help-19

[hearty nest]

.........

[reef belfry]

u type so slow

@high cave

[hearty nest]

shut up....

[high cave]

nah bro I quit

[reef belfry]

ok good

[cinder lava]

close the channel if doubts are resolved

[reef belfry]

paglu

[reef belfry]

accha bhai

[hearty nest]

i already did

[reef belfry]

wo kar chuki hai already

dumbass

[cinder lava]

real bhai

[high cave]

u sound like a villager from minecraft

[reef belfry]

u were the one who continued after closing

[odd edge]

Available help channel!

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[jovial sonnet]

Hello, I am quite lost here. I’m trying to figure out the upper and low quartiles but I can’t tell if I’m doing it right or wrong. Did I find the IQR correctly and Q1 & Q3? Also are the boundaries correct?

[odd edge]

@jovial sonnet Has your question been resolved?

[jovial sonnet]

<@&286206848099549185>

[fair olive]

u need help with what?

[jovial sonnet]

erm if I found the IQR and upper + lower quartile correctly

and the boundaries

[fair olive]

iqr = q3-q1

what is erm bro say it out in full if u need help

[jovial sonnet]

are you reading what I’m saying

I didn’t know what to type back to u so I just stared at your text for a good 3 minutes

if I found the IQR and upper + lower quartile correctly

if I found the IQR and upper + lower quartile correctly

if I found the IQR and upper + lower quartile correctly

if I found the IQR and upper + lower quartile correctly

can you see now

[tacit wasp]

Calm down!!!

!volunteer

[odd edge]

Helpers are just people volunteering their time to help you. Be polite and patient.

[jovial sonnet]

ok ur right I shouldn’t have said that sorry

nvm I figured it out

.close

[odd edge]

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[odd edge]

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

https://www.desmos.com/3d/eimtttshan

[frozen musk]

vector u <1,1,0> and vector v <1,1,sqrt(2)>

idk how to use right hand rule to find direction

whole concept confuses me

book shows this but I still don't get it

.close

[odd edge]

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
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[solid briar]

why is the range 1 to inf?

[ember oak]

No. x

[tacit wasp]

I think his y stood for "why"

[solid briar]

mb, meant why?

[solid briar]

yea ha

[gilded bear]

Because demonstrate (x^2-1) = 0 and x =1,-1 you want positive 1. Therefore range from 1 to inf. It's also indicated that's where limited hit.

[solid briar]

why wont it take -ve values less than -1 as well?

[odd edge]

@solid briar Has your question been resolved?

[orchid torrent]

No, because that would require you to "cross" $x=1$, which is not possible.

[clever fjord]

Civil Service Pigeon

[solid briar]

oh interesting, does that have something to do with"existence of a unique solution theorem ", are we taking the +ve side of the range instead of -ve bc it's y(2) = 1/3?

[orchid torrent]

The existence and uniqueness theorem does guarantee a unique solution on some open interval containing the initial point. This open interval is $(1, \infty)$. And we choose this interval because it contains $2$.

[clever fjord]

Civil Service Pigeon

[odd edge]

@solid briar Has your question been resolved?

[odd edge]

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[mystic saffron]

bro

[mystic saffron]

question

a dot product of 2 vectors produces a scalar

right? right

wait nevermind

[safe vapor]

yes

[edgy remnant]

I don't think that first vector is orthogonal to u

[safe vapor]

,dot <4, -3, 7>, <1.5, 5, 3>

,w dot <4, -3, 7>, <1.5, 5, 3>

[clever fjord]

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Requested by stitches3356

[safe vapor]

orthognal means 0 dot product

[mystic saffron]

i set y = 5 and z = 3 and solved for x

[safe vapor]

show your work

[mystic saffron]

btw i know i forgot a negative but even when i added it it was wrong

[edgy remnant]

That's probably because they don't consider (0, 0, 0) to be in the opposite direction to your first vector

[mystic saffron]

oh

should i scale v by -1

then

yeah i totally missed that part

i instead thought they wanted 2 orthogonal vectors

in general

[edgy remnant]

Yep, that should work 👍

[mystic saffron]

i have a general question and it might be silly, intuitively how does something * a negative give us a negative

in the real world

[edgy remnant]

Hmm... not sure I know how to answer that, but in a sense negative numbers are created to model negative "things" in the real world

[mystic saffron]

like 2 * 3 is 2 three times or 3 twice of whatever we quantified, but whats 2 * -3? 2 negative 3 times? whats that

what does -3 times of something look like

[edgy remnant]

You can't always find a good visualization of stuff in maths. You think of it as debt, or as reversing a vector I guess. But if you give 2 and 3 units, there might be easier ways to think of it

[mystic saffron]

but a negative doesnt exist irl right

whats a negative apple for example

[edgy remnant]

2 * -3 mph means you double the speed in the backwards direction

[rapid reef]

what's a positive apple?

[mystic saffron]

1 apple

more than 0 apples

anything more than 0 apples are positive apples

or not positive apples but apples that can be quantified as positive cause they represent what positive means

idk 😂

[rapid reef]

then a negative apple is an apple you are yet to obtain

[mystic saffron]

but we have negative quantifies but based off of what though

like if we are yet to obtain it then howcome we do negative operations

[edgy remnant]

That's a philosophical question, but some mathematicians are of the opinion that math is just a model of reality, so a negative number cannot exist by definition. You can only use it to model things irl

[mystic saffron]

i see

i think i might get it

[edgy remnant]

But personally I don't see why we need to ask whether math is real or invented; it works, that's what matters

[mystic saffron]

well because we are using applications right

like the a wall problem with vector forces as ropes or a spring mass system

[edgy remnant]

Yep 👍 there are also people who study math for math's sake, but surprisingly often it turns out that even obscure areas of math end up having some applications

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
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[loud rune]

I’m taking the derivative of a function and need assistance with recalling how to manipulate fractions in order to remove h from the denominator

[graceful viper]

was the original function f(x) = 5?

[loud rune]

i’ve tried a few manipulations with factoring out 1/h and cross-multiplying, but I know there’s some shortcut

[vivid wyvern]

Js simply the numerator ig

[loud rune]

the original function was f(t)=19+(20/(t+1))

i used (f(a+h)-f(a))/h to get here

[graceful viper]

can u show ur working? how did u get that limit?

[loud rune]

sorry forgot to include that t=3

[graceful viper]

okay combine the fractions in the numerator

[loud rune]

okay, so need gcd

which would be 16+4h

one moment

[vivid wyvern]

If by Gcd u mean the common denominator when u combine the numerator, then no,it should be 4+h itself

[loud rune]

how does that work tho?

sorry for sluggishness and messy handwriting — in the car 😭

[vivid wyvern]

$\frac{20}{4+h} - 5 =\frac{20-5(4+h)}{4+h}$

[clever fjord]

Victimizer

[loud rune]

ahhh okay

didn’t need to keep it over 4

[vivid wyvern]

Yea but urs is right too

Cancel out the h

[loud rune]

i thought you couldn’t cancel out a variable in a binomial tho

so i can just cancel and end up with 20/20/1

which is 1

[vivid wyvern]

I meant to cancel these two h out

[loud rune]

ah, okay and then direct sub it

so the key was GCD

[vivid wyvern]

Yes

[loud rune]

okay cool, thanks for your help

so, the rule is you can snag variables by themselves

if that had been ((4+h)/h)/h

it would have just resolved to 4+h

but if it was something like ((4+h)/x)/h i can’t cancel the h in the highest numerator, right?

[vivid wyvern]

Yes

[vivid wyvern]

U cant directly, yes

[loud rune]

okay, cool. thanks again for your help!

[vivid wyvern]

Nw

[odd edge]

@loud rune Has your question been resolved?

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[quartz shuttle]

how is the answer to this DNE instead of -9? when plugging in 8.9, 8.9999 etc i get that it's approaching -9? so why isn't the answer -9 but dne instead?

[atomic hornet]

u sure when you plug in 8.99999999 you get -9?

[quartz shuttle]

am i just really ass at using ti30 🥀

2nd pic is ans used for 1st pic

[mint mirage]

You forgot brackets in the denominator

,w 0.0001/(9-8.99999)

[clever fjord]

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[quartz shuttle]

ohhhhh that makes a lot of sense now

im too used to ti84 but im only allowed a ti30 for this class 🥀

thank you, idk how to close the channel

[short terrace]

!done

[odd edge]

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

[quartz shuttle]

ty 😭 🙏

.close

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[mint mirage]

Just remember to use brackets for the denomintor

[odd edge]

Available help channel!

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[raven sun]

hey guys so this is a random question but if anyone here knows sudoku i got a real hard one.. im trying to deduce which character is in each box with these hints. only ratty bear, moldy muffin, cerberus, peanut, and chestnut are left

[raven sun]

literally have a damn chart

[graceful viper]

locked in for popmarts

[raven sun]

100%

ive had friends help buy from that box to deduce the characters

i have the link if anyone else wants to help

if we have more ppl buy we can trade amongst each other and it will help us know more

but ive already spent over $80 today 😭 so has my friend

if any of u wanna buy and help deduce i'll send the link. yall gotta let us guess the boxes though so we can get more hint points

[odd edge]

@raven sun Has your question been resolved?

[odd edge]

@raven sun Has your question been resolved?

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

[bold shard]

Trying to solve c, what do I do?

[safe vapor]

What do you mean by "solve"?

[tiny topaz]

The only thing left to do is removing the square root signs

[bold shard]

hm

[tiny topaz]

4m^2 is square of 2m

so the square root of 4m^2 is 2m

[bold shard]

yeah i see that

mhm

you multiply 81 x 2?

2m

[tiny topaz]

yes

[bold shard]

that would get rid of the radical

[tiny topaz]

yes

[bold shard]

same for the right one?

i think the middle one you can't simplify more

actually you can

[tiny topaz]

the middle one can be written as 4.3.m^2

[bold shard]

?

[tiny topaz]

then it is 24m. radical 3

[bold shard]

4.3m^2?

[tiny topaz]

12 is 4 times 3

and 4 is the square of 2

[bold shard]

yeah

mhm

[tiny topaz]

so you multiply 12 with 2m

[bold shard]

its 72

not 12

btw

heh

[tiny topaz]

oh sorry

[bold shard]

al lgood

[tiny topaz]

the method is the same though

finding factors that can be simplified

[bold shard]

yeah

i think the left one

is simply

9 x 4

but im assuming you multiply 3 by 16

so 48m sqrt of 4

or is it sqrt of 2

?

[bold shard]

just this last one

@vernal yacht if u avail, it'd be great to do it now

if not i guess that's it for tdy

[bold shard]

^^ when you can!

[vernal yacht]

Nah, buying lunch

[bold shard]

okay!

[vernal yacht]

Will be available 40 minutes later

[bold shard]

hmm i think that will work

that would be 940 for me

[vernal yacht]

Alright !

[odd edge]

@bold shard Has your question been resolved?

[atomic hornet]

Alright, we're back into some bussiness

@bold shard u there?

[bold shard]

yeah

[atomic hornet]

What's the current progress?

[bold shard]

i think we finished 3 question out of the rest

other than that i got taught how to solve one question

i was stuck on

[atomic hornet]

Do you have the questions?

I mean the ones I assigned last time

[bold shard]

i can look for it

there

[atomic hornet]

We should move on from 4 right?

[bold shard]

yes

[atomic hornet]

Alright, have you given it a try yet?

[bold shard]

will do now

[atomic hornet]

Decent

[atomic hornet]

Alright

[bold shard]

is it wrong

[atomic hornet]

lemme check

[bold shard]

ok

[atomic hornet]

that's correct

[bold shard]

ain't no way

i conjugated it

i think]

[atomic hornet]

no cap

lol

[bold shard]

did i conjugate it

[atomic hornet]

You're correct

[bold shard]

ok

next oned

[atomic hornet]

it depends

[bold shard]

alright

[atomic hornet]

on the last question, no

You have to be creative

[bold shard]

ok

[atomic hornet]

Let's go, 5th question

[atomic hornet]

yep, that's it

[bold shard]

can the 2 on the denominator cancel both of those twos

on the num

i thought you could do one only

[atomic hornet]

how come?

lemme demonstrate

$\frac{x+y}{2} = \frac{x}{2} + \frac{y}{2}$

[clever fjord]

This is sad 😢

[atomic hornet]

$\frac{2x+2y}{2} = \frac{2x}{2} + \frac{2y}{2}$

[clever fjord]

This is sad 😢

[atomic hornet]

$= x + y$

[clever fjord]

This is sad 😢

[bold shard]

WAIT THATS WRONG

[atomic hornet]

unfortunately, this one is very tricky

[bold shard]

i think i messed up on the 4

[atomic hornet]

You cannot do like this

[bold shard]

ok

i see

explain

[atomic hornet]

No, you messed up the entire concept, you cannot autopilot that method

[bold shard]

i think the cube root is what messed me up

[atomic hornet]

Have you ever seen $x^3 + y^3$ formula?

[clever fjord]

This is sad 😢

[atomic hornet]

yeah...

[bold shard]

no

[atomic hornet]

oh, then this question would be too advanced for you

Do you wanna learn more?

or just...nah

[bold shard]

i think it would be on the quiz though

i assume

[atomic hornet]

u need this to solve that

[bold shard]

my math teacher puts like weird ahh stuff on the quiz without letting us know

[bold shard]

is it that hard?

[atomic hornet]

not really, it's just a formula

[bold shard]

hm

you can help me with these two math problems that my teacher said will be on the quiz

[atomic hornet]

$x^3 + y^3 = (x+y)(x^2 - xy +y^2)$

[clever fjord]

This is sad 😢

[bold shard]

so she told us to practice

[bold shard]

i mean long

yeah

[atomic hornet]

Sure, do you want to solve the last question in the end? It's a matter of choice

It's okay to not solve it

[bold shard]

we can do this one last

[atomic hornet]

Alright

[bold shard]

let me send those two questions, they are pretty tricky to me

teacher told me to remember the pie r^2

[atomic hornet]

Sure, I'll pin them to this channel just in case

[bold shard]

im pretty sure my teacher wrote a big amount on these

[bold shard]

a big amount of solving

like 10 equal signs

[atomic hornet]

oh, it's not that difficult

But I'd still like you to share what you've known so far with me

[bold shard]

okay

lets do 15

let me try to solve it

[atomic hornet]

yep

[bold shard]

actually i can't solve the first one

let me show my prog

[atomic hornet]

Alright

[atomic hornet]

What's Hc

[bold shard]

Ac

just shortened the circle

into Ac

[atomic hornet]

oh

[atomic hornet]

Can you find the radius of the circle with the given information?

[bold shard]

im not sure how to :/

[atomic hornet]

How do you find the area of a circle?

[bold shard]

wait

2 pie r

[atomic hornet]

nah, that's the wrong one

area formula

might consider look it up rq

[bold shard]

wasn't it pie r square

[atomic hornet]

yes

it's $\pi r^2$

[clever fjord]

This is sad 😢

[atomic hornet]

So you get the area, can you find the radius with it?

[bold shard]

3.47

i think

[atomic hornet]

nah, gimme the sqrt form

[bold shard]

now thats where im stuck

[atomic hornet]

wait, do you use calculator to get this?

[bold shard]

google

[atomic hornet]

no wonder

[atomic hornet]

😭

[atomic hornet]

Fine, I'll demonstrate how to do it

and we'll also have exercise later

[bold shard]

ok

!

[atomic hornet]

So we have $\pi r^2 = 38\pi$. First of all, we divide both sides by $\pi$ and we'll get $r^2 = 38$. Next, we take square root from both side and we get $r = \sqrt{38}$

[clever fjord]

This is sad 😢

[atomic hornet]

Is that clear?

[bold shard]

lemme see

what would taking away the square root look like

[atomic hornet]

$$r^2 = 38$$
$$\sqrt{r^2} = \sqrt{38}$$
$$r = \sqrt{38}$$

[clever fjord]

This is sad 😢

[bold shard]

okay

[atomic hornet]

oh wait, r can also be -sqrt 38 if its not a radius

Do you know why?

[bold shard]

hm

nope!

[atomic hornet]

Because $$-\sqrt{38} \times \left(-\sqrt{38}\right) = \sqrt{38} \times \sqrt{38} = 38$$

[clever fjord]

This is sad 😢

[bold shard]

m

[atomic hornet]

minus times minues = positive basically

[bold shard]

ooh

[atomic hornet]

yk minus times minues = positive right?

[bold shard]

mhm

[atomic hornet]

I don't want mhm 😭
It makes me think I'm a suck explainer 😭 😭

[bold shard]

no i understand

lol

[atomic hornet]

You should say: YES SKY, YOU'RE FANTASTIC

[atomic hornet]

Alright !!

[bold shard]

im too nonchalant to say that

[atomic hornet]

So basically r can have two possible values if it weren't for the fact that it's a radius

So far so good?

[bold shard]

yeah

[atomic hornet]

Cool, so now we all agree that the radius is sqrt38

[bold shard]

yeah

[atomic hornet]

Can you do part a from here?

[bold shard]

doing it

now

the diagonal part of the square

should mean the diamater of the circle right

[atomic hornet]

yes

there you go, that's part a

[bold shard]

opk

[atomic hornet]

Ya know how to solve b?

[bold shard]

i don't know the perimiter

[atomic hornet]

Have you heard of pythoa theorm?

$a^2 + b^2 = c^2$

[clever fjord]

This is sad 😢

[bold shard]

yes

are we slicing the square

[atomic hornet]

yep

[bold shard]

ok

[atomic hornet]

cut it in half

[bold shard]

ok

its a triangle now

[atomic hornet]

Yes

It's a right triangle

[bold shard]

nvm

[atomic hornet]

ya need help?

[bold shard]

yes

[atomic hornet]

yeah, just ping me if you need help next time. I dun bite

Can you draw me a right triangle?

[bold shard]

ok

done

[atomic hornet]

pic

I need the pic

[atomic hornet]

Alright, let me demonstrate

[bold shard]

ok

[atomic hornet]

So according to the therom we mentioned earlier, we know that $$x^2 + x^2 = k^2$$

[clever fjord]

This is sad 😢

[atomic hornet]

Agree?

[bold shard]

yes

[atomic hornet]

So we've known that $k = 2\sqrt{38}$. How would you find x?

[clever fjord]

This is sad 😢

[bold shard]

help i dont know

[atomic hornet]

$2x^2 = 4 \times 38$

[clever fjord]

This is sad 😢

[atomic hornet]

Is it solvable for you?

@bold shard sry, I got to go. Tell the next helper that you're working on the first question's part b if they ask.

[bold shard]

ok

byeeee!

[atomic hornet]

Question 15, part b

[atomic hornet]

Take care

[shell haven]

Hallo so how far are you on b

[grand halo]

is this a doubt still

[shell haven]

So we know the diameter is

$2\sqrt{38}$

[clever fjord]

denzio321

[shell haven]

Since the square is inscribed in the circle

The length of the squares diagonal is equal to the diameter of the circle

Since the one internal angle of a square is 90°

Then the triangles formed by splitting the square by it's diagonal are right triangles

With equal sides(since the sides of the square are equal

Denoting the length of one side of the square x

By the Pythagorean theorem

[bold shard]

yeah

[shell haven]

$x^2+x^2=(2\sqrt{38})^2$

From there solve for x

To get the the length of a side of the square

Then finding the perimeter is easy from there

[bold shard]

mhm

[shell haven]

Ye

[clever fjord]

denzio321

[bold shard]

can you do 2x^2 = 2sqrt38

then divide by 2

[shell haven]

I actually a typo

It's 2x^2=(2√38)^2

Which would be
2x^2=4•38

But sure yea divide by 2 on both sides

To isolate x^2

[bold shard]

that leaves with x^2

= sqrt38

im not sure what to do from there

[snow rune]

2x^2=(2sqrt(38))^2 so sqrt(2)x=2sqrt(38 so x = 2sqrt(19)

[bold shard]

could you make it visibe

visible

<@&286206848099549185>

[vivid wyvern]

Which step r u at

[bold shard]

doing B right now

[vivid wyvern]

Alr

[bold shard]

did all this

2x^2=(2sqrt(38))^2 so sqrt(2)x=2sqrt(38 so x = 2sqrt(19)

but i don't know this thing

i wished it was represented in an imaged

the brackets and everything makes it weird

[vivid wyvern]

Wait

$2x^2=(2\sqrt{(38)})^{2} so \sqrt{(2)}x=2\sqrt{(38)} so x = 2\sqrt{(19)}$

[clever fjord]

Victimizer

[vivid wyvern]

Lemme type step by step

$2x^2=({2\sqrt{38}})^{2}$

[clever fjord]

Victimizer

[vivid wyvern]

So putting sqrt over both sides

U get,

$\sqrt{2}x=2\sqrt{38}$

[clever fjord]

Victimizer

[bold shard]

ok

[vivid wyvern]

$x=\frac{2\sqrt{38}}{\sqrt{2}}$

[clever fjord]

Victimizer

[bold shard]

what if i just simply

divided 2 sqrt 38 and 2x^2 by 2

[vivid wyvern]

Yes u can do it

What do u get after doing it

[bold shard]

can we go that way

[vivid wyvern]

Oh wait

[bold shard]

that path is a bit easier for me to understand!

[vivid wyvern]

U r planning to dividde both sides by 2 here

[bold shard]

yes

[vivid wyvern]

So dont forget the square over 2sqrt38

[bold shard]

yes

[vivid wyvern]

Alr

$\frac{2x^2}{2}=\frac{4 \times 38}{2}$

[clever fjord]

Victimizer

[vivid wyvern]

Did u understand this

[bold shard]

hm

what happened to the radical

[vivid wyvern]

When u square the radical it goes away

[bold shard]

you only squared the radical

?

[vivid wyvern]

$({2\sqrt{38}})^2= 2^2 \times {\sqrt{38}}^2$

[clever fjord]

Victimizer

[bold shard]

What I was thinking of was

Having it into x^2 = sqrt 38

[vivid wyvern]

x² would give u only 38 ×2

That would be ur answer

[bold shard]

But it’s wrong?

[vivid wyvern]

They asked for area of square right?

x²=2×38

[bold shard]

Yeah

[bold shard]

Perimeter

[odd edge]

@bold shard Has your question been resolved?

[odd edge]

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[shell haven]

Can you draw the line 2x and -x+3

Like on their own

[rich vapor]

I don’t know how to

[shell haven]

So you don't know how to graph lines in general

[rich vapor]

I know a bit but I don’t see a b

[shell haven]

Ok so I guess we can start from there

[shell haven]

If you can't see a b

The b is 0

2x=2x+0

[rich vapor]

Does that mean I start from the origin and use 2/1

[shell haven]

Yes

Then -x+3?

Is that fine

[rich vapor]

I don’t know how to do -X

[shell haven]

-x is -1x

That means m = -1

[rich vapor]

How many times horizontal and vertical is that

[shell haven]

Remember m=rise/run

If rise/run=-1

If run=1

[halcyon needle]

May I help anyone

[shell haven]

Then rise/1=-1

So rise=-1

So when we move 1 unit right horizontally

We move down 1 unit vertically

[rich vapor]

Breh I point 3 and dunno what to do

[tacit wasp]

But haven't you been taught to draw 2 (or 3) points and connect them? @rich vapor

Because imo that's a bit easier than finding a and b each time (at least for the drawing part, but otherwise you indeed must know what a and b are)

[shell haven]

You can pick 2 integer x values to plot as points

And if the coefficients are integers too you should get a nice integer y coordinate

Giving you two integer points to draw the line through

[rich vapor]

Dang this is hard I give up

Nvm I think I’m right for the first one

Ima go sleep to recharge

.close

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[pale iron]

What are the formulas for:
(a+/-b)(c+/-d)

[low locust]

learn how to multiply this out

dont just remember some formula

[pale iron]

My teacher wants to see the formula

Because i know how to do it without it

[low locust]

then whats the problem if you know how to do it

just multiply it out and the result is the "formula"

[pale iron]

Uhm okay

Thank you!

.close

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[quasi shard]

why is desmos having trouble adding?

[signal yacht]

uh what

[quasi shard]

it is the same equation all I did was simplify the first two terms but now desmos says it is two different values

[royal dragon]

if you want ppl to understand your question, you'll have to provide a complete context

[shell haven]

Without context I'd say you should double check your simplification

[signal yacht]

doesnt look the same to me honestly

[shell haven]

Yea but maybe it's because we can't see the other terms or something

So we kinda need the context frfr

[stoic cloud]

in one there’s a fraction, in the other there’s a cosine

So yeah, context plx!

*plz

you know you can stretch the desmos equation side to be bigger right?

[quasi shard]

ok I am doing this question. I am trying to simplify the integral. I wanted to make sure my algebra was right so I created m and n as test values in the desmos and started simplifying. the step before the error has this. $\frac{2m-2n}{2m+2n(2m-2n)} -\frac{2m+2n}{2m-2n(2m+2n)} - \frac{(2m-2n)cos(2\pi m +2\pi n)}{2m+2n(2m-2n)} + \frac{(2m+2n)cos(2\pi m - 2\pi n)}{2m-2n(2m+2n)} $

[quasi shard]

didn't know that

[stoic cloud]

thank you!

[quasi shard]

ignore the last cell

actuallyt nvm I looked again thats this issue

[golden sonnet]

maybe we couldve broken the integral to somehow follow walli's theorem, or am i wrong?

ig i am wrong

[quasi shard]

idk what that is

[golden sonnet]

walli's was for smth else

[signal yacht]

this is insanity lol

[golden sonnet]

it is an indefinite integral for sin^mx cos^nx b/w 0 and pi/2

[quasi shard]

why?

[stoic cloud]

TBH it could just be value error, and it really is 0.

[low locust]

why are you even doing all of these steps

from int sin((m+n)x) dx its already clear that this is zero

[quasi shard]

sometimes I use desmos to make sure my algebra is all good because then it should instanly pop up

[quasi shard]

one sec let me see I guess I missed that then

[low locust]

I dont care about you using discord to verify the steps

I am talking about doing these steps in the first place is pointless

you were supposed to use part (a) of the question

[golden sonnet]

no he couldve multiplied and divided this by n and proceeded

atleast that would've been over yet

[quasi shard]

ah I see let n = m+n

[golden sonnet]

ah no

shit

[low locust]

no...

in programming speak, sure

[golden sonnet]

applying sinxcosy formulae?

[low locust]

that step was legit

I didnt say anything about that

[low locust]

and m+n is not zero because ... ?

[golden sonnet]

whatever

[quasi shard]

because of the n^2 doesnt equal m^2

[fringe cradle]

jus multiply and divide by 2

the term inside the integral becomes sin((m+n)x) + sin((m-n)x)

u can jus integrate now

[low locust]

good job for reading the conversation

[signal yacht]

well

lmfao

[odd edge]

@quasi shard Has your question been resolved?

[odd edge]

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[toxic rose]

.close

[odd edge]

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[frigid canopy]

Let $a_1>0; a_{n+1} = a_n + \frac{1}{a_n}$. Prove $(a_n)$ diverges

[clever fjord]

wai

[signal yacht]

huh

[frigid canopy]

As $a_1>0 , \frac{1}{a_1}>0$. So $a_1+ \frac{1}{a_1}>0$
\
Let $(a_n)>0 $, so $\frac{1}{a_n}+a_n=a_{n+1}>0$.
\
Thus the sequence is strictly positive.
\
$a_{n+1}-a_{n} = \frac{1}{a_n}>0$.
\
Thus the sequence is strictly increasing.
\
Let the sequence converge to $L$ via MCT
\
Then $a_n≤L \forall n \in \N$
\
Then $a_{n+1}≤L \implies a_n+ \frac{1}{a_n}≤L \implies a_n ≤L - \frac{1}{a_n}$.
\
This means $L$ is not the supremum, which is a contradiction. Thus $(a_n)$ doesn't converge

[clever fjord]

wai

[low locust]

thats not a contradiction to L being a supremum

[frigid canopy]

hmm

right

[frigid canopy]

Am I on the right track though

[low locust]

yes

[fervent hawk]

i was thinking about this too, but for the last part, i would use limit to make something like
L+1/L = L
but not sure if its valid

[low locust]

shhhh

[fervent hawk]

sorry

[frigid canopy]

Yea, that's what I thought now

$a_n≤L; \frac{1}{a_n}≤L \implies a_n+\frac{1}{a_n}≤ L+1/L$

[clever fjord]

wai

[frigid canopy]

right so $lim(a_n) \to L$ and $lim(\frac{1}{a_n}) \to 1/L$ , so by algebra of limits $lim(a_n + 1/a_n) = L+1/L$ which would require $L=L+1/L$ which is false

[clever fjord]

wai

[tacit elm]

I found something that might work. Have you studied series ?

[frigid canopy]

no.

[tacit elm]

Ok so only sequences

[frigid canopy]

yea

[tacit elm]

Too bad I relied on series 🙁

Is this good ?

[odd edge]

@frigid canopy Has your question been resolved?

[frigid canopy]

Hmm?

[tacit elm]

You don’t like it ?

[odd edge]

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[sturdy root]

In a raffle, there are 24 prizes out of 100 tickets in the pool. How many times would you have to draw from the pool at least to have a 97% probability of winning at least one prize?

[sturdy root]

so how do i do this

so i got 24/100 and 0,97%

[atomic hornet]

What's the win rate if you pick one time?

@sturdy root

[sturdy root]

24/100

[atomic hornet]

How about two times?

[sturdy root]

24/100

[amber veldt]

it can't be the same if you have two chances

[sturdy root]

but i think

we’re supposed to think

that it doesnt like

im not sure tho

okay so 23/100

[amber veldt]

not quite

first of all, youre only thinkiing about the second time

we want you to think about both times

[atomic hornet]

Look, there are two possible win conditions. The first one is you get the prize on your first try, the other is you grab it on the second

[sturdy root]

oh

uhh

23/99

or

[amber veldt]

I see numbers that are part of the correct answer!

that's good

if you have a chance of 24/100 the first draw and 23/99 the second draw

[sturdy root]

im so confused

[sturdy root]

yes

[amber veldt]

whats the probability of winning at least once when you try 2 times

I think it will be easier

to consider the probability of not winning

and then take 1 minus that as the probability

because there are 3 ways to win here : the first try, the second try, and both tries

and there's only one way to lose

so it's less math

make sense?

[sturdy root]

wait

why 3 tried

2 tries

[amber veldt]

you can win the first time, the second time, or both the first AND second time

that's 3 possibilities

[sturdy root]

but how do you know about the 97% like

[amber veldt]

find out how many draws until the probability of losing is 3%

[sturdy root]

yes so how do i calculate that

sorry

[amber veldt]

Im not saying you HAVE to do it this way

just that it's less math to consider

what's the probability of losing if you draw twice?

[sturdy root]

and win?

[amber veldt]

no

I recommended

[sturdy root]

76

%

[amber veldt]

instead of focusing on the 97% chance to win

you focus on the 3% chance to lose, do you understand why i am recommending that?

[sturdy root]

uhh

no man not rly

[amber veldt]

then just go bck to the chance to win then

[sturdy root]

oka

[amber veldt]

chance of winning when you draw once?

[sturdy root]

24/100

24%

[amber veldt]

chance of winning when you draw twice?

[sturdy root]

23/99

[amber veldt]

it cant be LESS LIKELY to win if you draw twice than it is to draw once

think about it

that's like saying buying two lottery tickets is less likely to win than buying one

[sturdy root]

yes but its 1 less

1 less ticket

[amber veldt]

23/99 ~= 23.23%

24/100 = 24%

[sturdy root]

ok so wtf

[amber veldt]

here's the part you're missing

you need to add them

wait

yes

but not straightforward how to do so

because you cant just add them until you get to more than 100% that makes no sense

[sturdy root]

yeah

so thats why i said i think its with puttin back

wait let me check the result

like my teacher put some in there

[amber veldt]

oh that would make it easier

[sturdy root]

the fuck is this

bro yk ill just

memorize it

and not understand it

[amber veldt]

she's doing the other way I was saying

no, hear me out

ill explain it again differently

[sturdy root]

okay

[amber veldt]

theres a complication in calculating how many ways there are to win

for example

if you draw twice

you have to consider:

the chance you lose on both

the chance you win on both

the chance you win on the first and not the second

the chance you win on the second and not the first

that's complicated, and we'd prefer not to do that

[sturdy root]

oh

yeah

[amber veldt]

it's not WRONG

but it's extra work

so what your teacher did, and what I talked about a bit at first

was instead consider the different ways to LOSE, because if you dont lose you win

if you draw 2 tickets

the only way to lose is if you lose on both

there are 3 ways to win and 1 way to lose, if you draw twice

[sturdy root]

ohhh

yeah

whats with the infinity n tho

[amber veldt]

she's not saying n is infinity

she's treating it as a variable and taking logs

[sturdy root]

oh

[amber veldt]

she's also assuming you put the ticket back each time before you draw again. even though that's not realistic, it makes the math easier

[sturdy root]

i think its bcs we didnt do that yet

[amber veldt]

probably

so the probability of losing if you draw once is?

1-0.76

right?

wait I mean

0.76

1-0.76 is the chance of winning

[sturdy root]

yes

[amber veldt]

yes

chacne of losing if you draw twice?

[sturdy root]

1-0.76

[amber veldt]

it cant be the same

it has to be LOWER

if you buy more tickes, the chance of losing goes down

it's (0.76)(0.76)

[sturdy root]

how

u just lost tho

it gotta be the same

[amber veldt]

because if you buy two tickets, you have more chances to win

than if you buy just one

[sturdy root]

yeah but thats obvious

its still 0.24 tho

[amber veldt]

you just said it has to be the same

[sturdy root]

to win

[amber veldt]

and that it's obvious it's lower

[sturdy root]

ok mb

but its still 24 to win and 76 to lose tho

[amber veldt]

that's if you play once

if you play twice, it's different

[sturdy root]

what is it then

[amber veldt]

(0.76)(0.76)

multiplied

[sturdy root]

oh

okay

so now

[amber veldt]

for three times it's

(0.76)(0.76)(0.76)

[sturdy root]

0.76 0.76 0.76

[amber veldt]

yup

or can say

$0.76^3$

[clever fjord]

gfauxpas

[sturdy root]

yeah

ohh

so u use n as unit

to get to 0.97

yeah ive been there

[amber veldt]

so your teacher wrote

[sturdy root]

but im confused on when the < flips

[amber veldt]

$1-0.76^n \ge 0.97$

[clever fjord]

gfauxpas

[amber veldt]

because she's writing it as, when does the probability of winning exceed 0.97

you could also write

$0.76^n < 0.03$

[clever fjord]

gfauxpas

[sturdy root]

yes but the symbol flips sometime

i remember there was some rule

[amber veldt]

if you multiply or divide both sides by a negative number

it flips

[sturdy root]

ohh

[amber veldt]

think about it

1 < 3

-1 > -3

[sturdy root]

oh

and it cant be like

unlogicsl

like -1<-3

when i have this then ik i did something wrong?

[amber veldt]

presumably

or someone did somehting wrong

[sturdy root]

okay so

yk what my problem is

i use all the exercises to understand the concepts but i never have any exercises left

and like i cant think on my own

but when you have the probability of 1/100 winning something

then it doesnt change when you pull a 2nd time

[amber veldt]

sure it does

[amber veldt]

it's 1-99^2/100^2

[sturdy root]

oh okay

[amber veldt]

khan academy can make exercises

for you to do

you just have to find the section matching what youre learning

[odd edge]

@sturdy root Has your question been resolved?

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[sturdy root]

wait bro

[sturdy root]

There are 6 black and 4 white balls in an urn. It will be pulled 3 times with placing back
pulled.
a) Explain whether this is a Laplace experiment

so ik it isnt a laplace experiment but

[odd edge]

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[tulip oyster]

Can anyone reccomend a good channel that helps you understand general topics in math a?

[odd edge]

Please don't occupy multiple help channels.

[signal yacht]

.close

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[final flume]

I need help with this question to get a specific answer . I have attached my working and the desired answer

[faint knot]

write ln(sqrt(t)) as ln(t^(1/2)) then simplify

[glass vault]

integral of lnx is xlnx - x

[faint knot]

this is not necessary, FTC can be used here as shown in the work

[final flume]

Still not getting that

[faint knot]

which part of it

[final flume]

The simplification like how are they getting that as answer where is the 4 coming from

[faint knot]

did you try this?

[final flume]

I did

[faint knot]

what does ln(t^(1/2)) simplify to?

[final flume]

1/2 ln t ?

[faint knot]

yes,

[final flume]

Oh wait

I got it

Thank you very much

@faint knot

[faint knot]

np

[final flume]

.close

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[outer wadi]

kinda stuck on what’s to do after
i’m proving order of a equal order of inverse of a

[outer wadi]

I want to show n less than equal to m and the converse

[merry finch]

suppose m < n

[odd edge]

@outer wadi Has your question been resolved?

[outer wadi]

is this ok

[merry finch]

yup

exacrtly

i was gonna give you more but i thought it probably was enough

[merry finch]

this i was gonna do with contradiction

[outer wadi]

I see thanks for the help!

illl leave this open if i have trouble with infinite case

.close

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[desert dome]

how do i find c with only a

[forest sky]

if an object started out at the terminal velocity, then it would stay at that terminal velocity forever

so the terminal velocity should be a constant solution to the equation of motion

[odd edge]

@desert dome Has your question been resolved?

[sour storm]

/bnnnnnnjggggggjjhujgvbcdxszdxzx M=]ik9l0t6yrq3ww1``233E45R6TY7UO8IP[=][\v v n m

[stoic cloud]

i can feel my braincells slipping away

[sour storm]

i can feel i have no braincells

[odd edge]

@desert dome Has your question been resolved?

[odd edge]

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[odd edge]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[feral mirage]

i came across a geometry question in my past year papers that i do not know how to solve

Uhh the translation is:
In the diagram, what is ∠a + ∠b +∠c + ∠d + ∠e + ∠f

lemme know if this image is not clear enough

[red perch]

Oh hold on is

Yea ik this

[feral mirage]

i have the answer

but i just need to know how to solve

[red perch]

Is the answer D 360 degrees

[atomic hornet]

你有先試過什麼東西嗎？

Assume that your mother tongue is also Mandarin

[feral mirage]

i speak simplified

[red perch]

额 这样你知道解题方向怎样做吗

哪里部分不会的

[atomic hornet]

好，那我開本帳來回，IOS有一鍵翻譯

[feral mirage]

yea

it is 360

[feral mirage]

isnt this traditional?

[red perch]

注意到 有两个 直角三角形吗 在中间

角C 和 F

[feral mirage]

isnt 直角三角形 right angled？

like 90 degrees?

[red perch]

对

[feral mirage]

i cant see it

[red perch]

If you're ok I can stick to en

[feral mirage]

en