help-43

[random mica]

ok

[heady spruce]

im kinda confused

[random mica]

so AD is the angle bisector

[heady spruce]

yea

angle BAD = andle CAD

[random mica]

correct

we are given that B > C

now look at the other angles

[heady spruce]

hmm

[random mica]

is angle ADC bigger or smaller than ABD

(hint sum of 2 opposite interior angles = exterior angle)

[heady spruce]

bigger

[random mica]

Yes, as ADC = ABD + BAD

so ADC > ABD = B

but we know B is greater than C

so ADC > C

so what can you conclude about the side inequalities for triangle ADC

[heady spruce]

wait

wym

[random mica]

where are you stuck

[heady spruce]

the ADC>C part

waht is c

[random mica]

sorry its the shorthand they used

angle C

basically angle ACD

should I explain again

[heady spruce]

no need

so in ADC

AD>CD

[random mica]

uhm not quite

whats the side opposite to angle ADC

[heady spruce]

AC

[random mica]

yes and whats the side opposite to angle C

[heady spruce]

AD

[random mica]

yes so AC > AD

right?

and thats one of the options

[heady spruce]

oh ye

[random mica]

what else can we definitely conclude

[heady spruce]

CD>BD

[random mica]

yes but why

its the angle bisector ratio right?

[heady spruce]

ye

[random mica]

BD/CD = AB/AC

[heady spruce]

im preety sure that is it

[random mica]

yes

the other ones cant be properly concluded

[heady spruce]

trye

true*

[random mica]

you can play around with moving the point A in your head

and see how AD changes

or in geogebra

[heady spruce]

ok

tysm btw

[random mica]

welcome

[heady spruce]

.close

[compact pewter]

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[compact pewter]

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[vernal jacinth]

hi

[vernal jacinth]

this question is wrong, right ?

[kind viper]

why so

[vernal jacinth]

we dunno the exact value of lix x to 1

-2 as well

soo ?

this one is also wrong , right ?

hey guys ... pls help

[feral flicker]

Who made these questions? 🤯

[vernal jacinth]

dunno

[vernal jacinth]

they are wrong , right ?

[feral flicker]

I think they don't contain enough information to answer them. The latter question has a problem at x=1 where it is discontinuous and therefore does not exist. You would have to stipulate if the limit is coming from one direction or the other.

[vernal jacinth]

plus we dunno the value at f(0)

thats also a problem i guess

[feral flicker]

You can infer that all the values at x=0, lim x->2 f(x), and lim x->1^+ f(x) are the same.

And that's where the lack of information comes into play.

[vernal jacinth]

what about this ?

[feral flicker]

It doesn't even show f(x) at x=-2.

Perhaps they meant x->2.

[vernal jacinth]

and it doesnt exist

🙁

[feral flicker]

These questions are not written well.

[surreal shard]

u need to take into account that the slope is constant

[vernal jacinth]

yes i did that part , after writing to u

but can we infer that for x = -2 it will be -1

i mean , we dont see that part of the function

[vernal jacinth]

aaa lim to k is 0 always

isnt it ?

[vernal jacinth]

so limit exists only at x = 0 , x = k , x = 2

at x = k answer is 0

at x = 0 and x = 2 answers are the same

soo 2l = 4

l = 2

f(0) = 2

and if we make the equation of line then k is -2

[compact pewter]

@vernal jacinth Has your question been resolved?

[vernal jacinth]

.CLOSE

.close

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[delicate pulsar]

Need help with number 13

[vague sparrow]

I can't see. Maybe take it clearly

[delicate pulsar]

As part of Kayla's exercise program, she either runs 6 miles/day or rides her bike 10 miles/day. Her new goal is to cover a minimum distance of 200 miles, with at least 15 of the days running. She would like to determine the number of days it would take to accomplish this.

[vague sparrow]

so we're using the x y coordinate

right

[delicate pulsar]

i made the equations x>_ 15
and
6x+10y >_ 200

[delicate pulsar]

yeah

idk what to do now

but i decided to make y=mx+B out of the 6x+10y equation

[vague sparrow]

alright so we know that the slope of the line is $\frac {y_2 - y_1}{x_2 - x_1}$

[boreal girder]

Average_X^2

[delicate pulsar]

its system of equations

i mean systems of linear inequalities

[vague sparrow]

so we're using $y = mx + b$

[boreal girder]

Average_X^2

[delicate pulsar]

heres the equations i made out of the word problem:

6x+10y > 200
x > 15

turn it into y + mx+b ill end up with

y> -3/5x + 20
x>15

[vague sparrow]

ok

so its $y> = -3/5x + 20.$ where b is 20.

[boreal girder]

Average_X^2

[delicate pulsar]

b is 20

i just dont knows how to graph the -3/5

(its not a full graph)

[vague sparrow]

alright so,

Lets see if we can graph $-3, -5$

[boreal girder]

Average_X^2

[vague sparrow]

will that work?

[delicate pulsar]

yeah

Also this is the graph they gave me

@vague sparrow

oh wait now i get how to graph

alr im good now

[compact pewter]

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[grizzled geyser]

I'm wondering whether a set $X \subseteq R^k$ can be both open, closed and bounded. I think it should be impossible. My attempted proof is as follows. $X$ is bounded, so take $q \in X$. Let $h = inf{d(p,q) | p \in X^{\text{c}}}$. There are two possibilities:

If there is a point $p \in X^{\text{c}}$ such that $d(p,q)=h$, then $p$ is a limit point (otherwise, there is a neighborhood $N_p$ such that $N_p \cap X = \emptyset$. Taking a $r \in N_p$ between $p$ and $q$ produces a point with shorter distance to $q$ than $p$, which is a contradiction). Thus $p$ must be an element of $X$ because $X$ is closed, violating the construction of $p$

But, I'm stuck on the case that there is no such point. I think there should be a point, since $X$ is open.

[boreal girder]

Xwtek

[untold cairn]

There are exactly two subsets of R^k that are both open and closed: the empty set and R^k itself. This is because R^k is connected

[south sage]

Well if it's closed and bounded it's compact which means that it can't be open

[untold cairn]

There are some topological spaces where some non-empty sets are both compact and open though. The reason you can't in R^k is basically because R^k is connected I think

[compact pewter]

@grizzled geyser Has your question been resolved?

[grizzled geyser]

Wait, give me time to think

[grizzled geyser]

I'll look at the connectedness of R^k.

[south sage]

Proof?

[trim isle]

no

[untold cairn]

Here are some examples: https://math.stackexchange.com/questions/1403826/a-compact-open-set

[grizzled geyser]

Okay, I found the proof here. Thanks.

.close

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[cerulean thorn]

yes and whats the side opposite to angle C

[compact pewter]

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.

[proven wren]

do you have a problem?

[bleak dock]

it's a bot that spams random messages

mods are aware of it

.close

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[ionic ermine]

when will PA.PB be minimum? Any concept here that I'm not aware of?

[compact fractal]

does PA.PB mean PA * PB

[ionic ermine]

yea

[compact pewter]

@ionic ermine Has your question been resolved?

[ionic ermine]

i guess I'll ask a bit later, try and see if you find something

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[open plaza]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[open plaza]

@ionic ermine is trig allowed

[ionic ermine]

yea sure

[open plaza]

here's our figure

[cursive harbor]

Wait why does the question give you P

I just noticed that

[open plaza]

P is already given

[cursive harbor]

Yeah why lol

Oh nvm A and B are not

I'm dumb sorry

[open plaza]

so from sine theorem, PA = PO*sin(POA)/sin(A)

likewise for PB

[ionic ermine]

ok

[open plaza]

and because PA.PB needs to be the minimum, sin(A).sin(B) needs to be the maximum

[ionic ermine]

but

POA and POB also vary

ok

yea i see

[open plaza]

$sin(A) \cdot sin(B)$ can be rewritten as $\frac{1}{2} \cdot [cos(A-B) - cos (A+B)]$

[boreal girder]

(De)Carbonized

[tiny prawn]

What is this channel for

[open plaza]

@ionic ermine this should be enough hints I suppose

[ionic ermine]

yea

[open plaza]

there are prob easier approaches

[ionic ermine]

really good idea though that the lines can be seen as origin and thier angles

.close

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[old flame]

simple Algebra II question, working on a practice assignment.
I've been trying to figure out a step in the factoring of this polynomial,
How does x^2 - x + 6 turn into (x^2 + 2x) + (-3x - 6), where do the 2x and -3x come from?

[mild sky]

2x-3x = -x

[old flame]

okay gimme a second

where do you get that from

[south kindle]

...

We turn -x into 2x-3x

[old flame]

based on...

[south kindle]

To factories it

Ize

Ex nihilo

Idk whats the word in english

[topaz cargo]

Educated guess based on Vieta's

[south kindle]

無中生有。

[old flame]

confused

[south kindle]

Its hard to explain ;-; its just u need to change stuff based on what u have

[old flame]

i have the knowledge base of an american hs junior

lol

[old flame]

is vietas a math thing or a name

[topaz cargo]

Vieta's theorem

[old flame]

alright let me look into that

wasnt covered by name

[south kindle]

.

Do you have a calculator

[old flame]

yes

im just struggling with theory here

[south kindle]

You can do it with cross method or

[old flame]

so you just find this by plotting it

is that the simplest solution

oh im an idiot

alright thank you

.close

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[old flame]

How would I properly write the linear factors of (x^2 -4) (x^2 -9)

[mild sky]

Do you know how to factor x^2-4 and x^2-9

[old flame]

yes but my practice assignments answer checker is saying its wrong

[mild sky]

What was your result

[wooden slate]

the expansion using algebraic identity a^2 - b^2 was wrong?

[old flame]

give me a second to type it out

(x + 2) (x - 2) (x+ 4.5) (x- 4.5)

did i misplace a negative

[mild sky]

4.5?

[old flame]

oh man

would it not be 9/2

[strange ermine]

Nope, what formula are you using?

[old flame]

well there it is

[mild sky]

This is the identity

You are using

[old flame]

oh its the square not the half

boy am i dumb

[strange ermine]

Square root, not square

[old flame]

ykwim

[kind viper]

no, don't say the opposite of what you actually mean

[old flame]

alr

mb

[mild sky]

Accuracy is important

[kind viper]

if you must shorten it to one word then say root

[old flame]

i was doing it backwards in my head when i said it

mistake

anyway

.close

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[wispy warren]

please check if the range is correct

[viscid shard]

why is positive 1 excluded from the range?

[wispy warren]

because it doesnt include it from -x wherein x is between 0 and 1?

[balmy dawn]

It should be included.

[wispy warren]

I think I got it now

[viscid shard]

you have three cases which produce 1 though

[wispy warren]

so I should base on the cases that it does?

[viscid shard]

yes, because it is indeed possible for f(x) to be equal to 1

[wispy warren]

okay thank you

.close

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[wispy warren]

@viscid shard uhm wait is the domain correct tho does it affec

[wispy warren]

since theres a hole in 0 shoud I also restrict it in my domain?

[viscid shard]

!noping please

[compact pewter]

Please do not ping individual helpers unprompted.

[wispy warren]

I forgot to ask this

[viscid shard]

but let's see

[wispy warren]

oh sorry

[grizzled geyser]

$x \leq 7$?

[boreal girder]

Xwtek

[viscid shard]

I don't see anything wrong with the domain as written, but what's that at the top?

[wispy warren]

the scribbled part?

[viscid shard]

(as in the line pointed out by Xwtek)

[wispy warren]

thats 1

[grizzled geyser]

That still doesn't make sense since it overlaps with second, third, and fourth condition.

[viscid shard]

you mean **-**1?

[wispy warren]

oh sorry its -1

sorry its a mess when I write

[viscid shard]

it's fine. so what's the issue again?

[wispy warren]

my question basically is or does the restriction of range affect the domain or its independent of it

[viscid shard]

independent of it

I mean, you can have f(x) = 5 for all real x

the domain is still (-infty, infty)

[wispy warren]

oh okay thank you, and sorry for pinging u

.close

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[foggy wing]

2/x+3=-1/x

[chilly basalt]

$\frac{2}{x} + 3 = -\frac{1}{x}$

[boreal girder]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[chilly basalt]

or

$\frac{2}{x+3}= -\frac{1}{x}$

[boreal girder]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[viscid shard]

!original, please.

[compact pewter]

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.

[foggy wing]

This

[dense dagger]

You want to solve for x?

[foggy wing]

Yes

Teacher told us about proportion way but what if there's negative number😨

[dense dagger]

start by bringing all the x terms to numerators. So multiply both sides by x+3 to eliminate the x+3 in the denominator of the LHS. Repeat with x.

[foggy wing]

Ok i will figure it out

.close

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[old mica]

Let $M$ be a smooth manifold with or without boundary and $p$ be a point
of $M$. Let $\mathscr{I}_p$ denote the subspace of $C^\infty(M)$ consisting of smooth functions that vanish at $p$, and let $\mathscr{I}_p^2$ be the subspace of $\mathscr{I}_p$ spanned by functions of the form $fg$ for some $f, g \in \mathscr{I}_p$.
\begin{enumerate}
\item Show that $f \in \mathscr{I}_p^2$ if and only if in any smooth local coordinates, its first-order Taylor polynomial at $p$ is zero. (Because of this, a function $\mathscr{I}_p^2$ is said to \textbf{vanish to second order}.)
\item Define a map $\Phi: \mathscr{I}_p \to T_pM$ by setting $\Phi(f) = df_p$. Show that the restriction of $\Phi$ to $\mathscr{I}_p^2$ is zero, and that $\Phi$ descends to a vector space isomorphism from $\mathscr{I}_p/\mathscr{I}_p^2$ to $T_p^*M$.
\end{enumerate}

[boreal girder]

higher!

[old mica]

I'm trying to do part 1, the <- direction

i.e. first order Taylor vanishes -> f is in I_p^2

here's what I have so far

Conversely, suppose that the first order Taylor polynomial of $f$ at $p$ vanishes in any smooth chart $(U, \varphi)$ containing $p$, and use Taylor's theorem to write
\begin{align*}
\hspace{-20px} f(x) & = \underbrace{f(p) + \sum_{i = 1}^{n} \pdv{f}{x^i} \left(p\right) (x^i - p^i)}{= \ 0} + \underbrace{\sum{i, j = 1}^{n} (x^i - p^i)(x^j - p^j) \int_{0}^{1} (1 - t) \pdv[2]{f}{x^i}{x^j} \left(p + t(x - p)\right)}{R_1(x)} \ \
& = \sum{i = 1}^{n} (x^i - p^i) \sum_{j = 1}^{n} \left(x^j - p^j\right) \int_{0}^{1} (1 - t) \pdv[2]{f}{x^i}{x^j} \left(p + t(x - p)\right)
\end{align*}
for any $x \in \varphi(U)$. Now set [f_i(x) = (x^i - p^i), \quad g_i(x) = \sum_{j = 1}^{n} \left(x^j - p^j\right) \int_{0}^{1} (1 - t) \pdv[2]{f}{x^i}{x^j} \left(p + t(x - p)\right).] Then $f(x) = \sum_{i = 1}^{n} f_i(x)g_i(x)$, and both $f_i$ and $g_i$ vanish at $x = p$.

[boreal girder]

higher!

[old mica]

so now I'm stuck: I've obtained the result (f = sum of functions f_ig_i that vanish at p) within the chart U, but I don't know how to extend this to all of M

[sullen merlin]

Without thinking too hard ||partitions of unity go brrr?||

[old mica]

I thought so, but then f wouldn't equal the sum anymore

right?

oh wait

I was thinking of bump functions, but those would decay and it'd be doomed

but partitions of unity are always 1 everywhere

okay lemme try this

[sullen merlin]

In general whenever you have a construction locally you want to extend to a global one, in partitions of unity you trust

[old mica]

I dunno why I cast them out of my mind

ig I just assumed they wouldn't work without thinking harder

[sullen merlin]

||Also as an aside: This result you're proving is actually how cotangent and tangent spaces are defined for more abstract spaces||

[old mica]

yeah ISM has a remark on that

[sullen merlin]

Yeah the main benefit I got out of befriending a lot of DG people before taking it was a short cut on getting the useful intuition

[old mica]

apparently people define the cotangent space first as I_p/I_p^2

and then set the tangent space to be the dual

this sounds kinda cursed to me since idk anything other than the very basic smooth setting, but I imagine it's useful when people leave this little bubble

[sullen merlin]

very very useful in this case

[old mica]

hm, what sort of thing does this definition allow us to do, just out of curiosity?

you say we can define the co/tangent space on more abstract spaces with this

wdym by that?

[sullen merlin]

Its more natural when you're working say with schemes and want to define a cotangent and tangent space but you don't necessarily have a smooth space, (schemes allow singularities, e.g. a figure eight)

meanwhile in the algebraic geometric setting one does have ideals

and here J_p and J_p^2 are two ideals

so taking their ring theoretic quotient is very natural

[old mica]

interesting

I have much to learn, it seems

anyhow, thank you for the hint lance

I shall close this now

.close

[compact pewter]

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[sullen merlin]

don't we all

[sullen merlin]

nw, any time :)

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[short ferry]

<@&268886789983436800>

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[solemn matrix]

I needed help with this question for $x \in \mathbb R$. I tried letting $a = \sqrt{x}$ and $b = \sqrt{x - 1}$, this results in $(a + 1)(a + b) = \frac{9}{2}$. This is neat but the fractions are annoying to deal with. What can I try?

[boreal girder]

1 divided by 0 equals Infinity

[open plaza]

what other equation do you get

[solemn matrix]

a^2 + a + b + ab = 9/2
a^2 + ab + a + b = 9/2
a(a + b) + (a + b) = 9/2
(a + 1)(a + b) = 9/2

the most annoying thing to deal with is the 9/2

somehow, the calculator can't solve this equation too

[open plaza]

^

[solemn matrix]

$b = \sqrt{a^2 - 1}$ and if you want other equations, i can't really get anything else that makes a difference

[boreal girder]

1 divided by 0 equals Infinity

[open plaza]

in other words a^2 - b^2 = 1

now multiply 1 to the right hand side (yes it is a legitimate hint)

[solemn matrix]

i see

[solemn matrix]

i see where this is going

a = -b is one possible case

a^2 + a + b + ab = 9/2
a^2 + ab + a + b = 9/2(a^2) + 9/2(b^2)
(a + 1)(a + b) = 9/2(a - b)(a + b)

?

isn't a + b = 0 then a = -b?

[boreal dawn]

That would contradict a^2-b^2=1

[solemn matrix]

i'll handle that later

thanks for reminding me of that

so $a + 1 = \frac92(a - b)$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

$a < b$ $\forall a, b \geq 0$

$\Leftrightarrow a - b < 0$ $\forall a, b \geq 0$

$\Leftrightarrow RHS = \frac92(a - b) < 0$ $\forall a, b \geq 0$

but $LHS = a + 1 \geq 1$ $\forall a, b \geq 0$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

so this is not possible

[open plaza]

√x <= √x-1 wa lao eh

[solemn matrix]

<

but you get my point

[boreal girder]

1 divided by 0 equals Infinity

[open plaza]

you just made a big mistake there lmao

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

lmao

thanks for reminding me of that lol

[boreal girder]

1 divided by 0 equals Infinity

[open plaza]

what is this??

[solemn matrix]

...

im going to think on this question

[open plaza]

is this for some olympiad stuff

[solemn matrix]

toan chuyen

[open plaza]

chuyên rách

ko chơi

[solemn matrix]

homework

[open plaza]

trường rách

[solemn matrix]

a = sqrt(x - 1)
b = sqrt(x + 1)
=> b^2 - a^2 = 2

(b^2 - a^2)(a + b) = 9a^3
(a + b)^2(b - a) = 9a^3

[boreal dawn]

Solve for b

[solemn matrix]

something like this

[solemn matrix]

inequality proves should do the trick

[open plaza]

dont you realize your mistake here?

[solemn matrix]

i saw that

√x > √x-1, not <=

[open plaza]

đòi học chuyên

tuah

[solemn matrix]

ko thi chuyen toan

hell nah

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[grizzled elk]

so this might be bad, but i only know the first step of this

so take a point 0,0 , transforms to 0 0

then lets take the left most point -2,2

transforms to 1,3

then 2,2 transforms to-2,2

[compact pewter]

@grizzled elk Has your question been resolved?

[compact pewter]

@grizzled elk Has your question been resolved?

[rotund sphinx]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/05%3A_Linear_Transformations/5.02%3A_The_Matrix_of_a_Linear_Transformation_I

This means that you should figure out how T acts on $(1,0)$ and $(0,1)$

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

problem is that you're not given those

however, recall that because $T$ is linear, $T(\alpha \mathbf{u}+\beta \mathbf{v})=\alpha T(\mathbf{u})+\beta T(\mathbf{v})$

[boreal girder]

Civil Service Pigeon

[rotund sphinx]

think about how you can use that combined with these pieces of info to find $T(e_1)$ and $T(e_2)$

[boreal girder]

Civil Service Pigeon

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[west pulsar]

Hey guys, I’m working on problem one right now

[wide dune]

what've you done so far?

[west pulsar]

Literally nothing

Would the first step be getting rid of the parentheses?

Idk my teacher didn’t tell us what do when there’s parentheses

We’re just finding asymptotes

[wide dune]

yes, try distributing

[west pulsar]

Just start with that?

Ok well
I have 24x^4 -5x^2-36

What now

<@&286206848099549185>

[shadow abyss]

which question

[west pulsar]

1

I’m finding asymptotes

[shadow abyss]

tbh couldnt you just square root the entire thing

[west pulsar]

What

Why would you

[shadow abyss]

what is an asymptotes

[shadow abyss]

less exponents

[west pulsar]

No

[shadow abyss]

what is an asymptote

[wide dune]

did you distribute the denominator?

[west pulsar]

No but

Is that

Necessary?

[wide dune]

dude you gotta just try random shit sometimes and see what happens

you can't know it's necessary until you actually try to solve the problem

[west pulsar]

Yeah ig but like

Blech idk we never had to do that in class but at the same time it’s like

We never did it in class

[wide dune]

just distribute the denominator for now and see what happens

[west pulsar]

Ok what now

[wide dune]

now you have an x^4 term in the numerator and denominator

[west pulsar]

So

Multiply by 1/x^4?

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[proper forum]

I need help

[proper forum]

I'm not sure how to solve this problem

It seems easy, but the word "rotate" is confusing me

[proven wren]

that's a convoluted way to ask for intersection of two line segments

[proper forum]

um..?

what is convoluted?

🙁

[wide dune]

you have 2 line segments AB and A'B', the point where they intersect is the axis about which they rotate

[proper forum]

ohhh

Do they intersect?

[proper forum]

I surely know that, but.. I don't think they intersect

[open plaza]

incorrect in this case

[wide dune]

uhh look at the full line whoops

[proper forum]

what is full line whoop

actually the two lines don't intersect each other at all

[open plaza]

clearly the intersection isnt the center of rotation

[proper forum]

oh

if we prolong the line

I mean expand

[proper forum]

wait, then is that the rotating point?

[open plaza]

there are 2 points that satisfy this

[proper forum]

wat are they>.?

[open plaza]

1. The intersection of the perpendicular bisectors of AA' and BB'
2. The intersection of the perpendicular bisectors of AB' and BA'

[proper forum]

ohh

then what is the right point?

[open plaza]

both points should be right

[open plaza]

so this problem has 2 correct answers

[proper forum]

no

see the question

[open plaza]

I checked the coordinates of the correct points

and found both of them are included in the options

[proper forum]

wtf

[open plaza]

I don't know if the question specifically wants A -> A' and B -> B' after rotation

[compact pewter]

@proper forum Has your question been resolved?

[proper forum]

I will just see the solution

gg

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[slate socket]

I have a math test on 25th this month, can anybody guide me how to study and some tips as well please, I can let you know my syllabus

[dusky nymph]

do practice exams if you can find any

[slate socket]

i do have but how? with the solution right below it?

[dusky nymph]

cover up the solution, do the problem on separate paper

[wanton ferry]

please write something specific

[dusky nymph]

try to do it in the same conditions (closed book, calculator or not, etc) as the real exam

and under the same time limits

[slate socket]

Oh

[slate socket]

Wdym specific?

[wanton ferry]

like the syllabus

[wide dune]

which class is the exam even for?

[slate socket]

this is the syllabus

[wide dune]

if you need study tips in general, this isn't really the place for it

[slate socket]

grade 10

[slate socket]

yeah I thought about it

[slate socket]

can you see the syllabus dude?

[wanton ferry]

yeah

and i still dont know how to help you review 20 subjects

[slate socket]

cool

maybe tell me about few

geometry part mainly

Similarity, Circles, Section Formula, Equation of a Straight Line

[slate socket]

mango can you help me now?

[wanton ferry]

be specificccccccccccc

[slate socket]

these topics: Similarity, Circles, Section Formula, Equation of a Straight Line

[wanton ferry]

thats still too generalllllllllllllllllll

[slate socket]

wait imma let you know more

[slate socket]

if i let you know my chapter's summary will that be good enough??

obv not all

just few

[wanton ferry]

maybe just lemme help you on one topic and open a new channel for another topic

[slate socket]

Sure

[slate socket]

Topic: Circles

Course in the attachment, obviously you don't have to read everything just read the bold headings ig that can you make you understand on whats coming

[wanton ferry]

can you hang in there i gotta help another channel xd

[slate socket]

just one topic as you asked

[wanton ferry]

v quick

[slate socket]

Yes sure

no problem, I'm glad that you are helping me

[wanton ferry]

bad news: you will have to memorise nearly everything here

but theres still intuition to help

yeah this is so dense

exactly which parts do you not understand

[slate socket]

The properties or theorems where to apply on which question and stuff

[slate socket]

I can do that, any tips on how to?

how should I or would you study this chapter (for the first time)?

or maybe even suggest me a youtube video on how to study math, if you have or think that can help

[wanton ferry]

so if you see angles whose vertexes lie on the circle, theres a high chance you should use one of the angle properties. if you see a quadrilaterall whose angles satisfy those conditions you should draw a circle that passes through all points. if you see two chords passing through a single point you should use the power of the point theorems. if you see a single tangent, you can probably draw a line from the center to the point of intersection. if you see two tangents passing through a point, you should know that they are equal

now, this may seem that im restating the theorems but the key takaway is that if you see any of these things theres a very high chance that the theorems will work

also sorry, this is still very general for us to help

[slate socket]

Yeah sure I totally get it

Thank you so much

[wanton ferry]

and the very important thing is to do problems

and also

reverse thinking

like, when you see the thing you prove, think what should you prove in order to prove that. you can also rewrite a strange answer using similarity identities to make it less weird

algebraically

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[compact pewter]

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[sand magnet]

two second order ODEs (dx^2+a(x)dx+c(x))psi=0 y (dx^2+d(x))psi=0 can have the exact same set of solutions?

[noble sorrel]

can u send image of the question for clearer understanding

[gaunt lynx]

Do you wanna prove this or you want the answer for this?

[sand magnet]

I want to prove it

okay, I send the picture, one moment

I mean, I want the answer and a proof of why yes/no

[gaunt lynx]

Got it

[gaunt lynx]

Yeah pls do

[sand magnet]

these* equations

ah psi depends on x

[hushed magnet]

if a(x)=0, c(x)=d(x), sure

[sand magnet]

well yes

but not doing that

I cannot do a=0

[hushed magnet]

well psi could satisfy (a d/dx+c)psi=d(x)*psi

[sand magnet]

yep

[hushed magnet]

which is just another ode which could have a solution

[sand magnet]

I've arrived to a compatibility condition, yeah

but this does just mean the intersection is not the vacuum

[gaunt lynx]

For them to have exact same set of solutions then the fundamental solution of one equation must also be the solution for other, so ig you gotta so psi1 and psi2 and find solution

[sand magnet]

of the solutions

[gaunt lynx]

You cannot do a(x) = 0 but you can prove it right ?

[sand magnet]

what do you mean?

My theory forbids a(x) from being zero

physical theory*

[gaunt lynx]

.

[sand magnet]

Yes, that's the proof I would like to have

[gaunt lynx]

If you find solutions for that, you’ll eventually end up having a(x) = 0 if the answer is yes

You’ll have to assume psi1 and psi2 for fundamental sets and then find solution for both the conditions

[sand magnet]

can you prove it by first solving a solution and then the other?

[gaunt lynx]

Yeah maybe you can

Try it the way you think, and if doesn’t work, try it the way i suggested

[compact pewter]

@sand magnet Has your question been resolved?

[compact pewter]

@sand magnet Has your question been resolved?

[sand magnet]

yep but I'm not able to prove it

[rotund sphinx]

No. From the second equation, we have that
$$\dv[2]{\psi}{x}=-d(x)\psi$$
Substitute this into the first equation:
$$-d(x)\psi+a(x) \dv{\psi}{x}+c(x) \psi=0$$
$$a(x) \dv{\psi}{x}+(c(x)-d(x)) \psi=0$$
$\psi$ can be described as a linear combination of two linearly independent basis solutions $\psi_1$ and $\psi_2$. Thus, $\psi_1$ and $\psi_2$ are solutions to the same first-order ODE, which is impossible as they must be linearly independent.

[boreal girder]

Civil Service Pigeon

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@sand magnet Has your question been resolved?

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[sand magnet]

thankss

[compact pewter]

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[fallen frost]

Calculus -- In the following type of limit, does dt have the same value as h?

[mild sky]

dt is not a number

its notation that is part of the integral

[fallen frost]

hmm, my current impression is that this is basically 'f(t) * h', which amounts to the area of the rectangle sliver being integrated.

[kind viper]

dt is not any sort of number

it is just a notational device

which NOTIONALLY stands for the width of a tiny increment on the t-axis and is sometimes illustrated as such

however it must not be taken literally

[fallen frost]

This comment serves enough for my purpose. Thank you.

.close

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[honest merlin]

im so confused whats happening in part b please help

what's the formula for curvature?

[kind crane]

https://en.wikipedia.org/wiki/Curvature

[honest merlin]

okay I guesssssss I'll read into it

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[paper tendon]

hi, i have an exercise:

How many triangles can be created if each side can be one of the following lengths: 4, 5, 6, 7, 8, 9 (+ no two triangles are congruent)

My logic is

The lengths can be used multiple times and the order doesnt matter (467 is same as 476), so we use combinations with repetition + there are 2 combinations from which triangles cant be created (4,4,9 and 4,5,9)

My result: (8 choose 3) - 2

Is my logic correct ?

[weak cobalt]

What about 4,4,8

[paper tendon]

Ye forgot that one

So 3 combinations

[weak cobalt]

Not sure how (8 choose 3) answers that

[fluid stag]

Is 4,4,8 considered a triangle?

[paper tendon]

no thats not a valid triangle

[weak cobalt]

yeah I think you're correct

[remote current]

I don't think the choose function is correct if you allow repeat lengths

maybe I am wrong though

[paper tendon]

i just used the equation for combinations with repetitions -> n+k-1 choose k

[weak cobalt]

Right, right, 5+3 choose 3

[remote current]

ah mb, I think you are right

[paper tendon]

ye results say its correct thx all

.close

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[paper tendon]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[paper tendon]

if someone has more intuitive way to think about this, let me know, i just used the equation but im not really sure why it works, maybe i just need to go over the logic of the combinations but not sure

[weak cobalt]

It's about putting things in boxes

You have 6 lengths, those are your boxes

[paper tendon]

ye i know the boxes and divisors logic

[weak cobalt]

You must select 3, with repetitions, so you're putting 3 things in those 6 boxes

[paper tendon]

ill just go over it again

[weak cobalt]

That's 5 dividers + 3 things, so 8 things total, and you need to choose 3

[paper tendon]

ye this makes sense, maybe im just struggling with transforming it into the combination number

[weak cobalt]

You can also consider cases: 6 equilateral triangles, 6*5 isosceles triangles (the 6 is for the length of two sides, the 5 for the third length), and 6*5*4 / 3! other triangles (/3! to remove duplicates)

[paper tendon]

ye makes sense

why cant it be just 6 * 5 * 4

oh

nvm i see

[weak cobalt]

lengths:  4 | 5 | 6 | 7 | 8 | 9
example:    | x |   | xx|   |       (5,7,7)
5 bars, 3 "stars" -> 8 things, choose 3 to be the stars (or choose 5 to be the bars)

xxx||||| would be choosing the first 3, x||x|||x would be choosing the 1st, 4th, and 8th

Once you can visualize examples, it should be clear how to get the formula

[paper tendon]

ok, thx a lot

i will post couple more exercises to verify my logic if u dont mind, if someone else wants to correct me im down, if not i will just close this channel and look at the results

a library has 5 shelves and each can hold 20 books, how many ways can we place 20 books

my logic

there is total 100 places at the beggining to place a book so my result would be: V(20, 100) = 100*99*98*...*81

[strong crest]

I don’t think this is what theyre looking for unfortunately

[paper tendon]

ye seemed too good to be true, the results say 24!/4!

[strong crest]

Right, well it’s a slightly ambiguous question but

[paper tendon]

there was also a hint that i forgot to mention: "you can think of shelves next to each other where each two shelves are divided by same item"

[strong crest]

Sure, I don’t think (I might be wrong) they want you to think of each shelf as something with 20 slots

[paper tendon]

so its permutations with repetitions where each book is different and we have to divide by all possible permutations of divisors

[strong crest]

I think they want you to just think of the shelf as something you can put up to 20 books on, so the real problem is how many ways can we distribute the 20 books among 5 shelbes

[paper tendon]

is my guess

[strong crest]

In other words, any 5 numbers that add up to 20

But it also depends on whether the books are identical or not

[paper tendon]

probably not

[strong crest]

Mmm well if they are distinct then the answer is kind of easy to calculate but it’s not 24!/4!

[paper tendon]

we can probably think about it like BBBBBBBBBBBBBBBBBBBBDDDD where B is a book and D a divisor

[strong crest]

Yes so the books are all alike

And you’re right yes

[paper tendon]

oh ye true they are the same actually

[strong crest]

If they were distinct the answer would be

5^20

[paper tendon]

how come

variations make more sense to me

[strong crest]

There are 20 books, each one you have 5 shelves to choose from

[paper tendon]

without repetition

[strong crest]

So itll be 5x5x5…..20 times

But that’s only if they’re distinct

[paper tendon]

ok i guess it makes sense

thx

[strong crest]

Have you done stars and bars?

[paper tendon]

ye, Nel mentioned it 20 messages above

[strong crest]

You might notice it’s a stars and bars problem

[paper tendon]

yep makes sense

another exercise:

a selfcheckout cofee shop has 4 types of coffee, each has 50 grams. how many combinations of 250g of coffee can we buy
a) if there is enough coffee of all types

my logic

stars and bars again, 3 divisors (bars) and 5 coffees (stars), so result 8!/(3!*5!)

b) if 2 types of coffee have 10 packs of coffee each and rest of the coffee types only have 4 packs

my logic

we take result from a) and decrement 2 for the two coffee combos where we cant create 5 of the same type

both are actually correct, these were easier for me so i just checked results

thx all

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[rotund ether]

Determine all positive integers n such that there exists a positive integer m with the property that n is divisible by m and n² + m² is divisible by m²n + 1.

hello can someone help me with this ?

[open girder]

What have you tried so far?

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[rotund ether]

wait sorry ill ask again later thank uu

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[trail yew]

What is the definition of a definition in geometry as in definition of angle bisector

[grizzled geyser]

[trail yew]

What don’t you understand about your question

My*

I just realized I mistyped

It should make sense now

[kind crane]

what is the definition of definition?

[grizzled geyser]

In mathematics, a definition is used to give a precise meaning to a new term, by describing a condition which unambiguously qualifies what the mathematical term is and is not. Definitions and axioms form the basis on which all of modern mathematics is to be constructed
According to Wikipedia

[trail yew]

Oh ok

I have a few more questions if you don’t mind be asking

Hello are u still here?

[kind crane]

you don't need to ask if they're here. it's faster if you just ask

[trail yew]

Ok

[kind crane]

for this one: there's no widely accepted convention. you just figure it out from context

not really enough context to answer this one:

and the rest are just done with practice. do more actual problems

[trail yew]

There’s the problem where I had the question

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[soft sundial]

Proof that angular displacement AOT = -OAT

Hey! I’m working on an engineering project with angular displacements - it’s basic trig with the caveat that I need angles to preserve their sign as they’re displacements and i’m using a polar coordinate system.

Can anyone perhaps give me a few slight hints as to how I could elegantly prove the relationship I demonstrate in this picture? I think this is a great opportunity for me to practice my proofs so the vaguer you can start off the more awesome 😁 thanks!

[soft sundial]

I could use the coordinates of A and T but I feel like I should be able to do it just from this 🤔

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[soft sundial]

Basically if we “pivot” at O and sweep from A to T the “angle” is negative because that’s a clockwise sweep which is a negative angle vector in a polar space :)

But if we pivot at A and sweep from O to T the angle vector is counterclockwise, which is positive in polar coordinates

Unfortunately the classic engineer’s vibes-based proof aint gonna cut it 😭 i’d like to prove this mathematically

[soft sundial]

.solved

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[stiff flume]

not sure how to simplify the derivative in terms of dx and dy

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[tall pecan]

hi

you there

first you have to differentiate the y = wx to make a differential equation

so that you can substitute parts of the differential equation given

then you will get a short equation, which has to be integrated

then with the point A(e^2, 0) you will get the value of the constant

then finally put the value of constant

@stiff flume

[stiff flume]

hm

[tall pecan]

don't have to integrate the function, proceed with differentiating y = wx

you seeing?

hey ping me when u r free

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[rocky berry]

How would I go about finding if something is differentiable on an open interval

[rocky berry]

Do I derive the function then set it equal to 0 to solve for x then put the numbers of the interval back into the original function to see if it is equal to what I got for x

[stark marsh]

It’s is differentiable if the left limit of the derivative of the function is equal to the right limit of the derivative of that function

[rocky berry]

So like how does this show that

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[rocky berry]

Is it where they solve for x on the derivative and find it equal to -3

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[leaden leaf]

Whats the best way to cram for PSAT math?

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[flint cliff]

Would the top view or plan view of the screw be a circle with diameter 23mm

[hard isle]

.

need help?

[steep crypt]

Me

[hard isle]

say

what q

[steep crypt]

What q?

[upper bane]

!occupied, sorry

[compact pewter]

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

[steep crypt]

What di you mean

[hard isle]

oh

i meant what question

this is occupied

🥀

[upper bane]

if you want to ask a question, please use an available channel, like #help-29

don't use a channel with a | and a person's name in the title

[hard isle]

#help-1 #help-3 #help-4 #help-5 #help-12

ping me

@steep crypt

[steep crypt]

Yo

[strange ermine]

!noping

[compact pewter]

Please do not ping individual helpers unprompted.

[upper bane]

don't troll in help channels please.

[hard isle]

vro im asking him to like choose a channel

[upper bane]

and you didn't notice OP has already opened a channel?

[hard isle]

um no

[upper bane]

plus, there are already people helping OP

now, let's stop flooding this channel

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[flint cliff]

.close

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[tall lantern]

Can someone help me?

<@&286206848099549185>

[kind crane]

read #❓how-to-get-help

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[noble copper]

does PA.PB mean PA * PB

[noble copper]

?

[thorny urchin]

yes

[noble copper]

oh

[fossil finch]

ig yes

[noble copper]

ok

[strange ermine]

!done

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[noble copper]

!done

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[noble copper]

.close

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[versed wraith]

.close

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[sacred wave]

Hello I need help

[sacred wave]

I’m trying to calculate the effective rent on a loan but i keeps getting 0%

Its in Danish so it hard to understand

<@&286206848099549185>

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[glossy sentinel]

Question: If A is an n × n matrix with entries in R satisfying (A^T)A = 2In, show that A is
always invertible.

Do I multiply each side by 1/2 to isolate the In matrix so 1/2(A^T)A = In. After this I am lost.

[open girder]

You can just analyse the determinant

[glossy sentinel]

The determinant of 2In? or what? I'm not given the A matrix.

[open girder]

(A^T)A = 2I_n ==> det((A^T)A) = det(2I_n)

[open girder]

You should know what the determinant of 2I_n is.

[glossy sentinel]

ok, ty. I got it.

.close

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[raw dawn]

This look correct

[dusty harbor]

why you got it in blue light

but yeah looks good i think, id have to solve it to really know

[raw dawn]

Too bright for my regular light so I used my led’s

[raw dawn]

Would you

preferably using the same methods I did

I had a teacher solve it but she got a slightly dif answer that idk if it’s equivalent

[dusty harbor]

yeah hang on

it looks like you messed up somewhere

gimme a sec to finish my q rq

[raw dawn]

Alr thanks

[dusty harbor]

wrong

place

sorry

[raw dawn]

lol

[raw dawn]

I remember ts 🥹

What a throwback

[dusty harbor]

can you help me w it then

[raw dawn]

Lemme try

it’s been a min

[dusty harbor]

why use trig sub

[raw dawn]

Like it hetter

Also didn’t think of the u sub

lol

[dusty harbor]

[raw dawn]

What lol

Bro don’t mess with trig sub

It’s my favorite

[dusty harbor]

yeah

thats crazy

ive not done trig sub since i had to 😭

im struggling over here

[raw dawn]

Do you want my answer

@dusty harbor

[dusty harbor]

yeah

[raw dawn]

I split it up like CRAZY

If u want my work

lol

30π

[dusty harbor]

hm

yeah show me

please

[raw dawn]

I did the u sub for cos2theta but js didn’t show it, it equals 0 anyway

[raw dawn]

After doing mad problems w it it gets kinda fun

[dusty harbor]

ive not a clue what i did wrong

but thank you abhsbshah

[raw dawn]

U subtracted

Not multiplied

[dusty harbor]

also sorry to say, i cannot help you

atl w trig sub

forgot that long ago

[raw dawn]

lol

I try to stay updated

[dusty harbor]

thats cuz i got 15[theta - cos u +sin u]

[raw dawn]

I haven’t been w calc 3 lol

[dusty harbor]

from 2pi to pi

[raw dawn]

Erm idk

I lowk js split it up a shit ton

[dusty harbor]

its easy peasy youll do great

[raw dawn]

I alr took it lol

[dusty harbor]

if you take it anyway

oh

[raw dawn]

I mean keeping up with it

[dusty harbor]

ohhhh

type

for why

[raw dawn]

If you gave me a spherical triple integral idk if I’d be able to solve it lol

[dusty harbor]

you in uni or no?

[raw dawn]

Nah

Senior yr hs

[dusty harbor]

damn

another child prodigy

sick of this world

[raw dawn]

Why

Wbu*

[raw dawn]

I’m not that good lol

[dusty harbor]

2nd year of uni

[raw dawn]

my friend who’s the same age is a soph in college rn

😭

[dusty harbor]

bro some 14 year old was helping me w this not too long ago 😭

i cant handle this no more

[raw dawn]

What’s ur major

It’s also def a lil easier to get ahead in hs

[dusty harbor]

biochemistry

last year of math 🙏

[raw dawn]

I went from alg 2 (the end) to calc 3 and Lin alg (the beginning) in a year

[raw dawn]

U need calc 3? 😿

[dusty harbor]

yeah but honestly its not that bad

im just brain dead

and i think also linalg

but thats cake

[raw dawn]

Yea I took ts in 2 weeks and it was easy asl

It was in my calc 3 over the summer which was a 6 week course

[dusty harbor]

yeah kudos

6 week courses are brutal

you're dual enrolled?

[raw dawn]

It was fun ash lowk

Nah

Not offered at my school

[dusty harbor]

interesting

[raw dawn]

I’m ahead of evb else