help-23

[dull sequoia]

Oh yeah you can do that

[torn vessel]

so 1/cscx should not exist when y =0 ?

[dull sequoia]

Well it can’t ever be 0

Since the numerator is 1

[torn vessel]

ah ok makes sense

but where should it not exist ?

[dull sequoia]

Precisely this

csc has holes in its domain

[torn vessel]

but u said it never is 0

[dull sequoia]

That’s when it doesn’t exist

1/cscx = 0 is never true

[tidal imp]

I tried graphing this and desmos doesn’t even show the holes lol
So boooo.

[torn vessel]

where should the holes be ?

[tidal imp]

But yeah, if the denominator is undefined, the whole thing is undefined

[dull sequoia]

Where are the holes in 1/x

Or rather, discontinuities

[torn vessel]

at 0 ig ?

[dull sequoia]

At what 0

Be more specific

[torn vessel]

at x = 0

[dull sequoia]

Why x = 0

[torn vessel]

cuz the whole functions becomes undefined

[dull sequoia]

Ok

So for 1/cscx

Where will our hole(s) be at

[torn vessel]

cscx never becomes 0

so nowhere ?

[dull sequoia]

No, whenever cscx = 0

,w graph cscx

[flat frigate]

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[dull sequoia]

Ah wait

Whoops

[torn vessel]

but is has constraints for x right ? x should be R-npi

[dull sequoia]

Yeah

That’s the other thing

So whenever cscx = 0 then 1/cscx is undefined

In this case that’s never bc cscx doesn’t equal 0 anywhere

But csc also has a domain problem

And 1/cscx also has that

Cos if csc(a) is undefined then so must 1/csc(a)

[torn vessel]

yes

so when we compare it to a sinx graph , why is it similar even tho we have different domains ?

[dull sequoia]

It’s close

But it’s not the same

[tidal imp]

It looks similar because the only differences are in tiny holes

[dull sequoia]

1/cscx has holes

It has holes where cscx is undefined

[torn vessel]

where are the holes and how do we find them ?

[tidal imp]

Practically, 1/csc(x) is equivalent to sin(x)

[dull sequoia]

It’s not

[tidal imp]

But mathematically, wherever csc(x) is undefined

1/csc(x) will also be undefined

And that’s where the holes are

[torn vessel]

where is it undefined ?

at npi ,right ?

[tidal imp]

ye

[torn vessel]

so that means whenever x = pi , 1/cscx should not exist ?

[tidal imp]

yes, but not only pi

[torn vessel]

yea like all the pis

[tidal imp]

and zero

n can be any integer

[torn vessel]

yea

so there should be a hole on every pi and 0 ?

[tidal imp]

Yep

Basically wherever it “touches” zero on the graph, there’s a hole there in reality

[torn vessel]

but desmos doesn't show that ?

cool i understood

thanks everybody

[tidal imp]

Yeah, sorry for being disjointed

[torn vessel]

.close

[tidal imp]

Took me for a spin

[safe radish]

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[short epoch]

Hi so im a bit confused on what exactly im supposed to do to solve the problem asking for a value of c? like i know the IRC and ARC equation but im not exactly sure if using one or the other is actually needed to find c if c is part of the indefinite form of integral? some clarifying and guidance will help, ty!

[tidal imp]

No integrals here

Let’s start with ARC

[short epoch]

ok

[tidal imp]

If you have a closed interval [a,b] and you want to find the ARC of f(x) on that interval, it’s a matter of slope basically

[short epoch]

ohh so find values of when x = 2 and x= 0 then plug into arc?

$f\left(2\right)=\frac{1}{6}\ and\ f\left(0\right)=\frac{1}{4},\ \frac{\frac{1}{6}-\frac{1}{4}}{2-0}=-\frac{1}{24}$

[flat frigate]

Eggy 🍳

[tidal imp]

Yep

(The key word here btw is because of “rate of change”, hence we’re using slopes)

If we were asked for the average value, then we would use integrals

Now the next part is the IRC

How can we express the IRC of this function?

[short epoch]

we use the function place into the irc equation

[tidal imp]

Ye, and that entails finding what

[short epoch]

that shows the slope at a point (c?)

[tidal imp]

ye

You’d need the derivative

Evaluate it at x=c, and set it equal to what you got in your ARC

[short epoch]

ok so $I\ got\ lim_{\left(x\to c\right)}\ \frac{8+c}{c\left(x+4\right)\left(x+c+4\right)}$

[flat frigate]

Eggy 🍳

[short epoch]

would this be right format?

[tidal imp]

Hmm

You need the derivative of 1/(x+4)

[short epoch]

ohh

[tidal imp]

Yeah, whenever you see “instantaneous rate of change”

That should make you think of slope of a tangent line

And that should alert you to taking a derivative

[short epoch]

aa ok

i got $\frac{-1}{\left(x+4\right)^{2}}$

[flat frigate]

Eggy 🍳

[tidal imp]

Yeah

So what’s the IRC of the function at x=c?

[short epoch]

$\frac{-1}{\left(c+4\right)^{2}}$

[flat frigate]

Eggy 🍳

[tidal imp]

Yes

But there’s another thing we know about the IRC

at x=c

[short epoch]

yes value of c is in the interval [0,2]

[tidal imp]

And?

There’s a stated relationship between this particular IRC and the ARC

[short epoch]

and the irc and arc is identical

[tidal imp]

Yes

Which means?

[short epoch]

we can set both irc and arc equal to each other to find value c

[tidal imp]

Yep

[short epoch]

that makes sense ok

[tidal imp]

Now, what must c be?

[short epoch]

$sqrt(24)-4=c$

[flat frigate]

Eggy 🍳

[tidal imp]

Yeah

[flat frigate]

CST

CST

[tidal imp]

Now finally, we go to the last part of that question

And this is 90% conceptual knowledge

[short epoch]

i see yea the information threw me off so i wasnt sure how and now i know, ty!

[tidal imp]

No problem

[short epoch]

.close

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[lean otter]

The motion of a circular elastic membrane, such as a drumhead, is governed by the two-dimensional wave equation in polar coordinates [u_{rr} +\f1r u_r + \f1{r^2} u_{\theta \theta} = a^{-2} u_{tt}
]
Assuming that$\map u{r,t,\theta} = \map R r \map T t \map \Theta \theta$. Find ODEs satisfied by $\map R r, \map T t, \map \Theta \theta$

[lean otter]

I don't quite see how to do this just yet. Any nudges?

[plucky elk]

plug u into your PDE

[lean otter]

Oh I see now lmao

Ty

.close

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[eager niche]

Hello I'm working through Dummit & Foote's Abstract algebra and I'm stuck on this problem from Section 1.6, number 23 (which bodes well for the next 800-odd pages lmao)

[eager niche]

Let $G$ be a finite group which possesses an authomorphism $\sigma$ such that $\sigma(g)=g$ if and only if $g=1$. If $\sigma^2$ is the identity map from $G$ to $G$, prove that $G$ is abelian (such an automorphism $\sigma$ is called $fixed\ point\ free$ of order 2. [Show that every element of $G$ can be written in the form $x^{-1}\sigma(x)$ and apply $\sigma$ to such an expression.]

[flat frigate]

SherlockSage

[eager niche]

Basically I'm not sure how to 1) turn every element into $x^{-1}\sigma(x)$ and then 2) compute $\sigma(x^{-1}\sigma(x))$

[flat frigate]

SherlockSage

[feral linden]

You have

I use f for σ

Wrote something wrong, one min

Oh I am dumb. Okay now:

G->G

Mapping x to x^-1f(x)

This is injective:

If x^-1f(x)=y^-1f(y)

Then yx^-1=f(yx^-1)

Then yx^-1=1, y=x

Now any injective map from a finite set to itself is bijective

So this mapping is bijective

Therefore any g there exists x such that x^-1f(x)=g

Then f(g)=f(x^-1f(x))=f(x)^-1x=(x^-1f(x))^-1=g^-1

So f(a^-1b^-1)=f(a)^-1f(b)^-1 we have ba=ab

QED

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[eager niche]

Thank you very much!

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[lean otter]

How do I do this?

[safe radish]

@lean otter Has your question been resolved?

[median vigil]

we have a periodic function (sine) here. What is the average value of sin(x) and when does sin(x) reach that value?

[lean otter]

List out the infromation you know

What information don't you know

What equations can you use to figure out the missing information

[lean otter]

0, pi over 2..?

[lean otter]

I know midline, period, amipitude phase shift, and can find B

[median vigil]

so to find when sin(x) is at its average value we have to set up an equality of when the input will equal π

[lean otter]

What is B?

[lean otter]

Is it pi?

[lean otter]

Why though? Sorry I don't understand

[median vigil]

sin(x) = 0 when x = π right? so if we substitute x = π(t+0.23), we still find when x = π to find when sin(x) = 0

[lean otter]

Ohh

So we basically substitute it to find when sin of x is equal to zero

[median vigil]

yes, since that's its average value

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[wraith prism]

When 31513 ad 34369 are divided by a certain three digit number, the remainders are equal, then the remainder is

[wraith prism]

Is it solvable by modular arithmetic?

[cold aurora]

Yes.

Well, kind of. Modular arithmetic here also seems like normal algebra.

But yeah, in modular arithmetic, just keep going by the rules. Lol

[wraith prism]

ohh

i got now one thing

34369=r mod (abc)
31513=r mod(abc)

2856=0 mod (abc)

[cold aurora]

Yes. Nice.

Although it's probably bad notation to write a three digit number as abc.

,w 34569 - 31513

[flat frigate]

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[cold aurora]

@wraith prism

[wraith prism]

34369

[cold aurora]

Ah. You wrote wrong here.

Anyway, now just factorise 2856 and there's your answer.

[wraith prism]

Can you explain why we are doing it?

This is the gap between numbers? Which should be divisible by a three digit number perfectly

[cold aurora]

Do you know what 2856=0 mod (abc) means?

That number abc should divide 2856.

So, abc is a factor of 2856.

[wraith prism]

2^3 ×3×7×17

Options are
95
96
97
98

@cold aurora

[cold aurora]

Did you find the three digit number?

[wraith prism]

There are many

[cold aurora]

Yes. List them.

Are you sure that your question gives nothing else?

[wraith prism]

136 168 357 408 952

Yes. Question is completed

[cold aurora]

289?

[wraith prism]

It is up to us to use any theory

[cold aurora]

That shouldn't be there.

[cold aurora]

Ah. Makes sense. Actually, take any number, they'll all give the same remainder.

So, just do with any one of them.

[wraith prism]

I found a different solution on google

@cold aurora

I guess all the 3 digits factors will give the same remainder

[cold aurora]

Already said that here.

[safe radish]

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[wraith prism]

a^2-4<0

[wraith prism]

a^2<4

a<+-2

a<2

and a<-2

[calm bridge]

(a+2)(a-2)<0

[wraith prism]

yes

[calm bridge]

wavy curve?

[desert pasture]

for such a problem, just graph it instead

[wraith prism]

I have solved it by desmos I am thinking about it more

[desert pasture]

don't use desmos, plot it manually

[wraith prism]

I don't know why I am stucked today on this little thing

[cold aurora]

Thank goodness. At first i thought, you were suggesting desmos or sth.

[wraith prism]

I have plotted by random numbers

[calm bridge]

just plot y = x^2

and observe when is it less than 4

[wraith prism]

I want to learn that inequility sign where i am confused

[fiery merlin]

You can do it without graphing as well:

a² - 4 < 0

Solve a separate problem to see which a values are boundary values where it switches from being true to being false.
a² - 4 = 0
(a + 2)(a - 2) = 0
a = -2 or a = 2

We have three ranges: (-infinity, -2), (-2, 2), and (2, infinity). Try a number from the middle of each of those three ranges to see whether they make the inequality true. Then, try the boundary values -2 and 2 and see if they're true or false. Then, write inequalities for when they're true.

[wraith prism]

yeah i did the same

[calm bridge]

wavy curve^^

[wraith prism]

thanks chai

[fiery merlin]

No problem.

[wraith prism]

i need to learn it

[desert pasture]

This might help you:- https://brilliant.org/wiki/wavy-curve-method/.

[calm bridge]

what rex did is what you do in wavy curve (but skipping these steps)

[wraith prism]

got it

+-+-+-+-wavy curve

just watched an awesome video

which says just watch the powers and write critical points

[calm bridge]

those are not critical points

[wraith prism]

so if powers are even then don't cross if odd then cross the cruve

above line > below <

so this is the whole story of wavy bhai

[lean otter]

Yeah

[wraith prism]

so let us back to our main sign thing
a+2(a-2)<0
a<2

a<-2?

inquility failed?

[lean otter]

Yes

Therefore a belongs to (-2,2)

[wraith prism]

hmm true

.close

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[gentle zodiac]

can someone explain

[gentle zodiac]

so an orthogonal linear transformation is a matrix that preserves

length

and angle of vector

right?

[light shoal]

yes

[gentle zodiac]

then how is

the question false exactly

🤔

[obsidian oracle]

We can't see the full statement 3

[gentle zodiac]

it is the full statement

it's a true/false question

[obsidian oracle]

is an orthogonal matrix?

[gentle zodiac]

yes

it's a typo im guessin

isn't an orthogonal linear transformation and orthogonal matrix?

how is the 2x2 matrix provided orthogonal exactly?

[obsidian oracle]

wait a bit

[gentle zodiac]

they're not right?

[obsidian oracle]

that's the point

it's not orthogonal

[light shoal]

the statement is true for any orthonormal basis

not necessarily any arbitrary basis

[obsidian oracle]

a reflection is an orthogonal linear transformation

[gentle zodiac]

i think my prof wrote it wrong then

cuz

the 2x2 matrix is not orthogonal

[obsidian oracle]

that's the point

[light shoal]

he' showing you a counterexample

[gentle zodiac]

o wait

ok

wait

[obsidian oracle]

so where the statement is wrong is "any basis",

the statement becomes true if you add orthonormal basis

[gentle zodiac]

ok

[light shoal]

the change of basis matrix is $P = \begin{pmatrix}1 & 1 \ 0 & 1 \end{pmatrix}$, try computing $PMP^{-1}$ where $M$ is the given matrix

[flat frigate]

Bungo

[light shoal]

the result is an orthogonal matrix

[gentle zodiac]

what's the inverse of p again

[1, 0] [-1, -1]?

[light shoal]

>> inv(P)
ans =
     1    -1
     0     1

i'm lazily using matlab haha

[gentle zodiac]

o yea mb

ok ty

.close

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[lean otter]

Given the function f(x) = 1/3x^2 - 6x + 5, determine the inverse relation.

[lean otter]

determine the inverse relation? how do i do that

.close

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[junior atlas]

mg-\left(B+D\right)\frac{dx}{dt}=m\frac{d^{2}x}{dt^{2}} how to solve this second order DE to get x(t)

[junior atlas]

Where m is mass, g is a constant, B is a constant, D is a constant, x is displacement and t is time.

[worthy hemlock]

$mg-\left(B+D\right)\frac{dx}{dt}=m\frac{d^{2}x}{dt^{2}}$?

[flat frigate]

dldh06

[pseudo scroll]

,w y'' = ay' + b

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[umbral swan]

@junior atlas you first solve for x', which you can do by first dividing by m, thereby leading to the form mentioned by Neon. That way you get an exponential function plus a constant for x'. Then you can integrate x' for x

which yields the result wolfram spits out

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[torn vessel]

won't the greatest integer function be an odd functions always ?

[torn vessel]

instead of writing cases like [x] + [-x] = 0 when x belongs to I and [x]+[-x] = -1 when x doesn't belong to I , can't i just say that its an odd function ?

[plucky elk]

,w ceil(pi)

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[plucky elk]

Or is GIF the floor?

[torn vessel]

i think it is ciel , the function which goes like :

[1.2] = 1
[-1.2] = -2

[umbral swan]

I'd presume ceil, either way it isn't odd

it's only odd if f(-x)=-f(x)

you've just provided a counter-example

[torn vessel]

but if we see it's graph it is odd right ?

[umbral swan]

how is [1.2] = 1 for ceil

[umbral swan]

isn't ceil(1.2) = 2 and ceil(-1.2) = -1

that'd already show it's not odd

[torn vessel]

idk 😭 my teacher said its a GIF function

then it must be floor

[torn vessel]

i am sure that it does this

[umbral swan]

yeah looked it up as well, it's floor, but still not odd

floor(1.2) isn't -floor(-1.2)

[torn vessel]

but floor(1) = -floor(-1)

right ?

[umbral swan]

yes

[torn vessel]

so what is it ?

even or odd ??

the whole floor function

[umbral swan]

if you refer to the floor function only for whole numbers, it's odd

[torn vessel]

i c i c

[umbral swan]

whole*

[torn vessel]

but when you look at the graph of the floor function , it resembles that of any odd function . So can't i say its always odd ?

ik numerically it won't be but what about graphically ?

[stoic dune]

One implies the other. Either your graph is not odd, or our argument is wrong.

[safe radish]

@torn vessel Has your question been resolved?

[torn vessel]

i c

thanks guyys

.close

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[lean otter]

I have already solved this question but it would be nice if I get some verification that my answers are correct

[covert yoke]

Ok, could you show your answers and work then?

[lean otter]

My handwriting is very bad and I did it with h3 pencil cuz I don't have a pen right now💀

But for first part I needed to find

P(w>12.4)

In standard normal distribution thats

Let z be variable of standard normal distribution

Then I need to find

P(0.8<z<inf) where inf is infinity

Then I used stand normal distribution table

So after using the table and doing some manipulation

The answer I got is 0.2118

For the first part

[covert yoke]

I can confirm this is correct

@lean otter

Or at least if we made a mistake we both made the same mistake

[lean otter]

Thanks sir

Similarly for the second part the answer is (1÷1.88)

Which is equal to 0.53 approximately

Similarly for the third part the answer is 11.657

[covert yoke]

I have confirmed your second answer

Working on the third

My answer is very close to yours for the third, but it disagrees in the first decimal place

Ah, I made a transcription error

Now it's closer but disagrees in the second decimal place

I get 11.6776

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[lean otter]

.reopen

[safe radish]

✅

[covert yoke]

Hello again

[lean otter]

Its probably cuz my prof told me not to interpolate and just take the closest value that corresponds to the value so it's cool

[covert yoke]

Oh great

Each using tables can be tricky

I just use a calculator

[lean otter]

So our answers are probably right

Ohh I see

Thank you very much sir

.close

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[covert yoke]

,wa solve[1/2 erfc((x-12.5)/(sqrt(2) 0.5)) == 0.95, x]

[safe radish]

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[blazing dew]

Whats the difference between big N and small n?

[split ether]

The small n is a running index, it goes through every integer from 0 to N

So N is the highest value n takes

[blazing dew]

so technically its small n

[split ether]

? No

N is a constant and n is a running index

[blazing dew]

i dont understand

[split ether]

How can N be the same as small n if n could be any integer from 0 to N?

[blazing dew]

if n is a running index, what is the point of N? is it just neglected

[split ether]

No, N is the highest value for n

[blazing dew]

let me actually put this in context of a question

[split ether]

n starts at 0, then it's 1, then it's 2

And it stops at N

[blazing dew]

im trying to answer 1b

this is what ive gotten to so far

and im not exactly sure how to insert e^(2i/n) into the summation formula

[split ether]

You rewrite it as (e^(2/n))^i

And now it's a geometric series

[blazing dew]

an infinite one right?

[split ether]

No, you first take it for some n and then let n go to infinity

It would be the same yeah but there's 2/n in front of the sum so

[blazing dew]

im not exactly sure what to do with this though

[split ether]

Alright, so, given the formula [ \sum_{i=1}^n r^i = \frac{r^n - 1}{r-1} ] What is the following equal to? [ \sum_{i=1}^n\left(e^{\frac2n}\right)^i ]

[flat frigate]

A Lonely Bean

[blazing dew]

hm well if r=e^2/n

[split ether]

Yeah you can let r = that

[blazing dew]

however, my issue is that i have no idea how the i is carried

unless the i becomes n

[split ether]

Again, i is a running index, i could be any integer out of 1, 2, 3, ..., n

[blazing dew]

so n is just the highest value just like how N was before

[split ether]

Yes

[split ether]

Btw I meant to put i = 0 in the first sum here

[blazing dew]

why is it i=0 instead of 1

[split ether]

I mean it's just what the formula says, should start at i = 0

You want to go over the proof?

[blazing dew]

uhhh no

lmao

[blazing dew]

the highest value of n is infinity

[split ether]

That does not change what I said

[blazing dew]

no it doesnt

[split ether]

Infinitely does not have a value btw, it's not a number to begin with

[blazing dew]

yeah

so if I input infinity here

i get an exponent of 0

[split ether]

Yeah but you aren't yet supposed to do that

[blazing dew]

what

[split ether]

The thing is that what you wrote is is $\sum_{i=0}^n\left(e^{\frac2n}\right)^i$

[flat frigate]

A Lonely Bean

[split ether]

But we had $\sum_{i=1}^n\left(e^{\frac2n}\right)^i$ there

[flat frigate]

A Lonely Bean

[blazing dew]

true

[split ether]

So you are gonna have to subtract 1 from there

[blazing dew]

from the running index?

[split ether]

No I meant from the whole expression

[blazing dew]

why the whole expression?

if there is a difference of 1 in the index

[safe radish]

@blazing dew Has your question been resolved?

[safe radish]

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[marsh oracle]

Hello

[lean otter]

Hello

[marsh oracle]

Can someone help me with my math here

[lean otter]

Yes

[marsh oracle]

Ok

So im at the last question and it says what is the x equal to in this equation

x² + 5x + 10 = 0

@lean otter

[lean otter]

Do you know how to complete the square?

Do you know the quadratic formula?

[marsh oracle]

No 😔 im bad at math i need to solve this question fast

I missed many days in school #¥

So ye

Noo did u go

[lean otter]

You will not be able to solve this problem fast

[marsh oracle]

Oh dang

[lean otter]

Your first step is finding a book for learning algebra

Do you already have one?

[marsh oracle]

i have my math book with me

[lean otter]

How much of it have you read and fully understand?

[marsh oracle]

Well our 9th grade classes started like yestr day so ig just afew pages lol

[lean otter]

algebra 2?

[marsh oracle]

Btw our teacher is the same for the 3 grades so they teached us 9th grade stuff in 8th grade but i forgot

Lol

[lean otter]

Well, if this is the work on the first day

All you have to do is read the textbook, I should be in there

shouldn't take long to find

[marsh oracle]

Oh ok

[lean otter]

don't skip pages

[marsh oracle]

Im tryna catch up to the other guys but its so hard 😔 since i kept skipping school in every grade oof

@lean otter btw did you learn math in school or did you just watch online videos to learn it

[lean otter]

Khan academy then textbooks

[marsh oracle]

Oh wow

Ohh wait i know the quadrtic formula when i saw it i remembered it i acttualy learned it i just didnt know what it was called so i was confuaed

Or acttualy i guess just a little bit lol i saw a lot of equations that i didnt know there xd

[tranquil timber]

do you know which formula you have to use

if quadratic, a =1, b = 5, c=10

b is before the x coefficient

a before the x^2

and c is a constant

[marsh oracle]

💀

[tranquil timber]

plugging this into the formula should give you the solution

[marsh oracle]

Translating and talking english is hard oof

Lemme translate all the math things you said rq

[tranquil timber]

do you know how the quadratic formula looks like

[marsh oracle]

I searched it up

So ig ye

[tranquil timber]

yes

that formula is for equations such as ax^2+bx+c=0

[marsh oracle]

Ye

[tranquil timber]

so for x^2+5x+10=0

(1)x^2 + (5)x + 10 = 0

now you know a = 1, b = 5, and c =10

[marsh oracle]

Oh

@tranquil timber are you at 9th grade too

[tranquil timber]

8th actually

[marsh oracle]

Lol ur smart

[tranquil timber]

thanks

hope this solved your problem

[marsh oracle]

Ye ima try to learn things i missed in 8th and 7th grade lol but ye thanls

Thanks"

[tranquil timber]

anytime

i recommend using khan academy to catch up

[marsh oracle]

Ye that was the first thing that popped up when i searched quadratic formula lol

[tranquil timber]

yeah

[marsh oracle]

Ima look at theyre videos and stuff

[tranquil timber]

good idea

[marsh oracle]

😁

[safe radish]

@marsh oracle Has your question been resolved?

[safe radish]

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[inland salmon]

I have a question

[inland salmon]

A gambling game involving two players.

Player 1 starts with a d100 and rolls it on the first turn, and we can call the outcome of this dice “x”. If x=1, then player 1 wins. Else, Player 2 then rolls a dx and we can call the outcome of this roll “y”. If y=1, player 2 wins, if not, the cycle repeats.

What is the probability that player 1 wins?

[safe radish]

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[safe radish]

@inland salmon Has your question been resolved?

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[slow ivy]

someone please help me

insert two arithmetic means between -4 and 5

[lean otter]

What have you tried?

Do you know like the formula for an arithmetic mean

[slow ivy]

yes

i got the answer -3 😭😭 but it seems to be wrong

should i send my solving?

[lean otter]

Yeah of course

[lean otter]

Ur algebra here is like

Wrong

[slow ivy]

does it turn into positive? 😭

[lean otter]

Yeah

[slow ivy]

Okay ty

[lean otter]

\env{align*}{
5 &= -4 + 3d \\
5 +4 &= +3d
}

[lean otter]

Sorry for late responses btw

[slow ivy]

its alr

[lean otter]

Discord on phone mieeries

[slow ivy]

thanks for the help!!

[lean otter]

Nw

[safe radish]

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[latent heath]

I was watching a video , in that video he turn square into root in opposite side , please see image for clarification , how that possible ,
I'm 21 years ago , i remember I used to do that in school , turning + into - and × into ÷ when bring number opposite side of equal ,
My question is how square turn into root , what if it as cube or 4 or 5 ,,,

[fluid token]

square root of both sides

[latent heath]

okay

.close

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[steep vessel]

factorizing 16x^2 -1

[stable inlet]

hint: 16 = 4^2

[steep vessel]

(4x)^2

-1

[dull sequoia]

What does that look like

[steep vessel]

a^2-b^2

yay

that's all i need to know thank you

.close

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[steep vessel]

factorize x^2+x+1/4 i dont understand where is the relation between these

[safe radish]

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[steep vessel]

factorize x^2+x+1/4 i dont understand where is the relation between these

[stable inlet]

hint: 1/4 = (1/2)^2

[steep vessel]

ok so now i see the a and b but how about that isolated x

[stable inlet]

$a^2 + 2ab + b^2 = (a+b)^2$

[flat frigate]

Toblerone

[steep vessel]

2*x.4/2 is not x

so it doesnt work

[stable inlet]

huh?

2* 1/2 = 1

[steep vessel]

oh

lol

thx

[stable inlet]

np

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[steep vessel]

(2x-4)(x+1) -2+x factorize

[steep vessel]

do we have to factorise or expand first

[full oar]

expand brackets/parenthesis, simplify then try to factorise

[upper mulch]

Also a method to use is to change -2 + x 's position to $(2x-4)(x+1)+x-2$

[flat frigate]

Good

[steep vessel]

so its 2x^2-x-6

[full oar]

yep

then you can simplify

[steep vessel]

does the change of position do something? like it might help us find the identitie?

identities

[upper mulch]

If you factorize the expression $$(2x-4)(x+1)+x-2$$ You might find something in common.

[flat frigate]

Good

[steep vessel]

so we shouldn't expand

[upper mulch]

Expand is a way, this is also a way.

[steep vessel]

the question says that i should factorise, if i expand i'll have 2x^2-x-6, how can i factorize this?

[upper mulch]

A way is to use the quadratic formula, or factorize it.

You'll be focusing on the -x.

[steep vessel]

i didnt study it yet

[upper mulch]

Try to change the expression to -x, so that you'll have something like a(x-b)+c(x-b)

[steep vessel]

the book i took this from is just some simple math lol

[upper mulch]

So you're going to factorize it, right?

[steep vessel]

so technically it shouldn't be complicated right?

yes

[upper mulch]

Then look at the message I sent. Expand -x to something else, so that you have $$a(x-b)+c(x-b)$$

[flat frigate]

Good

[steep vessel]

so $$(2x-4)(x+1)+x-2$$ would become $$(2x-4)(x+1)-x+2$$

[flat frigate]

Jasblox

[upper mulch]

The hint that I gave you was to change the expression to $$(2x-4)(x+1)+x-2$$

[flat frigate]

Good

[upper mulch]

Now, is there something to factorize?

[steep vessel]

yes i did in my paper

[steep vessel]

i dont think so

[upper mulch]

Take another look.

[steep vessel]

is it the third identity

?

i mean the 1st

[upper mulch]

Could you write it?

[steep vessel]

a^2+2ab+b^2

[upper mulch]

Not quite. I'll make the range smaller $$(2x-4)(x+1)$$

[flat frigate]

Good

[upper mulch]

There, is there something we can factorize?

[steep vessel]

*2

[upper mulch]

Very good.

What is the expression after?

[steep vessel]

how abt that 4?

x-2

[upper mulch]

Now, you got a factorized expression $$2(x-2)(x+1)+x-2$$What is something in common.

[flat frigate]

Good

[steep vessel]

ooooh

x-2(2+x+1)

$x-2(3+x)$

[flat frigate]

Jasblox

[upper mulch]

Not quite

[steep vessel]

oop

[upper mulch]

So the common factor is (x-2), factorize it.

[steep vessel]

?

[upper mulch]

You made a small mistake, look over again.

[steep vessel]

oh cuz nothing multiplies the second x-2

[upper mulch]

Something here is wrong

[steep vessel]

$(x-2)(2+x+1) +x-2$

[flat frigate]

Jasblox

[steep vessel]

but 1 multiplies the second x-2

so $(x-2)(2+x+1+1) $

$(x-2)(2+x+1+1)$

[flat frigate]

Jasblox

[steep vessel]

$(x-2)(x+4)$

[flat frigate]

Jasblox

[upper mulch]

You'll have $$(x-2)(2(x+1)+1)$$

[flat frigate]

Good

[steep vessel]

i dunno why but now i gtg

Tysm for ur help!!

[upper mulch]

You're welcome.

[steep vessel]

if anyone can make me understand, leave it here, and i'll come read them after 🙂

[safe radish]

@steep vessel Has your question been resolved?

[steep vessel]

<@&286206848099549185>

.close

[safe radish]

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[steep vessel]

.reopen

[safe radish]

✅

[steep vessel]

$(2x-4)(x+1) -2+x$ factorize

[flat frigate]

Jasblox

[steep vessel]

i came to this deduction $(x-2)(2+x+1+1)$

[flat frigate]

Jasblox

[steep vessel]

but i was told it was wrong

[obsidian oracle]

so the good first reflex is spotting that x-2 will be the common factor

in that case, $(2x-4)(x+1) = (x-2)(?)$

[flat frigate]

rafilou2003

[steep vessel]

2

$2(x-2)(x+1)+x-2$

[flat frigate]

Jasblox

[steep vessel]

hello?

[fresh owl]

you want to factorise that is that correct?

@steep vessel

[steep vessel]

that's exact!

[fresh owl]

alright. Do you see a common factor?

[steep vessel]

yes

x-2

[fresh owl]

which is..?

correct

so what will this become

[fresh owl]

^

[steep vessel]

$(x-2)(2(x+1)+1)$

[flat frigate]

Jasblox

[steep vessel]

thx

[fresh owl]

ya got it

so the final form is $(x-2)(2x+3)$

[flat frigate]

xtra

[fresh owl]

do you understand what is being done here?

[steep vessel]

we expand the 2

i got it ty

.close

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[halcyon lance]

h..how do I solve this

[halcyon lance]

where the results will be these

[stable inlet]

use a calculator

[halcyon lance]

yea I have the 991EX right besides me

but

what next

[stable inlet]

that should give you the roots

[halcyon lance]

you mean using the EQUATION/FUNC

didn't get same numbers

😦

[stable inlet]

yeah

[stable inlet]

these aren't the correct roots

your calculator should be giving you the correct ones

[stable inlet]

yeah, looks right

its one real, two complex

[halcyon lance]

I hate it when this happens
professors should take a penalty for flawed material 😦

anyway, thanks

[safe radish]

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[bleak wedge]

1. 90

2. 10

I seem to be stuck on the third one

I don't really get the multiplication part

[thin bridge]

in this context * isn't multiplication
rather its a self defined operator

[bleak wedge]

ah

so 6 . 8 is 24

and then 24 . 3 is 36?

[thin bridge]

yes

[past birch]

yep

[bleak wedge]

how do I solve the 4th one?

[cold aurora]

Try to write each term in such a manner that they have same powers.

[thin bridge]

assuming calcs aren't allowed?

[bleak wedge]

calcs are not allowed

they used estimation in the video

https://youtu.be/h3nWz-DN1V8

[harsh parcel]

All of the powwe

Powers

Are multiples of 83

So make the exponent 83

[bleak wedge]

thanks!

so i should get something like: 5 to the power of 83 x 2

right

?

OOHHHH

I THINK I GOT IT

[bleak wedge]

is it 2 to the power of 415?

the answer for q4

[cold aurora]

Yes

[safe radish]

@bleak wedge Has your question been resolved?

[safe radish]

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[safe radish]

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[timber garnet]

A region in the xy-plan is bounded by the following lines:
y = x, x +y=6 , y = 0
i) Find the area of this region.

I keep getting stuck and I just need a step by step guide on how to do this

[steep pewter]

have u tried graphing?

or sketching

[timber garnet]

No I just wanna be able to do it straight up😭

But I can't figure out where the actual part that's being integrated comes from

[steep pewter]

can u visualize those lines in ur mind?

[desert pasture]

When it comes to finding areas, usually grphing is important

[steep pewter]

ahh ur integrating

[timber garnet]

Not really, the way my prof has taught it which is probably very backwards is without any graphing at all

[unique bison]

to do it "straight up" you need to be able to visualise the region

[unique bison]

he prolly visualised it in his head

[timber garnet]

He said I didn't need to graph anything🥲 but the graph isn't provided for these questions

[steep pewter]

its pretty simple since there y=0 but for a general solution u kinda need to know where they are i.e. visualize them

[steep pewter]

yeah just a sketch should be enough

[unique bison]

yea cuz u can visualise it

if you can't, graph

[desert pasture]

alternatively, if the region is a triangle , find the vertices, and use the determinant to find the area

[timber garnet]

The thing I'm struggling with is if the area is between the curves of f(x) or f(y)

[unique bison]

bruh

name checks out

this seems too complicated

[steep pewter]

wow, what?

[timber garnet]

I have no idea what that is😭

[timber garnet]

Hold on I'll take a pic of what I learned looks like cause I don't even know how to explain it

[wet burrow]

and unnecessary

[steep pewter]

u guys remember that without looking it up?

[steep pewter]

😮‍💨

[desert pasture]

If you're using integrals, IMO, just graph it

and then integrate

[timber garnet]

I mean there is a graph, but he hasn't told me anything about graphing or how to graph it and I specifically asked if it needed to be graphed and he said no

So Idk what I'm doing in conclusion

[steep pewter]

ok can u sketch linear graphs considering constant and gradient

[timber garnet]

Idek what those r🥲I pretty much didn't take any hs level math in hs I'm jumping into the deep end here

[plucky elk]

https://www.khanacademy.org/math/precalculus

[wet burrow]

Yeah you need to integrate f(y) = ((-y+6)-y) from 0 to 3

[timber garnet]

Yes okay that's what I did😭but chatgpt told me something else for the part I'm meant to integrate

[wet burrow]

Chatgpt??!

[plucky elk]

don't use chatgpt for math

[timber garnet]

I don't know where to get the answer key for this from and my test is on Thursday 🥲

I have limited options I don't even think my teacher has the answer key

[desert pasture]

GPT is a LLM, not a math AI

[timber garnet]

Well could you point me in the direction of an accurate calculator that can help with this

I'm desperate 🥲

[wet burrow]

Wolframalpha

[plucky elk]

,w int from 0 to 3 of (-y+6)

[flat frigate]

Results provided by WolframAlpha

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

Requested by rie.mann

[plucky elk]

oh i missed the other y

[timber garnet]

I'm in the dark so sorry for the piss poor lighting but wouldn't it be zero?

Wait no 9

,w int from 0 to 3 of (6-2y)

[flat frigate]

Results provided by WolframAlpha

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

Requested by andrometa

[final halo]

You really should take the time to figure out/learn how to graph these lines

[timber garnet]

I really would but my test is on Thursday so I don't have much time🥲

[plucky elk]

takes like 30 minutes or less to figure out how to graph

[safe radish]

@timber garnet Has your question been resolved?

[safe radish]

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[supple shore]

I have an equation : $cos(\theta) - cos(2\theta) = sin(3\theta)$ I end up with $2sin(\theta/2)sin(\frac{3\theta}{2}) = sin(3\theta)$ I don't know how to continue, does someone have a int ?

[flat frigate]

phoestaclies

[feral linden]

then use sin(2a)=2sin(a)cos(a) for a=3θ/2

[supple shore]

yes but with that what can I do, I want my result in form of : $sin(3\theta) = sin(y)$ no ?

[flat frigate]

phoestaclies

[supple shore]

ok thanks I solve it ty 🙂

.close

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[umbral swan]

For Q1 you seek the clockwise angle from North from A to B

Which will be 180° - 60°

since these two angles must add up to 180°

yes that's your answer

k, if you can't visualize it in your head it helps drawing it out, but not the exact angles and speed to scale, just for visualization

Plane A will have travelled 4800km and Plane B 6400km, now, given the angles you can calculate the third side of the triangle

because they come from the same origin (airport), you can picture it as a triangle

where you know two sides

= their travelled distance

and one angle

= their bearing difference

It doesn't get deleted

wdym by with you :D you can keep it open for as long as you're not done with it if you want

keep this channel occupied

[rocky beacon]

well you can close the chat then search your username to go back

[umbral swan]

once you want to let this channel be occupied by someone else again, you can use .close, once you've fulfilled the task or the question is answered

there are many channels so you don't need to worry about ppl waiting

np, try solving for the third side of the triangle and if you're stuck just come back here

[safe radish]

@lean otter Has your question been resolved?

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[proud dagger]

Hi everyone, i was wondering how i'm supposed to know which angle they want me to find?
For example, my original thought was they wanted me to find "A+B+D" or "AxBxD"
I'm just not sure which angle "ABD" is reffering to? Does it mean to add them? Is it the middle one?

[proud dagger]

Thank you in advance

[mossy fractal]

imagine a line going from vertex A to B to D

what would be the angle between those 2 lines

[proud dagger]

The vertex meaning the rotational point, correct?

So angle A?

[mossy fractal]

the corner that A is labelled at

[proud dagger]

Right

So it's angle A they want?

My apologies by the way, i'm completely new to this

[mossy fractal]

show me what angle you think it is, you can draw it on paint if you want

[proud dagger]

Sure one sec

thank you btw

dodgey drawing but hopefully you get the point lol

[mossy fractal]

not quite, that would be the angle BAD or DAB

lemme show you something real quick

[proud dagger]

Right

[mossy fractal]

it wants the angle ABD so we want to go from A to B and from B to D

[proud dagger]

ohhh i seee

[mossy fractal]

so what i have drawn in red is the line from A to B

[proud dagger]

Rightttt

So it would be the angle which is made when you draw a line from a-b-d?

which in this case is 90

it's a right angle

[mossy fractal]

yes

[proud dagger]

And that works consistently across all shapes/angles?

[mossy fractal]

yeah

[proud dagger]

thank you so much

[mossy fractal]

eg angle BCD is the angle on the line B to C and C to D

[mossy fractal]

np 👍

[proud dagger]

.close

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[lean otter]

I think i'm stuck on this problem

[lean otter]

Which part of the question are you having problem with?

Here is clearer picture

[lean otter]

Umm

||Use relative acceleration with mgh , that would be m(g+a)h||

i do not think the velocity part is correct

[lean otter]

Sub questions, I meant, a b c or d

Or all of them

c

i meant

d

Speed

d

Yeah

You found c?

yeah

c is just the work done

So

Yep

Work Done by conservative forces = Change in kinetic energy

Good

Then just use 1/2 mv²

Oh you did

yeah

Did you not get a numerical value for kinetic energy??

Idk

haven't put it into a calculator yet

Hmm

My answers look right?

I'm trying to figure out

What if F(h) ??

Force of helicopter

Uuh

I don't understand what you did

But it's simple

let me

Yeah

You go first then me

I just used relative Acceleration of astronaut which comes out to be (g + g/10),
used m(11g/10)15 = 1/2 mv²

Simple

We did same thing

Correct I assume

D is wrong

[lean otter]

It was harder for me to think it in force

[lean otter]

Why tho

Lemme calculate the value

It isnt 70.35 something??

the velocity?

no

oh wait

I actually did D right

I just messed up while putting it in desmos the first two times

Yeah

https://www.desmos.com/calculator/5zz5wayea7

TY for help

.close

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[flat frigate]

Şêro

[median vigil]

the slope of the tangent line gives you information about the first derivative, and the fact it's not a min/max gives you info about the second derivative

yes

[placid oak]

if a stationary point is not a local extrema, the second derivative is zero

[median vigil]

generally, we can say that if the second derivative at a point where the first derivative is 0:

• is positive, the point is a minimum
• is negative, the point is a maximum
• is 0, is neither (inflection point)