help-13

[shut wadi]

all the other sas angles I've ever encountered look like this

where the 2 sides are surrounding the angle

but in this instance

only one side is near the angle

if it was sas

criteria

shouldn't the bottom angle be shown as congruent

bottom side*

and not the left side

of the triangle

BRUH

[hollow trail]

you're right, actually

[shut wadi]

you guys wer wrong

😭

I already sumbitted the homeowrk

rip

it is ok

[shut wadi]

?

[hollow trail]

the issue is that you could also draw it like this, and geometry diagrams are under no obligation to look right

[shut wadi]

yes

because we don't know the length of bottom side

and we don't know the angle at the top

so the left side can swing over

[cedar kiln]

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[oblique tundra]

(New account. Old one hacked))

Ive learned a thing or two about linear equations and linear equation systems..

But..

How do i do inequalities and ratio/proportions?.

[oblique tundra]

Questions: 3, 5, 6 to be specific

Ive heard we treat the <, and >. Same as the = in linear equations

Does that mean i solve lets say..

A) like this?

X + 3 < -x + 7.
. - 3. .- 3

X < -x + 4....wait..im stuck

...uhhhh

Yeah i dunno what to do tbh

Any tips?

Hmmm

X < -x + 4.

I think this task is done?.

I know i gotta display it in a graph, if -x + 4 is bigger than X.

X = 1.

-1/0/1/</2/3/4_-

((Weird graph i know-))

[oblique tundra]

<@&286206848099549185> help would be appreciated, i just wanna learn for the big test tomorrow.

[cedar kiln]

@oblique tundra Has your question been resolved?

[oblique tundra]

Anyone?

...i guess i gotta close then

.close

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[austere pawn]

the < and > are pretty much the same, but you need to be careful when doing operations on both sides, which can change the symbol

[austere pawn]

so like:

a) x+3 < -x +7
x+3+(x-3) < -x+7 +(x-3)
2x < 4

x < 2

lets say we have like: $-x < -8$

[wraith dagger]

clonesolopros

[austere pawn]

then we cant say: $x < 8$; by multiplying by -1

[wraith dagger]

clonesolopros

[austere pawn]

you have to invert the symbol when multiplying by a negative number

a.k.a: $x > 8$; correct solution

@oblique tundra

.close

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[spiral marsh]

hi

[cedar kiln]

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[wicked path]

how do i find the area using horizontal elements?

[cedar kiln]

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[coarse pulsar]

Can someone help me with this function question?

[floral salmon]

supposing that $h^{-1}$ means inverse function, and that it is defined for $h$, then $hh^{-1}(x)=x$ by definition.

[wraith dagger]

Crystopher

[floral salmon]

now, I am not sure what they mean by $h(p-2)^2$ is it $h(h(p-2))$, $[h(p-2)]^2$ or $h((p-2)^2)$. I am guessing it is the first alternative but not sure tbh.

[wraith dagger]

Crystopher

[coarse pulsar]

This is the solution apparently

[floral salmon]

yes, makes sense, they meant $h((p-2)^2)$

[wraith dagger]

Crystopher

[floral salmon]

are there any doubts about the solution?

[coarse pulsar]

Doubts?

[floral salmon]

yes like, you posted a question in this channel, do you have any questions or anything? Do you need some clarification?

[coarse pulsar]

I need clarification

[floral salmon]

what do you need clarification about?

[coarse pulsar]

I don't understand the second line of the answer

[floral salmon]

This one?

[coarse pulsar]

Yes

[floral salmon]

On the left side they used the property I talked about inverse functions. The first step would be to prove that the inverse $h^{-1}$ exists, but it is implicitly given it exists.

[wraith dagger]

Crystopher

[floral salmon]

This follows from the definition of inverse function

in our case $f$ can be $h$ and $g$ can be $h^{-1}$

[wraith dagger]

Crystopher

[floral salmon]

so it says in this case that for all $x$ it holds that $h(h^{-1}(y))= y$ for all points $(x,y)$ on the line given by $h(x)=4x+5$

[wraith dagger]

Crystopher

[floral salmon]

meaning that $h(h^{-1}(4p^2+9)) = 4p^2+9$

[wraith dagger]

Crystopher

[floral salmon]

on the right side they evaluated $h(x)$ on $x = (p-2)^2$. That is to say, in $h(x) = 4x+5$ they replaced $x$ by $(p-2)^2$, giving $h((p-2)^2)=4[(p-2)^2]+5 = 4(p-2)^2+5$

[wraith dagger]

Crystopher

[coarse pulsar]

Can you explain more about the inverse function because there are symbols I have forgotten the meaning of

[floral salmon]

You can see inverse functions as functions that 'restore' the input from a function. One way to visualize them is like in this image

you have $f$, it 'pairs up' the inputs $a,b,c$ as follows: $(a,1), (b,2), (c,3)$

[wraith dagger]

Crystopher

[floral salmon]

$(a,1)$ can be read as "the function gives 1 if $a$ is 'fed' into it"

[wraith dagger]

Crystopher

[floral salmon]

the inverse $f^{-1}$ does the opposite. It pairs up: $(1,a), (2,b), (3,c)$.
Here $(1,a)$ can be read in this context as "$1$ is the output of $f$ when $a$ ins 'fed' into it", or some equivalent formulation.

[wraith dagger]

Crystopher

[floral salmon]

In this case we have $h(h^{-1}(4p^2+9))$, which could then be read as "The output of $h$ when feeding it the input $x$ which gives $4p^2+9$", which is a convoluted way of saying just $4p^2+9$

[wraith dagger]

Crystopher

[floral salmon]

For a function to have an inverse it needs to fulfill two properties, it has to be injective and it has to be surjective.

[coarse pulsar]

Injective and surjective?

[floral salmon]

They can be somewhat complicate to understand. But imagine an injective function as one where each output is only given by a specific $x$, that is to say, there are no two different values $x$ that give the same output

[wraith dagger]

Crystopher

[floral salmon]

Formally this is written as: $f(x_1)=f(x_2)\Rightarrow x_1 = x_2$

[wraith dagger]

Crystopher

[floral salmon]

or some equivalent formulation

[coarse pulsar]

And surjective?

[floral salmon]

Surjectivity means that for each possible output $y$ there is at least an $x$ such that $f(x)=y$

[wraith dagger]

Crystopher

[coarse pulsar]

Okay

I think I still don't understand injective

[floral salmon]

Alright, lets give a concrete example

you see, the first one is injective since each output $y$ ($1,2,3,4$) is given by at most a single $x$ ($a,b,c,d$)

[coarse pulsar]

So, no same codomain for two domains?

[wraith dagger]

Crystopher

[floral salmon]

but on the upper right we have the pairings $(c,3)$ and $(d,3)$, meaning that there are 2 different inputs $x$ ($c$ and $d$) such that $f(x)=3$

[wraith dagger]

Crystopher

[floral salmon]

A common example of a non-injective function is $f(x)=x^2$

[wraith dagger]

Crystopher

[floral salmon]

for any $y>0$ you see there are two different $x$ values that give $f(x)=y$, these are $x=\sqrt{y}$ and $x=-\sqrt{y}$

[wraith dagger]

Crystopher

[floral salmon]

Domain refers to the 'set of inputs' of the function, the codomain is the 'set of the possible outputs', two different domains may as well have the same codomain.

[coarse pulsar]

Okay

[floral salmon]

for instance $f(x)=x^2$ has the domain $\mathbb R$ and codomain $\mathbb R^+$ and $g(x)=\sqrt{x}$ has domain $\mathbb R^+$ and codomain $\mathbb R^+$

[wraith dagger]

Crystopher

[coarse pulsar]

If all the info about injective you said is that's all, then I think I got it

[floral salmon]

alright, injectivity and surjectivity together give bijectivity

which is a must have for a function $f$ to have an inverse $f^{-1}$

[wraith dagger]

Crystopher

[floral salmon]

In short, bijectivity implies that for each possible output $y$ there is an unique input $x$ such that $f(x)=y$

[wraith dagger]

Crystopher

[coarse pulsar]

Okay

[floral salmon]

the function in the problem, $h(x)=4x+5$ is intuitively bijective.

[wraith dagger]

Crystopher

[coarse pulsar]

Is there a way to instantly know if a function is bijective?

[floral salmon]

to prove that a function is bijective imples proving that it is both injective and surjective. There are certain types of functions that can easily be proven bijective.

For instance strict monotone functions (functions that only either decrease or increase)

like a linear function

such as $h(x) = 4x+5$

[wraith dagger]

Crystopher

[floral salmon]

but for more complex functions it may not be as straightforward. And then there may be cases where a part of the function can be invertible. and other parts are not.

[coarse pulsar]

Okay

[floral salmon]

if you look at a graph you can put a ruler parallel to the x-axis and go upwards, if two points intersect the ruler you know it is not bijective in the interval shown on the graph (since it is not injective).

This is only valid for real-valued functions of course.

[coarse pulsar]

So, can we go back to the second line of the answer?

[floral salmon]

sure, are you wondering about the right side?

[coarse pulsar]

Yeah, and how the left side lost h and h-¹

[floral salmon]

the left side lost the $h$ and $h^{-1}$ since they are inverse of each other, as I told you.

[wraith dagger]

Crystopher

[coarse pulsar]

That made sense

[floral salmon]

so they 'cancel out' each other

[coarse pulsar]

Okay

Now the left side

[floral salmon]

you mean right side?

[coarse pulsar]

Yeah, my bad

[floral salmon]

ok, apparently $h(p-2)^2$ here actually means $h((p-2)^2)$

[wraith dagger]

Crystopher

[floral salmon]

what they did there is just evaluate $h$ on $x=(p-2)^2$

[wraith dagger]

Crystopher

[floral salmon]

very similar as when finding $h(5)$, you just replace $x$ by $5$ in $h(x) = 4x+5$, giving $4\cdot 5 + 5 = 25$

[wraith dagger]

Crystopher

[floral salmon]

in this case you just replace $x$ by $(p-2)^2$ in $h(x) = 4x+5$, giving $h((p-2)^2) = 4(p-2)^2 + 5$

[wraith dagger]

Crystopher

[floral salmon]

in other words, you set $(p-2)^2$ as the input of $h$ and work out its output

[wraith dagger]

Crystopher

[coarse pulsar]

Okay, I understand now. Thank you so much

.close

[cedar kiln]

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[crimson sedge]

Gallileo Gallilei stated that motion in mutually perpendicular directions are independent of each other.\
$v_y$ will denote velocity in y direction and $v_x$ will denote velocity in x direction.\
intially a particle has $v_y = 2.5\cross 10^8$ and $v_x=0$. by an external force, i manage to increase $v_x$ to $2.5\cross 10^8$.\
now, the magnitude of both the velocities will become $\sqrt{12.5}\cross 10^8$ which is more than the speed of light. \
what is wrong here?

[wraith dagger]

deltaG

[crimson sedge]

my explanation: since there's already $v_y = 2.5\cross 10^8$, i'll require infinite amount of energy to increase $v_x$ such that the magnitude just reaches the speed of light. which implies that there has to be some sort of "threshold velocity" for x direction.

[wraith dagger]

deltaG

[crimson sedge]

but i dont really like this explanation

[mental trail]

if you do manage to get v_x to 2.5*10^8, what's to say v_y stays the same

[crimson sedge]

Gallileo's state says v_y remains the same

statement*

[mental trail]

well it's just the limit of classical physics

[crimson sedge]

since any acceleration in x direction wont affect velocity of y

[mental trail]

that you can't "add" two velocities together

[crimson sedge]

sorry i didnt understand you here

[cedar kiln]

@crimson sedge Has your question been resolved?

[mental trail]

in classical physics for example, if a train moves at 30m/s and looking inside the train, a person moves at 2m/s, then we can say that outside the train, the person moves at 30m/s + 2m/s = 32m/s

Doesn't work in Relativity

see that we dont have u = v + u'

we have a correcting factor 1/(1+vu'/c^2)

[crimson sedge]

oh i did not know that, tyvm

.close

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[outer void]

can someone help me factorize x^3-y^3+1+3xy?

[cedar kiln]

@outer void Has your question been resolved?

[sly pike]

Ello

hmmm

Ok

So it's x^3-y^3+1+3xy
So we consider it as a polynomial over the x variable.

So we change the eq to x^3+3yx-y^3+1

Lemme think

So I THINK (Take this with a grain of salt), You Find on Factor of the Form x^k+m where x^k divides the monomial with the highest power of x^3, so m divides the constant facor -y^3+1.

So one of the factor's is x-y+1.

I forgot the last part.

hmmm

Need a bit of help

[livid hound]

try starting with the factorisation of x^3 - y^3

[sly pike]

yea

[livid hound]

and from there complete rectangles, split terms

[sly pike]

thanks

Aight

[cedar kiln]

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[sly pike]

.reopen

wawa

So we Factor the polynomial by dividing it with the Revised EQ. (x^3+3yx-y^3+1)
And we get, this is a long LONG one
(x - y + 1)(x^2 + xy - x + y^2 + y + 1)

Am I correct?

Did I make a oopsie

[cedar kiln]

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[civic coral]

ayy

[civic coral]

if I get sin(theta)=2/sqrt(13)

[nova snow]

PLEASAEEE CORRERECTO MY GRAMMMER

[civic coral]

waterbeammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[nova snow]

PCMG

[civic coral]

can I determine the value of cos(theta)

[fast lotus]

Did they give which quadrant thetha belongs to?

[civic coral]

no

[nova snow]

i think you can use the pythagorean identity but there might be a problem with the +- sign when u take the square root

[civic coral]

so the sign cannot be determined without more infos

[nova snow]

im not 100% sure

$\cos^{2}\theta=1-\sin^{2}\theta$

[wraith dagger]

water beam

[nova snow]

maybe use this tho

[civic coral]

you're 200% sure

[fast lotus]

This works

[civic coral]

really

[nova snow]

1000000% sure

try it yourself

[fast lotus]

But yea ull be getting 2 values of cos tho

[civic coral]

what

I'm trying to get the sign

[nova snow]

are u given if its like obtuse or something

[crimson sedge]

water beam

[fast lotus]

U cannot unless u have which quadrant thetha belongs to

[nova snow]

yajat

[crimson sedge]

helper role era?

[fast lotus]

This also works

[nova snow]

like 999999.999% of the people here are smarter than me so its gonna take 90000000 years

[crimson sedge]

am i in those 999999.999% of ppl

[nova snow]

u should just create an alt account and ask rlly easy questions and let me answer all of them for real

[nova snow]

im 200% sure

[crimson sedge]

waittttt a min i never thought of this before

we could actually do that

shi

[nova snow]

ong

[civic coral]

aren't my question easy enough for you to answer.

[crimson sedge]

but now u have said it in public

so

F

[nova snow]

im gonna guess that the answer is

[crimson sedge]

he didnt answer ur question yet?

[nova snow]

$\cos\theta=\sqrt{1-\frac{4}{13}}$

[wraith dagger]

water beam

[nova snow]

thats what i think

[civic coral]

what about the sign

[civic coral]

what about the sign, it is illegal

[nova snow]

https://tenor.com/view/let-it-go-elsa-disney-disney-princess-queen-gif-3597315

i dont know about the sign tbh

[civic coral]

what a reckless young people

[nova snow]

maybe if there was more information

@crimson sedge what about the sign

[crimson sedge]

wait i didnt see the question yet

lemme read

[civic coral]

maybe there are extraterrestrial lives in the moon.

[nova snow]

im gonna correct your grammar

[civic coral]

my grammar is perfect

[nova snow]

PCMG!

yajat got this

he does jee

hes 999999999 intelligence level

[civic coral]

I'm so talented.

at everything.

[nova snow]

At*

but akshually

should be one whole sentence

yknow

[crimson sedge]

i mean really dont see any problem with it

[crimson sedge]

this is correct

and 2/sqrt{13} is

0.55

so u know theta

[nova snow]

if yajat says its correct then its correct 🥳

[crimson sedge]

is from first quadrant

and cos is +ve is frist quadrant too

so

[crimson sedge]

nah 💀

[nova snow]

https://tenor.com/view/ladacarpiubellachece-gif-25096196

[civic coral]

is he dababy.

[nova snow]

suiiii

[cedar kiln]

@civic coral Has your question been resolved?

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[dusk finch]

?

.close

[cedar kiln]

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[stoic perch]

I have the following problem. Let's say a seller has a t-shirt available in 7 sizes and 12 diffenret colours?
a)In how many different ways can we buy 4 t-shirts of different size in white colour.
b)In how many way can we buy 5 t-shirts of different size and in different colours?
c)In how many ways can we buy 6 t-shirtss of the same size(they can be the same or different colour)
d)In how many ways can we buy 2 white t-shirts and 3 red t-shirts (They can be the same or different size)

[cedar kiln]

@stoic perch Has your question been resolved?

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stoic perch]

2

[wicked mantle]

Show your work

[austere pawn]

what are you doing in c)???

[stoic perch]

for c and d i dont know

that's where i'm stuck

[austere pawn]

you were on the right path until c 🙂

[austere pawn]

you have to "choose" 6 t-shirts of different color

and for d) you have to first: "choose" 2 t-shirts of a color, then: "choose" 3 t-shirts of a color

does that help? @stoic perch

[stoic perch]

yes but i am not sure how to handle the colors bit

[austere pawn]

just think of it the same way as the size

[stoic perch]

because for c i thaugh the answer was 1 * 12 * 1* 12... etc, where 1 represent the one choice of the size and colours thr 12 different colours

[austere pawn]

like choosing 2 different sizes of 10, is the same as choosing 2 colors out of 10

[stoic perch]

so it should be something like C(12,6)?

that now just represent 6 t-shirts of different colours

[cedar kiln]

@stoic perch Has your question been resolved?

[austere pawn]

yes

[stoic perch]

but how about the fact that in bracket it says it can be the same or different colours

do i need to use the formula for combinations with repetitions then?

[austere pawn]

I think it means the color can change, but size is the same

[stoic perch]

so the you think the right answer for c is simply C(12,6)?

[austere pawn]

yea

what I meant before is that it doesnt what thing you are choosing, as long as its the same thing

[stoic perch]

ahaa i understeand.. so then for the last one it would simply be C(7,2) * C(5,3)?

[cedar kiln]

@stoic perch Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[bright ruin]

I understand why this works after seeing the solution

but not how to come up with it

[silver tangle]

It may be easier with contradiction perhaps.
Assume root xy greater than RHS
As both positive, we can Square both sides
Multiply by 4.
Bring to other side
Factorise
0 >( x-y)^2
Contradiction

[bright ruin]

hmm

yes this is easier to understand

thank you

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[silver tangle]

No problem

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[drowsy rapids]

not sure if i should solve with e differently

[cerulean sail]

How did you get that 0.523? I have no issues with your working out (albeit tiny ones, make it clear in each of these that the integral limits refer to x, not u, or even better, change the limits and use the changed ones!) but the latter, I do not get 0.523 from that-

[drowsy rapids]

i was using e = 2.72

idk what im supposed to do with it(with e )

if i need decimal form'

[cerulean sail]

How are you working this out? In a calculator?

[drowsy rapids]

ye

[cerulean sail]

(if so, can you not use the e button there?)

[drowsy rapids]

i only have e to the exponent tho

[cerulean sail]

There should be an option to get e as the value by itself, similar to pi, no? And if that is the case, then just do e^1

[drowsy rapids]

i might just be blind

im using a TI-30Xa

[cerulean sail]

(and really, if you simplify stuff, there's only one place where you actually need the value of e tbf)

[solid juniper]

moment

[wicked mantle]

Sponge bob is having a good time Patric

[cerulean sail]

Oh, so it doesn't damn, well, as above, you can just use e^1 in the one place you need e

(in particular, what's ln(1), and can you rewrite ln(6e)?)

Anyways, besides-

[drowsy rapids]

i got 1.459

says no

[cerulean sail]

,calc 8 * ( (log(6 * e))^7 - (log(1))^7 )/(7 * (6e - 1))

[wraith dagger]

Result:

98.666905508694

[drowsy rapids]

its right

wait what did i do wrong

then

i integrated it correctly no?

[cerulean sail]

You did, as before, and-

,w int 8/(6e - 1) * (ln(x))^6/x, x from 1 to 6e

[nimble mountain]

bit!

[wraith dagger]

Results provided by WolframAlpha

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

Requested by chartbit

[cerulean sail]

more likely you're not putting it into the calculator right yours is likely to be very picky about how you type stuff out

[drowsy rapids]

mayybe

well thank you anyways

[cerulean sail]

(and hiii )

[nimble mountain]

hiii!

[drowsy rapids]

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[cloud field]

Someone help question number 13

I tried solving it and even google didn't work

[cedar kiln]

@cloud field Has your question been resolved?

[cloud field]

<@&286206848099549185>

[signal lynx]

For question 13, using a simple proportion equation can be used. Saying that 1/1.18 = x/17,700 is the proper set up. The top is the original number we are trying to find and the bottom number is the relationship to that number based of the numbers in the equation. The 1.18 comes from the 18% increase of the original cost of the article to the selling price.

@cloud field If you need more help, I would be happy to answer any more questions

[cloud field]

Wait lemme read

[cloud field]

This is what i did

Now i dont know what to do

[signal lynx]

The setup looks perfect and you have the understanding down. Remember that in fractions, we can't cancel numbers that are added to others. You have to cancel on all terms in a fraction.

[cloud field]

Ok gimme 1 min

The correct answer came
Thx

[signal lynx]

You're welcome

[cloud field]

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[marble sluice]

How to solve $$ \int{\sqrt x ln x dx} $$

[sinful bay]

Presumably there's a dx after that?

[marble sluice]

Oh yea uhh typo

[wraith dagger]

Nerdy_Coder

[sinful bay]

Are you familiar with integration by parts

[marble sluice]

Yea

I just sucked at it

[sinful bay]

Well, what does it look like we should make v and what do you think should be du

[marble sluice]

I did the problem multiple times…all in failure

As you can see

[sinful bay]

Ok so it seems like you're not totally comfortable with how to use IBP, which is fine

All IBP really is is $\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$

[wraith dagger]

خرشوف

[marble sluice]

I know

It’s the integration product rule

[sinful bay]

Right

[marble sluice]

Wdym?

[sinful bay]

Oh I think you can disregard that, I misread the work

[sinful bay]

We want to figure out what we can set f and g to so that f'(x)g(x) is easy to integrate

[marble sluice]

I was given choices of u and dv by my textbook

[sinful bay]

Sure, they mean the same thing

[marble sluice]

In this case it was u = ln x and dv = sqrt(x)dx

[sinful bay]

Right, but do you see why this is?

[marble sluice]

Yea I may or may not suck at math…

[marble sluice]

They are easy to integrate and differentiate

[sinful bay]

Well specifically, that choice of u and v makes vdu easy to integrate

If we flipped that choice, we'd get something that is harder to integrate than what we had originally $\left(\frac{\ln x}{\sqrt{x}}\right)$

[wraith dagger]

خرشوف

[marble sluice]

Okay…so what am I doing wrong?

[sinful bay]

Oh it seems like you just messed up uv

[marble sluice]

So my ability to do algebra is bad?

[sinful bay]

In the first attempt, uv shouldn't be what you have there

[sinful bay]

Or maybe you lost track of which thing you were differentiating? Not sure

[marble sluice]

Probably algebra

Sigh…

[sinful bay]

Well as they say, calculus is just an algebra check

[marble sluice]

People told me to do integration problems rather than go back and study algebra…well here I am I have needing help.

[marble sluice]

My algebra knowledge is good but…my ability to keep track of stuff is kinda mid.

[sinful bay]

I mean, that isn't necessarily bad advice

Many people find they're better at algebra after doing calc

[marble sluice]

I know it’s just annoying needing help.

[sinful bay]

Even math PhDs need help sometimes, that might not go away

[cedar kiln]

@marble sluice Has your question been resolved?

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[jagged holly]

In this function, why is the domain ] −∞,2] and not ]−∞,2[

[jagged holly]

Im talking bout why is the 2 included

[raven shard]

what happens when you plug in x = 2?

[jagged holly]

oh I see

you get sqr root of 0

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[tender frigate]

Gambling! :D

how to do basic statistics and get odds?

Let's say you have a 1:100 chanche to win the main price. But you got 90% ROI. so for example if you gamble 10$, you get back 9$, then 8,1 and then 7,29, then that times 0,9 and going on like that. I am pretty sure this is simple logarithmics but can't figure out a generic formula (not even one where the main price is not included in the ROI)

[wraith dagger]

Levgu

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@tender frigate Has your question been resolved?

[quasi merlin]

What probability are you trying to calculate?

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[south terrace]

Should the wind be at 250 degrees in quadrant 3 or is it supposed to be in the opposite direction and in quadrant 1?

[hushed garnet]

Yeah that seems right. If a problem only states direction then most likely it's just counterclockwise from quadrant 1.

[south terrace]

In the lesson, they do it CW from the vertical

[hushed garnet]

Huh that's pretty wack. Do it that way then I guess.

Wait is that not CCW from the quadrant 1 horizontal?

No it isn't... weeeeeird

Ah because it said bearing is why. So that means starting from North CW.

[south terrace]

Is there a different between bearing and heading?

[hushed garnet]

No, doesn't seem like
https://en.wikipedia.org/wiki/Bearing_(navigation)?wprov=sfla1

[south terrace]

ohh okay

thank you so much :D

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[oblique leaf]

beginner stats question:

say I have n people split into two categories A and B. if I choose 3 people at random, how do I find the probability that at least two of them are from category A?

my intuition was
((#A choose 2) + (#A choose 3)) / (n choose 3) but it doesn't feel right 🤔

[wicked mantle]

You didn’t mention the proportion of A and B

[oblique leaf]

why does that matter?

I'm using #A to represent the number of people in category A

[mental trail]

when you do "#A choose 2", you didn't choose the 3rd person

[oblique leaf]

so are you saying the numerator would need to be (#A choose 2)*(#B choose 1) + (#A choose 3)?

[mental trail]

yes

[oblique leaf]

hmm trying to understand why

[mental trail]

how many ways to choose 3 people at random so that exactly 2 of them are in A?

well you choose 2 people in A

and then you choose the remaining person in B

similarly, how many ways to choose 3 people at random so that exactly 3 of them are in A?

well you choose 3 people in A

and that's it

so how many ways to choose 3 people at random so that at least 2 of them are in A?

(#A choose 2)*(#B choose 1) + (#A choose 3)

how many ways in total to choose 3 people at random?

(n choose 3)

uniform probability -> [(#A choose 2)*(#B choose 1) + (#A choose 3)]/(n choose 3)

[oblique leaf]

okay so say there were 3 categories A, B, C, and i want to find the probability that if i choose 3 people, there is one from each category:

(#A * #B * #C) / (n choose 3) ?

it's the same idea as the previous one but i dont know how im supposed to figure out the # of groups that contain 1 a 1 b 1 c

[mental trail]

it's exactly (#A * #B * #C) / (n choose 3)

because you pick one person from A

one person from B

one person from C

(k choose 1) = k

of course we're supposing A,B and C don't overlap

[oblique leaf]

yeah they don't

okay I think I see

thanks for the help

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[uncut phoenix]

can someone go through the solution and tell me if something's wrong with it or not?

[uncut phoenix]

the question is to find the min and max of f(x,y,z) given 2 constraints g(x,y,z) and h(x,y,z). and also stating the points where the minima and maxima occur

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@uncut phoenix Has your question been resolved?

[uncut phoenix]

<@&286206848099549185> ?

[cedar kiln]

@uncut phoenix Has your question been resolved?

[cedar kiln]

@uncut phoenix Has your question been resolved?

[uncut phoenix]

bruh

[cedar kiln]

@uncut phoenix Has your question been resolved?

[uncut phoenix]

...

<@&286206848099549185> ?

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[gusty plume]

didn’t have time to do it myself, from chat gpt 4 plugin (wolfram alpha)

Solution is the same as your original

[uncut phoenix]

ok makes sense thanks

[tranquil nebula]

hello @gusty plume

[cedar kiln]

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[zinc mist]

am trying to find what is the minimum velocity we'll need to get a projectile to pass through a specific point. However am trying to do this using only parabolas and well am stuck

[subtle hinge]

!original

[cedar kiln]

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.

[mighty shuttle]

could you show the context of the problem

that will help

that determines which stategy we will use

[zinc mist]

i assumed the point to be (a,b) and the equation of the parabola to be y=-x(x-p)
where p is the range of the projectile, or the zero of the equation

[zinc mist]

well the exact problem is:
'determine minimum velocity for a projectile to pass through (30,40)'

[carmine kraken]

That only works if b is 0

[zinc mist]

so am trying to derive it for a general point (a,b)

[carmine kraken]

Could you just use the system you had before and shift it up by b

So like -x(x-p) + b

[zinc mist]

but how would i get the value of velocity from that? velocity will be defined by dy/dx

[zinc mist]

wouldnt i get the same value for both of those equations?

[carmine kraken]

Do you mean with the shift and without it

[zinc mist]

yes

with and without the shift, we'll get the same velocity right

but wait

we can't use the shifted equation

the range is no longer 'p'

b has to be zero for that to work, which is not a general case

moreover we can find the value of p by subbing a,b in the equation, not like that's leading to anywhere helpful

[carmine kraken]

But don’t you want it to just pass through the point

[zinc mist]

indeed

[carmine kraken]

So why not just shift it to pass through a and b

[zinc mist]

there will be infinite parabolas that will pass through the point, we're only interested in those that also satisfy the point, along with the range of the trajectory

by shifting it, we do not satisfy the range

[carmine kraken]

What do you mean by range of the trajectory

In this context

[zinc mist]

well u throw something at an angle to the earth and the net horizontal distance it travels is the range

i took the range to be 'p' ie it travels from origin to (p,0)

so 0 and p are zeroes of the equation

[carmine kraken]

Okay

[zinc mist]

that's why i made the equation to be y=-x(x-p)

-x so that the curve is concave downwards

[carmine kraken]

Yes

In that case to find how much you need to shift it to maintain the range I suggest setting x to be a and y to be b and simplifying

To find the upwards translation

[zinc mist]

actually i was wondering if there were a way to find the components of the velocity

if u threw it with the minimum velocity, it should just touch that point and come back

ie at that point, the value of net velocity should be the minimum

[carmine kraken]

Oh

[zinc mist]

means the vertical velocity is 0, and such a point is the highest point of the trajectory

[carmine kraken]

I see what you mean

So the derivative is 0 at that point

[zinc mist]

yep

iguess that should be it

[carmine kraken]

Okay

[zinc mist]

but the velocity is never 0 in trajectory motion right

[carmine kraken]

Yes

It would be more like a vector

[zinc mist]

only the vertical component can get to 0, the horizontal is never zero-

[carmine kraken]

Yeah

Technically

[zinc mist]

even though the answer we get is correct

like if you equate the derivative to 0, and find the value of velocity

it solves the question that was asked in the class

but that doesnt make sense since that says that the net velocity became zero which is well not what occurs

sooo uhhh

what should i do now

the reason why i was thinking to somehow break it into components is because then i could represent the velocity as a function;
sqrt[(v1)^2 +(v2)^2]

and then find the minimum value of it

there will be a relation between v1 and v2 (horizontal and vertical components)

which should help me to do the same

[split viper]

is ur question solved? if not whatr is it

[cedar kiln]

@zinc mist Has your question been resolved?

[zinc mist]

it isnt actually :<

i was trying to solve a kinematics question using only parabolas; the question being:
minimum velocity to shoot a projectile such that it touches a point (a,b)

[split viper]

what is the question

[zinc mist]

now i thought of it by taking the equation as
y=-kx(x-p) where k is a positive reak number and p is the range of the projectile

essentially 0 and p are the zeroes of the equation

the velocity will be defined as sqrt[(dy/dt)^2 + (dx/dt)^2]

am improvising my answer a bit from the earlier discussions

[split viper]

see the velocity in x remains same throughout the journey

also you need to throw a projectile at 45 angle for it to reach maximum distance

[zinc mist]

actually there is no regard for the angle in this case

whatever the angle maybe, we want the minimum velocity for it to touch the required point

[split viper]

do you see where i am going with thus

ok.. lemme think

[zinc mist]

and well you can intuitively tell that the point will be the highest point

because at any other point than the highest point, the velocity is greater

if it werent the highest point, we couldve had another which when treated has the highest point wouldve needed a lesser velocity

[cedar kiln]

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[zinc mist]

.reopen

[cedar kiln]

✅

[zinc mist]

so once u assume the given point to be the highest point, you do indeed swiftly arrive at the answer but

i wanted to do that mathematically

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[weary lintel]

can anyone help with this

[cedar kiln]

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[fleet inlet]

Suppose I have a random event that on average occurs every minute, and must occur 7 times before it 'succeeds'. Suppose I now want to slow down how quickly that rate of success happens by 32x (so it takes 7 x 32 = 224 minutes to 'succeed' on average). Would it be sufficient to simply make the event only actually occur 1/(32/7) times when it is triggered? How could I generalize how much I have to reduce the chance?

[cedar kiln]

@fleet inlet Has your question been resolved?

[fleet inlet]

<@&286206848099549185>

[raven forge]

how do you not know this

[fleet inlet]

@raven forge diff person bro

[raven forge]

no

correct person

[slate lintel]

yeah that seems fine

well

it would be 1/32 times

[solid juniper]

:waves:

[fleet inlet]

Ah okay

Great thank you

That’s all

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[quick fulcrum]

yo

[quick fulcrum]

Im not sure whats that even supposed to mean can someone help me with that?

I know this summation symbol and know what it does

I also thinkt htat I know what the n and what k is just dunno what the other stuff is

[floral salmon]

I'm guessing $O$ refers to the Ordo-notation

[wraith dagger]

Crystopher

[quick fulcrum]

Could u explain what that is?

[floral salmon]

It is a way to describe how fast a function $f(x)$ changes/behaves as its argument $x$ goes to infinity.

[wraith dagger]

Crystopher

[floral salmon]

For instance, $f(x) = x^2 + x$ is $O(x^2)$

[wraith dagger]

Crystopher

[floral salmon]

since when $x$ grows bigger and bigger $f(x)$ approximates the function $g(x)=x^2$

[wraith dagger]

Crystopher

[quick fulcrum]

so it ignores the other stuff thats not the biggest?

[floral salmon]

Yes, in a way it describes how the function behaves for big input values. This is used in fields like computer science to analyze an algorithm's time complexity (how the execution time varies with the size of input).

[quick fulcrum]

Ok thank you.
So how can I prove that?

[floral salmon]

good question, they want you to use linearity of summations, which entail the following properties

unless you have some definition of linearity I'm unaware of, I myself am not that sure if those properties I showed are the linearity they talk about here.

[quick fulcrum]

I dont see one this are a few pages Ill look if I find smth

[quick fulcrum]

here smth I found.

[floral salmon]

At most this has to do with the lower sums in the image, but it is almost the same as what needs to be proven.

[quick fulcrum]

yeah thats also what I thought

and there doesnt seem to be any definition there

bruh

math

[floral salmon]

the only difference is that in this problem we are summing over different functions $f_k$ evaluated on the same input $i$.

[wraith dagger]

Crystopher

[quick fulcrum]

anyways lets just do another problem where I have trouble with.
Im not what evaluate here means or what exactly im supposed to do here

[floral salmon]

looks like a power series?

[quick fulcrum]

yea looks similar I guess. Something I can do with that?

[solid juniper]

hint: antiderivative

[quick fulcrum]

For the first problem or the second problem?

[solid juniper]

this one

[quick fulcrum]

ok

Yo Ive tried bit from my understanding the solution is just 0?

since one is like infinitely large and the other think is like infinitely small so it just approaches 0

[solid juniper]

huh? no

[quick fulcrum]

:<

[quick fulcrum]

Could u help me with that?

Please

[solid juniper]

ok

write $f(x) = \sum_{k=1}^\infty (2k+1)x^{2k}$ (for $|x| < 1$)

[wraith dagger]

chmonkey #1 simp

[solid juniper]

now my suggestion is to find an antiderivative of f

[quick fulcrum]

ahm

[quick fulcrum]

k^2+c x^k(^2)+c

?

[solid juniper]

no

[quick fulcrum]

:<

[solid juniper]

it's not much different than finding an antiderivative of a polynomial

[quick fulcrum]

bro first time Im doing that just watched an 3minute video from khan

[solid juniper]

you've never learned how to take an antiderivative (or integrate as some may say)?

[quick fulcrum]

yea

just derivative

[solid juniper]

hmm ok maybe there is a different intended solution then

[quick fulcrum]

khan did that so I tried to do the same do I also need to antiderivative the sum symbol?

hm the + 1 prob doesnt go away

[solid juniper]

i'm not sure what question means

[quick fulcrum]

hm

[solid juniper]

anyway maybe we shouldn't do this

[quick fulcrum]

yea

math

math gone

.close

[cedar kiln]

Channel closed

Closed by @quick fulcrum

Use .reopen if this was a mistake.

[solid juniper]

[cedar kiln]

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[crisp dagger]

this is a bit of a weird one…
given a line y=mx+c, find a second line which is parallel to the first, where the distance between the two lines is k
that isn’t the weird bit ofc, the weird bit is i’ve already solved it:
the equation of line 2 should be

y = m(x - a) + ga + c
where g = -m^-1 and a = 2k/sqrt(g^2+1)

but I’ve forgotten how i’ve solved it(and i’m pretty sure there was a large amount of guessing and desmos involved in my finding this solution) so I haven’t got a clue why the solution is what it is
so, I suppose, I need someone to explain to me why this is the solution, and how one might solve this without guesswork

[civic magnet]

ok so

first of all

if it's parallel

what's the m of the second line

[crisp dagger]

the same

[civic magnet]

yes

therefore

[crisp dagger]

i meant more how tf did i find a

[civic magnet]

yes im getting to it

[crisp dagger]

because especially the 2 seems out of nowhere

[civic magnet]

follow my steps

so we know our second line is of the form y=mx+k, where k is different from q

not 1,q

my bad i keep mistyping today

[crisp dagger]

lmao same

[crisp dagger]

yes

[civic magnet]

ok

now what would you do

[crisp dagger]

set simultaneous equations and solve

[civic magnet]

that wouldn't really work

how do you find the distance between two lines

[crisp dagger]

solve the two lines seperately with y = -m^-1 * x and find the distance between the points

[civic magnet]

well that would work

i wasnt thinking of that but it works

so now we apply your formula

we will set -x/m=mx+c and -x/m=mx+k

actually sorry

dont use k

use c+k

because then you'd get problems

y=mx+c+k is the one you're trying to find

if you're struggling with this method i had in mind another

yeah let's use my method, i think you'll undestand it more

[crisp dagger]

lets try yours
rn all i’ve got is x = -c/(m^2 + 1) for the first one

which seems to look like something similar

[civic magnet]

ok so

if we have two lines

parallel

there is a easy way to find their distance

since their distance is always the same

we can choose a random point on one

then calculate the distance of that point to the other line

so we get our starting line y=mx+c

lets choose x=0

we get y=c

A(0,c)

right?

[crisp dagger]

yeah

[civic magnet]

now we use the formula for the distance between a point and a line

which is.....

[crisp dagger]

….not something I know
the best thing I can think of is solving for r for x^2 + (y-c)^2 = r^2; y = mx + k
but I don’t think that yields a constant solution

[civic magnet]

with the equation you dont know the problem becomes extremely easy

the distance between a point and a line is = absolute value of the line with instead of x and y the coordinates of the point, divided by the square root of the coefficient of y squared + the coefficient of x squared

is it clear

[crisp dagger]

kinda

hang on

|c - 0 + k| / m^2 + 1^2
= | c + k | / m^2 + 1

[civic magnet]

yes but there is a square root on the bottom

is square root of that

[crisp dagger]

oh yes

[civic magnet]

and since the distance you wanted was k

you set that equal to k

and you might be like, why is there an absolute value

it's there because you clearly have 2 lines that have the same distance from that line

one above, one below

and so that gives you the double solution

[crisp dagger]

which is…different to the k in that expression?

or same?

[civic magnet]

uh wait

what are your two equation

just making sure

write them

yeah my bad, i chose bad letters

the k you have in the line equation is a k which makes your line go up and down right?

and clearly it is not the same k as the distance between the two lines

so call the distance a different way or rename the other k

[crisp dagger]

yeah

[civic magnet]

sorry for the incomprehension

[crisp dagger]

lets say distance is d then

[civic magnet]

sure

but you made a mistake i think

in the absolute value

look at it again

you have y=mx+c+d

if you sub y=c and x=0

the c actually cancels out with the other one

and only d is left

so this is independent of the initial c, as it should be

[crisp dagger]

are you sure this is d though

[civic magnet]

so the answer is k=|d|/sqrt(m^2+1)

[crisp dagger]

but yeah I get what you mean

[civic magnet]

oh you called the other one d

my bad

d=|k|/sqrt(m^2+1)

read what i had said but with k

is it clear? i know it was a bit wonky

always hard to explain typing

[crisp dagger]

so k = +- d * sqrt(m^2 + 1)?

[civic magnet]

no the distance is the d

[crisp dagger]

yeah and we solve for the k surely

[civic magnet]

we're getting confused with the variables

[civic magnet]

doesnt really matter but sure do it

we get that as a result, which looks similar to yours right?

[crisp dagger]

kinda
mine is really weird though; its 2d/sqrt(m^-2 + 1)

but

i multipy it by both -m and -m^-1

so

i’ll mess around with it a bit

[civic magnet]

sure thing

check that distance formula you didnt know about, here it is very useful, it basically solves the problem by itself

[cedar kiln]

@crisp dagger Has your question been resolved?

[crisp dagger]

wait @civic magnet
i’m not tripping, right?

our results are different?

c is 0 here, not that it should matter

[civic magnet]

let me write it down myself

mine is correct, look

i set the distance between the lines as 2

you can check those two points i highlighted have indeed distance 2

[crisp dagger]

oh yeah
that explains the random 2k, the explanation being that the 2 was wrong

[civic magnet]

and also yours didnt have the plus-minus solution

which wouldnt have made sense

[crisp dagger]

before i go, isn’t +- sqrt(d^2(m^2 + 1) the same as +- dsqrt(m^2 + 1)?

[civic magnet]

yep, you can see i wrote that in my desmos

[crisp dagger]

huh
but wolfram gives it in the latter form

weird

[civic magnet]

i only used the case with + as the other was irrelevant

[civic magnet]

if they are equal it doesnt matter

also dont always trust what a computer says

[crisp dagger]

not really
I just usually expect wolfram to give me the most simple form

[civic magnet]

sometimes wolfram overcomplicates stuff

i had some examples in the past

anyways, you have your solution

[crisp dagger]

indeed

thank you very much

.close

[cedar kiln]

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[cedar kiln]

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[pearl vigil]

Can someone help me with this

I have this question

What is wrong

Or we just dont have functions with such property that works in the same time for n=1 and some n>=2

[pearl vigil]

So ?

[cedar kiln]

@pearl vigil Has your question been resolved?

[pearl vigil]

I just want to know if there is such function

<@&286206848099549185>

Yes

For every n

Because I have a proof i think above why this such function does not exist

Ok thanks

.close

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[strong merlin]

Hi! Can someone help me figure out how to solve this?

[restive pilot]

u want to find the function?

[humble lintel]

What are the properties of limits do you have?

[restive pilot]

like
f(1) = 4

it gonna have a lot o f soluctions

[strong merlin]

this what my prof gave us

[strong merlin]

the instruction is to find each limit if it exists

[humble lintel]

Yep, now the limits for g and f exists at 1. So now use those properties

[restive pilot]

i think this doesnt import

cause limit isnt a operator

its like a ideia

[strong merlin]

like this?

[humble lintel]

Yep

[strong merlin]

IT WAS THAT SIMPLE??

you're a lifesaver

my teacher explained it so complicated

mb 😭

tyty

.close

[cedar kiln]

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[restive pilot]

how i open channel?

.open

[humble lintel]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[restive pilot]

.reopen

[cedar kiln]

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[marble sluice]

What is wrong with my answer to solving the integral of arcsin?

[cedar kiln]

@marble sluice Has your question been resolved?

[cedar kiln]

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[storm otter]

can someone tel me if i got the right answer

[storm otter]

i got 75.7

[prime sun]

👍

[storm otter]

its right?

i got 15

is this right

can someone hekp

[prime sun]

sorry dude

not good with shapes

[cedar kiln]

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[cold geyser]

How many ways can 4 boys and 4 girls be arranged in a circle if the boys and the girls are in distinct groups?

[scenic stirrup]

BBBBGGGG and GGGGBBBB are the same

that makes sense right

[cold geyser]

yes

[scenic stirrup]

so find the number of ways you can order 4 boys and 4 girls

[cold geyser]

is it 4!x3! because someone has to sit down first?

[scenic stirrup]

oh

i was just thinking (4!*4!)/2

[cold geyser]

.close

[cedar kiln]

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[scenic stirrup]

wait nvm mb

[cedar kiln]

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[sour wave]

hi

[sour wave]

A block with mass 3 kg is held at rest against a vertical wall by a horizontal force. If the coefficient of static friction is 0.45. Determine the horizontal force to hold the block at rest against the wall.

[flint plinth]

what's your question?

[hushed garnet]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sour wave]

I know that Fs = mg

since the static force counters the force of gravity

Fs = mew(FN)

wait

Fn counters Fa

if we know Fn then we know Fa

[flint plinth]

what is Fa?

[sour wave]

Fn = Fs/mew

[hushed garnet]

Fs=mg? how?

[sour wave]

the horizontal foce applied

[flint plinth]

that one that's just labeled F in the diagram?

[sour wave]

isnt the static force supposed to counter Fg

[sour wave]

yea i mis drew

[flint plinth]

nw

[hushed garnet]

Yeah you're right I forgor

[flint plinth]

this is correct since there's no vertical motion

[sour wave]

so I know Fs = Fg = (3)(9.8)

and we can use Fs = mew(Fn)

Fn = Fs/mew

[flint plinth]

yes

[sour wave]

oh dam i solved me own questin

[flint plinth]

nice job haha

and yea F = Fn since there's also no horizontal motion

[sour wave]

29.4/0.45 = 65 N or something

ok and I remember my teacher got the same too

[flint plinth]

,w 3*9.8/.45

[wraith dagger]

Results provided by WolframAlpha

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

Requested by bungo.the.original

[flint plinth]

yep looks good

[sour wave]

,w 3*9.8/.45

oh dam

[wraith dagger]

Results provided by WolframAlpha

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

Requested by hugoisme

[sour wave]

thanks

[flint plinth]

cheers

[cedar kiln]

@sour wave Has your question been resolved?

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[limber snow]

How would u do this

[tulip tinsel]

u need the same final combination of steps every time

[vagrant elbow]

How many times do you have to move up, and how many times right?