sets, but the cartesian product between two intervals

[fathom onyx]

Hence why we're drawing it

[fathom onyx]

That's what I've been trying to tell you

An interval is a set

$[a,b]$ is shorthand for ${x \in \mathbb{R} | a \leq x \leq b}$

[tidal turret]

\leq

but yes

[fathom onyx]

yh ik

[grand pond]

Waes (Wires)

[fathom onyx]

But you can see, an interval is a set

[tidal turret]

ok

A = . . .
i) -2 <= x <= 3
ii) 1 <= y <= 4

I think I lowkey get it tho

A = {(x,y) € R^2 | -2 <= x <= 3 and 1 <= y <= 4}

and its represented by this rectangle

but the cartesian product is what complicated things for me @fathom onyx like the cartesian product between two sets / intervals

[fathom onyx]

tbh you didn't need to go down an entire rabbit hole

[tidal turret]

what rabbit hole?

[fathom onyx]

All that was needed to make sense of it is that it was shorthand for the set A as described in the original question

Because it's faster for me to write

[fathom onyx]

ik

I wrote A in shorthand using a cross-product, because it's smaller

"A = [-2, 3] x [1, 4] ( \subset R^2)" is a much smaller sentence

Fundamentally this doesn't change the outcome of how the question is handled

Not to be dismissive but it's getting late over here so

[tidal turret]

bro please just stick with me for a couple more minutess

are you from spain?

[fathom onyx]

no

[tidal turret]

from where?

[shadow schooner]

idk tbh he speaks a lot of language

[tidal turret]

wdym? u know him

[fathom onyx]

puedo leger un poco de español pero soy del reino unido
/ I can read a bit of Spanish but I'm from the UK

asi que no estaba tan difícil leger la pregunta
/ so it wasn't that difficult to read the question

[shadow schooner]

i’ve seen him help other peoples

anyhow back to your question

[tidal turret]

ok

[tidal turret]

R and Q are in A n B

P is not

[fathom onyx]

@shadow schooner Would you mind taking over if you could?

[tidal turret]

good night waes

love you bro, xoxo

[fathom onyx]

https://tenor.com/view/mario-sleep-gif-13382784005848704668

[tidal turret]

how did you found A U B? btw, any idea?

[shadow schooner]

give me a sec eating the pizza right now

[tidal turret]

yep, no worries

[shadow schooner]

this is the question

[tidal turret]

A ∩ B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) and (0 <= x + y <= 5)}

[tidal turret]

yes

A ∩ B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) and (0 <= x + y <= 5)}
A U B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) or (0 <= x + y <= 5)}

how to plot A U B tho?

[shadow schooner]

For the union it geometrically contains points that are either in A the rectangle or B

[tidal turret]

oooooooookay

[shadow schooner]

maybe a drawing will be clearer

this basically

[tidal turret]

so P, Q and R are in A U B

[shadow schooner]

you could also explicitly verify each point satisfies either the condition for being in A or being in B

but in R^2 you can levereage geometric intuition

[tidal turret]

P (purple) is in B (green)
Q (black) is in A(red rectangle) and in B (green)
R (red) is in A(red rectangle) and in B (green)

[shadow schooner]

yep but if P is purple it seems like it’s only in the green region

ah yep

[tidal turret]

yes is in green which is B

and not in A

so its in A U B

[shadow schooner]

I think you did A inter B with Waes so that should be it for a) imo

[tidal turret]

recall that

A ∩ B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) and (0 <= x + y <= 5)}
A U B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) or (0 <= x + y <= 5)}
A = {(x,y) € R^2 | -2 <= x <= 3 and 1 <= y <= 4}
B = {(x,y) € R^2 | 0 <= x + y <= 5}

[tidal turret]

and we found which points are in A U B aswell

what about b)?

[shadow schooner]

What’s C geometrically in R^2?

[tidal turret]

do you want me to plot it?

is a line I think with slope 2

[shadow schooner]

yep plotting it could help confirm things

[shadow schooner]

i mean with the other sets as well

[shadow schooner]

Then what is A inter C geometrically?

[tidal turret]

A ∩ B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) and (0 <= x + y <= 5)}
A U B = {(x,y) € R^2 | (-2 <= x <= 3 and 1 <= y <= 4) or (0 <= x + y <= 5)}
A = {(x,y) € R^2 | -2 <= x <= 3 and 1 <= y <= 4}
B = {(x,y) € R^2 | 0 <= x + y <= 5}
C = {(x,y) € R^2 | y = 2x}
A ∩ C = {(x,y) € R^2 | -2 <= x <= 3 and 1 <= y <= 4 and y = 2x}

[tidal turret]

give me a second to think bro

[shadow schooner]

i’m still eating the pizza

[tidal turret]

C is blue line

red rectangle is A

[tidal turret]

the portion of the blue line that intersects with the red rectangle bro

[shadow schooner]

yep

[tidal turret]

exactly

is not entirely contained

is only a proper subset

no?

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[shadow schooner]

indeed for instance (2,4) is not in B but in A inter C

without the visual you could check with the equation it’s not in, but in desmos it’s a bit obvious once you have the pictures

that should be it imo, you good?

[tidal turret]

I dont get it

the question is asking if A n C is a proper subset of B

no?

is not using \subseteq is using \subset @shadow schooner

[shadow schooner]

oh maybe it’s a conflicting definition on my part then, for me A \subseteq B indicates that all that is in A can be in B and A could be equal to B, and A \subset B indicates that all that is in A can be in B and A \neq B

[shadow schooner]

So A inter C \subset B is false because there are some point like (2,4) in A inter C but not in B

we don’t have that for every (x,y) in A inter C that this same (x,y) is in B

[tidal turret]

can you elaborate?

[shadow schooner]

I can write it out in full quantifier if you prefer that

[tidal turret]

in predicate logic

?

[shadow schooner]

here from my point of view we don’t have A inter C \subseteq B and we also don’t have A inter C \subset B as the later is stronger

[shadow schooner]

yeah

[tidal turret]

yes, this is proper subset

[shadow schooner]

then if our definition match i argue that since it would need to be a subset to be a proper subset and that it’s not even a subset then it can’t be a proper subset

“it” being A inter C

anyhow i have to go take my bath

i’ll comeback in ish 20 min

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[marble trellis]

I forgot how to do dis

[floral apex]

you just use the compound interest formula

[midnight plank]

@marble trellis Has your question been resolved?

[marble trellis]

I forgot how to use it

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[steady hill]

is this correct? i need to get this right

[slender walrus]

did you check for whether there's a second possible triangle?

[steady hill]

i got it wrong.

is this one right?

<@&286206848099549185>

[jagged sandal]

@steady hill Angle C is close, but is incorrect. Please show me your work

[steady hill]

idk how to type my work onto discord

[steady hill]

a/sinA= c/sinC
sin C= csinA/a
sin C= 54xsin(29 degrees)/37
sin(29degrees) ≈ 0.4848
54×0.4848=26.1792
sinC= 26.1792/37
sinC≈0.7075
sin −1 (0.7075) ≈ 45.1
​

[jagged sandal]

Huh

[steady hill]

idk how to show

thats what i did tho

[jagged sandal]

Everything went well until the final answer

The answer is 45.0318707…

Rounded to 45

[steady hill]

so rounded to the nearest tenth its just 45.03?

i mean

45.0

[jagged sandal]

Yeah

[steady hill]

and the second C is correct?

[steady hill]

so is this correct?

[jagged sandal]

Why are there two solutions? (Just making sure)

[steady hill]

bc of the ambiguous case in the Law of Sines.

[jagged sandal]

Ah…

[steady hill]

sin(C)=sin(180degree−C)
This happens when
sin(𝐶)<1sin(C)<1. The calculator will give you the acute angle, but the supplementary obtuse angle is also a valid solution for
𝐶
C if it still allows the triangle's sum of angles to be `80

180

[jagged sandal]

In that case, just fix the second C, both B’s, and recalculate both b’s just to be safe (Case-sensitive)

[steady hill]

ive got the same answers

[jagged sandal]

You’re supposed to use the new C angle

[steady hill]

im stupid

[jagged sandal]

It’s all good

[steady hill]

okay how does this look

[midnight plank]

@steady hill Has your question been resolved?

[jagged sandal]

180 - 37 =143

[steady hill]

okay so i passed that module im on my last one and i really struggle with this

[jagged sandal]

Oh, ok

[steady hill]

do you know how to do this?

[jagged sandal]

Angle c is definitely easy to solve

[steady hill]

okay i got that one

last problem and ill be fully done

i cannot get this wrong

[jagged sandal]

You got this!

[steady hill]

is it 2.3?

[jagged sandal]

How did you solve it?

[steady hill]

top angle = 180 degree - (70 degree + 50 degree) = 60 degree
AB/sin (60 degree) = 2/sin(50 degree)
sin(60 degree) ≈ 0.8660
sin(50 degree) ≈ 0.7660
AB = 2x0.8660/0.7660
AB ≈ 2.26 miles
rounded to 2.3

[steady hill]

is that correct?

[jagged sandal]

Wrong

Sin(50) should be sin(70)

Check Angle B again

[midnight plank]

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[jagged sandal]

Bro, what?

.reopen

[steady hill]

.reopen

[midnight plank]

✅

[jagged sandal]

@steady hill The channel closed by mistake

[steady hill]

its all good

okay i got 1.8

[jagged sandal]

Correct

[steady hill]

bet so thats for sure correct?

[jagged sandal]

Absolutely

[steady hill]

i appreciate your help very greatly. thank you.

[jagged sandal]

You’re welcome!

[steady hill]

.close

[midnight plank]

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Closed by @steady hill

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[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[small zenith]

If this question is asking whether E is the circumcentre of triangle ABC, can i simply say that since |EA| ≠ |EB| so it’s not?

[lyric charm]

can we see the original problem just in case

[small zenith]

cuz for E to be circumcentre we must have EA = EB = EC

[lyric charm]

yeah true

...why are they even concerning themselves with whether EM is perp to AB i wonder

oh i guess

ah

no ok their way works and so does yours

and imo there's less overhead with just finding the lengths of EA and EB

[small zenith]

that step is worth 1 mark tho

idk but the past paper marking scheme says that “any acceptable answers should be considered”

[lyric charm]

that sounds reasonable

[small zenith]

so ig mine makes sense if i just stop after saying EA ≠ EB

[lyric charm]

,calc 20^2 + 18^2 + 4^2

[grand pond]

Result:

740

[lyric charm]

,calc 10^2 + 6^2 + 8^2

[grand pond]

Result:

200

[lyric charm]

ok yeah way different

[lyric charm]

yes it does

[small zenith]

thanks!

.close

[midnight plank]

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[midnight plank]

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[steel rivet]

Find deriviate of given function w.r.t the independent variable y = (x-1)(x²+x-1)

[steel rivet]

Product rule is not working

What else is applicable here?

[deep oriole]

product rule should work, but you could also consider just distributing out that product

and then differentiating term by term

[steel rivet]

Do I have to open brackets first then dy/dx

[steel rivet]

Lemme send u my solution

[deep oriole]

you need to multiply these whole bracketed sets of terms by the (x-1), as written

[steel rivet]

Oh

👍

Got 3x²

[deep oriole]

yep thats the right answer

[steel rivet]

Thanks

.close

[midnight plank]

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[fringe karma]

how to start solving for x?

[astral terrace]

Firstly assume 2x^2 -5x -2 = t

then you get sqrt(t) - sqrt(t-7) = 1

take sqrt(t-7) to rhs

then you gotta solve its pretty standard

lmk if you need further help

[fringe karma]

ok

wait

done

[fringe karma]

thanks..

.close

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[solemn bear]

Pls explain me the concept of remainder theorem and factor theorem
Like p(x) ÷ (x-a) = p(a)
I know doing ques using this but my concept is not clear , how we get remainder without dividing . We just pu value in place of variable and get remainder how explain concept behind like how this is division

[lyric charm]

Like p(x) ÷ (x-a) = p(a)
this is very very dodgy notation

anyway, do you know how to do polynomial long division in general?

[astral terrace]

Yo ann how do you do that

[lyric charm]

do what

[astral terrace]

like

the reference

[lyric charm]

> message goes here

message goes here

[astral terrace]

Alr

Like p(x) ÷ (x-a) = p(a)

oo

ale

back to the problem

[lyric charm]

@solemn bear are you still here?

asks for explanation of concept
disappears

[midnight plank]

@solemn bear Has your question been resolved?

[solemn bear]

Am back

[solemn bear]

Yes

[lyric charm]

click ❌ on this so the channel doesnt close on you

[solemn bear]

@lyric charm am asking like how putting value

Is same

As division

[lyric charm]

"putting value is same as division" prime example of few word go bad

specifically, the thm states that the remainder of p(x) on division by (x-a) is equal to p(a)

[solemn bear]

Yea I know this but how I want concept clarity I am in class 9

[lyric charm]

ok right

so let's try to think about it now

let's take p(x) = x^2 as example

[solemn bear]

Ok

[lyric charm]

can you work through the polynomial division of x^2 by (x-a) in full and show me your result

[solemn bear]

Ok w8

[lyric charm]

there will be more after this but i want you to do this for now

as a starting point

[solemn bear]

Remainder is -ax

[lyric charm]

nope

you made a sign error and didn't carry the division process to completion

the remainder must have strictly lower degree than the divisor

your remainder has degree 1 and (x-a) also has degree 1 so you aren't yet done

and also fix your sign error:

x^2 - (x^2 - ax) = +ax and not -ax

[solemn bear]

But we have to multiply x by x to get x² so we do x(x-a) = x²-ax

And x² got cancelled

[lyric charm]

yes, and then you need to subtract that from x^2

x^2 - (x^2 - ax) = +ax and not -ax

[solemn bear]

Oh

I didn't change sign

[lyric charm]

you forgot what you were actually doing then lol

[solemn bear]

Sry I learnt division just now

[solemn bear]

K what next

[lyric charm]

well you have one more step to do

x times what gives ax?

[solemn bear]

X times a

[lyric charm]

ok

so we need to subtract a(x-a) too

[solemn bear]

?

[lyric charm]

in the 1st division step you subtracted x(x-a) to cancel the x^2 term

[solemn bear]

We doing division further?

[lyric charm]

now you subtract a(x-a) to cancel the ax term

[solemn bear]

Oh ok

a² remainder ?

[lyric charm]

indeed

so we have verified that this sort of thing works for the polynomial x^2

do you have paper by the way

[solemn bear]

Yes

[lyric charm]

ok

do the division thing for p(x) = x^3

again divide by (x-a)

and do it in full

[solemn bear]

Ok

[lyric charm]

and show me your result on paper

[solemn bear]

Done a³ remainder

[lyric charm]

ok right

so maybe you can see a sort of pattern here

when you divide a pure power of x, i.e. just x^n, by (x-a)

the remainder will be a^n

[solemn bear]

Yes I see

[lyric charm]

you can play the same game if it's multiplied by some coefficient, e.g. 2x^n rather than just x^n -- the two will just stick around and nothing will happen to it

(or any other constant)

that make sense?

[solemn bear]

Yes but

How dividing and putting value is same

[lyric charm]

patience

i am trying to explain step by step

[solemn bear]

Ok

[lyric charm]

what we've got thus far is that the "remainder of division by (x-a) = value at a" thing is true when p(x) consists of just one term like cx^n

i was going to say this now before you interrupted me:

when you do it with a general polynomial,
it is basically like you are doing it for each of its terms but in parallel,
and the intermediate results kind of snowball until all the x's become replaced with a's like you saw with the x^3 example (x^3 -> ax^2 -> a^2x -> a^3)

do you understand now?

[solemn bear]

Let me understand this message

[lyric charm]

if you want, i can give you some more examples to work through

[solemn bear]

I understand somewhat

[lyric charm]

try p(x) = 2x^3 + 5x

[solemn bear]

At?

I will ask if I have doubt cya

.close

[midnight plank]

Channel closed

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[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[cursive swan]

I have a question.

[cursive swan]

Just a minute. Let me send the image.

[exotic stratus]

!da2a

[midnight plank]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[cursive swan]

'tis a geometrical question.

Figure is not to be scaled.

[lyric charm]

,rccw

[cursive swan]

,rotatex

[lyric charm]

not to scale, you mean?

[cursive swan]

Yes.

Not to be done by that.

,rotate

[lyric charm]

...

[cursive swan]

,rotate

[cursive swan]

😦

[lyric charm]

do you want to rotate this picture through 17 more full turns

[cursive swan]

,rotate

[lyric charm]

also

[cursive swan]

Done.

[lyric charm]

where are the vertex labels

and also what's the question

[cursive swan]

Find alpha.

[lyric charm]

and why didn't you give the vertices any names?

[cursive swan]

,rotate

[lyric charm]

@cursive swan what have you tried?

[cursive swan]

Oh.

Let me send the picutre.

I have also marked the intersection of the diagonals.

O.

And a point P (just for fun).