help-27

[atomic drum]

oh got it

so the left is x(x-1)

am i able to do it to the right side too?

[heady plinth]

Get to a common denominator bro

It’s essentially the same as normal fractions

[atomic drum]

ok so can i multiply the numerator by x for the fraction on the right

[heady plinth]

Yes

[atomic drum]

ok

[restive river]

Example 16

Explain me 🥹

[stark anvil]

Open a new help channel

[restive river]

Why?

[stark anvil]

This channel is already occupied

By someone else

[atomic drum]

is it -6?

[heady plinth]

That doesn’t seem right

What did you get once you got common denominators

[atomic drum]

x+2x-2=x^2

here's my work for it

sorry if u can't read it

[restive river]

Want to see the answer

[atomic drum]

sure

i also wanna know what i did wrong tho

[restive river]

Bro wait a second

That was a prove question

What the heck 😭

How did u get -6 in prove question

[heady plinth]

The equation looks right but I can’t read anything else lol

[atomic drum]

what's a prove question

[restive river]

Read question carefully

You have to prove LHS = RHS

-6 😭

[atomic drum]

how do i prove it 💀

[atomic drum]

there's no question just that equation

[restive river]

what's your academic year

[atomic drum]

junior ha

hs *

[restive river]

😭

[heady plinth]

What the fuck are you talking about bro

😭😭

[atomic drum]

we are doing rational equalities or something

[heady plinth]

You have to solve for a

X

I’ll do it rq

[atomic drum]

that's what i thought

[heady plinth]

Just to see the problem

[restive river]

there's no X in the question tho

[heady plinth]

Are you blind or what

This is the problem

You might be in the wrong channel

[atomic drum]

💀

[heady plinth]

@atomic drum make sense?

[restive river]

Bro

[atomic drum]

nope

[heady plinth]

Which part

[atomic drum]

the whole bottom part

where you were solving for x

[heady plinth]

Once you have three fractions with the same denominator you can multiply both sides by the denominator

And it cancels out with everything

Does that part make sense

[atomic drum]

yes

why did you make it equal to 0

aren't you supposed to keep x on one side

[heady plinth]

Oh that was just solving the resulting quadratic

Do you know how to solve quadratics

[atomic drum]

no

oh wait

factoring?

[heady plinth]

You need to learn that before you can do this

Yes

[atomic drum]

yeha i know how to factor

[heady plinth]

Or quadratic formula

[atomic drum]

i see what you did now

i just was confused

[heady plinth]

Do you get why x can’t be 1

[atomic drum]

because uhh

wait

you set the terms in the denominator equal to zero and what you get are the holes?

or is it a different term

[heady plinth]

Yeah basically

[atomic drum]

ok i got it

lemme do one more so i get it 100%

@heady plinth for this one would i square root both sides

or do i add 4

[heady plinth]

If you really want to square root both sides you can add 4 and then do it

But it’s slightly better practice to do like this

One sec

[atomic drum]

ok i got it

wait

why is it -20

why isn't it x^2+20=0

@heady plinth u there?

[heady plinth]

Because you subtract 16 from both sides

How would you even get to +20 lol

[atomic drum]

u right

ok so

i have one more problem

ok so i created a common denominator and stuff

so all i have to do is

subtract the whole fraction

from the right

to the left

[heady plinth]

Sure that would be really clean

But you can skip a step here and just cross multiply both sides from the start

Just make sure the denominators are never 0

[atomic drum]

ok let me try that

just to make sure, so i put a negative sign in between them and set it equal to zero

wait that seems weird tho

because i could combine like terms but idk if i'm supposed to

do i factor them

WAIT

is it

-4 and 5

wait

it's 11

-11

NO ITS 14

IM THE SMARTEST PERSON IN EXISTENCE

[devout snow]

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[devout snow]

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[thorn sage]

I came too late

[thorn sage]

Shouldn’t I use this somehow?

[spare crypt]

you can rewrite all the equations from j-i to L instead if you want

it just means the induction step lets you use j-i to prove the j-i+1 case

[thorn sage]

Well, if I don’t need to it doesn’t matter

Yeah makes sense

But so the one you’re talking about is the final one, right

[spare crypt]

yea I was working from there

[thorn sage]

Why can I turned it into two times the sum of two to the power of K from etc. etc.

[spare crypt]

I looked at these in terms of the summation bounds

turns out the second one will just be the first terms but backwards

like the terms are {2^1, 2^2,...2^(j-i)} for the first box and the second is {2^{j-i), ... 2^2, 2^1}

[thorn sage]

Okay I must admit I don’t fully see that

k-i+1 and j-k

How are they the same?

Or the same backwards

[spare crypt]

the start of k-i+1 is 1 and the end is j-i right?

and it's incrementing by 1 each time to get from start to end

[thorn sage]

Hm yea

[spare crypt]

and if you look at j-k it's just the reversed start and end, and going down 1 each time

[thorn sage]

Ye it becomes 2^1

In the end

[spare crypt]

right, so you can reorder those terms to get the same things in the first exponent

meaning it's just twice the first exponent

[thorn sage]

So two times that

[spare crypt]

right

and there's a bunch of -1's to deal with

[thorn sage]

But shouldn’t we sum over the -2 as well

[spare crypt]

yea -2 shows up j-i times

[thorn sage]

Yeah, there’s two-ones which becomes -2

j-1?

This is a one by the way

i to j-1

[spare crypt]

yea finding how many terms total that is is sort of annoying
I do {i,...,j-1} is the same number as {i+1,...j} same as {1,...j-i} so j-i terms total

like how the number of terms from 3 to 7 is 7-3+1

[thorn sage]

Aaaah

I see

So, j-i*2

-2

Not 2

[spare crypt]

right that's the constant for the whole thing

[thorn sage]

Yea

But what I don’t understand is that the first one is supposed to be going from I to K and the next one is supposed to be going from K plus1 to j

So why are they the same just backwards?

When they’re going for completely different ‘areas’

Anyway, I typed it

Even though it seems weird to me all of this😂

I feel like there should be an easy way to prove that this is true

[spare crypt]

t_i,k means a certain t at those 2 parameters, it's not going from some place to another

[thorn sage]

I see it as it needs to cover from I to j

So it deals with two sub problems instead

I to k and k+1 to j

And thereby the whole solution I to j is covered

But is it wrong to think that way?

[spare crypt]

yea that's the wrong picture, it's like a 2d grid and wherever k is that's the coordinate that's ranging

like t_i,k is going from t_i,i to t_i,j-1 when you plug in k

so the first coordinate is just fixed

idk here's a bad chart

[thorn sage]

Yeah, okay thanks♥️

[devout snow]

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[rain bluff]

how do i solve this?

[crisp trench]

This is biquadratic

Aka $$ax^4 + bx^2 + c$$

[woven radish]

casework

[rain bluff]

ooohk

[crisp trench]

Good thing about it is that it is very easy to solve as you can just use substitution

Substituting what will give you a quadratic?

Or i can just make it a little more obvious $a(x^2)^2 + b(x^2) + c$

[woven radish]

casework

[rain bluff]

wait so what are the steps

[crisp trench]

Substitute $t = x^2$

[woven radish]

casework

[crisp trench]

Solve for t

Then solve for x

[rain bluff]

ooooh

lemme try

[devout snow]

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[modest ermine]

I have just seen a proof that if A, B are sets, then A - (B - A) = A - B. This is since A - (B - A) = everything in A that is not in A-B, which is everything in A that is not in B, which is A - B. I follow every line of this proof. However I really am failing to understand why it is that when I draw venn diagrams, I see the sets as not equal.

Additionally, Aren't A and B-A disjoint? So A intersect (B - A) = empty set, so A - (B - A) = A? Where is the flaw in my logic? And why are these venn diagrams not equal? See screenshot attached.

[wicked turtle]

A - (B - A) is just A. The thing you're subtracting (B-A) doesn't contain any elements of A, so subtracting it from A has no effect

[modest ermine]

yeah so doesnt that mean that A - (B - A) = A, which is NOT A - B?

[wicked turtle]

correct, they are not the same

[modest ermine]

but

[wicked turtle]

the figure is correct

[modest ermine]

where is the flaw here then?

[wicked turtle]

the second line is not the same as the first

the second line could be removing things that are in A

[modest ermine]

ok well what about this:

$$A - (B - A) = { a \in A | a \not \in B - A } = { a \in A | a \not \in B , or, a \in A } = { a \in A | a \not \in B } = A - B$$

[woven radish]

bot

[modest ermine]

idk why its cut off but the last part is supposed to say A-B in my typesettin

[wicked turtle]

"(a not in B) or (a in A)" is not the same as "a not in B"

[modest ermine]

ohhh rihttt

ok

okay so my answer key was just wrong than lmao

i was being driven insane

thanks for the help

[wicked turtle]

yea sounds like it

sure

[modest ermine]

fuark dude

currently in honors topology and just got driven crazy by a intro to logic error

[wicked turtle]

haha, it happens

easy to get your brain twisted by stuff that is not in complements, too much negation

[modest ermine]

yeahhh

thx

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[blissful heron]

Hi, I need some help with this question

[blissful heron]

this is my attempt so far at it and it is incorrect

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@blissful heron Has your question been resolved?

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[vast spoke]

this was the proof that I followed but I don't think its right

[devout snow]

@vast spoke Has your question been resolved?

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@vast spoke Has your question been resolved?

[urban jungle]

This looks right. A good way to test it would be to try the same method of proof on sqrt(4), as if it works on sqrt(4), there's something wrong since sqrt(4) is rational. I would recommend specifying that 5|a^2 implies 5|a is a result of the fact that there are no integer square roots of 5. In fact that's the reason we can extend this proof to all non-integer square roots but not to integer square roots like sqrt(4).

[vast spoke]

hmmmm okay I am going to try and do that too ! as long as it looks right lol

[urban jungle]

Yeah, the proof is definitely sound though

[vast spoke]

I put a pic of the new proof tho

[urban jungle]

still looks good

[vast spoke]

its longer but more organized i think

WOOOOOOOOOOOOOOOOO

[urban jungle]

a good practice when a proof feels fishy is to try to use it to prove a similar but false statement

if you can prove something false with it there's obviously a problem

[vast spoke]

damn wtf

i never thought to do that lol

thank you XD

[urban jungle]

you're welcome

[vast spoke]

THANK YOU SIR

now i turn in thank

https://tenor.com/view/snow-pepe-pepe-snow-dancing-snow-pepe-pepe-dancing-in-snow-happy-snow-pepe-gif-19261612

have a good day !

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[vital cradle]

hey guys, can some one please explain this process to me, I'm really confused by this

[restive river]

do you know about u-substitution to solve integrals

[vital cradle]

yes i do

[restive river]

alright then you can solve this integral by substituting x=sin¢

[vital cradle]

im sorry, but like sin what?

[restive river]

anything sint, sinu sin(theta) whatever variable you want to use

[vital cradle]

oh,I see

Thank you!

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[restive river]

if a = (4i + 5j) and b = (2i + 5/2j), is a parallel to b

[severe nest]

they're parallel if they are multiples of each other

[restive river]

Bard AI is an idiot

thank you

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[restive river]

can anyone help me with 7 number?

[vapid reef]

do you understand the question?

[restive river]

yeah but I am facing difficulties making the formula

[vapid reef]

ok then consider H to be helens age and D to be fathers age
then you have H +D = 51
and H = D/2

now substitute the values
you'll get the result

[restive river]

THANKS

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[vagrant cedar]

why isn’t the first one right

[strange arch]

But is says 20 in the text

[vagrant cedar]

i know

so what’s wrong w it?

[strange arch]

I think they just wanted the unit after the value

20 meters/min

instead of 20

[vagrant cedar]

oh okay

but

u did

i

in the right bottom corner

i said slope as rate 0 = 20 meter/min

i think i did something wrong with the interpretation?

[strange arch]

ah thought that was the answer key, yeah then idk what they thought was wrong

[vagrant cedar]

who do i ask?

[strange arch]

No interpretation is fine

[vagrant cedar]

why is it marked wrong i’m so confused

[strange arch]

maybe ask your instructor but looks all correct

[vagrant cedar]

.close

[devout snow]

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[vagrant cedar]

he

i wanna reopen

can someone else check

just in case

[strange arch]

.reopen

[devout snow]

✅

[strange arch]

sure, but am certain

[smoky gyro]

nice handwriting

[vagrant cedar]

thank you

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@vagrant cedar Has your question been resolved?

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[forest stirrup]

can someone help me understand how they got multiply by 2 in this equation? (line 4)

[lavish nimbus]

Could you also post the question?

[forest stirrup]

i think i get it now, they have a +1, and they could express that as log2(2) = 1.

and logx + logy = log(xy), we just multiply everything together?

thanks,

they prob shouldve put it as (2)(x-9)(x+53) to make it more clear

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[limber quartz]

4 Four boys and four girls are arranged in a circle. In how many ways can this be done:
a if there are no restrictions,
b if the boys and the girls alternate,
c if the boys and girls are in distinct groups,
d if a particular boy and girl wish to sit next to one another,
e if two particular boys do not wish to sit next to one another,
f if one particular boy wants to sit between two particular girls?

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[visual prairie]

How to solve f(x)≤g(x)

They didn't give f(x) nor g(x) in that question

<@&286206848099549185>

[restive river]

I think you have to assume it's a parabola with roots -1 and 3

And that's a straight line with the intercepts as you can see

[visual prairie]

After using a ruler which is not something you should do to calculate this I got that
U->(1/2)

What is mean is that it's 1 and alpha is 2

Not used to English statements

I don't study in english

That's for g(x)

After some rules I got that g(x)=2x-3

But

f(x) is just not gettable

I mean yeah I could say that it's just a square fonction which has a degree of
1, alpha=-4 but that's not how it should be solved

[restive river]

x² -2x -3 ≤ 2x - 3

Solve this

[visual prairie]

x²≤4x

[restive river]

x € [0,4]

[visual prairie]

Hm

How did u get x²-2x-3

They gave us f(x) after this but it's not allowed to use it ig

We have to get it ourself or not use it

[restive river]

Roots of quadratic -1 and 3

Look in graph

[visual prairie]

Yeah I don't think I studied that tbh

Wdym -1 and 3

Like the cut in x->?

We didn't study that

[restive river]

Yes

[visual prairie]

Intresting

Ig I didn't do too bad on the exam but I was genuinely confused Abt this

How do u get x²-2x-3 from
-1 and 3

[restive river]

If roots are a and b

[visual prairie]

How I solved it is I said
The fonction was taken
U->(1,-4)
Then I took the rule and got
That -a=1
a=-1
b=-4
Since this is a square fonction
Which is f(x)=(x+a)+b
I got to
f(x)=(x-1)²-4

[restive river]

Quadratic polynomial is (x-a)(x-b)

You have to learn it first

Then

[visual prairie]

We didn't do it in school

[devout snow]

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[hard knot]

Quick Question

[hard knot]

I did this

dy/dx C = dy/dx 3x+4

X = ln(2) /4

to find Y, I can plug in either one right?

Ohh i guess not

Bc tangent to C

.close

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[urban harbor]

still just not sure how to go about finding these tangent equations for polar eq

[twilit comet]

yoooo soosh

hmmm

so basically -pi, 0

this is weird

thinks

[urban harbor]

@winter torrent (sooory to ping, but any chance you'd know how to do this one? by any method)

[winter torrent]

lol I was trying to do this one myself earlier to prove I knew how

[twilit comet]

i thought we weren't supposed to ping mods/helpers

huh

[winter torrent]

i couldn't figure it out

[twilit comet]

soosh im sorta new to this so i have one question

is the tangent line expressed in terms of theta by any chance

like rotation per dx

or smth

[urban harbor]

i don't know thats the thing, my book is kinda awful for this chapter, there is literally not a single example of where they find a tangent line of a polar graph in the polar graph chapter haha

[candid lance]

r = theta

is that curve

[twilit comet]

hmmm

[candid lance]

like what sorta equation is that

[urban harbor]

ye

[twilit comet]

this, snow

[urban harbor]

archimedean spiral i think, just spirals out

[twilit comet]

yes

soosh btw what course is this

[candid lance]

well Im outta here I never heard abt it

[twilit comet]

cya

[urban harbor]

im not in school, just reading

[twilit comet]

oh, graduated already?

xD

okay

hmmmm

[urban harbor]

going through an old calculus book and brushing up on stuff that i don't feel solid on

[twilit comet]

i see

hang on, imma search up on what tangents in polars are even

[urban harbor]

i think ill just have to look at polar stuff in a diff book though

[twilit comet]

ok apparently

the trick is to think of r as f(theta)

so r = theta

dr/dtheta = 1
so at pi, pi, it's pi

[woven radish]

ren

[urban harbor]

at pi, pi, what is pi?

[twilit comet]

dr/dtheta

[urban harbor]

and yeah dr/dtheta is 1, but i don't even know what that is geometrically in terms of the tangent line, it definitely isn't slope...
i think a big part of the problem is i just dont really have familiarity with equations of lines in polar coordinates

[twilit comet]

this

[twilit comet]

the radius changes pi times as fast as the angle at theta = pi

[urban harbor]

wut? dr\dtheta = 1

constant

but how does that help with coming up with an equation for the tangent line anyway?

[twilit comet]

bc we set f(theta) = r

but here f(theta) = theta

think of it as another coordinate system

[urban harbor]

you're just confusing me

no idea what youre saying 😐

[twilit comet]

hmmm

okay

basically, dr/dtheta measures how fast r changes in relation to theta at a certain point/angle, right??

[smoky willow]

You're essentially representing a polar curve in polar coordinates to a parametric curve in Cartesian coordinates

[urban harbor]

ofc but how does that relate to the slope of the tangent

[twilit comet]

it isn't a SLOPE

it's how fast the radius changes in relation to the angle

[smoky willow]

With the standard conversion (x,y) = (r cos(theta), r sin(theta)), but now considering r as a function of theta, you obtain a parametric curve with parameter theta in the Cartesian plane, and here slopes/tangents work like you're used to

[twilit comet]

from that, we can find the slope i believe

(which, again, is pi)

[smoky willow]

The slope/tangent/rate of change, whatever you like to call it, of a parametric curve (x(t),y(t)) can be calculated as dy/dx = (dy/dt) / (dx/dt), so you just apply that here

[urban harbor]

so we need everything in terms of just theta?

since r = theta, can we write
x = theta cos(theta), y = theta sin(theta) ?

[smoky willow]

Yes

[urban harbor]

just gonna say t for theta
so dy/dt = t cos(t) + sin(t)
dx/dt = -t sin(t) + cos(t)
and now find dy/dx with that?

[twilit comet]

what the hell--

@urban harbor write that in r and theta form in latex bro i can't understand

[smoky willow]

Yes, then just this

[urban harbor]

so i think dy/dx evaluated at t = pi is dy/dx = pi, but once that's obtained, how do you find the equation of the line? just cartesian representation and then convert to polar with x = r cos(t) etc.?

or is there a more clever way

since problem asked for the line eq. in polar representation

[smoky willow]

That should work, yes

[urban harbor]

hm i seem to just be of by a negative from the answer

oh nm, my mistake, ok yeah that worked, thanks @smoky willow

.close

[devout snow]

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[urban harbor]

i guess you were right about slope being pi, but yeah i didn't understand where you were coming up with that

[devout snow]

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[cerulean helm]

yo what does this mean

[covert root]

mean do you not know what e and ln are?

[twilit comet]

don't judge ok

[covert root]

I didn't not mean come off aggressive

[cerulean helm]

nevermind..

[twilit comet]

basically if you have sm number "a"

and you want to know what number "e" must be raised to in order to get "a"

that number is ln a

[cerulean helm]

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[twilit comet]

well, that was quick'

[blissful herald]

lmfaoo

[cerulean helm]

yeah someone told me it

Lol

i just got stuck on it

[devout snow]

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[still crown]

anyone pls help me with this

is it d?

[versed hawk]

Think

It is a,ar,ar^2,ar^3 and so on

l = ar^k

ar^(k - 1)

[still crown]

yes

[versed hawk]

ar^(k - 2)

and then ?

[still crown]

so the answer is d right?

[blissful herald]

what do you think

[still crown]

i think d

is the correct answer

[blissful herald]

yes

it is

[still crown]

can you explain ?

[blissful herald]

so

take any gp

lets say 2,4,8,16

and reverse it

[blissful herald]

what is the common ratio of this?

[still crown]

its 16,8,4,2

1/2

is the common ratio

[blissful herald]

yes

yes

so nth term of a gp

is first term times the (n -1)nth power of the common difference

and here

l is the last term of the original gp

which becomes the first term

when you reverse it

[still crown]

ok bro

[blissful herald]

got it?

[blissful herald]

the common ratio becomes 1/r

[still crown]

ok bro

understood

[blissful herald]

alr

[still crown]

I thank thee for thine's assistance.

[blissful herald]

Be thou welcome

@still crown close the channel

.close

[still crown]

i thanketh thee, I appreciate your warm welcome

.close

[devout snow]

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[devout snow]

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[sturdy bobcat]

what kinda bridge would be strongest structurally

[strange arch]

are you sure that's not purely physics related

[sturdy bobcat]

[strange arch]

🐣

[dusk cedar]

lattice

[sturdy bobcat]

really, that would be a big pain to do

[dusk cedar]

r u making these bridges?

[strange arch]

do physicists not yearn pain?

[sturdy bobcat]

yeah

[sturdy bobcat]

yeah but from ourselves

[devout snow]

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[devout snow]

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[terse canyon]

I have some math problems I need assistance with?

[blissful herald]

what is it that you need assistance with

[solar goblet]

!da2a

[devout snow]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[terse canyon]

I need help with 3.

[terse canyon]

Number 3

[blissful herald]

the question is not visible properly

[terse canyon]

Here I’ll zoom it in

Is that better

[blissful herald]

yes

[terse canyon]

Ok

[devout snow]

@terse canyon Has your question been resolved?

[devout snow]

@terse canyon Has your question been resolved?

[devout snow]

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[grand meadow]

if P(A)=3 and P(A|B)=0.1 how do i solve this?

[devout snow]

Please don't occupy multiple help channels.

[dense lynx]

@grand meadow u need to close one of ur help channels

[grand meadow]

hey how do i close this one?

[dense lynx]

oh

.close

[devout snow]

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[devout snow]

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[restive river]

What's wrong here

<@&286206848099549185>

[drowsy peak]

The integral of -1/x is -ln(x)
You're missing the minus sign on the third line maybe

[devout snow]

@restive river Has your question been resolved?

[restive river]

Oh yeah

Thank you

.close

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[restive river]

This image is from my teachers answer key. I got -45 as an answer, but she got 216. I believe it's because she did 19+29 instead of 19-29, but I might be wrong.

[devout snow]

@restive river Has your question been resolved?

[stark anvil]

-45 looks correct.

[stark anvil]

The common difference is negative, so the nth term will also be in this case.

[restive river]

wouldn't that only apply if n is odd?

[stark anvil]

Can you give me an example?

[restive river]

I can't think of one on the pot, but if the number of terms is odd, and the common difference is odd, wouldn't the answer be odd?

[stark anvil]

What do you mean by answer?

[restive river]

the sum

[kindred whale]

Is it correct to say Sn = 25n - 6(n(n+1)/2)

Which gives S9 = -45

[stark anvil]

It is

[stark anvil]

1,4,7 but 1 + 4 + 7 = 12

Where did you get that

[restive river]

Oh, i'm stupid. For some reason I was thinking that all odd numbers are negative

[kindred whale]

Bold assumption

Lol

[restive river]

It's wednesday brain

I'll go ahead and close

.close

[devout snow]

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[placid narwhal]

Are those 2 midpoints?

So only the image?

Well we need more info

Knowing 2 sides isnt enough

Especially since we dont know how those lines are defined

[devout snow]

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[devout snow]

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[fluid jewel]

How do I solve this?

[dense jay]

well, what is a function

[fluid jewel]

no x repeats

[dense jay]

that all?

[fluid jewel]

Im not sure

[dense jay]

'each input has exactly one output'

[fluid jewel]

yes

I forgot that

[dense jay]

so, what if an x doesnt have an output

[fluid jewel]

then it is not a function

[dense jay]

correct

[dense jay]

so, what are your thoughts

[fluid jewel]

well I am confused on where it says y as a function

of x

[dense jay]

what confuses you about it?

[fluid jewel]

I havent done functions with equations

Usually just on a table or graph

[dense jay]

ah, it just means the value of y is dependent on x

like y=f(x)

f is the function

[fluid jewel]

ah alright

so is there something I need to do to find y as a function of x

or just by looking at it I should know

nvm I got it

.close

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[copper spoke]

Hi

[copper spoke]

guys

why does out of a = sup {a_n | n in N} follow that there must be an N in N such that a_N > a-epsilon?

[restive river]

the supremum is the smallest upper bound

if there were no such a_N then there would be an upper bound smaller than a

[copper spoke]

wait

so a = sup

so is a the lowest upper bound right?

so a - epsilon

could you explain in more detail

somehow dont get it

[restive river]

say there are no a_N > a - epsilon

then a - epsilon is an upper bound of the sequence

and a - epsilon < a

so a cannot be the smallest upper bound

[devout snow]

@copper spoke Has your question been resolved?

[devout snow]

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[cursive magnet]

hello

[cursive magnet]

i need help with geometry problems

can someone help

<@&286206848099549185>

please

i need help

[devout snow]

@cursive magnet Has your question been resolved?

[cursive magnet]

no

can you please send help

[devout snow]

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[drowsy thorn]

Hi! I'm taking real analysis rn and am stuck on this problem. To my understanding, the distance between any p with any q is r p,q, so couldn't i say that the "r" we're looking for is just half of this distance? i think i'm oversimplifying this problem

[drowsy thorn]

i'm going to see what i can come up with please ping me if you're able to help

[devout snow]

@drowsy thorn Has your question been resolved?

[devout snow]

@drowsy thorn Has your question been resolved?

[devout snow]

@drowsy thorn Has your question been resolved?

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[spare gyro]

Hi I’m trying to solve this with partial fractional decomp but I think I’m doing something wrong because I feel like my values aren’t correct when setting them to the coefficients

Was it a specific one or all of them?

Oh yeah I missed the x^2

Thanks I’ll try fixing that and seeing what happens!

.close

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[restive river]

what is the limit of 1/sqr of x as x approaches 0

[restive river]

@fervent swallow

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[restive river]

okay l will wait

[dense jay]

what do you think

[devout snow]

@restive river Has your question been resolved?

[restive river]

it doesnt exist

[quaint citrus]

,w 1/(sqrt x)

[woven radish]

Results provided by WolframAlpha

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

Requested by northsteve

[restive river]

.close

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[cunning roost]

whatd they do here to find the derivative?

[dense jay]

implicit differentiation

collecting dy/dx terms on one side

then making dy/dx the subject

[cunning roost]

did he do power rule here?

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[dense jay]

product rule

[cunning roost]

oh ok

[devout snow]

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[runic rose]

how to find sine cosine and tangent of 13pi/12

[runic rose]

i get all the way to sqrt2/4 -sqrt6/4

[graceful cosmos]

Which one are you doing

[runic rose]

the first

[graceful cosmos]

So hopefully you did:
sin(9π/12 + 4π/12)

Or something similar, it's just gotta add to 13π/12

[runic rose]

i used the sum and difference formulas

yes

i did

3pi/4 + pi/3

[graceful cosmos]

This looks like a solid answer. What's up?

Then again, it shouldn't be negative

[runic rose]

yeaha

the answer in the book says

sqrt2/4(1-sqrt3)

i dont know how to get there

[devout snow]

@runic rose Has your question been resolved?

[devout snow]

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[magic hollow]

the question was "Write the equation represented by the graph below."

my answer was sin^-1(x)

[round jacinth]

sin pi/2 =1 and sin-pi/2 = -1 so yea that's right

[magic hollow]

I figured

for some reason the website says its wrong

[round jacinth]

do arcsinx

[magic hollow]

It wont let me answer again but Ill ask my teacher about it

thanks

[round jacinth]

that is the graph of sine inverse/arcsinx

maybe it wants it typed a certain way

[magic hollow]

thats probably it

thats what I typed

[graceful cosmos]

You're mathematically correct

[magic hollow]

except in previous problems the website uses that version of inverse sin

[round jacinth]

maybe it wanted parenthesis around the x

[magic hollow]

not arcsin

I havent had to do that for previous problems

I think the website is wrong

anyways, thanks

.close

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[sonic smelt]

What's your question?

[hasty flame]

i was going to ask if i was right

obviously uh no

so

how does one solve this problem correctly?

[graceful cosmos]

This is just checking that you know the factor theorem

[hasty flame]

oh no

what is that??

[quaint citrus]

i put an X cuz f(-2) isnt 5

[hasty flame]

yeaah

theres only one answer left

i know its not A, since x+2 is in fact a factor

[steady forge]

well what do you get when you plug in -2

[hasty flame]

2^3 -2+1

[quaint citrus]

no

its

-(-2)^3 - 2 + 1

[steady forge]

no they're right

[hasty flame]

doesnt the negative on x cancel

[steady forge]

which is?

[hasty flame]

like

9

wait

7

[quaint citrus]

lol i think they were right for the wrong reason

[hasty flame]

what reason?

[steady forge]

they realized the negatives cancel?

[steady forge]

yeah so you can determine the answer now

[hasty flame]

yipee!!

C (right?)

[steady forge]

yep

[hasty flame]

woooohooo

[steady forge]

congrats lol

[hasty flame]

this was not as hard as i thought it would be

my brain is fully mush ive been working for like 5 hours

while im here anyone wanna make sure i didnt write something stupid here

optional lol yall have been patient enough

[steady forge]

point of inflections are whenever concavity changes

wait

you're right but you should specify in the explanation that there's two changes, from up to down then down to up

[hasty flame]

oh yeah that omission might get me dinged

thank you!

.close

[devout snow]

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[devout snow]

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[restive river]

Can someone help me draw this

I dont know how to draw the solid

[graceful cosmos]

It's kind of like a cone, except hollow. The inside is cone-shaped

Well, of course

I imagine they don't expect you to be Michelangelo

[restive river]

Let me try again

I dont think i understnad this conceptually

[graceful cosmos]

Okay I see what you're up to. Yeah that looks pretty good

[restive river]

The third one is the best so far?

I think this is what it is supposed to look like

<@&286206848099549185>

Is this third drawing an accurate sketch of the solid

Okay

I renovated my drawing

I think I am content with how it turned out

Now time to solve the problemo

.close

[devout snow]

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[rain bluff]

how is my explanation for ((a^3)^4)^-2

?

[devout snow]

@rain bluff Has your question been resolved?

[rain bluff]

<@&286206848099549185>

[wanton yacht]

You want to do ((a^3)^4)^-2 = (a^12)^-2 =a^-24

[rain bluff]

oh crud

is my wok botched

can anyone else confirm

[wanton yacht]

Looks like your first step you made an error

[rain bluff]

which makes the rest of it bad too

[wanton yacht]

Think of (a^3)^4 as (a•a•a)^4 = (a•a•a)(a•a•a)(a•a•a)(a•a•a) = a^12 if that helps

[rain bluff]

hm

thanks!

Ive got another question if youv time

[wanton yacht]

Sure

[rain bluff]

i cannot for the life of me find anything wrong with this

[wanton yacht]

I think it’s the x^3 term missing

I’m assuming you’d have to put a zero for that place

Like the top row is 3 0 -8 11 1

[rain bluff]

yeah i mean

wait

what x term

[wanton yacht]

Like the polynomial is missing a x^3 term so for the long division method you’d want to write it as 3x^4 +0x^3 -8x^2 +11x+1

Because that top row in the problem would correspond to 3x^3-8x^2+11x+1 which isn’t what you want

[rain bluff]

oh yeah

i guess that must be it

synethetic division is so silly

thanks

[wanton yacht]

Yeah haven’t seen it in awhile lol np

[rain bluff]

.close

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[restive river]

In this exercise I have to find the red area

We're given P(4,5|3) and f(x)=1/2x^2

So I thought f'(x)=x --> f'(4,5)=4,5=m

So 4,5x+c

But that's incorrect

Why?

How do I find out the tangent

Nvm I got it

.close

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[swift sage]

Could I get some help regarding understanding greedy algorithms? specifically exchange argument

[swift sage]

Given an alternative solution to compare the greedy to, its supposed to be a general solution right?

not a specific one

[devout snow]

@swift sage Has your question been resolved?

[graceful cosmos]

"Greedy algorithm" = "always make the best short term choice"

Notable greedy algorithms will also give the best long term result as well

[swift sage]

right, but the exhange arguement to prove the greedy algorithm

is that comparing to a "general" alternative solution

to say that for any solution, the greedy algorithm is at least as good as it

[devout snow]

@swift sage Has your question been resolved?

[devout snow]

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[hollow mantle]

$\frac{\cos^2(x) - \sin^2(x)\cos^2(x) - \sin^4(x)}{\sin^2(x) + \sin^2(x)\cot^2(x)} = 1 - 2 \left(\sqrt{\frac{\tan^2(x)}{1 + \tan^2(x)}}\right)^2$

[woven radish]

Alex

[hollow mantle]

how do i begin here?

i have to prove they are equal

i tried taking common factor but im stucked

[vestal dirge]

maybe start by writing everything in terms of sine and cosine

[radiant dune]

Show your work, where are you stuck

[hollow mantle]

$\frac{\cos^2(x) (1 - \sin^2(x)) - \sin^4(x)}{\sin^2(x) + \sin^2(x)\cot^2(x)} = 1 - 2 \left(\sqrt{\frac{\tan^2(x)}{1 + \tan^2(x)}}\right)^2$

[woven radish]

Alex

[hollow mantle]

i am messing up right?

[safe knoll]

doesnt the sqrt and square cancel each other ? or am i dumb

[feral agate]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[eager nova]

Yes, but i imagine is there to give a hint

[hollow mantle]

This is the original, i have to prove lhs = rhs

[radiant dune]

Id recommend working on the rhs first

To get a decent idea of what you want to do with the lhs

[feral agate]

Both sides need to be worked on

[radiant dune]

Yes ofc but simplifying the rhs will give him a big lead

[feral agate]

The lhs is also pretty simple to simplify

[eager nova]

Rhs last is just sin^2x

Last part i mean

[radiant dune]

Go ahead 1-sin²x=cos²x and also do similar factoring in the denominator

[eager nova]

Rhs would be 1-2sin^2(x)

[radiant dune]

And use identities like a⁴-b⁴=(a²-b²)(a²+b²)

[eager nova]

Ok i know already which identity is this xd

[hollow mantle]

mmm

i could do denominator like 1+cot^2(x) and divide numerator bu sin^2(x)?

[radiant dune]

Why are you stopping? Go ahead and keep using those identities lol

Proofs aren't supposed to be done completely mentally

[hollow mantle]

sorry i am blocked

$\frac{\cos^2(x) (1 - \sin^2(x)) - \sin^4(x)}{\sin^2(x) (1 + \cot^2(x))}$

[woven radish]

Alex

[hollow mantle]

Wait cos^2(x)/sin^2(x) is cot^2(x)

i could expand the numerator and divide

$\frac{\cos^2(x)}{\sin^2(x)} - \cos^2(x)\sin^2(x) - \sin^4(x)$

[woven radish]

Alex

[hollow mantle]

$\cot^2(x) - (1 - \sin^2(x)) - \sin^2(x)$

[woven radish]

Alex

[hollow mantle]

$\cot^2(x) - 1$

[woven radish]

Alex

[hollow mantle]

$\frac{\cot^2(x) - 1}{1 + \cot^2(x)}$

[woven radish]

Alex

[hollow mantle]

@radiant dune is this correct?

[devout snow]

@hollow mantle Has your question been resolved?

[devout snow]

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[frank zinc]

Two pair of dice are rolled simultaneously, what is the probability the product obtained is exactly divisible by 3

[frank zinc]

I think 20/36?

Am I wrong?

[neat solstice]

looks good

[frank zinc]

So it is 55%, thank you

[neat solstice]

you can think about this as prime factoring

for z=a*b to be divisible by 3, it must have a factor of 3
since 3 is a prime, either a or b have to have a factor of 3

so a or b have to be 3 or 6

for a=3 we have 6 options for b
for a=6, we have 6 options for b
for b=3, we have 4 options for a (a=3 and a=6 are already taken)
for b=6, we have 4 options for b (same logic)

[frank zinc]

.close

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[alpine hull]

help im lost

[full flume]

What's the question

[alpine hull]

everything

in the picture

[full flume]

Youre taking the derivative in respect to x