help-10

[inland pine]

well that proves it right

[royal basin]

no

for the inductive step you do not put n as any value

[inland pine]

ah ok i got it now

thanks

.close

[obtuse pebble]

Channel closed

Closed by @inland pine

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

can anyone help me with this?

so far i've got

1. [p && ] - Assumption
2. p - & Elimination (1)
3. q - & Elimination (1)

[weary delta]

What rules do you have in the book?

[timid silo]

@weary deltaany idea?

[weary delta]

Sorry, I’m not familiar with this sort of logic

Maybe modus tollens to get a double negation?

Assume p \land q. By modus ponens this statement is nothing but not(p \implies not(q)). By initial assumption + and elimination p. Thus by modus ponens the statement is not(not(q)), by double not elimination we reach q. By initial assumption + and elimination q. Thus 1?

[stable rain]

woah woah

u cant just do double elimination like that

[weary delta]

I guess you’d also need to show Assume not(p \land q). No idea. Was just guessing since noone else seems to be responding 😄

[timid silo]

these proofs will be the end of me

@stable rainAre yuo familiar with these?

[stable rain]

a bit

but

like i said

double elimination is sus

[timid silo]

would it be possible if we could hop in a call and i could share my screen and you could give me a little guidance?

i'm really struggling with these

[stable rain]

u cant really prove

p from ~~p

[timid silo]

well assumptions are allowed if you didnt already know

[stable rain]

@timid silo

i kinda forgot how u would go about working

can u show how u would show

p or ~p is a tautology

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

@stable rain

here's an example proof my prof did

[stable rain]

yea i get that

but in this case the assumptions is nothing right

so how do we start

(p and q) -> ~(p -> ~q), start, 1
(p -> ~(p -> ~q)) or (q -> ~(p -> ~q)), from start, 2

can we do this

[timid silo]

well i think we need to start with just

p and q

and work towards p and q -> ~(p->~q)

im so confused bro

[stable rain]

can we start with p and q?

if we start with p and q then we are taking p and q as our premises right

[timid silo]

yes

we need to eventually workout way from:
p and q
.
.
.
(p and q) -> ~(p->~q)

[stable rain]

oh

yea i get it now

so u are saying instead we can equivalently prove

p, q |= ~(p->~q)

yea?

[timid silo]

kinda yeah

idk if that's 100% accurate to say it like that

but we start with the left hand side and assume that

and then work our way down

[stable rain]

yea

i feel its equivalent

but idk if we can start with that

do uve an example of a tautology being proved

[timid silo]

lemme c

[stable rain]

other than a when a is true or tautology itself

[stable rain]

ok i guess they are saying

when trying to prove

a->b

we can assume b?

idk a bit confused

idk looks weird

they assumed p then later assumed not p

[timid silo]

so the idea is

you can assume anything

thats what the brackets denote

[stable rain]

and

whats that conditional on

im pretty sure u had to resolve what u assumed somewhere right

else u can just assume the tautology is true

[timid silo]

so for example here

everytime u assume something

you indent one inwards

and it changes the scope of the variables you can use

by the end of the proof you need to be back to the leftest indent

but you can assume things to help you introduce stuff you didnt have before

[stable rain]

ok so

back to

p or ~p

[timid silo]

i'll just go to my professors office hours and try to get extra help

[stable rain]

lol

yea

i think im understanding a bit more now

lol

so the table u showed earlier

if weve a->b

we can assume anything

like we can assume c

[stable rain]

but later on we have to add in a

c->(whatever we got)

is that right

so for

p or ~p

we can assume p

from there we get ~p

going back assumption we get

p-> ~p

[timid silo]

well

you have to "leave an assumption world"

like you have to leave that assumption's scope

that's done through something like

-> introduction

or Reduction to absurdity

so if you see these two rules

they assumed [a]

and they have variable b

through -> intro they can leave that assumption world and say that a->b

[stable rain]

oh yea i get it now

ok i get how to do

wait

its still feels weird

p or ~p

[timid silo]

well p or ~p is just = to p

[stable rain]

[p]

[~p]

p and ~p is false, resolving ~p, we get p

[timid silo]

hm

circle would you be willing to call and discuss this i think that might be easier

[stable rain]

sure

wait actually nvm

my mic is bad

[timid silo]

that's fine tbh

i can deal with that

i just wanna understand this stuff 😭

[stable rain]

o ok

[timid silo]

sent u a friend req

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[past rivet]

hi I need help with proving that cos(α −β) = cos(α) cos(β) + sin(α) sin(β) when α = 2π, α = β and β = 0 (not at the same time, but substitued in one at a time they should keep the equality of the equation)

[kind hawk]

well what can you use?

do you know complex exp?

[past rivet]

complex exp?

[kind hawk]

exp(z) for complex z

although I guess that's overkill if you only have to show it's satisfied for those few values

[past rivet]

yea

[kind hawk]

cos(2pi), cos(0) etc are special values

what are they

[past rivet]

cos(2pi)=1 (cos0)=1 no?

[kind hawk]

good and what about sin of these values?

[past rivet]

sin(2pi)=0 at least

sin(0)=1?

[kind hawk]

no

[past rivet]

0

it equals 0

[kind hawk]

yes

so when yyou plug these values in, which equalities do you actually have to show

[past rivet]

ill try to do it one moment

i just dont know what to do with cos(α −β)

isn't it cos(1-beta)?

[kind hawk]

only if alpha=1

[past rivet]

alpha is 2pi for the first value

which is 1

[kind hawk]

2pi is not 1

[past rivet]

cos(2pi)?

[kind hawk]

well that's different

[past rivet]

true

i just end up with cos(2pi-beta)=cos(beta)

which is not the same

i don't know where im doing something wrong

[kind hawk]

it is

which properties of cos do you know

[past rivet]

uhh the ones we discussed above at least

how is (2pi-beta)=cos(beta)?

[kind hawk]

do you know that cos is 2pi periodic?

[past rivet]

I don't know what that means sorry

[kind hawk]

it repeats after 2pi steps

cos(x)=cos(x+2pi)

[past rivet]

oh right, yea that i understand

[kind hawk]

ok

so cos(2pi-beta)=cos(-beta), yes?

[past rivet]

I suppose

[kind hawk]

so now we need to show that cos(-beta)=cos(beta)

do you know that already?

[past rivet]

but what about the 2pi part, it's still 1 or..?

which gives cos(1-beta) or am i wrong here?

[kind hawk]

which 2pi part

you can't evaluate functions like this

[past rivet]

oh

I mean yeah cos(-beta)=cos(beta) cause it's the same axis

[kind hawk]

good. so we are done with the first one

[past rivet]

but I still don't understand what happens to cos(alpha-beta)?

[kind hawk]

well in general cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)

which we are here proving for a few special cases

[past rivet]

ahh so then it would be the same as 1*cos(-beta)?

[kind hawk]

that's what we are showing

[past rivet]

yeah I mean, that makes sense

wait give me a moment to try to solve for 2pi

alright I got it

nice

so you end up with cos(-beta)=cos(beta) which is the same

@kind hawk

[kind hawk]

yes

[past rivet]

nicee

ok so i'll try for alpha=beta

ok so i get it down to cos(beta)cos(-beta)=cos(beta)sin(beta)+sin(beta)

[kind hawk]

no

not sure how you got that

[past rivet]

ill try again

with just replacing alpha for beta

ok so it gives us:

cos(beta-beta)=cos(beta)cos(beta)+sin(beta)sin(beta)

[kind hawk]

yes

on the LHS beta-beta=0

so we are just left with ?

[past rivet]

0=0

[kind hawk]

no

cos(beta-beta)= ?

[past rivet]

cos(0)?

[kind hawk]

yes

which is?

[past rivet]

its.. 1?

[kind hawk]

yes

[past rivet]

ah

😄

[kind hawk]

so LHS is 1

lets do the RHS

[past rivet]

okay

[kind hawk]

we notice that the inputs are teh same

so it's cos^2(beta)+sin^2(beta)

do you know this expression

[past rivet]

looks like the pythagorean theorem

[kind hawk]

it is

[past rivet]

so can you just take the squareroot then?

[kind hawk]

square root of what

what does pythagoras say

[past rivet]

side^2+side^2=hyp^2

sorry we don't use it in english 😄

[kind hawk]

good and to which triangle can we apply that here?

[past rivet]

uhmm

i dont know tbh

[kind hawk]

do you know the unit circle?

[past rivet]

yea

[kind hawk]

so where in the unit circle could we find all these values? which triangle?

[past rivet]

cos^2(beta)+sin^2(beta)?

[kind hawk]

of which triangle are cos(beta) and sin(beta) the sidelengths

[past rivet]

1st quadrant?

and then you would have a right triangle there i guess

where the hypothenus has a radius of 1

[kind hawk]

where the hypothenus has length 1, yes

so what does pythagoras tell us about that triangle then?

[past rivet]

cos^2(beta)+sin^2(beta)=1^2

[kind hawk]

yes

and how does that help us?

[past rivet]

its explains why both sides must be equal

no wait

yeah just cos^2(beta)+sin^2(beta)=1^2

[kind hawk]

yes

so we are also done with this case now

[past rivet]

amazing

so if beta=0

I simplified it down to cos(alpha-0)=cos(alpha)*1

not completely sure what to do with the lefthand side

ohh no nvm I think I know

cos(alpha-0)=cos(alpha)cos(0)=cos(alpha)*1 correct?

[kind hawk]

yes

[past rivet]

alright so that are all the calculations which shows that all the different values still keeps the equation equal

[kind hawk]

yes

[past rivet]

thank you very much!! I really appreciate the time and effort you took to help me with this ❤️

[obtuse pebble]

@past rivet Has your question been resolved?

Channel closed

Closed by @past rivet

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[foggy jewel]

Hello

[foggy jewel]

Can you figure out the sequence/progression of the white circles?

[flint sundial]

that is why it is called identity,It holds true for all values

[past rivet]

ahh right that makes sense

[obtuse pebble]

@foggy jewel Has your question been resolved?

[obtuse pebble]

@foggy jewel Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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 turtle]

let a , b , c and d be natural numbers such that :1<a<b<c<d show that 1/a+1/b+1/c+1/d+1/abcd ≤31/24

[hybrid turtle]

Any help guys ??🙏

[haughty coyote]

You can actually maximize this quite easily

Just find the 4 numbers that give the highest value

[obtuse pebble]

@hybrid turtle Has your question been resolved?

[obtuse pebble]

@hybrid turtle Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

The amount of chemical in the i-th shell is not approximately k/(1+r_i^2) since the radius can vary does it not?

ahh it was a very shallow pool

.close

[obtuse pebble]

Channel closed

Closed by @sleek solstice

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

Can someone please help me?

I had to prove

-(a+b) =(-a)+(-b)

The solution given started with
(-1)(a+b) and then used left distributive property.

I don't understand how they turned -(a+b) into (-1) (a+b)
Can we do it simply? I'm confused because we only have to use field properties
Can I say that's multiplicative identity?

[restive ridge]

,, -n = -1 \cdot n

[warm shale]

illuminator3

[timid silo]

I don't think we can do it randomly
Because for another question
(-1)a =-a they asked for a proof

[timid silo]

But that is not a field property, is it?

[restive ridge]

no but if you already proved it in a previous exercise then you can just refer to that proof

[timid silo]

Ah okay makes sense. Thanks a lot!

.close

[obtuse pebble]

Channel closed

Closed by @verbal 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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[pale stag]

hey im looking for some help on a trig substituion problem, im working on the problem, ∫1/(x√ (x^2+25)

[pale stag]

so i started off by setting x = 5tan(θ)

so dx = 5sec^2(θ)dθ

and then i set √(x^2 + 25) = 25secθ

then i substitute in to get (1/25)∫(secθ/tanθ)dθ

and no matter where i go from here my solution is not getting accepted

can anyone show me if im doing something wrong in this setup?

[cedar lichen]

√(x² + 25) is not 25 sec θ

[obtuse pebble]

@pale stag Has your question been resolved?

[pale stag]

could you elaborate as to how?

[cedar lichen]

First, elaborate how it is

[pale stag]

because i set up (x^2+25) =√( (5tanθ)^2+25)

and then i get √(25tan^2θ +25)

and i pull out a 25

[cedar lichen]

√(ab) ≠ a √b

Otherwise we could take any square root easily by saying √x = x √1 = x

[pale stag]

oh i see

so that should leave 5secθ?

[cedar lichen]

Yes

[pale stag]

oh ok thanks for showing me that

so just after that i set it up using that and the dθ correct?

[cedar lichen]

Yep

[pale stag]

∫(5sec^2θ)/(5tanθ5secθ)dθ ?

is that set up right?

[cedar lichen]

Looks like it

[pale stag]

ok so then i get 1/5∫cscθdθ

giving me (-1/5)ln(abs(cscθ + cotθ)) + c

and then using x = 5tanθ

i set up a right triangle to find cscθ = √(x^2 +25)/x

and cotθ = 5/x

since it was beginning to get difficult to type these all out

however that answer is still not being accepted

i suppose i could simplify a bit further

but it still won't take that even

-ln(abs((x^2+25)^(1/2)/x))/5 +c

i just can't figure out what im getting wrong

[obtuse pebble]

@pale stag Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[grizzled bay]

Assume that you have a committee of 14 members, made up of 7 men and 7 women.

In how many ways can the 3 tasks be assigned so that both men and women are given assignments?

[true rain]

which 3 tasks

[grizzled bay]

its imaginary

[true rain]

wow!

I dont think I understand what this question is asking

men and women for each task? any task? all 14 must be assigned? any 14?

[drifting wraith]

agreed

[grizzled bay]

Ok

[drifting wraith]

oh it's like

3 people are chosen

except you can;t choose 3 men or 3 women

that's the solution too

[true rain]

3 people are chosen for what

[drifting wraith]

to each get one task

[true rain]

every task?

[blissful bane]

I think it's asking, how many ways we can arrange 3 from 14 remember

given, 2 must be men and women

[true rain]

Hmmm if every man and woman must be unique

I think you should consider removing the number of unique selections that choose only men or only women

or answer, choose 2 men and 1 woman, and choose 2 women and 1 man

Do you know the equation for that?

[blissful bane]

from 14 (7M+7W) people, select 3 ( 2 Men and 1 Women)

i guess so

is it 7C2 * 7 C1 ?

[true rain]

yeah that would be for 2m and 1w

[blissful bane]

so 7C2 * 7 C1 + 7C1 * 7 C2 ?

[true rain]

looks right to me

[blissful bane]

I guess so, not sure

[true rain]

I was also thinking 14c3 - 2* 7c3

if they're the same then I'm confident

[blissful bane]

I'll verify that

[blissful bane]

shouldn't it be only one way ?

that is just for 2m and 1 w?

[true rain]

should be either or

2men 1 woman or 2 women 1 man

[blissful bane]

right

[blissful bane]

=294

[blissful bane]

=294

noice, it's same for both of us

@true rain

[true rain]

thumbs up emoji

i need to leave now

[blissful bane]

have a good one

@grizzled bay

[grizzled bay]

What is it?

Oh

I tried 294 and I think it sisd it was wrong

Becuase I also looked it up online and that’s what it said

[blissful bane]

@grizzled bay i mean, that could possibly be the answer

[grizzled bay]

Let me try it again

[blissful bane]

we both did in different ways

[grizzled bay]

294 didn’t work

[blissful bane]

${n \choose r}$

[grizzled bay]

Was there another one?

[blissful bane]

screenshot the question

and post here

[blissful bane]

@grizzled bay

[grizzled bay]

Yes

[blissful bane]

I think, i understand what the mean

[grizzled bay]

14 people

3 tasks

7 of each gender

[blissful bane]

Total ways to arrange will be $14 * 13 * 12$

[grizzled bay]

What

[warm shale]

CapitalW

[blissful bane]

nice

see, here first any of that 14 can go into task 1

then 13 can be only chosen for task 2

then 12 of them can go to final task

so by multiplication theorem 14 * 13 * 12= 2184 ways

what we need to find is both men and women choosing for tasks

If only males can fill the tasks= 7* 6 *5=210

If only females can fill the tasks = 7* 6 *5 =210

[grizzled bay]

Still says incorrect

[blissful bane]

so both male and female = 2184-504-504=1176 Ways

try this

[grizzled bay]

Still no

[blissful bane]

i made a lil mistake

try 2184-210-210=1764 ways

@grizzled bay

[grizzled bay]

Correct

[blissful bane]

great

[grizzled bay]

Thanks for all your help

[blissful bane]

wait what?

don't you want to know?

[grizzled bay]

I was going to read up

I figured your thought process was above

If not, I’m all ears

I guess I shouldn’t have assumed

[blissful bane]

if you understood my thought process through text

it will be great and good to go

[grizzled bay]

So from what I see

The three tasks have to be delegated individually so to speak

Then the overall number of people goes down

To 13

And then 12

[blissful bane]

ie, if you select from the entire 14 without filtering

if I subtract arrangements of only female and only men from the total arrangements , I'll get both men and female

pretty intuitive

[obtuse pebble]

@grizzled bay Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sonic turtle]

im doing a practice sat questions for math and how would i solve this very quickly

[fierce lagoon]

I mean this kinda requires you knowing what that function looks like

If you think about it

@sonic turtle is x > x-1 (for all x > 0)?

Bruh

@sonic turtle yo you there?

[sonic turtle]

yo one sec

[fierce lagoon]

Ping me when you answer this

[obtuse pebble]

@sonic turtle Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[fierce lagoon]

L

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[obtuse pebble]

Please don't occupy multiple help channels.

[sleek ocean]

need help solving the meters part :)

[obtuse pebble]

@sleek ocean Has your question been resolved?

[obtuse pebble]

@sleek ocean Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

i dont understand anything desmos is confusing and this as well

[obtuse pebble]

@delicate karma Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

how will you solve this one

[timid silo]

y can’t you multiply and divide and then get the answer

[pine sail]

$\f13 \times 3 \times 3^5 = 3^n$

[warm shale]

What the hell am I doing here?

[pine sail]

Yeah?

[timid silo]

this one

[pine sail]

Simply multiply, yes that is what you have to do.

Or use Exponent laws to write each side as an exponent of 3.

[timid silo]

but there comes a rule in your mind that if same bases are being multiplied you add the powers

[pine sail]

Yeah then do that...

[timid silo]

but here that isn’t making sense

[pine sail]

It is.

It clearly is.

[pine sail]

What do you mean by that?

[timid silo]

can you make 3/3 * 3 ^ 5 = 3 ^ n

[timid silo]

ive learnt a rule in indices that when bases are same you add the powers and if division, you subtract it

[pine sail]

Yes.

You've learnt correctly.

[pine sail]

You can.

[timid silo]

this one

[pine sail]

You can do it so many ways, which are all the same, just a different way to look at it.
If you want to use the rule that states you can add exponents when the base is same. Then note 1/3 is just 3^(-1) and 3 is just 3^1 so you have 3^(1-1+5).
Another way is to first simplify 3/3 which is just 1 so you have 1 * 3^5. Whatever you do, you end up with a 3^5 in the LHS.

[timid silo]

if you just cancel three and three but there is a multiplication sign in the middle

[wild swallow]

3/3 = ?

theres no more multiplication

if you just cancel the 3s in 3/3

[timid silo]

so you get like 3^5 = 3^n

[wild swallow]

yes

[timid silo]

and now you cancel both bases

and so you get 5 = n

five should be the answer

could have done 9/3 but division was first so had to cancel

[timid silo]

ive done this without the rule

[pine sail]

And it's correct still.

5 is your answer.

[timid silo]

it says 9 is the answer

how will nine be

[pine sail]

I think this is where you share an image of the original question.

[timid silo]

right ill add the question

this one

[pine sail]

Basically this.

[timid silo]

yes this one

the answer comes as nine

[pine sail]

The answer isn't 9, but a 5. If you are convinced 9 is the correct answer, then you should share an image of the original question from the source you're doing this. That is a book, or whatever.

[timid silo]

this

my bad

it’s not 1/3, it’s 3/3

[pine sail]

It's not even 3/3, I don't even see a /

It's poorly written.

[timid silo]

it should have that division line

but because of some sort of misprint it has got removed

[warm shale]

What the hell am I doing here?

[timid silo]

it’s like 3^3/3

[pine sail]

I think, this is what they want you to solve.

The thing I've written.

[timid silo]

i think this one is making sense here

[pine sail]

Yes, try solving that one.

[timid silo]

right

3^3 + 1 + 5 = 3^n

[pine sail]

Use parentheses

[timid silo]

3^9 = 3^n

[pine sail]

Yes.

And that gives you the desired answer.

[timid silo]

the bases gets canceled and nine is the answer.

is indices the topic whole based on practice?

because everytime you come across some new questions

[pine sail]

Probably.

[timid silo]

right

thanks

for the help

.close the channel

[obtuse pebble]

Channel closed

Closed by @buoyant zephyr

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[mighty ether]

do i need to use squeeze theorem for this problem?

[mighty ether]

i think i might be overcomplicating it?

i feel like there is something here to trip me up... I was planning on just calculating the double integral

and taking it from there but idk if that would be enough

[wild swallow]

maximise the numerator

minimise the denominator

that puts an upper bound on the integral

[mighty ether]

oh okay, and thats it? because i don't have a lower bound

so i wouldn't need to worry about that right/

?

[stable rain]

hii

[mighty ether]

hi

.close

[obtuse pebble]

Channel closed

Closed by @mighty ether

Use .reopen if this was a mistake.

[mighty ether]

what am i doing wrong here? the answer looks different from the calculator?

[stable rain]

finally lol

why wouold u

close and ask a question

lol

[mighty ether]

i was like i'll just move on to a different q

cause i didn't think the helper had anything more to say?

sorry, is it against the rules?

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[warm plover]

I dont understand where to start

I tried different things but it doesnt work

[obtuse pebble]

Channel closed

Closed due to the original message being deleted

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[stable rain]

ive a question

[stable rain]

say weve f(x) = 1+O(x)
can we or how do we show
f(x) = (f(x))^-1

i know we can show it for f(x) = (f(x))^n, where n is N>0

but idk how to show it for negative

wait it was x->0

does it hold otherwise

doesnt seem like it? idk

[wild swallow]

wtf

[wild swallow]

very bad

[wild swallow]

very bad

[stable rain]

$$[1+O((z-n)^2)]^{-1}=1+O((z-n)^2)$$

[stable rain]

OH WAIT

[wild swallow]

binomial

[stable rain]

in this case its

x->0

oops

ok then what about binimial

bino

[wild swallow]

binomial theorem

[stable rain]

nvm but

first

the cond must be

x->0 right

[wild swallow]

yeah

[stable rain]

i thought

binomial was for

n being natural

[wild swallow]

might work if not can't remember

[stable rain]

0

[wild swallow]

extended binomial theorem

the one with the negative binomials

[stable rain]

like this? idk

[wild swallow]

pochhammer symbols

[stable rain]

confused

[wild swallow]

gamma functions

[stable rain]

i have

never heard of those

well i heard of gamma but nvr touched it

[wild swallow]

anyway i'll be back later

[stable rain]

:c this any relavant

ok got it

.close

[obtuse pebble]

Channel closed

Closed by @stable rain

Use .reopen if this was a mistake.

[stable rain]

ty

https://brilliant.org/wiki/negative-binomial-theorem/

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[timid silo]

Hello 👋

[timid silo]

I Neeed help with the following question

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[toxic crescent]

Hey can someone help me with this

[toxic crescent]

Here's my working out so far

[kind hawk]

do you want to solve for x?

[toxic crescent]

Yes

[kind hawk]

add 48 on both sides of the first equation. what do you get

[toxic crescent]

Wdym both sides

[kind hawk]

left-hand side and right-hand side

[toxic crescent]

Someone said to me there could have been a misprint on the paper

Because we were meant to factorise the trinomial but I don't seem to see how

[kind hawk]

why are you expected to multiply this out and then try to factor again

if instead of 48 the number was 49 then all the numbers involved would be a bit nicer

but it's still possible like this

[toxic crescent]

Yeah

I'll retry with 49, if I get stuck I'll come back

[trail musk]

Why would you expand it back out

Lol

Lets say you had the equation:
u^2 - 48 = 0

How would you solve?

[obtuse pebble]

@toxic crescent Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sage hawk]

please help me find x,y,z

[small thicket]

what r u stuck on

[stable rain]

i dont think u can find y or z

[proven zephyr]

you can

[stable rain]

oh nvm

[proven zephyr]

wait... i dont think you can

[sage hawk]

i've found x, only z and y to go

[proven zephyr]

wait.. can you?

[stable rain]

lol

idk man

[sage hawk]

i was trying to use substitution but it doesn't work

[stable rain]

idk gl

[sage hawk]

my teacher told me to look at the triangles

or interior angles

[proven zephyr]

or does the line divides the angle to 2 parts...

[stable rain]

er if u can prove it

idk is that a property of parallelogram

oh it is

wait nvm

idk

ignore me

[proven zephyr]

i dont think it is

[stable rain]

ye they dont

oops

[proven zephyr]

try asking your teacher for the answer

im curious too

[sage hawk]

if i solve this i get $5, used to be $10 but then i asked for the hint and it got reduced

[small thicket]

dont the diagonals bisect

[proven zephyr]

yeah, i forgot if there is a law for that

[snow pewter]

you can

[proven zephyr]

no i cant

[snow pewter]

x is easy, is 40°

[proven zephyr]

you can, but i cant

[snow pewter]

then you consider the sum of angles in a triangle

[sage hawk]

180 = z + y' + 40

[snow pewter]

hold o lemme try on paint

[sage hawk]

z + y' = 140

if we try substitution

[small thicket]

the two should be equal

z and y'

[sage hawk]

y' = 140 - z

z + (140 - z) = 140

[small thicket]

bc. diagonals bisect

[sage hawk]

the z's cancel out

[wild swallow]

y and z are not uniquely determined by the diagram

[sage hawk]

z = y because of alternate angles

it shouldn't be solvable

[snow pewter]

it should

there's only one paralelogram obtainable with x angle

therefore there has to be only one solution for z = y

[wild swallow]

this is incorrect

[snow pewter]

why is that?

[wild swallow]

there are infinitely parallelograms obtainable with x=40

[snow pewter]

sorry

i should mention

that they're all similar

infinitely many parallelograms

[wild swallow]

no

[snow pewter]

however all similar

[wild swallow]

that is still wrong

[snow pewter]

how?

i have this angle

[wild swallow]

observe

[snow pewter]

okay true

so this is not solvable

[sage hawk]

so there's an infinite amount of parallelograms

that are similar

[snow pewter]

yeah

[sage hawk]

given that x = 40

so the value of y and z are different

[snow pewter]

they're governed by the equation z+w = 140

i use w for the unmarked angle

[sage hawk]

in every similar infinite parallelogram

[snow pewter]

not similar

[sage hawk]

no similar okay

[wild swallow]

no y=z that we know

the value we do not

[snow pewter]

yeah

[sage hawk]

z + w = 140

z = 140 - w

substitute

140 - w + w = 140

yeah with substitution it could not be solvable

[obtuse pebble]

@sage hawk Has your question been resolved?

[stable rain]

yay

fun

[stable rain]

[obtuse pebble]

Channel closed

Closed due to timeout

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hoary widget]

Find out what a+b+c+3p is if a+b+c=p and the pic down below

[hoary widget]

The one it says 1/a...

[thick oracle]

well

[stable rain]

wao

[hoary widget]

Did u understand?

[thick oracle]

do you know what -1 exponent does to a fraction?

[hoary widget]

Idk

[thick oracle]

it flips it

[stable rain]

wao

[hoary widget]

Oh yea

But how do I get the answer

It's eating my brain

[thick oracle]

well

a+b+c=p

[hoary widget]

Yes

[thick oracle]

but you know a+b+c=2

[hoary widget]

Yes

[thick oracle]

so what is p?

[hoary widget]

2?

[flint sundial]

substitute them

[thick oracle]

yes

[flint sundial]

correct!

[hoary widget]

But how did you know that?

[flint sundial]

because he said

a+b+c=2

a+b+c=p

so i subt

[thick oracle]

from this

you flip the fractions

[hoary widget]

How

OH

[thick oracle]

1/a flipped is a

[hoary widget]

I understand

Let me do this and see what I get

The answer is 8?

[flint sundial]

correct

you get 2 + 2 times 3

which is 8

[hoary widget]

Yes

Thanks for the help

.close

[obtuse pebble]

Channel closed

Closed by @hoary widget

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[short warren]

there's this question :
The line 7x+2y= -20 intersects the curve x²+y² +4x+6y-40 at the points A and B. Find the length of AB.

[short warren]

I tried solving the question by making x in the first equation the subject

so x= (-20-2y)/7

then i substituted this equation into the second one

[devout sable]

Yes

[short warren]

then idk where to go from there

wiat

something like this

[last falcon]

thats right

[short warren]

i simplified it further

wait

[last falcon]

i would try to open the parentheses

since u got rid of x now

[last falcon]

correct

u have a clue for the next step?

[short warren]

then idk what to do

how do i get rid of the fraction?

[last falcon]

find common denominator

[short warren]

oh

then let me try it

[last falcon]

sure

[timid silo]

how is x²+y² +4x+6y-40 a curve

[short warren]

idk man, thats what my book says

[last falcon]

wouldnt that be a circle with a center?

[timid silo]

no

[last falcon]

oh

[timid silo]

that isnt equal anything i mean if thats equal 1 or 2x or anything else then that makes sense

[short warren]

i feel like i messed up here

[last falcon]

i dont see no problem

add them up now

wait

[timid silo]

.... its like saying $x^2+y^2$ is a curve but only something like that $x^2+y^2=1$ is a curve

[warm shale]

does calc even end(cat gang)

[timid silo]

or is this multivariate

[last falcon]

i get your point

im not sure about this tho

can u assume its =0?

[timid silo]

i think its meant to be something like $f(x,y)=x^2+y^2+4x+6y-40$ not sure tho

[warm shale]

does calc even end(cat gang)

[short warren]

i got 4y^2+367y-2120

also i used a graph thingy

for the "curve"

how is that a curve

[timid silo]

thats =0

in the graph

which u didnt wrote in the question

ig the graphing calculator or whatever just assumes it

[last falcon]

hiba check ur book do u have a =0 or is it just like u wrote it?

[short warren]

its =0

[last falcon]

Oh

[last falcon]

so u got this?

[short warren]

let me send a pic of the question form the book

[last falcon]

wait

how did u get 49y

[short warren]

where?

[short warren]

it was y^2

[last falcon]

yea

[short warren]

to find a common denominator, i multiplied it by 49

[last falcon]

y^2*49=?

[short warren]

you said i had to make all the terms have the same denominator

so i multiplied the top and bottom of y²/1

[last falcon]

i meant to find a least common multiplier of 49 and 7

which is 49

now in order for y^2/1 to get to a denominator of 49

u gotta also multiply y^2

[short warren]

why tho?

i dont understand

[last falcon]

okay wait a sec

[short warren]

y² is rlly just y²/1

so, to make it have 49 as the denominator, i have to multiply both the numerator and thw denominator with 49.

[last falcon]

yes

[short warren]

i cant just multiply the denominator and leave the numerator alone.... right?

[last falcon]

thats right

[short warren]

oh

[last falcon]

bc then it wont be equal to each other

y^2/1 is not equal to y^2/49 right

[short warren]

49y²/49 is equal to y²

[last falcon]

which is equal to y^2 and y^2/1

[short warren]

yes

[last falcon]

u just expanded the fraction to make it easier for u to work with

[short warren]

i just realised

[last falcon]

say u had 2/3 + 4/6

how would u solve this

[short warren]

its 49y²/49 and not 49y/49

[last falcon]

yea

[short warren]

i forgot to put the ²

Sorry i messed up 😓

[last falcon]

its okay

[short warren]

i was in a rush

[last falcon]

no problem

fix it up and wel solve it

[last falcon]

also

can u spot another mistake?

that 80-8y/7 should have been 80+8y/7

u still there?

[short warren]

uhhhhhhh

[grizzled stream]

yeah

[short warren]

OH RIGHT

[last falcon]

mhm

[grizzled stream]

it should be (-80-8y)/7

[last falcon]

yea but since he put a minus there

it became -80y+8y/7

[short warren]

SO THAT'S WHERE I WENT WRONG

[last falcon]

yea no worries just change it up

should be easy from there

[short warren]

i kept getting the right answer but in negative

[short warren]

yeah i can do the rest

[last falcon]

after u get that u should solve it with the quadratic formula

will be easier

oh and dont forget to find x after u found y

[short warren]

get the values of x and then use the formula to find the distance between the two points

right right

thanks for the help

[last falcon]

no problem :)

[short warren]

that was long sheesh😣

[last falcon]

phew at least we got it

[short warren]

rlly appreciate it man 🙏

[last falcon]

my pleasure

do ,.close if u dont have any more questions

[short warren]

.close

[obtuse pebble]

Channel closed

Closed by @short warren

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[hexed tide]

How do I do this question?

[hexed tide]

I don’t understand what I’m supposed to do

[grizzled stream]

you're supposed to simplify the expression

[royal shard]

$x^{a}\cdot x^{b}=x^{a+b}$

[grizzled stream]

try using the exponent laws

[warm shale]

~Martin

[royal shard]

$\frac{1}{x^{a}}=x^{-a}$

[warm shale]

~Martin

[hexed tide]

The answer I got for it was (10p^2)/p

I don’t understand how to continue

[fierce lagoon]

Simplify

Because p^2 = p • p

So you gotta recall that $\frac{p^a}{p^b} = p^{a-b}$

[warm shale]

Umbraleviathan

[hexed tide]

I don’t understand

Wait so would it be 9p^2?

Please help, I still don’t understand

<@&286206848099549185> I don’t know what the correct answer for this is. Can you explain how to do the last step please?

.close

[obtuse pebble]

Channel closed

Closed by @hexed tide

Use .reopen if this was a mistake.

[obtuse pebble]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[tall arrow]

find the exact value for sine 45
I'm new to trigonometry, and from what I know, in order to solve this problem you would have to draw an actual right triangle with an angle being 45 degrees. This is easy for 30-60-90 degree triangles, but I can't really find/draw sides of a triangle when angle is not equal to 30 or 60

[tall arrow]

How should I approach this problem?

[thick oracle]

sin 45 degrees?

[tall arrow]

yeah

[timid silo]

bisecting an angle of 90 degrees gives you 45

[tall arrow]

could you clarify?

I know what all the triangle's angles are, just not the sides

[timid silo]

there is a known method in geometry to draw a bisector of an angle. You apply it to the 90 degree angle in a right triangle

[timid silo]

you should draw a right triangle where you know its sides

try for example a right triangle with sides 3, 4, 5 (we know it's possible because 5^2 = 4^2 + 3^2)

then bisect the right angle

[thick oracle]

i've done it

so

we need a right and isosceles triangle

[tall arrow]

the term "bisect" is new to me

[silver fox]

You dont need to draw a correct triangle

[tall arrow]

so I can simply do the 30-60-90 one?

[silver fox]

What do all the angles add up to?

No

[tall arrow]

180

[silver fox]

U have to reason on the correct one

But u dont have to draw it perfect

Ok so whats the other angle gonna be

[thick oracle]

seed

can i explain?

[tall arrow]

It would be 45, too, since all right triangles have 90 degree angles, and sine (45) tells you that the second side is 45 degrees, making the third one 45 degrees too, since a right triangle has 180 degrees total
the thing that i can't understand is how you can use that information to find the sides of triangle

[thick oracle]

or do you want ninja to explain

[silver fox]

Well given that 2 angles are the same you will have symmetry

[tall arrow]

Both your explanations are valuable to me

[thick oracle]

ok ninja explain first

[silver fox]

So call one of the smaller sides x

U cant actually figure out what the sides are although u can figure out their ratios

If all you know are the angles

[tall arrow]

Okay, with only angles in our hand, how do we do so?
SOH-CAH-TOA won't be applicable in this situation

[silver fox]

Yes it will just apply pythagorean to the triangle to get all the sides in terms of x

Both of the smaller sides are the same length