help-0

[stuck otter]

Out of previous answers to this question, it can not be a pool paradox

Did, but it was not a aplyable

So wrong

Out of the task we can not start in a corner

I tried to apply pool but it doesn't work in this particular case

[lone heart]

@stuck otter Has your question been resolved?

[stuck otter]

As I figured out from task starting point could be were ever on x,y

<@&286206848099549185>

[stuck otter]

<@&286206848099549185>

[stuck otter]

<@&286206848099549185>

[bleak creek]

1. The number of distinct pairs (d,a), you need to count how many distinct corners can be reached by traveling a distance d ≤ D. The distance to each corner (or reflected corner) can be calculated geometrically, and each such corner gives a unique pair of (d,a).
2. The number of distinct pairs (n,a) is determined by how many distinct corners (including those from reflected rectangles) can be reached after exactly n reflections. Each such corner corresponds to a valid pair.

[stuck otter]

There is one complication, d (vector length) can not be bigger than D (we don't know but suppose to pretend like we know it)

And for 1. Task, we have a start angle of 30°

[stuck otter]

Do you count with a following things or could it be like that even after thouse things pop up?

<@&286206848099549185>

[bleak creek]

figure that out

[stuck otter]

How 😢

[bleak creek]

Be better

[stuck otter]

I really need a deep explanation about this case

[stuck otter]

Wow, and what i suppose to improve

[bleak creek]

school

[stuck otter]

Get out, door is there, I'm asking for help couse I got lost in it and can not figure it out during a week, in case you don't whant to explain just get out and let someone else help me

[bleak creek]

do you want an answer or explaination

[stuck otter]

Explanation how to get to the answer, just in witch way I suppose to move and what to use

[bleak creek]

okay so instead of thinking about bouncing, imagine the rectangle is repeated like tiles in an infinite grid. The satellite just moves in a straight line

[stuck otter]

Okeeey

But how in this case satellite going to get into corner, if it moves straight and out of the task we can not start on edges

Like only thing that comes into my mind is make infinite vector but I don't have a single idea how it could help

[bleak creek]

you're right

[bleak creek]

the satellite reaches a corner because its path involves reflections off the walls. Think of the rectangle as being repeated in an infinite grid, you can treat the satellites movement as a straight line path that eventually intersects a corner. That is how the satellite can reach a corner even though it starts inside the rectangle

[stuck otter]

Okey, make sense, but there a problem with vector length can pop up

Or ami getting something wrong?

[bleak creek]

yeah that's also right

just use rational proportions for the satellites movement and carefully consider reflections. you can avoid problems with vector lenght and ensure the satellite reaches a corner after a certain distance or number of reflections

[stuck otter]

And also we can not use only one infinite vector , cause we are not going to reach corners in direction of vector

[stuck otter]

Make sense

[stuck otter]

In case of only one bounce is allowed

[bleak creek]

its rare

[stuck otter]

How u mean it?

[bleak creek]

youre correct that using just one infinite vector wont work unless its perfectly aligned with a corner which is rare. THe satellite must rely on multiple reflections to adjust its path and eventually reach a corner. Each reflection changes the satellites direction, allowing it to correct its trajectory and intersect a corner. So use this piecewise linear approach, you can ensure that the satellite reaches a corner after some distance or number of reflections

thats all I know

[stuck otter]

Ooooooh , now i got it

This really helped much, thanks a lot !

.close

[lone heart]

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[gritty mesa]

Given:

ABCD - Rectangle
BC - 64 cm
AB - 34 cm
O - Bisects KL and AC, and is also the center of ABCD.
AC - Diagonal ABCD
Angle AOL = AOK = KOC = COL = 90

Find:

Area ABKO

Also ABKO = LOCD and AOL = KOC.

[small lance]

Hint: CKO is similar to ABC(AA similarity)

[gritty mesa]

And how to solve this?

[gritty mesa]

Sorry!
BC - 10 cm
AB - 5 cm

[alpine sable]

find ac by pythgoreas

and use similarity to find others

since you mentioned angles are 90

each

[gritty mesa]

AC = √(AB^2 + BC^2) = 11,1

AO = OC = 5.59

[gritty mesa]

Knowing that AOL = AOK = KOC = COL = 90, how can you find other angles?

[alpine sable]

use similarity

and get the sides

and obtain the area

[gritty mesa]

Need to find AL to find the area AOL through the height, and then multiply it by 2, subtract it from the area ABCD and divide by 2.

Knowing AL, we can solve the problem like this:
Let's draw the height OJ from the corner AOL.
And then the area of ​​AOL will be equal to 0.5*OL*OJ

Then everything is simple: Multiply the area AOL by 2, subtract it from the area ABCD and divide by 2.

But the question is: How to find AL?

[gritty mesa]

OR need to find KL

I don't know how to solve this

Help me!

@alpine sable @small lance

[alpine sable]

wait

i will send you the solution

[gritty mesa]

okay

[alpine sable]

send me the question again

properly

[gritty mesa]

In what sense?

[gritty mesa]

Here

[alpine sable]

like properly

wait

[gritty mesa]

The source text of the task:

1. He takes the opposite corners of the sheet of paper diagonally and aligns them with each other.
2. The resulting shape is then folded exactly in half.

Now, Georgy wants to know what the area of the resulting figure is.```

Note

In other words, you need to find the area of ​​the figure highlighted in dark color if you unfold a sheet of paper to its original size and draw lines along the folds.

[gritty mesa]

Also BK = LD

[lone heart]

@gritty mesa Has your question been resolved?

[gritty mesa]

<@&286206848099549185>

[twilit frigate]

nope no idea mb g

[gritty mesa]

?

[opal ridge]

thats a fun one

[gritty mesa]

How to solve this???

[opal ridge]

I m working on it

I think I got it

[opal ridge]

nvm

I am going crazy with it

[lone heart]

@gritty mesa Has your question been resolved?

[gritty mesa]

Me too. This task is very difficult.

[opal ridge]

Its a think with trigonometry

I dont know how I got that KC = 32 And it needs to be a bad answer

[gritty mesa]

Why 32?

[opal ridge]

I made a mistake

If my calc's are good KC = 40

and the BKAO = 704

[gritty mesa]

How did you get this?

[opal ridge]

Wait I will draw it in paint

[gritty mesa]

Draw with length 10 and height 5

I have the answer for this height and length

But the main decision

[gritty mesa]

With BC = 10 and AB = 5 the answer is 17.187500

[gritty mesa]

COS a or d on start 2-3 lines?

[opal ridge]

and with AL you can calculate the tringle and get the area

[opal ridge]

cos alpha

alpha = OAL

do you see it?

[gritty mesa]

It's a little unclear what is written here

[opal ridge]

e = AC = 32 sqrt(5)

cos alpha = 64/32sqrt(5) = 2/sqrt(5)

cos alpha = (e/2)/a = (16sqrt(5))/a

and when you have a you can calc the OAL triangle and with it calculate the area of BKAO

[gritty mesa]

e is the AC diagonal?

[opal ridge]

yes

[gritty mesa]

Why sqrt(5)?

Where does 5 come from?

AO?

[opal ridge]

x^2 + (2x)^2 = e^2

x = 32

x^2 + 4x^2 = e^2
5x^2 = e^2
e = x*sqrt(5)

[gritty mesa]

$e = x * \sqrt{5}$

[ocean seal]

vo0ov

[opal ridge]

yup

Pythagoras

[opal ridge]

AO is e/2

[gritty mesa]

$AC = \sqrt{BC^2+AB^2}$

[ocean seal]

vo0ov

[opal ridge]

exacly

but when proportin bc to ab is 1:2 the ac = bc sqrt(5)

[gritty mesa]

Let's say BC = 10 and AB = 5. Now I'll try to solve with these numbers.

[opal ridge]

it will be 5 sqrt(5)

[gritty mesa]

AC = 11,1

[opal ridge]

its a aproximation

[gritty mesa]

AO = 5,5

[gritty mesa]

yes

[gritty mesa]

I keep deciding

[opal ridge]

do you understand the whole thing with a?

[gritty mesa]

$COS(α) = \frac{BC}{AB * \sqrt{5}}$

[ocean seal]

vo0ov
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty mesa]

COS a

[opal ridge]

if yes you can calc the triangle

h = 32/2 = 16
a = 40

P = ah/2
P = 4016/2
P = 40 * 8
P = 320 cm^2

[gritty mesa]

And again the same question. In sqrt(5), 5 is AO?

[opal ridge]

where is AO?

[gritty mesa]

Here

[gritty mesa]

AO is 5,5

[opal ridge]

dude pls dont do aproximations

AO = 2.5 sqrt(5)

A0 = AB/2 = (5sqrt(5))/2

[gritty mesa]

$AO = \sqrt{5}{5}/2$

[ocean seal]

vo0ov

[gritty mesa]

Error

I wrote it down incorrectly

[gritty mesa]

But I still don’t understand what needs to be written in line 2

[opal ridge]

do you know about triangle funcs?

I used cos alpha

for cos alpha I need AB (its a AD = 64 here) and BC (its a AC = 32 sqrt(5) here)

[gritty mesa]

hm

[opal ridge]

did got it?

but if we are working with 5 and 10

AC is 5 sqrt(5)

AO is AC/2 and its 5sqrt(5)/2

and AD is 10

so cos alpha = 10/5sqrt(5) where you get a 2/sqrt(5)

bc triangle funs are proportions

[gritty mesa]

$AO = \sqrt{BC^2+AB^2} / 2$

[ocean seal]

vo0ov

[gritty mesa]

Right?

[opal ridge]

yea

[gritty mesa]

I'll upload it again so I don't have to scroll through it

[opal ridge]

np

[gritty mesa]

To get COS(LAO) we need to know AL

[gritty mesa]

.

[opal ridge]

no

we need to know AO

wich we know

[gritty mesa]

We already know.

[opal ridge]

yea

[gritty mesa]

What action is next?

[opal ridge]

cos(LAO) = cos(CAD)

cos(CAD) = AD/AC

[gritty mesa]

I think I understand

[opal ridge]

cos(LAO) = AO/AL

did you had cos, sin, tg, ctg in school?

[gritty mesa]

$COS(LAO) = \frac{AD}{AC}$

[ocean seal]

vo0ov

[opal ridge]

yea

[gritty mesa]

COS(LAO) = 10 / 11.1...

[opal ridge]

10 / 5sqrt(5) = 2/sqrt(5)

really dont aproximate

[gritty mesa]

Now we know COS(LAO). What does this give us?

[opal ridge]

COS(LAO) is AO/AL

and COS(LAO) is AD/AC

becouse its propotions

so AO/AL = AD/AC

and only thing we dont know is AL

so we can calc it

[gritty mesa]

I understand!

[primal vessel]

Hello is this for all grades

[opal ridge]

what do you mean by for all grades?

[boreal epoch]

but have you tried #❓how-to-get-help

[primal vessel]

I am new but I did read it.

[gritty mesa]

AL = AD/AC/AO?

[primal vessel]

Any one can help with grade 9 work ?

[opal ridge]

I think you should go to open help https://discord.com/channels/268882317391429632/903486481929224232

and clame it

[primal vessel]

ok thank you so much.

[primal vessel]

Will do.

[gritty mesa]

?

[opal ridge]

AL = (AO*AC)/AD

but I think its the same thing

but I m not sure

[opal ridge]

bro sad read the docs

[gritty mesa]

Aera AOL = 0.5 * (AB/2) * AL

Correct?

[opal ridge]

yea

[gritty mesa]

Aera ABKO = (AB * BC - (Aera AOL * 2)) / 2

Right?

[opal ridge]

exacly

[gritty mesa]

$AO = \sqrt{BC^2+AB^2} / 2$ \ AL = (AO * AC) / AD \ Aera(AOL) = 0.5 * (AB / 2) * AL \ Aera(ABKO) = (AB * BC - (Aera(AOL) * 2)) / 2$

[ocean seal]

vo0ov
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty mesa]

@opal ridge Right?

[opal ridge]

Yea

[gritty mesa]

Now I will substitute the values 8 and 8

[opal ridge]

AB = 8 and BC = 8?

[gritty mesa]

Yes

AO = 5,6...
AL = 7,9...

[opal ridge]

do you understand why we got this equations?

[gritty mesa]

In which line?

[opal ridge]

like all of it

[gritty mesa]

I roughly understand

Solution:

AL = 7,9...
Aera(AOL) = 15,8
Aera(ABKO) = 15,8```

When 8 8

[opal ridge]

You have a problem with AL?

[gritty mesa]

I don't understand AL = (AO*AC)/AD a bit

Why it works?

[opal ridge]

bc of triangle functions

we used cosinus

bc we needed a AL

and we used cosinus bc AL is touching the angle

and cosinus is a proportion

proportion between the side of triangle touching alpha and the side of triangle not touch the 90 degrees

and bc we know that proportion is 2/sqrt(5) for AD and AC we know that its the same for AC/2 and AL

[opal ridge]

https://cdn.discordapp.com/attachments/490557019623915520/1290733866541125784/iu.png?ex=66fd8916&is=66fc3796&hm=2f6262c85020d98ec5e2b7fb0001418b7620ae77f2a7bcaec52249be56832cf1&

[opal ridge]

I expliened it wrong

our AL is BC

our AD is AB

our AC/2 is AB

our AC is BC

I know its not the best explanation

try to ask your teacher about it they will answer it better

[gritty mesa]

Thank you so much! The five of us solved this problem for 2 days, but we still haven't solved it!

[opal ridge]

lol

[gritty mesa]

We haven't been through this at school yet.

[opal ridge]

is that was a homework?

bc if it was I have no idea of how to solve it with out cosinus

[gritty mesa]

No. This task is from the programming championship for grades 8-11. Alas, there may be problems that cannot be solved in the 8th grade.

There is solution on Python:

def calculate_area(ab, bc):
    ac = (bc**2 + ab**2) ** 0.5
    ao = ac / 2

    al = (ao * ac) / bc

    area_aol = 0.5 * (ab / 2) * al

    area_abko = (ab * bc - (area_aol * 2)) / 2
    
    return {
        'AO': ao,
        'AL': al,
        'Area_AOL': area_aol,
        'Area_ABKO': area_abko
    }

result = calculate_area(*map(int, input('AB BC: ').split()))

print(f"AO = {result['AO']}")
print(f"AL = {result['AL']}")
print(f"Area(AOL) = {result['Area_AOL']}")
print(f"Area(ABKO) = {result['Area_ABKO']}")```

Something like this.

[opal ridge]

<3 python

I think you should use math lib here (IT WRONG I KNOW IT NOW)

import math

def calculate_area(ab, bc):
    ac = math.sqrt(bc**2 + ab**2) ** 0.5
    ao = ac / 2

    al = (ao * ac) / bc

    area_aol = 0.5 * (ab / 2) * al

    area_abko = (ab * bc - (area_aol * 2)) / 2
    
    return {
        'AO': ao,
        'AL': al,
        'Area_AOL': area_aol,
        'Area_ABKO': area_abko
    }

result = calculate_area(*map(int, input('AB BC: ').split()))

print(f"AO = {result['AO']}")
print(f"AL = {result['AL']}")
print(f"Area(AOL) = {result['Area_AOL']}")
print(f"Area(ABKO) = {result['Area_ABKO']}")

[gritty mesa]

It seems like i can't even use the built-in libraries at this championship

[opal ridge]

really???

woah

so you need to make root algoritm for it

[gritty mesa]

But this is not accurate.

[opal ridge]

?

[gritty mesa]

I'll rewrite this code better later.

I think the issue has been resolved. If something is unclear to me, I will write in private messages. Thanks again!

Accept dm pls

Close tiket

.close

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[buoyant fractal]

yall are dumb

[buoyant fractal]

this is so easy

[alpine sable]

uno reverse card

[modern sedge]

ok

Anyway, what's your question?

[opal ridge]

https://tenor.com/view/exploding-car-explode-cat-boom-bah-gif-5387928273689188008

[tiny sky]

<@&268886789983436800> they seem to be trolling in #discussion too

[wide tartan]

don't troll in the help channels please.

.close

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[swift saddle]

Why is 2pi in the formula for a Gaussian distribution? What does it have to do with soicles

[fallen verge]

https://youtu.be/cy8r7WSuT1I?si=tq3a-n2AJuP7kEgq

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@swift saddle Has your question been resolved?

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[coral stone]

hi, how do i find a sup for x/(n(1+x^n))?

[coral stone]

for x>=0

and n>=1

n is an integer

[tacit arch]

have you learned derivatives

[coral stone]

yes

but i dont think it works

[tacit arch]

show your attempt

[coral stone]

im an undergarde

ok

[tacit arch]

?

[coral stone]

it is positive when

[tacit arch]

a fraction is zero when the numerator is zero

[coral stone]

in the derivative there is this x power of n

[coral stone]

and what do i do next

can you pls show me why the sup is 1/n?

[tacit arch]

so?

,w diff x / [n(1+x^n)]

[ocean seal]

Results provided by WolframAlpha

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

Requested by rie.mann

[tacit arch]

set the numerator to zero and solve for x

[coral stone]

it is x = n-th root of 1/(1-n)?

[tacit arch]

yes

plug that into your function

[coral stone]

ok but then i need to find the limit when n --> infinity

mmh wait

[tacit arch]

well when did you say taht

[coral stone]

no i mean the og question asked me to show that it's uniformly convergent when x >=0 but i guess it works

but then how do i find the limit if the n-th root of 1/(1-n) when n --> infinity?

uh it's 1 then

[tacit arch]

it's a limit problem

yes 1

[coral stone]

yes the limit is 1 actually im dumb

or?

0^0?

it's not 1

[tacit arch]

????

[coral stone]

i tried lim (1/n)^(1/n)

and both 1/n approach 0

[tacit arch]

why are you changing the limit problem

what inequality are you using

[coral stone]

no im not changing it i just want to know this limit

[coral stone]

this one

[coral stone]

so its 1

ok thanksss

[alpine sable]

you can use logs to figure this one out, if you want to know why it's 1.

[coral stone]

ok so i's not the 0^0 thingy

[alpine sable]

not really no

[lone heart]

@coral stone Has your question been resolved?

[lone heart]

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[alpine steeple]

dont rlly understand this because theres nothing to base it off of and i know its 4 units to the left but i only got partial credit

[prime badge]

it's maybe 2,3,4

[teal moat]

I’m working on a problem where a jar contains 4 red marbles numbered 1 to 4 and 8 blue marbles numbered 1 to 8 and when I was asked what the chance of the drawn marble being blue given it was odd numbered I got 8 over 6 (8/6). How can I find where I went wrong?

I don’t trust my answer if the numerator is bigger

[prime badge]

90° rotation plus the translation is one way to do it

the reflection is the second way

depending on which one is the original it's also −90°

so maybe it's only 4

but you said 3 is right so i guess it's not just 4

[alpine steeple]

yeah i think it could be all three, but i have issues understanding like reflections etc in general since graphs aren't my strongsuit

[teal moat]

Oh wait nvm

I’m overthinking this

I should divide the number of odd blues that I know by the total of odd blues and reds

[alpine steeple]

aoh also this aswell?" i think it might be d but i could be wrong .. i hate shapes and i struggle with them

[prime badge]

yeah d has no second line

B has 0 I think

I mean A

the last one

the circle has more than eighteen

[alpine steeple]

oo okay

i was curious about this one aswell ? i think its either b or a but it misclicked and im not rlly sure where to go with it either

[alpine sable]

B

It will need to look exactly like the first figure

Just flipped

[lone heart]

@alpine steeple Has your question been resolved?

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[inner haven]

How do i do this

[buoyant saddle]

integration

[inner haven]

yeah im having trouble setting it up

[inner haven]

not sure how to format the integral tho

[tacit arch]

where did -1.94591 come from

[inner haven]

intersection

[tacit arch]

the answer probably should also include e^(-x)

[inner haven]

how do i format the integral?

[tacit arch]

use math to find the intersection

not desmos

[inner haven]

ok but how do i find the intersection without demsos

[tacit arch]

https://www.cuemath.com/algebra/logarithms

[charred pilot]

I put this into desmos, I would be inclined to rewrite the functions as negative and integrating from the x axis

[inner haven]

what does this mean

[buoyant saddle]

ok han solo

[charred pilot]

Integrating using the bottom two functions

(Purple and black)

[inner haven]

Whys it +7 and -e^x?

[charred pilot]

Do you notice anything similar about the area between purple and black and the area bounded by the first 3 functions?

[inner haven]

flipped no?

[charred pilot]

The area

[inner haven]

I mean

theyre the same

[charred pilot]

Yes they are the same

[charred pilot]

What could this possibly mean

[buoyant saddle]

idk that’s what riemann called you

like a week ago

[charred pilot]

I don’t know riemann

[buoyant saddle]

🤷🏼‍♂️

[charred pilot]

Interesting..

[buoyant saddle]

🤔

[inner haven]

?

This doesnt work tho

[charred pilot]

Don’t use decimals

[inner haven]

So i should integrate by hand like this?

[buoyant saddle]

i’ll never understand the decimal fascination

symbolic answers are much nicer

[inner haven]

😓 just easier

[buoyant saddle]

why would you want the 1.982736$:&3’shdjx instead of ln(7)

[inner haven]

😭

Good point

[charred pilot]

Good job

Do you see how I got those functions?

[inner haven]

yeah

i was just slow

but u did top - bottom function right

[lone heart]

@inner haven Has your question been resolved?

[lone heart]

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[ornate iron]

how would i go about solving d-f 😔🙏

[hybrid inlet]

so for the matrix part

you simply take constants that are attached to x y and z

and create a matrix using that

[alpine sable]

do you know how to write this as an augmented matrix?

[ornate iron]

r u talking about the equation to matrix part

yeah

i just have no idea how to find a and b

[alpine sable]

ok

for d

do you know what results in an infinite number of solutions

a free variable

so set up an equation with the augmented matrix. with a and b as a 4th row. and try to find values that would create a free variable

[ornate iron]

uh i found that z is arbitrary if that what u mean but other htan than i havbe no idea

ouh

[hybrid inlet]

ok I did not read your post at all

so sorry 😭

did not see the d-f part

[ornate iron]

nooo its okay 😭

[alpine sable]

for the next part. the one solution

the reduced row echelon

should be like 1 ... number and a diaganol matrix

that follows that pattern

like 1 00 ... number

then next row

010 .. number

and so on

for no solutions the easiest way would be having a row 0 0 0 then augment it with a number

[ornate iron]

ohhhh

[alpine sable]

hope this makes a bit of sense

im not the greatest at explaning

[ornate iron]

i think i kinda get it for the no solution part

[alpine sable]

for the first part

try to get a free variable

[lone heart]

@ornate iron Has your question been resolved?

[ornate iron]

okieokie :,) ill try that thank uuu

[lone heart]

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[tulip marsh]

I’m given a matrix $A=LL^T$, where L is lower triangular and I have to prove A is both symmetric and positive definite.\
I did $<L\vec{x},\vec{x}>$ and simplified but im not sure where to go from there

[ocean seal]

KySquared

[lone heart]

@tulip marsh Has your question been resolved?

[tulip marsh]

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[paper fulcrum]

.close

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[agile heron]

Hello! I'm doing this extremely simple math question and I feel like I'm going insane. This is what I've calculated:

Start: 10,000
End of first year (+45%): 14,500
End of second year (-55%): 6,525
Overall investment change: Loss of 34.75%

That's just not an option, so I have no clue where to go from here

[agile heron]

It's just simple percentiles, I don't even know why we're spending so long relearning this, but I guess if I'm missing something it makes sense, I just can't see how it's anything different.

<@&286206848099549185>

[alpine sable]

10,000 * 1.45 * 0.45

[agile heron]

Second decrease is 55% not 45%, so that would be *.55, right?

Or wait no

[alpine sable]

no, it will be *(1-0.55)

[agile heron]

Yes, that's what I got before

6525

So you agree?

[alpine sable]

yes but it doesn't math any of the options

[agile heron]

Nice pun lol

Yeah this class is becoming my personal hell

This will be I think the 5th or 6th time I've had to correct the professor on a problem

God save me

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[agile heron]

Thank you @alpine sable

[alpine sable]

np

the overall is a decrease in approximately 35%

[agile heron]

Yep

Specifically 34.75

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[torpid dirge]

associativity of natural numbers. where could i find the list of axioms for this? what are they called?

[tiny sky]

peano axioms i believe

[torpid dirge]

yea tt

ty

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[tiny sky]

nw

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[gaunt falcon]

I do not grasp how the 2e^-t comes or how the number 2 comes in cos(2t)? Could somebody demonstrate the steps in between that got skipped?

[unkempt pollen]

um did they give any relation between s and t?

[lone heart]

@gaunt falcon Has your question been resolved?

[lone heart]

@gaunt falcon Has your question been resolved?

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@gaunt falcon Has your question been resolved?

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[indigo lotus]

how did they solve that integral on RHS

nvm they used partial fractions

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[steep wing]

Hi, how can i calculate that argumnent? What is my mistake?

I got it

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[steep wing]

.reopen

[lone heart]

✅

[steep wing]

I do not got it

Ik what i did wrong the first time

But i did smzh wrong the second time too

<@&286206848099549185>

[rustic coral]

$-1-4i$ is in the third quadrant of the complex plane, so $$\arg(-1-4i)=\pi+\arctan 4$$

[ocean seal]

Civil Service Pigeon

[steep wing]

Sorry, what do dollar signs mean?

[rustic coral]

Read this.

[steep wing]

No it's wrong 😦

Or at least thats not what belongs here

[rustic coral]

Yeah cause z = (-1-4i)^5, not just -1-4i

[unkempt ravine]

i need help

dm me pls

[steep wing]

So what do i do?

[rustic coral]

De Moivre’s

[steep wing]

😭😭😭

You think if i knew what that meant id ask for maths help on discord

[rustic coral]

$z=re^{i \theta} \implies z^n=r^n e^{n i \theta}$

[ocean seal]

Civil Service Pigeon

[rustic coral]

Here, n=5 and z=-1-4i

[steep wing]

Ok

Thanks

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[latent cosmos]

I'm trying to find the Magnitude for velocity:

[thin lance]

Does anyone know how to do 2.24

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[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

[latent cosmos]

I think 1. Was trying to explain what I tried

the 2nd post distracted me 😅

realizing what I did probably doesnt make sense now though

I did this with each v to find it's their position

but can I just add the velocities of x and y and just do ?

[rustic coral]

This formula is fine, your word choice is questionable

[latent cosmos]

Thats what it gave me for my answer so thought maybe I was on the right track

(for 8.52)

[rustic coral]

Use the values stored in your calculator for v_x and v_y

I bet the issue is that the intermediate rounding is fucking you over

[latent cosmos]

Ah, alright then. That was it

[desert sonnet]

hey can somone help me with somthing i have no idea how to solve it

[latent cosmos]

.close

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[latent cosmos]

#❓how-to-get-help use one of the channels here

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[tight laurel]

We have that the function f(x): R -> R it is such that f(2)=4 and f(0)=2. Consider the function g(x)= f(2x-1). Which of the following equalities are definitely true? g(2)=7 ; g(3/2)=4 ; (f o g)(1/2)=4 ; g(1)= f(1)

[tight laurel]

please help🇪

[tacit arch]

can you find a value for g(?)=? when f(2) = 4

[tight laurel]

mhh

how?

[tacit arch]

similarly g(?) = ? when f(0) = 2

use

[tight laurel]

so like I put 2 in the place of g(x)

[tacit arch]

no

g(2) = f(2 * 2 - 1) = f(3) and you don't know f(3)

[tight laurel]

and what can I do with f(3)

[tacit arch]

that means this is the wrong approach

find another value instead of 2

f(2) = f(2 * ? - 1)

[tight laurel]

can you make me please I don't understand

[tacit arch]

fill in the question mark

that question mark will go into here

[tight laurel]

really I'm not undestanding

🫠

[tacit arch]

i can't understand why not

? is a number here

[tacit arch]

.

[tight laurel]

can we restart please

so what we have to find

[tacit arch]

just go up and re-read

[tacit arch]

.

[tight laurel]

g(?)=? when f(2) = 4. Yes for example 4=f(3)??

so we put four in g(x) and 2 in the place of x

[tacit arch]

no. f(2) = 4

you don't know f(4)

[tight laurel]

f(3) sorry

[tacit arch]

you also don't know f(3)

i don't know why you think you know that value

[tight laurel]

bhu

[tacit arch]

these are the only two values of f(x) you know

that gives you two values for g(x)

[tight laurel]

can you please solve me, i'll try to understand so i don't take your time

[tight laurel]

okok

[tacit arch]

no do your own work

[tight laurel]

but how can I find the values of g(x)

by g(x)= f(2x-1)

so

[tight laurel]

we put 2 in the x of g(x) and f(2x-1)?

[tacit arch]

.

[tight laurel]

please can you say me only the next step

[tacit arch]

.

[tight laurel]

ahhhh

we have to put a number, but a number that we to obtein 2 or 0

???

[tight laurel]

so x=3/2and 1/2

??

[tacit arch]

,calc 2 * 3/2 - 1

[ocean seal]

Result:

2

[tacit arch]

yes this corresponds to x=3/2

[tight laurel]

and for 0 to 1/2

[tacit arch]

so f(2 * 3/2 - 1) = f(2)

[tight laurel]

so now we know that values

and we kwon that g(3/2)=2

[tight laurel]

so the option 2 in my first message is false

no but that's imbossible because the book sais that th correct answer are b,c,d

you solved it? @tacit arch

[tacit arch]

.

[tight laurel]

yes, i would like if possible if with this method the answer is correct

i don't want to know the answer

[tight laurel]

because for what I'm doing right now the book says the opposite

[tacit arch]

this is wrong

connect those 3 to find g(3/2)

[lone heart]

@tight laurel Has your question been resolved?

[lone heart]

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[burnt reef]

Hello,

How should I pursue with this limit?

[burnt reef]

I know the outcome is 0. But I need to derust my algebra hehe. 😄

[celest stump]

Idk if this works

But you can root n on the top and bottom

Then you have 2/(n^1/n)(n-1)!

Then that equals 2/(n-1)!

[burnt reef]

I think I need to use the stirling formulae.

Otherwise they (the class) wouldn't use it in a similar homework.

[celest stump]

Idk what that is that's just what I got from intuition

[zealous ingot]

another method you can use is that, by simplifying the numerator to (2e)^n, and then basically ignoring the sqrt(2pi), you get (2e)^n/n^(n+1/2), where you can convince yourself, as n goes to infinity, the denominator grows faster than the numerator, so the limit goes to 0

something like asymptotic analysis

[burnt reef]

My class is specifically about this analysis

Asymptotic analysis hehe.

[celest stump]

Oh wait the reason why my method is complicated is that you'd have to prove that raising it to 1/n would maintain the limit

So it depends on the teacher

[burnt reef]

But

I do need to reach the divide by infinity by applying the limit.

To reach lim = 0

Here is a similar problem with the solution:

But mine is 2^n, not e^n.

[vale crag]

you can use essentially the same trick

[burnt reef]

I'm trying haha.

PS: Hello again ahah 😄

[vale crag]

thing you have inside the limit is $$\frac{1}{\sqrt{2\pi n}} \left(\frac{2e}{n}\right)^n$$

[ocean seal]

aPlatypus

[vale crag]

hi^^

if you manage to show (2e/n)^n goes to 0 you won

now use the x = e^ln(x) trick for with x=(2e/n)^n

[burnt reef]

Where does this equation come from?

I'm not seeing it in my helper.

[vale crag]

imitating what they did here

[burnt reef]

Yes, where does that come from.

[vale crag]

e and ln are inverses of each other, essentially by definition depends how you defined e and ln tbf

it's weird it's not in your helper

maybe they find this fact too basic idk

[burnt reef]

L'aide mémoire n'a pas de dérivées non plus. Je vais devoir réviser les dérivées que j'ai apprise il y a 12 ans. 😢

But alright I'll try.

[vale crag]

pas besoin de dérivée là

[burnt reef]

Non en effet.

[rain urchin]

ola

[vale crag]

alo

[burnt reef]

Mais disons que pour mon cours... J'en aurai besoin ahah.

[rain urchin]

jajaaj

[vale crag]

c'est pas une ytp ici

[rain urchin]

j3a3

[burnt reef]

Aaah I think I got it. Hold on.

Like this?

Or did I mess up some algebra?

I still have 2^n that is problematic.

[vale crag]

oui bha 2^n c'est juste e^(ln(2^n)) = e^(n*ln(2))

always the same trick

[burnt reef]

Aaaah wait

This one?

[vale crag]

it's a special case of this one yes

[burnt reef]

where b is e

b = e?

[vale crag]

right

[burnt reef]

Aaah I see, now I can jsut throw it on the bottom?

Then voila?

[vale crag]

yeah merge the two e^blabla

[burnt reef]

Yeah I see the rest of the sequence.

Wow that was kinda annoying.

Can't believe I may have to do that on a exam 😢

(My handwritting sucks)

[rain urchin]

so now niamey=p

k

[vale crag]

went through there :/

[burnt reef]

Hahaha. 😄

[vale crag]

at least you can write as shit as you want on brouillons

[burnt reef]

Yes!

[vale crag]

my math teacher in terminale really encouraged me to take my time with writing for the bac

small details like cross your z's so that you don't mismatch them for a 2, etc...

[burnt reef]

That makes sense.

My issue is more about the size ahah.

Or being aligned to the lines.

Either I go fast, or extremely slow.

[rain urchin]

if niamey + aplaytypus = p

[vale crag]

my uni doesn't have lines on the paper for exams most of the time, it's just blank

[burnt reef]

Mine had some lines, luckily! But... if I try to be aligned, I'm slow.

[vale crag]

how slow are we talking, maybe you can try to get extra time if it's really problematic

[burnt reef]

And the time limit can be tight.

Well

The class is not a math exclusive. So I don't expect to be that tight.

But back when you helped me in linear algebra, I had to be very very fast. Time was kinda tight (I'm the kind of student to double check (or even triple check) my calculations)

Alright, off to eating, I'm hungry 😄

[vale crag]

aight it's a good time as any other to eat

[burnt reef]

Yes!

Have a nice end of day

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[rain urchin]

test

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[final badge]

I need some help understanding the notes, where did the x-2 come from?

[final badge]

and the -12x

[tacit arch]

f(2)=0 implies x-2 is a factor

[final badge]

f(2)=0 ?

[tacit arch]

can you read your problem again

[final badge]

ohhh

im blind bruh

ok

thanks

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[gritty violet]

I wanna know if c is the correct answer for ex 9

[tight pier]

why

[arctic lintel]

its not

[tight pier]

i would reconsider the argument geometrically

7pi/4 is just too off

[slim jolt]

Sigma

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[modern sinew]

Is y=2x+3 a direct variation

[modern sinew]

I dont think it might be

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@modern sinew Has your question been resolved?

[modern sinew]

<@&286206848099549185>

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@modern sinew Has your question been resolved?

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[rain spindle]

Just 13. I know how to do the transformations its just i don’t know how to execute it bc of yk the equation (hope that make sense)

[rain spindle]

oh and this is transformations of functions

[buoyant saddle]

,rotate

[buoyant saddle]

@rain spindle

[lone heart]

@rain spindle Has your question been resolved?

[rain spindle]

i tried doing it without simplifying it (not sure if i should do that first) and it didn’t look right

[buoyant saddle]

what did you get?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[rain spindle]

.reopen

[lone heart]

✅

[rain spindle]

i erased it so hold up

i think my main issue is idk how to really start tackling it. bc once i get the flow of it i got it

do i simplify the equations first?

[buoyant saddle]

well in terms of a general function f(x) how can we write a horizontal shift right 4 units

[buoyant saddle]

how would you simplify 13a?

[rain spindle]

get rid of the square root and then transform it into (3-x) (3-x) -4

[buoyant saddle]

get rid of it how?

[rain spindle]

hold up nvm if i simplify it i’ll be stuck with y^2 bc to get rid of the swuare root u gotta square it

[buoyant saddle]

yea

[buoyant saddle]

^

[rain spindle]

it would be (x-4)

[buoyant saddle]

yes f(x) -> f(x-4)

can you do that for this one

what would it look like

if $f(x) = -\sqrt{9+x^2} - 4$

[ocean seal]

knief

[rain spindle]

do i just apply it to that equation?

[buoyant saddle]

yes

tell me what f(x-4) is

[rain spindle]

\sqrt{x^2-8x+25} +4$

[buoyant saddle]

use a dollar sign at the beginning

too

[rain spindle]

$\sqrt{x^2-8x+25} -4$

[buoyant saddle]

yes except it should be -4 yes?

[rain spindle]

yeah mb its -4

[buoyant saddle]

edit it

the bot will fix it

[ocean seal]

emss

[buoyant saddle]

ok that’s f(x-4) now we need to transform that again by compressing it by a factor of 1/3

do you know how we might do that

[rain spindle]

from what i remember replace x with 3x?

[buoyant saddle]

yes

[buoyant saddle]

do that here

[rain spindle]

$f(3x) = -3\sqrt{1 + x^2} - 4$

i used the og equation btw^

[buoyant saddle]

what

[buoyant saddle]

nono

no idea what happened

[rain spindle]

OH WAIT mb mb that was for smth different

[ocean seal]

emss

[buoyant saddle]

hmm no we’re applying consecutive transformations

we first apply the shift

then compression and so on

[buoyant saddle]

it’s really f(3(x-4))

[rain spindle]

oh mb i have to do the stretch and then the reflections first

[buoyant saddle]

i mean it’s convention to do that

but i’m reading from left to right

i assumed they meant apply the transformations in that order

from left to right

[rain spindle]

$f(3(x - 4)) = -3\sqrt{1 + (x - 4)^2} - 4$

[ocean seal]

emss

[buoyant saddle]

why the -3?

we go from here

$-\sqrt{9x^2-24x+25} -4$

[rain spindle]

ohh okay

[buoyant saddle]

then vertical translation up 5 should be quick and easy

[rain spindle]

yuh got it

$\sqrt{9x^2 - 24x + 25} + 1$

[ocean seal]

emss

[buoyant saddle]

yup

now vertical flip

then horizontal flip

(reflection)

oh wait there was a - sign missing

[buoyant saddle]

here

[ocean seal]

knief

[buoyant saddle]

then

[rain spindle]

instead of 4 its one right?

[buoyant saddle]

$-\sqrt{9x^2-24x+25} + 1$

[ocean seal]

knief

[buoyant saddle]

i was just editing the one from before

[rain spindle]

oh got it

[buoyant saddle]

ok reflection about the x axis

what is it

[rain spindle]

that but remove the negative before the square root

[rain spindle]

$\sqrt{9x^2 - 24x + 25} - 1$

[buoyant saddle]

what about the +1?

[rain spindle]

oh

[buoyant saddle]

reflection over x axis of f(x) -> -f(x)

[ocean seal]

emss

[buoyant saddle]

mhm

then do reflection over y axis

or f(-x)

[rain spindle]

$-\sqrt{9x^2 - 24x + 25} - 1$

[ocean seal]

emss

[rain spindle]

might have done that wrong

[buoyant saddle]

that’s the just the vertical reflection

you need to reflect over the y axis now

[rain spindle]

u mean x?

the x-axis?

[buoyant saddle]

a horizontal flip is over the y axis

[alpine sable]

Hey guys what’s a postulate in geo

[buoyant saddle]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[buoyant saddle]

it’s just an assumed fact that’s so trivial you can use it in proofs

without having to justify it

[alpine sable]

yea I’m learning proofs rn

[rain spindle]

alright so i did everything in order aka compression then reflection and then translations

$f(x) = 3\sqrt{1 + (x - 4)^2} + 8$

[ocean seal]

emss

[rain spindle]

this is my final answerf

[buoyant saddle]

huh

[buoyant saddle]

oh you did the other order?

[rain spindle]

yes

[buoyant saddle]

again idk if that’s what they want

do they specify which order you should use

or is it ambiguous

[rain spindle]

eh idk the question is written poorly, i was taught to always do stretches and reflections first

ill probs just ask my teacher

ty for the help i appreciate it <333

[buoyant saddle]

you’re welcome

[rain spindle]

!close

.close

[lone heart]

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[alpine sable]

@buoyant saddle would this be correct

[buoyant saddle]

is this a test

[alpine sable]

No

Hw

[buoyant saddle]

do you see the big blue button that says "check answer"

!noping

[lone heart]

Please do not ping individual helpers unprompted.

[alpine sable]

If I enter that in then it will finalize the grade

I don’t wanna do that without clearifying first

[buoyant saddle]

sounds like cheating sir

[exotic sun]

[alpine sable]

How?

[alpine sable]

It’s homework

[buoyant saddle]

^

[alpine sable]

I just wanna make sure if I got it correct

[buoyant saddle]

well if that was true then you’d hit check answer

you’re more concerned with the grade

hence why you aren’t hitting check answer

[alpine sable]

I can’t do it again tho that’s y

Look

It’s practice

Is it wrong?

Did I do something wrong in the question

bro

U just ghosted me

1. This is literally homework 2. How is this cheating if i already did it myself 3. Is it wrong to ask to clarify something before i submit 4. I genuinely wanna know if I did anything wrong

U guys just left

What kinda help is this

@buoyant saddle

@exotic sun

can someone just help me what did I do wrong in the question

[buoyant saddle]

the second half is wrong

[alpine sable]

Which part

The postulate?

[buoyant saddle]

"then the entire plane containing those points lies in that plane" like what

that’s saying the plane is itself

irrelevant

it’s "the entire line"

not plane

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Ohh

[lone heart]

Channel closed

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[wheat elk]

Can someone help me solve
A plane departs from an airport and travels at a bearing of N30°E at a speed of 25 knots for 2 hours. It then turns to a bearing of S45°E and continues at a speed of 30 knots for 1 hour. After that, the plane turns again to a bearing of S10°W and travels at a speed of 20 knots for 3 hours.
What bearing must be taken from the airport to reach the plane's current position?

[lone heart]

@wheat elk Has your question been resolved?

[midnight wharf]

you should add vectors

[wheat elk]

bruh

this is basic trig

i dont think

the teacher

expects

us to know vector

[midnight wharf]

vectors are things in 2(+)d

like with x and y positions

you don't have to call them vectors

but you need x and y positions

add x and y positions

[lone heart]

@wheat elk Has your question been resolved?

[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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• Be polite and have a nice day!

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[wicked idol]

Hello, I have a problem with a decomposition into single element (I don't know if this is the right term in English)

here is the equation:
$S(p) = \frac{E}{p(1 + 3RCp + (RCp)^2)}$

I find a decomposition in:

$S(p)= \frac{A}{p}+ \frac{B}{p-p_1}+ \frac{D}{p-p_2}$

$p_1$ et $p_2$ root of the polynomial $(1 + 3RCp + (RCp)^2)$

I have:
$\frac{E}{p(1 + 3RCp + (RCp)^2)}=\frac{A}{p}+ \frac{B}{p-p_1}+ \frac{D}{p-p_2}=\frac{E}{p(p-p_1)(p-p_2)}$

I am looking for the coefficients A,B,D
I want to do it by multiplying the coefficients
$\frac{A}{p}+ \frac{B}{p-p_1}+ \frac{D}{p-p_2}=\frac{E}{p(p-p_1)(p-p_2)}$
I multiply each part by $(p-p_1)$
so
$B=\frac{E}{p(p-p_2)}$ were $p=p_1$

wich give $B = \frac{E_0}{p_1 \cdot (p_1 - p_2)}$
I find the right term for A
but for B and D I don't find the right one for example for B
I find :
$B = \frac{E_0}{p_1 \cdot (p_1 - p_2)}$

by doing

But in my correction it is written

$B = -E_0 - \frac{E_0 \cdot p_1}{p_2 - p_1} = -\frac{E_0 \cdot p_2}{p_2 - p_1}$

I would like to point out that no, my value is not equivalent to that of the correction via Vieta's rule.

with the digital application I find 0.01708 and the correction gives 1.708

What is my problem ?

[ocean seal]

Silicium

[lone heart]

@wicked idol Has your question been resolved?

[lone heart]

@wicked idol Has your question been resolved?

[lone heart]

@wicked idol Has your question been resolved?

[alpine sable]

omg wow how does this bot work.

$f(x)= \frac{1}{h}\cdot \frac{1}{2-x^2}+ \frac{x}{2-x^2}$