help-38

[plain tusk]

like that?

[tidal igloo]

Exactly

[plain tusk]

cant you do 5/9 x t before u sub it in

[tidal igloo]

Now solve for t

[tidal igloo]

Sorry what do you mean

[plain tusk]

as in 5/9 x t would be 5t/9

[tidal igloo]

Yes that works

[plain tusk]

ok

[tidal igloo]

Multiplication is associative and commutative so it doesn't matter what order you do it in

7 * (5t) = (7*5)t = 7t)*5

[plain tusk]

um is this right?

that should be a plus symbol

sorry

[tidal igloo]

Looks right

Well done

[plain tusk]

got 54

[tidal igloo]

Can you find c now

[plain tusk]

yes

my final answers

[tidal igloo]

Now check if the inital equation holds

[plain tusk]

yep

it works

can you help me out with one more question if u dont mind?

@tidal igloo

its fine if u cant

[tidal igloo]

Go on

Sure

[plain tusk]

here

[tidal igloo]

Oh good god i haven't done that shi in so long lmao

Give me one moment lemme see

[plain tusk]

lol

i did annotate a bit

the Z is corresponding angle i think

[tidal igloo]

Yo

Sorry gimme 1

[plain tusk]

no worries

[tidal igloo]

Sorry mate i remever jack shit from hs geometry

[plain tusk]

oh lol no worries

[tidal igloo]

But it might be useful to construct BD then calculate ODB

By iscoleces triangle

Then u use alternate segment and inscribed angle thereoms?

[plain tusk]

o yea

[tidal igloo]

Something like that

Try doing that

[plain tusk]

whats inscribed angle theorem

[tidal igloo]

Angle at the circumference is half the measure of its intercepted arc?

Looked that up lol

Not sure if its true

It is teue

[tidal igloo]

That looks right

[plain tusk]

i know which one you're talking about but where do u see that

is this correct?

[tidal igloo]

Where do u get BAD being 32 from

[plain tusk]

i thought it was alternate

[tidal igloo]

If thats right then u can calculate the center angle

By angle sum

Do voa

Calculate base angle

Subtract

[plain tusk]

i got 39+32

71

wya

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[lunar stirrup]

can someone help w this summation

[trim lichen]

!status

[trim jolt]

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

[lunar stirrup]

1

[trim lichen]

ok

let $f(x) = (1+x)^n = \sum_{k=0}^n {}^nC_k x^k$

[solid kiln]

Ann

[trim lichen]

observe that $\sum_{k=0}^n {}^n C_k = f(1) = 2^n$ as a starting point

[solid kiln]

Ann

[trim lichen]

try looking at $f'(x)$ and $f''(x)$ and tying them back to your summation somehow

[solid kiln]

Ann

[lunar stirrup]

oh ok i tried f'(x) but i didnt get anything useful

let me try f''

[trim lichen]

you should get coefficients k * nCk from that

which is a step in the right direction

[lunar stirrup]

im not really understanding how to get our series from this

[trim lichen]

i mean you are making things difficult for yourself by not using sigma notation

[lunar stirrup]

oh this helps me think clearer

[trim lichen]

$f(x) = (1+x)^n = \sum_{k=0}^n {}^nC_k x^k \ f'(x) = \sum_{k=1}^n {}^nC_k \cdot k x^{k-1} \ f''(x) = \sum_{k=2}^n {}^nC_k \cdot k(k-1) x^{k-2}$

[solid kiln]

Ann

[lunar stirrup]

okay yeah isnt that what i got as well

[trim lichen]

yes but

$f'(1) = \sum_{k=0}^n {}^n C_k \cdot k \ f''(1) = \sum_{k=0}^n {}^n C_k \cdot (k^2 - k)$

[solid kiln]

Ann

[trim lichen]

(padding both sums with 0 terms to make the indexing go from 0 to n in both)

k^2 - k is k(k-1)

are you picking up what im putting down now

[lunar stirrup]

not really sorry....are u saying that adding both these will make the n C k .k term cancel?

[trim lichen]

yes

and you will be left with exactly the sum you were looking for

[lunar stirrup]

ohh ok got it thank you

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[latent jay]

The graph of the function y = x^2 was displayed symmetrically with respect to the abscissa axis, and then moved parallel 2 units up along the ordinate axis. What function did you get?

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[supple sun]

We need to calculate the volume of a figure bounded by the surfaces given by the equations:

1. (this defines a plane),

2. (this defines an ellipsoid).

[supple sun]

It is also limited by y>0 x>0 z>0

,rotate

4.39

[spiral ocean]

,rotate

[fierce rain]

,rotate

[fierce rain]

Which part

Am i looking at

[trim jolt]

@supple sun Has your question been resolved?

[supple sun]

4.39

but it is in polish

so let me explain

[fierce rain]

Sorry idk epsiloid

U can ping helpers

Since it's been 15mins

But right the explanation down

[supple sun]

<@&286206848099549185>

[fierce rain]

So when they come here

They can see the help

[supple sun]

so we have to calculate the volume of a solid that is created by surfaces given in exercise so x^2/a^2+y^2/b^2+z^2/c^2=1 ,x/a+y/b=1 and x>0 y>0 z>0

[fresh merlin]

ah yes texit light mode

pog

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@supple sun Has your question been resolved?

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@supple sun Has your question been resolved?

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[fast bronze]

hello

[fast bronze]

we have
A in (0;4)
B(-3;1)
now they told us to get the coordinates of E which is the point that is half of [BC]
then
they told us to draw D the picture of A with the rotation and the center of it is E and the angle of it is 180 degrees
then
they said get the coordinates of D
then
they said to probve that ABDC is a rectangle
if i can just prove that one of his angles is a right angle
i can prove that its a rectangle

i need help

with proving ABDC is a rectangle

i answer all the other questions

why s nobody answering me

.close

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[fast bronze]

Hi

[fast bronze]

i m here to get helped with how to prove ABDC IS A RECTANGLE

here is what i got

we have
A in (0;4)
B(-3;1)
C(5:-1)
now they told us to get the coordinates of E which is the point that is half of [BC]
then
they told us to draw D the picture of A with the rotation and the center of it is E and the angle of it is 180 degrees
then
they said get the coordinates of D
then
they said to probve that ABDC is a rectangle
if i can just prove that one of his angles is a right angle
i can prove that its a rectangle

oops wait

k done

so

can you help me please 😄

[acoustic path]

Sorry

[fast bronze]

?

ok ima wait

[vast egret]

$x_E = \frac{x_B + x_C}{2}$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[fast bronze]

no no ive don't this alr

it got me

(1;0)

and i got D at

(2;-4)

i need

to get help with

how to prove

its a rectangfle

ABDC

[vast egret]

Do u know scalar product?

[fast bronze]

wait ima translate that

[vast egret]

Produit scalaire

$\vec_{n}$ $\cdot \vec{v}$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vast egret]

That thing

[fast bronze]

ngl

ive never seen that

rule before

the laser n X the laser v

[vast egret]

I meant vectors

$\vec{AB} \cdot \vec{AC} = 0$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

That means angle BAC = 90°

[fast bronze]

sorry

i was disconected

anyways

oh

so

if i do that

[vast egret]

Did u learn that rule?

[fast bronze]

BAC is 90 degrees

ye but the teacher never said that that means an angle is a right angle

[vast egret]

It means that they are perpendicular.

Therefore, the angle is 90.

[fast bronze]

ok

[vast egret]

Since A is repeated, this means that its angle is equal to 90°.

(AB . AC) = 0

But, calcul vectors AB and AC first, then use scalar product

Or, u can use vectorielle product

$\vec{AB} \wedge \vec{CD} = \vec{0}$

[fast bronze]

ok

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

Alright

!done

[trim jolt]

If you are done with this channel, please mark your problem as solved by typing .close

[fast bronze]

.close

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[cinder elm]

goofy ahh problem but I'm not completely sure about the logistics of it

[wanton rune]

‎!status

hmm

Well, where are you confused

[cinder elm]

I'm not sure where I would even plug t in

[wanton rune]

do you know how to solve a separable differential equation

(c) doesn't necessarily need you to solve for y(t)

[cinder elm]

Yeah I just don't know how I would integrate it

Would it be like arctan?

[wanton rune]

you need to use partial fractions

https://tutorial.math.lamar.edu/classes/calcii/partialfractions.aspx

[cinder elm]

could you just show me where specifically I would decompose it tho

[zinc ginkgo]

show your separable form

[zinc ginkgo]

read example 1 here

and/or 6

[cinder elm]

and also how would I solve a

[zinc ginkgo]

put over a common denominator and match coefficients

or do "cover up method"

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

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[signal island]

hey so iget how lim as x approches 0 of sin(x)/x =1, but would the same thing for sin(x)/2x be 1/x, or would sin(2x)/2) be 1? or would sin(3x)/2 = 3/2 (all in the sense of the previously mention limit: limit of function as x approaches 0). Can someone help me expand on this?

[bright quarry]

$\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}$

[solid kiln]

knief

[signal island]

oh that makes a lot of sense

what if x approaches something other than 1?

would the behavior and limit be different?

[bright quarry]

it doesn’t approach 1

it approaches 0

[signal island]

thats what i meant

mb

[lilac flame]

obviously

[red loom]

Similarly:

[\lim_{x\to0}\frac{\tan(ax)}{bx}=\frac{a}{b}]

[solid kiln]

PajamaMamaLlama

[signal island]

how different?

[signal island]

neat

[signal island]

or different how?

oh ok

[lilac flame]

nvm

[signal island]

so if im approaching like 3 for example, how wou;ld the limit change

[bright quarry]

🤔

[lilac flame]

well if ur approaching p != 0 the limit is sin(ap)/(bp)

[bright quarry]

then it would be sin(3a)/3b?

[lilac flame]

just by continuity

yes

[signal island]

i dont get it

[lilac flame]

just plug in x=p

[red loom]

because the function is defined and continuous for all (c\neq 0) we can say:

[\lim_{x\to c}\frac{\sin(ax)}{bx}=\frac{\sin(ac)}{bc}]

However, when (c=0) we can see that:

[\frac{\sin(a\cdot 0)}{b\cdot 0}=\frac{0}{0}!!]

Thus more analysis is required and indeed it can be verified that:

[\lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}]

[solid kiln]

PajamaMamaLlama

[signal island]

makes sense

im done

somebody do that command thingy

[zinc ginkgo]

.

[red loom]

.close

[zinc ginkgo]

#help-38 message

[signal island]

!close

someone close this

[dull island]

.

not !

.close

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[quartz mauve]

Okay so I don't get how this distributes

I thought it would distribute the way I did it but it jsut pushed the a to the side?

[red loom]

idk what you mean by how this distributes but note that (f\left(-\frac{a}{5}\right)=-6\left(-\frac{a}{5}\right)+3)

[solid kiln]

PajamaMamaLlama

[ionic pendant]

multiplication doesn't "distribute over division"

e.g. $5 \times \frac 34 \ne \frac{5 \times 3}{5 \times 4}$

[solid kiln]

cloud

[quartz mauve]

yeah you plug

say 1x

and x=2

1*2

[red loom]

ohh I see what you mean by distrubite now, but yeah look at cloud's example, you multplied 6 to top and bottom

[jaunty mica]

hi

[red loom]

which is not proper multiplication of fraction and a number

[quartz mauve]

okay how do I do it ?

like -6/1 * x

woudlnt that still give 6a though

Oh I see that is 6a

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[proven jewel]

would someone help me understand each portion of the solution here

so i understand that its mostly

1/2 base times height triangles, a trapezoid and a rectangle

pls help me go over each triangle from left to right

[zinc ginkgo]

instead of trapezoid, split it into rectangle and triangles

[proven jewel]

yeah but im also trying to understand how they did the trapezoid

for the 3rd one,

how did they get

1 (3 +1/ 2) here

[proven jewel]

yeah its easier for sure

let me try

[vast egret]

Area of trapezoid is: h × (long of base + long of upside base) + 1/2

[proven jewel]

would this part be

[vast egret]

h × (a + b) × 1/2

[proven jewel]

like 1 times 1 for the first trapezoid (i made it into a rectangle)

-1 times -1

[vast egret]

First trapezoid = 2

-1 × (-1 -3) × 1/2

||Negative part||

❗️The area must be positive, too❗️

[proven jewel]

so

-1/2 + 1/ 2 ( 2) - 1/2 - (1)(1) - 1/ 2 + 1/ 2 (2)(2) + (2)(2) + 1/2 - 1/2 (2)

im not using the trapezoids

[vast egret]

-1/2 must be the first triangle, right?

[proven jewel]

yes its from left to right

[vast egret]

Area of triangle = -1/2

[proven jewel]

isnt it negative that part because its below 0

[vast egret]

Owh, we aren't calculating |h(x)|

Ok, let's check

[proven jewel]

i keep doing it wrong

not ending up with 9/2

[vast egret]

Before Last 1/2 (2)

[proven jewel]

huh

[vast egret]

That + 1/2 - 1/2(2)

It must be 1/2 (2) - 1/2(2)

[proven jewel]

owhh

ur right

ok let me simplify

[vast egret]

The two triangles are similar.

[proven jewel]

whatt im confused im doing simple math wrong

but the one on the very bottom is correct yes?

[vast egret]

Yes

Don't u use × ?

I see u using (2)

[proven jewel]

i do

[vast egret]

$-1 -\frac12 +\frac12 \cdot 4 + 4$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

U removed -1

Recheck

-1/2 -1/2 = -1
And 1/2 × 2 = 1
So -1+1 = 0

[vast egret]

$$-\frac12 + \frac12 \cdot 2 - \frac12 - 1 -\frac12 + \frac12 \cdot 4 + 4 +\frac12 \cdot 2 -\frac12 \cdot 2$$
$$= -1 + 1 - 1 -\frac12 + 2 + 4 + 1 - 1$$
$$= -\frac22 - \frac12 +6$$
$$= -\frac32 +\frac{12}{2}$$
$$= \frac92$$

[proven jewel]

yesss?

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

Idk how -2/2 -1/2 = -3/2 + 4/2
But u did it

Congratulations

[proven jewel]

-3/2 + 4/2

[proven jewel]

the next question relating to this is

what is the average value of h(x) on [ -6 , 6 ]

[vast egret]

Still wrong

[proven jewel]

whatttt

let me do it again

[vast egret]

Bruh

[vast egret]

Copy this

And study it

Until enlightenment comes to u

[proven jewel]

so

-1/2 + 1/2

is 0

whats left is 2

2- 1/2

[vast egret]

Wait

2 × 1/2 = 1

[proven jewel]

yes

[vast egret]

That's multiplication

[proven jewel]

-1/2 + 1

[vast egret]

Yes

[proven jewel]

is 1/2

[vast egret]

Continue

[proven jewel]

then

1/2 - 1/2

= 0

so we got -1 - 1/2

[vast egret]

Yes

[proven jewel]

-3/2

so its -3/2 + 1/2 (4)

and thats equal to 1/2

frick i got lost

[vast egret]

You are one step away from success

[proven jewel]

1/2 + 4

= 9/2

9/2 + 2/2 = 11 / 2

11/ 2 - 2/2 = 9/ 2

[vast egret]

Yes

My turn, lemme show u the fastest way

[proven jewel]

great thanks

i know there were some opportunities to cancel out alot

but thats how i got alot of mistakes

[vast egret]

If u got 1/2 × 2, make it 1

Then calcul

Dont use 1/2 × 2 × 2 - 1/2 × 2 while calculating

Make it 2 - 1

That's easier.

[proven jewel]

okay MOVINGG ONN

average value of h(x) on [ -6, 6 ]

[vast egret]

Average: $\frac{1}{b - a} \int_a^b h(x)\ dx$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[proven jewel]

it says here

9/2 / 12 = 9 / 24 = 3/8

[vast egret]

That's the rule

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

Use it

[proven jewel]

so the h(x) here

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

?

[proven jewel]

is the 9 /2

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

Yes

[proven jewel]

and that goes on top of the 1

[vast egret]

???

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[proven jewel]

my messeges are not sending

is 9/2

but

the h(x) is 9/2 and it replaced 1

bro my messeges are not sending?

then i just simplify

[vast egret]

1/(b-a) × int h(x)

[proven jewel]

i understand

okay i understand now thats easy enough

[vast egret]

What do u mean it replaced 1?

[proven jewel]

becusse u move whatevers the h(x) to the left of the integral sign

[vast egret]

U can do $\frac{\frac92}{b-a}$ or $\frac{1}{b-a} \cdot \frac92$

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

It's the same

[proven jewel]

yeah that makes sense

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

???

[proven jewel]

okay okay cool

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[vast egret]

I think it broke down.

[proven jewel]

might have did lol my messeges are delayes

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[proven jewel]

im gonna close this

[vast egret]

Do u know values of b and a in ur question?

[solid kiln]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[proven jewel]

.close

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[rocky arch]

Hello i dont know the steps for solving it but the answer should be x=0 and x=log4/log u btw

[zenith nexus]

Eh? I'm guessing it's supposed to equate to zero?

[rocky arch]

yes

youre right sorry i forgot that

[rocky arch]

Do you know how to do it? can you help me please

[whole coral]

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[high heron]

(moved to #help-3 )

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[copper sinew]

I feel like these values are not really close to what the answers are

Can anyone check me

[dusky thunder]

,r

.r

i do not know how to rotate an image

[hollow cypress]

,rccw

[dusky thunder]

thnak you

[copper sinew]

First try

[hollow cypress]

you have to specify the direction

cw or ccw

[copper sinew]

I got like 26 as the local min and 17.6 smt as the local max

[dusky thunder]

,w plot 1.07x^3-11.17x^2+23.71x+3.36 on -2<x<10

[solid kiln]

Results provided by WolframAlpha

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

Requested by _mia_34

[hollow cypress]

is it possible theyre asking for the turning point

[hollow cypress]

(still that wouldnt be correct)

[copper sinew]

Ya

[hollow cypress]

however it is closer to one of the options

[copper sinew]

True

[hollow cypress]

circle the BEST answer

[copper sinew]

I think he took this from a previous ap test

[hollow cypress]

the q never specifies correct 😭

[copper sinew]

With half changed number💀

[copper sinew]

Bro😭

Thanks

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[hollow cypress]

ay wait is that right?

do you have the answer scheme?

@copper sinew

@copper sinew

i think the qs is asking what's the most extreme points of the given that lies within the range

[copper sinew]

Yes

I checked that too tho and it’s again a bit off

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[dense zenith]

I need help with b

[dense zenith]

I just don’t really understand how you find RS or NS

[dusky thunder]

ah okay you actually need to write a line equation of MN and OR

in the form of
r(ß) = position vector + ß(direction vector)

[dense zenith]

ooo alr lemme try

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[chilly mulch]

Why is k, the scale factor, the square root of the determinant in this question?

I thought the scale factor was always just the determinant of the matrix

not the square root

later on in the following part of the question, we use the scale factor as 64, and multiply the area by 64 instead of 8

Which just makes me even more confused

my only assumption is they're using length scale factor for part a) i) but I don't know why they would do that?

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@chilly mulch Has your question been resolved?

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@chilly mulch Has your question been resolved?

[lime sphinx]

Oh i misunderstood

but I can say the area of the transformed square is det(M)(2a)^2 = 64*(2a)^2

because the determinant geometrically is how much multiplicatively area are scaled by a linear map

[lime sphinx]

tbh it depends how they defined enlargement. If their def is this

then it make sense imo. The scaling factor k is how much each axis is scaled not the area, but i agree this would be confusing

[chilly mulch]

Yeah that does make sense, but them saying scale factor without giving any information about the matrix feels a bit
... silly in my opinion. We work out the enlargement matrix later on and it makes more sense that way

Thanks 🤦‍♂️

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[split crypt]

I know the answer to this problem, but I don’t understand what steps I need to take to get there. I’m currently learning cos, tan, sin, and I know that those are used to figure this out, but I don’t know when and why to use those functions

(Both parallel lines are square)

[karmic acorn]

,rotate 270

[thorn bay]

what are the numbers on the diagram?

[split crypt]

Yeah sorry.
I meant it to be a coordinate plane. 5.76 is the length of the vector. The vector is (-5.44, -1.90)

[mystic iris]

heyyyy

[mystic iris]

what function lets you find the angle here?

[split crypt]

Yeah pretty much. I also know the hypotneus

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[rapid folio]

He asked which function will help u find the angle

[split crypt]

Oh sorry

[split crypt]

I believe it’s arctan..?

[mystic iris]

yeah

[split crypt]

Thank you!

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[wide narwhal]

Where did I go wrong

[wide narwhal]

I don’t understand how to take the integral of e^2x I thought it’s just e^2x

[versed jasper]

but what happens when you differentiate that

[wide narwhal]

2e^2x

[versed jasper]

yep

that's not equal to the original

but if you perform u substitution

[wide narwhal]

Um

[versed jasper]

you'll get a coefficient of 1/2 correct

[wide narwhal]

I don’t understand

Like I understand that 1/2e^2x is the correct answer

And derivitive if that is e^2x

But I don’t understand how to get that

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[smoky kite]

Why can’t the power be 1/alpha -1 instead

[whole coral]

I suspect that it should be are you told anything about alpha?

[pallid lake]

it's the same?

[whole coral]

Well actually, there’s a negative sign from flipping, but still-

[pallid lake]

should still be 1/(alpha - 1)

[smoky kite]

It’s a parameter between 0 and 1

[whole coral]

Yea it should be 1/(1 - alpha), I think they meant

[smoky kite]

Okay thanks

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[autumn violet]

Determine all positive integers $n$ such that $\phi(n) + \tau(n) > n$. I have arrived at the fact that $\phi(n) + \tau(n) = n+1$ under the constraints of $n$ and my conjecture is $n=1,4$ or $n$ is a prime number. Aops forms ain't doing shi so can I get some help?

[solid kiln]

ProjectTime

[stable ore]

Your conjecture is correct. I encourage you to think about the sets of naturals that correspond to \phi(n) and \tau(n).

[autumn violet]

That was my aops hint but I didn't get anything else besides the equation

[stable ore]

Well, what can we say about their union? Their intersection?

[autumn violet]

Their intersection is always one and their union at maximum can be n right?

[stable ore]

Right.

So what can we say about every number less than or equal to n?

[autumn violet]

Can you elaborate further on that?

[stable ore]

Well, call the set of numbers coprime with n P and the set of divisors T. You've said that P cap T is {1}, and that P cup T subseteq {1,\dots n}. This means that given |T cup P|>n, we can say what about every element of {2,\dots n}?

[autumn violet]

It has to be an element of either P or T?

[stable ore]

Yeah. Now say we have a composite number that isn't 4. What can we say about its prime factorization?

[autumn violet]

It is the product of at least 2 distinct primes?

[stable ore]

Right, but we can actually get a little more than that.

What has to be true of at least one of those primes?

[autumn violet]

It's not 2?

waot

wait

yeah

?

[stable ore]

Yeah.

[stable ore]

Now going back to this really quickly, this clearly means that we can't have some composite m<n such that m shares exactly one prime factor with n.

Why do this #help-38 message and this #help-38 message break that?

[sharp heart]

Are you leaving out composite numbers like 9?

[stable ore]

Oh, whoops! Not necessarily distinct primes, yes.

But that doesn't matter.

[autumn violet]

Yeah I guess but I proved that it can only be 4 with repeated primes

[autumn violet]

Also can you give a brief explaination of this my mind is fried after thinking about this problem for 4 hours

[stable ore]

Well being in P or T (and not 1) means a given m<n either shares no prime factors with n or all of its prime factors with n.

Why does that imply what I said?

[autumn violet]

Because if it has more than one factor, it either shares no prime factors or shares all of them. If it just has one prime factor then it's not composite?

[stable ore]

Yeah that's basically it.

Now why does this statement break if we have prime factors p,q with p>2?

[autumn violet]

You mean p,q prime factors of pq?

[stable ore]

p,q prime factors of n.

[autumn violet]

If n is even, then we have ${1,2,3, \dots 2k}$. Since there must exist at least one integer $m$ less than $k$ relatively prime to $k$, there exists an integer $2m$ in ${1,2,3, \dots 2k}$ for which $2m$ and $2k$ only share the factor two. If n is odd with a prime factor $p$, there exists a number less than $n$, $2p$, which only shares a factor $p$.

Oh god

May very overcomplicated

Brain can't handle contradiction

[solid kiln]

ProjectTime

[autumn violet]

Four passes this argument because there are no composite numbers between 1 and 4 and prime numbers pass this argument because $2p$ can't divide $p$?

[solid kiln]

ProjectTime

[stable ore]

This is true but you could also have just said (p-1)q shares only q as a factor and is composite

[autumn violet]

And that's general right for even and odd n?

[stable ore]

Yeah

Because we said p>2

[autumn violet]

Oh yeah so 4 works just because of how we set up the initial conditions?

[stable ore]

Mhm

[autumn violet]

And just one more thing, I proved it in my head but not rigorously, what covers the case where primes are repeated?

[stable ore]

There is no case where primes are repeated, actually.

We never used that p,q are distinct. In fact I didn't even realize you had said they were until Chai T. Rex pointed it out.

[autumn violet]

oh

Ohoo and we never used the fact that there's exactly two primes

Okay

I am gonna leave this channel open while I attempt to write the proof

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[autumn violet]

.reopen

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✅

@autumn violet Has your question been resolved?

[autumn violet]

@stable ore I was stuck on that for many hours thank you very much

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[vagrant imp]

how do i do this?

[vagrant imp]

I tried some stuff, but nothing worked, I did get b=-1 somehow tho

[empty orchid]

Quadratic formula?

[vagrant imp]

but like i get -b+-sqrt(b^2+4)=-1+-sqrt(1-4b)

oh wait i could just plug in the values and check

yeah I got (B), thanks a lot!

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[vagrant imp]

so i got 1+b^2=2bsin^2theta, how do i find theta?

[empty orchid]

With some inspection you can notice when the equation has only one value. Or you can just plug in the answer choices.

Or just use the discriminant.

[vagrant imp]

ohhh that makes sense, otherwise b would have 2 values which should not be possible

[vagrant imp]

thanks a lot!

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[tribal fractal]

im pretty sure 3 works

[tribal fractal]

but how can i prove that it is a prime in the quadratic field

in the field Q[sqrt(-1)]

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[tribal fractal]

<@&286206848099549185>

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[spiral ocean]

how do i visualize this

is it like

a smallest segment of X that contains o and the range of f(x) is still in the segment?

i cant really understand

like

if define ${f(x) = x + 1}$, then ${\text{clo}_f({2}) = ?}$

[solid kiln]

k

[spiral ocean]

is it ${2,3,4,\dots?}$

[solid kiln]

k

[trim lichen]

subset not segment

but yes clo_f(2) = {2,3,4,...}

[spiral ocean]

so like

to find it

o must be in it -> f(o) must be in it -> f(f(o)) must be in it -> ...?

so its like subset of X

that must follow have o, f(o), f(f(o)), f(f(f(o))), right?

[stark bison]

Not sure what you mean by X, but, yeah, the set of those (o, f(o), f(f(o)), …) is the closure

[spiral ocean]

i think of X like domain of f

[stark bison]

good point, they didn’t mention power set of what are we considering the f-closed sets from

It’s safe to say it’s the domain of f

So that makes the closure a subset of the domain like you said

[spiral ocean]

ahh ok

[stark bison]

Ah actually nvm it wouldn’t make sense for a set to be f-closed if it contains an element outside of the domain anyway

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[stark bison]

Forgot to say, your example is correct

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[vagrant kraken]

it is the same thing,

in the formula,
Esys is the energy LOST by the system
Esurr is the energy LOST by the system

thus Esurr is negative

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[turbid gazelle]

how to do part c?

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[lapis loom]

im not sure how i would do part c because i dont think u could do 1/2 x base x height as its not a right angle triangle

[humble forge]

herons formula?

[lapis loom]

whats that?

[vagrant prism]

i am not sure if that's the best way to do it but it works ig

[lapis loom]

whats a semi-perimeter??

[vagrant prism]

it says below there

it's the perimeter divided by 2

[lapis loom]

damn it does work :D, tysm

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[high flax]

i am at complete loss on this

[empty prairie]

!showyourwork

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[high flax]

not much work i just drew the triangle assumed that aplha must be near the vertex of 3rd quadrant because the horizontal distance is maximum there

i have no idea what else to do

[empty prairie]

You're given that point P is of the form $(\alpha, \alpha^2 - 2)$, think about which points will satisfy this property

[solid kiln]

@empty prairie

[empty prairie]

Hint: Co-ordinates of a point are of the format $(x, y)$

[solid kiln]

@empty prairie

[sacred sapphire]

maybe try linear inequalities

[high flax]

aplha = 1 will obv satisfy thats why there an exclusive bracket

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[chilly bobcat]

is there no way to do it other than manual addition?

[spiral ocean]

i suppose

splitting

[chilly bobcat]

hmm

[spiral ocean]

fraction?

[chilly bobcat]

.. then?

because you would reconsider after knowing the answer

[spiral ocean]

${\frac{1}{2} + \frac{10^5}{2^{r+1}}}$?

[solid kiln]

k

[spiral ocean]

does this converge?

[chilly bobcat]

well

of course

but it's still a large sum of some terms

[harsh cedar]

it's got to, the greatest integer function will only spew greater than 1 values til a certain r after which its all zero but yeah ull have to do it manually 'til there

[chilly bobcat]

well then

[spiral ocean]

oh wait

[chilly bobcat]

can you guess what the answer is

[spiral ocean]

[] is greatest integer function

😭

[chilly bobcat]

yeah

[harsh cedar]

even if that wasnt there, it was a geometric progression so it would converge regardless

[chilly bobcat]

to 1/2

sum is infinity

[harsh cedar]

huh

[chilly bobcat]

in that case

[spiral ocean]

if sum is infinite, it doesnt converge?

[chilly bobcat]

but here there's a gif

and the answer is...

10⁵

[harsh cedar]

oh i wasnt counting the having to add 0.5 infinite times i was only seeing the 10^5/2^(r+1)

[spiral ocean]

ye i think u have to do it manually

[spiral ocean]

@stark bison how is this statement wrong 😭

[stark bison]

Convergence means the limit is some real number

You may write the limit “= infinity”, but technically the limit still doesn’t exist, so it diverges

[chilly bobcat]

uh doesn't it converge to 0.5

i mean

the sum doesnt

[stark bison]

I am talking about the sum

[chilly bobcat]

what about this one then

[spiral ocean]

like sum = infinite, the sum doesnt converge?

[chilly bobcat]

it's of the same vibe

[stark bison]

Yeah

[spiral ocean]

maybe my phrasing is weird

[spiral ocean]

i think the idea is to consider the point where 2000/3^k = 1/6

[stark bison]

You may think of it as just convention, after all the limit being infinity simply tells you that the expression grows arbitrarily large, not that it converges to something

[spiral ocean]

cuz after 1/6, its just all 0

[chilly bobcat]

but it feels like unexpected number crunching

like why are the numbers 10 power multiples

[marsh forum]

[a+b]< [a]+[b] I beleive?

[stark bison]

[0.5 + 0.5] >= [0.5] + [0.5]

[chilly bobcat]

well it works the other way

[marsh forum]

you need a "revere triangle inequaliy" here imo

[stark bison]

There’s probably some way to count the number of terms equal to 0, 1, …; The rest is still arithmetic

Okay there’s definitely a way

[chilly bobcat]

i cant figure which one is easier

[stark bison]

Which one out of brute force and what I mentioned?

[chilly bobcat]

no like out of the two problems I just sent

the second one feels compressible

but idk

i tried brute force for the first one already and it worked

but it's like a sum of 20 numbers

with proper decimals

[marsh forum]

try writing out a few terms and see if there's a pattern

:D

[chilly bobcat]

there isn't

[true shuttle]

Who needed help

[chilly bobcat]

it just randomly increments 1 to random halved numbers

[true shuttle]

And wines

[chilly bobcat]

what

[true shuttle]

Wines is the integral part of that equation

[chilly bobcat]

i think your help would be even more integral

im underage too

[true shuttle]

Wdym

[chilly bobcat]

wait

are you being serious

[true shuttle]

Ye

[chilly bobcat]

wdym wines

[true shuttle]

I won’t tell you the answer but what will happen if you subsidies the integral part of the question into

It self

[chilly bobcat]

um you don't need to

i know the answer

[true shuttle]

What’s the answer

[chilly bobcat]

i got it using a painful method

10⁵

for the first one

[true shuttle]

Oh

My bad

I thought something else

[trim jolt]

@chilly bobcat Has your question been resolved?

[chilly bobcat]

also

[chilly bobcat]

another interesting lead on this one

if you substitute 10⁵ in the question with any arbitrary positive integer

you will get that arbitrary integer as the answer

i just tried it out

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@chilly bobcat Has your question been resolved?

[trim jolt]

@chilly bobcat Has your question been resolved?

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@chilly bobcat Has your question been resolved?

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@chilly bobcat Has your question been resolved?

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[leaden nebula]

i think my proof sucks help

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@leaden nebula Has your question been resolved?

[autumn violet]

<@&286206848099549185> (Sorry I can't help)

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@leaden nebula Has your question been resolved?

[thorn bay]

you wrote
"we know dim(W) = n"
but that's what you're trying to prove

what's the "trillium theorem for finite bases"?

[leaden nebula]

oh right

[thorn bay]

oh, I didn't know that theoremhad a name

[leaden nebula]

my prof came up with it lol

[thorn bay]

anyway, I think an easier approach is to use rank-nullity.
It's just, you have to think of a function that's appropriate to analyze with rank-nullity, any ideas what that could be?

do you know what rank-nullity looks like for linear functions, not necessarily defined by matrices? It's basically the same thing

[leaden nebula]

yeah rank(T) + nullity(T) = dim(V)

[thorn bay]

so, we need to think of a T that we can analyze with rank nullity to analyze the space W

[leaden nebula]

maybe a function that maps a polynomial in P_n(R) to f(a)? cuz then the kernel would just be W right?

[thorn bay]

yes!

[leaden nebula]

ohh

[thorn bay]

the map $f \mapsto f(a)$

[solid kiln]

gfauxpas

[leaden nebula]

and im(T) would be R.

[thorn bay]

so now you know 2 out of the 3 terms in rank nullity, you get the third with arithmetic

[leaden nebula]

tyy

[thorn bay]

https://tenor.com/view/youre-welcome-pleasure-keanu-reeves-its-a-pleasure-thank-you-gif-18395277

[leaden nebula]

.close

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[leaden nebula]

.reopen

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✅

[leaden nebula]

@thorn bay wait but the rank nullity theorem only applies to finite-dimensional spaces no?

[ionic pendant]

P_n(R) is a vector space of dimension n + 1

[leaden nebula]

ohh right. I was confused by the R

nvm haha

[thorn bay]

it applies to infinite dimensional spaces too if you use cardinal arithmetic, but, it's usually not very informative in that case.
but, yes, what cloud said

[leaden nebula]

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[wraith hinge]

whats the difference in tangent in these two unit circles one is straight goes to up while other goes round as angle increases touching the point

[wraith hinge]

sorry if my question is stupid i am just not able to understand tangent function geometrically

[violet gust]

If you look at the triangles OAB and OCD, these two triangles are similar (easy enough to prove)

So, you can write the ratio of $\frac{AB}{OB} = \frac{CD}{OD}$

[solid kiln]

Facter10Br4g

[violet gust]

At this point, can you compare this with the original unit circle?

[wraith hinge]

ok i got it that those triangles are similar but i was asking about how those tangents on both circle are similar

[violet gust]

Can you tell what are the lengths AB and OB from the unit circle diagram?

From the yellow bg figure

[wraith hinge]

ab is sinx and ob is cosx

[violet gust]

yea

and what about OD?

[wraith hinge]

1

[violet gust]

Great

So, can you put these back in the ratio and find CD?

[wraith hinge]

ok i know those tangent lines have similar legnths equals to sinx/cosx but how they are placed differently on the circles is confusing me

[wraith hinge]

wait

[violet gust]

Well, thats just the figure is mirrored

[wraith hinge]

cd = (od * ab)/ob

[wraith hinge]

oh

[violet gust]

well mirrored is wrong mb

its mirrored AND rotated

[wraith hinge]

is that allowed in geometry

[violet gust]

Well, yea. If you do everything correctly, why not

You can see how the orange line is now similar to the green line of this picture

[wraith hinge]

yeah

now i am getting a little bit. thank you so much

[violet gust]

np

[wraith hinge]

👍

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[hallow terrace]

.close

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[next gull]

Can I say that ABCD is a rectangle if there is 2 angles of 90? I don't know anything else about ABCD.

[trim lichen]

no

[next gull]

Why?

[trim lichen]

behold

[empty prairie]

!xy

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[trim lichen]

this look like a rectangle to you?

[next gull]

no.

It does make sense

but let say if we know the angles are opposite of each other does it make a change?

[next gull]

It just a problem that came across my mind

[trim lichen]

what angles

[empty prairie]

If the shape is a cyclic quadrilateral, in the Euclidian space, and two diagonally accross angles are 90 degrees then yes that shape is a rectangle

[next gull]

the 90 degree

[trim lichen]

wdym by regular...

[trim lichen]

ok then what do you mean by "opposite"

the 90° angles in my picture sit on non-adjacent vertices

what more are you saying

[next gull]

Let there at the oppoiste site of each other

[frail heron]

The angles that are not adjacent to each other?

[next gull]

eactly

[frail heron]

No

[empty prairie]

OH shit mb I meant to write cyclic

[frail heron]

Ann has already shown it

[frail heron]

[next gull]

Ok, thanks.

thats all

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[lethal bone]

How is this wrong?

[trim lichen]

what's the general equation of a circle?

[empty prairie]

Dang it sniped again

[lethal bone]

(x-x1)^2 + (y-y1)^2 = radius^2

[boreal apex]

no

[queen burrow]

it's radius²

[trim lichen]

https://tenor.com/view/are-you-sure-john-cena-ru-sure-about-dat-gif-14258954

[boreal apex]

It is not

[queen burrow]

lol

[lunar stirrup]

radius^2

[trim lichen]

r^2 yes

thats where the issue was and is

[lethal bone]

oh

so is it the square root of 9 then or 9^2 for the radius

it doesn’t have ^2 above it in the original equation it gave me so I thought it was already simplified from 3^2

[boreal apex]

r^2 = 9

[lethal bone]

Which is 3 right

[boreal apex]

r is 3 yes

[empty prairie]

$\pm 3$ but it can only be three in this case

[solid kiln]

@empty prairie

[lethal bone]

Fire, thanks guys

[empty prairie]

!done

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[lethal bone]

.close

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[blazing geode]

I need a term

[blazing geode]

(x+3) is called monomial

right?

what is (x^2+3x+4) called?

guys

throw me a term

[supple copper]

quadratic

[violet gust]

tbf, x+3 is not a monomial, its a binomial

[supple copper]

that too

[violet gust]

and the quadratic is a trinomial

[gaunt rain]

is x+3 not a linear expression/factor