help-0

[woeful pulsar]

and the definition of derivative?

[sage lynx]

they are the same thing

[woeful pulsar]

erm... not really

[sage lynx]

i thought a function is differentiable if its continuous

[woeful pulsar]

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

[sage lynx]

so when something is continuous everywhere its differentiable nowhere

[woeful pulsar]

no it doesn't generalise

[sage lynx]

oh

then how do i find out if the function is continuous, differentaible, or if it does not exist

[radiant anvil]

@woeful pulsar true, so in the case of sum of averages of time per step it's a bit "unfair" to sum the averages in order to get the total time. I mean, the average per step is good information for the time per step, but in the case of the time for the complete flow, as you said, it's better to sum the total time for all stages and then average those total times.

[woeful pulsar]

use definition

[sage lynx]

how do i find that

[woeful pulsar]

definition of continuity has an epsilon delta definition
definition of derivative has a limit definition

should be able to easily find it online

[sage lynx]

alr thanks

i think its f is differentiable at x=2 then

[buoyant vale]

why does this have two answers

?

i can only think of 36.87

[woeful pulsar]

is it just differentiable at x=2?

[buoyant vale]

got that using 0.5absinc=15

[woeful pulsar]

is there an obtuse solution?

[buoyant vale]

whats that?

[woeful pulsar]

is there a configuration such that theta is obtuse?

[sage lynx]

i guess it could be continuous at x=2 too, but are they mutually exclusive? If not then it would be both, because the derivative clearly does exist

[woeful pulsar]

differentiability implies continuity

[sage lynx]

oh

alright thanks, so it would be continuous and differentiable

i'll keep that in mind for future problems

[buoyant vale]

i think thereis but im not sure how to make it

[woeful pulsar]

you can calculate sin theta in that case

but what does it tell you about the value of cos theta?

[buoyant vale]

i dont know

[woeful pulsar]

cos theta of an obtuse angle is negative?

[buoyant vale]

wouldnt that just take me back a step from how i found the acute angle?

[sage lynx]

if im doing this correctly, the slope would be 0 and f(x) would be increasing at the interval of 0 to infinity in the original equation correct?

[woeful pulsar]

well the obtuse angle would have the same sin theta

[buoyant vale]

how can i use that to find the obtuse angle

?

[woeful pulsar]

no for a) the slope of f at 0 is not 0, (THIS IS NOT THE ORIGINAL FUNCTION)

[woeful pulsar]

you know the sin of the obtuse angle

[sage lynx]

oh alright, how do i find the original function?

[woeful pulsar]

do you have to?

[sage lynx]

hmm

probably not

[buoyant vale]

yes but what do i do with that to find the obtuse, sin^-1 of 0.6, which is sin of the acute angle just gives me the acute angle, i dont know what to do with this information to find the obtise angle

[sage lynx]

i just dont know how i'd find the slope of the original function

[woeful pulsar]

try visualising the angle on the unit circle, the sin is the y-coordinate

[woeful pulsar]

f' is slope

[sage lynx]

at x=0 how is it not 0 because its a flat line at 0

there is no rise or run

[woeful pulsar]

what function is being plotted?

(THIS IS NOT THE ORIGINAL FUNCTION)

[sage lynx]

f'(x)

oh yeah

[woeful pulsar]

and it's asking about f

[sage lynx]

so how do i find the slope of f from a f' graph?

[woeful pulsar]

f' is slope

[sage lynx]

so slope = 1?

[woeful pulsar]

remember your parentheses

(1 - cos^2(x))/cos(x)

you are trying to combine the two fractions

yeah you are just doing things to the left side

[tawny lion]

fair enough, i see

[sage lynx]

@woeful pulsar so is the slope = 1 then?

[woeful pulsar]

no

read the graph

[sage lynx]

im confused as to how to get the slope value

all i see is a flat line at x=0

[carmine kayak]

helppppp meeee i need area

[woeful pulsar]

decompose into simpler shapes

[sage lynx]

can you tell me what the slope is so i can bracktrack and see how to get the slope from the actual value

[woeful pulsar]

read the graph to find the slope

don't look at slopes, the value is the slope you are looking for

[sage lynx]

ohhh its 4 i think

[woeful pulsar]

-4

[sage lynx]

ah

thanks

[stable dune]

can someone help me get the answer

[woeful pulsar]

do you know pythagorean theorem

[stable dune]

yeah

[woeful pulsar]

just apply it

[stable dune]

how?

[woeful pulsar]

find the lengths of the sides

[stable dune]

idk how im rlly confusd

[woeful pulsar]

which sides of the triangle can you find?

[stable dune]

12^2+5^2=c^2

[woeful pulsar]

yeah you can finish that

[stable dune]

okay

[still crest]

@stable dune did you finish? making sure

[stable dune]

yes

i got 13

[still crest]

Could someone help me with this?

<@&286206848099549185>

[strong furnace]

can you show some work or some idea that you might have?

[still crest]

I make as always a graph

Let me send

[strong furnace]

so you know about diameter of parabola right?

[still crest]

It's parallel to the symmetry axis of the parabola, correct?

[strong furnace]

yes

I would say start by making some assumptions

use those initial assumptions to get other initial conditions to be true

[still crest]

Assumptions like?

[strong furnace]

point B(h,k)

[still crest]

About the angles?

[strong furnace]

taking assumptions about angle would make it more complex I guess , if we just assume a point we get the angles by this assumption itself

[still crest]

Is there a reason you haven't named B(x,y)?

Or h and k stand for something

[strong furnace]

when we write equations for lines and curves we use x and y for general coordinate if we have a constant point with x and y it just makes it confusing to keep track of which one is which

[still crest]

I see

[last sinew]

Hi is this question channel currently open?

[still crest]

Nope

[last sinew]

Oh ok my bad

[still crest]

no worries^^

Is naming the coordinate B(h,k) an assumption?

I'm not exactly sure what an assumption is tbh

[strong furnace]

we were not given the coordinates for point B so yes

[still crest]

Okay

Could we name our angles too?

[strong furnace]

we can name angles

but the names won't be significant

[still crest]

Let me draw what I thought

We could say that those are 90 degrees each

Although I don't know how would that help

[strong furnace]

the thing we need to prove is that our normal bisects that angle

we only need 2 line equations to prove that here

[still crest]

What does bisect mean?

[strong furnace]

cut in half

[still crest]

So we need the equation for e and z

[strong furnace]

can you think and point out which two lines would be sufficient

in the diagram

[still crest]

I'm guessing B, where they intersect

[strong furnace]

B is a point

[still crest]

Ohh my bad

Wait

e and z?

[strong furnace]

nope

think about it like this

they mention the reflective property of parabola

[still crest]

So one side reflects the other

[strong furnace]

you know the law of reflection?

[still crest]

I can guess what it means but not the law

[strong furnace]

never mind then

just to check can you point out which angle is being bisected in the diagram

[still crest]

The diameter one?

[strong furnace]

can you mark the angle in your picture or use two lines to describe the angle? diameter one does not really describe an angle :/

[still crest]

Okay, just a moment

[slow mason]

can someone help me here?

[strong furnace]

channel is busy try another one

[slow mason]

kk

[still crest]

Is this correct?

[strong furnace]

yes

if you want to prove this angle is bisected how would you go about doing it ? what would you want to find out first?

[still crest]

Could I get the equations of each line?

Or try to find an angle

[strong furnace]

this

if you had the values for those 2 angles made by bisecting our angle

and they were equal

that would prove that the angle was bisected right?

[still crest]

Correct

So can we assume from now that they are equal?

Or that's wrong

[strong furnace]

no

I mean we can

but that would be a different approach

we are going to find the angles

and we are going to prove they are equa

[still crest]

What's an initial condition?

[strong furnace]

its a weird habit I have of calling given condition initial conditions

ignore that for now

[still crest]

Okay

So what's the other approach we could take

[strong furnace]

I have not really thought about it

but if I wanted to simplify it

its kind of like going from RHS to LHS in a proof

[still crest]

Angles right?

[strong furnace]

assuming what we want to prove is true and then showing that this would happen only if we had the given conditions

what we were doing was assuming there exists a point and then finding the angles and proving them to be equal

that would be assuming angles are equal and showing that the line between them is the normal

[still crest]

Wait, when I asked this, I meant the first approach that we would go for

[strong furnace]

we will find the 2 angles which are part of the bisected angle and prove they are equal to show that the angle is bisected

[still crest]

I see

[strong furnace]

how would you go about finding these angles?

[still crest]

I mean whichever is more familiar to you

So the explanation is easier

So

Let me think

[strong furnace]

I meant what would you do to find these angles?

[still crest]

I realised after 😅

I'm pretty bad with angles, EBx looks like 90 degrees

[strong furnace]

we cannot get the exact value of this angle

[still crest]

Right we can't

[strong furnace]

we assumed our point to be (h,k)

which can be any point on the paraboa

[still crest]

Could we find the slope?

[strong furnace]

yes

[still crest]

With E and B, right?

[strong furnace]

the fact that diameter is parallel to x-axis is very convenient

[still crest]

My slope is k/(h-a)

And we plug that to find the equation of the line?

[strong furnace]

all we need is slope we don't need to find the line

but we need 1 more slope which one is that?

[still crest]

Okay

Ohhh

The one that is parallel to the symmetry axis

Right?

[strong furnace]

no

[still crest]

Hmm wait

Okay I think I got it

[strong furnace]

symmetry axis is x-axis so we know its slope is 0

[still crest]

The opposite/reverse of this is the slope of line z?

Since it's perpendicular

[strong furnace]

how is that perpendicular

[still crest]

It's not

[strong furnace]

you have the slope of line EB

you have the slope of line Bx

[still crest]

Yes

[strong furnace]

which slope are you missing

which includes the 2 angles

[still crest]

Tangent?

[strong furnace]

that we are planning to compare

[still crest]

Of the parabola

[strong furnace]

look carefully

[still crest]

I mean one remains because I failed all ahah

Line z?

[strong furnace]

true and yes

BxBE is the angle that has been bisected so we want to prove angle between BE and (z) and the angle between Bx and z is equal

[still crest]

Clear

[strong furnace]

or you could prove BEBx angle is double the angle between Bx and Z

[still crest]

How could we find the slope of line z

[strong furnace]

I am assuming you know some calculus?

[still crest]

If there's a formula I could recognise it

[strong furnace]

we can also use algebra and coordinate but it takes longer than just differentiating

[still crest]

Wait

Could we ydy/dx = 4a?

[strong furnace]

yeah

[still crest]

And solve for dy/dx

[strong furnace]

dy/dx = 2a/y for tangent to parabola

that is the slope of line (e) which is perpendicular to z

[still crest]

Why 2a/y?

Isn't it 4a

[strong furnace]

y^2=4ax
differentiate 2y dy/dx = 4a

[still crest]

I forgot the 2 yep

[strong furnace]

so the slope to e is 2a/k

what is the slope of z then

[still crest]

-k/2a

[strong furnace]

we have our 2 slopes that we need

[still crest]

Wait

[strong furnace]

just use tan identities to prove the relation between these angles

[still crest]

Okay we substituted y with k

[strong furnace]

if you have to prove BeBx is double the angle between z and Bx we have the tan of both of these angles we just use tan(2(theta))=2tan(theta)/(1-tan^2(theta))

[still crest]

I forgot the formula. Isn't it something like tan = (slope1+slope2)/2 * slope1 * slope2?

[strong furnace]

acute angle between two lines having slope m1 and m2 is

arctan(mod(m1-m2/(1+m1*m2)))

[still crest]

I've never seen this

Wait

I may have seen this but is this transformed?

[strong furnace]

transformed ?

[still crest]

Let me find smth

I found the formula we have for finding an angle between 2 lines

Is this correct for here?

tanx = (m2-m1)/(1+m1*m2)

[strong furnace]

yeah

[still crest]

How did we get arctan?

[strong furnace]

if tanx=(m2-m1)/(1+m1.m2) , x in (-pi/2,pi/2)
x=arctan((m2-m1)/(1+m1.m2))

[still crest]

Okay let me do the operations

Wait, if we have 1+(two slopes that are perpendicular) that makes it 1-1 which is 0

We can't have 0 in the denominator

[strong furnace]

we can't have pi/2 inside tan() either

[still crest]

Am I wrong with my operations?

[strong furnace]

show your work

[still crest]

,rotate

[strong furnace]

which angle are you trying to find ?

[still crest]

Oh that's what I'm doing wrong.

So I could pick up EB slope with z

[strong furnace]

did you get it right?

[still crest]

Yep, I'm doing it right now

Let me finish in a moment

I found tanx = (ak+kh)/(2ah-2a^2-k^2)

[still crest]

ty

[trim hull]

my bad

sorry

[strong furnace]

use the fact that h,k lies on parabola

[still crest]

no worries^^

I could substitute h and k, so it's k^2 = 4ah?

[strong furnace]

yeah

[still crest]

And solve for k?

[still crest]

Is this right btw?

[strong furnace]

not necessarily just substitute h in the equation with k^2 to eliminate h from that angle and you would prove what you wanted

[strong furnace]

will have to check i used tan(2(theta)) approach

[still crest]

I will do it again in case it's wrong

[strong furnace]

k^2 has a positive sign I think

[still crest]

If we multiply negative * positive (the slopes) and and it's 1 + (-)

I think it's negative

[strong furnace]

my bad yeah

it is -k^2 so that is correct

[still crest]

Yep it's correct

I'll substitute parabola in too

I found tanx = (2k^2+2ak)/(ak-4a^2-2k^2)

So it would be x = arctan((2k^2+2ak)/(ak-4a^2-2k^2))

Right?

[strong furnace]

we don't have to find the angle

we just have to compare it to the other angle

so keeping it as tan is alright

[still crest]

Okay

We check line z with Bx now right?

[strong furnace]

we don't need to find the tan of angle between them since Bx is parallel to x-axis the slope of z is the tan of that angle

[strong furnace]

all you need to prove is that this is equal to that slope

[still crest]

How do I see if those are equal

Do I equal them?

[strong furnace]

that expression is incorrect I think you made a mistake while replacing h with k^2/4a

[still crest]

This?

[strong furnace]

nah you have to show that expression is equal to -k/2a

simplest idea is to take -k/2a common from numerator and then it should balance itself

[strong furnace]

yeah

[still crest]

Let me correct my mistake

I mean it's only the operations now, I can't find my mistake. Thank you very much

And when we find that the tans are equal we have proved it right?

[strong furnace]

Yes I think that should be sufficient

[still crest]

Okay, thank you for your time

^^

[golden willow]

can someone help me do repeating decimal to fraction

[gloomy lintel]

Sure, I haven’t done this since gcse but let’s have a go I guess haha

[golden willow]

how do i do 0.083 (3 is repeating) as a fraction

[gloomy lintel]

So your first step is let x = 0.083333333333

Recurring

[golden willow]

ok

[gloomy lintel]

Then let 100x = 8.3 recurring

[golden willow]

where does 100 come from

[gloomy lintel]

Now we need to work out 1000x, because we want the reoccurring bits to cancel

We need to move the decimal place 2 points initially to get to the reoccurring bit

That’s why it’s 100

[golden willow]

ok

[gloomy lintel]

So once you have 1000x, subtract 100x from it to get 900x=

So it turns out that it’s (83.33333-8.3333333)

So 75

If 900x=75

X=75/900

And that’s your answer

Should work out if you type into a calculator

[golden willow]

should i simplify it

[gloomy lintel]

I never do if the question doesn’t ask, it’s unnecessary

[golden willow]

ok

[gloomy lintel]

Otherwise you might get it all right then mess the simplifying up and it’ll sort of undermine what you just did but idk that’s just me

Yeah so that’s it

Always get the inital multiple to the reoccurring bit

Then the next multiple in order for the trailing numbers to cancel

[golden willow]

ok

thanks

[keen wasp]

how do u do this? answer is c, but i get cube root 24

[harsh swallow]

to change the limits from x to t

you have to solve t = x + 1

basically f(x) is the integral with those limits

actually i don't really know how to explain

😄

[keen wasp]

thats how i got cube root 24

second fundamental theorem of calc

[gray isle]

although unsimplfied your result is correct

[harsh swallow]

i like did it quickly and got the answer and now trying to justify it i am blank

😄

[keen wasp]

but cube root 24 is 2* cube root 3, is it not

is the book just incorrect

[gray isle]

based on their intended answer, it seems like there may be a typo and the upper bound was supposed to be x-1 instead of x+1

[keen wasp]

oh thanks

[harsh swallow]

that would make sense

it definitely would make sense so much more if it was x - 1

[coral ridge]

anyone knows why is this answer not correct for this

isn't this correct

[ocean seal]

KRoNlC
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ornate harness]

sigh, sorry. fixing tex and re-writing.

[wanton turtle]

c = (ba - b^2)/a

c = b - (b^2/a)

b^2/a = b - c

therefore a = b^2/(b-c)

[ornate harness]

I'd appreciate any help correctly propagating the errors for this calculation:
$x = \frac{10^{(-0.4)(15.162 \pm 0.108)}}{10^{(-0.4)(14.606 \pm 0.055)}}$
edit: <@&286206848099549185>

[ocean seal]

KRoNlC

[ornate harness]

Bonus (optional): I also need to code this error propagation into Python! So I'll be writing each step in a Python script.

[light sleet]

How does terminating and repeating decimals work?

[golden willow]

terminating means the decimal only goes a certain amount. Repeating means one number or multiple numbers in the decimal are repeating

[light sleet]

Can you give a equasion example?

[golden willow]

so 2/11

it would be 0.18

but 1 and 8 keep repeating forever in the same order

[light sleet]

but like if you do it on calculator it'll round to the 10billionth place because it has to stop somewhere right?

[golden willow]

on a calculator it would either round or just stop at one point

[light sleet]

ohh okay

but what im not understanding is

how do you get 0.18 from 2/11 how does that work?

[golden willow]

2 divide 11

[light sleet]

Okay so division, got it

so that's all of it?

[golden willow]

yup

unless you want an example for terminating

[light sleet]

I think that'd help clear the fog as well for me

[golden willow]

terminating is any number that doesn't repeat

like 2 divde 10

[light sleet]

okay

[golden willow]

its just 0.2

the 2 doesnt repeat

[light sleet]

ohh so when it's terminating

it doesnt repeat

[golden willow]

yup

[light sleet]

but when it's not terminating it DOES repeat

got it

now I understand

thank you

[golden willow]

ok

your welcome

[stiff moth]

the product that i chose is a scrub daddy and the price i chose is $8, i dont know how to incorporate, the 15 cent increase. to be frank im not even sure if i understand what it mean s

[alpine sable]

You sold 500 items at the price of $8

Whats the revenue?

[stiff moth]

i would subtract my cost function from it right?

C(x) = 48.3475(x) + 6000

so r(x) - c(x) ?

[alpine sable]

Is the cost function from something previous?

[stiff moth]

it was this question

[alpine sable]

Subtracting cost from revenue gives you profit

Or loss if costs are more than revrnue

Revenue

[stiff moth]

i see, so in this case R(x) = 500(8) - 48.3475(x) + 6000

where would the 15 cent come into play

[alpine sable]

So 500×8 gives you revenue
If you increase the price by 0.15 you sell 10 items less:
(500-10×1) × (8+0.15×1)
Where 1 represents the number of increases

If we had 0 increases, you get (500-10×0)×(8+0.15×0)
Which is just 500×8

[wanton edge]

hey, does anyone know this formula????

[stiff moth]

@wanton edge sorry this channel is occupied

[wanton edge]

mb

sry

[stiff moth]

no problem

would it be important to add the x 1?

[alpine sable]

So R(x) = (500-10x)×(8+0.15x)

[stiff moth]

oh i see

[alpine sable]

Absolutely

Im just showing you how to model the function

[stiff moth]

mmhm okay makes sense

[alpine sable]

One more thing

R(x) - C(x) = (500-10x)×(8+0.15x) - (48.3475x + 6000)

Be careful with the - sign on the right

It will give you -6000

[stiff moth]

we dont want that right?

[alpine sable]

Not sure if you need it for another problem

But for the revenue, you dont need r-c

[stiff moth]

thats profit, right okay

[alpine sable]

Yep

[stiff moth]

i have a quick question, so this was in the composite functions unit but im not too sure how its related

[alpine sable]

Im not sure

Doesnt seem related

I guess you can view the increases as a composite

[stiff moth]

yeah thats what i thought as well

how so

[alpine sable]

The increases can be viewed as a constant function

So for R(0) you get the revenue for 0 increases

R(1) for one increase, etc

[stiff moth]

hmm okay

[alpine sable]

You can ask your professor

[stiff moth]

yeah thats what im going to be doing now. thank you !

[faint tiger]

is this channel still busy?

[ornate harness]

@faint tiger I asked a question one hour ago and pinged Helpers but no response yet. I'll wait a bit longer. You can take it over.

[faint tiger]

appreciate it,

"For every function f, the Maclaurin series generated by f converges for all values of x in the domain
of f."

how can i prove that the statement is false?

<@&286206848099549185>

[shrewd oracle]

@alpine sable

[alpine sable]

@ornate harness not sure what you're asking

the range of the error?

[pliant helm]

@faint tiger maclaurin series of 1/(1-x) diverges for x > 1

[ornate harness]

Just solve for x basically, but make sure to propagate the error through correctly.

Should we wait for the current discussion to finish or?

[alpine sable]

We can move to dm

[ornate harness]

okay thanks

[worldly breach]

expression are positive?```

[strong furnace]

lets say P(x)=x^2+Bx+24=(x+a)(x+b) = x^2+(a+b)x+ab , we know about ab they want us to find all the possible values of a+b

[native temple]

how to solve this?

[strong furnace]

x^2+1>0 for all x in real numbers , so you can multiply both sides by 1+x^2 to get x^3-x>0 which should be easy to solve?

[native temple]

I tried to get the factors

which is x(x+1)(x-1)/(x+1)(x-1)

[oak chasm]

@native temple The denominator is wrong, but you don't need it according to nyann's advice.

[strong furnace]

denominator has complex facctors

[native temple]

wait what

it does?

[strong furnace]

but you don't need to factorise

[thorny terrace]

Can someone help me with basic math 😰😰

[oak chasm]

@thorny terrace Sorry, channel busy.

[thick yoke]

$x^2+1$ doesn't have any real solution, only i and -i

[ocean seal]

AR13L

[thorny terrace]

Alright

[shrewd oracle]

yeah, you cannot factorise is like you did, its solutions are i and -i

[oak chasm]

@thorny terrace #help-1 is open if you hurry.

[native temple]

okay

Then ummm what van I do

My teacher told me to solve this like a quadratic inequality

so I was

[thorny terrace]

Ok so like im afk farming at a game and I get 10 gems every 8 seconds and I'm gonna afk for 7 hours hoe many will I get in total

[native temple]

but I couldn’t find any rational factors even used p/q

[oak chasm]

@thorny terrace Please switch to #help-1.

[thorny terrace]

Oh ok

[native temple]

I thought that factos were x(x^2 + 1)

and X^2 +1 could be equal to ( x+1)(x-1)

[strong furnace]

x^2-1 is equal to (x-1)(x+1)

[native temple]

okay then how to solve a rational inequality

[next void]

I am learning Markov Chains and could use some guidance on solving this problem. I am not exactly sure what to do.

[native temple]

Do I cancel

@next void bye

hello to solve my problem wasn’t in my lesson

it gave fa ctors usausally

hello

[native temple]

Can you show that in texit?

[alpine sable]

$x^2+1>0$ for all x in real numbers , so you can multiply both sides by $1+x^2$ to get $x^3-x>0$ which should be easy to solve?

[ocean seal]

sxkura

[native temple]

seriously go in anouther channel

dtop

stop it

[covert wind]

How can x squared plus y squared also be x times y

[native temple]

no

[covert wind]

why should i

[native temple]

STOP IT

[covert wind]

Why

[native temple]

<@&268886789983436800>

[covert wind]

What

Am i gonna be banned for asking a question?

[native temple]

#❓how-to-get-help

[mossy tinsel]

what is 1 + 1

[shrewd oracle]

this one is being used

[sly mantle]

@covert wind please move to an unoccupied channel

[covert wind]

ok

Ok

nobody was talking

but alr

[shrewd oracle]

no but rules are to wait or go to another channel when one is being used

[native temple]

like you mean canceling

[slow current]

4 dumbass

[native temple]

The Fractions

[shrewd oracle]

depends on the space dont it

[alpine sable]

idk i just put what the other person sed in texit so u could read it better

[strong furnace]

can you elaborate what you mean here?

[native temple]

simplifying the fraction is that how you solve?

equation?

[strong furnace]

no

do you know how to multiply on both sides in inequalities?

[native temple]

you just multiply both sides

[strong furnace]

its not that simple , we have to take the sign of the number we are multiplying on both sides into consideration as well

[native temple]

yea it’s positive so it’s just multiplication

[strong furnace]

yeah

when we multiply 1+x^2 on both sides we remove the denominator from the LHS and since RHS is 0 we don't get anything in RHS

[upper spruce]

I need help

[strong furnace]

reduces the problem to x^3-x>0

[strong furnace]

use a different channel this one is busy

[native temple]

rhs

[upper spruce]

No one is replying in other channels

[native temple]

I got this

[strong furnace]

just post the problem some1 will reply sooner or later

[strong furnace]

not here

in the other channel

[native temple]

please no not here @upper spruce

[upper spruce]

Sorry

[strong furnace]

by multiplying on both sides I meant LHS and RHS not numerator and denominator

[native temple]

LHS RHS?

[gray jungle]

wouldn't it be 2(x*y)?

[strong furnace]

Left hand side and right hand side

[covert wind]

thats the question

i just did integer only questions

so i thought this was an integer only

so its a decimal

cuz it cant be an integer

[upper spruce]

Is this channel free?

[gray jungle]

i guess

[upper spruce]

Pls help

[haughty marsh]

Pls I need help

[upper spruce]

@haughty marsh it's too simple .....u just have to equate

[strong furnace]

some1 responded to your problem in question 8 stay there

[haughty marsh]

Please answers

[haughty marsh]

Pls give us the answers

[strong furnace]

this is not supposed to be a place where you get answers , you should show your work and then ask for ways to figure out the answer

[haughty marsh]

Please I beg you

[native temple]

SO I CAN

T just multiply both sides of a fraction

I have to multiply the otherside by 0

[strong furnace]

(x^3-x)/(1+x^2) > 0 now multiply both sides by 1+x^2 (since its positive) what do we get ? x^3-x>0

[twin pagoda]

Please see #rules

[native temple]

ooh wait I didn’t think about this

[haughty marsh]

Come on please you have got to make an exception if not I will get beaten

[glass lichen]

No one's going to just give you answers.

Also dont spam your question in the channels

[muted raft]

Maybe stop playing minecraft and do your work? lol

[haughty marsh]

I’m not that is my brother

[warm marsh]

I need to calculate the operator norm(i think this is the terminology in english) of this linear application. Can someone verify if my attemp to find the maximum is correct? I would be really thankful

Does it exist other methods to compute the norm?

[alpine sable]

(x-y)^2 = x^2-2xy+y^2

[oblique comet]

as multiplication is to the ! operator
is there an equivalent addition operator?

[glass lichen]

$\sum_{i=1}^n i$?

[ocean seal]

moshill1

[warm marsh]

I need to calculate the operator norm(i think this is the terminology in english) of this linear application. Can someone verify if my attemp to find the maximum is correct? I would be really thankful
@warm marsh
Can somebody help?

[coral ridge]

can someone help this too

<@&286206848099549185>

[paper ibex]

Can someone help me with this problem

[past lichen]

Lol I can’t figure out what am I missing

[still crest]

ty

[past lichen]

Sorry I was replying, my fault

[still crest]

no worries^^ @paper ibex is in queue

[strong furnace]

you're always ready with that

[still crest]

i'm in business rn, so much chaos

[ashen wave]

can anyone explain to me what outliers are?

[strong furnace]

what do you understand by similar shapes?

[strong furnace]

odd ones in a data set ? we like to remove outliers from a data set because they cause discrepancies if I am correct

[ashen wave]

im confused-

what are you removing?

[strong furnace]

lets not talk about removing for now

outliers are datapoints that are away from other datapoints

[last patio]

bakery 1 sells a dozen cupcakes for 21 dollars. bakery 2 sells half a dozen for 12

[still crest]

ty

[last patio]

oh

sorry :3

[still crest]

no worries^^

[strong furnace]

I think he is le epic troll

[last patio]

ill go now i have no life

[ashen wave]

could you show me like a visual representation? Im a bit confused about what your talking about-

[strong furnace]

straight out of google

[ashen wave]

oh

ok

Erm could you explain this problem to me? @strong furnace

i dont know what the 1.5*IQR rule is

[limpid spade]

Neither do I

[strong furnace]

write the numbers in ascending order

[ashen wave]

okk

now what?

[strong furnace]

you are supposed to find Q1 Q2 and Q3 I am trying to figure out if these are terms in the dataset or average between two successive terms for even number of terms

Q1 is the median of first half

Q3 is the median of second half

you only need these for now

so split the dataset in two

for even numbers median is the n/2)th term in this case 2nd term

[still crest]

ty

[strong furnace]

use a different channel this is busy

[alpine sable]

k

[ashen wave]

okk let me write this down

[ashen wave]

how did you determine what the quartiles were?

[strong furnace]

looked up the definitions

[still crest]

ty

[ashen wave]

is quartile 3 the median?

[still crest]

@next void deleting the image would be helpful^^

[next void]

sorry

[still crest]

no worries^^

[ashen wave]

is quartile 3 the median?

[strong furnace]

quartile 2 is the median of the whole dataset

quartile 3 is the median of the other half of dataset

last 4 terms

[ashen wave]

But? When you look at a box plot like this one the 3rd quartile is the median @strong furnace

[trail garden]

can i get help?

[still crest]

ty

[ashen wave]

This channel is open for others to use :))

I have to go anyways

[lime glacier]

Could I send my q ?

[still crest]

Yes

[lime glacier]

Alright

[still crest]

No one is in queue

[lime glacier]

so i found the derivative

but then i got stuck af

like i plugged in the value pi/3

x

Here is my working out

idk how the answer is e^-pi

i got -2e^-pi

i was so happy for once i thought i got it right then bam i checked the ms and i fucked up somehow

wait is it cause cos(0) = 1 ?

and sin(0) = 0

if so then im a dumbass

cause the -3e^pi wouldnt apply if sin(0) = 0

[still crest]

Someone will arrive in a bit to help you

[lime glacier]

alr

i got it

dw

i just want reassurance

by someone haha

[still crest]

no worries^^

[strong furnace]

they take the first and last terms as points as well

[still crest]

ty

[lime glacier]

:/

i asked my question dude

[midnight cloud]

oopsie

[lime glacier]

haha all good

@still crest

Can i ask another question in lieu of my previous question?

since i think i got it

or shall i mvoe to another channel and let k/ have their answer answered

[still crest]

You shall ask

[lime glacier]

Alright

So I set this equal to 0

then idk what to do xD

3cosx=4sinx

[limpid spade]

3cosx-4sinx=0

[lime glacier]

yeah

then do i just plug

that into a calc

and find the roots ?

or is there a way to compute it algebraically

[limpid spade]

Is it allowed

[oak chasm]

$$1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}$$

[lime glacier]

thats the thing

[ocean seal]

Chai T. Rex

[lime glacier]

idk

how would i do this algebraically

[limpid spade]

Idk

[lime glacier]

cause

like

i thought of the possibilities

3cos(pi) = - 3

and 4sin(2pi/3) = 3.46

so we wont get 0

exactly

[limpid spade]

Has to be same x value

[lime glacier]

oh fuck

that doesnt make sense then

[limpid spade]

,w 3cosx-4sinx

[ocean seal]

Results provided by WolframAlpha

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

Requested by The Fractalogist#2374

[lime glacier]

wtf

this needs a calc

100%

[limpid spade]

Lol

[lime glacier]

right ?

can u solve this algebraically

[limpid spade]

Idk

[lime glacier]

as a good maths person

xd

[wild talon]

I need help with this one

x = pi/2 is all i got

[still crest]

@lime glacier have you finished?

[limpid spade]

Yes

[strong furnace]

it is called trigonometric equation , lookup solutions to trigonometric equations

[wild talon]

there needs to be 3 more answers

for it

[strong furnace]

how did you approach this problem?

[wild talon]

no idea how to approach it, thats what im looking for help on

[strong furnace]

here's a tip for such problems whenever you have different angles inside trigonometric functions try to reduce them to one common angle so you have 2 different approach available for this problem

[wild talon]

like try to get it all to sin or cosine?

[limpid spade]

Sin

[strong furnace]

try to get 2x to x

or x to 2x

[wild talon]

oh

[civic crypt]

Both that and the same trig function.

Both same angle and same trig function.

[wild talon]

alright lets say im at cos2x + sin2x = -1

oh

[strong furnace]

you have an identity for cosx+sinx you have to multiply and divide by 1/sqrt(2) to get it

in general for asinx+bcos we multipy by 1/(sqrt(a^2+b^2)

[civic crypt]

Let's step back a bit. Why would you change x to 2x.
You change sin2x to 2sinxcosx.
To have: 2cos^2(x) + 2sinx cosx = 0

You divide by 2 and factor out cos(x):
cos(x)[cos(x) + sin(x)] = 0

At this point you have 2 solution groups:
1° cos(x) = 0 2° cos(x) + sin(x) = 0

[wild talon]

wait sin2x = 2cos^2(x) ?

[woeful pulsar]

R formula hmm?

[civic crypt]

(btw, my hint to convert to the same trig function was bad. the only procedure you have for sure is to convert to the same angle. That nyann suggested.)

[civic crypt]

No.

[wild talon]

oh wait im blind

whoops

[woeful pulsar]

Here's a link to read up for the R formula thing
https://brilliant.org/wiki/trigonometric-r-method/

[civic crypt]

This is not needed here. As far as I can tell.

[woeful pulsar]

oh yeah

[strong furnace]

this is not needed if he reduces it to x

[woeful pulsar]

Just realised you can use tan x in cos(x) + sin(x) = 0

[alpine sable]

I have done this https://bit.ly/Maths-Solution

[strong furnace]

but it helps to learns about the R method

[civic crypt]

Yea, it is kind of advanced tecnique. I think at this point it might be too much for him.

[wild talon]

uh im looking at it

hmm

[civic crypt]

First solve it by reducing to x

[strong furnace]

yeah I believe that is the approach they want him to take as well

[wild talon]

i mean i can just remmeber asinx + bcosx = sqrt(a^2 + b^2)sin(x + a)

or something

wait why would I use R method

to get it into sine?

[strong furnace]

sure you can do that

r method is just how you get that

[wild talon]

alright ill try to use it

[strong furnace]

sin(x-arctan(b/a)) btw

[strong furnace]

go through @civic crypt's approach first

[civic crypt]

The formula is too much to remember. If you remember this you forget in 3 days. There are many other more basic formulas to remember

[wild talon]

i just put it into a doc

[strong furnace]

the fact that it has a very intuitive and easy to remember derivation gives us even less incentive to remember it

[civic crypt]

Yea. I just remember that I can transform them into one function. Then derive it:
asinx + bcosx = A sin(x+phi) = Acos phi sinx + A sin phi cosx

[strong furnace]

there is also a method of deriving it using phasor diagram and vector addition

[civic crypt]

a = A cos(phi) b= A sin(phi)
And divide the two you get b/a = tan(phi).
Sum and square and you get a^2 + b^2 = A^2

[wild talon]

we havent learnt phi yet :/

[civic crypt]

It is just the name for the angle 😐

[strong furnace]

phi is just a symbol lol

[wild talon]

oh

i thought it was the ratio thing

[civic crypt]

But as I said, this is too advanced for you at this point.

[strong furnace]

google will say golden ratio because it is represented by phi as well

[wild talon]

mm

okok wait

so

for the question

all i got to do is get everything to the same angle ?

and then use formulas and try to reduce it

[civic crypt]

yes. and you basically always want to have this angle as "x". Rarely "2x" or "x/2"

[wild talon]

alright

[alpine sable]

<@&286206848099549185>

I needs help and an explanation

[still crest]

ty

[alpine sable]

oh ok sorry

[still crest]

no worries^^

[strong furnace]

<@&268886789983436800> epic troll

[night geyser]

thanks

[woeful pulsar]

can you tell us how you arrived at D?

[alpine sable]

honestly i have no idea

[woeful pulsar]

okay, how does the derivative and second derivative related to increasing/decreasing and concave up/down respectively?

[alpine sable]

so ou basically find point

s

and testif their negative or positive

for both first and second

you can find inflection points and where the funtion is increasing

[woeful pulsar]

for x>1, is the function increasing or decreasing, and how do you tell that?

[alpine sable]

i test points less than and greater than 1 in the first derivative?

[woeful pulsar]

okay, so what does it tell you?

[alpine sable]

It tells me where it increases?

so its increasing?

@woeful pulsar

[woeful pulsar]

yeah, it's increasing for x>1

how about for x>e, is it concave up or down?

[alpine sable]

Its increasing so concave up?

[strong furnace]

what is the definition of concave up?

[alpine sable]

It opens ups?

[strong furnace]

a function is concave up in an interval (a,b) if its slope is increasing in a,b

[alpine sable]

ok

[woeful pulsar]

wait, what's increasing?

how did you get concave up?

[alpine sable]

Yeah what do I do

I dont get it

I plugged in e for the second derivative

then i pluggled in something higher than e

[woeful pulsar]

for >e what did you get?

[alpine sable]

I got .877

[woeful pulsar]

what did you plug in?

[alpine sable]

3

[woeful pulsar]

so you are telling me $\frac{1-\ln 3}{3^2}\approx0.877$

[ocean seal]

Element118

[alpine sable]

yeah

[woeful pulsar]

,w (1-ln 3)/3^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by Element118#2437

[woeful pulsar]

that doesn't look right

[alpine sable]

I typed it in google

[woeful pulsar]

did google use log base 10?

[alpine sable]

Oh it formatted ti wrong

your right

[woeful pulsar]

what's on my right?

[alpine sable]

it didnt put the denomiator across the entire thing

[woeful pulsar]

creepy noise intensifies

[strong furnace]

the result would still be different but continue

[woeful pulsar]

so it's negative

that's a lot of difference

[alpine sable]

yeah

[woeful pulsar]

so what can you conclude from here?

[alpine sable]

So its concave down from e to 3

[woeful pulsar]

and after 3?

[alpine sable]

same

its still concave down

@woeful pulsar right?

[woeful pulsar]

yeah

so from there you should be able to figure out which one is the answer

[eager fern]

Each problem has 6 parts. Specify the limit next to the problem. Give the answer the best describes the graph.
I’m not sure how to do this

[woeful pulsar]

okay start with 1a)

[woeful pulsar]

do you know what it is asking for?

[dusky quarry]

anyone know how you get -2cov(x,y)?

[woeful pulsar]

Var(X-Y)=E((X-Y)^2)-E(X-Y)^2
Cov(X, Y)=E(XY)-E(X)E(Y)

do you use these definitions? if not how do you define them?

[dusky quarry]

yes

we use those definition

[eager fern]

i think the function. Those were the exact instructions

[woeful pulsar]

yeah, can you calculate limit to 1-?

[woeful pulsar]

then you should be able to do the algebraic manipulations necessary? given those definitions?

[dusky quarry]

yep thank you

[honest orbit]

Hey, not really sure what this problem is asking me to do

for context, this lesson has been all about normal distributions and the z-table

[glass lichen]

what's the probability of an exact specific value occuring in a probability distribution?

[honest orbit]

honestly, im not sure

i can't find anything about it in the notes

[glass lichen]

Well how do you find P(x<a) in a continuous variable for example?

[honest orbit]

z=x-mean/standard deviation

[glass lichen]

Oh wait are you doing calculus probability?

[honest orbit]

i mean the course im taking is algebra 2, but it could be a calculus lesson

[ionic jewel]

its not

[glass lichen]

Or look up the number in s table?

[ionic jewel]

its supposed to show a specific thing

here

between x=0 and x=1 there are infinite possible values you can pick

the chances you pick specifically x = 1 is a 1/infinity chance

and the limit of that is ...?

[honest orbit]

infinity right?

[glass lichen]

0

1/big number approaches 0

[ionic jewel]

0 is right

yes

[honest orbit]

oh

alright, thanks

[ionic jewel]

i hope you see why

[honest orbit]

yeah it approaches 0, right?

[dusky quarry]

cant seem to manipulate it

[woeful pulsar]

where were you stuck?

[dusky quarry]

from the very beginning, i dont know where i start off from those definition

[alpine sable]

this is suppose to be tan^-1 but I get it wrong

[alpine sable]

what have you tried?

not sure, I wasn't completely taught on this stuff and the school expects me to do it, I tried something like 120*atan(800)

idk what I'm doing

[alpine sable]

you know SOH CAH TOA right

yes

first step is to label the opposite, adjacent, and hypotenuse

that's opposite and adjacent so it's tan

but you're trying to find an angle so it'd be tan^-1 aka atan

so lets look at angle x, we know tan(x)=opp/adj

so tan(x)=120/800

then apply the arctan function to both sides

ah so you're just switching the variables around

arctan( tan(x) ) = arctan( 12/80 )
x=arctan(12/80)

x=arctan(3/20)

see another problem I did which I had help on was sin(63) = 120/x and the x was the denominator

just solve the equation for x

right right

@alpine sable but a question, how do you know it's tan(x) and not tan(120) or something like that

[woeful pulsar]

tan is only applied to angles

[alpine sable]

because x is the angle

ohhh so the (x) is where the angle is, even if it's a number

tan(55) would mean the angle is 55 degrees?

yes

makes sense

[amber cedar]

Could I have some help with this? I'm just really unsure what sort of other "values" I can find here.

[amber cedar]

<@&286206848099549185>

[woeful pulsar]

$a^b=e^{\log(a)b}$

[ocean seal]

Element118

[woeful pulsar]

what values of log are there?

[amber cedar]

Sorry I just saw this. What do you mean?

[woeful pulsar]

what values x are there such that $e^x=1$?

[ocean seal]

Element118

[amber cedar]

0 right?

[whole palm]

Whats 2 plus 2

[glass lichen]

1

[whole palm]

Big brain

I now have the cheats neccasery to pass my test!!11

[glass lichen]

cool

[woeful pulsar]

not just that

there are other values too

[amber cedar]

Like what?

[woeful pulsar]

2pi i

[amber cedar]

Oh.

[wraith locust]

how am I wrong??

[woeful pulsar]

wait how are you right?

how did you come up with -1?

[wraith locust]

smhhh I wasn't paying attention lmaoo

i did it for t^2 -1 nvm

[woeful pulsar]

whoops

[amber cedar]

So just being sure, I got to 1^i=e^(ilog(1))

[full wasp]

it says to write in the form x + iy though, like sure you have many values in cis form but those are all gonna be 1 + 0i

[woeful pulsar]

yeah, what sort of values do you have for log(1)?

[amber cedar]

log(1) is 0 right?

[oak chasm]

,calc log(1)

[full wasp]

joy

[oak chasm]

@amber cedar $\log_b(1)$ means solve $x$ in $b^x = 1$.

I guess the bots are dead.

[amber cedar]

Lol

I kinda understand it what's the b in my case?

[woeful pulsar]

e

[amber cedar]

Okay.

But in my case it's e^ilog(1)

[woeful pulsar]

Yeah so you need to find all the possible values of log(1)

[amber cedar]

Yes. Log(1) is 0.

Right?

Channel is occupied

[keen wasp]

do points of inflection not include end points, while critical points do?

[woeful pulsar]

a point of inflection is essentially a place where it changes from concave to convex

... so if it's an endpoint it can't change from concave to convex

[full wasp]

lmao ok

[sly mantle]

👢

[keen wasp]

ok thanks

[alpine sable]

i don't understand why i'm wrong

so something starts off at 2000 and it doubles every 30 minutes, so the function should be 2000*2^(x/30)

and then it asks when it'll reach 500,000

and it says to use log10(2) = 0.3, but when i do 2000*2^(x/30) = 500,000, i encounter with log10(5) and it's supposed to be a whole number

[full wasp]

are u allowed to use a calculator

[alpine sable]

no

[full wasp]

oh interesting