help-36

[fathom kayak]

i honestly have no clue if there any books like that

[terse crypt]

wait

[fathom kayak]

but the solutions would definitely point out the links between those

[terse crypt]

sorry

its like this

find the sum of the roots over interval [0, pi/2]

dawg

i dont understand anything

[fathom kayak]

man

is that tamil

[robust minnow]

Stole from ChatGPT

1.    “Thomas’ Calculus” (14th Edition): This is a classic textbook that covers foundational and advanced calculus topics, along with sequences, series, functions, and transcendental functions. It also includes chapters on analytic geometry and vector calculus, making it a comprehensive choice for calculus and analysis. Some editions provide introductory statistics and probability as well .
2.    “Schaum’s Outline of Advanced Mathematics for Engineers and Scientists”: This outline-style book covers a wide array of topics including calculus, linear algebra, complex numbers, differential equations, probability, statistics, and an introduction to linear programming. It’s a useful book for quickly covering topics with practice problems, though it may be less detailed than other textbooks .
3.    “Mathematical Analysis” by Tom Apostol: This textbook offers extensive coverage of set theory, logic, calculus, series, and vector analysis, with detailed sections on sequences, series, and functions. Though it doesn’t go in-depth into linear programming or statistics, it is very thorough in other areas such as number theory and functions .

4. “Advanced Engineering Mathematics” by Erwin Kreyszig: Known for its wide coverage, this book is ideal for advanced students and covers topics like linear algebra, calculus, complex analysis, Fourier series, partial differential equations, vector calculus, probability, and statistics. It includes sections on linear programming and is practical for applications in science and engineering .
   5. “Cambridge Mathematical Textbooks”: Cambridge has a series of texts that cover a range of mathematical topics in individual books, from complex analysis to abstract algebra and statistics. These books may be more specialized but offer a thorough and rigorous approach to each subject, suitable for upper-undergraduate or graduate-level study .

For a complete coverage in one textbook, “Advanced Engineering Mathematics” by Kreyszig and “Thomas’ Calculus” would provide broad coverage. However, for deep dives into specialized topics like number theory or logic, consider pairing a primary text with a more focused book from a series like Cambridge.

[terse crypt]

ok

ill look into thomas calc first

since i have that

[robust minnow]

For the rest, search “(title) free pdf download”

[terse crypt]

alrighty

ill start doing practice problems and then cram then

thank u @robust minnow @fathom kayak 🙏

[robust minnow]

You know the basics at the very least right?

[terse crypt]

yeah

other than linear programming

and number theory

[robust minnow]

Yeah then it’ll be possible to do well by Sunday, just grind and minimise distractions

[terse crypt]

👍

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[echo seal]

i got maximum, stuck finding minimum

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[honest egret]

Could someone just explain to me what I did wrong.

[trail mango]

x^3 * x^2 is not x^6

[honest egret]

Erm wait

Lemme use my slow ass thinking

[distant sleet]

Same with y^2*y^3

[honest egret]

Wait so when I’m confused so when should I multiply exponents vs adding them 🥹

[tranquil pine]

Factorization was not necessary since the exponent of the radicand is divisible by the index of the root.

[honest egret]

Wait what 😭😭😭

[honest egret]

I get this

[honest egret]

^^^

[honest egret]

How is (x^3)^2 not x^6?

[trail mango]

that’s different

[honest egret]

Is that not what that is?

What I wrote

[trail mango]

x^3 * x^2 = x^5

(x^3)^2 is x^3 * x^3

[honest egret]

my question isn’t this thou

[honest egret]

It’s this one

[trail mango]

but look

[honest egret]

OH

bye

[trail mango]

you replaced x^6 with x^3 * x^2

[honest egret]

Wait wait stay with me ☝️

[honest egret]

X^2 y^2 is the final right

[trail mango]

yea

[honest egret]

Omg I’m such a Baka tysm

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[chrome hemlock]

hii

[chrome hemlock]

y is this wrong

[tranquil pine]

It loses 63 percent

So you would be left with 37 percent of the original kinetic energy

[chrome hemlock]

oh

ok thank u!

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[shell storm]

I don’t understand the blue part

[opal osprey]

yea, dont worry, Do you want it exactly as it is, or do you want an easy way?

ill elaborate it to you

So, you know how we factorize X^2 -10X + 9 =0?

[shell storm]

Exactly

Thank you

[shell storm]

Yes I know

[opal osprey]

Mhm. You do the same here, so take the coefficient of the c^4 term and constant and multiply them together alright?

that is 8 x -3

its -24 right?

[shell storm]

But why 8 disappear

[opal osprey]

Yea,

The equation is just equivalent to the other one we saw

But i can explain you without the 8 disappearing

[shell storm]

Sorry I don’t understand my English is not good😭😭could you explain why 8 disappearing

[opal osprey]

I get it… umm

Lets do it without the 8 disappearing ok?

[shell storm]

Ok

[opal osprey]

Ill send you a pic

[shell storm]

Ok!! Thank you so much 😭😭😭

[opal osprey]

here

do you get it?

[shell storm]

Ahhhh I understand now, bc I used to use this way to solve equations, but my teacher gave the answer that is like the pic that I sent so I don’t understand

Thank you🥹🥹

[opal osprey]

i get it

[opal osprey]

youre always welcome

[opal osprey]

Always trust you. if teacher feels wrong, try the problem in your way

[shell storm]

Okay 🥹🥹thank you so much!!!❤️

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[warm python]

I'm totally lost here

[warm python]

I think I'll have to use the rank nullity theorm somewhere here, not sure how

[inland kettle]

physics u remember navier stokes man

he's back

(sorry for off-topic)

[warm python]

I don't

[inland kettle]

o

ok

[warm python]

The only thing that comes to mind is starting off with basis of ker(t)

[scarlet sequoia]

rank(g o f) <= min(rank g, rank f)

[warm python]

Hmm

Okay

[warm python]

I don't see how that helps

So I prove that 2 \leq rank(g), rank(f)

Hmm

Well, $5-rank(TS) \leq 5 -rank(S) + 5-rank(T) \implies rank(TS) -5 \leq rank(T) -rank(S)-10$

[soft zealot]

A dense set

[warm python]

$rank(TS)+5 \geq rant(T) +rank(S)$

[soft zealot]

A dense set

[warm python]

$5-rank(TS) \leq 5- ran(k(T)+ 5-rank(S)$
\
\
$rank(TS) \geq rank(T) + rank(S)-5$
\
Let rank(TS) =3 (say)
\
that would mean $ 8 \geq rank(T)+ rank(S)$
\

[soft zealot]

A dense set

[scarlet sequoia]

you should probably reason on the dimensions of ker(T), ker(S)

recall the relation you found between the dimensions of ker(T), ker(S) and ker(TS)

[warm python]

I have used that here

oh

[warm python]

I don't follow

I've used that here

[scarlet sequoia]

yeah but keep reasoning with the kernels

instead of the ranks

[warm python]

Hmm

but

I have this result

which uses ranks

[scarlet sequoia]

we already used it didn't we

so rank(ST) <= min(rank(S),rank(T))

and now we wanna show one of rank(S), rank(T) is <= 2

[warm python]

Well, from here that's simple enough

[scarlet sequoia]

edited

[scarlet sequoia]

how would your last line show that?

[warm python]

8 = rank(TS)+5

[scarlet sequoia]

so lemme get this straight

you supposed rank(TS) = 3 (which is supposed to be a contradiction)

[warm python]

Yes

[scarlet sequoia]

and you found... no contradiction?

[warm python]

yes

[scarlet sequoia]

so...

[warm python]

I've made a mistake somehwere

[scarlet sequoia]

I'm just telling you to stay with kernels

[warm python]

Okay

will do

$Ker(TS) \leq ker(T)+ ker(S)$

[soft zealot]

A dense set

[scarlet sequoia]

try to show null(T) + null(S) >= 5

[scarlet sequoia]

you mean in dimension right

[warm python]

ye

[scarlet sequoia]

well we got that now right

so at least one of them is >= 3...

[warm python]

yes, so the image will have a dimesnion less than or equal to 2

[scarlet sequoia]

<= 2*

[warm python]

yes

$2 \leq min(dim(im(T)),dim( im(S)))$

[soft zealot]

A dense set

[scarlet sequoia]

?

you mean the opposite

[warm python]

no

becasue we'e assuming the nullity to be 3, the kernal has dim=2

[scarlet sequoia]

?

the nullity of either T or S is at least 3

[warm python]

yes

[scarlet sequoia]

so the rank of either T or S is AT MOST 2

so the minimum of rank(T), rank(S)

is at most 2

not at least 2

[warm python]

right

so this proves the desired result

does't it

[scarlet sequoia]

yes

[warm python]

cool

thanks

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[drowsy ember]

hi

[drowsy ember]

might be silly but I dont understand the difference between nCr and nPr

<@&286206848099549185>

[pallid birch]

Permutation is used when order of arrangement does matter.
While in combinations, the order of arrangement does not matter.

Let me explain you through an example

[drowsy ember]

how to understand if the arrangement is needed in the question

[drowsy ember]

Ye go ahead

[pallid birch]

Let's say you have five letters a,b,c and d.

Now, you wish to create a word of length two.
How many possible arrangements are there?

[drowsy ember]

four?

i mean there should be four letters

[pallid birch]

Yeah four letters

[drowsy ember]

i think answer should be 4!

[pallid birch]

Correct.
Here are the arrangements
ab ac ad ba bc bd ca cb cd da db dc

[drowsy ember]

yes

ik arrangement is permutation and selection is combination

but i don't understand which to apply in question

[pallid birch]

In combinations of words,
ab and ba are same.
While in permutations they're different

As b should come first is different arrangement then a comes first

[drowsy ember]

4C1

is 4! = 4P1

how dude

[drowsy ember]

whaat

maybe ur right

[pallid birch]

What if I tell you to select any character from a, b, c, d in how many ways you can choose?

[drowsy ember]

4C1

4!/3!

so thats just 4

abcd bcda cdab dabc

in permutations

its also the same

?

@pallid birch ??

[pallid birch]

Now, What if I ask you select two characters from a, b, c, d. In how many ways you can choose.

[drowsy ember]

4C2

12

[pallid birch]

Which is 6, not 12

[drowsy ember]

4!/2!

4×3?

wait

sorry

i forgot 2!

[pallid birch]

Yeah, in here ab and ba is considered as one combination

[drowsy ember]

ok

in combination its 12

[pallid birch]

4c2 is 6.

Possible selection is 6, while possible arrangement is 12

[drowsy ember]

ye

So

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[chrome hemlock]

excuse me how would i do this with logs

[chrome hemlock]

this is half liofe formula

[inland kettle]

to solve for t?

i mean just take log_0.5 of both sides?

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[chrome hemlock]

can i use log base 10?

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[reef canyon]

Sorry for the mess,

basically integrated the first part with respect to x

and then found the derivative of that with respect to y

now do I just plug in the condition y(1)=1 into that to find C(y) ?

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[reef canyon]

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[reef canyon]

<@&286206848099549185>

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[twin beacon]

looks right to me

[twin beacon]

the x at the very front would go away

unless you’re taking an implicit derivative, then every x needs to have its derivative taken

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[stone wagon]

can someone walk me through why this is true

as in the equality

i don't get how these two are the same

the first sum looks like this

$\left\lfloor\frac{x}{1+\lfloor x / (\ell_0 + 1)\rfloor}\right\rfloor + \left\lfloor\frac{x}{2+\lfloor x / (\ell_0 + 1)\rfloor}\right\rfloor + \cdots + \left\lfloor\frac{x}{x+\lfloor x / (\ell_0 + 1)\rfloor}\right\rfloor$

[soft zealot]

artemetra

[ocean lintel]

if they didn't totally change the indices I could say things about it being telescopic
but with the index change I'm kinda confused

[stone wagon]

it is a telescoping sum indeed but the indices make it confusing

context
https://math.stackexchange.com/a/4995164/1153468

[ocean lintel]

oh it's still that
making any progress?

[stone wagon]

yes, i'm trying to understand the answer

but it's REALLY cool if correct

the second sum looks like

[ocean lintel]

oh

I see

it's summing by grouping by the value of [x/n]

[stone wagon]

oooooooooooooooooohhhhhhhhhhhhh

[ocean lintel]

it equals l for any n in the interval, whose length is the difference in parentheses

[stone wagon]

wait elaborate

[ocean lintel]

need more elaboration?

[stone wagon]

OOOOOHHH

i get it

this does not have to be this terse

[stone wagon]

thank you so much

[ocean lintel]

welcome to math literature

[stone wagon]

yea i should start getting used to this

thanks again

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[full halo]

Why - and not ±

[waxen lintel]

because y is negative from the initial equation.

[full halo]

But then its 5y²=16

[waxen lintel]

yes, it is. And?

[full halo]

So the solutions are about 5y²=16

[waxen lintel]

yes, -4/sqrt(5) satisfies the equation. And 4/sqrt(5) does not

[full halo]

Why it doesnt?

[waxen lintel]

substitute each one into the initial equation and check which one is ok

[full halo]

The initial equation has y² and substituting each for y² gives the same result

[waxen lintel]

but it has y without a square on the right.

[full halo]

Oh u meant that, let me see

Why y<0

4-y² -2<y<2

Not y<0

[waxen lintel]

bc square root is non-negative. If multiplied by -2 you get negative. And it equals y. So y must be negative

[full halo]

Yea i see that

Im trying to see where I failed on just calculating it

Probably need to put a ± somewhere

Pribably when I did everything ^2

[waxen lintel]

Squaring the equation is exactly the place where you get excessive roots

[full halo]

How to avoid excessive roots

[waxen lintel]

just check them all at the end and throw out the ones which aren't ok

[full halo]

What if its a complex long equation

Which i cant tell has a defined sign before squaring it

[waxen lintel]

or. at the beginning mind that squaring adds roots

[full halo]

Yeah and how do I tell which is the excessive one

[waxen lintel]

For equations it is easier to sub them all at the end into the initial equation

[full halo]

Isnt there a way to just tell which root is wrong?

When squaring

Without looking the sign

[tawny forge]

Just set conditions at the start

[waxen lintel]

As I said mind your squarings. Each time you square equation f(x)=g(x) you add roots of the equation f(x)=-g(x)

[tawny forge]

There cannot be a negative number under the square root if calculating in Real numbers

[full halo]

What if its fx=-gx

[waxen lintel]

If your initial equation was that, then squaring add roots of f(x)=g(x)

[full halo]

Right, what if there was no root?

[waxen lintel]

Squaring may add some. You should remove them at the end

[full halo]

I meant, what if its an equation which has a fx that has no defined sign

And you have to square it

Both signs are ok?

[waxen lintel]

I dont know what is defined sign.

[full halo]

I think I get it bro

Ty

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[rugged merlin]

Hiii can someone please help me with e and f? i am so lost and i have an exam soon 😭

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Please don't occupy multiple help channels.

[rugged merlin]

<@&286206848099549185>

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[acoustic finch]

This can't be right? Where is the mistake

[elfin tulip]

If I'm interpreting what you've read correctly, I think the issue is that you're assuming it's the same fixed a' in S for all values of epsilon. 2) is trying to tell you that there are points in S, arbitrarily close to u.
When you note that which a' you're talking about depends on your value of epsilon, the deduction that u < a' + eps for all eps implies u <= a' makes no sense.

In the set S = (0,1), the point a' = 0.995 would work for epsilon = 1/2 or 1/10 or 1/100, but would not work for epsilon = 1/1000. You'd need to switch to a different a' at that point, which is why you can't deduce 0.995 = 1.

If there was a single point that was within epsilon of 1, for any value of epsilon (0.99...) then of course that would have to just be 1.

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[acoustic finch]

Thank you

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✅

[acoustic finch]

Then is this question valid? It's what made me think of that "proof" in the first place

[elfin tulip]

Yeah, w is always in the closure of A. It's just not necessarily in A itself.

For closure, we need to argue that any neighbourhood of w contains some point of A. This does not need to be the same point for every neighbourhood. It just means any neighbourhood of w intersects A somewhere.

[acoustic finch]

Ah is that what the bar above A means? lol I haven't done this yet and looking at it made me come to the conclusion above

[elfin tulip]

Yeah, the bar means the closure. It is the set of all points x such that any neighbourhood of x contains a point of A. Equivalently, it is the union of A with the limit points of A. Also equivalently, it is the intersection of all closed sets containing A, and thus the smallest closed set containing A (hence the term "closure").

[acoustic finch]

Thank you

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[sour fiber]

I don’t get this…

[raven marsh]

what

[sour fiber]

1/4 is 25%

Isn’t it

[raven marsh]

yea

[sour fiber]

And

5*5 is 25

[raven marsh]

wait so whats the quesiton

[sour fiber]

To solve that

[raven marsh]

oh

[trail mango]

<3

[sour fiber]

Hiiii

[raven marsh]

wait so your answer is $\sqrt{\frac{1}{4}}=5$

[soft zealot]

Death.cv

[sour fiber]

Yes

But apparently that’s not right

[long flame]

It should be 0.5

Not 5

[raven marsh]

try to check

[sour fiber]

Wait I’ll try

[raven marsh]

is 5^2 = 1/4 ?

[sour fiber]

Oh yeah now it worked

[raven marsh]

fr

💀

[sour fiber]

Don’t judge I just started this chapter of math..

[raven marsh]

no is ok

[sour fiber]

.close

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[raven marsh]

bro didnt even thank me

[sour fiber]

Thank you🫶

[inland kettle]

yeah that's fairly rare with helpees

at least this one was nice

[raven marsh]

np 🙂

[sour fiber]

.reopen

[final saddle]

✅

[sour fiber]

Hi im back…

I’m confused again

[inland kettle]

,tex .exp rules

[soft zealot]

ren

[raven marsh]

hello

[sour fiber]

I still don’t get it

[raven marsh]

note that $\sqrt{\frac{9}{49}} = \frac{\sqrt{9}}{\sqrt{49}}$

sry

[soft zealot]

Death.cv

[sour fiber]

ohhhhhhhh

So

Top is 3

Bottom is

Uh

7

So

[raven marsh]

yep

[sour fiber]

3/7

[raven marsh]

mhm

🙂

[sour fiber]

Okay let me try it

[inland kettle]

here you can sub a = 3 and b = 1/7 for the second-to las one

[sour fiber]

Lhhhh it was right okay wait I’ll write this down

[raven marsh]

yes this might be helpful at times 🙂

[sour fiber]

Yeah I’ve wrote them down in my note book

Thank uuuu agaunnn

[raven marsh]

thats a good habit

[raven marsh]

np

[sour fiber]

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[jagged flare]

let a and b be positive integers such that a+b|ab, prove that a=b

[jagged flare]

im not really sure what to try to do here...

[raven marsh]

what

[swift nebula]

hello sv

[raven marsh]

all i can think of is let ab = ka + kb

then $(a-k)(b-k)=k^2$

[soft zealot]

Death.cv

[jagged flare]

18,9wait what

this isnt even true wtf

erm

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[finite root]

Which answer is right, the one in red or black?

[finite root]

(I used multiple ai software to check, they are agreeing with both answers)

If someone could tell me where i went wrong, i would appreciate it a lot

[final tangle]

what software are you using that's giving different results

[finite root]

gpt, and one given by school

[winged anvil]

!nogpt

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[final tangle]

is that initial expression **exactly **how the question was presented to you?

[finite root]

Yes, the very top line

I think i rewrote it correctly on the second line

[final tangle]

seeing as they had () around the other denominators,
it is not entirely clear whether they intended
6y/(2y) or 6y/2 * y

based on your interpretation, its fine,
the red is just wrong regardless

[finite root]

huh, ill have to ask my teacher about that

thanks for the help

[balmy holly]

black is the correct answer from what I can understand.

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[oblique turret]

I understand everything there but why is there a 2k pi added at the end

[severe canyon]

Because of the periodicity

If you don't put it, you wouldn't be able to find all the four 4th roots

[oblique turret]

so if it was z^3 you would still put the 2k pi there

[severe canyon]

Yep

[oblique turret]

so essential its there no matter the z^n because thats the period of cos and sin

[severe canyon]

Exactly

[oblique turret]

what didnt make sense later in the video they work out when k=-1,0,1,2 whats the need of this?

is the final answer whats seen on bottom right in that case cant you just give your answer when k=0?

[severe canyon]

I admit I'd never seen negative values of k before

Usually k = 0, 1, 2, ..., n - 1

[oblique turret]

i mean since thats the period part (2kpi) it doesnt matter what k is

[severe canyon]

Where n is the root index (in your case 4)

But any "sequence" of consecutive integers with length 4 works
So you could have picked k = 6, 7, 8, 9 for instance

[oblique turret]

is it because its to a power of 4 so it has to have 4 roots?

[severe canyon]

Yeah

[oblique turret]

so essential k=0,1,2,3

becuase first 4 e.g

that makes sense thank you

[severe canyon]

You're welcome

[oblique turret]

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[rotund dagger]

How to find h

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[vagrant wraith]

the ( ) here is jacobi symbol

[fallow wren]

maybe ren will arrive soo

[vagrant wraith]

Im not sure how they get S(m+Np)=S(m)

[atomic spindle]

He told me my photo was illegible

Very helpful

[inland kettle]

sorry but it honestly is

[atomic spindle]

😭

[inland kettle]

i can't exactly help you if i don't know what you've written

[atomic spindle]

It’s fine

[fallow wren]

maybe he was a little bit rude ...but he work a lot here ...it is indeed a good guy

[inland kettle]

everg i'm sorry but we're clogging up poor @vagrant wraith's channel

(thanks for the compliment though)

[fallow wren]

you are rright , my dear ren

[atomic spindle]

It was typed

[vagrant wraith]

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[inland kettle]

bruh

sorry man

[fallow wren]

np

[inland kettle]

:(

no i was talking to somethingwrong

[vagrant wraith]

oh no dw about it, i figured out the question haha

[inland kettle]

ah nice

[fallow wren]

ohh I see...

[inland kettle]

just looked at the jacobi symbol

seems very interesting

what course is this from?

[vagrant wraith]

an introduction to number theory

[inland kettle]

ah

didn't see this in the one i did

[vagrant wraith]

yes the topic is very wide afterall haha but its very strongly related to solutions to quadratic residues if you worked with that b4

[inland kettle]

ah i haven't done those

thanks!

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[gilded schooner]

Can someone tell me how this is possible lol or if my reasoning doesn’t work

[gilded schooner]

Specifically talking about c=0

Ignore the =9 I wrote out the full sum of angles somewhere else

Actually I just didn’t use that to solve for x

[woven ledge]

Whats the given

[gilded schooner]

Everything but x and the shape drawn in red I inputted

The answer I got for x is correct using the fact that the sum should be 540

Using my logic there

So given angle 85, 85, 95, and 155

Goal is to find x and ik I kinda didn’t show that well

[woven ledge]

ok so x is not the angle of the red cyclic

[gilded schooner]

No

[woven ledge]

but you calculated x as if it is in the red cyclic

[gilded schooner]

I meant yes sorry lol

[woven ledge]

(x=180-85=95)

oh

[gilded schooner]

I got x for the 5 sided shape equals 120

But x in the red shape is 95

Sorry let me just redraw

[woven ledge]

How did you get x?

was the value of d given?

[gilded schooner]

No

I used cyclic quadratic theorem to get y and the 95

[woven ledge]

Nono like

you said x = y + d. (correct)

you got y as 95 (correct)

so u said x = 95 + 25

but that would mean d = 25

but we never really got d seperately. We can only say that x = 95+d

c+d = 25 but not d specifically

in fact, d can be any value

where 0<d<25 . same for c (0<c<25)

@gilded schooner

[gilded schooner]

But we know the total angle where that 95 is on the left is the total for the 5 angled shape

It somehow equals the angle in the red shape

Which means angle c would be 0

[woven ledge]

i couldnt understand could u please elaborate

cus the sum of angles for a quadrilateral is 360

and sum of angles for a pentagon is 540

they are not the same

and if u mean sum of two internal angles = opp external angle

that does not apply here

cus the line is bent

[woven ledge]

If its given that 95 is a + c

then a is 0

cus c is also95

[gilded schooner]

Why wouldn’t it be a+c?

[woven ledge]

because it just isnt

what logic did u use to arrive at a+c =95?

also d is not = 25-c

[gilded schooner]

I meant 25-a

[woven ledge]

its also not equal to 25-a

[gilded schooner]

Because 95 is made up of a +c, I cut 95 into c and a

[woven ledge]

yeah then it is

i thought 95 was part of the quadrilateral

[gilded schooner]

No

Well it is

But that’s cuz a is 0 which makes no sense

[woven ledge]

Is this from a specific problem>?

can u send the problem if so?

[woven ledge]

yeah a is 0 if a+c us 95

and u are saying thats what they wrote

So what we can summarise

IS

[woven ledge]

I think whats happening is that if these are the angles for this pentagon, then it will not fit on a circle

it just isnt cyclic

[gilded schooner]

We are given this without the angles being split up into a b c or d and there is no imaginary line, those equations at the top I put

We are given the circle

[woven ledge]

yeah i understand

read what i said above

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[boreal thicket]

A survey was conducted to figure out the percentage of high school seniors who had 5 or more drinks in a week. 13,000 seniors participated. Based on the data from the survey, 48.3% of males had 5 or more drinks and 28.7% of females had 5 or more drinks. What percent of both males and females had 5 or more drinks?

[boreal thicket]

been getting into an argument with my HS health teacher over this question, i argue its 38.5%. she says its 77%

[tiny gorge]

how could it be 77%?

P(5 or more drinks | male)P(male) + P(5 or more drinks | female)P(female)

is what you want

if you assume that P(male) = P(female) = 1/2 then your answer is correct

[boreal thicket]

thank you

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[wise plume]

How can I use L'Hopital's rule to evaluate the limit as x approaches 1 from the RHS of 1/lnx - 1/(x-1)

[hybrid heath]

$\frac{1}{\ln x}-\frac{1}{x-1}$?

[soft zealot]

SWR

[wise plume]

Yes

[vital crag]

put over a common denominator

$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$

[soft zealot]

riemann

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[strange moth]

If someone asks me a question and gives me 4 possible option and one is correct, my chance of guessing the answer is 1/4 or 25%, but what would be the change of winning the whole show if it costs of 8 rounds, and each round has different available answers?

1-4th round - 4 answer options
5-7th round - 3 answer options
8th round - 2 answer options

What’s the chance of me winning the show if I only guess the answer?

[rocky tusk]

assuming these events are independent you can just multiply the probability of getting it correct in each round

[strange moth]

Joe

[rocky tusk]

facts Joe

[short talon]

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[jolly tapir]

hello
for this question, I'd need to write f(x) as a piecewise function, then find f_Y(y) using those intervals, would that be correct?

[jolly tapir]

eg f(x) is defined as the following
$$x+1, x\in [-3,0], y\in[-2,1]$$
$$-x+1, x\in[0,1], y\in[0,1]$$
$$x-1, x\in[1,3], y\in[0,2]$$

so for the interval of say $y\in[0,1]$, would i need to take into account the contributions of all three $f_Y(x)$s?

[soft zealot]

jack

[jolly tapir]

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[unique estuary]

can someone help with c)

[boreal locust]

whats the position at t=0

[unique estuary]

10

[boreal locust]

whats the position at t=2

[unique estuary]

0

[boreal locust]

0 no?

[unique estuary]

typo my bad

[boreal locust]

cool now whats the position at t=7/2

[unique estuary]

-9/4

[boreal locust]

ok so now what you need to understand is

the paricle starts somewhere

goes somehwerre negative

then goes to x=0 position as well

ur considering distance travelled

[unique estuary]

yeah not displacement

[boreal locust]

right so like consider the distance where it hits 0 the first time

it should be some posiive displacement

(which is at t=2)

so x_0 - x_2

now after t=7/2 it again goes to 0

since a parabola is symmetrical

just consider distance from t=2 to t=7/2

and multiply it by 2

[unique estuary]

so would the textbook be wrong when it says it is 14.5m?

[boreal locust]

uhh lets see

9/2 + 10

14.5

its correct

[unique estuary]

cool thank you

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[compact horizon]

Is there a way I can learn math through my passions ? Psychology, astrology and fashion design, I know I can apply the inherent skills that math gives u, like critical thinking, and patterns identification.But not the hard contents ,coz I really like the conceptual part and not the monotone practice part ( doing ton of exercises), I want to identify this concepts in the world , extrapolate systems and use them to improve my life , understand the world and it’s habitants better, apply knowledge to substantial things

[sonic crystal]

there is no math or science in astrology

[compact horizon]

There is some math

[barren hound]

i think there's a little math in astrology

some geometry

[compact horizon]

Yeah but is not the foundation

[barren hound]

psychology i would expect to see some, especially statistics

[sonic crystal]

do you mean astronomy or astrology

[compact horizon]

Astrology

Well ignore Astrology what about the other stuff like understanding the world better

?

[barren hound]

well, there's a lot of pop math channels

numberphile, 3blue1brown, mathologer

they do a good job explaining things (usually) without requiring lots of background

[compact horizon]

How can I practically consciously do that fun stuff , and not just staying on the abstract and the repetitive exercises but seeing the abstract on the practical and material world

[barren hound]

well, if you really want to understand something, the easiest way really is repetition and practice

it's a skill just like playing an instrument or drawing a dress

[compact horizon]

Oh okey just push through till I can do the cool stuff

[barren hound]

what i find works well is spaced repetition

seeing things for the first time, then again in a month really getting into them

[compact horizon]

What do I need to learn before I can do all the things I listed before

?

Idk anything ( I ditched school for like 4 years)

The fundamentals of algebra ?

[barren hound]

i guess it depends on what math you run into but algebra is pretty fundamental yeah

but before then it'd be good to memorize times tables, i know it's a pain but it's so helpful to be able to look at 6x8 and not have to think about it

[compact horizon]

Oh I know timetables

Idk the quadratic function or like the roots properties

And radicals

[compact horizon]

Like consciously using the skills math gives u

[barren hound]

definitely you can learn as you go and as you run into stuff, this is what i mean by the fundamentals of algebra though

you don't have to do these right now :) but that's the idea i had in mind

[compact horizon]

Is the book from Harold Jacob’s good ?

[barren hound]

never heard of it

it's probably fine

[compact horizon]

Elementary algebra

[barren hound]

seems expensive

but it has good reviews

shrug

[compact horizon]

But do you recommend another algebra book for complete beginners

?

[barren hound]

i don't have any on hand :/ if you can afford it (and find it, it seems to be out of print?) then it's probably just fine

but there are much cheaper algebra books that i'm seeing on amazon that seem to have good reviews

[compact horizon]

I found it as a website but the images are kinda blurry, and having to study with the computer is distracting

[barren hound]

physical books are really nice

[compact horizon]

But my sister has the physical baldor algebra book

[barren hound]

oh are you a spanish speaker?

[compact horizon]

But I have really loved the book

[compact horizon]

Yeah how did you know

[barren hound]

pq lo googleé

(baldor algebra)

there's a channel that i really like for spanish math instruction https://www.youtube.com/@MateFacilYT

pa algebra hay https://www.youtube.com/playlist?list=PL9SnRnlzoyX1sF5fX83CleyK_SATfbhia

[compact horizon]

Hablas español?

[barren hound]

sip

[compact horizon]

Wow

[barren hound]

(mal jaja)

[compact horizon]

Tranquila mi inglés tan poco es perfecto

[barren hound]

no he dado cuenta hasta que dijiste baldor

[compact horizon]

Oh thanks sweetie

I think the resource doesn’t matter but the student

[barren hound]

a lot of it is that you'll want multiple modes

and yeah one of those modes being writing will help a lot, at least it did for me

[compact horizon]

But I loved the Harold Jacob’s book it was so fun

Okey

[compact horizon]

Like writing proofs?

[barren hound]

but like debes utilizar la que te interese pa determinar que estudiar

[barren hound]

sure, or just writing exercises, if you don't have to think about the easy things you can look at more interesting things

[compact horizon]

Yeah I have to push through the fundamentals

To be able to do the things I want to do

Thank u very much

[barren hound]

x nada

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@compact horizon Has your question been resolved?

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[ocean merlin]

2:1?

Because

lets say the radius is like 1

of the smaller circle

so then the radius of the larger circle is √2

√2^2/1=2

[ocean merlin]

wtf

[short lake]

.

[ocean merlin]

m8 get ur own help channel

[short lake]

ok

[fast creek]

its best to use letters like a or α to call the length of the radius

avoid numeric values, always generalize

anyways, yes ur answer is correct, its 2:1

[ocean merlin]

alr nice

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[subtle urchin]

Is the domain of this

[-4, inf)

[atomic moon]

Domain ?

[subtle urchin]

yes

domain

[atomic moon]

That is [-2,7] - {3} or im missing something

[subtle urchin]

ohh

yeh i got it wrong

[wind glen]

[-2, 3) U (3, 7]

[subtle urchin]

i kinda forgot how to do piecewise functions

[subtle urchin]

yes this makes sense

thanks

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[bleak island]

#help-4

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[red field]

I have two lines in 3d space , they are in the pic,
I have found and verified that the minimal distance between them is 17/sqrt(29)
I want to find the point C (in l1) which is the nearest point to l2
how can I do that?

[final saddle]

@red field Has your question been resolved?

[edgy burrow]

are you familar with vector forms of the line and dot product

[red field]

yes

[edgy burrow]

i think you can use something to do with the fact the line between the two points is perpednicular to either line so when dotted with either line it will equal 0

but its been a while since ive done this so i cant remeber exactly sorry

[red field]

ye I get where ur coming from'

maybe I can let A be a point on l1, B point on l2
and the line in the direction of AB is perp. to both lines\

[edgy burrow]

yeah thats what im trying to do now

it feels like there should be a faster way though

[red field]

yes I think i knew one and forgot it lol

[edgy burrow]

do you think this could be used?

[red field]

I used this actualy to get the 17/sqrt(29)

maybe we can use the distance between two points

but the problem is we have 2 variables

one for each point

[red field]

waittt

maybe we can use the one u sent

and we have only 1 var

[edgy burrow]

yeah i think we can find the point on l2 with it

[red field]

like deriviative and find minimal point

[edgy burrow]

would this not work?

however this only gets the point on l2

[ocean relic]

Uhhh

Idk what any of this is

👍

[red field]

ye it should work

[red field]

ye the other one we can find if we just shift the plane to l2

like chnge D in Ax+By+Cz+D=0

ty for help

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[edgy burrow]

no problem

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[subtle urchin]

How do I find the zeros for X of F(x)

that's the question

i dont get it

[drowsy epoch]

Just look where sqrt(x+4) = 0 and x²-9 = 0 on their given intervals

[subtle urchin]

i mean

the top one equals 0 when x is -4

but that's not on the interval

[drowsy epoch]

yup

so we can now consider x^2-9 = 0

[subtle urchin]

and bottom one when its 3 or -3

none of them are

on the interval

[urban burrow]

yeah

[drowsy epoch]

yup

so what can you deduce?

[subtle urchin]

there are no zeros for f(x) in the given interval

?

thanks

.solved

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[tranquil pine]

Consider only the centre and S. The centre is fixed. This is vertical. We release a ball tied to the string which is fixed at the centre. We also give it a velocity of 5 m/s. What will be it's speed when it has travelled through an angle which is less than 90 deg as shown. The mass of the stone is m and the lenght of the string is l.

[tranquil pine]

Consider K too.

Haha

[barren hound]

complex

i'd say look at potential energy and kinetic energy

[inland kettle]

lagrangian?

[tranquil pine]

no

[inland kettle]

lagrangians are closely related to potential energy and kinetic energy is what i mean

and iirc pendulums are famously easier to solve using lagrangian mechanics

[tranquil pine]

oh

[granite heath]

isnt like the euler-lagrangian equation useful here?

[tranquil pine]

Idk

[granite heath]

i think ive seen it used for double-pendulum

[inland kettle]

it has

[granite heath]

i think i saw it in a veritasium video lmao

it was a good video ngl

[tranquil pine]

thank you,

got the correct answer

[granite heath]

oh i see

just find vert distance from S-K and then get pe and solve for velocity

[tranquil pine]

yup

law of conservation of energy

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[tranquil pine]

.reopen

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✅

[granite heath]

what happened

[tranquil pine]

what about this

LOL

[granite heath]

lmao

what was that?

[fallow wren]

it was like a virus..donno

[tranquil pine]

everg just challenged picasso with that one

he just turned in his grave

[inland kettle]

??

[tranquil pine]

something not to your liking ren?

[inland kettle]

no i just wanna know what happened lmao

i wasn't here

[tranquil pine]

oh

[tranquil pine]

@fallow wren sent a spin-off of the smiley face on the problem i'm replying to right now with a frown thingy

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[vivid cloak]

is this not suppsoed to be 1/2 - 1/2cos2theta

[drowsy epoch]

they pulled out 1/2

[vivid cloak]

it would still be 1/2 - 1/2cos2x

so 1- cos2x

not 1-2cos2x

[drowsy epoch]

yea

that right

[vivid cloak]

so the mark scheme is wrong

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@vivid cloak Has your question been resolved?

[vivid cloak]

ig?

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[twin oriole]

I'm struggling with this proof:

$$ if |f| \leq g, then

\int_a^b |f| \leq \int_a^b g
$$

[twin oriole]

$$ if |f| \leq g, then

\int_a^b |f| \leq \int_a^b g
$$

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[drowsy epoch]

I'm struggling with this proof:

[ \abs{f} \leq g \Rightarrow \abs{\int_a^b \abs{f}} \leq \abs{\int_a^b g} ]

[twin oriole]

The integrals should be in absolute values as well, I just messed up my code

[soft zealot]

bacc (unhelpful)

[twin oriole]

Yeah

[twin oriole]

Pls help

It says: "proof is left as exercise" but idfk what to do 😭

.reopen

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✅

[drowsy epoch]

The only way what I am thinking of is to start with the definition of integral, the riemann sum

Despite it being so obvious i am also a bit lost

[deft ravine]

Why do you need the absolute values? Both integrals are positive or 0 anyways?

[twin oriole]

It's for both a < b and a > b

I'm allowed to use these tho

[deft ravine]

Doesn't this cover the case b > a immediately?

[twin oriole]

Yeah, but what about a > b?

[deft ravine]

Then you switch the bounds and introduce a - which cancels with the absolute value

[twin oriole]

Like so?

[deft ravine]

Yes

[twin oriole]

Damn, I feel so dumb

Thanks

[deft ravine]

You're welcome

[twin oriole]

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[acoustic oar]

Hello I need help with this limit: $a_0 > -1$ and $a_n = \sqrt{2^{n+1}a_{n-1}+4^n}-2^n$ compute $\lim_{n\to\infty}a_n$

[soft zealot]

ω

[acoustic oar]

I also have to prove it exists as thats non-trivial

I proved that this sequence is decreasing

also via induction $a_n > (-1)2^n$

[soft zealot]

ω

[acoustic oar]

we can rewrite the expression to $a_{n-1} = a_n + \frac{a_n^2}{2^{n+1}}$

[soft zealot]

ω

[fallow wren]

also from here it can be seen the decreasing property with ease

[acoustic oar]

yeah thats how I got that

[fallow wren]

so now you only have to apply the limit with n approaches infinity of the last thing you wrote

[acoustic oar]

but how idk what $\frac{a_n^2}{2^{n+1}}$ is convergent to

[soft zealot]

ω

[acoustic oar]

and even if it is at all

okay its either convergent or divergent to -infty since its decreasing

suppose it is convergent

so can I say that this term $\frac{a_n^2}{2^{n+1}} \rightarrow 0$?

[soft zealot]

ω

[acoustic oar]

I guess so

[fallow wren]

right

[acoustic oar]

or not