help-13

[tiny socket]

ty

.close

[cedar kiln]

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[glacial locust]

Is the graph correct?

[cosmic steppe]

No

Your first partition isn't

[glacial locust]

the first "condition" or piece?

[cosmic steppe]

Piece

The 1/2 + x

[glacial locust]

ah ok

[cosmic steppe]

I think you mistaken that as 1/2 x

[glacial locust]

what should it be

[cosmic steppe]

Well it's just in point intercept form

x + 1/2

[glacial locust]

oooops

[cosmic steppe]

If I were to rewrite it

[glacial locust]

that too

damn

ill post the corrected one asap

Took too long

@cosmic steppe

[cosmic steppe]

GeH

Yeah*

[glacial locust]

phew

alright, thanks

.close

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[reef glacier]

https://cdn.discordapp.com/attachments/903481106819612714/1054124141189533726/Screenshot_134.png

[reef glacier]

i need explanation how to do this math

[cedar kiln]

@reef glacier Has your question been resolved?

[reef glacier]

explain

[cedar kiln]

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[plucky dagger]

how D1 is find out?

I tired to solve it....

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[zenith ridge]

Hi I have no idea how to do question 15 can someone help me

[cedar kiln]

@zenith ridge Has your question been resolved?

[elder slate]

what is the formula for compound interest

[cedar kiln]

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[coral valley]

I've been staring at this for 30 min and cannot figure out where
(1,1,0) and (-2,0,1) comes from

any1 have any idea?

[scarlet garnet]

look where $t_1$ and $t_2$ are in the equation

[wraith dagger]

Jigglyproff

[scarlet garnet]

and remember that the vertices stand for (x,y,z)

[coral valley]

.close

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[graceful hinge]

Hi, can someone help me solve this problem? I know it can be solved using Laplace transform, but I don't know how to start

[latent bloom]

There is a relationship that links the Laplace transform with derivatives

[cedar kiln]

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[loud wyvern]

What formulas do i use for 7

[loud wyvern]

I got 7i)

[stone lake]

Hey guys, I have really been enjoying Discovery Pre-Calculus as a free EdX course with the Uni of Texas at Austin - does anyone know of any similar courses where the emphasis is on problem-based learning and exploration, perhaps on statistics?

[red pumice]

#books-old

stats is great but you wont get very far in stats without knowledge about probability and more calculus

[inland ocean]

You should first find the slope of the both lines

[stone lake]

Cool - are there any nice courses on probability that you know about?

[inland ocean]

Then use tanΘ = m2 - m1/1 + (m1m2) to find the angle

[red pumice]

could you open your own channel so we can discuss this more

[loud wyvern]

Sorry its messy

M1 = -3
M2= -1

Got 45*

[inland ocean]

That's correct

[loud wyvern]

Do you know the formulas for the next Qs

Parts*

[inland ocean]

How do you find the point where two lines intersect

Do you know how to?

[loud wyvern]

Ax1 +bx1 +c

Wait

[inland ocean]

I meant the point

[loud wyvern]

Ermmm i forgot

[inland ocean]

You can solve both the equations to find x and y

To find a point (x,y)

That is the intersection coordinate

[loud wyvern]

Kk

[inland ocean]

That's the formula for the perpendicular distance from a point

[loud wyvern]

This?

[inland ocean]

We have been asked to find the distance from the origin to the point of intersection

[inland ocean]

Is that correct

Hmmm

I got 3,-5

[inland ocean]

y = 3

Not -3

5y = 15
y = 3

[loud wyvern]

Ohhh

[inland ocean]

Good

That's correct

[loud wyvern]

Ayyy

[inland ocean]

Now use the distance formula

Origin is 0,0

And you have the point of intersection -5,3

[loud wyvern]

What is the formula

Wait

This?

[inland ocean]

Yep

[loud wyvern]

How do i get the other point x2 y2

[inland ocean]

We already have two points

[loud wyvern]

Oh kk

[inland ocean]

Let me draw the thing to make you understand

[loud wyvern]

Though there was another

[dire geode]

,rotate

[inland ocean]

Oh yeah this is correct

[loud wyvern]

Also tysm for helping me

[inland ocean]

Np

This is just an example

The two lines l1 and l2 meet at point (x,y) and d is the distance from origin

[loud wyvern]

Kk

[inland ocean]

here the two lines are

3x + y = -12

[loud wyvern]

L and k

[inland ocean]

And x + 2y -1 = 0

Yes

L and k

and we got the x,y as -5 and 3

[loud wyvern]

Yes

[inland ocean]

and d is √34

So if you understood this one

Shall we move to the next part

[loud wyvern]

Yes

[inland ocean]

Its better to draw the lines

Do you know how to draw the line for the equation

[loud wyvern]

Lemme try i dont think so

[loud wyvern]

Ohh im kinda close

[inland ocean]

If we have 3x + y = -12

Put x = 0 and solve for y for first time and then do y = 0 and solve for x to find the two points and join the two points

To get the line

Same process for x + 2y = 1

[loud wyvern]

Is l= (-4,-12)
K= (1,1/2)

[inland ocean]

0,-12 and -4,0 for L

[loud wyvern]

Okay

[loud wyvern]

L = (0,-12) (-4,0)
K= (1,0) (0,1/2)

[inland ocean]

Yes

[loud wyvern]

Kk

[inland ocean]

Now can you tell me which points make a triangle

With the x-axis

[loud wyvern]

Is it (0,-12 ) (-4,0) (0,1/2) idk how to find which points to use

[inland ocean]

Look at these two

[loud wyvern]

Ohhh

[inland ocean]

Its not 0,1/2 since they dont meet at x - axis

We look at points at x axis

And the point of intersection

[loud wyvern]

Okay

[inland ocean]

So the points are

(-5,3),(-4,0) and (1,0)

[loud wyvern]

Then we make 1 (0,0)

[inland ocean]

Its not 0,-12 its -5,3

[loud wyvern]

Kk

[inland ocean]

Wdym

[loud wyvern]

Make 1 point origin

Move

And the rest aswell

[inland ocean]

I dont understand can you explain what exactly you mean

[loud wyvern]

Sorry bad explaining xd

[inland ocean]

Its finee dw

[loud wyvern]

-1 not +1 at the top

Oop

Wait i did it wrong

Then the 1/2 | x1y2 - x2y1| right?

[inland ocean]

Hmmm

[loud wyvern]

And i dont think that works

[inland ocean]

We only need three points for a triangle

[loud wyvern]

Im getting confused w another q

Sorry

[inland ocean]

Its alrr

We already have three points

Which make a triangle

[loud wyvern]

Yep

[inland ocean]

This is the formula

To find the area

[loud wyvern]

Gimme a sec

[inland ocean]

Sure

[loud wyvern]

So the formula i was using worked

The answers at. back of the book had this aswell

[inland ocean]

I got 5

And this looks

Incorrect

Since

[loud wyvern]

We never did that formula

Hmmm

We can skip that part

[inland ocean]

Sure

[loud wyvern]

I got that

Accudently brought 9 over the equals

So i brought it over at the end

[inland ocean]

Ohhh yeah

This is correct

🤦‍♂️

[loud wyvern]

The one above this part?

[inland ocean]

This one

[loud wyvern]

Kk

[inland ocean]

The 4th part

[loud wyvern]

How do i do part 5

[inland ocean]

I was confused since i later realized the line l is 3x + y = -12

[loud wyvern]

Ohhh

[inland ocean]

Do you know the formula for the perpendicular distance

[loud wyvern]

Is it

,rotate

[inland ocean]

Yep

[loud wyvern]

Kk

Lemme do it

,rotate

[inland ocean]

Ohh wait yeah

1/√10

Sorry

🥲

[loud wyvern]

Its fineee

The book said √10/10

[inland ocean]

Yeah 1/√10 is √10/10 after rationalizing

They rationalized it

[loud wyvern]

Hows that

[inland ocean]

By multiplying √10 to numerator and denominator

[loud wyvern]

Oh kk

Tysm for your help

[inland ocean]

Np

[loud wyvern]

https://tenor.com/view/love-thank-you-thanks-crypto-nft-gif-26347302

[inland ocean]

And i think the triangle one is correct

[loud wyvern]

Kk

Ill see tmr

[inland ocean]

You should ask your teacher

[loud wyvern]

Ye i will

[inland ocean]

About it too

[loud wyvern]

Im gonna go cya have an amazing day

[inland ocean]

You too

Cya

[loud wyvern]

Any mods or admins seeing this you can close now ty

Oh wait i think I know

.close

[cedar kiln]

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[molten wave]

which area should be shaded?

[molten wave]

as in, what is the region that the expression asks for?

[cedar kiln]

@molten wave Has your question been resolved?

[molten wave]

i get it, and i've solved for the area as well

.close

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[violet mountain]

How do we solve for the volume of this solid of revolution?

[crimson sedge]

Revolved around x = -2?

[violet mountain]

I’m not sure, I think so

[dire geode]

follow example 3
https://tutorial.math.lamar.edu/classes/calci/volumewithcylinder.aspx

[violet mountain]

That did the trick! Thanks

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[zenith sail]

not really any work to show

What do you know about sec?

as it relates to cos?

[zenith sail]

ok

[pine estuary]

That’s what I thought too but if you had to show work, what would you write?

[zenith sail]

Maybe just say cos(theta) = 1/sec(theta)

then plug in 6/5

even that's kinda overkill though lol

[pine estuary]

Dope thanks

[zenith sail]

np 👍

[cedar kiln]

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[cosmic bramble]

hello

[cosmic bramble]

anybody could help me?

[velvet mortar]

Well, find the ratio of sugar to water for both

[cosmic bramble]

i don’t under stand the lesson this is my first time doing it

[velvet mortar]

I don't know what the RACE strategy is. Do you know about it?

[carmine bronze]

Restate the question in the form of a statement.
Aswer the question.
Cite the evidence.
Explain the answer.

The question is Which season uses a greater sugar to water ratio? How could you rephrase that question into a statement?

[cedar kiln]

@cosmic bramble Has your question been resolved?

[cosmic bramble]

I don’t need help with RACE i need need a good enough answer with a explanation

<@&286206848099549185>

[carmine bronze]

Have you come up with some type of answer yet?

[cosmic bramble]

Yes

[carmine bronze]

Let's see it and we can offer suggestions for improvement.

[cosmic bramble]

I want to add extra work on this slide but i don’t know what to write.

[carmine bronze]

How did you come up with 3/2 divided by 2?

[cosmic bramble]

27/18= 3/2

Since 3/2=1.5

and 3/2 / 2 is 1.5

[carmine bronze]

$\frac{3}{2} \text{ divided by } 2 = \frac{\frac{3}{2}}{2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$

[cosmic bramble]

What is this answer for

which one

because there are 2 of them

[wraith dagger]

Kookiemon

[cosmic bramble]

close

.close

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[eager pier]

I'm using the lagrange method

idk what im supposed to do next

[upper abyss]

Solve the system. You can get a lot of mileage doing line 2 - line 1, for example

[eager pier]

so itll be the difference equal to 0?

[upper abyss]

λ - λ is equal to 0

[eager pier]

yeah so y = x

then we can see that 2x+z=100000

what do I do now?

@upper abyss

<@&286206848099549185>

[cedar kiln]

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[eager pier]

.close

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[fringe anvil]

how do i find antidervative of this

[fringe anvil]

and these

for the first one this is whta solotuion say

but i dont understand why the g'(x) dissapears

[red pumice]

they substituted u = g(x)

du = g'(x)dx

dx = du/g'(x)

so you get g'(x)/g'(x) = 1

[cedar kiln]

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[fringe anvil]

ty

[cedar kiln]

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[dim copper]

how do i calculate the near and far plane of a camera

[dim copper]

tbh i dont even know what they heck they do or what they are but i need them to project 3d point to 2d

[cedar kiln]

@dim copper Has your question been resolved?

[craggy oriole]

Not really a math question, isn't the near and far planes the boundary of what's to be rendered by computer graphics?

[cedar kiln]

@dim copper Has your question been resolved?

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[upper abyss]

Not doing anything wrong. The answer is given by a double integral with integrand 0, over the rectangle.

This is, of course, 0.

[cosmic steppe]

... why would you use greens theorem

The question's making you suffer

[upper abyss]

Yes. Umbra raises a good point that this is naturally 0 by the fundamental theorem of vector calculus. Using Green's to get the result is weird.

But whatevs!

[cedar kiln]

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[crimson sedge]

,rotate

[crimson sedge]

Just want to know is c) Loga (2)(3xy)-loga(5)(z)

[gusty forum]

no, because only y is squared

[crimson sedge]

Z⁵ on bottom

[gusty forum]

-5log(z) is correct

[crimson sedge]

Where'd a go

[gusty forum]

im lazy and there's no other bases in the question :p

-5log_{a}(z) it is

oh actually now that i look at it, that's not the correct application of log laws in your answer

[wraith dagger]

IV

IV

[crimson sedge]

I'm getting really confused

I used rule 2 and then 3

[wraith dagger]

IV

[gusty forum]

rule 3 indicates you should bring the exponent outside and in front of the log

[crimson sedge]

Oh my bad

Can you teach me how to do 4 a or b

.close

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[strong silo]

so for polynomial graphing with multiplicity how do you tell the multiplicity of a y intercept like how can i tell if i should touch or cross on the y int

[gusty forum]

usually you won't worry about a y-intercept's multiplicity

(assuming your polynomials have the form y=f(x) for some function of x)

[strong silo]

are the y-int always going to cross and not touch/swoop in graphing then?

[gusty forum]

if it swoops then for some value of x you get 2 different y values

then you don't have a function but instead a relation

[strong silo]

alright! thank you!

.close

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[crimson sedge]

what am i do wrong

[violet night]

have you got working??

looks like you raised a 1 to a common denom

$a_n=(3a_{n-2}) \cdot (2a_{n-1})$

[wraith dagger]

Clarkie

[violet night]

in the third case , $a_3$

[wraith dagger]

Clarkie

[crimson sedge]

okok

[violet night]

your n value is 3,

so you get something like looks like $a_3=(3a_{1}) \cdot (2a_{2})$

[wraith dagger]

Clarkie

[crimson sedge]

yes thank you

6(1/24)(1)

sometimes my brain no work

.close

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[violet silo]

how to solve this ?

[violet silo]

using complex numbers

english ?

[thick kraken]

id imagine you could just write cosec(a+ib) = 1/sin(a+ib)

then expand with the sin addition formula using the givens and it should just like pop out

[violet silo]

okay trying it

[crimson sedge]

Lol

[solid juniper]

haha that’s just unsightly typesetting

[cedar kiln]

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[inland ocean]

What's the difference between cosh and cosx tho

[cedar kiln]

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[glacial locust]

This is the given limit, I'm testing if it's continuous or not

[gilded elm]

x->5?

[glacial locust]

Now I can do the whole right and left hand limit thing to do it

Or just substitute x=π

[gilded elm]

oh pi

[glacial locust]

Oh yes

[gilded elm]

$\lim_{x \to \pi} (\sin x + \cos x)$

[wraith dagger]

DerpZ

[clear sage]

do you know the requirements for a limit to be continuous

[gilded elm]

f is continuous at x if ...

[glacial locust]

How do I know when to just x=π or do this

[clear sage]

yuo do x=pi when you are trying to evaluate the limit

but in this case, you are trying to see if the limit is continuous

[manic acorn]

For a limit to be continuous?

[glacial locust]

The book did this for a few provlems

[glacial locust]

Yeah

[gilded elm]

wait, you want to prove continuity?

[glacial locust]

Shit I'm not expressing my problem right

[glacial locust]

Is there something wrong with doing this, other than it being taking longer to solve

[gilded elm]

are you trying to prove continuity or solve the limit?

[glacial locust]

I'm trying to prove continuity of the function (sinx + cosx) at x=π

[gilded elm]

if you want to find the limit you can just plug in sin(pi)+cos(pi)

[glacial locust]

Yes

[gilded elm]

ok so let epsilon > 0 be arbitrary

[glacial locust]

Alright

[gilded elm]

so now you want the delta such that |x-pi| < delta

implies

|f(x)-f(pi)|<eps

where f(x) = sinx+cosx

[glacial locust]

I'm not getting

[gilded elm]

wait

uh

is this like pre-university math?

[glacial locust]

Yup

High school

[gilded elm]

ahh

Ok ok

[glacial locust]

Pre calc

[gilded elm]

So continuous just means like

You can draw the function without lifting your pencil

[glacial locust]

yea

[gilded elm]

So

[glacial locust]

i have trouble with limits

[gilded elm]

Is sin(x) + cos(x) defined for the entire domain?

[glacial locust]

its not given

[gilded elm]

Correct

You don't need it to be given

Alright what's the domain of sinx

[glacial locust]

im so bad

i forgot

[gilded elm]

Okay uh

[glacial locust]

its R

right

[gilded elm]

Yep

And what about cosx

[glacial locust]

R

[gilded elm]

great

Now are there any numbers

from R that can't go into sinx

or cosx

[glacial locust]

nope

[gilded elm]

great so

what about sinx+cosx?

[glacial locust]

R

[gilded elm]

great

[glacial locust]

is the domain

[gilded elm]

so are there any numbers from R which

again can't go into sinx+cosx?

[glacial locust]

nope

[gilded elm]

now at x=pi

can you see it's continuous?

[glacial locust]

yes

[gilded elm]

alright

then you're done

[glacial locust]

yes but that doesnt sort the problem i have tho

its more to do with limits i think

[gilded elm]

could you like

post the exact question please

[glacial locust]

yea ill just post the ques

yea

[gilded elm]

Oh ABSOLUTE value

[glacial locust]

I have no trouble understanding how they did this part othe than that how they just substituted x=π in the limit. I thought continuity required LHL=RHL=f(π)

[gilded elm]

Oh you're confused why you can just

g(pi)?

[glacial locust]

No not the function itself

The limit

I was proving continuity using this method

I hope it's legible

this is just for the right hand limit, similar for lhl, i can do that too if you want

@gilded elm sorry for the ping but help

[gilded elm]

I'm confused on what you are confused about

[glacial locust]

oh

[gilded elm]

The solution seems perfectly reasonable to me

[glacial locust]

i see

[gilded elm]

as in

[glacial locust]

can it be done via this way

[gilded elm]

this

Yeah

[glacial locust]

oh alright

ig the book way is just shortened of what i did

[gilded elm]

but notice why this works

[glacial locust]

the book sued 2 methods and what confused me is can both be used or not

used*

[gilded elm]

could you show both?

since sin(pi)+cos(pi) is defined it's faster to go with that

[glacial locust]

this is one of them

[glacial locust]

wdym defined

[gilded elm]

defined meaning (in a simple way) that

you can evaluate that

for example 1/0 is not defined

[glacial locust]

oh yes

[gilded elm]

say you want to evaluate $\lim_{x \to 0} 1/x$. Then you can't just calculate 1/x

[wraith dagger]

DerpZ

[glacial locust]

This is the method i used in my "solution"

[gilded elm]

Then you'll need to do the

0+ and 0-

[glacial locust]

Ah ok

[gilded elm]

Yeah your method works

[glacial locust]

So we cant substitute where there are constraints

[gilded elm]

You can only substitute when the limit is defined

I.e. you can only compute the limit when it's defined

[glacial locust]

Wdym by the limit is defined

When the functions defined right?

At the value where x is tending to?

[gilded elm]

for example

lim (1/x) is not defined at x=0

[glacial locust]

Yes

[gilded elm]

in an intuitive way if you imagine the graph

you can see it's shooting off to infinity

[glacial locust]

Yup

[gilded elm]

so that limit isn't defined

[glacial locust]

I see

[gilded elm]

but lim at x=pi of sinx is defined

[glacial locust]

And for cos x too so we can substitute it

[gilded elm]

Correct

[glacial locust]

Got it

[gilded elm]

So if you visualize the graph of the function

[glacial locust]

Limits are really confusing haha

[gilded elm]

You can see it

Well you can usually draw a graph and have a look

[gilded elm]

lmao wait till you get to real analysis

[glacial locust]

True

[gilded elm]

some of the functions are even weirder

[glacial locust]

Yea no I'm good lol

Alright

Thanks a bunch

.close

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[brave frost]

Y = (3x/2) - 2

State the rise of the line if the run is: 2, 4, 7
State the run of the line if the rise is: 3, 9, 4

[brave frost]

pls help

nvm

Im gucci, gang gang

.close

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[crimson sedge]

My guy @brave frost

What the hell is ur name 💀

[violet night]

[cedar kiln]

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[exotic field]

Slove in Laplace transform

[cedar kiln]

@exotic field Has your question been resolved?

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[nocturne oriole]

Anyone can help me with the 9th question?

Any help????

[lusty grotto]

so ice is at temp 0C

[nocturne oriole]

That's wt confusing I don't know the temperature of ice lets assume it to be as 0°C

[lusty grotto]

well ice is 0C u dont have to assume it

[nocturne oriole]

Ok then..

[lusty grotto]

anyway

so how much heat is lost by the ice in melting to water at 0degC

[nocturne oriole]

Q=Lm

[lusty grotto]

right

and here that qty is equal to?

[nocturne oriole]

Heat gained by water ?

[lusty grotto]

no

okay

so ice loses a heat Lm

[nocturne oriole]

S

[lusty grotto]

in turning to water at 0degC

[nocturne oriole]

S

[lusty grotto]

in doing so

it absorbs heat from the water

but it continues to heat up

right

[nocturne oriole]

S

[lusty grotto]

it wont just stop at 0degC

[nocturne oriole]

Yep

[lusty grotto]

by S do u mean yes?

[nocturne oriole]

Yes

(S)yes

[lusty grotto]

when does the heat exchange stop?

[nocturne oriole]

When the water and ice both become same temperature

[lusty grotto]

yeah

so we dont know what temp this will be right

so take a variable for it

[nocturne oriole]

T

[lusty grotto]

call it T

right

[nocturne oriole]

Ok

[lusty grotto]

so the ice first melts to water at 0degC

and then this water heats up to T

[nocturne oriole]

Yep

[lusty grotto]

and the water in the beaker is what must give out the energy required for all of this

[lusty grotto]

so the water in the beaker starts at 50degC

and cools down to T

[nocturne oriole]

So it's Q = mct

[lusty grotto]

the heat the water in the beaker loses in cooling down from 50deg to T, is what the ice cube takes to melt into water at 0deg + the energy that water at 0deg takes to heat up to T

now u must be able to write an eqn

[nocturne oriole]

Heat energy gained by ice at 0°C to become 0°C of water then to T°C of water = Heat energy lost by water to reach T°C from 50°C

[nocturne oriole]

Is it correct?

@lusty grotto what's the equation?

[lusty grotto]

yeh

[lusty grotto]

this is correct

[nocturne oriole]

200×4.2×(50-T)=42000-840T

[nocturne oriole]

Amt of heat lost by water

[lusty grotto]

yes

[nocturne oriole]

Amt of heat gained when 40g of ice at 0 °C converts into 0 ° C water

40×336=13440

[lusty grotto]

yes

[nocturne oriole]

Now 0°C to T°C water

[lusty grotto]

yes

[nocturne oriole]

40×4.2×T

Which is 168T

[lusty grotto]

yes

[nocturne oriole]

After this ?

[lusty grotto]

put those values here

[nocturne oriole]

13440+168T=42000-840T

Is this correct

[lusty grotto]

yes

[nocturne oriole]

Yaa got it

[lusty grotto]

👍

[nocturne oriole]

T=28.33

Tku

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[high marten]

Hello.
I have to prove this for any natural m>1.

[high marten]

I tried to use induction, now I'm stuck tried to prove this

[crimson delta]

try converting it into exponential form

[high marten]

like this?

I've got it, thanks

.close

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[proud monolith]

having trouble trying to prove this identity

[manic acorn]

Ur numerator looks wrong

Shud be sin(-x) at the end

Yea and then youre done

[proud monolith]

Did I get one of the rules wrong?

[manic acorn]

Yea

[spice prairie]

I remember doing that

Wait some seconds

[manic acorn]

The last one is cos C - cos D = 2 sin((C+D)/2)sin((D-C)/2)

[proud monolith]

oh explains why its wrong, appreciate the help <3

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[cedar kiln]

@meager flare Has your question been resolved?

[ancient talon]

maybe something like a_n+2 = 5a_n+1 - 6a_n + 2n(-2)^n
n = 0

a_2 = 0

everything is 0

wait that cant be right lol

idk im bad at seq someone else can help

[dire geode]

don't do people's work for them

[peak lynx]

👀 sorry idk

[cedar kiln]

@meager flare Has your question been resolved?

[cedar kiln]

@meager flare Has your question been resolved?

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[maiden cloak]

Hello, how would I solve for A and B if

A*cos(x)^2 + B*cos(x) - 1 = 0

Where two possible values of x are (pi/3) and (5pi/3)

[maiden cloak]

Thanks

[dire geode]

quadratic formula

[maiden cloak]

yea I tried that and it didn't work

did my calculator just break?

I guess so then

[dire geode]

show your work

[maiden cloak]

so I would solve for A then substitute that back into the quadratic formula to solve for B?

[dire geode]

$\cos(x) = \frac{-B \pm \sqrt{B^2 - 4A(1)}}{2A}$

[wraith dagger]

riemann

[dire geode]

now plug your two values and get 2 equations for 2 unknowns

[maiden cloak]

wait why is it cos(x) = ...

I thought it would be just x?

[dire geode]

follows from your equation

if it helps, rewrite your equation using the variable

$y = \cos(x)$

[wraith dagger]

riemann

[maiden cloak]

ok thanks

I thought quadratic formula solves directly for x but I guess not

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[lapis mantle]

How would I be able to equate this answer to [(-2)sqrt3]/3?

[lapis mantle]

I tried getting it down to secx, but without it being a special angle I can’t equate it further

[zealous compass]

is it 7pi/6 insid the brackets on the last line

[lapis mantle]

Yea

[soft owl]

if you can give us the original question

[gaunt hamlet]

7π/6 is a special angle, no?

[zealous compass]

^

7pi/6 is just pi/6 in the 3rd quadrant

^

[lapis mantle]

Basically the original question was to sub in -pi into g(x)/f(x), where f(x) = sin(x-pi/6) and g(x) = tan(x-pi/6)

I meant special angle as in like this special triangle

[gaunt hamlet]

It's on the unit circle

[gaunt hamlet]

.

If you know which trig functions are positive/negative in the 3rd quadrant, you can just use π/6

Then apply the right sign

[lapis mantle]

Ahhh okay I think that makes sense

Let me try it

Got it

Thank you all so much for the help ❤️

[cedar kiln]

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[dapper imp]

"The duration of different electronic components of a fabric follows a normal distribution of N(500,σ). How much is the value of σ, if the probability of having an electric component of a duration between 450 and 550 is 0,75?"

• I tried to solve this probability exercise but I keep getting stuck at the end and I can't find the value of "σ". The exercise says that the solution is "σ=43,48" , but I can't find a way to get there, I don't really know how to continue or solve this, I tried to search this online but I found no sources or resolutions to this problem.

[lethal jackal]

imagine the probability density function of the normal distribution

it looks somewhat like this

Let F^-1 be the inverse cdf of the normal distribution

F^-1(0.125) is the z-score corresponding to 450 in your distribution

but also, the z-score corresponding to 450 is calculated as (450-500)/sigma

and you can simply set these equal and solve

or another way of looking at is

F((450-500)/sigma) = 0.125

and solve for sigma

[cedar kiln]

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[dapper imp]

I see now, the drawing makes it clearer, thank you

[cedar kiln]

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[earnest pendant]

Could anyone please explain why in the integrals there is npix/L by cos

[upper abyss]

It looks like we're missing context. Was anything before this? What was the question?

[earnest pendant]

It’s the integral, I just wanna know the integral

Idk how npix/L is coming

[solid juniper]

probably integration by parts

with the computation totally omitted

[earnest pendant]

That doesn’t make sense

Could you please elaborate

Idk why it’s in the numerator

I am really confused by the integration by parts here

[solid juniper]

what doesn't make sense? i don't think it should that clear until you do or see the computation, anyway

like unless you can find an antiderivative of this just in your head

you kinda just have to compute it lol

[earnest pendant]

What do you mean?

As in here

This is what I am getting it doesn’t look the same

[solid juniper]

$\int x \sin\left(\frac{n\pi x}{L}\right) dx$

oops didn't mean to reply

[wraith dagger]

💜𝓁𝒶𝓎𝓁𝒶💜

[solid juniper]

this is something to compute

and you can do it with integration by parts i imagine

[earnest pendant]

Yes that’s what I tried doing

[earnest pendant]

As in here

I am not getting the same thing

[solid juniper]

probably made an error then

:c

on your side i mean

OR it's the same thing but in a different form, or just not done yet

[earnest pendant]

.close

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[crimson sedge]

any good online calculators for parametric equation to cartesian equation

i dont have time to do all of these by hand

cant find one

[crimson sedge]

Nope, sorry! If you're not getting an answer it's because we don't do that stuff here. If you don't have enough time, then come back when you do. Sorry that was not the answer you expected, it's the best you'll get on this server for questions like this :/

its ok i get it

[cedar kiln]

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[wild stone]

this says that the second derivative being 0 makes the theorem inclusive

so when does the second derivative being 0 indicate an inflection point

or was that just some false thing i learned during high school?

using x^3 as an example theorem 16 checks out because 3rd derivative would be the first non vanishing derivative at x=0

and k would be odd so it's a point of inflection at 0

[gusty forum]

that's insufficient and the reason that's why you need the generalization

inflection point => second derivative = 0, but the other direction doesn't work, consider y=x^4

[wild stone]

ahh yea i see true for the first derivative and how it equaling 0 doesn't imply a local min/max

also

[gusty forum]

pretty sure that actually does hold

[wild stone]

in theorem 16 there's a part that says suppose the k+1 derivative of f exists, i don't understand the relevance of it

it can be a point of inflection but the converse is true that a min/max => derivative = 0 , it was something they pointed out in this pdf

[gusty forum]

oh right i misread

oh hey riemann

i guess you can construct a grotesque counterexample to abuse the lack a of continuous derivative but i cant do it off the top of my head

[wild stone]

imma just ignore it and proceed lmao

thank you for your help

.close

[cedar kiln]

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[hidden radish]

Is chi square test technically always a onesided test so you never have to multiply the p-value by 2?

[cedar kiln]

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@hidden radish Has your question been resolved?

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[fiery berry]

how can i find derivative of arcsec and arccosec ?

[fiery berry]

i found arcsec like that

[crimson sedge]

You can derive them using the derivative of the inverse

Unless you are talking about how to apply the formulas you know of those

[fiery berry]

what is my mistake ?

this is derivative of arcsec

i would like to apply formula

i applied this formulas

but something wrong

[cerulean sail]

How comes you did $\frac{8}{1 \cos^{-1}(5v)}$ if you don't mind me asking?

[wraith dagger]

chartbit

[fiery berry]

arcsec = 1/cos

i made it with this formula

[crimson sedge]

No that's not how it works

[fiery berry]

oh

[crimson sedge]

[
\arcsec(x) = \arccos(\frac{1}{x})
] is the correct statement

[wraith dagger]

♡LexQa♡

[fiery berry]

when arcsec = 1/cos

never ?

[sacred grail]

when x is approximately 7.07891760755

[cerulean sail]

That's effectively saying arcsec = sec

[solid juniper]

arc = 1

[sacred grail]

arcsec is arcsec

[fiery berry]

wow

[sacred grail]

,w plot arcsec(x)

[wraith dagger]

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[sacred grail]

,w plot 1/(cos(x))

[wraith dagger]

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[solid juniper]

and dividing by arc gives us sec = sec

[sacred grail]

ah yes ofc

[cerulean sail]

Mind you, thought the original thing written was arcsec = 1/arccos as well

[crimson sedge]

Layla inventing the Laylan lemma I see

[solid juniper]

my favorite lemma, sec = sec

ok sowwy this is too many stupid jokes

[sacred grail]

anyway arcsin is arcsin, arccos is arccos, arcwhatever is arcwhatever

no rewriting them as 1/sin or 1/cos or 1/whatever

[solid juniper]

the confusion is some people write cos^{-1} for arccos or whatever

[sacred grail]

the confusion is probably more that people think sin(x) means some symbol sin multiplied by x

which is just like completely not what it is

[solid juniper]

https://tenor.com/view/why-not-both-why-not-take-both-gif-11478682

[sacred grail]

together its a whole mess

just confuses the hell out of people

[bold lotus]

huhh really

[fiery berry]

i found 40/(u^2)(1-25u^2)^1/2

[bold lotus]

i thought everyone knows its s times i times n times x

[sacred grail]

[fiery berry]

i found [40/(u^2)(1-25u^2)^1/2
]

[sacred grail]

for c)?

[fiery berry]

its wrong why

[cerulean sail]

How did you find it?

[fiery berry]

firstful i made

[fiery berry]

yes

i made -1 *8 *-5u^2/(1-25u^2)^(1/2)

[
i made -1 *8 *-5u^2/(1-25u^2)^(1/2)
]

i made
[
-1 *8 *-5u^2/(1-25u^2)^(1/2)
]

[cerulean sail]

$$

• \frac{ (1)\cdot (-8) \cdot 5u^2}{(1-25u^2)^{1/2}}
  $$

[fiery berry]

not 18 -8 it was 1*-8

[sacred grail]

where are you getting the 5u^2 from

[cerulean sail]

Keep telling me if I'm typesetting what you mean?

[fiery berry]

no 1-8 inside the dominator

[cerulean sail]

Hmm but why?

[fiery berry]

numerator sorry

[wraith dagger]

chartbit

[fiery berry]

8arccos(1/5u)

[crimson sedge]

\dv looks better dc

[fiery berry]

then i made arccos derivative

yes i made like that

5u^2

5u^-2

[cerulean sail]

Hmmm, that's better, but not quite there I think?

[sacred grail]

[cerulean sail]

Though of course, they do give you the derivative for arcsec right there anyway, why not just use it?

[crimson sedge]

[
\dv{x} >>> \frac{d}{dx}
]

[wraith dagger]

♡LexQa♡

[fiery berry]

in exam i cannot regonize all formulas i just tried solve with arcosx derivative

[sacred grail]

i dont like the upright d

[fiery berry]

is it possible to solve with arccosx derivative

[cerulean sail]

...anyways, believe what you should be getting is something like
$$
-\frac{8 \cdot (-1/5u^{2})}{\sqrt{1 - (1/5u)^{2}}}
$$
no?

[wraith dagger]

chartbit

[cerulean sail]

Before simplification

[fiery berry]

and

[cerulean sail]

At least what I've typeset doesn't seem to match what you have...

[fiery berry]

yes it doesnt

[cerulean sail]

Do you agree though that this is correct?

[fiery berry]

im stuck at third step

This is my work but i cannot understand why my answer not correct

@cerulean sail

[cerulean sail]

,w diff 8*arcsec(5x)

[wraith dagger]

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Requested by chartbit#3847

[cerulean sail]

How did you input it - did you use u's rather than x's?

[cerulean sail]

This is equivalent btw

[fiery berry]

calculator wrong

oh ok

this respect to u

[cerulean sail]

Actually ignore

Hmmm, I don't know, what you have at the end should be right, unless I'm missing something clear...

Might be the way you're typing it in, maybe?

[fiery berry]

i mean i dont understand third step on this picture

[cerulean sail]

Oh they're just weirdos for that

[fiery berry]

?

[cerulean sail]

They've done some weird form there - it's still the same at the end of the day, but I don't even see why they won't accept yours

...they probably have some formula they're using, I'm guessing?

[fiery berry]

i didnt send my answesrs yet

[cerulean sail]

,w diff arcsec(x)

[wraith dagger]

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[fiery berry]

not same

[cerulean sail]

Ah, they're taking this to be the "formula", strange people there

[cerulean sail]

The one they have here is better looking imo

[fiery berry]

they re bastrds they re manipulating students not fair

[cerulean sail]

Actually strange people they have (and tbh I don't even see why Wolfram writes it like that either)

[cerulean sail]

Makes no sense to me, this is a more simple form

[fiery berry]

so you mean its correct

thank you so much

[cerulean sail]

Might be that they're taking this one I guess...

[cerulean sail]

Yeah, it's the same thing at the end of the day - personally I would rather accept your version than theirs, but that's just me(!)

[cedar kiln]

@fiery berry Has your question been resolved?

[cedar kiln]

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[crimson sedge]

@austere zealot hey sry last question

[austere zealot]

Yes?

[crimson sedge]

So I am still struggling with the objective function

I have this now:

But this is just revenue

But I want to maximise the profit

this would be the profit one for book 1 (I think)

[austere zealot]

I believe that is correct.

You just need to calculate that value for all 5 books and sum it all up. I presume you know how to get Excel to find optimal solutions?

[crimson sedge]

Yes I do

So the objective function

would be:

• and then the same but for book 2,3,4, and 5?

[austere zealot]

I believe so.