help-0

[thick lynx]

So then the only property that we know is pretty much $\langle \varphi(v), w) \rangle = \langle v, \varphi^(w) \rangle \iff \langle A \cdot v, w \rangle = \langle v, A^ \cdot w\rangle$

So $\beta(\varphi(v), w) = \beta(Av, w)$

[ocean seal]

Kepe

[marsh rapids]

in the standard dot product case

so half in general

[thick lynx]

Let $\lambda$ be an eigenvalue of $A$. Then $\beta(Av, v) = \beta(\lambda v, v) = \lambda \beta(v, v) = \lambda ||v||^2$

[ocean seal]

Kepe

[marsh rapids]

calm down buddy, no eigenstuff needed here

[thick lynx]

Alright

[marsh rapids]

this is also intended from elementary stuff

yeah but I said standard dot product

[thick lynx]

So..?

You mean I should rather write it as < , >?

[marsh rapids]

yeah, beta is unnecessary

[ocean seal]

Kepe

[thick lynx]

So $\langle A \cdot v, w \rangle - \langle v, A^* \cdot w\rangle = 0$

[ocean seal]

Kepe

[thick lynx]

Now how do we write this as one scalar product?

[marsh rapids]

the point of it being the standard one is you can use its definition much more simply

[thick lynx]

Ah

[ocean seal]

Kepe

Kepe

[thick lynx]

Is this what you wanted?

[marsh rapids]

yes

this wouldn't be so recognizable without the standard dot product

[thick lynx]

I compared both sides, though not sure if that's a very rigorous argument

[marsh rapids]

hence the orthogonal group is the matrices whose inverse is their transpose (no matrix in between)

[thick lynx]

Maybe we should argue by plugging in unit vectors for v and w or something like that?

[marsh rapids]

you've just proven a special case of a theorem you've seen

yeah proving they're equal because they generate the same function is a necessary step
Choosing smart vectors is a way

[thick lynx]

Thanks a lot Bezier

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[marsh rapids]

there's more theory about the orthogonal group but ig I'll stop here for now

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[fervent crystal]

theres a second part to this question
b. Calculate an estimate of the lower boundary of the masses of the heaviest 50% of these animals.
a is 159 and b is 636
could someone explain to me how and why the answer to this is 23.5kg
like i first did 50% of 2226 which is 1113 so we know we get heaviest starting from number 1114
but afterwards then i calculated cumulative freq which showed me that 13-32 has 1590
but if we bring that in half to 13-22 and 23-32 which is a freq of 530 and 530 each
then we get to now 1060 which is even closer
then finally using that i got 23kg
but how do u get 23.5

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<@&286206848099549185>

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@fervent crystal Has your question been resolved?

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@fervent crystal Has your question been resolved?

[fervent crystal]

@craggy dagger

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@fervent crystal Has your question been resolved?

[fervent crystal]

<@&286206848099549185>

[fervent crystal]

<@&286206848099549185>

its been 6 hours :(

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[paper fulcrum]

is this not considered the same answer??!?!

[paper fulcrum]

is my professor crazy or are these not the same answer

Did I do grouping wrong?

[quasi vector]

maybe try writing it like $-24x^2\sin\left(2x^3+28)^4\right^3$

[ocean seal]

kheerii

[paper fulcrum]

hmm ok thanks I assume the answer would've been correct if I've written it with the brackets like this lol I didn't realize it made a difference

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[grim pagoda]

I am not sure what the relative minium is

[sour verge]

To get a relative minimum you need the function to go from decreasing to increasing. Looking at your intervals can you see where this happens?

[grim pagoda]

(3/4,1)?

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@grim pagoda Has your question been resolved?

[sour verge]

Well this is just a part where it's increasing.

Find a point where it's decreasing right before it and increasing right after.

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[worldly sequoia]

Hi

[worldly sequoia]

Help with this

.close

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[jaunty sierra]

is the answer 8+2pi?

[exotic canopy]

no

[exotic canopy]

what is the integral of this function from -2 to 2 ?

[jaunty sierra]

8?

[exotic canopy]

you sure?

$\int_{-2}^2 x dx$

[ocean seal]

artemetra

[exotic canopy]

what do you get if you evaluate this by hand

[rustic ruin]

Wait its not 8+2pi? Woah

[jaunty sierra]

0

wait

[exotic canopy]

yep

[jaunty sierra]

yeah 0

[exotic canopy]

OH

sorry

[knotty tapir]

I'm getting that too

[exotic canopy]

it was me who was wrong

[knotty tapir]

(I put the 2 above instead of beneath the curve by accidnet)

[exotic canopy]

yeah it is 8+2pi

sorry lol

[rustic ruin]

Exactly

[exotic canopy]

yes 8+2pi is correct

[jaunty sierra]

okay thanks

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[errant gull]

In a triangle a / b = sin(alpha) / sin(beta) then a - b / a + b = ?

[native cloud]

!status

[lone heart]

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

[errant gull]

1

[lone heart]

@errant gull Has your question been resolved?

[errant gull]

<@&286206848099549185>

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@errant gull Has your question been resolved?

[errant gull]

<@&286206848099549185>

[real gazelle]

is that (a-b)/(a+b)? or a - (b/a) + b? and also what is the meaning of "in a triangle"?

can you just send the original question

[errant gull]

sorry

but i can not send i am on laptop

it is ( a- b) / (a + b)

[real gazelle]

what are alpha and beta then?

[errant gull]

angles

[real gazelle]

are they two angles in a right triangle?

[errant gull]

in a triangle

i do not think right

i think as according to sin law

a / sin(alpha) = b / sin(beta)

[real gazelle]

oh, a is the length of the side opposite to alpha?

this is all information that's necessary to solve the problem 😭

[errant gull]

a / b = sin(alpha) / sin(beta)

[errant gull]

??

you mean it is incomplete

[real gazelle]

I'm saying we can't guess what you mean

[real gazelle]

is this right

[errant gull]

ya

i think

cuz by using sin law

[real gazelle]

are you sure you can't take a picture or something

[errant gull]

sorry man

[real gazelle]

okay well one thing you could try is dividing numerator and denominator by b

then you get (a/b - 1)/(a/b + 1)

[frank briar]

Isnt this true in every triangle?

[real gazelle]

and then you can substitute in a/b = sin(alpha) / sin(beta)

[real gazelle]

yeah

[errant gull]

ya

[frank briar]

Then basically there is no given information you just have to find (a-b)/(a+b)?

[errant gull]

right

btw i have an idea

correct me if i am wrong

What about we use Commando and dividendo theorem

[frank briar]

Yeah thats correct

[errant gull]

ok

[frank briar]

I just looked that theorem up

[errant gull]

let gooooo

[frank briar]

Never heard of it

[errant gull]

Thanks for the help

Bye

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[alpine sable]

m,n,p are roots

find...

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[alpine sable]

<@&286206848099549185>

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[unborn scarab]

<@&286206848099549185>

[glass surge]

?

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[gloomy holly]

I join

[alpine sable]

hhi

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[left bridge]

Hi

Guys how i ask for help with my sum?

@here

<@&286206848099549185>

[unkempt wind]

Literally just write the question here and then someone can try to help

[bright tundra]

Or make your own channel for it

[unkempt wind]

Oh this isn't their channel, oops

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[distant heath]

gi

[distant heath]

help me how to sketch a quad graph from the given equation fx = -2(x-1)^2 + 3 from -4 <= x <= 0

i know that the shape of the graph is downward

i know that the vertex is x = (1, 3)

how do i shape it halp

and how can i start

[open urchin]

try finding two other points and connecting them

[distant heath]

;-;

how

[open urchin]

for example plug in x=2 and x=0 into f(x)

[distant heath]

why that number tho

[open urchin]

any number works but those are easier to draw since 2 and 0 are close to 1

actually nvm those numbers aren't good for the range

pick three numbers between -4 and 0

[distant heath]

-3 and -1?

and -2

[open urchin]

yeah try that

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@distant heath Has your question been resolved?

[earnest juniper]

can anyone explain me how integration came ?

[lone heart]

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[zenith latch]

Prove, without Fermat's and with induction only, that any power of 17 is expressible as the sum of two squares.

[zenith latch]

Fermat's trivializes this so it's not allowed

[slate vortex]

Notice that $17^{n+1} = 4^2 \cdot 17^n + 17^n$

[ocean seal]

Ari

[zenith latch]

oh

[slate vortex]

I'm not sure if that's the correct way but it looks like it could be useful

[zenith latch]

wait

no

i tried this

nvm

sp $17^n = a^2 + b^2$. Then you get $(4a)^2 + (4b)^2 + a^2 + b^2$

[ocean seal]

RedJive

[zenith latch]

<@&286206848099549185>

maybe i should've asked in the elementary number theory channel

[real gazelle]

Perhaps use 17^(n+2) = 17^2 17^n instead

[slate vortex]

that's smart

[real gazelle]

You'll need two base cases tho, one for n=0 and one for n=1

[zenith latch]

right

hm

ok thanks

i'll go ahead an try that

much appreciated to both @slate vortex@real gazelle

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[real gazelle]

you're welcome!

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[warped topaz]

The white is the question and the green is the solution

Im confused about something in the solution, why is it 1/2sqrt(x+1) ? why not x/2sqrt(x+1) ?

[exotic canopy]

the derivative of $\sqrt{x+1}$ is $\frac{1}{2\sqrt{x+1}}$

[ocean seal]

artemetra

[exotic canopy]

so we pick 1/2sqrt(x+1) for h'(x) because it has a known and easy antiderivative

there's no rule to this; with integrals you just have to be very observant but over time something like this becomes very clear

[exotic canopy]

you can try going with your option and you [might] arrive at the result but it will definitely take more work

[warped topaz]

Ah yeye I think I get you

Thank you!!

❤️

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[kind sparrow]

stuck with /(x+2)(x-4) +1

[kind sparrow]

not sure where to go next

im trying to find the equation

got that wrong

?/(x+4)(x-1)

idk about the +1

[hardy prawn]

(x+2)(x-4)/(x+4)(x-1)

(the first two are from x intercepts, last two are from vertical asymptotes, i know that there is no constant multiplying it as if you expand the brackets and compare the coefficient of the highest power it should be 1)

[kind sparrow]

i thought x intercepts where from the denominator

[hardy prawn]

x intercepts are where the f(x) = 0

[kind sparrow]

ah

[hardy prawn]

while the y intercept is f(0)

[kind sparrow]

yeah thanks

hole doesnt matter

its pods

[hardy prawn]

yea i realize what hole means now

[kind sparrow]

which there arent in your equation

[hardy prawn]

i've never heard of it called that but yea

i thought it meant undefined value for a second and got so confused

lmao

[kind sparrow]

the whole time they called it pods until this question lol

thanks mate

[hardy prawn]

weird

no problem :)

[kind sparrow]

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[fickle tinsel]

I think it's easier to simply post the link here:

[fickle tinsel]

https://desmos.com/calculator/lsqndbe1w5

I'm looking to find the area of the spaces between the circles

I was thinking I could theoretically find the area(?) under the spiral by integrating with respect to the angle

But that number is way too big

Also I just noticed that the spiral doesn't go through the tangential points of the circles anyway so its useless

[lone heart]

@fickle tinsel Has your question been resolved?

[fickle tinsel]

I have an idea but I'm not 100% sure

https://desmos.com/calculator/9yixwfrhdo

I could construct triangles from the centers of circles n, n+1, and the origin

And stopping at the triangle with the 7th and 8th circles

Then making a triangle between the 8th and the 1st

I calculate the area of that, calculate the area of the circles partially included, then geometric series the ones that are fully included

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@fickle tinsel Has your question been resolved?

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@fickle tinsel Has your question been resolved?

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@fickle tinsel Has your question been resolved?

[fickle tinsel]

real

new link https://www.desmos.com/calculator/bb72g4sppv

<@&286206848099549185>

I'm trying to find the areas of the gaps inbetween the circles

My current strategy is taking the triangle formed by the centers of $C_n, C_{n+1}$, and $C_{n+8}$ as well as $C_n, C_{n+7}$, and $C_{n+8}$

[ocean seal]

7aman

[fickle tinsel]

You can play around with the T parameter under the "sectors" folder to visualize this

$C_n$ is externally tangent to $C_{n+1}, C_{n+7}$, and $C_{n+8}$

[ocean seal]

7aman

[fickle tinsel]

C_0 is a unit circle a distance d away from the origin

and given $C_n$, $C_{n+1}$ is created by scaling the radius of $C_n$ as well as the distance of it away from the origin by $S$, where $S \approx 0.906331406149$

[ocean seal]

7aman

[fickle tinsel]

(yes I need the precision)

[fickle tinsel]

After that, I subtract the sectors of the circles that are shaded in the triangle

But the problem is that the sectors are somehow calculated to be bigger than the actual triangle

this is all in the "explicit area" folder

(I ran out of variable names so please excuse the arbitrary numbers and letters)

Q is for the triangle formed by $C_n, C_{n+1}$, and $C_{n+8}$ while q is for the triangle formed by $C_n, C_{n+7}$, and $C_{n+8}$

I've triple checked this and I can't seem to find the mistake

And there's no one nearby whose awake, able, and willing to help me

<@&286206848099549185> save me please

[ocean seal]

7aman

[fickle tinsel]

Here's the screencap of the original problem

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[median blaze]

help

[real gazelle]

this channel is already taken, please open your own help channel (see #❓how-to-get-help )

[median blaze]

ok

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[verbal kindle]

I simply can not picture this in my head

how is it the top half of a cone?

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[jaunty mulch]

Im a bit confused

is the reason I can't integrate this funciton using the ln function

immediately

because the denominator is a product expression

[subtle light]

mhm

partial fractions would do the trick

[jaunty mulch]

ok thanks

[rustic ruin]

You can write the numerator as m+(1-m)

[jaunty mulch]

omg i wrote 1-m the wrong way round ha

but yeah i got it now thanks you

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[upbeat escarp]

wait

[gritty bramble]

what is i

[upbeat escarp]

i sent the wrong quesiton

[upbeat escarp]

the square root of negative 1

[upbeat escarp]

so basically what i did is

i first tried to make the fraction easier to work with

by multiplying it by 3+7i/3+7i

coincidentally it became 58(3+7i) / 9(58)

so that was good

so its 3+7i/9

so now i have z^2 = 1 + i + ((3+7i)/9)

[arctic lintel]

yep

[upbeat escarp]

i don't know where to go from there

i could turn the 1 into 9/9

[arctic lintel]

start by splitting that frac

[upbeat escarp]

3/9 + (7/9)i

[arctic lintel]

yeah

and combing the imaginary and real parts

[upbeat escarp]

so like

12/9

• (16/9)i

or ig 4/3 + 16/9 i

[arctic lintel]

from there u just have to find the roots of this

there are a few ways to do that

[upbeat escarp]

uhh

how do i get the roots of

(16/9)i

the real part would just be like

2 root 3 / 3 yea

or 2/root3

[arctic lintel]

u dont just square root each term

try subbing in a+bi for z

[upbeat escarp]

a^2 + 2abi - b^2

[acoustic hull]

@paper portal I have the solution for your last problem

[upbeat escarp]

so (a^2 - b^2) = 4/3

and 2ab = 16/9

?

[arctic lintel]

yeah

which simplifies to ab=8/9

[upbeat escarp]

ye

a = 8/9b

[arctic lintel]

u can do this by inspection i believe

[upbeat escarp]

what's inspection

[arctic lintel]

just looking at it

2 numbers that multiply to 8/9 and the difference of their sqaures is 4/3

[upbeat escarp]

a is -4/3 and b is -2/3

[wanton wigeon]

Here for help
!original

[upbeat escarp]

this is the

q

[arctic lintel]

yeah this looks fine

[upbeat escarp]

how would i get that with a formula

[arctic lintel]

but remember there should be 2 different roots

[upbeat escarp]

yes true

its +-

[arctic lintel]

doing this

by substitution

[upbeat escarp]

substitution?

oh like

[arctic lintel]

sub this for a in a^2-b^2

[upbeat escarp]

a^2 - (8/9b)

[arctic lintel]

yeah

[wanton wigeon]

What about expressing in Euler form?

[upbeat escarp]

= 4/3

[upbeat escarp]

idk that

[wanton wigeon]

Ok imma check

[upbeat escarp]

so if i have a^2 - (8/9b) = 4/3

wait whoops

[arctic lintel]

oh that sub is wrong

[upbeat escarp]

(8/9b)^2 - b^2

= 4/3

[arctic lintel]

yeah

[upbeat escarp]

lemme write it down

would it be fair to say

[wanton wigeon]

can we not just express z=x+iy and square it and compare real and complex parts both sides?

[upbeat escarp]

(64/(81b^2)) - b^2 = 4/3

[arctic lintel]

what we are doing

[wanton wigeon]

Okok

[upbeat escarp]

i've been trying to do that

[arctic lintel]

yep

[upbeat escarp]

so what do i do now

multiply by

81b^2 ?

[arctic lintel]

b^2 should suffice

[upbeat escarp]

true

64/81 - b^4 = (4/3)b^2

[arctic lintel]

almost

4/3 * b^2

yeah

[upbeat escarp]

i have never worked with exponents higher than two lmao

[arctic lintel]

its just a quadratic

[upbeat escarp]

fr?

[arctic lintel]

yeah try letting x=b^2

[upbeat escarp]

bet

64/81 - x^2 = 4/3x

-x^2 - 4/3x + 64/81 = 0

[arctic lintel]

yeah

[upbeat escarp]

can i multiply by -1

negatives confusing

[arctic lintel]

yeah

[upbeat escarp]

so x^2 + (4/3)x - 64/81 = 0

can the last term be divided

nvm

it can't

[arctic lintel]

its not nice

quadratic or completing the square here

[upbeat escarp]

quadratic

is better imo

what do u think

[arctic lintel]

cant remember what i used to do

i think quadratic is just simpler

[upbeat escarp]

its

-4/3 + or - the square root of (((4/3^2 - 4(-64/81)/2)

[arctic lintel]

i think that notation is confusing

[upbeat escarp]

yea

hmm

well when plugging it into my calcl

i have

x = 4/3 and x = -16/9

[arctic lintel]

$\frac{-\frac{4}{3} \pm \sqrt{(\frac{4}{3})^2-4(-\frac{64}{81})}}{2}$

[ocean seal]

Galaxy

[upbeat escarp]

exactly

[arctic lintel]

i think thats good

[upbeat escarp]

so its

b^2

[arctic lintel]

yes

[upbeat escarp]

since earlier we said

x = b^2

[arctic lintel]

correct

[upbeat escarp]

so b^2 = (4/3) and b^2 = -16/9

b = 2/root 3

[wanton wigeon]

bro listen

[upbeat escarp]

and b = -4/3

?

[upbeat escarp]

listening

reading*

[arctic lintel]

nah this is off

[wanton wigeon]

Im getting z as +-√(20/9) * e^i(53/2)

[arctic lintel]

its negative

so its square is imaginary

[upbeat escarp]

???????

[wanton wigeon]

53/2 is in degree

[upbeat escarp]

right

[upbeat escarp]

what does that mean

[upbeat escarp]

2/3 i

[wanton wigeon]

bro listen e^ix=cosx+isinx

So the answer is

[arctic lintel]

yes but remember z is a+bi where b and a are real numbers

[upbeat escarp]

z = 4/3 + (2/3)i

or the negative version of that

[wanton wigeon]

+-√(20/9)[cos(26.5°)+isin(26.5°)]

[upbeat escarp]

when did trig come in

[arctic lintel]

mod arg form

[wanton wigeon]

U all talkin bout this right?

[upbeat escarp]

yess

[wanton wigeon]

If yes then this is the answer

[arctic lintel]

we had the solution ages ago

[upbeat escarp]

oh

u solved it faster

but i dont understand that solution at all

[wanton wigeon]

So what you all were discussing imma getting confused now

[arctic lintel]

fastest is way is by inspection

[upbeat escarp]

yea

[upbeat escarp]

just solving the

system of equations

[wanton wigeon]

Okok

Okok

I was getting a^2-b^2=12/9 and 2ab=16/9

[upbeat escarp]

i wanted to get it without inspection

[upbeat escarp]

hmm

eys

[wanton wigeon]

But Instead of solving I used euler

[upbeat escarp]

yes

why'd you use euler

[upbeat escarp]

is that easier?

[wanton wigeon]

Yes

Direct answer

[upbeat escarp]

oh

how do i do it

[wanton wigeon]

Listen bro every complex number can be expressed as

[arctic lintel]

finding the modulus and arguments of the complex number

[upbeat escarp]

ok nvm

[wanton wigeon]

z=mod(z)*e^i(arg(z))

[upbeat escarp]

i don't know what any of those are yet

[arctic lintel]

u would have needed to cover demoivres

and mod arg form

[wanton wigeon]

Fr bro is correct

[upbeat escarp]

ah

is that in the

dp aa hl course

[arctic lintel]

idk what that ccourse is but you should get taught it

[wanton wigeon]

I don't know bro I studied complex numbers separately

[upbeat escarp]

ah

[upbeat escarp]

oh its this curriculum called IB

i will be taking the highest level of math so im tryna be prepared beforehand cause im not that good in math but enjoy it a lot

thanks guys

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[iron pendant]

A factory that gives rewards and monthly commission . if the employee are 3 types according to their time of work . Workers that works at night receives double monthly commission to those who works at day , Workers who have part-time receives a commission that is 1/3 the commission of who works at night . If the day workers is paid 3600 pounds for each as a commission . Then find the commission of those who have part-time .

[iron pendant]

my aaah alwaays gets it 2400

XD

even though it should be simple

[empty cedar]

!status

[lone heart]

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

[iron pendant]

3. I got an answer but I was told that it's wrong.

[empty cedar]

you know the day workers are paid 3600

[iron pendant]

yes

[empty cedar]

how much are the night workers paid

[iron pendant]

*2 is 7200

*1/3 = 2400

idk really

[empty cedar]

Is this the exact problem?

[iron pendant]

yes

[empty cedar]

Where did you get this problem from?

[iron pendant]

the exact problem

eeh

asignment

[empty cedar]

Do you have a picture of the problem from the assignment?

[iron pendant]

nvm of the logo

[forest marsh]

It should be 2400 indeed

[lone heart]

@iron pendant Has your question been resolved?

[iron pendant]

well idk

[forest marsh]

Mistake from book

[iron pendant]

i dont think its a typo in the book

imma tag a helper

its in the rules

<@&286206848099549185>

[wanton wigeon]

Yo

[iron pendant]

yo

[forest marsh]

The answer should be 2400, but its not in the choice, wanted the opinion of a helper

[iron pendant]

weird that there are no <@&286206848099549185> to be found

[ornate ginkgo]

It should be 2400

[iron pendant]

then a book typo then

and i was right

you are pretty helpful ty

.close

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[hazy bay]

First things first, you need to have decent mental calculation skills

what im typing might be a bit long

try bear with it

@alpine sable
Break Down Problems:
For addition and subtraction, break numbers into parts. Add or subtract hundreds, tens, and ones separately. For example:
(712 + 281): Think “700 + 200,” “10 + 80,” and “2 + 1.” Then combine to get 993.
(37 + 45): Think “30 + 40 = 70” and “7 + 5 = 12.” Add 70 + 12 to get 82.

Adjust problems to work with round numbers, then correct at the end. For instance:
(596 + 380): Round 596 to 600, add 600 + 380, and then subtract 4 to get 976.

Learn to Add Many Numbers at Once:
Reorder numbers to create convenient sums. Look for pairs that add up to 10 or other round numbers. For example:
(7 + 4 + 9 + 13 + 6 + 51): Reorganize as ((7 + 13) + (9 + 51) + (6 + 4) = 90).

Multiply from Left to Right:
Keep track of hundreds, tens, and ones places. Start with the leftmost digit:
(453 \times 4): First, (400 \times 4 = 1600), then (50 \times 4 = 200), and finally (3 \times 4 = 12). Add them to get 1812.
For larger numbers, break it down further. For example, (34 \times 12):
((34 \times 10) + (34 \times 2) = 408).

[lone heart]

@alpine sable Has your question been resolved?

[hazy bay]

its like that for mental calculations

thats why we have a calculator

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[sonic otter]

I wanted to solve this for fun but I realized it's not as easy as I thought so like where do I even begin? 💀

[forest marsh]

Take it geometrically

[west girder]

dx under root 💀

[forest marsh]

First distribute

[dim garnet]

What does that even mean

[sonic otter]

Didn't think of that

[gray isle]

consider odd/even identites for definite integrals

wait

typo?

[forest marsh]

Dx is not suppose to be under root

[gray isle]

looks like that meme problem, and someone fked up trying to copy it

[forest marsh]

Yeah

[forest marsh]

Anyway what do you get after distribution ?

[lone heart]

@sonic otter Has your question been resolved?

[sonic otter]

Sorry, my pc took a while to start.

I think I can figure it out from here, though. 🤔

[forest marsh]

Can you tell what is the parity of the function at right

?

[lone heart]

@sonic otter Has your question been resolved?

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[carmine solstice]

Wasn't negative values unaccepted?

[carmine solstice]

was it due to finding the intersection point, which requires 2 x values, then comparing to find the intersection point

[lone heart]

@carmine solstice Has your question been resolved?

[rustic ruin]

,rotate

[carmine solstice]

why is c -1/2

nvm both 1 and 2 are correct

figured it out myself

.close

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[crude flint]

Can someone help me with (ii) please?

[torpid oasis]

i just wanted some clarification if it is D

Oh sorry

[crude flint]

Np

[merry depot]

Hint: Think about the original triangle inequality and the fact that a = (a - b) + b

[crude flint]

I don't get how that helps 😢

[merry depot]

the triangle inequality says |x+y| <= |x| + |y| yeah?
What if x = a-b and y = b?

[crude flint]

Why is it ||a|-|b|| instead of just |a|-|b|?

$|a|-|b|$

[ocean seal]

DapperJaguar197

[crude flint]

🤣 With the modulus around both

[merry depot]

because you need to make an argument that the modulus is still less than. Should either be a cases situation a<b and b<a, or playing with the order (i think, i haven't written the proof up)

[crude flint]

What does that mean?

[lone heart]

@crude flint Has your question been resolved?

[lone heart]

@crude flint Has your question been resolved?

[crude flint]

.close

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[fallow path]

Is this true?

Is the pink area=blue area?

[severe portal]

Well, ever heard of pythogoras' theorem?

[fallow path]

A bit but not a lot

[severe portal]

Right, it's not much. Essentially, for any given right angled triangle, let the hypotenuse be c, and the two sides be a and b

Then this equation holds true

$$ a^2 + b^2 = c^2 $$

[ocean seal]

StrangeQuarkAL

[fallow path]

Ohh

Kind of get it now

[severe portal]

1. The triangle in the image is always a right angled triangle as it's formed by a diameter and a point on a circle

[fallow path]

Yes

[severe portal]

2. the pink area(semicircle)'s diameter is always the hypotenuse of the triangle

[fallow path]

Oh yeahhhhhh

I get why it has the same area now

Thx a lot

.close

[severe portal]

No problem

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[elder stratus]

Is it correct to use average and standard deviations if you have data where A + B + C = 100% and you know triplicate data was used for (A, B, C)? To me it feels wrong to indicate stdev because I presume adding or removing ±σ randomly to the average results in A + B + C =/= 100%. Is there a different way to handle them?

The specific data I'm working with is the following:

[lone heart]

@elder stratus Has your question been resolved?

[lone heart]

@elder stratus Has your question been resolved?

[mossy field]

just normalize, then +1/3 for A B C

[lone heart]

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[hazy nexus]

can someone explain to me how i would do this?

[jagged cobalt]

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

if that helps

[hazy nexus]

i may be stupid but whats a b and c here

[jagged cobalt]

theyre just arbitrary numbers

you can relate them to what you have here

[hazy nexus]

i know that

but im saying

which number represents a, which represents b and which represents c

[jagged cobalt]

have a guess

[hazy nexus]

2/3 is a

x+4 is b

x is c

?

[jagged cobalt]

not quite

[hazy nexus]

😭

[jagged cobalt]

when i said the rule i was more just focussing on solely the first term

after you can factorise

[hazy nexus]

ok thanks i should b good

i got it correct ty

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[vocal tapir]

Vectors, colinear with a line, form a 1D real linear space

[vocal tapir]

our lecturer said that on a line exists a single linearly independent vector

but how?

[summer dirge]

what's the definition of linear independence?

[vocal tapir]

the linear combination for a certain amount of vectors is the 0 vector only and only when the scalars are 0

[summer dirge]

right

if we have two collinear vectors, what happens?

[vocal tapir]

they are linearly dependent

[summer dirge]

right, why?

[vocal tapir]

they can be expressed as one another?

they are collinear*

[summer dirge]

they can be expressed as a scalar multiple of the other vector yes

so if we have vectors v and w which are collinear

then by definition, v = kw for some nonzero k in F

so v - kw = 0

this is a nontrivial linear combination of v and w that equals the 0 vector

therefore v and w are not linearly independent

so on a line, we can have only one linearly independent vector

does that make sense?

[vocal tapir]

right yeah

[summer dirge]

[vocal tapir]

but

what about more

[summer dirge]

what do you mean more?

[vocal tapir]

ah nvm

nothing

ty

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[summer dirge]

like if we had three collinear vectors?

you can just set the coefficient on one of them to be 0 and do the same thing we just did in that case

still a nontrivial lin combo that hits 0

so not LI

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[alpine sable]

was helping a friend out with this porblem and this is a new topic for me too was hopign someone could just make sure i am not leading them in the wrong direction

[cinder sundial]

Grammar sir

[alpine sable]

also im not rly sure how to progress with this

[cinder sundial]

What is “D60s”?

A 60-sided dice?

[alpine sable]

yeah

same with d100

100 sided

[cinder sundial]

That’s insane

Fk

I think I got it

@alpine sable Sailor

Could you give me the correct answer please

[alpine sable]

uh

gpt says53,970,463,281.

i dont know how it got there

[cinder sundial]

Let me check

[alpine sable]

i only checked gpt to make sure i was doing somethign right

[cinder sundial]

No

It is too big to be the answer

[alpine sable]

is my work so far wrong

[cinder sundial]

I got (30+31+…+89) + 29

[alpine sable]

i like

genuinely do not know

how to go further with this

or if im even doing this right anymore

[cinder sundial]

Set D60s as some number

[alpine sable]

i set them as x

and the d100s as y

and set them up as an equation where if you added all 8 it equals 120

where x is 1-60

and y is 1-100

and then i made them a fucntion

by subtracting by 1 so it would be easier to make it

so it became x!1 where X1 is 0-59

and y!1 etc

where y! 0-99

[cinder sundial]

Emm

I thought you was unable to solve it?

[alpine sable]

well

after i made the function

and grouped the 2

thats where i couldnt

thats just what went in my head

at the time

[cinder sundial]

I didn’t make any function tho

I just sort of operate it in my mind and get the number

Let A to be the D60s

B and C to be the two D100s

Note that B and C are identical.

Then for the number of solutions where A=[1,60]

You there?

[alpine sable]

Yes

[cinder sundial]

the number of solutions where A=[1,60] should be (30+31+…+89)

[alpine sable]

How did you get here

[cinder sundial]

That’s a flawed claim tho

[alpine sable]

<@&286206848099549185> can we have someone check both of our statements I am lost rn…

[cinder sundial]

You could actually set A as 60 and operate it to get the number of solutions within such limitation.

You would get there are 30 ways to distribute 60 to B and C.

[wanton wigeon]

Ho

Yo

[cinder sundial]

Ho

[wanton wigeon]

Yo

Here for help

[cinder sundial]

YoYo

[alpine sable]

Hi

[wanton wigeon]

Hi

[cinder sundial]

Check the answer

[wanton wigeon]

Ok

Mark the msg

[cinder sundial]

Luv u

[wanton wigeon]

Aw 💞

[cinder sundial]

Here

[alpine sable]

Here

[alpine sable]

And then this is what I did

[alpine sable]

And idk this to

[wanton wigeon]

Aah

Probability

we have to add to 120 rigjt

100 and 60

[cinder sundial]

It is permutation, no?

[wanton wigeon]

Oh yes PNC they only asked ways

Like take d 60

[alpine sable]

With 6 d60 and 2d100yeah

[cinder sundial]

My bro is on something strong

[wanton wigeon]

Yes

Bro just woke up

Yea took strong black ☕

Tell me

How u all approaching this one

Basic counting or making cases individually

[cinder sundial]

Sure

[alpine sable]

I did this

[cinder sundial]

That’s my approach

[wanton wigeon]

My guy

Let's go with it

[cinder sundial]

So what’s the answer you got

[wanton wigeon]

Bro imma start doing now

[cinder sundial]

Cool

I solved it within like 4 minute

I wish you could outstrip me

[wanton wigeon]

Bro imma not so good at combinatorics

Trying but

Na bruh

[cinder sundial]

Show me your answer

[wanton wigeon]

I am stuck bro

Me not so good

[cinder sundial]

It is fine

[wanton wigeon]

I took 6D60 as X and 2D100 as Y

So X lies from 6 to 360

And Y from 2 to 200

We need solution for X+Y=120

Help me proceed

[cinder sundial]

I think there’s no need to model it as some function?

[wanton wigeon]

But I think it will get simplified what u think?

[cinder sundial]

No.

The case is already simply

[wanton wigeon]

Ok bro what answer did u come up with

[cinder sundial]

^

I split it into 2 cases

[wanton wigeon]

Yea bro listening

[cinder sundial]

The first is A=[1,60] and B=[30,89].

[cinder sundial]

^

[wanton wigeon]

Okok

[cinder sundial]

Then you got (30+31+…+89) for the number of solution in such limitation

[wanton wigeon]

Bro listen the problem is that I am getting stuck let the sum of D60s bes X then the sum on D100s should be 120-X

Now X has minimum value of 6

So 120-X=114

[cinder sundial]

That’s the problem

[wanton wigeon]

Now on D100s there are many ways for 114

[cinder sundial]

You are overcomplicating it

When you model it as a function

Like these algebraic assumptions and operations is unnecessary

[wanton wigeon]

Listen

If X is 6

Listen bro pls

[cinder sundial]

No

[wanton wigeon]

There are I think 85 ways for 120-X to 114

[cinder sundial]

Im quit bro

[wanton wigeon]

Ok bro you tell

[cinder sundial]

I love you

[wanton wigeon]

Sry bro imma out

Me not good at combinatorics

[cinder sundial]

But you are good at treating me right

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Do I just ping helper again then

[cinder sundial]

Sure

[alpine sable]

<@&286206848099549185>

[alpine sable]

<@&286206848099549185>

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

@craggy dagger

[rigid dew]

stop pinging helpers

[alpine sable]

Yeah my bad for asking for help

Sorry!

Guess my shit is too hard

[rigid dew]

are you being sarcastic

[rigid dew]

https://discord.com/channels/268882317391429632/488120190538743810

[lone heart]

@hollow aurora Has your question been resolved?

[lone heart]

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[worldly jungle]

I have got the formule = 3x - 5y = 10. and i want the 5y infront so -5y= 3x +10

[worldly jungle]

But my aswer book says its -5y = -3x + 10

Why is it -3x and not 3x?

[hollow vapor]

youre changing the side the 3x is om

thats like saying 3-2=1 —> -2=3+1

doesnt make sense

[worldly jungle]

I see, but why does the 10 stays a 10?

[hollow vapor]

cause we dont touch the 10. its already on the other side

an example, 3-2=1, if we change that its -2=-3+1

you see how we only moved the 3

the 1 stays there so it stays pos

[worldly jungle]

oh like that

I understand

thnx m8

.close

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[devout cypress]

a little lost on how to proceed here, what steps can i take?

[subtle light]

see how you have a x^{1/3} dx already in your integral

youre just missing that 4/3 to replace it with du

so you can multiply by 4/3 and divide by 4/3 to do that

[harsh rain]

constants don't matter in substitution, you just need the derivative in your integral

[devout cypress]

Oh shit yeah

is this process correct

[subtle light]

looks good to me

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@devout cypress Has your question been resolved?

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[alpine sable]

why is there an exclamation mark at the end

!status

[lone heart]

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

[limber ember]

1

[alpine sable]

oh ok 😭

you know how to find the projection of F1 vector on x axis?

[limber ember]

nah we didnt learn that

i know what a projection is

[thorn heron]

Is it option C

?

[limber ember]

yeah lol

wow

[thorn heron]

Have you learnt resolution of vectors

?

[limber ember]

just theory

not really in practice

[thorn heron]

Like what theory?

[limber ember]

components of vectors can be added up using the triangle or parallelogram rule

how did you get C

[somber crest]

consider just the vector F1

draw a right triangle, with legs parallel to the axes and F1 its hypotenuse

that should also contain the 60 degree angle

ok better description, draw a vertical line from the tip of vector F1 to the x-axis

now that's a right triangle lol

[limber ember]

yeah i got it

[somber crest]

the (magnitude of) the x-component of F1 is now just (the length of) the bottom leg

[limber ember]

right

[somber crest]

and we know from trig tricks that the cosine of an angle is the same as the lengths of the adjacent side over the hypotenuse

call the adj side length x momentarily, we have that the hyp length is 10 and the angle in question is 60 degrees

let me try that again

it's adj length over hyp length

hence cos(60) = adj / hyp = x/10

thus x = 10cos(60)

and cos(60) is discernible

[limber ember]

yeah 1/2

[somber crest]

and thus the horizontal component of F1 has magnitude 5

the same procedure can be done for F2

then the resulting force is just the vector sum of F1 and F2, having horizontal component just the sum of F1.x and F2.x

[limber ember]

and then they just add up?

[somber crest]

yup

[abstract bone]

i love physics vectors ❤️

[somber crest]

they're like real vectors so it's nice

[limber ember]

makes sense, i tried adding F1 to F2 and drawing triangles but kept getting nowhere

[somber crest]

yeah it's difficult because the resulting force may have a more annoying angle

it's easier to determine the angle from the components, rather than the reverse

at least I think so

*of the sum of two vectors

[limber ember]

i see, thanks

.solved

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[brazen wadi]

how should i do this

the z component is here

[alpine sable]

dot product

[brazen wadi]

ohhhh

right

.close

[lone heart]

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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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[glossy niche]

the question needs me to find two pairs of coordinates and graph them and write where they intersect, finding the intersection point is easy but I can't graph it correctly

[glossy niche]

I tried substituting 0 into x and y for both equations and got x = 2.5 and y = 5 for the 1st equation, and x = 0.8 and y = -4 for the second equation

[simple loom]

I feel like you'd have to divide 4 with -5 to find x in the second equation

X should be negative afterwards