help-13

[crisp flint]

How do i find the least area of line when there's 3 pillars with r = 5 for every pillar and minimum required line width when tying pillars with line

[cedar kiln]

@crisp flint Has your question been resolved?

[cedar kiln]

@crisp flint Has your question been resolved?

[subtle harbor]

!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.

[cedar kiln]

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[raven sandal]

I have a factor graph and am trying to calculate the probability P(x2=1) with the given probabilities

[raven sandal]

Any help would be greatly appreciated

[cedar kiln]

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[raven sandal]

<@&286206848099549185>

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@raven sandal Has your question been resolved?

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@raven sandal Has your question been resolved?

[worn bramble]

what should we find?

[cedar kiln]

@raven sandal Has your question been resolved?

[warped coyote]

Is this the full question? It looks like you’re missing some information. What is f_a … f_f ? And how does x x’ relate to x y ?

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[wraith dagger]

Ransik

[cloud quartz]

that’s not a line thats just a point, or its two perpendicular lines

ah they’re probably referring to the actual lines x = 2 and y = -5, two separate lines

[bold hinge]

Nope

x=2, y=5 is a line

r(t) = (2,5,0)+(0,0,1)t

[cloud quartz]

Oh woah

[bold hinge]

This is the vector form

(Position vector) + (direction vector)×(free variable)

Because z varies freely

[cedar kiln]

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[near aurora]

Cos=-0,8 need to find sin and tg

[subtle harbor]

,tex .unit circle

[wraith dagger]

Moosey

[silent finch]

Unit circle won’t help

This just Pythagorean identit

[subtle harbor]

ah

ye

$(\cos(\theta))^2+(\sin(\theta))^2=1$

[near aurora]

Sin^2=1-0,8^2?

[wraith dagger]

Moosey

[near aurora]

Sin^2 is 1,64??

[subtle harbor]

no...

cos(theta)=0.8 yes?

[near aurora]

-0,8

[subtle harbor]

-0.8, alright

$(\sin(\theta))^2=1-(\cos(\theta))^2$

[wraith dagger]

Moosey

[subtle harbor]

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

[wraith dagger]

Moosey

[near aurora]

Will the - change to + ??

[subtle harbor]

,calc (-.8)^2

[wraith dagger]

Result:

0.64

[subtle harbor]

,calc 1-.64

[wraith dagger]

Result:

0.36

[subtle harbor]

,calc sqrt(.36)

[wraith dagger]

Result:

0.6

[subtle harbor]

technically, sin(theta) will have two answers since we were squaring before.

and so tgnt will have two answers

[near aurora]

I need the minus one

For both

So sin is -0,6?

[subtle harbor]

ye :)

[near aurora]

What about tg

[subtle harbor]

we can verify that sin(theta)=-.6 with Pythagorean identity

[subtle harbor]

for that, just use definition of tg

[near aurora]

Sin/cos ?

Would tg be 0.75?

@subtle harbor

[subtle harbor]

,calc (-.6)/(-.8)

[wraith dagger]

Result:

0.75

[near aurora]

Can you help me with a few more of these so i can get the hang of it

Ctg is. -sqrt3

Need to find sin cos tg

<@&286206848099549185>

[subtle harbor]

use pythagorean identity involving ctg

alternatively draw out the triangle representing ctg. Also you need to specify which of sin and cos are negative

[near aurora]

ctg =1/tg?

is this right?

ik tg and ctg now how do i find sin and cos

@subtle harbor

[subtle harbor]

$(\cot(\theta))^{2}+1=(\csc(\theta))^{2}$

[wraith dagger]

Moosey

[subtle harbor]

cot=ctg

csc=1/sin

[near aurora]

i havent lerned thoes ye cot and csc

so can i use 1+ctg^2=1/sin^2

[subtle harbor]

yeah :)

[near aurora]

soo sin is 1/2?

[subtle harbor]

it comes from this identity when you divide both sides by sin(theta)^2

ye :), or -1/2

sin and cos have to have opposite sign tho

since ctg is negative

[near aurora]

always?

[subtle harbor]

only since ctg is negative

yea

if your question mentions something about quadrant that's important to include as well

[near aurora]

i need help with one more and thats it

i need to find cos from

2tg^2a+3tga+1=0

@subtle harbor

[subtle harbor]

let tga=u

[near aurora]

ok

2U^2?

[subtle harbor]

so you get 2u^2+3u+1=0

apply quadratic formula

[near aurora]

ok now what?

ok wait

got -1

and -2/4

now what

[subtle harbor]

now since you know tga=u

tga=-1 and tga=-1/2

[near aurora]

ok so what now?

[subtle harbor]

there is another pythagorean trig identity involving tan and sec

sec=1/cos

tan=tg

$(\tan(\theta))^{2}+1=(\sec(\theta))^{2}$

[wraith dagger]

Moosey

[near aurora]

i havent lernd sec and thoes stuffso is it posseble to do it other way

tg^2+1 eaquels 1/cos^2

???

[subtle harbor]

you can also draw the right triangle that corresponds to tg=-1 and tg=-1/2

[near aurora]

no i need to do it with thes formulas

[subtle harbor]

alright, then yeah, use that formula

[near aurora]

ok what do i do now with this formula

[subtle harbor]

solve for cos^2

[near aurora]

do i substitute tg?

[subtle harbor]

ye

[near aurora]

-1 or -2/4

[subtle harbor]

both is good, since both are valid for tg

unless your teacher put any specification

[near aurora]

so cos is 1 and 2 is this correct?

[cedar kiln]

@near aurora Has your question been resolved?

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[loud stump]

how do I do this? I've used every standard convergence test I know

[raw gulch]

use criterion of cauchy

$\lim_{n \to \infty } \sqrt[n]{\left| a_{n} \right|}$

[wraith dagger]

Joanna Angel

[raw gulch]

and then, when you show the convergence of this series, using the theorem on the differentiation of power series, you will calculate the sum of the appropriate power series

[loud stump]

Ok I think I was able to prove convergence, but I have no idea how to use power series to find the sum, could you explain what you mean?

[raw gulch]

yes sure

look:

yoru series can be writtien in a scuh form:

$\sum_{n=0}^{\infty }\frac{\left( -1 \right)^{n}\cdot 3n}{4^{3n}}=3\sum_{n=1}^{\infty }n\left( -\frac{1}{64} \right)^{n}$

[wraith dagger]

Joanna Angel

[raw gulch]

do you agree on it ?

[loud stump]

yes

[raw gulch]

ok to evaluate its sum we have to replace

-1/64 with x

and we get power serries of the form:

$3\sum_{n=1}^{\infty }nx^{n}=S\left( x \right)\text{ }\text{ if}\text{ }\left| x \right|<1$

[wraith dagger]

Joanna Angel

[raw gulch]

since center of the series is x = 0

and

Radius R = 1

that si easy to find

so to find oru sum we ned to find S(x)

we start form geometric series

[loud stump]

I see

got it

thank you

[raw gulch]

$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\text{ }\text{ if}\text{ }\left| x \right|<1$

[wraith dagger]

Joanna Angel

[raw gulch]

yw )

etc

[loud stump]

.close

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[undone crystal]

sorry need help with this again

[undone crystal]

does anyone here know how to map 3d points onto a plane

(theyre all coplanar)

i found the normal vector

and established 2 perpendicular vectors along the plane

to serve as x and y

idk what to do from here tho

@old ridge

sorry for ping

i rlly wanna get this over with tho

😭

[waxen mortar]

Can you be more specific about what you mean by "map" in this case?

[undone crystal]

yea

so i have a bunch of points with x y and z

they all lie on the same plane

i want them to only have 2 coordinates

[waxen mortar]

Oh, that's a fun thing to do. You've got a few things to do. You've already found the plane in the original coordinate system, but you need another coordinate system to map them into.

Does this plane intersect with the origin of the original space?

[undone crystal]

it can but not necesarily

im doing this all with variables for a general case

[waxen mortar]

If it doesn't, does your use-case allow you to translate the plane and points such that it does?

[undone crystal]

use-case?

[waxen mortar]

You're asking for a reason, right? (personal interest, a project, a class, ...etc)

[undone crystal]

math paper

my math paper is

finding the area when a cube is sliced by a plane

of the polygon formed

[waxen mortar]

Oh, neat!

[undone crystal]

yea its alot of fun so far

i assigned the back right corner to be the origin

so now i have like

6 points in space

now i need to map them onto a single plane

so i can use gauss method

to find area

😄

[waxen mortar]

You'll need to decide on an origin for the coordinate system of your plane in some concrete way - could be a point. You'll need to establish basis vectors for your coordinate system - could use two other points on the plane. You may want to correct that to be an orthonormal basis - not too bad hard from there. If you do all of these things, you can probably produce a linear transformation that encodes all of that work.

[undone crystal]

ok since all the points are coplanar i can justy make r_1 the origin right?

then from there i have my concrete point (r1) and i have 2 perpendicular vectors

[waxen mortar]

Sure, just making any one of them the origin is a start.

[undone crystal]

i also have the nromal vector

so i have my plane fully established

but now how do i make it into x', y' ( ' = new plane)

rather than x, y, z

[waxen mortar]

So think about that point you made the origin. Is there another point on the plane towards which you could draw a vector? If so, normalize that vector. You've got your x' basis. Then, take the cross product of that and your plane's normal. Now, you've got a y' basis.

[undone crystal]

i already have my x' and y'

i took a random vector between 2 points

then crossed that with normal

so i got x' and y'

[waxen mortar]

Alright, fair enough. That'll do it. Now that those are real vectors and you have an origin, you need to set up a linear combination of those vectors that reach each other point.

Those linear combinations will be your coordinates in the plane's coordinate system.

[undone crystal]

uh what wopuld that look like

so like 6 individual vectors?

[waxen mortar]

Great question.

So

Take the vector from your origin (r1) to some other point (r2)

That's a perfectly valid vector, right?

[undone crystal]

ya

[waxen mortar]

Now project it onto x'

There's the coefficient you get is the x' part of your coordinate.

Do the same again for y'

and you have the y' part of your coordinate

[undone crystal]

ok

i think i get that

[waxen mortar]

If your x' and y' are normal vectors in the original 3d coordinate system, then your area won't need to be scaled!

[undone crystal]

awesome

[waxen mortar]

There's also another trick for getting that area, come to think of it...

[undone crystal]

im using shoe string method

but im open to sduggestions

[waxen mortar]

That's what I was going to suggest.

Nice.

[undone crystal]

also general question for my paper

do you think i need to delve into each case

[undone crystal]

a

cus thatll take so long 😭

i was hoping to make a general formula but i doubt thats possible

[waxen mortar]

That depends on your target audience, honestly.

[undone crystal]

yea your right

IB is so unclear

when it comes to ia's

ok thanks for all the help!!

.close

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[storm crescent]

is y=1/2(x+1)^2+5 a vertical compression by a factor of 1/2?

[cedar kiln]

Please don't occupy multiple help channels.

[storm crescent]

.close

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[hot rampart]

How do I find how many sequences of 100 consecutive numbers containing exactly 25 primes?

[hot rampart]

Not sure how to start

[amber ridge]

well, prime numbers are basically random, they don't have a pattern to them

are you allowed to write a computer program or smth

[hot rampart]

No you’re not.

[potent fractal]

I think it's useful to note that having 25 primes among 100 consecutive integers is very restricting

[hollow minnow]

Key observation:||for n>11, there are at least 76 numbes in the seqeunce n,n+1,n+2,...,n+99 divisible by 2,3,5,7 or 11 and also note that the sum of 2,3,5,7,11=2310.||

[hot rampart]

This seems quite random…

How do you prove that?

[potent fractal]

You first eliminate all even numbers - there're 50 of them

Other numbers are of form 2k+1

They can be ±1, 0 mod 3

[hot rampart]

Oh yeah I see now

Okay so

[potent fractal]

When k = 1 mod 3 we eliminate the number

Etc.

[hot rampart]

Okay lemme see

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[urban ermine]

how do i set this up?

[urban ermine]

i did $183=0.7x+0.3y$ and $93=0.1x+0.6y$

[wraith dagger]

Joshii

[urban ermine]

and im currently at the matrix $\begin{bmatrix}
0 & 1 & 120\
1 & 6 & 930\
\end{bmatrix}$

[wraith dagger]

Joshii

[urban ermine]

but i dont really understand what this would imply here

[cedar kiln]

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[cedar kiln]

@urban ermine Has your question been resolved?

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[split pier]

So my approach in this problem is:
Let c be in R. We construct a sequence x_n of rational numbers in (c, ∞) that converges to c. Then limf(x_n) converge to c² + 1.
Now i construct a sequence y_n of irrational numbers in (c, ∞) that converges to c but then limf(y_n) converge to c. Therefore we get two different sequences in (c, ∞) that converges to different limits by sequential criterion. So right hand limit at c doesn't exist of f(x). Hence second kind discontinuity. Is this right ?

[dire geode]

What does discontinuity of second kind mean

[split pier]

That left hand and/or right hand limit of the function doesn't exist at a point

[dire geode]

Seems fine overall. Maybe you have to cite density of rational and irrational in R

[split pier]

Uhh yeh that's right i think even my prof mentioned it at a slightly similar problem before

Thanks for checking

.close

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[latent totem]

Anyone?

[potent fractal]

You can use the summation we discussed yesterday

The one with exponents

[latent totem]

The problem not with the summation

It's the alpha and beta

Generalization for the roots

[drifting marlin]

,rcw

[latent totem]

To find some suitable n for that

[potent fractal]

The sum will be some function g(x)
g(x)=0 has a finite amount of solutions on the specified interval

Yeah g will depend on n

[latent totem]

I used the sin AP series though

[potent fractal]

A pair of those solutions should satisfy the last relation
Hopefully it gives a strong condition on n

[latent totem]

First I took alpha to be pi/n and 2pi/n as it's standard procedure
Then it didn't work

[potent fractal]

If you want to guess alpha then maybe you should search among alpha such that $e^{i \alpha}, ..., e^{n i \alpha}$ form a regular polygon on the complex plane

[wraith dagger]

EQUENOS

[potent fractal]

2pi/n was a good try but there are also 4pi/n, 6pi/n, etc.

[latent totem]

There would be many though

And n varies too

Brute forcing would extend it
I tried with n

[vocal crescent]

One second

[potent fractal]

It's reasonable to calculate the sum

So we get an explicit expression for g(x)

[latent totem]

The relevant terms of sum are
Sin(nx/2) sin ((n+1)x/2)

The numerators

For alpha and beta should satisfy these

[royal loom]

!occupied

[cedar kiln]

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).

[latent totem]

Hey I tried this

Alpha = a pi/n , beta. = bpi/n

Got me n = 37.5
Close in our range

If we do
Alpha = pi/n, beta = 4pi/n
We get n = 15

Can we manipulate it in some way to move n closer to our range?
@potent fractal

[cedar kiln]

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[latent totem]

.close

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[tawny drum]

Does anyone have any resources that can help me train for tests like this

[surreal karma]

Hi! For all?

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[tawny drum]

hi

[tawny drum]

yes

scroll up for my question btw

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[thin galleon]

How to approach proving:

[thin galleon]

$\lim_{n\to\infty}\frac{f(n)}{g(n)}= 0$

[wraith dagger]

kytsu1

[astral bay]

...depends what f and g are

[thin galleon]

🧐

[silver fable]

u can use big O definition

[thin galleon]

It is a task related to understanding the o expression, I am trying to figure out what the functions are.

[astral bay]

...ok wait can you just post the original task

[thin galleon]

The task is to prove

[astral bay]

...where does it say that that's the task

[thin galleon]

Sorry, this is the correct task:

[silver fable]

What is this book called ? please

[astral bay]

ok that makes a lot more sense

so probably the first step would be to look at the definitions of $\Theta$, $O$, and $\Omega$

[wraith dagger]

bee [it/its]

[thin galleon]

Introduction to Algorithms by Cormen, T

[silver fable]

ty

[cedar kiln]

@thin galleon Has your question been resolved?

[thin galleon]

O(g(n)) = the output of f(n) is never greater than cg(n)
Θ(g(n)): ~∃ f(n) outside the bounds set by c1g(n), c2g(n)
Ω(g(n)) means f(n) cannot be under the curve of cg(n)

c is a positive constant, n0 is the smallest meaningful value - I don't know why n0 is the smallest value

ln x has negative y values when x<1, so perhaps n_0=1 for ln x

O(g(n))={f(n) | f(n)∈∫_n0^∞{cg(n)}}

I think an analogous problem is to prove that A ⊂ B <=> A ∩ B = A

This is fun! 🙂

@astral bay do you think this is a good solution?

[astral bay]

...is what a good solution

as far as i can tell none of what you've said here is a proof of anything, just restatements of the thing you've been asked to prove

[thin galleon]

The Venn diagram shows the proof, that x is in A if and only if it is in B and C.

[astral bay]

ok so here's a proof that every number between 0 and 100 is prime

is this proof correct?

@thin galleon

[thin galleon]

B and C should be prime numbers

[astral bay]

why?

the venn diagram shows the proof, x is in A if and only if it is in B and C
i'm doing the exact same thing you did

your "proof" doesn't show that B and C are prime numbers

[thin galleon]

This is a good analogy, thank you

[thin galleon]

[astral bay]

[thin galleon]

The O is for O(g(n)), not zero

[astral bay]

yeah i know

i'm just giving you a proof that all numbers between 0 and 100 are prime

[astral bay]

so... is this proof correct?

the point i'm making here is
a "proof" of a statement isn't just a pile of symbols and various different obviously equivalent formulations of the statement

part of the point, the reason that proofs matter at all, is that you can only prove things that are true

so if you have a "proof" that works equally well at proving something false, it must not be valid

what a correct proof looks like is a series of steps where each step is necessarily true if the previous steps are true

producing out of nowhere the claim that $\Theta = O \cap \Omega$ is not a step that immediately and necessarily follows from nothing

[wraith dagger]

bee [it/its]

[astral bay]

if $\Theta$ is the set of primes, $O$ is the set of positive integers, $\Omega$ is the set of integers less than 100, then it's false, for example

[wraith dagger]

bee [it/its]

[astral bay]

and in fact nothing you can do by just symbol manipulation can ever prove that \Theta is the intersection of O and \Omega, because you'd always run into the fact that that's not always true

this is why i suggested looking at the definitions - you need to reference the definitions in the proof and deduce things from them

[astral bay]

?

[thin galleon]

I am trying to translate this problem to the prime number example, I find it difficult, but it is interesting. Thank you for the explanations, they make sense.

[cedar kiln]

@thin galleon Has your question been resolved?

[cedar kiln]

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[formal pasture]

how do i do 7

i dont get how to start it

[cedar kiln]

@formal pasture Has your question been resolved?

[formal quest]

let the circle be a unit circle, which is $x^2+y^2=1$. then how can you express Q's coordinate, using $\theta$?

[wraith dagger]

이재현

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[dreamy totem]

how is acceleration m/s^2. I understand that it’s messaging the change is meters per second per second, so it would look like 5m/s/1s. However, wouldn’t 1s jump to the top, making it 5m•1s/s canceling out the seconds. Pretty sure the last part is where I’m making a mistake.

[crimson delta]

(m/s)/s is different from m/(s/s)

you are thinking of the second

[dreamy totem]

true

but what is m/(s/s)

[crimson delta]

would just be m

[bronze hazel]

its a sign convention, it means: change of velocity with respect to time
as velocity is written as m/s and time is written as s, change in velocity with respect to time will be m/s per second, which makes it m/s^2

[dreamy totem]

So if a car can go from 0 m/s to 60 m/s west in 1 second then its acceleration would be 60m/s^2?

[bronze hazel]

yes

as the change in velocity per second is 60m/s, so acceleration will be 60m/s^2

[dreamy totem]

okay thank you

[bronze hazel]

welcome

[crimson delta]

well, average acceleration over that second. probably not the same during the whole second

[dreamy totem]

isn’t the average acceleration always different in between the start and end time

Also how can you find instantaneous velocity, position, and acceleration using algebra I thought that required calculus

@crimson delta

[crimson delta]

yes that usually requires calculus

except for some very simple cases I guess

[dreamy totem]

ok thanks

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[latent totem]

.rotate

help please

?

[cedar kiln]

@latent totem Has your question been resolved?

[latent totem]

.close

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[true galleon]

How would you solve this?

[inland sky]

first: there's a common factor in every term. what is it?

[true galleon]

X

[inland sky]

Ok, so that's factored out

[true galleon]

We can pull x out

[inland sky]

now we have x(x^3 + x^2 + 4x + 4)

Now: attempt to factor out something from x^3 + x^2 and 4x + 4 separately

[true galleon]

From x^3 and x^2 we can factor out x^2

[inland sky]

ok so we have x^2(x+1) and then for 4x+4?

[true galleon]

4

[inland sky]

and the other factor?

4 times

[true galleon]

But wait, what happens to the x we factored out beforehand too?

[true galleon]

2

[inland sky]

no, 4*x + 4 does not equal 4 times 2

[true galleon]

Oh hang on…

4(x+1)?

[inland sky]

Yes, so now we have x^2*(x+1) + 4(x+1)

do you see what we do next?

[true galleon]

(x^2 + 4) (x+1)?

[inland sky]

Yep, now you're done

so the final answer is x(x+1)(x^2+4)

[true galleon]

Thank you!!!!

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[inland sky]

since it says linear you might want to put (x+2i)(x-2i) for the x^2 + 4

[true galleon]

Really? Ok

[cedar kiln]

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[weak ore]

how is −4(1+k)(9) = -36-36k and not just 36k?

[jade charm]

Multiply

[weak ore]

my logic is that -4*(1+k) = -4k

[jade charm]

That is false

[weak ore]

and -4k*9=-36k

[jade charm]

You need to distribute

[weak ore]

okayy yeah i get it now

[jade charm]

If you're ever unsure about that sort of stuff, try plugging in a concrete value of k

For instance k=-1

[weak ore]

yeah thanks

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[formal pasture]

how did they get cos2x at end agaib

again

[glacial heron]

can you show the whole image

what was the question?

[formal pasture]

11a

[glacial heron]

so sin4x

[formal pasture]

yes

i got the first setp

but when they factored 2 out

i got lost

[crimson sedge]

hey,, can i get some help w math?..

[glacial heron]

2 * sin2x * cos2x?

[glacial heron]

#❓how-to-get-help

[formal pasture]

but when u take 2 out

wouldnt it be 2(sinxcosx)

[glacial heron]

no

[formal pasture]

hm

[glacial heron]

we can so a substitution method at first to make it easier to understand

let’s say u = 2x

[formal pasture]

okay

[glacial heron]

you then have sin(2u)

then use the double angle formula

and you get 2 * sinu * cosu

[formal pasture]

yes

[glacial heron]

replace u with 2x

[formal pasture]

wdym again

[glacial heron]

2 * sin2x * cos2x

mb

[formal pasture]

4x?

[glacial heron]

?

[formal pasture]

isnt that

sin4x

[glacial heron]

yes

[formal pasture]

okay

so what is next

i get those steps

[glacial heron]

just do the double angle formula again

for both sin2x and cos2x

[formal pasture]

how did they get (2)(2sinxcosx)(cos2x)

[glacial heron]

sin2x = 2sinxcosx

[formal pasture]

i dont get where thr last came from

[formal pasture]

yes

[glacial heron]

so 2(sin2x)(cos2x) becomes 2(2sinxcosx)(cos2x)

[formal pasture]

why didnt cos2x

become smthn else

[glacial heron]

they were probably focusing on sin2x first

[formal pasture]

ok

yo but how did 2sin2xcos2x

[formal pasture]

become

that

like they just subbed in

the doublr angle

[glacial heron]

yes

[formal pasture]

formulas?

ou

[glacial heron]

yup

that’s all

it’s still mathematically correct

[formal pasture]

bruh they couldve just said that

sob

[glacial heron]

oh well, now you know

[formal pasture]

okk tysm

:DD

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[manic cipher]

Goooodddd afternoon

[manic cipher]

I'm solving a question in logarithms that says i need to find the domain for f(x)=ln x^2-x-6. While solving, i ended up with the answer (-oo, -3)U(2, oo) as i used (x-3)(x+2) when factoring the problem. However, when using mathway to confirm my answer, it gives me (-oo, -2)U(3, oo)

I was wondering if the reversal of my answer actually matters or if there's something i'm doing wrong

When factoring x^2-x-6, -3 and 2 both multiply to -6 and add to -1, so i'd assume my answer was correct. But if the true answer is what math way gave me, what am i doing wrong?

here's the photo of the question

[glacial heron]

the domain for a polynomial is all real numbers

ah

[manic cipher]

I forgot to put ln my apologies

[glacial heron]

since the polynomial's domain is all real numbers, you just have to worry about the domain for the natural log function

[manic cipher]

what do you mean?

[mortal hemlock]

for what values is $ln(x)$ defined

[wraith dagger]

Flappie

[manic cipher]

what do you mean defined?

i'm sorry if i'm a little slow, it's been hard trying to understand this

[mortal hemlock]

what is $ln(-1)$?

[wraith dagger]

Flappie

[manic cipher]

-4? -1^2 is 1, so 1+1= 2 and 2-6= -4

so ln(-1)= -4

[mortal hemlock]

ln(-1) actually is not defined

[manic cipher]

Ohhhhh i see my mistake there

[glacial heron]

basically negative numbers

and 0

[manic cipher]

I mixed up ln with f(x)

okay so ln(-1) isn't defined

[mortal hemlock]

yes, and for what other values for x is ln(x) not defined

[manic cipher]

0?

[glacial heron]

that's one value, yeah

[mortal hemlock]

0 is one of them yes, but which other ones

[manic cipher]

Any negative value im assuming?

[mortal hemlock]

yes

[manic cipher]

Any value less than 0?

Ah okay

[mortal hemlock]

for any x<=0 ln(x) is not defined

[manic cipher]

Okay got that

[mortal hemlock]

what does this mean for the domain of ln(x)

[manic cipher]

I don't know

[mortal hemlock]

do you know what the domain is

just the term

[manic cipher]

wait the domain for ln(x)?

is that what you mean by the term?

[mortal hemlock]

if i say the domain of some function f(x), what do i mean by that

[manic cipher]

All i know is that the domain is the x's

[mortal hemlock]

but if x<=0

we concluded a moment ago that ln(x) is not defined for those x's

so would those x's be in the domain?

[manic cipher]

No?

[mortal hemlock]

correct

so then what is the domain of ln(x)

if x<=0 is not in it

[manic cipher]

(0, oo)?

[glacial heron]

bingo

[mortal hemlock]

indeed

basically, any positive number

[manic cipher]

Ah okay now i understand that

Okay

[mortal hemlock]

so then what if we have ln(x-3)

hwo does this change the domain

[manic cipher]

well then it'd be 3 and any positive number above it right?

(3, oo)?

[mortal hemlock]

is 3 included?

[mortal hemlock]

this is correct

[manic cipher]

I don't know

[mortal hemlock]

then what if we have ln(g(x))

what does g(x) need to be for this to be defined

[manic cipher]

I'm using parenthesis so i'm assuming not

[manic cipher]

Greater than 0?

[mortal hemlock]

if you have () brackets it means its not included, the [] means it is included

[mortal hemlock]

exactly

so now what if we look at the original question

what does that ask

[manic cipher]

it asks for the domain of f

[mortal hemlock]

if we look at the function

what is f(x)

[manic cipher]

and f(x)= ln x^2-x-6

[mortal hemlock]

*with an ln

[manic cipher]

Right sorry

[mortal hemlock]

so with what we have concluded earlier

what is in the case the g(x)?

[manic cipher]

I don't know

[mortal hemlock]

what did you say here

[manic cipher]

Oh right greater than 0

[mortal hemlock]

exactly

so then what about ln(x^2-x-6)

what needs to be greater than 0

for this function to be defined

[manic cipher]

x?

[mortal hemlock]

not quite

if x=-10

then ln(x^2-x-6)=ln(100+10+6)=ln(116)

so this is defined

[manic cipher]

That's because 116 is a positive number

[glacial heron]

hint hint interval test

[manic cipher]

if the number is negative, it would be false, if it was positive, it'd be true

[mortal hemlock]

how would you check if its positive or negative

[manic cipher]

Well, i know that in the number line after i factor x^2-x-6, i pick any number in each section (in beween -oo and -3, -3 and 2, etc.) so i would replace x with that number and solve the problem

then the answer i get should be either negative or positive

that's how i learned it

[mortal hemlock]

x^2-x-6 = (x-3)(x+2), right?

[manic cipher]

Yes

[mortal hemlock]

so, ln(x^2-x-6)=ln((x-3)(x+2)), right?

[manic cipher]

Yes

[mortal hemlock]

and we need this function to be defined, so whatever is inside the brackets of the ln needs to be greater than 0

[manic cipher]

Are you saying that whatever is in the domain has to be greater than 0? Is that what you mean by inside the brackets?

[mortal hemlock]

ln(stuff inside the brackets)

[manic cipher]

OH okay

[mortal hemlock]

this is what i mean by 'stuff inside the brackets'

x^2-x-6 is a parabola and by factoring it, we can easily see what the intersections with the x-axis are

[manic cipher]

Okay yeah i get that now

Okay

[mortal hemlock]

(x-3)(x+2) is that factorization, so what are the intersections?

[manic cipher]

-3 and 2?

[mortal hemlock]

close

[manic cipher]

wait

3 and -2?

[mortal hemlock]

yes

you can see it as x-3=0 or x+2=0

[manic cipher]

Yup

[mortal hemlock]

so if we take the point x=3

if we go up, does it go positive or negative

if we go down, does it go positive or negative

same for x=-2

[manic cipher]

Well, for -2 if it were to go up it would become positive right?

[mortal hemlock]

so, if i fill in x=-1 it will be a higher value then when i put in x=-2?

Might be easier to graph it so you can visualise it

[manic cipher]

Oh wait. If i'm replacing the x's with -1, the answer would come out to be -4. But if i replaced with -2, the answer would be 6.

So if we go up, it will become negative

[manic cipher]

i think so

[mortal hemlock]

this is what you get

[manic cipher]

I still haven't come to understand how the answer would be (-oo,-2)U(3, oo) and not (-oo, -3)U(2, oo). But i think im getting somewhere

Ah okay

[mortal hemlock]

we can see our points x=3 and x=-2

from (x-3)(x+2)

[manic cipher]

Okay

[mortal hemlock]

in this graph, for what intervals is this function positive

[manic cipher]

(0, oo) i'm guessing

cause anything below that is negative

[mortal hemlock]

for what values of x, do we get a positive value for y

[manic cipher]

Oh 3 and anything above

[mortal hemlock]

maybe phrased like this its better

yes, (3,$\infty$)

[wraith dagger]

Flappie

[mortal hemlock]

and what else?

[manic cipher]

and -2 and anything below

[mortal hemlock]

exactly

so we have the two intervals (-inf,-2) and (3,inf)

for which our function x^2-x-6 is positive

together with our knowledge of the domain of ln(g(x))

we know that ln(g(x)) is defined for g(x)>0

in our question, g(x)=x^2-x-6

with all of this combined

what is the domain of ln(x^2-x-6)

[manic cipher]

(-oo, -2)U(3, oo)

[mortal hemlock]

boom

[manic cipher]

OHHHHHH

I get it now

[mortal hemlock]

thats fantastic

do you also know why?

[manic cipher]

Why what?

[mortal hemlock]

why this is the solution

[manic cipher]

😅

[mortal hemlock]

if i gave you ln(x^2-5x+6) would you be able to give the domain aswell?

[manic cipher]

Probably yeah

I could give it a try

give me a min

[mortal hemlock]

or ln(x^2+3)

you can also just write your thought process here

[manic cipher]

Okay so i factored the first one you gave me. I factored it into (x-2)(x-3)>0

[mortal hemlock]

• =0, but yes

[manic cipher]

Oh in a video i watched they just put >

But = yeah that's what i mean

[mortal hemlock]

actually, that also works

mb

[manic cipher]

For the numbers in between and above/below, i chose -4, -2.5, and -1

since -2 and -3 are right next to each other, could either or just be used instead of a decimal?

like just use -2 instead of -2.5?

[mortal hemlock]

at -2 and -3 its 0

so it wouldnt give you much information

[manic cipher]

Oh okay

[mortal hemlock]

what you could do, is take the derivate and evaluate them at x=-2 and x=-3

[manic cipher]

After replacing x with -4 i got 42

[mortal hemlock]

42?

oh right, mb

ignore that

:)

[manic cipher]

Haha okay

[mortal hemlock]

so 42 is positive

and -2.5 and -1?

[manic cipher]

solving that now

[mortal hemlock]

oh wait

[manic cipher]

?

[mortal hemlock]

are you sure its at x=-2 and x=-3?

i see why i was confused now

[manic cipher]

well yeah, -2 + -3 = -5 and multiples it gives 6

[mortal hemlock]

if i fill in -2 at (x-2)(x-3)

i get (-2-2)(-2-3)=-4*-5=20

thats not 0

[manic cipher]

okay wait now i'm confused

[mortal hemlock]

yes (x-2) and (x-3) to get the factors

but -2 and -3 are not the roots

(roots are where the fucntion is 0)

the roots are x=2 and x=3

[manic cipher]

What?

Oh

Ohhh

Okay

[mortal hemlock]

because (2-2)(2-3)=0*-1=0 and (3-2)(3-3)=1*0=0

[manic cipher]

Okay okay

This is so much to think about...

[mortal hemlock]

it'll be muscle memory once youre used to it

[manic cipher]

I hope so

[mortal hemlock]

so back to ln(x^2-5x+6)

what is the domain

[manic cipher]

Wait before i get the domain, do i need to change the sign of the numbers in the line? Cause i used -2 and -3 originally so i should change that to positive numbers?

[mortal hemlock]

you want the roots

which are 2 and 3

so, yes? if trhats what youre asking

[manic cipher]

Oh okay

Is it (2,3)U(3, oo)??

[mortal hemlock]

almost

is (2,3) the correct interval?

if necessary, look at its graph again

[manic cipher]

(-oo, 2)?

[mortal hemlock]

yes

so its domain is (-inf,2)U(3,inf)

correct?

[manic cipher]

Yeah

[mortal hemlock]

so now what about f(x)=ln(x^2+3)

what is its domain

[manic cipher]

Would the factor for x^2+3 be (x-3)(x+3)?

[mortal hemlock]

no

if you work out those brackets you get x^2-9

dont try to factor this one

[manic cipher]

Oh

[mortal hemlock]

you can see at a glance

[manic cipher]

Okay

[mortal hemlock]

when is this function positive

[manic cipher]

I dont know

Is it positive at 3?

[mortal hemlock]

is x^2 ever negative?

[manic cipher]

No

[mortal hemlock]

so is x^2+3 ever negative?

[manic cipher]

No

[mortal hemlock]

so where is ln(x^2+3) defined?

[manic cipher]

any number greater than 0?

x>0

[mortal hemlock]

we just concluded that x^2+3 is never negative

[manic cipher]

So it's never undefined?

[mortal hemlock]

exactly

[manic cipher]

So if it's never undefined

[mortal hemlock]

the whole domain is just R

[manic cipher]

Would the domain just be -oo, oo?

[mortal hemlock]

yep

thats R

[manic cipher]

Oh okay

[mortal hemlock]

idk how to do the latex cool R thing

one sec

[granite knoll]

\mathbb{R}

[mortal hemlock]

$\mathbb{R}$

[wraith dagger]

Flappie

[mortal hemlock]

here

[mortal hemlock]

thanks

[manic cipher]

Oh

[mortal hemlock]

idk if you've had those numbers

anyway

idk if i explained it properly

the whole thing

im new to this

[manic cipher]

It's a bit difficult to understand but it's not your fault. I'm a little bit tired and have been studying for a while trying to master logarithms

[mortal hemlock]

were the small steps useful or annoying

[manic cipher]

Domain of logarithms is what i'm just having most trouble with

They were useful

[mortal hemlock]

glad to hear

[manic cipher]

I was able to identify some mistakes i've made

I'll just need to practice more. Thank you though for the help i really appreciate it

[mortal hemlock]

no problem

if your question is solved you can close the channel

[manic cipher]

Actually, if you don't mind, would it be okay if i can dm you if i need help with anything else? Despite my misunderstandings, you were really helpful and i'd like to learn a bit from you

[mortal hemlock]

yes sure, i dont mind

[manic cipher]

Oh cool thanks! I really appreciate it

.close

[cedar kiln]

Channel closed

Closed by @manic cipher

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[modest shadow]

how do i grpah this without a calculator

[cedar kiln]

@modest shadow Has your question been resolved?

[modest shadow]

<@&286206848099549185>

[nimble veldt]

It depends on how accurate the graph needs to be.

[modest shadow]

oh

well its like this

[nimble veldt]

well then just think how the function behaves. eg. if x grows -> 0.2/x goes to 0 -> e^(0.2/x) goes to 1 and f(x) goes to 2. and so on. what happens if x goes to 0 from the left, then from the right, ...

[modest shadow]

what abt the asymptotes

[nimble veldt]

i am not sure what you mean. what should be about asymptotes?

[modest shadow]

like how to find them

wihtout a calculator

[nimble veldt]

just as i said before: think how the function behaves. e.g. what happens to the graph if x grows?

what is with x = 1000, x = 10000, x = 100000, ...

[modest shadow]

oh ok

thanks .close

[cedar kiln]

@modest shadow Has your question been resolved?

[modest shadow]

didnt close it right 💀

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[empty berry]

Hi

[crimson sedge]

hi

[empty berry]

How r u

[upper garnet]

Good and u?

[empty berry]

Can anyone here teaches me grade 9 math

[crimson sedge]

probably

[empty berry]

Im tense about math

[royal loom]

If you have a specific question you need help with, ask it

Otherwise the point of these channels are not to teach you entire subjects

[empty berry]

Im not getting even a single thing

[crimson sedge]

about what

[empty berry]

yeah right

so let me give you question

[empty berry]

Algebraic expression and Formulae

[velvet hemlock]

ask ur question

[velvet hemlock]

are you familiar with laws of power?

[hollow minnow]

,,\expolaws

[empty berry]

Yeah kind of

[wraith dagger]

Pure

[velvet hemlock]

Ty pure

ye read it i guess

[empty berry]

yeah I know thanks

[royal loom]

Well if you know, these are all you need to solve that question

So where are you stuck?

[empty berry]

ok

[velvet hemlock]

maybe in (ii) u can use the fact that:
(x-a)(x+a) = x^2 - a^2

[empty berry]

I just dont know how to solve or go further like where to start

[velvet hemlock]

if ur not familiar with

lets start with (i):
u agree that 120/30 = 4?

[empty berry]

yeah

[velvet hemlock]

aight you agree that x^2/x^3 = x^-1?

which is also eqaul to 1/x

[empty berry]

yes

[velvet hemlock]

aight u agree that y^3/y = y^2?

[empty berry]

how

[velvet hemlock]

y^3 / y = y^3/y^1 = y^3-1 = y^2

Quotient of power

[velvet hemlock]

.

[empty berry]

oh ok

[velvet hemlock]

can u try now to complete it?

[empty berry]

ok

[velvet hemlock]

u can tag me if u need help with others

[empty berry]

is it 4x^-2y^2z^3

Is my answer correct @velvet hemlock

[velvet hemlock]

u sure x^-2?

[empty berry]

oh it should be 1 / x^2

but how will it fit in equation

Ok its 2:24AM

Gn

see ya'll later

It would be nice @velvet hemlock if you finish answering me

[velvet hemlock]

its x^-1

not x^-2

[empty berry]

oh

yeah my mind

ok Thanks

see ya later

bye

Gn

[cedar kiln]

@empty berry Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[graceful rose]

Suppose Lisa has a collection of 20 necklaces. Twelve are made of pearls, 6 are made of gold, and 2 are made of both pearls and gold. If she randomly selects 5 necklaces to wear to a gala event, find each probability.

P(2 pearls and 3 gold or 1 pearl and 4 gold)
P(at least 1 necklace made of pearls or gold, but not both)

[cedar kiln]

@graceful rose Has your question been resolved?

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

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[reef hare]

For this question, I think I got a) and b), in the sense that a) is just P(AandB) = P(A)P(B), and b) is just when they are mutually exclusive P(AandB) = 0 so it's minimu, but what's the condition for the maximum?

[cerulean sail]

Hint: if both A and B have happened, then A has to have happened, and B has to have happened...

[cedar kiln]

@reef hare Has your question been resolved?

[reef hare]

So it's... 1?

😭

[cerulean sail]

Not quite

You have to have A happen (which has probability 0.3) and you have to have B happen (which has probability 0.8) for A and B to happen both

[reef hare]

So that's the same as a)? assuming that they're indepdent and just using the multiplication rule

[cerulean sail]

Not quite either

[reef hare]

😭

[cerulean sail]

It's not that they're independent that you want to look for, but rather, something else (not mutually exclusive either)

[reef hare]

Ohhhhh

B is dependent on A?

or something like that

Actually A is dependent on B

like if B happens A is guaranteed to happen

or something like that

Those cat emojis are so cute

[cerulean sail]

That's the idea

I love the cat emotes

[reef hare]

So practically P(AnB) = P(B) because if B happens A is guaranteed to happen, as such P(AnB) = 0.8

[cerulean sail]

But for example, it's more likely I'm typing this in bed than it is that I'm typing this in bed and drinking hot chocolate, that's the idea

[cerulean sail]

Not quite P(B) either, almost though

[reef hare]

Wat

Wait

I'm

[cerulean sail]

Note that P(B) is bigger than P(A)

[reef hare]

Given that P(B) has happened...