help-39

[short orchid]

I get it now

Thank you so much

Thank you everyone

Will close the channel

Wait i forgot how to close

.close

[pearl pond]

Channel closed

Closed by @short orchid

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusk edge]

I need help drawing the diagram for this problem. I can’t seem to visualize it

[pearl pond]

@dusk edge Has your question been resolved?

Channel closed

Closed by @dusk edge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[void estuary]

can somebody please tell me how to solve these type of questions where there is exponential function too

[toxic lichen]

dear god this is ugly

but my first idea is integration by parts somehow

[inner cloud]

∞

[void estuary]

the whole square thing and the e^x seperately?

[toxic lichen]

i guess?

though i think maybe uhhh

let's see

[void estuary]

ig it will get more ugly

[toxic lichen]

$\frac{(x+2)^2}{(x+4)^2} = \frac{x^2+4x+4}{x^2+8x+16} = 1 - \frac{4x+12}{x^2+8x+16}$

[jolly parrot]

Ann

[toxic lichen]

is this cooking at all

maybe put u := x+4?

[void estuary]

e^x is still in it's multiple. What about that?

[severe quarry]

You can "long-divide" that fraction (i.e. just write x + 2 = x + 4 - 2) to get (1 - 2/(x + 4))^2, multiply that out and now Ann's IBP idea should work because the derivative of 1/(x + 4) is -1/(x + 4)^2 so you can view the last two terms as what the product rule gave you for (e^x/(x+4))'

Let's see if that works out

[\int \l(1-\frac{4}{x+4}+\frac{4}{(x+4)^{2}}\r)e^{x}dx] is what you have after multiplying out, i.e. [\int e^{x}\dd x-4\int \frac{e^{x}}{x+4}\dd x+4\int \frac{e^{x}}{(x+4)^{2}}\dd x] and now if you set $g(x) \coloneqq \frac{1}{x + 4}$ you notice that those last two terms are [-4\int e^{x}f(x)\dd x-4\int e^{x}f'(x) \dd x=-4\int e^{x}(f(x)+f'(x))\dd x]

[jolly parrot]

Kepe

[severe quarry]

and that thing in the integrand is what you get after using the product rule on [...]

and then you are done

[void estuary]

what are after the last step btw?

like that f(x)+ f'(x)

[severe quarry]

e^x * f(x) + e^x * f'(x)

[void estuary]

yea

[severe quarry]

Do you notice what derivative evaluates to that?

think of the product rule

[void estuary]

left d-right + right d-left

I get it

[void estuary]

so integration of this thing is e^x*f(x)

[severe quarry]

Yes

[void estuary]

i seee

[severe quarry]

But also you can just IBP on the second line as Ann said and then you don't have to do this

IBP is just based on the product rule

IBP'ing this thing should probably also work

the thing that you subtract should exactly cancel that third integral and you remain with your uv-expression

[void estuary]

ohh and what about the third thing then?

ohh it will cancel out

i get it

[severe quarry]

that third integral is cancelled yes

[void estuary]

damn i am not used to these exponentials but thanks

[severe quarry]

when there is e^x you kind of automatically think of IBP

[void estuary]

I see. Will remember that from next time onwards

.close

[pearl pond]

Channel closed

Closed by @void estuary

Use .reopen if this was a mistake.

[plush vessel]

my first reaction was g(x)=x/x-4

@void estuary

g(x)+g'(x) = ((x+2)/(x+4))^2

*e^x

=(g(x)*(e^x))'

answer can concise

[void estuary]

wait just a min

[void estuary]

how?

ohh

so you manipulated this into (g(x) + g'(x))

right?

[plush vessel]

yep

[plush vessel]

It looks like a differential equation

[void estuary]

you did well

[plush vessel]

the answer is concise

[void estuary]

but as a person who is not much friendly with exponentials i had to go step by step and see how things work

[plush vessel]

That's no problem, I'm just providing a better solution

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shut flicker]

how do we know that all terms in the Sigma are intergers

[rustic atlas]

because the power of q is positive

n-1-k >= 0

[shut flicker]

ahh tyty

ig thats the reason why we have to multiply by q^n-1 at the start makes more sense now thanks

.close

[pearl pond]

Channel closed

Closed by @shut flicker

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight haven]

hello

[midnight haven]

in this definition of a sub field

for the first condition

k' + sub group of k +

shouldn't a and b be in K' not in the whole K?

[lilac obsidian]

guys help 1 plus 1 how much does it come out

[midnight haven]

$\forall(a,b) \in \mathbb{K}^2, a-b \in \mathbb{K}'$

[jolly parrot]

Drk

[midnight haven]

$\forall(a,b) \in \mathbb{K}'^2, a-b \in \mathbb{K}'$

[jolly parrot]

Drk

[midnight haven]

basically is it the first or second one?

[midnight haven]

please move to another channel

[golden crag]

Drk

[brisk scarab]

I agree

I think it's just a typo but it should be (K')^2

[midnight haven]

i mean the whole point is closure and the inclusion of an inverse

[golden crag]

u might beat all of chatgpt, gemini and gauth

i havent do this type of question yet

[midnight haven]

thanks edward i thought i was bugging out

this professor is incompetent

shm

.close

[pearl pond]

Channel closed

Closed by @runic shore

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scarlet trellis]

I think I need to relate radii & distances to tangent length

but I don't know how to do that

once I have that I just run it for [1, 18] and sum

I drew a diagram but I see nothing useful from that

[glass meadow]

Is w supposed to be an integer?

I guess that's the "simplest radical form" thing

[scarlet trellis]

Probably

[glass meadow]

The circles are x units apart, does that mean their centers is x+12+9 units apart?

[scarlet trellis]

Yes

Or at least that's how I understood the question

[glass meadow]

No sure there's a better way than calculating the tangent equations for both circles and solving for a common one, something like that: https://www.desmos.com/calculator/xpvlic6zmm

[scarlet trellis]

There has to be another way

this is a geometry competition, the assumption is that you have 1 semester of precalc knowledge

so i doubt it would have a derivative

also, this is question 6/20

it isnt always perfect, but test designers are supposed to put easier questions first, with 20 being the hardest

so this should be a really easy problem...

but it sure doesnt seem like it

[glass meadow]

Ok but in competitions, most questions require knowing some kind of trick...

I know some but clearly not enough for this one

You may have better luck in https://discord.com/channels/268882317391429632/433980473367330826

[scarlet trellis]

I just tried googling "relate distance between circles and their radii to external tangent length"
and this result seems promising...I'll try it out. Maybe I should've checked google in the first place lol

[glass meadow]

Ah yeah that makes sense

[scarlet trellis]

ok so answer key says 16

I find the values of x and 7 & 17

so i think the answer should be 24

unless i made a mistake here

[glass meadow]

It's not x^2

[scarlet trellis]

oh wait

yeah its x+12+9

^2

[glass meadow]

,w sqrt((x+21)^2-3^2) for x={1,2,3,4,5,6,7,8,9}

[jolly parrot]

Results provided by WolframAlpha

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

Requested by nelfeal

[scarlet trellis]

yep i get 16 now

2, 4, and 10

[glass meadow]

,w sqrt((x+21)^2-3^2) for x={10,11,12,13,14,15,16,17,18}

[jolly parrot]

Results provided by WolframAlpha

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

Requested by nelfeal

[glass meadow]

Yeah

[scarlet trellis]

thank you for your help

im going to put this formula in my calculator now

.close

[pearl pond]

Channel closed

Closed by @scarlet trellis

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hybrid ivy]

. uh?

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[inner spear]

[robust kraken]

[robust lynx]

[cobalt hinge]

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scarlet trellis]

I have spent so long on this

Let h = initial hour hand in hours

Let m = initial minute hand in minutes

then $h_\text{final} = h + \Delta h$

[jolly parrot]

UCYT5040

[scarlet trellis]

and $m_\text{final} = m + \Delta m$

[jolly parrot]

UCYT5040

UCYT5040

[scarlet trellis]

also, $\frac{h_\text{final}}{12} = \frac{m}{60}$

[jolly parrot]

UCYT5040

[scarlet trellis]

and $\frac{m_\text{final}}{60}=\frac{h}{12}$

[jolly parrot]

UCYT5040

[scarlet trellis]

i also notice that $\frac{\Delta m}{60} + \frac{\Delta h}{12} = 1$

[jolly parrot]

UCYT5040

[scarlet trellis]

but i really only have 3 equations here, with a 4 variable system

what am i missing?

[pearl pond]

@scarlet trellis Has your question been resolved?

[scarlet trellis]

<@&286206848099549185>

[pearl pond]

@scarlet trellis Has your question been resolved?

[dense jasper]

wait why are there four equations

let $t$ be the number of minutes after 7:00 now and $\Delta$ be the number of minutes requested in the answer (so $\Delta<60$)

[jolly parrot]

Civil Service Pigeon

[dense jasper]

Then, $$6(t+\Delta)=210+\frac{1}{2}t+360n$$
for some integer $n$ (since the minute hand travels $6$ degrees in every minute and the hour hand travels $\frac{1}{2}$ a degree per minute)

[jolly parrot]

Civil Service Pigeon

[dense jasper]

you can use similar logic to get an equation for the hour hand

it should finish nicely from here

I'd be shocked if it didn't

[scarlet trellis]

1 sec lemme read this

ah ok i understand

but i prefer the idea that there are 60 minutes on a clock rather than there are 360 degrees

[scarlet trellis]

thats confusing

$t$=$\Delta$ then

[jolly parrot]

UCYT5040

[scarlet trellis]

it is currently between 7:00 and 8:00 (or, it is currently 7:t)

the answer is requesting how many minutes past 7:00 it currently is

do you mean $t$ is the number of minutes after 7:00 after the swap takes place?

[jolly parrot]

UCYT5040

[dense jasper]

degrees gives you a way to relate the changes of the minute and hour hand under a common unit though

[dense jasper]

$\Delta$ was supposed to be the number of minutes after the swap took place

[jolly parrot]

Civil Service Pigeon

[dense jasper]

mb

[scarlet trellis]

ok

[scarlet trellis]

same with minute ticks tho

[dense jasper]

eh to each their own

[scarlet trellis]

1hr = 5 minute ticks

1 min = 1 min tick

ok im going to attempt this whole problem now

sorry i got distracted

actually attempting it now

$t$ number of minutes past 7 $\$
$\Delta$ number of minutes that have passed after the swap

[jolly parrot]

UCYT5040

[scarlet trellis]

$\frac{7\cdot 60+t+\Delta}{720}$ represents the position of the minute hand before swap

$\frac{7+\frac{t}{60}}{12}$ pos of hour hand before swap

[jolly parrot]

UCYT5040

UCYT5040

[scarlet trellis]

$\frac{7 \cdot 60 + t}{720}$ pos of min hand after swap

[jolly parrot]

UCYT5040

[scarlet trellis]

$\frac{7+\frac{t+\Delta}{60}}{12}$ pos of hour hand after swap

[jolly parrot]

UCYT5040

[scarlet trellis]

$\frac{7\cdot 60+t+\Delta}{720} = \frac{7+\frac{t+\Delta}{60}}{12} \ \frac{7+\frac{t}{60}}{12} = \frac{7 \cdot 60 + t}{720}$

[jolly parrot]

UCYT5040

[scarlet trellis]

@dense jasper i dont think im doing this right

how did you get n in there?

[dense jasper]

angles are equal modulo full turns

[scarlet trellis]

thank you for your help, but honestly my brain is too fried today to attempt this further

i think i know what you mean though. ill try again tomorrow

.close

[pearl pond]

Channel closed

Closed by @scarlet trellis

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fast heron]

u = lambda * x

can someone help me simplify the first integral?

help pleasse

[pearl pond]

@fast heron Has your question been resolved?

[fast heron]

.close

[pearl pond]

Channel closed

Closed by @fast heron

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wind lagoon]

I have perhaps a very elementary question. Let $|\cdot|$ be the Euclidean norm in $\mathbb{R}^d$ and $q$ a fixed point, $x,y$ two other arbitrary points. If $|x-q|\neq |y-q|$, is it true that $y\neq x$?

[jolly parrot]

psie

[robust kraken]

you can see it by the contrapositive of this statement i think

if y=x, then |x-q|=|y-q| which is obviously true

[wind lagoon]

.close

[pearl pond]

Channel closed

Closed by @wind lagoon

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vivid sky]

x = y+3

[vivid sky]

How would I plot this

For example id do x = 1…

so y = 4

@tulip ore

[tulip ore]

1 isnt 4 + 3

try that again

[vivid sky]

Oooh

So 1 = y + 3

y = -2

@tulip ore

[tulip ore]

yep

another way you can do this is to solve for y

so from x = y + 3, simplify this to get y = (...)

[vivid sky]

Okay so y = -2

How to get coordinates

(-2,0)?

@tulip ore

[tulip ore]

remember coordinates are (x, y)

not (y, x)

also, (-2, 0) isnt correct

-2 isnt 0 + 3

[vivid sky]

Oh so it’s

(1,-2)?

@tulip ore

[tulip ore]

yes, that would for example be a point

if you keep putting in x and y, you can plot the graph out

another way you can do this is to solve for y

so from x = y + 3, simplify this to get y = (...)

[pearl pond]

@vivid sky Has your question been resolved?

Channel closed

Closed by @vivid sky

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gray valve]

So im just doing some exercises in my school book: Derive a formula that can be used to calculate the volume of a sphere from its diameter.

[gray valve]

I lowkey dont know how to proceed/ where to even begin

[plush bramble]

Are you supposed to use calculus

[gray valve]

Dont think so

Matter of fact idk what calculus is ngl

[slow grove]

i think maybe it just wants you to convert the volume of a sphere formula from using the radius to the diameter

so to start, do you know what the formula is for the volume of a sphere with radius r ?

[gray valve]

4/3 x pi x r^3 right?

[slow grove]

yep!

and do you know what the relationship is between radius and diameter

[gray valve]

Not really

Its kind of a weird exercise

The ones before was just calculating the volume

Now im supposed to think of a formula

[slow grove]

this is not to do with this specific question

there is a formula relating radius r and diameter d

[gray valve]

Well yea radius is the half of a diameter

[slow grove]

yep perfect

can you write that in symbols

[gray valve]

1/2 d = r?

[slow grove]

yep

now plug that into your volume formula from before

[gray valve]

Just add it?

Like at the end?

[slow grove]

no, substitute this formula into the other one

[gray valve]

So uhh d = 1/2 x pi x r^3?

[slow grove]

not quite

what you need to do is replace any rs with 1/2 d

[gray valve]

So V = 4/3 x pi x 1/2d^3

[slow grove]

almost, we cant forget brackets

[gray valve]

The 1/2d^3 in brackets?

Never understood how they work

[slow grove]

yes, so (1/2 d)^3

think about writing out r^3 the long way

so r x r x r

[gray valve]

Yea

[slow grove]

then replace each of those with 1/2 d

[gray valve]

No like i know the purpose

[slow grove]

i think this will help illustate how this works

humour me

[gray valve]

Wdym

[slow grove]

as in just go along with it

[gray valve]

So just do what u Said earlier

[slow grove]

yes

[gray valve]

1/2d x 1/2d x 1/2d

[slow grove]

and what does that simplify to

[gray valve]

Which is (1/2d)^3

[slow grove]

yep

but thats not totally simplified

[gray valve]

So what would i have to do to simplify it

[slow grove]

its easier to simplify from here

[gray valve]

Maybe uhh

Also get the d out of the brackets?

Nvm that makes no sense

[slow grove]

forget about any brackets for now, we're just looking at 1/2 d x 1/2 d x 1/2 d

[gray valve]

3 x (1/2d)

But thats just the same thing

Ohhhhh im dumb asl

Maybe 1/6

[slow grove]

if i said simplify 1/2 x 1/2 x 1/2 what would the answer be

[gray valve]

1/8

[slow grove]

if i said simplify d x d x d what would that be

[gray valve]

d^3

[slow grove]

ok so what should 1/2 d x 1/2 d x 1/2 d simplify to

[gray valve]

So its not 1/2 but (1/8d) ^3

[slow grove]

not quite

because (1/8 d)^3 would be 1/8 d x 1/8 d x 1/8 d
which i think you can agree is clearly different

[gray valve]

Yea

[gray valve]

1/2d^3

[slow grove]

still not quite

[slow grove]

think back to this

[gray valve]

1/8d?

[slow grove]

no bc now youre missing this

[gray valve]

Now you lost me completely

Would that just be 1/8d^3

[slow grove]

yes

[gray valve]

So just without the brackets

[slow grove]

yes

(1/2 d)^3 = 1/8 d^3

[gray valve]

So thats the difference

Ohh

[slow grove]

because (a b)^n = a^n b^n

[gray valve]

So what do I put in my formula

[slow grove]

so applying that here
(1/2 d)^3 = (1/2)^3 d^3 = 1/8 d^3

so we have V = 4/3 pi r^3

[gray valve]

So is my formula V = 4/3 x pi x 1/8d^3

[slow grove]

yes perfect

[gray valve]

Fuck that was hard to think of

[slow grove]

then that can simplify just a bit more

[gray valve]

To?

[slow grove]

we have a 4 in the numerator and an 8 in the denominator

[gray valve]

Nvm 1/2

[slow grove]

yep

so now whats the formula

[gray valve]

V = 4/3 x pi x 1/2d^3

[slow grove]

you kept the 4

[gray valve]

In the 4/3?

[slow grove]

yes

thats where the 4 in the numerator is coming from

[gray valve]

You got me confused

So is it 1/3 now?

[slow grove]

$\frac{4}{3} \cdot \pi \cdot \frac{1}{8} d^3 = \frac{1}{3} \cdot pi \cdot \frac{4}{8} d^3 = \frac{1}{3} \cdot \pi \cdot \frac{1}{2} d^3$

[jolly parrot]

Acman

[slow grove]

does that make sense ?

[gray valve]

Trying to think right now

You just multiplied both numerators right?

[slow grove]

i just moved some stuff around

maybe it would be easier to see if we got rid of some stuff that isnt changing

$\frac{4}{3} \cdot \frac{1}{8} = \frac{1}{3} \cdot \frac{4}{8} = \frac{1}{3} \cdot \frac{1}{2}$

[jolly parrot]

Acman

[gray valve]

I mean I got the formula now

[slow grove]

well it still needs to be simplified

[gray valve]

Why is there always something to simplify

[slow grove]

well not always, we're almost there now

[gray valve]

Dont tell me that i gotta take the 3rd square root now

Does that even count as simplify

[slow grove]

we dont need to do that

were just doing arithmetic at the moment

multiplying these fractions

[gray valve]

Maybe multiply again to 1/6d^3?

[slow grove]

bingo !

[gray valve]

So its V = 1/6d^3 x pi

[slow grove]

yep

[gray valve]

Gotta take it down step by step now

Thank you for the help

Atleast you didnt give me the answer right away

Made me use my brain instead

[slow grove]

yes no problem, happy to help

[pearl pond]

@gray valve Has your question been resolved?

Channel closed

Closed by @gray valve

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[proper nova]

I needed help proving M, K, N all lies on a straight line.
Question: Given $\triangle ABC$ has heights of $BD$ and $CE$ intersects at $H$. The angle bisector of $\angle ABH$ and $\angle ACH$ intersects at $K$. Let $M$ be the midpoint of $AH$ and $N$ be the midpoint of $BC$. Prove that $M$, $K$, $N$ all lies on a straight line.

[jolly parrot]

1 divided by 0 equals Infinity

[rugged raven]

do you know determinants?

[proper nova]

?

im doing normal geometry here, not coordinates-related

[rugged raven]

yea but you can still assign their x and y coords

[proper nova]

i don't think that im allowed to do that here

[rugged raven]

Why? is that not part of your syllabus?

tho you may be right, it might not be possible to assign them exact coords

[proper nova]

yeah i know, im supposed to do it like, not related to coordinates and descartes plane and stuffs

[rugged raven]

you should be able to assign them some variable coords tho

[proper nova]

i mean, my curriculum didn't learn about coordinates in geoemtry yet

[rugged raven]

ah

then nevermind

[proper nova]

<@&286206848099549185>

[earnest escarp]

btw 1 divided by 0 does not equal infinity

[proper nova]

get out

[earnest escarp]

i aint helping you with your problem btw good luck

[proper nova]

idc

[rugged raven]

wow I suck at geometry without numbers

[earnest escarp]

same

[pearl pond]

@proper nova Has your question been resolved?

[pearl pond]

@proper nova Has your question been resolved?

[dusty jungle]

hint BKC = 90

@proper nova

[proper nova]

if i wanted to prove BKC = 90, then i have to prove KN = BN = NC?

[dusty jungle]

bro we have all the angles

since we can calculate BKC, we have all that

from which we can calculate KNB

do a lil angle chasing

[proper nova]

alr

[rugged raven]

interesting that forms a circle then

[proper nova]

let me try something:
EBH = DCH (there are many reasons to say that)
since BK and CK are angle bisectors of these 2 angles then KBH = KCH

[dapper kraken]

ok from what ive got alr: -||prove K lies on (BEDC)||
-||K is the midpoint of some arc||
-||K lies on perp bisector of some segment, and so is N||
then something something nine point circle?

[proper nova]

didn't hear the term midpoint of an arc before

[dusty jungle]

i mean if you have that, we know that MN passes through the midpoint of ED

(both perpendicular)

so we're done

but how did you get that

[dapper kraken]

which part

[dusty jungle]

midpoint of arc

[dapper kraken]

BK is angle bisector of BED

[dusty jungle]

OH NVM IM DUMB

yeah we're done

[proper nova]

lmao

[dapper kraken]

wait how did you get this

[dusty jungle]

just use this to finish

[proper nova]

only thing is that im trying to see why BKC = 90

[dapper kraken]

i did some funky and finnicky stuff with the nine point circle trying to prove this

[dusty jungle]

both are centres of circles, and ED is a common chord, so if the midpoint is L, ML is perpendicular, NL is perpendicular, so MNL is collinear

[proper nova]

what the 💀 is even the midpoint of an arc

[dusty jungle]

EK = KD

[dapper kraken]

ohhhhh

[proper nova]

my brain hurts when i encounter geometry 😭

[dapper kraken]

i was thinking of constructing (AEHD) but i didnt realize M was the center of it

[dusty jungle]

yeah lol no matter how much i try i still suck at it

[proper nova]

NE = ND is easy to prove since BC is the diameter of (BEDC)

the hard part is ME = MD and KE = KD

[dusty jungle]

like did you not understand or are you remarking

[proper nova]

im just remarking

[proper nova]

and trying to see why this is true

[dapper kraken]

(AEHD) is concyclic. M is the midpoint of the diameter of it (AH), thus M is the center of (AEHD)

ME and MD are radiuses

[proper nova]

oh thanks!

didn't see that

[proper nova]

i can't seem to see why KE = KD

[proper nova]

i only got KE = KD left and im trying to see why that is true

[dapper kraken]

thats a consequence of angle bisectors and arc bisectors (no idea which theorem states it tho)

[dusty jungle]

chords that make equal angles have equal length

[proper nova]

can you elabortate more on that

?

[proper nova]

.close

[pearl pond]

Channel closed

Closed by @proper nova

Use .reopen if this was a mistake.

[proper nova]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[proper nova]

i actually have one more question i needed help with: Given $\triangle ABC$ which is an acute triangle, three heights of that intersects at $H$. From $A$, draw tangents $AP$ and $AQ$ of the circle with diameter $BC$. Prove that $P$, $H$, $Q$ both lie on the same line

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

in the diagram above, i let $O$ be the midpoint of $BC$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

.close

[pearl pond]

Channel closed

Closed by @proper nova

Use .reopen if this was a mistake.

[elfin compass]

hey so, i'm just curious about a concept of math... If someone wants to chat about that

i'm new to this server and i'm trying things out

[ripe lantern]

This channel is occupied, you can take your discussion in #math-discussion

[elfin compass]

Thank you.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[elfin compass]

Hi, i'm with the topic, "equations with radicals" you see, sometimes you get two diferents values for x, like "x¹=4, x²=6" but why can a solution be "fake" (in mathematical terms) when you do the comprobation? (When the result does not satisfy equality)

[inner granite]

A solution is deemed false only when the logic building up to the result is contradicted by the answer.

[compact ridge]

because when you square both sides, you create solutions that aren't part of the original

if $a = b, a^2 = b^2$ yes

[jolly parrot]

south

[compact ridge]

but $a^2 = b^2$ can also come from $a = -b$

[jolly parrot]

south

[inner granite]

Like the following example, when the squares are taken off from both sides, it doesn't guarantee that both sides must be equal.

Unless there is some conditions sit upon them.

For example, if $a, b > 0$ and $a^2 = b^2$, we can be certain about $a = b$

[jolly parrot]

Erebus

[inner granite]

(Because they both need to be > 0 and a = -b doesn't satisfy the condition)

[elfin compass]

Ohh right

[inner granite]

Another time is when you get contradictory results.

$x > 0$ and you get $x = - 4$

[jolly parrot]

Erebus

[inner granite]

Then you can reject this case.

[elfin compass]

But like i was asking, mathematicaly talking, why can a solution be "false" if it does not satisfy the equality? How can you get to that result

[compact ridge]

do you understand how squaring both sides creates extraneous solutions?

[elfin compass]

Yes, because i they wouldn't satisfy the equality?

[inner granite]

By equality, do you mean why a solution might be false although they fit in an equation or another thing?

[compact ridge]

I don't understand where you're confused on

[inner granite]

Mostly, I'll give you another example.

$x^2 = 4 \implies x = \pm 2$

[jolly parrot]

Erebus

[inner granite]

Here, non of the solutions are false because there is no other restrictions on it.

$\sqrt{4x - 3} = x \implies 4x - 3 = x^2 \iff x^2 - 4x + 3 = 0 \iff (x -3)(x - 1) = 0$

[jolly parrot]

Erebus

[inner granite]

Wait bad example

[elfin compass]

Wait

It a good one

Or no

?

[compact ridge]

there's even $\sqrt{x} = -2 \implies x = 4$

but when you check $x = 4$ by subbing in the original equation, it doesn't work ($\sqrt{4} = 2$ only)

[jolly parrot]

south

[compact ridge]

so the conclusion is that there are no (real) solutions

[inner granite]

In the above case, no square roots could be negative.

This leads to a contradiction.

[elfin compass]

So, what happens?

[inner granite]

Well, Assuming $\sqrt{x} = -2$, then $x = 4$ (squaring both sides.)

[jolly parrot]

Erebus

[inner granite]

But $\sqrt{4} \neq -2$

[jolly parrot]

Erebus

[inner granite]

So, our previous assumption is false.

[elfin compass]

Yeah, so what's the solution then?

[inner granite]

There is no solution.

The solution simply does not exist.

It is somewhat of a law that square roots *aren't negative.

$\sqrt{x} \ge 0$

[jolly parrot]

Erebus

[elfin compass]

Yes, i understand. Thank you.

❤️

I love you both

[compact ridge]

no worries!

[pearl pond]

@elfin compass Has your question been resolved?

[latent quail]

.solved

[pearl pond]

Channel closed

Closed by @latent quail

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sharp smelt]

I;d like to solve this using convolusions

[sharp smelt]

$h(z)= \int_{a}^{x} \frac{a}{(z-y)^2} \frac{a}{y^2} dy$

would this work?

[jolly parrot]

wai

[sharp smelt]

my only concern here is finding the domain of z

I don't follow

[elfin stirrup]

the probability density of a sum of independent random variables is the product of the probability densities

[sharp smelt]

so just a^2/x^4

[elfin stirrup]

thats not correct

[sharp smelt]

oops

my bad

[sharp smelt]

why is this wrong ?

[elfin stirrup]

no worries we are all here to learn :)

[hallow timber]

$f_Y$ is nonzero on $[a, \infty)$ and $f_X(z-y)$ is nonzero on $(-\infty, z-a]$

[elfin stirrup]

what is z ?

[jolly parrot]

flynger

[sharp smelt]

the variable in terms of which I'll express the final answer

I also just realised I assume both RVs are the same

[elfin stirrup]

you have to take to different ones

but with the same distribution and densities

[sharp smelt]

this is the formula

so I tried using the last one

[hallow timber]

I think this implies that the integral is between [a, z -a] right

[sharp smelt]

so their intersection

[elfin stirrup]

and again what is z in theorem 10.1 ?

thats important to understand

[hallow timber]

yeah

[sharp smelt]

this is thm 10.1

[sharp smelt]

10.3

you mean

[elfin stirrup]

yup so what would Z be in this case?

maybe that question is too trivial for you and if so Im sorry, but its important to understand

[sharp smelt]

the RV?

X+Y

so I have $a≤y≤x$;
$a≤z-y≤x$?

[jolly parrot]

wai

[sharp smelt]

( to find the bounds of the integral?)

[elfin stirrup]

you can insert the functions even with the cases

[sharp smelt]

I don't follow

I'm just unsure of how to do this using convolusions

(which is what I wish to use here)

[elfin stirrup]

okay first of all what formula do you want / need to use?

[elfin stirrup]

from the pictures that you presented

[sharp smelt]

the last one

[elfin stirrup]

okay what is f(z-y) and g(y) in this case?

[sharp smelt]

$\frac{a}{(z-y)^2}; \frac{a}{y^2}$

[jolly parrot]

wai

[hallow timber]

From direct substitution you have

$h(z)= \int_{-\infty}^{\infty} \frac{a}{(z-y)^2} \frac{a}{y^2} dy$

[jolly parrot]

flynger

[elfin stirrup]

thats not correct

you are both missing the cutoff

[hallow timber]

oh wait sorry

yea thats invalid

[sharp smelt]

the what?

[elfin stirrup]

\[
f(x)=
\begin{cases}
\dfrac{a}{x^{2}}, & x \ge a,\\[6pt]
0, & x < a.
\end{cases}
\]

[jolly parrot]

tobi

[hallow timber]

From direct substitution you have

$h(z)= \int_{-\infty}^{\infty} f_X(z-y)f_Y(y) dy$

[jolly parrot]

flynger

[elfin stirrup]

this is only valid in its given domain

[hallow timber]

$f_Y(y)$ is nonzero on $y \in [a, \infty)$ and $f_X(z-y)$ is nonzero on $y \in (-\infty, z-a]$

[sharp smelt]

ye

[jolly parrot]

flynger

[sharp smelt]

oops, right

[elfin stirrup]

so yes from direct substition you would first get this

so thats what I meant with inserting the function even with the different cases

[hallow timber]

By this you have

$h(z)= \int_{a}^{z-a} f_X(z-y)f_Y(y) dy$

[jolly parrot]

flynger

[sharp smelt]

got it

tq

[elfin stirrup]

I hope you dont have to integrate that, that would not be fun (but possible)

[sharp smelt]

I think my course would require me to, but we usually don't get such problems on the exam

such computational ones

[elfin stirrup]

use partial fraction decomposition, if you wish to evaluate the integral

[sharp smelt]

one more question

[elfin stirrup]

sure

[sharp smelt]

one min

I'm actually unsure of which method to use here

oh

nvm

[elfin stirrup]

have you worked with probability measures?

[sharp smelt]

this course doesn't do measure at all :(

It's done alongside RA1

[elfin stirrup]

sorry I only know how to do it with probability measures

[sharp smelt]

I do have a formula in mind

the pdf of $X$ is $ \frac{4h^3}{\sqrt{\pi}} v^2 e^{-h^2v^2}$

[jolly parrot]

wai

[elfin stirrup]

yeah that works here

[elfin stirrup]

(very important formula in physics btw!)

[sharp smelt]

what is T here

like T looks like it's x^2

[elfin stirrup]

Y = X^2 = T(X)

[sharp smelt]

So W(y)=√y

[elfin stirrup]

well

yes but why

look at the definition of the r.v.

[sharp smelt]

Z=1/2mV^2

[elfin stirrup]

Why does your example satisfy this assumption?

[sharp smelt]

the RV increases on the domain

[elfin stirrup]

f(x) = x^2 is not an increasing function

yeah it would not be on whole R

but on [0,inf) it is

[sharp smelt]

right

dx/dy=1/2√y

and $f(W(y))= \frac{4h^3}{\sqrt{π}} ye^{-h^2y}$

[jolly parrot]

wai

[elfin stirrup]

you are missing the 1/2 * m

but otherwise your done

1/2 * m is just a constant and therefore independend so you can just mulipliy it to the density of V^2

[sharp smelt]

Yea

Tq

I think that's all for now , thanks so much!

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[sharp smelt]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[sharp smelt]

@elfin stirrup you were saying something?

[elfin stirrup]

I asked what Tq means but I looked it up

[sharp smelt]

thank you

[elfin stirrup]

:)

[sharp smelt]

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[urban crypt]

Hello, I am in quite the bind at the moment because I need to get to calculus undergrad levels of math.. and currently I can barely do 5th grader math. That in itself isnt that much of a problem...
The problem is that I start school 13th of January.. and I need to get up to a reasonable level by then. I dont particularly want to cram but I dont think I have any choice? Are there any good tools I can use to help me along the way? The software my school uses is ALEKS but of course its paid software so I dont wish to spend too much time with it. I spent 13 years in school yet I remember literally nothing. I am a bit stressed by now..

Sorry if this question is too generic..

[hallow timber]

Work from the ground up and understand the fundamentals. Don't be scared to stay on the beginner material until you fully grasp it. Once you learn a topic, push yourself with harder questions to test how strong your understanding is.

[rotund ferry]

I think the main problem here is the time limitation

It's already November, and school will open on January?

I think that skipping a few topics is a must atp

Just learn what is more important, try to watch YouTube serieses like Organic Chemistry Tutors

And watching a few 1-2 hour "intro to alg", " Algebra review", "basics of trig"

After watching those 1-2 hour videos, use apps like Khan Academy to strengthen your understanding by practice

[hallow timber]

They said they can barely do 5th grade math, so I'm gonna assume they could need to start even earlier

[rotund ferry]

True

[umbral osprey]

yeah and to kinda add onto this, don’t worry about going too deep into it because usually the intro calculus classes will kind of guide you through the problems so you can learn any gaps if you need

[rotund ferry]

Only having 2 months to re-study 13 years of math is just... Insane

[pearl pond]

@urban crypt Has your question been resolved?

[urban crypt]

Thing is that this seems a bit scary 💀

[umbral osprey]

to be honest a lot of it is just going to be repetition

[urban crypt]

They expect me to get up to that class before I start if I read that right? English isnt my first language

that means I need a score of 76% or better

I did it once already and I got 18% 😭

[umbral osprey]

yeah you need to place into it

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shut flicker]

i was trying to appraoch this question with a stars and bars method, i think i got there would be 97 choose 3 ways if we ignored the odd condition, but im not entirely sure how to appraoch it with the odd condition (maybe might have to use a different method), i was thinking maybe it involves reducing the previous 97, maybe dividing it in half or something similar, as we need odd gaps between each bar?

[tropic saddle]

x_i = 2a_i+1 because they are odd

[pearl pond]

@shut flicker Has your question been resolved?

[shut flicker]

so does it become like this

[clear shore]

The concept is right but you're overlooking one thing

[shut flicker]

can u hint where that may be

[clear shore]

What's the least value a_i can have in your method

[shut flicker]

ah is it that i should have done 2a-1 instead, since x_1 can be 1, so if 2a_1 - 1 = x_1, then 2a-1 can equal 1, so a = 1, whereas with 2a+1, i can't get x_1 = 1 unless a is non positive

[clear shore]

Yup

[shut flicker]

so its 50 choose 3

yippee

thanks for ur help

.close

[pearl pond]

Channel closed

Closed by @shut flicker

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shadow cloak]

What are some applications for the fundamental theorem on homomorphisms? And why can you say that the diagram you draw for it commutes?

[cobalt hinge]

Same as first isomorphism theorem correct?

[shadow cloak]

Nope

[cobalt hinge]

this right?

[shadow cloak]

Yes

[cobalt hinge]

Ok so just slightly different format

[shadow cloak]

? That's not similar to the first isomorphism theorem, is it?

That's 1st iso

[cobalt hinge]

Wait lemme check my textbook

I might be tripping

This is what I have in my book, its just not labelled

[shadow cloak]

yes

[cobalt hinge]

Ok phew

So what I mainly use it for is like elaborating on "induced" homomorphisms whenever I'm creating a proof

A lot of the time if you have these conditions satisfied then its easy to prove that \tilde{\phi} is bijective since according to this theorem its already an isomophism

There are a few examples I have

[shadow cloak]

How do you remember it? Like can you describe it in one sentence intuitively or something

[cobalt hinge]

Well its the diagram

[shadow cloak]

You just think of the diagram?

[cobalt hinge]

Yep

Once you start working with it a little bit it gets easier to remember

[shadow cloak]

But (i) is not in the diagram, is it?

[cobalt hinge]

Hmm

[cobalt hinge]

Well I guess you can memorize that little part

Its not in my book for some reason

[shadow cloak]

True

Ok and one more thing:

Do you remember all the three iso theorems?

[cobalt hinge]

Uh no

[shadow cloak]

haha

[cobalt hinge]

Partly because I haven't gone in-depth and partly because my book doesn't name the theorems 🥲

[shadow cloak]

Alright

[cobalt hinge]

I gotta go eat

[shadow cloak]

Thanks

[cobalt hinge]

You can keep this open to see if anyone else can help you with memorization

[shadow cloak]

So the new question is: How do you usually remember the three isomorphism theorems? Do you have like a one-sentence-intuition you keep in mind?

[shadow cloak]

Also by the way, apparently yes, it is the first iso theorem, we just named it a little differently and that's why we don't call it that for some reason

.close

[pearl pond]

Channel closed

Closed by @shadow cloak

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quick venture]

hey im unsure why these angles are the same

[shadow stag]

they don't have to be the same, that's just given

[quick venture]

Ik but how would I know that the top angle is 30

[warm tiger]

Nc is perpendicular to that inclined

[quick venture]

can you show me a diagram?

[warm tiger]

As normal force acts perpendicular to the plane

@quick venture

[quick venture]

Thanks

.close

[pearl pond]

Channel closed

Closed by @quick venture

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[robust lynx]

it’s closed

[solar fern]

Ok

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[kindred forge]

#19

Do I have to calculate the area of both circles and subtract their intersection?

So I’m assuming the circles have radial bounds of 0 to their respective equations, and angles 0 to pi for sin and -pi/2 to pi/2 for cos

I can do that all well and dandy, what about their intersection? How can I calculate that

?

<@&286206848099549185>

.close

[pearl pond]

Channel closed

Closed by @kindred forge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[kindred forge]

17

[kindred forge]

.close

[pearl pond]

Channel closed

Closed by @kindred forge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[edgy laurel]

I dont really understand the answer

[sharp vigil]

,rccw

[boreal nexus]

a-b the whole square = a² - b² - 2ab

that should be about it yeah

don't look at it differently just because of the -x

[sharp vigil]

what is the question?

[edgy laurel]

But we have two specific formulas
a^2+2ab+b^2 for when (a+b)^2
And a^2-2ab+b^2 for (a-b)^2
When i asked chatgpt he said the answer is x^2+2x+1
But i get the answer in the picture

[inner spear]

,rotate

[boreal nexus]

right so the 'a' being -x, doesn't make it a + b the whole square.

you compute it just with the formula of a - b the whole square

it just gives you a funny 'a' that you don't expect to be +/positive (while doing the computation)

[edgy laurel]

Ohhh so -x-1 is treated as a a+b so i have a positive 2 when multiplying

[boreal nexus]

no no, -x - 1 (the whole square), is treated as a-b (the whole square).
the "a" Here being -x (negative guy) doesn't change the fact that it's gonna be a - b the whole square.

[edgy laurel]

So the formula is a2-2ab-b2?

[boreal nexus]

yes

and not the +/positive formula

might have gotten confused with the -x squaring to positive x²

but you should not really ask chat gpt for answers, as it has a lot of errors/mis understandings and you won't get what you're doing

[flint fulcrum]

There's no two formulas

It's simple (a+b)^2= a^2+2ab+b^2

[boreal nexus]

there is, he means the a - b the whole square and a + b the whole square formula.

[flint fulcrum]

If you sub b for -b'

You get second formula

[boreal nexus]

yea there's 2 formulas then, aren't they!

[flint fulcrum]

Say you wanna evaluate (-x-1) ^2

Think like this a=-x , b=-1

Then do (a+b)^2

If you wanna do by (a-b)^2

You can think a=-x , b=1

Or you could be smart and think (-x)^2 =x^2

[boreal nexus]

I mean... do we really wanna confuse someone who's just learning this?

it's simple enough either way

[edgy laurel]

Ok im genuinely confused now
Our proffesor told us to just fit in the numbers into the formula depending on the +/- beetween a and b

[boreal nexus]

Yes

[edgy laurel]

Is the -x treated as x ?

[boreal nexus]

nope

it's treated as -x, but there's a difference between treating it as -a and +a

so when I speak about -a or +a, I'm speaking about the left side of the sign in between

[boreal nexus]

let's not confuse you and take 55 + 45 here for example

when I say +a, here I mean 55 because it's left to here

but if I say -55 + 45

I take "a" as -a

get the point? it can be -a or +a depending on the sign of the variable/constant

[edgy laurel]

Ohh i see

[primal pasture]

(−x−1)square2 =(−x−1)(−x−1)= xsquare 2+x+x+1=xsquare2+2x+1

[boreal nexus]

now think about this, -a doesn't matter here really because you compute two numbers hmm and the second one matters more
so, what matters is the b sign

[edgy laurel]

Ohh ok

[boreal nexus]

now here, the b is "-1" so it has a negative sign

[edgy laurel]

Ye

[boreal nexus]

that means, you use the formula for (a-b) ²

[edgy laurel]

Yep

[boreal nexus]

the a being -x or +x, does not matter

you can even confuse it by saying the "a" as -x² or x²

but it's just gonna be "a"

the "b" matters majorly here

[edgy laurel]

So in this scenario how would i fit the numbers into the formula?

[boreal nexus]

right, so the formula results in a² + b² - 2ab

for (a-b) ²

[boreal nexus]

now we plug in what's "a" and what's "b" here

[edgy laurel]

Shouldnt it be a2-2ab+b2

[boreal nexus]

yea it is, we just flipped the order

changing -2ab to the right edge doesn't matter

because we brought +b² to left without changing anything

[edgy laurel]

Ohhhh

Ok ill try to solve now

[boreal nexus]

it doesn't matter if you do it that way as well

[boreal nexus]

yeah, now plug in a and b to the formula

and check what's happening

be careful with the cancelling after you do it

[edgy laurel]

Im stuck at the -2 times -x times -1 since its
-2 times -x is 2x and 2x times -1 is -2x
But in the original answer it should be +2x

[boreal nexus]

mhm

[vernal bear]

,rotate

[boreal nexus]

now think of it like this, what would be -2 times positive/+x?

[edgy laurel]

-2x

Then that times -1 is +2x

[boreal nexus]

right? now what if the x was also negative? the answer changes it's symbol right?

[boreal nexus]

no no dont move on yet

we have -2 times x, resulting in -x.

[edgy laurel]

U mean -2-x?

[boreal nexus]

no, I mean -2 × X being -2x, then what does -2 × "-" X be?

[edgy laurel]

+2x

[boreal nexus]

right?

now, multiply that with 2 times -1

what is 2 times -1?

[edgy laurel]

-2

[boreal nexus]

so you get the 2 times a and 2 times b, now our step is to multiply those both:)

so what is -2 times 2x ?

[edgy laurel]

-4x

[boreal nexus]

exactly!

that's 2 times a × b, don't get confused if "a" is something hard or weird!

[boreal nexus]

and lemme just point out the idea here, you multiply 2 times "a" properly, and 2 times "b" properly.
before you can multiply those both together

[edgy laurel]

Omg i know the problem
I was solving this thinking x2+2x+1
Is the answer what i got from Ai and photomath

[boreal nexus]

they may be correct

but they most likely mis understood your wording

[edgy laurel]

So the answer is x2-4x+1

Is 2ab
2 times a and 2 times b

[boreal nexus]

mmmmm it would be if something was different, but I told you wrong somewhere, where is it?

[viscid shale]

are we still on the (a+b)^2 / (a-b)^2 thing?

[boreal nexus]

(Hint: I did no calculating error, something went wrong in calculating the 2ab)

[edgy laurel]

Idk im genuinly confused
Since i done all my other with 2 times a times b rather than 2 times a and 2 times b

[boreal nexus]

scrape out everything, think of -2ab when a is -x and b is -1

[viscid shale]

Are we still stuck on this?

[edgy laurel]

Ye

[viscid shale]

okay, recap, cause the explanation is pretty easy

[boreal nexus]

we're not stuck really I'm just trying to get him though it so that he understands it completely

[viscid shale]

You have to know the following properties of arithmetics and thats kinda it:

$$a\cdot (b+c) = ab+ac$$
$$(ab)^c = a^cb^c$$

[viscid shale]

Do you remember these two?

basically its distribution over one lesser operation

[edgy laurel]

Uhh i dont think so
I may have but i wasnt a real math person trough 5-7 grade
Now i want to lock in

[boreal nexus]

Okay, to ease out your confusion. For this time I'm telling you where I went wrong,
when I tried to calculate 2ab , I thought it was 2(a) × 2(b). Something that is only correct if it was 2 (ab) [ with ab in bracket]

[boreal nexus]

it's a common error, and you only had to be multiplying it once

[edgy laurel]

Ohh so its 2 times a and the answer of that is ans times b

[boreal nexus]

yes exactly

thinking you should multiply 2 times a and 2 times b, then multiply it again, is a common error

[edgy laurel]

Oh ok

[viscid shale]

Okay, well, lets recap on how distributive property works:
Do you agree that something like:

$3 \cdot (5+2) = 3 \cdot 7$

[vernal bear]

What is original que?

[viscid shale]

?

[viscid shale]

this is just pemdas applied

[boreal nexus]

now let's not get too complicated of ourselves^ and see what 2 times -x is, and then the answer of that to times -1.
So 2 × -X × -1

[edgy laurel]

Yes

Guys its hard keeping up with two people

[boreal nexus]

mate appreciate you, but this is in a complex language for him right now

need to help him with pure basics, nothing else

[boreal nexus]

@edgy laurel think of this

and tell me the answer what you think is correct. (Nothing too complex, just ease out your mind)

2️⃣ ✖️ ➖ ❌ ✖️ ➖ 1️⃣

[boreal nexus]

Red X is x as in x mass, and black x is multiplication sign

[edgy laurel]

My original answer

[boreal nexus]

correct!

[vernal bear]

One mistake

[edgy laurel]

Oh
So my problem was checking with AI

[edgy laurel]

Where

[vernal bear]

When you write b take as +1

[edgy laurel]

But isnt it -1 since u need to take the - if its infront of a number?

[vernal bear]

-2(-x)(+1)

[edgy laurel]

Could u explain why +1?

[vernal bear]

Yes but here we applying (a-b)^2
a-> -x
b-> +1
-x - (+1)

= -x-1

[edgy laurel]

Problem is i dont understand why -1 would turn into +1
If so
Then AI was indeed correct

[viscid shale]

Might i try to take a jab at it? I think distribution is easy enough of a concept.

[vernal bear]

Go on

[edgy laurel]

If +1 then answer
x^2+2x+1

[edgy laurel]

sure

[vernal bear]

A bit off

[viscid shale]

Using pemdas, we got this results previously, which, mind you, is also = 21.

Well, distribution basically tells you that the order of operation doesnt really matter
So, you can multiply and then add
or add and then multiply.

going back a step

$$3 \cdot (5+2) = 3 \cdot 7$$
but also:
$$3 \cdot (5+2) = 3\cdot 5 + 3\cdot 2$$

[edgy laurel]

Ohh i know about that
We learned distribution + commutitation or commutation

[viscid shale]

You can multiply term-by-term

You can check on your own, both expressions are equal to 21.

[edgy laurel]

Oh i understand that part

[viscid shale]

Without too much on the details of why
we say that multiplication distributes over addition (what we just did)
And that exponents distribute over multiplication

[viscid shale]

This formulas should make some sense now

[edgy laurel]

Yes

[boreal nexus]

mmmm I see, the -1 does not turn into +1 .
-1 is actually - times +1 , which turns it into a -1 ( - times + is -) .
We just take it as +1 and leave the minus symbol left of it because the formula we're using, already has in the account of it being ("a" '-' "b")².
Suppose, it was actually -x + 1 instead of -1 ,

We would use the other formula wouldn't we? so when we consider the b, we take it as + and not negative. But when we think of a, we take it as whatever it is no matter what it is. Because the formula only changes because of it being - or + , and the formula has taken into the account of the number/variable without it being positive or negative.

[viscid shale]

Well, we can use these to easily solve for why we get x^2 + 2x + 1

[edgy laurel]

OMG ik
Bc of the minus in the formula we remove the - from the 1 since its alr in the formula by default

[boreal nexus]

yeah

[viscid shale]

Consider a small fact: We usually call "factorization" to the process of doing """inverse""" distribution

sooo, as you just said

$$(-x-1)^2$$
$$\left((-1)(x+1)\right)^2$$
$$(-1)^2(x+1)^2$$
$$1\cdot (x+1)^2$$
$$(x+1)^2$$

[edgy laurel]

Oh yes now i understand this

YALL ARE LIFESAVERS🙏 🔥

[boreal nexus]

took a bit too long to tell that, but I hope you've learnt common errors and mistakes along the way.
the level you're in, it does not matter in a bad way for you to do mistakes.
Learn from your inability of knowledge. The lesser things you know, the more you can learn and you aren't compulsioned to be perfect. So don't take it too hard, and be in a stage where you can sleep a good night even if you hadnt studied for the exam in the morning

[edgy laurel]

I think i understand other stuff
Dont mind the questions in croatina
I will hit yall up if i need sum more help

U can make this channel available now

[boreal nexus]

Yeah 👍 glad to be of help

[boreal nexus]

you actually need to do it by typing ".close"

[edgy laurel]

Oh

Bet

.close