help-10

[wheat sonnet]

i will try blender

thanks

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[jolly agate]

could someone help me with the process of finding the length of BC

I dont know how to find it with 2 angles

and ive searched youtube and theres nothing

[blazing maple]

Angles in a triangle sum to 180

[potent badger]

use this
https://en.wikipedia.org/wiki/Law_of_sines

[jolly agate]

I know that

[jolly agate]

this scares me

ill look at it

[dreamy solar]

its simple sine rule

[jolly agate]

yeah ive never seen this before

okay so can someone still walk me through the question

i dont even know what im looking for on wiki

[jolly agate]

bhidu ur the most fantastic person on the planet

this is what I got so far

[dreamy solar]

the angles are bad so you may have to use calc

[jolly agate]

is this correct

what im doing

[dreamy solar]

werent the angles 74 degrees and 64 degrees?

[dreamy solar]

though its correct

[jolly agate]

ooh

I was waiting for u to answer so I moved to the next question

but yeah ill go back to it

thank you

ugh I have to cross multiply

[dreamy solar]

just apply sine rule in prevs q

yeah the angles are bad you have to use calculator I suppose

to obtain their values

[jolly agate]

thanks dude

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[dreamy solar]

dm if you still dont get the answer

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[urban kiln]

but it’s wrong for some reason and I have no idea why, it looks all good to me

or I guess I could write the last thing as h/a^h+1= 0+

[polar fossil]

the last part is easier if you just take a^x - 1 / a^x + 1 and substitute x=0 into it

you get 0 pretty quickly

[urban kiln]

yeaa I think I jus did what I did instinctively

but for some reason 0 is the wrong answer

The answer is negative one-

wait

[urban kiln]

oh yeah ull get -1 too w that

but then again, all I did was multiply and divide by x which is approaching zero

[polar fossil]

you shouldn't get -1 there, a^0 = 1

[urban kiln]

how do I end up getting the wrong answer

[polar fossil]

so it's 0/2

[urban kiln]

nvm yea I shd be asleep rn but am not 😭

[polar fossil]

,w graph (2^√x - 2^√{1/x})/(2^√x + 2^√{1/x})

[warm shale]

Results provided by WolframAlpha

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

Requested by kitten.in.a.teacup

[urban kiln]

is their solution wrong?

[wooden cipher]

remember a^(bc) is not equal to a^b(a^c)

[polar fossil]

oh thanks

days since hayley was bamboozled by a student = 0

but yea $a^{\f1{\sqrt x}}(a^x) \neq a^{\sqrt x}$

[warm shale]

hayley 🥥 🌴

[urban kiln]

oh crap

yeah I just realised what I did too 😭

wow

anyways thanks a lot<3

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[crude lotus]

Can someone explain why we do it like this, when dividing: Say -3y = -x + 6

[crude lotus]

We divide by -3 on all

So it's y = 1/3x -2

Is it because there's an invisible 1 infront of the x?

[dark stirrup]

That's an acceptable way to think of it, yes

x=1x

[crude lotus]

Thanks. What if I had 6/x times x, do they cancel eachother out, so I'm left with 6?

[dark stirrup]

Yes

[crude lotus]

What is the other or more correct way to think of it?

[dark stirrup]

$\frac{6}{x}\times x=\left(6\times\frac{1}{x}\right)\times x=6\times\left(\frac{1}{x}\times x\right)=6\times 1=6$

[warm shale]

SWR

[crude lotus]

What's the rule for placing a 1 instead of the 6 in its place?

[dark stirrup]

There is no "more correct" way. Your way is perfectly valid, it just relies on the more fundamental properties of algebra, which is fine when starting out

[crude lotus]

It's that only when moving something infront of a fraction from the fraction?

[dark stirrup]

From definition of division: $\frac{a}{b}=a\times\frac{1}{b}$

[warm shale]

SWR

[dark stirrup]

Not really the "definition of division" in the elementary sense, but in the algebra sense.

[crude lotus]

Thank you for explaining it all. I really do appreciate it 🙂

[dark stirrup]

no prob. I know a lot of people get confused when first learning algebra

A lot of symbology to absorb

[crude lotus]

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[grand relic]

If the area bounded by f(x)= (x³/3)-x²+a and the straight line x=0,x=2, x-axis is minimum then find a

[rancid sky]

derive it to find the turning points

max at x=0, min at x=2

sub 2 into f(x)

to get a

[grand relic]

Turning point is x=1?

but i dk what that means

[rancid sky]

$$\frac{x^3}3-x^2+a$$

[warm shale]

Crissyboi_15

[rancid sky]

this is the equation right?

deriving it gives

$$x^2-2x$$
$$x(x-2)$$

[warm shale]

Crissyboi_15

[rancid sky]

turning points at x=0,2

[daring sorrel]

It is the minimum of the area below the curve, not the minimum of the function

[rancid sky]

ohh mb

[daring sorrel]

Create a function for the area between the bounds then take the derivative if that function

With respect to a

[grand relic]

Differentiate w.r.t a??

I mean i know the result

[daring sorrel]

Yeah, the area changes depending on your choice of a

So we need to differentiate with respect to a because that is the variable here

[grand relic]

If f(x) is a strictly monotonic function in [a,b] then the area bounded by y=f(x), x=a, x=b, y=f(c) c€[a,b] is minimum when c= (a+b)/2

But im having problem applying it here

[daring sorrel]

Hmm I don’t think you can apply this here. This theorem is about having control over one of the bounds, but in the original problem you cannot change the bounds, only the function

[grand relic]

He just directly wrote f(1)=0

Ik its minimum at x=1 but how did he figure out that its equal to 0

[daring sorrel]

I dont get what you mean by its minimum at x=1. Why are you picking an x value at all

[grand relic]

a+b/2

Bounds are 0 and 2

2/2=1

[daring sorrel]

Ok im going to ignore the x=0 and x=2 constraints temporarily

The problem you originally sent mentions area bounded between y=f(x) and y=0. These bounds do not change.

The theorem you sent above is talking about area bounded between y=f(x) and y=f(c). The y=f(c) depends on how you choose c, and the theorem tells you how to choose c, but this is not what we are doing in your original problem.

Actually maybe I see where your coming from let me think a bit more

[daring sorrel]

Yeah nvm I still don’t understand. Do you have a reason to believe you’re suppose to use this or was it just an idea that maybe you could use this.

[obtuse pebble]

@grand relic Has your question been resolved?

[grand relic]

It was the first qn our prof made us write after that theorem

[obtuse pebble]

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[loud iron]

not help on any math question, but how do i remember the content that i learnt today for a very long time?

[loud iron]

any methods?

[fathom flicker]

learn it again tomorrow

[loud iron]

LMAO didnt even think of that

thanks

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[solemn basin]

Practice questions

[loud iron]

ok

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[bitter terrace]

how do we know the two triangles are congruent?

[bitter terrace]

I can see how they have the same angles but how do we know they have the same size?

are they defining triangle xyz to have the same size as trangle pao?

[wraith hornet]

~ means similar, not congruent

[bitter terrace]

oh ok

oh ok that makes sense thanks

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[crude lotus]

How do I differentiate this beast

[plucky trail]

using the quotient rule might be the best way

you could also rewrite it and do the product rule

[plucky trail]

have you used the quotient or product rule before?

[crude lotus]

I don't know which one u refer to when u mean quotient role, but product rule is number 136, right?

[civic socket]

,tex .diff rules

[marsh geyser]

134

[warm shale]

Kokichi

[civic socket]

hope this rules helps

[crude lotus]

bruh where's this in my book xd

[civic socket]

maybe your book has some missing formulas

or rules

[crude lotus]

So quotient rule would look something like. (2x^2+3)´ * e^x - 2x^2+3 * (e^x)´ divided by (e^x)^2

[plucky trail]

yeah that looks right i think, except you would subtract instead of add

[crude lotus]

Why? There's an addition sign in my function?

oh

nvm nvm i saw

[plucky trail]

yeah like that

[crude lotus]

Are there other rules you could call so I could write it down in my book?

[plucky trail]

now that's a start, or, if you'd prefer, you could use the product rule if you rewrote the denominator in the original function as having a negative exponent

[civic socket]

those are self-study but most of them are from google

[plucky trail]

but both should work, i just know a lot of the time i find it easier to use the product rule

[civic socket]

i think those are all

[marsh geyser]

This is wrong

[plucky trail]

whoops

[crude lotus]

What part is wrong?

[plucky trail]

i also dont see which part is wrong but Samuel is usually right*

[marsh geyser]

Second part of numerator

Wasnt 2x^3?

[plucky trail]

missing parenthesis?

[marsh geyser]

Ah no, i invented that in my mind

[plucky trail]

no it should be 2x^2

[marsh geyser]

Lol

I thought the original was 2x^3

[plucky trail]

ah okay even someone who is often right can be wrong

yeah no problem

[marsh geyser]

Forget what i said

[crude lotus]

highlight usually 🍮

😄 d

[marsh geyser]

But yeah

Parenthesis

[crude lotus]

(2x^2+3)´ * e^x - (2x^2+3) * (e^x)´ divided by (e^x)^2,, right because of the subtraction sign, ty

[marsh geyser]

Take e^x as common factor in the numerator

And simplify

[plucky trail]

yeah that's a good idea do that shadoxxx

[crude lotus]

i saw that

[plucky trail]

https://tenor.com/view/big-smile-innocent-gif-11933775

[marsh geyser]

Lol

[crude lotus]

[plucky trail]

this is good for me because i haven't derived in months

[crude lotus]

its been 3 years for me

[plucky trail]

okay it's not a competition

[crude lotus]

but we've never learned this stuff

only simple ones like x^2 and such

[plucky trail]

damn that sucks

[crude lotus]

sorry for the rotated image

but like this?

[plucky trail]

yeah

that's correct

[crude lotus]

nice. thank you guys for helping

[plucky trail]

i tried to do it by rewriting it and using the product rule but i got it wrong, and np you were just missing a rule

now practice so you dont forget it like i have lmao

i will do the same

[crude lotus]

just work hard like a s

superman

[plucky trail]

https://tenor.com/view/predator-arnold-schwarzenegger-hand-shake-arms-gif-3468629

[crude lotus]

❤️

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[oblique oyster]

Can every even integer greater than 2 be written as the sum of two primes? help my professor sent me this

[plucky trail]

Two different primes or just any two primes even if it is the same one?

[vale pelican]

it doesn't say two distinct primes

[forest sinew]

cool professor

[high lily]

it's an unsolved problem

Goldbach's

[oblique oyster]

I dont know he said if i dont solve four problems he sent me im failing

[oblique oyster]

WAIT WHAT

[forest sinew]

lol

[plucky trail]

nah

[forest sinew]

you better solve it then

[vale pelican]

please share the whole list

[plucky trail]

b-rother

that's hilarious and horrible at the same time if what you're saying is true

[oblique oyster]

Can every even integer greater than 2 be written as the sum of two primes? Can every even integer greater than 2 be written as the sum of two primes? Does there always exist at least one prime between consecutive perfect squares? Are there infinitely many primes p such that p − 1 is a perfect square? In other words: Are there infinitely many primes of the form n2 + 1?

[plucky trail]

im, as per usual, out of my depth

[oblique oyster]

if he fails me

[forest sinew]

take the fail

[worn yoke]

well known unsolved problem https://en.wikipedia.org/wiki/Goldbach's_conjecture

[oblique oyster]

WHATTTT

[forest sinew]

also mersinne

twin

right

i recognize 2 of those others

[oblique oyster]

he said that test was part of our grade a big part he said but HOW BIG

did i fail the whole class

[worn yoke]

this is your professor's idea of a joke

[vale pelican]

you didn't take an exam?

[forest sinew]

its a funny joke

[vale pelican]

yeah it is good

[forest sinew]

u should tell him u solved one

[plucky trail]

XD

[vale pelican]

obviously we're missing context

[plucky trail]

lmao

[oblique oyster]

im going to go shout at him if hes in his office be back dm me for results

[forest sinew]

say u gave ur work to a bunch of randos online

[plucky trail]

"they worked out a solution"

[forest sinew]

bc they seemed really interested for some reason

[plucky trail]

"i swear they just gathered like 20 people in a chat room and we solved it"

[oblique oyster]

He said he was playing a joke

but i failed the test and have went from a b- to a d+

[vale pelican]

rip

[oblique oyster]

ill be studying now

[vale pelican]

enjoy

thanks for sharing

[obtuse pebble]

@oblique oyster Has your question been resolved?

[obtuse pebble]

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[broken river]

Hello

[broken river]

Can someone help me with trigonometry

[vale pelican]

what do you want to get out of this help?

[broken river]

Number 2 I don’t get what the square root do if they negative in and ()

[plucky trail]

when you say square root do you mean exponent?

are you wondering what a negative exponent means?

[broken river]

I think I got it right

[plucky trail]

I don't think that's quite right still but it is better than your first attempt

i will circle

[broken river]

Oh cause two negative make an positive

Right?

[plucky trail]

yeah yeah

negative times negative

or negative divided by negative

in this case we're multiplying them

[broken river]

I fixed it right?

[plucky trail]

yeah that looks correct i think but let me double check

yeah no that's right unless i am missing something obvious

[broken river]

Am confused on number 3 since I don’t know what would the i would do

[plucky trail]

you haven't gone over imaginary numbers?

sorry actually you might not need to have gone over imaginary numbers to simplify that

[broken river]

💀we just started school and I don’t remember if we did, I mean I know we delt with imaginary number but idk if we got that far with em

[plucky trail]

you can disregard i for this, sorry for confusing you

how would you begin or are you not sure?

also worth asking is if you're familiar with conjugates

$(a+b)(a-b) = a^2 - b^2$

[warm shale]

Neo

[broken river]

Yes

Sorry I don’t see how they are related

[obtuse pebble]

@broken river Has your question been resolved?

[broken river]

@stark ocean can you help (sorry idk who am supposed to ask I just saw @ helpers if it been 15 minutes without help)

@analog vault

[broken river]

Still need help anyone

[timid silo]

dang ive been there dw

[broken river]

Ight

Ngl got to go for an hour );

[stark ocean]

<@&286206848099549185>

[obtuse pebble]

@broken river Has your question been resolved?

[broken river]

@analog vault

Just need someone to assist

Me with 2

3

@ruby fulcrum

@obtuse pebble how do I get an helper to help

@warm shale how do I get an helper to help

💀 maybe it just time to give up

Oh they all busy I think

@ruby fulcrum

[broken river]

Yo

Close.

@ruby fulcrum

[obtuse pebble]

@broken river Has your question been resolved?

[near grove]

erm im not sure

but

i belive u need to multiply by the reciprical of the denominator

[oblique cedar]

multiply both the numerator and the denominator with (8 + 2i)

[near grove]

ya

and then simplify

so the answer should be uh

31+12i/34

[oblique cedar]

Sorry but you are incorrect. The answer would be 31/34 + 6/17i

[obtuse pebble]

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[spark horizon]

can anyone help me with 5b?

[high lily]

line of symmetry is directly related to the vertex form you found in part a)

that form also tells you the extreme point which allows you to determine the range

[obtuse pebble]

@spark horizon Has your question been resolved?

[obtuse pebble]

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[timid silo]

need urgent help ( i have 3 hours) to prepare for a test - my tutor is lame and didnt teach me

just help me for 3 questions and im good - voice chat would be nice

[high lily]

no vc here

just ask

and people will try to help to the best of their ability

[swift hull]

ok can you send the 3 questions?

[swift hull]

yeah people dont do voice chat here if someone did it would be in dm's and i dont think anyone would be comfortable with that tbh

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

YES sorry iu was eating

so bascially

im doing 2 units but in on eunit

its abt subsuituition grade 10

can someone teach me that

like

ik what to do

but some of them arent factors or multiples of each other

so it becomes a fraction

so wht i do

for example

2p + 3q = 5
8p + 6q = 0

tysm !!!

[high lily]

well here respective coefficients are multiple of each other

8 is a multiple of 2
6 is also a multiple of 3
so you have two options to start

[timid silo]

right so

i started with 8p + 6q

btw i had a test with these questions

and got

0/20

😭

[timid silo]

so i end up with

[high lily]

firstly do you want to do this using elimination or substitution

[timid silo]

im literally FORCED to use subsitution

😭

[high lily]

ok, just a little more algebra

can you show what you're doing

[timid silo]

oh i started with 2p + 3q =5*

[timid silo]

i can show u my test paper

but its rllyy messy

[high lily]

show it

[timid silo]

sorry took long!! i was recording rawrr 🦖

[high lily]

can you just take a pic of it

[high lily]

there doesn't seem to be anything wrong with your work

[timid silo]

if ur on pc i have it tilt

[high lily]

continue from:
$$\frac{40-24Q}{2} + 6Q = 0$$

[warm shale]

ℝαμΩℕωⅤ

[timid silo]

my tutlr said thats weo g

wrong

he said ur supposes to find the lcm of
8(5-3q)
/2

[high lily]

it'd be more efficient to simplify that way

but what you did isn't breaking any mathematical laws

[timid silo]

ok so we collect like terms and add 24q with 6 q

[high lily]

not so fast

[timid silo]

ok good bc idk lcm 😭

[high lily]

your terms don't have a common denominator so you can't combine them like that yet

you can start with simplifying that fraction

[timid silo]

oh

so we camt add 24q and 6q

then divide

okay then we divide 40, and -24q by 2

[swift hull]

By substituion do you mean like solving for p in terms of q then plugging that into one of the equations, getting the value of p, and solving for q using that?

[timid silo]

So is the the question 2q + 2q =5
8p + 6q= 0?

[timid silo]

yes

[high lily]

+3q in the first one

[timid silo]

Ah yes

So a simultaneous equation

Multiply 3q by 2 to get 6q

[high lily]

let them continue, decent amount of good work has already been done

[timid silo]

Okay

👍

[timid silo]

yes its this

[timid silo]

wait what

It’s a simultaneous equation so you would try to get the coefficient the same so you can add or subtract accordingly

They explained the solution already though right?

[swift hull]

well in that case its pretty simple i think
solve for p in the first equation to get
2p = 5 - 3q
so
p = 2.5 - 1.5q

plug that into the second equation and solve for q

[high lily]

let them continue, decent amount of good work has already been done

[swift hull]

then plug the q value into the equation p = 2.5 - 1.5q and that will give you the solution for both

[timid silo]

so we're restarting right

[high lily]

no

[timid silo]

do what they said first

[high lily]

continue from where you i told you to

m3nny is just repeating what you've already done

[swift hull]

ohh are you trying to approach it using elimination?
i think he said he HAS to solve it using substitution/its obligatory

[high lily]

rosy is trying to give the route of elimination

[timid silo]

yes

[timid silo]

rosy my opps ong

[timid silo]

Oh I thought it was a general approach question mb

[timid silo]

LMAO

[timid silo]

right so

my dad jsut wlaked past my smut but ok anyway so

[timid silo]

…

12

and

wtf

20

so like

[timid silo]

what.

right so then

wait so

12q right

OK WAIT

LEMMW REWRITE IT

[timid silo]

so we get 20 - 12q + 6q =0

so like WE ADD 12Q AND 6Q RIGHTTT

bc like bedmas

and bc like idk

right guys

hi guys

[high lily]

you add **-**12q with 6q

[timid silo]

and we get -6 ✌️😛

[high lily]

-6q

[timid silo]

AAAA

y3s so

yes so

-6q

and then

we

OOO

wait what

wait OH

wait no

ok so we get

20 on the

otherside

bc we want q

right

so like 20 - 20 -6q = -20

so like -6q = -20

OH

[high lily]

doesn't really matter which side things are on as long as they're separate

[timid silo]

OMG

[high lily]

you can swap them around later

[timid silo]

HOW ABT WE ADD 6Q

[timid silo]

oh ok

ok then whats next

[high lily]

having 20 = 6q is easier to work with since there aren't pesky - signs
but either way is fine

[timid silo]

yes so 20 = 6q

devide

divide

by

2

yes

rightt

[high lily]

you can be more efficient

[timid silo]

so 10 = 3q

[timid silo]

whats that

[high lily]

dividing by 2, you'll still have something multiplied to q

[timid silo]

oh

[high lily]

so you still need to do more stuff

[timid silo]

what do i do

[timid silo]

stop.

oh no

rlly

[high lily]

$\frac{6q}{\what} = q$

[warm shale]

ℝαμΩℕωⅤ

[high lily]

don't overthink this

[timid silo]

6Q DIVIDED BY 6

[high lily]

yes

[timid silo]

🦖

[high lily]

if you have a something multiplied to q, just divide by that directly and you'll get q alone

since you already divided by 2 to get

10 = 3q
which is fine

[timid silo]

but then we divide 20 by 6

[high lily]

and then apply what i mentioned

[timid silo]

so q = 3.3??

HI

HI GUYS

[high lily]

no

how are you getting 3.3

[timid silo]

bc 20/6

[high lily]

that isn't 3.3

[timid silo]

✌🏽 😜

[hidden compass]

This is not 3.3

[timid silo]

ok what is it gramps

BTW TYY HI

[hidden compass]

It's 10/3

[high lily]

leave answer as a simplified fraction ^

[timid silo]

oh

SO Q = 10/3

OK NOW WE GOT VALUE FOR Q SO WE REDO THE SAME EWUATION WITH THE VALUE OF Q

RIGHT GUSY

GUYS

@high lily hi sorry for pinging

[high lily]

keep showing your work

[timid silo]

so using the same equation we use the value of q
2p + 3(10/3) = 5

??????????

2 p + 30/3 = 5????

2p + 10 = 5????????????????

2p + 10 - 10 = 5 - 10????????

2p = 5 - 10???????

2p = 5 - 10
/2 /2???????????????????

p = 5 - 5??????

[rough stream]

Hii what is the question?

[timid silo]

HI

OMG HI

THE #5 ONE

[timid silo]

NBTW HII

AND TYY

[rough stream]

So you want to figure it out p and q right ?

[timid silo]

we are subsisutiting them yes

[timid silo]

so i did the 2p + 3q = 5 equation and the value for Q is 10/3

so now im using the same equation and inputing that value in

[rough stream]

How?

[timid silo]

does that make sense bc like i can elaborate yes

[rough stream]

Its Impossible for you to derive that Q is 10/3 just by the first equation

[timid silo]

i solved it

and got

[rough stream]

Ohhh ok

[timid silo]

yes

[rough stream]

You used the 2 equations and figure that Q is 10/3

Is that right ?

[timid silo]

no

im subsisutitng

so u solve one equation

(get a variable by itself)

and then

yes

[rough stream]

Wait a sec

Yeah ok Q is 10/3

So what is your question?

What is the value of P ?

[timid silo]

p

is

[timid silo]

this

[timid silo]

.

[timid silo]

so p = 0

☠️ .

WHY IS IT ZERO

it cant be zero

[rough stream]

P its not 0

2p + 3(10/3) = 5

So

2p + 10 = 5

[timid silo]

OH HELL NAWWW WHAT U DO

oh.

NVM

[rough stream]

What do you mean ?

[timid silo]

NVM.

CONTINUE

[rough stream]

2p = -5

p= -5/2

[timid silo]

what we do

with the

10

[rough stream]

2p + 10 = 5

2p + 10 - 10 = 5 - 10

Its an equation

[timid silo]

OH

right

i get it now

[rough stream]

So you would get 2p = -5

Now you divide each sides by 2

[timid silo]

why cant we dvide 2 by 2 and -10 by 2

??????????//

[rough stream]

2p/2 = -5/2

[timid silo]

OH BC WE'D GET ZERO

[rough stream]

p = -5/2

[timid silo]

okay

so

[rough stream]

Where ?

[timid silo]

now we know the value

for

p and q

[rough stream]

Yes

[timid silo]

whats next

[rough stream]

You figure it out

[timid silo]

2p = 5 -10

[timid silo]

no like

[rough stream]

You could

[timid silo]

do we reuse the same equation

[rough stream]

2p = 5-10

[timid silo]

???///

[rough stream]

2p/2 = (5-10)/2

[timid silo]

with the p and q

value

[rough stream]

p = 5/2 - 10/2

p = -5/2

The goal of this exercise is

To find

p and q

You did it

The exercice is done

[timid silo]

omg.

finally

holy

shomoly

ILY @rough stream @high lily TYSMM

.close

[obtuse pebble]

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[visual pulsar]

.close

[obtuse pebble]

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[austere wave]

Can anyone help me with the question 2?

I try to prove the blue want but I getting struck

the (2) i forgot to change to abs(a) = gcd(a,b)gcd(a,c)

<@&286206848099549185>

[solemn basin]

Gcd of bc is 1 so they are co prime so a must divide either b or c
U can take cases using this

[austere wave]

ahhh I see.

I frogot that I can use this fact

thank a lot

.close

[obtuse pebble]

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[solemn basin]

[obtuse pebble]

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[timid silo]

how do i find this? i cant find a good explanation online

[ripe flower]

dude

[timid silo]

sorry sorry

i was afk

i dont get it

wdym middle angles

yea

mhm

yes

yes

yes

wait wdym 180 is the angle of the middle

yes

yeah

like combined?

180 right

but

hold on

AOD

can you draw it for me

and MOB

i dont get how

its supposed to look

is this how its supposed to be or smn? lol

yeah the o right

so u just find the o's outside the triangle?

wait how is aob and doc 180, shouldnt it be 360? isnt there 180 in a triangle

😭

im so cooked bro

yk what ill jus leave this question

nvm

nvm

i understand now

my understanding was wrong

when it asks for AOD it asks for the O between A and D

i was thinking of it like

as if it was asking for all the angles for A and O and D combined

im sooo dumb

thanks bro

:)

.close

[obtuse pebble]

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[obtuse pebble]

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[tulip sail]

I dont really know what to do with this inf

[tulip sail]

PQ is root45 but what next?

i feel like this is a typo question

if it states that the width is 3cm wide then says tana = 3/4 then does that not mean 4 is the adjacent which goes against the q

[obtuse pebble]

@tulip sail Has your question been resolved?

[tulip sail]

.close

[obtuse pebble]

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[tulip sail]

,tex \arctan\left(\frac{1}{1+x}\right) + \arctan\left(\frac{1}{1-x}\right) = \arctan\left(\frac{6}{7}\right)

[warm shale]

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

[tulip sail]

,tex Let ( \alpha = \frac{x}{1+x} ) and ( \beta = \frac{x}{1-x} ).

[warm shale]

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

[tulip sail]

I did tan(arctan[alpha]) + tan(arctan[beta]) over tan(arctan(a))tan(arctan(beta))

then i cancelled out the tans so i get

alpha + beta / 1+ab = 6/7

[steady silo]

we need to solve for x right?

[tulip sail]

yea

i got this

[steady silo]

are you familiar with complex analysis?

[tulip sail]

like complex numbers

[steady silo]

yes

[median dome]

that doesn't seem necessary for this

do you know of the identity for $\arctan x+\arctan y$?

[warm shale]

kheerii

[tulip sail]

no i havnet done complex

[median dome]

i didn't ask you a question related to complex numbers..

[steady silo]

i think it could be faster because we're adding arctans which are just arguments of a complex number, so it's the same as (1+x+i)(1-x+i)=6+7i

[tulip sail]

no i dont know arctanx + arctany i only know up to compound angle formulas

[median dome]

what

that's not possible for real numbers x

[median dome]

okay well

it's not really necessary

you can just take the tangent on both sides

[tulip sail]

idk this is what i did

but ans is wrong

[median dome]

where did you get x/(1+x) from?

[tulip sail]

wait let me show working from beginnign

[ripe flower]

here its 1/(1+x)

[ripe flower]

here its x/(1+x)

[tulip sail]

oh mb

wrote the wrong number

srry

[median dome]

okay well that seems more legit

[median dome]

this simplification is wrong

[tulip sail]

oh wait

its minus on the bottom right?

[median dome]

it is

[tulip sail]

1- x/1+x ...

[median dome]

even if you take it as a plus the algebra is still wrong

[tulip sail]

this looks right ?

[ripe flower]

(maybe the algebra would be a bit easier when transforming the arctan to arccot first and then applying cot and using the cotangent sum formula)

[median dome]

yeah this is right

[median dome]

it's the same

algebra wise too

[tulip sail]

alr ill try do it again with - now

[ripe flower]

just a bit easier on the fractions

[tulip sail]

i think my algebra is wrong

nvm it works out

.close

[obtuse pebble]

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[obtuse pebble]

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[loud iron]

can anyone give me a hint

how should i represent the irational and rational number?

[marsh geyser]

This is very straightforward exercise

Not sure how to give a hint without spoiling out the solution

[proper kelp]

X is irrational y and z are rational , x+y=z, x=y-z

Which is rational

[nocturne bronze]

https://tenor.com/view/korean-penguij-annyeong-joeprocess-penguenito-gif-17326587

[nocturne bronze]

bro what

[timid silo]

u can prove using contradiction method

[nocturne bronze]

Bro contradicting the question itself

[royal swallow]

lol

[proper kelp]

No im saying

[timid silo]

you take a statement which is quite opposite to the thing u have to prove and let it be true
then u find a loophole and hence fidning a contradiction in your assumption

[proper kelp]

If you assume the sum is rational, u find x is also rational

Contradicting the statement of x being irrational

[royal swallow]

@loud iron

just do a thing

dont go for the question

just think, if you add a normal number, and then a root of a prime number

[loud iron]

so how should do i represent the irational and rational?

[royal swallow]

wouldnt that ROOT be still included there?

[loud iron]

as fractions?

[timid silo]

exactly
but the statements u gave were lets just say too much straightforward

[nocturne bronze]

rational number is just adding a distance to the number

[timid silo]

p/q thingy

[royal swallow]

just declare variables, or represent an example

[rocky glen]

idk man, if only the question mentioned that

[loud iron]

so a/b + c = d/e

[rocky glen]

im jus joking

[proper kelp]

Its a hint not a well articulated response

[covert pumice]

Guys somebody pm me the correct answer please(dont respond here i dont wanna disturb u guys)

[timid silo]

nah thats the thing youll have to think by your own

[loud iron]

alr i got this

thanks

[timid silo]

,rotate

[nocturne bronze]

!occupied

[obtuse pebble]

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

[loud iron]

.close

[rocky glen]

,rotate

[obtuse pebble]

Channel closed

Closed by @loud iron

Use .reopen if this was a mistake.

[nocturne bronze]

fetus deletus lmao

[covert pumice]

What happened 💀

[rocky glen]

wait

what is this

this isnt even the guy

[covert pumice]

Oh rotate okok

[rocky glen]

who asked the question

[rocky glen]

brudda get ur own channel and ask

[covert pumice]

ouh sorry okok im new

[rocky glen]

#❓how-to-get-help

[rocky glen]

bruh i thought this was the same guy who was asking a question my bad 😭

[nocturne bronze]

[royal swallow]

bro, this is a channel contradiction now, the guy's question's contradiction evolved into a channel contradiction, lol

[rocky glen]

life is contradiction

[royal swallow]

every one contradicts every one

[timid silo]

one of the factors of teh denominator of a raised to power 4 - 16 is 2
so you cna write it as (a - 2 )(some expression in a)
also -2 is also an fatcor so itd be better to wirte
(a-2)(a+2)(a quad eqn in a)

[rocky glen]

where's the guy who asked the question in the first place

bros gone

[royal swallow]

https://tenor.com/view/spider-man-no-way-home-spider-man-pointing-meme-spider-man-pointing-gif-25131254

[royal swallow]

he closed it and said he is done

we shouldnt talk here, bye

[rocky glen]

omg this is such a mess i thought we were still answering it

omd

[loud iron]

nah i got it now

thanks

[obtuse pebble]

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[silent nexus]

how to approach this question

[obtuse pebble]

@silent nexus Has your question been resolved?

[obtuse pebble]

@silent nexus Has your question been resolved?

[steady silo]

i think i might have got it

draw the diagonals of Q

and use the fact that for P the area (160)=base*height

@silent nexus

[hardy widget]

Another way to do it is to observe you have an angle bisector here and then do trig

[silent nexus]

ohhh

im getting m+n = 85

[steady silo]

yess

me too

so i think that's correct

this is how i did it btw

with A sub P and A sub Q i just mean area of p and q

btw dont forget to close the channel if you don't have any more questions

[silent nexus]

oh okii

[silent nexus]

thank youu so muchh!!!

.close

[obtuse pebble]

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[obtuse pebble]

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[broken mist]

At what value of n does using Stirling's approximation in equations involving factorials start to yield correct results when solving for n, where the resulting n is a rounded integer?

[obtuse pebble]

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[obtuse pebble]

@broken mist Has your question been resolved?

[obtuse pebble]

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[small shard]

I created some weird formulas that when integrated you get area and volume of circles and squares (spheres and cubes). What is even going on here, and why is it that when we increase the integration limit you get higher dimensions? Also I cant figure out how the integration limit relates to the volume of a sphere for f(x). https://www.desmos.com/calculator/cvozmrpdct

first picture is the circle one, second is the square one.

[worn yoke]

if you integrate over a multiple of 2pi, cos(x) integrates to 0, so you are just integrating 1/2 a^2 or 1/2pi a^2

[small shard]

How does that relate to the volume of a sphere though?

or like, how does 1/2 a^2 relate to a circle?

Im uneducated, so pardon me please ❤️

[worn yoke]

because if you integrate from 0 to 2pi, you get (2pi) * 1/2 a^2 = pi a^2

[small shard]

ah neat

so for the sphere its similar?

For even values of a, if b is like 2a/2 you get the volume

[worn yoke]

you can get the volume of a sphere of certain radii. e.g. for radius 3 sphere, the volume is 4/3 pi * 3^3 = 4pi * 3^2, which is 4x the volume of a circle of radius 3

[small shard]

dude, I put the formula for a sphere in desmos wrong xD

Thats neat though

I spent way too much time today with this.

[obtuse pebble]

@small shard Has your question been resolved?

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[vapid silo]

how do I solve this kind of inequality, when I'm working with two absolute expressions?

[grizzled shore]

You recognise it

[fossil crag]

I would split cases to get rid of absolute value

[grizzled shore]

Well you can do that too I guess

[grizzled shore]

But this form is something you can recognise

[fossil crag]

of course "you can visualize it"

I didn't even visualize it properly at first

[vapid silo]

I've gotten into this

But now I'm stuck

[grizzled shore]

Now check if x is in each of the combined regions what happens to your function

Also I think you wrote your inequality wrong

Some should be going the other way

[smoky vigil]

if x>2 what does this tell you about the sign of x-1 and x-2

[vapid silo]

Oh, I messed up the case part for the negative -(expression)

[grizzled shore]

Tbh I see it as a 1D ellipse, that’s what I was visualising when I see it

[marsh geyser]

Inspection is the easiest approach

[grizzled shore]

It looks a bit degenerate but I think it’s good to recognise it

[vapid silo]

I was using solely the module to solve, but for the simple ones (e.g. |x + 4| < 2)

the definition of module*

[obtuse pebble]

@vapid silo Has your question been resolved?

[vapid silo]

got some help from llm, but still don't know whether if It's correct or not:

[obtuse pebble]

@vapid silo Has your question been resolved?

[vapid silo]

hey, yea Im practising algebra rn

1st semester

doing pre-calculus

and being in uni doesn't mean you're good at math either :)

either way, managed to learn it

.close

[obtuse pebble]

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[indigo sierra]

Graph y = (1/3)*f(-2x+1) from y = f(x) for the function f(x) used in the image. Describe the steps they use, as in the notes in the image.

[obtuse pebble]

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[pseudo tulip]

Hi, Im a bit stuck on this question. Im not even sure where to start

[timid silo]

Believe in yourself leather dog

[pseudo tulip]

I thought the barycentre would be unchanged and remain at the origin but apparently its not right

[timid silo]

you are great

u can do this

okay buddy

[slim moss]

Most instructive advice ever 🔥

[pseudo tulip]

thanks for the encouragement but that doesnt really help

[timid silo]

the most important emal in the day is believing urself yk

[slim moss]

Yupp

[pseudo tulip]

right..

[obtuse pebble]

@pseudo tulip Has your question been resolved?

[obtuse pebble]

@pseudo tulip Has your question been resolved?

[tawdry lark]

Try using vectors

Wait no hes already using vectors

Mb

[obtuse pebble]

@pseudo tulip Has your question been resolved?

[void quartz]

try Newton's 2nd law of motion

not the ordinary one, ΣF=ma

try the rotational motion law

$\sum \tau = I , \ddot{\theta}$

[warm shale]

Emily

[obtuse pebble]

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[steady void]

Can someone explain the working of part b in this qn

[static furnace]

you should find the equation of the line that contains the point t=36

[steady void]

Wdym

[near kraken]

more specifically, the slope of that line

change in distance over change in time is speed

wait i thought it was a distance to time graph

[steady void]

It is

[near kraken]

Ok you know the acceleration is -1

It's not

v (m/s)

[steady void]

The deceleration is -1 from 30s to 50s

[near kraken]

yeah and the change in v over change in time is that acceleration

thats where (v-20)/(36-30) comes from

[steady void]

Wdym?

[near kraken]

the acceleration from 30 to 50 is constant

which means it's also constant from 30 to 36

[steady void]

Ok makes sense

Then what about the v-20?

[near kraken]

see where that comes from?

that's change in velocity

v at 36 - v at 30

v at 36, or v is what ur trying to find

[steady void]

So the v represents velocity?

[near kraken]

yeah

the graph is really messed up so it's hard to see where -1 comes from

[steady void]

Actually where is the v in the graph tho

[near kraken]

right above 20

[steady void]

So v is the max speed?

[near kraken]

v is used to label the y axis where it says "v(m/s)", and also to represent speed where it says "v" in the work under b

[steady void]

Oh I see

Thanks for helping

[near kraken]

no

np*