help-0

[soft bridge]

@still dove

well

use power of a point

[lone heart]

@still dove Has your question been resolved?

[lone heart]

@still dove Has your question been resolved?

[lone heart]

@still dove Has your question been resolved?

[lone heart]

@still dove Has your question been resolved?

[west girder]

This

Just show that H lies on the radax of (AE) and the circle centered at E with radius DE

It’s easy to compute the power of H wrt to each of these circles

[thorn yoke]

!help

[lone heart]

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

[still dove]

Could you explain this in more detail using a picture?

[west girder]

i dont ahve one

do you understand why H lies on the radical axis?

[still dove]

idk...

[still dove]

.

[west girder]

oh sorry, i meant do u understand why we need to show H lies on the radical axis

we dont know that yet, but it suffices to prove that

because radical axis is perpendicular to line through centers

[still dove]

I would like it to be proven at a middle school level

[west girder]

this is at a middle school level

radical axes and power of a point are important concepts

[minor imp]

Id like to learn math from 0 do you guys have any idea how to start

[still dove]

I think this might be a hint.
First, angle AEF and angle ECF are equal because they are inscribed angles subtended by the diameter CE of the circle.

Secondly, the intersection point G of the circle with diameter FC and the circle with diameter AC creates right triangle GAC with the common lines GF and GA.

[still dove]

You need to go to another channel and ask for help.

[west girder]

j compute the power of H wrt to both of these circles and show they’re equal it’s so much easier 😭

[still dove]

I’m sorry..., I’m not familiar with the Power of a Point. Could you explain me with a picture?

[west girder]

Oh ok

How did u know radical axis then?

I’ll try finding a synthetic solution later then

[still dove]

it's from perspective diagram, not math

https://www.youtube.com/watch?v=hOYjJibOOtE&list=PL4PCJHqcuSZuxZgMpKLBuorAzIagoZHDm&index=2

I didn't expect the problem to be so complicated 😦

[still dove]

Thank you so much for putting in so much effort for me.

[still dove]

I thought that knowing geometry at the middle and high school level would be enough, but I’m sorry. I didn’t realize the problem would be so difficult.

(for reference, I am Korean, we don’t learn these things in secondary education.)

[minor imp]

Which one

[winter light]

Read #❓how-to-get-help

[west girder]

hmm i think most good solutions will contain power of a point

maybe try to undersatnd it

[still dove]

I don't understand at all...

Even if I use the Power of a Point, I don't know how to relate it to the orthocenter.

[west girder]

The solution I was referring to also uses radical center theorem

[digital swallow]

Look orhtocentre has a definition

For a triangle ABC,

It's orhtocentre is the point of intersection of all altitudes drawn from vertices to opposite side

[still herald]

Ah, my first message.

In channel 0.

Noice.

[lone heart]

@still dove Has your question been resolved?

[ruby quail]

This is where I go to get help right? I have this question on GRE practice test. How can I solve this without this method and without a log or exponent function on the GrE calc. Focusing on the method being quick since it's for the gre

I'm confused I see someone say this isn't the spot earlier but the guide said to ask my question in help-0

[digital swallow]

Converting all numbers into 2s and 3s is actually the best and most accurate/clear method

The only way to fasten the process is able to do the same with bigger numbers,

Now the only other method left is log, which you don't want

What I would suggest is the go with the method of factorising those big numbers, i.e., converting them into 2s and 3s

If you unable to understand the solution you posted, read it calmly and thoroughly 2-3 times

At least this is what I can help

You may wait for other helpers if needed

[fallen plover]

ez fix

take log both sides

[still dove]

https://tenor.com/view/team-fortress2-engineer-tf2-wat-what-gif-21053319

[surreal chasm]

multiply by 0

[still dove]

Does solving this problem require infinite time?

https://tenor.com/view/infinity-loop-infinity-symbol-gif-16359183

<@&286206848099549185>

[still herald]

yes

[west girder]

Truth be told, I don’t think you’re ready for this problem

There are many solutions but they’re all power of a point

[still dove]

Even though it's difficult, I will choose that solution, sorry

Please don't use too much technical jargon instead, mix in simpler terms when explaining.

Should technical terms be used more frequently than simpler terms, please explain with the help of pictures, like I do.

I believe that pictures are the most effective way for me to comprehend the explanation, irrespective of the use of technical terminology or simpler language.

[livid sage]

look up power of a point / radical axis

[still dove]

Is knowing just those things enough to understand it?

Is there no information on how to apply these methods?

Could you provide at least a basic sequence or steps?

Why are the responses all so general?

[meager karma]

Choose basis such that:
A = (0, 0)
E = (1, 0)
D = (1, sqrt(d))
C = (1, 1/c)
By similarity of FCE and GFE
EG:FE = FE:CE
and since FE = ED
EG = (ED)²/(CE) = sqrt²(d)/(1/c) = dc

We can represent line EG as
y = dcx + 0
So gradient (aka rise/run) is dc

AED is similar to DEB, and by similarity
EB:ED = ED:EA
EB = sqrt²(d)/1 = d
B = (d + 1, 0)

We can represent line CB as
x + ky = j for some constant k and j
We know that C and B lie on CB, so their coordinates must satisfy this equation.
1 + k/c = j
(d + 1) + 0k = j
->
k = (j-1)c = ((d+1)-1)c = dc

Then the equation for CB is
x + (dc)y = (d+1)
y = (-1/(dc))x + (d+1)/(d/c)
negative reciprocal of the gradient is
dc
This is equal to the gradient of AG, which means that CB is perpendicular to AG.

Since G is the intersection of AG and CE (2 different heights), it must be the orthocenter.

[meager karma]

checkmate

[still dove]

https://tenor.com/view/leonardo-dicaprio-clap-slow-clap-good-job-nice-one-gif-14556567

I can't thank you enough

.close

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[west girder]

Well of course I was referring to synthetic solutions

[knotty helm]

Hi

[meager karma]

Say that AG meets BC at H.
By similarity of FCE and GFE
EG:FE = FE:CE
and since FE = ED
EG = ED²/CE
-> tan(HAB) = EG/AE = ED²/(AE·CE)
tan(HBA) = CE/EB
By similarity of AED and DEB
EB:ED = ED:EA
EB = ED²/AE
tan(HBA) = CE/EB = (AE·CE)/ED²
tan(HAB)tan(HBA) = 1 -> AHB = 90°

[west girder]

this is the same as power of a point.

[meager karma]

of course it is

[west girder]

which they couldnt understand

[meager karma]

i don't think 2 approaches being equivalent necessarily means they are equally easy to understand

[west girder]

this is true

[small lance]

no, this is not the place where you are supposed to get help

[lone heart]

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[junior escarp]

How do you acctualy find the value of pi?

[junior escarp]

i guess im asking how to find the value of pi.

[simple imp]

Yeah, i believe the most recent approximation was like 202 trillion digits and it used this algorithm.

they basically make a computer try and solve this series to the closest they can to find what 1/pi is

and from that easily find what pi is

[finite flax]

How quickly does that converge?

Wow

(Compared to antiquity)

[simple imp]

https://www.storagereview.com/news/storagereview-lab-breaks-pi-calculation-world-record-with-over-202-trillion-digits article on it

"team ran an almost continuous calculation for 85 days, consuming nearly 1.5 Petabytes of space across 28 Solidigm SSDs."

fr ramanujan is sadly forgetten about in western learning D:

one of my favourite things in maths is learning the history behind the mathematician but lots of people just skip it in teaching

well yeah i'd argue all arab and indian mathematicians

truly not enough recognition considering our numbers are literally yoinked from them

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[wanton lagoon]

i been at this problem for a little bit and i guess im wondering if my answer is right

[wanton lagoon]

i said 5 degrees based on inverse trig function of tangent

but i feel thats very wrong

[lethal belfry]

Do you mind showing your working?

[wanton lagoon]

sure

[neon mulch]

i mean if you round it to the nearest integer i believe that is that angle of elevation

[wanton lagoon]

for the angle i put a ?

just for reference that its the missing angle

[lethal belfry]

Huh. I had a different interpretation of the question

I thought the hypotenuse was 1 mile long

[wanton lagoon]

hypotenuse is not given for this scenario

and i dont think we supposed to calculate it

[lethal belfry]

They have said that a 1 mile stretch of road rises 450 feet.

[neon mulch]

i think by my math that regardless of the road being the longer leg or the hypotenuse, the angle of elevation rounds to 5 degrees

[lethal belfry]

,w arctan(450/5280) in deg

[ocean seal]

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[lethal belfry]

,w arcsin(450/5280) in deg

[neon mulch]

340?

[ocean seal]

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[lethal belfry]

hmm

They're close

[neon mulch]

what do you have to round it to

[wanton lagoon]

uhh

nearest degree

[neon mulch]

then ur chill

5

[wanton lagoon]

ok

ig the big numbers had me second guessing

a small angle

and i also had a second question bc i have no idea how to approach this

its from a study resource but it doesnt rlly give the answer

[neon mulch]

i think an 87 degree

i treated the tower as the hypotenuse

and the the rope as the longer leg

[wanton lagoon]

yeah thats what i did

so would the rope value

be

57 feet

[neon mulch]

yeah

[wanton lagoon]

ok

the 3 feet thing threw me off

[neon mulch]

mhm

[wanton lagoon]

so idk if thats relevant

[neon mulch]

kinda did for me too but wtv

anyways i gotta do my math now

cya

[wanton lagoon]

ok

cya

[wanton lagoon]

yo if ur still on

what is the hypotenuse length

bc i onyl found the

rope length

of 57

[neon mulch]

just do the pythagorean theorem

57^2 + 3^2

[wanton lagoon]

ok

thanks

[neon mulch]

alr

[wanton lagoon]

.close

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[cobalt token]

Looking for some help with an integration question, I was doing a line integral and did most of it but got to integral from 0 to pi of -4tcos(t)sin(t) dt and don't really remember how to integrate something like that...

[fallen verge]

u sub

oh

[tardy stag]

one thing you could try is using 2sinxcosx = sin(2x)

[fallen verge]

oh yeah then ibp from there

[cobalt token]

ah, okay, I always forgot like 90% of the trig identities lol

.close

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[gray pelican]

hey does anybody know significant figures?

[gray pelican]

I have a problem of 12.00 + 15.05 which equals to 27.05

and i have to go back 4 numbers left to right

so am i suppose to round the 5 or just leave it like this?

[primal bobcat]

I don’t understand this

[neat folio]

do you know how many S.F you are rounding to

[gray pelican]

its 4

[neat folio]

4 s.f

[gray pelican]

yes so its stops at 5

[primal bobcat]

27.05 already has 4 significant figures though

[neat folio]

remember that S.F means 4 significant figures starting from the first significant number

(so anything above 0)

hence the numbers both already have 4 s.f

[gray pelican]

so no rounding?

[primal bobcat]

Yeah

[gray pelican]

what about 5.7621 it has no math so am i suppose to leave it like it is?

[lethal belfry]

Well, the answer should have 4 siginificant figures afaik.

[gray pelican]

anybody know this one?

[neat folio]

if this is for 4 s.f

then 5.762 because

[gray pelican]

its a different problem

[neat folio]

that's 4 figures

[gray pelican]

ill send picture

[neat folio]

it would just be the same number

there's no need to round down and up

if it's been measured at 5.7621 then that's how many s.f

btw id probably delete this since it has your HS name on it.

[gray pelican]

thats jsut my teacher

but ight

[gray pelican]

so leave it as is?

[primal bobcat]

Yeah

[neat folio]

should be yes.

[gray pelican]

alright i should have no more questions thanks!

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[upper spire]

having trouble with 3y'+y=3

[lethal belfry]

$3\frac{dy}{dx}=3-y$

[ocean seal]

🏳🌈f(why am i here )= idk

[lethal belfry]

now what would you do?

[upper spire]

divide by 3 then solve by separation?

[reef pier]

the division isn't necessary but yeah separate

[lethal belfry]

Well, you don't NEED to divide by 3, but yeah, that works.

[upper spire]

so youd get 3y=3x-yx?

[lethal belfry]

uh, no

$\frac{3dy}{3-y}=dx$

[ocean seal]

🏳🌈f(why am i here )= idk

[lethal belfry]

Now $3-y=u$

[ocean seal]

🏳🌈f(why am i here )= idk

[upper spire]

got it appreciat it

[lethal belfry]

Awesome. :D

[upper spire]

its been pretty long since weve had a separable DEQ so i didnt even think about it

[reef pier]

you could technically try using integrating factor

but so unnecessary

[upper spire]

i was trying to solve it was 3r+1 = 0 but knew that it didnt make any sense

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[tepid ravine]

sorry, kinda lost ont his

[tepid ravine]

so we have the sample mean is 19.6

doesn't this fall below the null hypothesis of 20

so hence we reject it

and then consider the laternative hypothesis

[reef pier]

iirc just because the sample mean is below your hypothesized actual mean isn't enough to reject the null

[hushed locust]

whether or not we reject the null hypothesis depends on the test statistic. the sample mean is unlikely to be exactly equal to the null hypothesis just due to randomness, but the question is whether it's different enough

[tepid ravine]

how come?

[tepid ravine]

ok i see

i think this slide talks about it

also what does it mean by mean being greater than equal to 20

if it is greater than 20, wouldn' that also fail the null hypothesis

@hushed locust

[hushed locust]

yes, but the appropriate alternative hypothesis in that case would change

[tepid ravine]

oh nevermind, the data was suppose to say that if u > 20 then we walk

one last question

for this, why is it that alpa refer to null hypothesis being wrong?

also what part of the normal distributino does z_a refer to

i'm getting it confued if it is from left to right of the normal distribution

or right to left of the normla distribution

[hushed locust]

alpha is the probability that a given sample mean will fall into the "reject zone" even though the data was normally distributed about the null hypothesis is true, i.e. the probability that we will reject the null hypothesis even though it's true

[lone heart]

@tepid ravine Has your question been resolved?

[tepid ravine]

hmm i see

and for this right

why is it greater than alpha

[charred jewel]

the area greater than z_alpha is the rejection zone

[tepid ravine]

don't see the equality

also why did we go from -z_(1-a) = z_a

aren't they equivalent

so why no remaind as -z_(1-a)?

[lone heart]

@tepid ravine Has your question been resolved?

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[cerulean grove]

can anyone explain me like im five why dx=△x

[rare cradle]

its just a notation

in general

depending where you use it though

[cerulean grove]

Δx=dx+o(dx)

idk what's the difference between dx and △x

[rare cradle]

well there can be bunch of meanings

its just to shorten text though for this scenario I suppose so

its nothing to do with that you need to know what it is or something we want to shorten text

for example the finite change

in x

for example you wanna know the distance of change

[cerulean grove]

we use delta x

[rare cradle]

i mean yeah

[cerulean grove]

then how can dx=△x

[rare cradle]

thats when o(dx) = 0

remainder is 0

[cerulean grove]

o(dx)?

[zinc haven]

in my opinion, delta x refers to a change in x, while dx refers to an infinitesimal change in x

lets say you are traveling in a car, its displacement can be described as f(x), where x is time. at some point you want to find its velocity

delta x can be used to find the average velocity from x1 to x2. this assumes f(x) is linear from x1 to x2 when it, most of the time, really is not, but you want a fair estimate

now as you move x2 closer and closer to x1, the estimate's accuracy becomes better, where it is infinitesimally close to x1, but is not x1. the distance is now dx

in more formal terms, delta x represents a secant line, while dx represents a tangent line

they are not strictly equal, but definition-wise they are similar

[rare cradle]

but then because you know its not linear most of the time you have o(dx) which is the remainder or the error of dx so then you call it △x = dx + o(dx)

[lone heart]

@cerulean grove Has your question been resolved?

[cerulean grove]

Then why in the total dif, dx=Δx but dz≠Δz

[alpine sable]

can you give more context on where \Delta x and \Delta z occur? most of the time they are just syntactic sugar in the limit.

[lone heart]

@cerulean grove Has your question been resolved?

[lone heart]

@cerulean grove Has your question been resolved?

[tight pier]

like i am five ^

also dx = Δx is just wrong

[cerulean grove]

when dx is enough small, yes it will ig?

[tight pier]

You may say $\lim_{\Delta x \to 0} \Delta x = \dd x$ but $\Delta x = \dd x$ is very off because of how infinite small $\dd x$ is meant to be.

[ocean seal]

𝔸dωn𝓲²s

[cerulean grove]

When △x approaches zero, it can be considered infinitesimally small, and in this limiting case, dx ≈ △x

[alpine sable]

just a word of caution, don't be too serious about equality and approximately equal if you didn't define dx and \Delta x formally.

[cerulean grove]

shouldn't here be △x small? why here they said dx is small? im confused

[tight pier]

Usually you have an approximation Δy/Δx and their saying is as Δx approaches dx meaning gets also really small so does Δy meanin Δy ≈ dy

[tight pier]

This should help to visualize

[tight pier]

If Δx -> 0 we denote it as dx thats also why dy/dx is the notation for derivative because the derivative is the limit where the secant slope "became" tangent

[cerulean grove]

ok then why this ↑

[alpine sable]

according to your text, \Delta x = dx is by definition.

they also defined \Delta z, but what is your definition of dz?

[tight pier]

We defined the $\textbf{total differential}$ as $$\dd z = f_x \cdot \dd x + f_y \cdot \dd y$$ which comes from $$\Delta z \cong f_x \cdot \Delta x + f_y \cdot \Delta y$$

[ocean seal]

𝔸dωn𝓲²s

[alpine sable]

it begs the question what exact are dx and dy, differential 1-forms? 🙂

[cerulean grove]

ok but why they have to define dx=delta x, as a prequitst here?

ooh wait

i think i get it

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[sterile trench]

I am bad with rate changing problems

[sterile trench]

idk where to start with

[jade pond]

Try write the function of the area of both

Then differentiate them and see which is larger

For example the area of the circle is pi*x^2

And the rate of change is 2pix

[sterile trench]

lemme try

yea but

we need w.r.t t

[shadow saffron]

no conclusion it should be ig

[jade pond]

It's not no conclusion

[sterile trench]

reason?

btw answer is a (answer key)

[jade pond]

As it's with respect to time for both

[sterile trench]

option a

[jade pond]

We can multiply both by dt

[sterile trench]

oh that's true

yup got it

thanks

[jade pond]

No worries

[sterile trench]

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[tight pier]

You are given
\begin{align*}
&\mathbb{P}[A \setminus B] = \mathbb{P}[A] - \mathbb{P}[A \cap B] = 0.6 - 0.2 = 0.4 \
&\mathbb{P}[B \setminus A] = \mathbb{P}[B] - \mathbb{P}[A \cap B] = 0.5 - 0.2 = 0.3 \
&\mathbb{P}[A \cup B] = \mathbb{P}[A] + \mathbb{P}[B] - \mathbb{P}[A \cap B] = 0.6 + 0.5 - 0.2 = 0.9
\end{align*}

Let $C$ be another event with $\mathbb{P}[C] = 0.3$ and suppose $\mathbb{P}[C \cap (A \Delta B)] = \emptyset .\$
$\$
Show that $\mathbb{P}[C \setminus (A \cup B)]$ is uniquely determined by this information, and calculate this probability.

$\textbf{Solution.}$ $$\mathbb{P}[C \setminus (A \cup B)] = \mathbb{P}[C] - \mathbb{P}[A \cap B \cap C] = 0.3 - 0.2 = 0.1 \$$

I am trying to understand how
$$\mathbb{P}[A \cap B \cap C] = 0.2$$
was derived, knowing that
$$\mathbb{P}[A \cap B] = 0.2 \text{ and we know } \mathbb{P}[A \cap B \cap C] \subseteq \mathbb{P}[A \cap B].\$$

(Yeah I have drawn dozens of Venn diagrams by now but I don't get it, so don't ridicule me...)

[ocean seal]

𝔸dωn𝓲²s

[tight pier]

What you want first is open an own help channel (see #❓how-to-get-help ) 😄

brb gonna make something to eat

[lone heart]

@tight pier Has your question been resolved?

[fickle heath]

Well, P[A u B] = 0.9 and P[C] = 0.3 so P[(A u B) n C] >= 0.2

If P[C n ((A u B) \ (A n B))] = 0 then P[(A u B) n C] = P[(A n B) n C]

(and P[A n B] = 0.2 so P[A n B n C] <= 0.2)

[marsh rapids]

all that's left is to indeed verify the condition

[tight pier]

This is exactly what I don't get I just drew it again, and thinking where is the conclusion coming from that it's greater or equal to 0.2

[fickle heath]

P[universe] = 1.0

You can't have P[A u B u C] > 1.0

[tight pier]

yea ok

[fickle heath]

Well if P[A u B] = 0.9, P[universe \ (A u B)] = 0.1

So P[C \ (A u B)] <= 0.1

[tight pier]

oh shit

that 0.1 is in C \ (A u B)

and the 0.2 in inC and that A and B adn C

no not necessarily

[fickle heath]

C is the 0.1 not in (A u B) and the 0.2 in (A u B) which also happens to be all of (A n B)

[tight pier]

Ok but how do we certainly that the 0.1 is in C \ (A U B)?

it could also be outside

[fickle heath]

P[universe \ (A u B)] = 0.1

There is no outside

It's the rest of the universe

[tight pier]

i mean outside of (C \ (A U B)) but still in Omega

[fickle heath]

Yeah that's what I'm saying

[tight pier]

Ok then by that we know

[fickle heath]

P[A u B] = 0.9 implies P[Omega \ (A u B)] = 0.1

[tight pier]

(C \ (A U B)) <= 0.1

[fickle heath]

Yes

[tight pier]

ok and then?

[fickle heath]

And then that means P[(A u B) n C] >= 0.2

[tight pier]

hmm i think i am starting to get there

yeah that makes sense

[fickle heath]

Here's a poorly drawn Venn diagram

(somewhat to scale)

[tight pier]

i am slow

P(A U B) = 0.9

the moment we take C away it's at least 0.2

[fickle heath]

No that's useless

You want C n (A u B) (which is also C n (A n B))

[tight pier]

yes

sry

[fickle heath]

C \ (A u B) gives you the part of C that is not in (A u B)

C n (A u B) gives you the other part, the one that is in (A u B)

Add them together, they form C

If P[C] = 0.3 and P[C \ (A u B)] <= 0.1 then P[C n (A u B)] >= 0.2

[tight pier]

P[C \ (A u B)] + P[C n (A u B)] = P[C]

[fickle heath]

Yes

[tight pier]

P[C n (A u B)] >= 0.2 yes that makes more sense then what i wrote i confused myself

P[C n (A u B)] = P[A n B n C] right

[fickle heath]

Only because C n (A u B) = C n A n B

[tight pier]

yes

because the other "two parts" are empty

C n (A Delta B) = { }

[fickle heath]

Right

[tight pier]

ok now we need also to show that P[A n B n C] <= 0.2

then we showed it's equal right

[fickle heath]

Sure

[tight pier]

Oh

A n B n C ⊆ A n B

so that makes it

P(A n B n C) <= 0.2

[fickle heath]

Either use <= or don't use P[...]

[tight pier]

mb

right

i am just trying to connect the things

but it makes sense to say at least

thanks man

.solved

[lone heart]

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[tight pier]

life saver

[lone heart]

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[alpine sable]

I don't think in general there is a nice condition

there is a concept called "irrationality class" (I think?) though

[normal bone]

what kind of condition did you have in mind?

im not sure what you are asking

like display name said, there isn't a nice way of checking if something is irrational/rational

it depends on the number

[alpine sable]

No one knows if e+pi is a rational number, it could be, but possibly not. How do you prove it must not be a rational number? This is an open problem.

[normal bone]

i think you can just sum all n-1 sets together

then you will get that the sum of the numbers is rational

then subtract the integer n-1 sums to get that the numbers are also rational

first let the numbers be $a_1,\dots,a_n$, then you know that $(n-1)(a_1+\dots+a_n)$ is an integer

[ocean seal]

Bair

[normal bone]

you know that if $S=a_1+\dots+a_n$, then $S-a_1$, $S-a_2$, \dots $S-a_n$ are all integers, sum all of those together

[ocean seal]

Bair

Display name

[alpine sable]

Is this what you are thinking?

[normal bone]

this is what i mean

@hallow dome does that make sense

i think whatever you do will be equivalent to the computation i just laid out

[subtle parcel]

yo wsp i just joined

[pallid scarab]

you don't always have to reason by absurd when there is a direct proof

[normal bone]

what i said earlier, sum each of the integers $S-a_i$ to get $(n-1)S$ which shows that $S$ is a rational number, then $a_i=S-(S-a_i)$ which is a difference of rationals

[ocean seal]

Bair

[normal bone]

(n-1)S is an integer, so S isn't necessarily one

what is

say the set of rationals is 1/2, 1/2, and 1/2, then S=3/2

yes

[pallid scarab]

for any a_i yes

[normal bone]

why is that an issue?

yeah

[pallid scarab]

rational/rational

$\frac{\frac ab}{\frac cd}$

[ocean seal]

rafilou2003

[lone heart]

@hallow dome Has your question been resolved?

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[hoary herald]

How can I find a primitive kth root of unity and modulus n for powers of 2, k?
I want to find one of the answers, it doesn't have to be smallest or something

[limpid turret]

could you elaborate more what you mean, please?

[lone heart]

@hoary herald Has your question been resolved?

[hoary herald]

To phrase differently, given k, I want to calculate ω and n such that ω is primitive k-th root of unity under mod n.

[lone heart]

@hoary herald Has your question been resolved?

[lone heart]

@hoary herald Has your question been resolved?

[pallid scarab]

so you want a and n such that a^k = 1 mod n

well you need k = a multiple of phi(n) where phi is euler's totient

[hoary herald]

Yes
Thanks, let me search for that functions. Additionally if you have resources on actually computing these then lmk

[hoary herald]

Would that also be primitive or just kth root

[lone heart]

@hoary herald Has your question been resolved?

[hoary herald]

<@&286206848099549185> hi! If anyone can answer the followup question (1 and 2 message(s) above) lmk

[cunning needle]

Which follow up questions

[hoary herald]

Preferably both of them (as it says 1 and 2 messages above)

[cunning needle]

Can you reinput the questions to me as I can't find it, that's why I asked

[reef pier]

you might have better luck in #elementary-number-theory if you can't find too many helpers in the help channels who know how to help you

[cunning needle]

I see.

[hoary herald]

I'll rephrase the question as I might have found a better way to do so

Thanks btw

[hoary herald]

It's categorized under early university but I'm not in an university, this is for my hobby

Is it ok to post questions there even if so

[reef pier]

it should still be fine

[lunar urchin]

hi guys

[stone basalt]

Hi

[pseudo steppe]

Hello

[reef pier]

!redir

[lone heart]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[lone heart]

@hoary herald Has your question been resolved?

[hoary herald]

I asked in a different channel so let me close

[lone heart]

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[lone heart]

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[obtuse drift]

looking for help on setting this integral up

i have attempted rewriting in terms of x and then doing right minus left sqared times pi

but i am either doing something wrong or I am on the wrong track

@proud obsidian could you help me out on this one?

this is what i have right now, ad i just cant see what is wrong

this is the worksheet that this comes frm

[royal spoke]

can someone help me understand with the division?

[small lance]

!occupied

[lone heart]

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).

[obtuse drift]

yeah, i still havent been helped in this chat

.close

[lone heart]

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[glass plank]

Hi can you try find DEF, i feel like my answer of 67 degrees is wrong as when I use sine rule for DF its inconsistent on the side I use

[keen plinth]

i told you earlier that EDF is not 45 degrees

[glass plank]

so what do you suggest it is

[keen plinth]

thats something you'll have to figure out

its not necessary for finding DEF

[keen plinth]

why are you rotating it so that DF is horizontal

[keen plinth]

no

thats wrong

[glass plank]

can you explain why its wrong

[keen plinth]

this is the information you're given

[glass plank]

is my diagram not what the text descirbes

[keen plinth]

no

you've gone and rotated it

you've forced your incorrect assumption that DF is horizontal into your diagram, which has gone and made it incorrect

[lone heart]

@glass plank Has your question been resolved?

[glass plank]

so how would you find DEF

[keen plinth]

well if you ignore the entire bottom part of the diagram that you drew

the above is correct

[glass plank]

mhm so what would you do for the next step

[lone heart]

@glass plank Has your question been resolved?

[lone heart]

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[past escarp]

Hello...can somebody help me turn these into equations... I can solve for the rest,, please^^''

[jagged cobalt]

do you know the equation of a general circle?

[past escarp]

yeaa

[jagged cobalt]

what is it?

[past escarp]

Ax^2 + Bxy +Cy^2 + Dx +Ey + f = 0

oh wait

or is it x^2 + y^2 + Dx + Ey +F..

[jagged cobalt]

the form youre going for is
x^2+y^2+2ax+2by+c=0

but theres one thats a bit easier in my opinion

[past escarp]

what is itt

[jagged cobalt]

(x-a)^2+(y-b)^2=r^2
where the center is (a,b) and the radius is r

its what you get after completing the square on the other and simplifying

[past escarp]

ohhh the formula when the center isnt at the 0,0?

but how does that work...

[jagged cobalt]

this one isnt necessarily at (0,0)

a circle centered at the origin has the equation x^2+y^2=r^2

[jagged cobalt]

if youre aware of them, the (x-a) and (y-b) are translations for x and y respectively

which moves the center

[past escarp]

hold on..

wait lemme solve for both and imma come back here with my answers:-:

[jagged cobalt]

alrighty

[past escarp]

is this correct...

[jagged cobalt]

very nearly but no

[jagged cobalt]

your center is (-1,-4)
so its (x-(-1))^2+(y-(-4))^2=7^2

[past escarp]

OHHHH

so it makes it positive?

[jagged cobalt]

yup

[past escarp]

aaa ill brb

ohh so its just x^2 + y^2 + 2x +8y - 32?

im trying the other one rn

[jagged cobalt]

=0, but yeah seems alright

[past escarp]

aa i hate circles

but my friend says its x^2 + (y - 40)^2 = 625

[lone heart]

@past escarp Has your question been resolved?

[past escarp]

I need clarifications pls q^q

[proven forge]

sry wrong question hah

[past escarp]

😭

[proven forge]

i ll solve it fr y

[proven forge]

i am back, the wheeel question?

regarding the wheel question, take a look where your origin starts

at the ground, that means the centre of circle is going to be at 40m

r + the offset of the base = 25 + 15

how did we get the offset = total height - 2r

[past escarp]

offset of the base??

[past escarp]

so it is (0,40)?

but how did you get 40..

its 65 or the height, - 25?

[proven forge]

so the total height is 65m, and diameter is 50m, that means that something 15m in height is protruding from the top or bottom of the wheel, more reslistically a base on the bottom

[past escarp]

OH

So the answer is x^2 + (y-40)^2 = 625?

[past escarp]

HUH

the 15m wasnt mentionedd

does the diameter also apply vertically not only horizontally?

oh wait...

ok i think i get it now

[proven forge]

its not mentioned

but

its a text question

[past escarp]

is the origin still at (0,65) though?

[sleek girder]

that'd be a point way above-

[past escarp]

0,40?

0,45??

[sleek girder]

(0, 40) sounds like it

diameter is 50 so radius is 25
height is 65 so from top to origin is (65 - 25) because they're one radius away

[past escarp]

AAH

I GET IT NOW

[past escarp]

this is it?

[proven forge]

15 m is to the bottom of the wheel

the base

[past escarp]

65-15?

[proven forge]

the centre of wheel is 40m offseted from 0

[past escarp]

this isnt correct is it😔

[proven forge]

the equation is correct, if i see the number in bracket correctly, 40?

[past escarp]

yea

[proven forge]

on the drawing you marked the base as 15m tall but connected to the centre of the wheel

imagine it more as a circle that id standing on top of a block

[past escarp]

55 is the y?

i mean from the origin towards the centre

[proven forge]

the equation is correct

so 40

[past escarp]

oh..

thats it?-

[proven forge]

if you put in x = 0 you get two answrs

65 and 15

[past escarp]

wait i wanna get the two answers

aw

[proven forge]

that means the circle starts at height of 15 and ends st 65 (max height)

[past escarp]

Alr2

[past escarp]

but my first answer was correct?no?

[proven forge]

this one?

unfortunately no

[past escarp]

oh nonono

i meant the other one

[proven forge]

you made the centre of wheel be at height 65, meaning the max height would be 90m

[past escarp]

i just noticed that, sorry

i meant this one

[proven forge]

yeah i see it

[past escarp]

yipee

[proven forge]

close but your origin is on the wrong side

[past escarp]

wrong side??

[proven forge]

one sec

too lazy to count to 7 so i made radius 2

your equation is red, origin is mirrored through coordinate origin

[past escarp]

so what makes the origin wrong?

[proven forge]

the origin is supposed to be at -1,-4

your origins are at 1,4

[past escarp]

HUH

but thats the way our teach showed us to solveq-q

[proven forge]

try finding x and y that would make both of brackets 0

that is your circle origin

if you dont belive me you can graph your function at www.desmos.com

if your teacher says i am wrong she has beef with me 🖕👁️👅👁️🖕

[past escarp]

ok now im confused

why did the radius become 6.08 instead of 7

i just tried to convert it back

alr i think i got it now

thanks for the help everyone and sorry for the trouble q-q

.close

[lone heart]

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[ornate girder]

I have no idea how to go about this question; could someone help?

[still herald]

What is the sum of roots?

[ornate girder]

-b/a?

[still herald]

roots are clearly -d/c and -b/a

sum of roots = -k/6

k/6 = d/c+b/a

I also don't know what to do further

Lol

[ornate girder]

😭

[still herald]

wait

Leave this

[ornate girder]

mhm

[still herald]

you know d and b are constants

so the constant term in the quadratic = b*d right?

[ornate girder]

hm?

oh yes nvm

[still herald]

got it?

oh

[ornate girder]

yup

[still herald]

bd = -24
b is an integer.
so -24/d is an integer

[ornate girder]

ooohhh

omds thats all it was 😭

thank you so much!!

[still herald]

welcome

🙂

[ornate girder]

oh i have another similar question i didn't really get

i'm not sure what the two zeroes part means

[subtle sparrow]

2 zeroes means the function f(x) intersects the x axis at two points
so there are 2 roots, or 2 solutions to f(x)=0

[ornate girder]

ooh ok thats what i assumed

wait i'll try again

yeah i'm not getting anywhere 😭 ive used b^2 -4ac >0 and all i get is a>0

[south sierra]

Hep plsi dont understand position vectors

[ornate girder]

#❓how-to-get-help shows you how to get your own help channel!

<@&286206848099549185>

[subtle sparrow]

huh, none of the options actually seem valid

i'll go through my working again

oh wait, i got it

[proud obsidian]

have you learned differentiating yet..?

[ornate girder]

yup

but this is a SAT question so i don't think differentiation should be involved?

[subtle sparrow]

okay good, can you post your working?

[ornate girder]

i havent used differentiation in my working out 😭

[subtle sparrow]

there's a way to algebraicly derive the vertex of a quadratic

this is a common result so you're kind of expected to remember

[ornate girder]

ah ok

[subtle sparrow]

but essentially, u should get the x coordinate of the vertex to be -b/2a and the y to be -D/4a

once you get that, you can use the values given in the question to get 2 equations

[ornate girder]

wait im a bit lost where you got those from

[cedar kelp]

@ornate girder in this problem, you need to know what does Vertex actually mean

[ornate girder]

the turning point?

[subtle sparrow]

No

[cedar kelp]

it is all observation, no calculation

[subtle sparrow]

i think 77^2 will guide you from here

[cedar kelp]

maybe try to plot a parabola with that given vertex

it will be easier to visualize

[ornate girder]

where do I go from here?

[alpine sable]

Check out my soln...

[subtle sparrow]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[alpine sable]

Oops sorry..

But I hope @ornate girder understands...

[ornate girder]

I did the same working out as you at the bottom to get a>0

but I'm not sure how to get a+b+c

[alpine sable]

Yes so manipulate it to get a+b+c

[ornate girder]

oohh

ok wait I'll try that

[alpine sable]

See, u have to remember the formulae for the vertex of a general parabola

Which is (-b/2a,-D/4a) where D is the discriminant of the quadratic polynomial

[cedar kelp]

@ornate girder ayo sorry, I had to go somewhere

[ornate girder]

nw I'm still lost 😭

[cedar kelp]

you'll need a more relevant graph
since at x=1, they want negative value

now you'll probably wonder where does x=1 came form right?

[alpine sable]

77^2, pls let me guide him, I'm bored asf

[cedar kelp]

bro this questions doesn't involves any calculation at all

but if you think your way is good, then continue

[alpine sable]

Nope, we can I got the minimum value of a+b+c as - 6 so it helps

[cedar kelp]

@ornate girder ping me if you need any help later

[ornate girder]

I'd be interested in your methodtoo!! since this is a sat question so I don't think it should need too much calculating

[cedar kelp]

'kay

[cedar kelp]

see that thing

[cedar kelp]

and also this

[ornate girder]

mhm

[cedar kelp]

perfect

now from the given expression, what will be f(1}?

[alpine sable]

a+b+c is nothing but f(1)...

[ornate girder]

a+b+c

[cedar kelp]

since options for a+b+c are negative then f(1) should be negative right?

[ornate girder]

oohh

yes that makes sense

[alpine sable]

I found a very simple method to solve the question... 😆

[cedar kelp]

so considering all that, does this graph seems right?

[ornate girder]

yup

[cedar kelp]

like f(1) here will be somewhere under x-axis

[ornate girder]

yup that makes sense

[cedar kelp]

perfect

[ornate girder]

oh so it can't be below -6

and has to be -5

[cedar kelp]

yep

you got it my boy

[ornate girder]

wait that's way simpler than I thought 😭

thank you so much!!

[light aspen]

you can also use a>0

[cedar kelp]

thanks to jag and Raineifold too

[cedar kelp]

yeah that's a great point

[alpine sable]

Thanks 77^2 for appreciating me...

[light aspen]

since its a(x+2)^2 -6

[ornate girder]

.close

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[still herald]

. You know it bro

[cedar kelp]

.close

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[small lance]

Bruh, stop trolling

You might get banned for this

[still herald]

. I am very sorry.

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[alpine sable]

Hello, I have a question
what's the limit definition of the bigO notation??

[small lance]

!da2a

[lone heart]

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

[alpine sable]

sorry, accidentally sent it without completing

[small lance]

You’re good

[alpine sable]

<@&286206848099549185>

[real gazelle]

If f and g are both functions from R to R that are always positive, then lim x→∞ f(x)/g(x) being a finite number implies that f is O(g)

For example, x is O(x^2) because lim x→∞ x/x^2 = 0

Please only ping Helpers after your question hasn't been answered for 15 minutes tho

[alpine sable]

I see, sorry

& thx

.close

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[alpine sable]

So sin(2x) is not O(sin(x))?

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[limpid turret]

Predicting server load is still a very open question. If you look at cloud computing servies, you see that this is still a major point of contention. With that in mind, it's not "easy" to predict to extreme accuracy. But per hour, per day, and per year estimates can give you a lot to work with. Also I suppose per region.

Then you have to factor in major events that could lead to anomolous spikes that could skew your data

There's a lot to look into

but a simple approach would probably be a good essay topic

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[native temple]

So…according to Openstax Precalulus 2e..my answer to this question solve |x + 2| <= 6 is incorrect, but is this true if so why and what do I do better?

[sour verge]

Their answer would be wrong. Plugging in x=8 yields |8 + 2| = 10, which is not <= 6.

[pallid scarab]

x+2 is between -6 and 6 and your own solution is correct

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[high cypress]

not sure what the first step would be here

i know i have to work inside the bracket

but how could i possibly simplify this

hm

ah i see

i can do this inside the bracket:

log2(7x) - log2(10) x logx(16)

then i can

log2(7x/10) x logx(16)

maybe i can

turn it into

log2(7x/10) x 4logx(4)

ehh now

idk

hm

well now the whole problem is

log2[log2(7x/10) x 4logx(4)] = 3

hmm

multiply logs?

lets see...

turns into

idk

fawk

i forgot how to multiply logs

<@&286206848099549185>

help im stuck

i simplified it to

log2[log2(7x/10) x 4logx(4)] = 3

so fatr

far

whats next

!help

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

[high cypress]

<@&286206848099549185>

[proud obsidian]

first step would be to change the log 16 base x into base 2

[high cypress]

change of base?

so

logx(2^4)

then,

4logx(2)

then,

4 x 1/log2(x)

so now my equation is

log2[log2(7x - 10) x 4 x 1/log2(x)] = 3

[proud obsidian]

then, try to remove the other log2 function in the big brscket

[high cypress]

are you saying it is possible to cancel out the log2(x)

?

[proud obsidian]

just the log2 outside

[high cypress]

dunno how to do that

what can i do there

i forgot

[proud obsidian]

raise 2 to the power of both sides

you get log2(7x - 10) x 4 x 1/log2(x)= 8

[high cypress]

oh wait is that log10(2)

didnt notice

log2(7x - 10) x 4 x 1/log2(x)= 8

yes i see

i didnt know it was log10(2)

now

idk what to do from here

omg

can i make this turn into

7x - 10 x 4 x 1/log2(x) = 100000000

that looks wrong

yikes

but maybe it works

so now,

umm

[high cypress]

help plz

<@&286206848099549185>

[dawn spruce]

you can divide by 4 first

just to do a bit more simplification

[high cypress]

hm

oh to both sides?

[dawn spruce]

so now you have $log_2(7x-10)*\frac{1}{log_2(x)}=8$

[ocean seal]

AwesomeRat

[dawn spruce]

wait

htat

but on the other side is 2

[high cypress]

?

[dawn spruce]

I'm not looking at this problem wrong right

[high cypress]

$log_2(7x-10)*\frac{1}{log_2(x)}=8$

[ocean seal]

tokyomissionimpossible

[dawn spruce]

$log_2(7x-10)*\frac{1}{log_2(x)}=2$

[ocean seal]

AwesomeRat

[dawn spruce]

I suck at latex

but

[high cypress]

um

why 2

[dawn spruce]

because you divided by 4 on both sides here right?

[high cypress]

oh yeah

lol

ok so now

can i cancel out

log2

so then its

[dawn spruce]

how so?

[high cypress]

7x - 10

=2

7x = 12

12/7

[dawn spruce]

uh

[high cypress]

nope

frick

[dawn spruce]

I don't think you can cancel it out quite yet

[high cypress]

$log_2(7x-10)*\frac{1}{log_2(x)}=2$

[ocean seal]

tokyomissionimpossible

[high cypress]

so we are here

[dawn spruce]

yep

do you know the next step?

[high cypress]

can i multiply

[dawn spruce]

yea

yes you can

[high cypress]

but where

ohhh

log2(7x - 10)/log2(x) = 2

[dawn spruce]

yea

and then

[high cypress]

then,
7x - 10/x = 2

then,
7x - 10 = 2x

[dawn spruce]

uh

[high cypress]

omg

[dawn spruce]

one sec

hold up

I don't think you can do that yet I need to double check

a bit rusty on my log rules

[high cypress]

yeah i cant

[sour mica]

Do not divide anything by log2. Its like dividing the multiplication sign.

[high cypress]

nvm

[sour mica]

As long as there is a constant, you are stuck like that

make the log terms linearly separable

[high cypress]

?

[dawn spruce]

try multiplying by log_2(x) on both sides of this equation

(I think)

[high cypress]

ah

log2(7x - 10) = 2log2(x)

[dawn spruce]

yes

beautiful

now you're practically there

[sour mica]

Yea, so now you can undo the log

[high cypress]

log2(7x - 10) = log2(x)^2

[sour mica]

yep

[high cypress]

7x - 10 = x^2

x^2 - 7x + 10 = 0

[dawn spruce]

👍

[high cypress]

we did something wrong auugh