help-28

[tulip marlin]

I don't think it's a hyperboloid either

or is it

I think you're right here

this was incorrect

I'm not familiar with this equation either

I think c is related to the radius at the middle though. the slimmest part of it

x^2 + y^2 - z^2 = c^2

[stark wraith]

i just searched up a line rotated around an axis and got this

[tulip marlin]

but our line doesn't hit the z axis

you can see you can't solve for x = y = 0

[stark wraith]

yeah if it doesn't hit the z axis

at the origin there would be a big hole

something like this?

[tulip marlin]

yes

im still hoping you can just get to the answer parametrically / radially, without resorting to some knowledge

r^2 = x^2 + y^2

l = (1, 0, 2) + (1, 2, -1)t

l is parameterized by t

can you relate r and t somehow

or r and z

then you're done

[stark wraith]

i mean kinda

you can relate z to x or y then to r

but idk if its the same

[tulip marlin]

wait wait wait, this is just for a generic (0, 0, z) and (x, y, z)

r^2 = x^2 + y^2 doesn't have anything to do with the line l yet

you can use the equation of l now

suppose (x, y, z) is on l

then you have (x, y, z) = (1 + t, 2t, 2 - t)

and then use this to figure an equation for r

[stark wraith]

sry i have to be out for a few minutes

[tulip marlin]

this gives you an equation relating r and t

then, you can have t parameterize circles, rather than a point at constant z's, and that should get you to the final equation

previously, each t gives you a single (x, y, z) point, representing a point on the line, at a constant z.
for the rotation, instead have t give the equation of a circle for each constant z

[stark wraith]

hmm my brain was going in a different direction, we know
x = 1 + t
y = 2t
z = 2-t
then we can do t = 2 - z, so
x = 1 + (2-z) = 3 - z
y = 2(2-z) = 4 - 2z
since the equation we had was x^2 + y^2 - z^2 = r^2
and r^2 = x^2 + y^2
we do just plug in the values and ig the z doesn't matter

my prof gave these problems expecting us to collaborate with each other but idk any of the other kids
idk how much help we should get tho
now it just feels like your walking me thru the problem

[tulip marlin]

then get to know the other kids

well i tried to drop some hints, it's not like ive done this before either

[brisk obsidian]

What section of Calculus are you on?

[stark wraith]

like chapter? we are about to go to partial derivatives

[stark wraith]

how is it even possible in college bruh
all lecture we are supposed to listen to the prof then everyone goes their way

in highschool at least we had time to talk to each other in class

[tulip marlin]

form

a discord group

i dont know

[dark flume]

idk what uni is like but you could probably catch a few people and just ask if u wanna work on this together at some place

like library / cafe or something

[tulip marlin]

just hunt them down right before and right after class finishes

and INSIST they all join some group chat

[stark wraith]

lmao beyond my social capabilities maybe one day
turning a point into circles would be
x^2 + y^2 = (1+t)^2 + (2t)^2

that feels too wrong

[tulip marlin]

maybe, maybe not. then try solve for z

[stark wraith]

if i subsitute z in for t
it would become
x^2 + y^2 = (3-z)^2 + (4-2z)^2
expanding it would become
x^2 + y^2 = 5z^2 - 22z + 25

and it looks like an equation of a hyperbolic paraboloid of one sheet
wow this came together beautifully

how would someone even know if they reached the end and if they are right?

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[shrewd hawk]

not really sure where to start here on this question

[shrewd hawk]

.

[buoyant orbit]

it would be nice if we could find the velocity of the cruiser

[green iris]

draw the before and after of each scenerio

and label everything given

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@shrewd hawk Has your question been resolved?

[green iris]

ye im trying to do it rn n its tricky

this is the drawing i made

[cursive condor]

Make 2 different right triangles

Then use distance formula/d=rt

[shrewd hawk]

im up to 3²+(15x)²=10²

[green iris]

is this physics or geometry? i kinda wana use vectors for this

[shrewd hawk]

its from my maths textbook

i havnt learned vectors yet

[green iris]

ok so geometry

[shrewd hawk]

yea

[green iris]

ima try to use coordinate stuff

maybe something will pop out

[green iris]

i dont see where to go from there

rn im expressing the x and y coords of both ships, and seeing if i can do some sort of substitution to get the final d out

[shrewd hawk]

@green iris i got it

@green iris

[green iris]

ah ok then

glad u got it :>

[shrewd hawk]

tyy

close

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[honest quest]

Here what i understand: the beginning of the integration make sense. the limit from 0 to 2 make sense since that the area of that shape. the order of outer radius minus to inner radius make sense as well. what happen afterward doesnt make sense much to me. why are we finding the area of that? if anything shouldnt we find the area/vol of stuff from -1 to 0 and add it the other area

[quick stream]

what was the original question

[quick stream]

Well it specifically asks for the area from between x=0 to x=3

that is why they ignore the portion from x= -1 to x=0

they just mentioned that (-1,3) was an intersection point

however in this case it is not relevant to the problem

[honest quest]

reading comphesion went out of the window for me i guess

thank you!!

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[knotty bison]

wondering if someone could help me understand this

[cursive condor]

Draw a diagram first.

Nevermind

Note that $m<STV=m<UTV$

[glossy valve]

mathisfun

[knotty bison]

okay

so much to add up 😭

[cursive condor]

Show your work

[knotty bison]

okay give me a sec

srry for the sloppy writing

but how would I go abt solving this

[cursive condor]

Distribute the 4/5

a(b+c)=a*b+a*c

[knotty bison]

um

idk that term ur using

[cursive condor]

$\frac45(x+10)=\frac45x+8$

[glossy valve]

mathisfun

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@knotty bison Has your question been resolved?

[cursive condor]

$\frac45(x+10)=x-3$, $\frac15x-3=8$, $\frac15x=11$, $x=55^{\circ}$\
$m<STV=\frac45(55+10)=\frac45\cdot 65=52^{\circ}$, $m<UTV=x-3=52^{\circ}$
$m<STU=m<STV+m<UTV=52^{\circ}+52^{\circ}=104^{\circ}$

[glossy valve]

mathisfun

[knotty bison]

okay tysm! ill try to remember this

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[solid gorge]

lets assume there are two permutation groups

(1 2 3 4 5) (1 2 3 4 5)
(5 3 4 2 1) (2 4 5 1 3)

how can i make a product from them

[onyx glen]

those are permutations

not permutation groups

a permutation group is the set of all permutations on a given set of points/letters/whatever

[onyx glen]

do you want left * right or right * left? it might matter.

[solid gorge]

i want the first one

why would it matter

[onyx glen]

composition of permutations is not commutative

just to confirm, you want left * right?

[solid gorge]

yes

[onyx glen]

ok right

to figure out their product you need to run each number from 1 to 5 through one then the other

1 -> 5 -> 3
2 -> 3 -> 5
etc.

[solid gorge]

is their product

(1 2 3 4 5)
(1 2 4 3 5)

[onyx glen]

no, that's not what i'm getting

i mean, i just showed you that 1 gets sent to 3 but the result you just sent me sends 1 to 1

so you definitely got it wrong

[solid gorge]

i didnt see it before i said what i think

thank you

how do i close this

[arctic rapids]

Use .close

[solid gorge]

.close

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[jolly horizon]

Could anybody explain what the span of two vectors means geometrically?

[jolly horizon]

it would be the plane between the 2 vectors right?

or am I wrong

[hybrid zephyr]

Yep

[jolly horizon]

ok then I have a followup question

give me a sec

I feel like I'm misunderstanding the question

but how would the span of these two vectors be R2

R2 means the whole 2d plane right?

[hybrid zephyr]

Yes

[jolly horizon]

so I don't understand how the span of these vectors covers the entire plane

[hybrid zephyr]

If u want to prove, u can prove by

[jolly horizon]

I can understand it by algebraic proofs

just not geometrically

[hybrid zephyr]

Consider Ax = 0 is solution right

[jolly horizon]

yea

[hybrid zephyr]

Oh

So basically solve that

And ull get 2 3d lines

Those are x=0 and y=0

Uk that intersection of 2 3d lines gives a plane

And that plane is R²

Got it?

[jolly horizon]

sort of

ok

I understand

[hybrid zephyr]

Which part is ambiguous?

[jolly horizon]

The matrix A in Ax = 0 would just consist of the column vectors right?

[hybrid zephyr]

Yes

[jolly horizon]

k

[jolly horizon]

where does this come from

[hybrid zephyr]

Solve

This

1 1 vector and x y vector

x+y=0

For 1 -1 and x y

x-y=0

Solve those

x=0 and y=0

[jolly horizon]

yea

but I don't understand how that is 3d

instead of 2d

[hybrid zephyr]

Just consider z=0 or whatever

I can just represent a line like that

U can consider it to be any dimension line but taking 3d will make it easy for u

To understand

[jolly horizon]

ahhhh

I see ok

tysm

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[buoyant cradle]

I would like to know how to do a portrait of this outcome.

[buoyant cradle]

I don't know if it's a table or a mind map kind of thing.

<@&286206848099549185>

Hello?

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[wintry pelican]

Idk how to find the area of curved things.

[onyx glen]

they look like circular arcs

[wintry pelican]

I have never seen this in my life.

[onyx glen]

you have never seen a circle before?

[wintry pelican]

Not what I meant.

[onyx glen]

you have never seen a circular arc before?

[wintry pelican]

This specific shape.

[wintry pelican]

Maybe?

[onyx glen]

you're not expected to have seen this specific shape before, but surely you have seen bits of a circle?

this thing specifically is composed of 4 arcs of circles centered at (±3, ±4).

[wintry pelican]

I see.

[wintry pelican]

Yeah obviously I've seen a circle.

Anyways I still have no idea how to get the area.

The geometry teacher sucked last year so I learned nothing.

[onyx glen]

this is gonna take some familiarity with circle geometry on the cartesian plane

this thing specifically is composed of 4 arcs of circles centered at (±3, ±4).
like this point specifically

[wintry pelican]

I know radius and diameter???

That's the extent of my knowledge.

[onyx glen]

guh

well uh

this one's gonna be more than you can chew at the minute

[wintry pelican]

WHAT AM I SUPPOSED TO DO?!

JUST LEAVE IT THERE?!

[onyx glen]

https://www.youtube.com/watch?v=u_39J-syjB0 ig give this a watch for a start?

[onyx glen]

yes and go to sleep lmfao

[wintry pelican]

SHUT UP.

ITS 7 AM.

NO SLEEP ANYMORE.

ITS MORNING.

[onyx glen]

then at least go have breakfast? having pulled an all-nighter can't possibly be good for your brain function.

also don't yell at me please, what all did i do to deserve that?

[wintry pelican]

I apologize.

[onyx glen]

like, this problem can surely wait?

[wintry pelican]

This is how I talk.

[wintry pelican]

Ahgggg.

[wintry pelican]

This is sad. Ik how to graph a circle.

Just put the point and the little spinny thing.

[onyx glen]

what i was gonna say is this shape breaks down into 4 semicircles and a rhombus like so

[wintry pelican]

I see. So I would find the area of those individually?

I can maybe do that.

[clear jewel]

Hey @onyx glen is this circle equations?

[onyx glen]

sure, yes it is.

[clear jewel]

I mean is your topic is circle equation yk like ur whole topic is roaming around the specific circle equations?

[onyx glen]

??

what do you mean by "my topic"

i don't understand what you're asking here

[clear jewel]

Like your lesson is only targeted on circle equations

[onyx glen]

lesson?

[clear jewel]

Sorry I'm high rn my English ain't working somehow

[onyx glen]

i am barely engaged beyond giving OP a video to watch

[clear jewel]

Yeah does ur topic is specify only in circle equations

[onyx glen]

....

[clear jewel]

Alr nvm

But i can help u for somehow

$(|x| -3)^2 + (|y| -4)^2 \sqrt{25}$

[glossy valve]

nerfLeander

[clear jewel]

Alr

Basically the equation has an absolute value yk the thing which causes the circle like grumbling thing

|x| makes the equation symmetric accross the x - axis

I mean y - axis sorry

For |y| is the vice versa meaning it across in the x - axis

For that, it represents four identical circles as you can see from the graph

Which are (3,4), (-3,4), (3,-4), and (-3,-4)

And you have radius of 5

In order to solve the enclosed area, you need multiply it by 4 since the shape consists of four identical circular segment

You get it? @onyx glen

[wintry pelican]

It was my question. 😭

[clear jewel]

Oh shoot

My bad

[onyx glen]

you wrote this incorrectly

[clear jewel]

I told u im high rn

[wintry pelican]

But basically what you're saying is I have to find the area of one circle and multiply it by four?

[clear jewel]

How come it is wrong?

[clear jewel]

Yeah

Area of one segment

[wintry pelican]

Wtf that's so much easier than I thought.

[clear jewel]

Mhm

[wintry pelican]

TYSM.

[clear jewel]

Just use area of circle

Then ur good to go

Just use 5 as ur radius

[wintry pelican]

Mm. Yeah. Thank you. That'll be easy.

[clear jewel]

Wait no

I just said area of cirlce 💀💀

Don't use it

Ughhh

Area of one segment only

I'm sorry I'm high rn 😭🙏

[wintry pelican]

?

[onyx glen]

missing equals sign, and a square root that has no reason to be there

[clear jewel]

$A = \frac{(\pi )(r)^2}{4} (4)}$

[glossy valve]

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

[clear jewel]

Since it has 4 circles smthn

[clear jewel]

Mb mb

I told u I'm high asf 💀

I should not use carts rn 💀

[wintry pelican]

Am I crazy or wouldn't the ÷4 and *4 just cancel out?

[clear jewel]

No u need 4 to multiply it because u have 4 identical circles

[clear jewel]

Here

[wintry pelican]

Yeah but you put it over 4 meaning you're dividing too.

[clear jewel]

Kinda of but yeah

[wintry pelican]

I was right. It cancels out.

If you divide and multiply at the same time nothing is going to happen.

[clear jewel]

Mhm

[wintry pelican]

Why are you saying mhm as if you didn't suggest it. 😭

Do you want me to multiply by 4?

[clear jewel]

No

It's fine

It says here on my calc 79

(round off)

[wintry pelican]

Isn't that for one circle though?

[clear jewel]

Nope

Whole

79 square units

If u are like doubting, see the graph its like landing on ±6 and ±8 which the circle is small

Just logically

So it's 79 square units

[wintry pelican]

Huh?

[clear jewel]

See the graph

If u see it touches the -6 and 6 on x axis, it also touches -8 and 8 on y axis which is very small. The circle or the graph is small, so you can say the enclosed area is approximately 78.53

[wintry pelican]

Your logic is bc it's supposed to be a small number?

[clear jewel]

Mhm

[wintry pelican]

Thats not even anything solid.

That's just you speculating.

[wintry pelican]

If I use this method how do I find the length of the rhombus without putting it into a calculator. I can't figure out how to use the absolute value on my graphing calculator so I can't put it in.

Nvm.

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[raw island]

HI guys, for the 2nd question is the answer
y^6-3+16xy^2

[raw island]

so a is 1, and e is 16?

thanks, but wh ydo they write that thing under so weird? to promote confidence?

only reason i think they'd do that for

oh well, thanks.

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[olive olive]

who are you talking to

[raw island]

myself

lol

[onyx glen]

probably me also cause i was the one leaving those reacts?

[olive olive]

oh wait

now i see the reacts

nvm

[raw island]

yes

oh ann iremember u

hehe

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[ashen karma]

Good day! I need someone to explain me the concepts of Types of Continuities (Basic Calculus), please assume I have 0 knowledge about Continuities.. I just need to understand what was said on the pictures.. thank you!

[ashen karma]

If we're going about the specifics on what I do not understand it's:

On #1) I don't get the word "hole", and how it is indeterminate...

On #2) I don't get how it's literally a "jump" or not equal right n left

On #3) I don't get why cos, sin, tan, cot is also called infinite discontinuity...

[next gale]

on #1, its a hole because it is just a point on the line that isnt defined. just a point. its indeterminate because its not defined there

its called a jump because it is actually a jump

[next gale]

idk why cosine is in there, i think thats a mistake

the rest are called infinities because for some values the graph goes to infinity

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[sly trench]

for a hole discontinuitity limx->a- f(x) = lim x->a+ but f(a) does not exist right?

[next gale]

i guess f(a) can be defined if you want its just not equal to the limit from left and right

shure yah

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[raw island]

hi guys is the following correct ?

[raw island]

except 2nd last should be 5/4x^4

[lime ether]

,rotate

[frozen fiber]

looks fine

[raw island]

lets gooo

[lime ether]

good job sir

[raw island]

thanks

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[torn jolt]

This is only for calculations.

[torn jolt]

.commands

[onyx glen]

#bots

[torn jolt]

There is no guide on how to calculate with the Bot.

/calculate 1 + 1

?calculate 1 + 1

.calculate 1 + 1

calculate 1 + 1

,calculate 1 + 1

[glossy valve]

The following error occured while calculating:
Error: Undefined symbol ulate

[hot herald]

,calc for basic calculations
,w to call wolfram

[torn jolt]

I see Thank you.

,calc 1 + 1

[glossy valve]

Result:

2

[torn jolt]

,calc 1.25 / 0.0750

[glossy valve]

Result:

16.666666666667

[hot herald]

now that you know, do these in #bots

[torn jolt]

oh. I am sorry.

Thank you so much for the guide.

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[warm laurel]

hey

[warm laurel]

can I get some help on a work, energy, power question?

asking prior to giving the question because sometimes applied math/physics questions aren't welcome 😅

it's nothing too advanced, just a small question I have

if a pump is connected to the bottom of the well shown here, would the work done by it when pumping water up to the height h be mghx3/2 or mghx2

the argument of 3/2 comes from the water in the tank being at an height h, and that the gravity would do the work in bringing the water upto the level h and only the remaining work would have to be done by the pump

the argument of it being 2 comes from the pump being considered an independent system, where when it runs, it's pumping the water 2h upwards.

[wild sleet]

yeah up to h is free

[warm laurel]

so it's 3/2 mgh?

[wild sleet]

i don't know

i would say 1/2

[warm laurel]

hmm

i took the 3/2 as the displacement of the center of gravity

[wild sleet]

i'm assuming it's like an infinitely wide tank

[warm laurel]

yeah yeah

[wild sleet]

so there's no displacement

[warm laurel]

this is a part of a question but I want to get this small part covered

cause I have a doubt

[wild sleet]

hm

[warm laurel]

my argument was that the center of gravity lies at h/2. and it's brought up to 2h. So the displacement is h x 3/2

[wild sleet]

the center of the bold part rises by h

and that means m is 0 i see

[warm laurel]

wait shall I send the full question?

[wild sleet]

yeah i'm probably guessing wrong what the question is

it's not 1/2 or 3/2, it should be mgh

[warm laurel]

A water tank is located 8m below ground level is half filled with water. This water is pumped up by a water pump at ground level and expelled with a velocity of 2m/s. If it takes 10 minutes to empty the tank, if the power of the pump is 1kW, and the cross sectional area of the tank is 2m^2. Calculate the power required.

I'm not exactly sure what that means but I made sense of what I thought it means

[wild sleet]

right, not infinitely wide

[warm laurel]

the english is kinda terrible in the tute we have

[warm laurel]

yeah yeah

I'm wondering if the gravity reduces the amount of water that has to be done by the pump

[warm laurel]

Oh shoot just now I saw that it said the cross sectional area is of the tank 💀

[wild sleet]

it doesn't matter that you're pumping from the bottom

you can think of it as pumping from the top

it just depends on how much the level falls by the end

it would tell you the average height of where you pump from

[warm laurel]

oh, interesting

so for the question I have

would taking the displacement of potential energy of the center of gravity of the water be appropriate (6m) or would you have to multiply it by 8?

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[glad void]

Where

[glad void]

I don’t understand the first part of the mark scheme

How do they get yx-3y=x^2

?

[cursive condor]

Because $(x-3)y=x^2$

[glossy valve]

mathisfun

[cursive condor]

They multiplied both sides by (x-3)

[glad void]

Ohhh ya

Thank you 🙏

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[glad void]

.reopen

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✅

[glad void]

Wait I need help for another part😭

When u solve for y using the discriminant

I get y=12 and y=0

How do I know my y is in between those 2 values?

[cursive condor]

,w plot y^2-12y

[glossy valve]

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Requested by mathisfun111

[glad void]

Mhm

It’s ringing a bell but I forgot edy

Haven’t seen this in a while😭

[cursive condor]

Hm

If a>0 then it is conc up

[glad void]

Huh

I don’t understand😭

[cursive condor]

You can just use a number line

Plug in a value between 0 and 12 to see if it is positive or not

[glad void]

It’s negative

But like

Wait I forgot how discriminant works😭

So if <0 means no real roots

Then this graph is a minimum

And then I forgot how to determine if it’s the area inside or outside

[cursive condor]

Yes

[glad void]

How😭

Oh wait I get it now nvm

Thank you🙏

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[spring bay]

How would I graph a period of
y = -2tan(πx - (π/2)) with the relevant points being at 0, pi/4, -pi/4?

[eternal dock]

||using desmos||

[cursive condor]

Use integer x

[eternal dock]

what i find the easiest is graphing tan and slowly applying the changes 1 by 1 till i get what i want

[spring bay]

I graphed tan x but I just don’t know how to work with tan graphs

[eternal dock]

well - would be the opposite

of what u drew

obviously

[cursive condor]

Tell me the period of the graph at least

The one with algebraic stuff

[spring bay]

tan x is pi. -2tan(pix-(pi/2)) is 1

[cursive condor]

Good

Phase shift?

[spring bay]

-1/2

[cursive condor]

So plot it now

[spring bay]

wouldn’t I have to find the asymptote first? how would I find the asymptotes

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[cursive condor]

Which ones are between 49 and 59

Or, is 1/4>1/6, etc.

[versed urchin]

K

So is it all of them?

Ty

[cursive condor]

Ye

[cursive condor]

Same test

[versed urchin]

What?

[cursive condor]

Like

[versed urchin]

I don’t understand

[cursive condor]

4/5<8/9 etc

[versed urchin]

So like

Which one is less

Oh so it’s all of them

Thanks man, appreciate it.

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[ashen karma]

Good day! 😄 I need help with this one.. I absolutely do not get it 😭😭 Like I can't answer test 1, describing the normal curve... For test 2 I only managed to answer letter A, I hope it's correct

[ashen karma]

This is Statistics and Probability...

Is number 3 from test 1 mean?

[frozen fiber]

yes

[ashen karma]

Can you check if it's right?

1. Smooth
2. Vertical line
3. Mean
4. Standard deviation symbol
5. Mean + 3 standard deviation
6. Mean - 3 standard deviation
7. Standard deviation
8. Spread out
9. Peaked

This is just a guess and purely based on which fits the phrases more 😭

[frozen fiber]

2. no

[ashen karma]

That's the only thing wrong?

[frozen fiber]

7. no

1. ,8. ,9. no

[ashen karma]

Oh ):

[ashen karma]

How about now? @frozen fiber

[frozen fiber]

good

for future reference

mean is mu

standard deviation is sigma

[ashen karma]

Ohhh

[frozen fiber]

$\mu \sigma$

[glossy valve]

Xetrov

[ashen karma]

Can you explain to me why this is correct? Like a short explanation if you don't mind.. I'm still finding it hard to understand..

[frozen fiber]

explanation of what

[ashen karma]

Like how is a general normal curve smooth.. 😭

[frozen fiber]

how is it not smooth?

or why would it not be smooth?

In a large population, do you think the height of people suddenly tip off at 5ft 9?

[ashen karma]

Idk what makes it smooth.. is it because it has no vertices? And just a curve? Sorry my brain is slow 🥹

[frozen fiber]

a curve is smooth yes

in maths something which is not smooth is something which has corners

or jumps

[ashen karma]

Ohhh

[frozen fiber]

hopefully that makes sense?

I'd say read a primer on normal distributions

[steel solar]

Is this a scaled version of e^{-x^2}

[frozen fiber]

probably will get a better intuition

if it's truly normal then yes

[ashen karma]

Then what about number 9? What makes it peaked? I don't get how if the standard deviation is small the curve will be more peaked..

[frozen fiber]

I'll just give you the formula lol

[ashen karma]

Okay I'll take your advice on that 😔

[frozen fiber]

$f(x|\mu, \sigma) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$

(let me check this first)

[ashen karma]

Wth is that

[glossy valve]

Xetrov

[frozen fiber]

that's the general formula for a standard distribution

plug it into desmos, see what you get

(replace sigma with s and mu with m)

[ashen karma]

Rn?

[frozen fiber]

general formula for a distribution is usually referred to as a probability density function

[frozen fiber]

if you're interested go ahead

but don't worry about that

you're only asking about the shape

[ashen karma]

Ohh.. I'm still unfamiliar with that.. all we do is substitute and solve that's about it.. then graph.. then z score

[ashen karma]

Thanks a lot! Though I still need help with test 2

[frozen fiber]

interesting

[ashen karma]

Yeah our educational system sucks ass 😔 I'm in shs

Grade 11

[frozen fiber]

basically, if you decrease sigma to almost zero, then you should see that $f(\mu) = \frac{1}{\sqrt{2\pi}\sigma}$ which is very big

[glossy valve]

Xetrov

[ashen karma]

Ohhh

[frozen fiber]

which is why it peaks higher

[ashen karma]

I'll check that out once I'm home. Thank you, I appreciate it 😄

[frozen fiber]

don't know how I'd explain how it corresponds to its spreadoutness but you get the gist

[ashen karma]

Yeah

[frozen fiber]

this is a class test?

[ashen karma]

Nope homework

If it were one I would have never been allowed to take a picture

[frozen fiber]

anyways i require sleep

good luck and honestly just use the drawing in the first picture

[ashen karma]

Oh I still need help with test 2 though.. but if you really need sleep then I don't mind.. Good night to you!

<@&286206848099549185> please help me on test 2 🥺

I already answered letter A, I hope it's correct...

[clear jewel]

Stats and prob?

[ashen karma]

Yess

I can't answer b, c and d

[clear jewel]

Filipino arent u?

[ashen karma]

Yeah 💀

[clear jewel]

Sabi na e

[ashen karma]

LMAO

[ashen karma]

Patulong po bigyan kita c2 na red

[clear jewel]

Tignan ko I'll try my best

[ashen karma]

Absent ako nan kuya/ate/lolo/lola 😔

[ashen karma]

Thank you 🫡💪🔥🗣️💯

[clear jewel]

Yung sa part na "thus" is it conclusion?

Basically lumabas is 19 right?

[ashen karma]

Opo

[clear jewel]

Makikita mo nmn sa graph mo na nagtouch ung curve don sa 19

I would say its still the same

[ashen karma]

Bakit

Bakit po same ung area sa area between ng 35 at 19 😭 di ko naiintindihan kung anong "area" kinocompare nila sa area between ng 35 at 19

[clear jewel]

Ang interval na 19 hanggang 35 ay mula mu - 2 standard deviation hanggang mu .

Nakalimutan ko latex sorry 😭

[ashen karma]

Iz oke

[clear jewel]

Kapag tiningnan natin ang normal distribution curve, ang normal distribution ay symmetrical, ibig sabihin, ang probability na makuha ang isang value sa left side which is equal ng probability sa right, kung parehong distansya mula sa mean.

Gets mo?

[ashen karma]

Bat equal?

Himdi

[clear jewel]

Equal sila kasi nga sa mu - 2 times standard deviation

[ashen karma]

Ksi alam ko left ng mean is -sigma then right ng mean +sigma

[clear jewel]

Yes

Pero u can see na nag stop din sya don sa 19 and 35

Which is equal ung part B don sa graph

[ashen karma]

Di ko gets ung -2 times standard dev... San nanggaling un sa emperical rule ba un?

AHAHAHAJJS

[clear jewel]

Focus muna tayo sa main prob mo baka san pa toh pumunta eh

[ashen karma]

Absent ako neto 😭🙏

Di ko tlga gets bat same ksi di ko alam kung anong area kinocompare sa area between 35 & 19..

Un ba ung area sa letter A na pinapashade? @clear jewel

[clear jewel]

Hold on

[ashen karma]

Ge lng

[clear jewel]

As much possible let's chat in English so that anyone can understand us and also help you

Hold on

I'm tryna make simple an explanation

[ashen karma]

Okay

[clear jewel]

$\mu - 2\sigma = 35 - (2 \times 8) = 35 - 16 = 19$

So, the area between 19 and 35 is the same as the area between $\mu - 2\sigma$ and $\mu$ .

[glossy valve]

nerfLeander

[ashen karma]

Yeah okay that makes sense lol

How about letter c

[clear jewel]

Basically, when we look at the normal distribution curve, the normal distribution is symmetrical, meaning that the probability of obtaining a value on the left side is equal to the probability on the right, given the same distance from the mean.

[ashen karma]

How th is 35 to 27 = 34%, while 27 to 19 = 13.5%

[clear jewel]

For C it should be 47.5%

[ashen karma]

Bruh... WHY

😭 wait

[clear jewel]

27?

Where'd you get that?

There's even no 27 on the prob

Just focus on your problem

[ashen karma]

I'M SO STUPID 😭😭😭😭😭😭😭😭

😭😭😭😭😭😭 I was wondering how tf is it 34% and 13.5%

[clear jewel]

The possible you can think of this is make it logical thinking

[clear jewel]

It is given i think

[ashen karma]

Ye 😭 I thought if I knew it was from emperical rule I'd be able to get the answer but it was more simpler than that 😭😭😭😭

[clear jewel]

Are you familiar with 68%, 97%, 99% pattern?

Wait hold on

[ashen karma]

YEAH THAT

[clear jewel]

To make clear things for u

[ashen karma]

Emperical rule the old one

[clear jewel]

This is where 34 and 13 gets

[ashen karma]

Ywaah that's what I was wondering about 😭

[clear jewel]

Gets mo na?

Madali n lng ung last part

[ashen karma]

Opo

[ashen karma]

Paturo pa rin po 🥺🙏

[clear jewel]

Final answer sa last part is 47.5%

Since the total probability (area under the normal curve) is 1, the probability of the yield being between 19 and 35 bushels is 0.475 (or 47.5%).

I wonder what to put on the phrase (last part) "it is ___"

Maybe ask your seatmate

[ashen karma]

Di ko gets ung sa 1 💀 100% ba? Ksi tanda ko 1 un

[clear jewel]

Yes 1 = 100%

[ashen karma]

Yay 😁🙌👍🎊🎊🎊👍👍👍👍👍👍🎊🙌🎊🙌

[clear jewel]

Goodluck sainyo

And have a great day ahead

[ashen karma]

Kayo rin po kuya/ate have a great day 😁💗

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[limber crag]

gonna send my work

#1

lmk if thats hard to read

opening in browser makes it easier

also this one

[cursive condor]

$\lim_{n\to\infty}\sum_{i=1}^n\frac{b-a}{n}f\qty(a+\frac{b-a}{n}\cdot i)$, so $b-a=1$, and you can deduce that $a=?$

[glossy valve]

mathisfun

[limber crag]

a = 1-b

[cursive condor]

The exact value.

Note what you have on the inside of the ^3 expression.

[limber crag]

pasting just so i can seen again

ah wrong problem

I have

1+i/n inside

a should be 1

I think i figured that out at some point just didnt write it down

[cursive condor]

$u=2x$, $du=2dx$, $\frac{du}{2}=dx$, so we have $\int\frac{\frac{u^2}{4}}{\sqrt{5-u^2}}\cdot\frac12 du$

[cursive condor]

Yes

[limber crag]

b-1=1 so b should be 2

[glossy valve]

mathisfun

[limber crag]

oh is it a

[cursive condor]

Good

Nah it is not a)

Look at function again

[limber crag]

xi = 1 +i/n

we figured that out

oh oh

shit

c

[cursive condor]

W

[limber crag]

I always mix that up

[cursive condor]

Here now

[limber crag]

lemme past that one again

[cursive condor]

$\frac18\int\frac{u^2}{\sqrt{5-u^2}}du$

[glossy valve]

mathisfun

[cursive condor]

I gotta go man I'll brb

[limber crag]

oh okay

ill try to figure out from what u said before

oh

did i forget the 1/2

[hot lily]

@limber crag do you know about trigonometric substitution?

[limber crag]

i might the name doesnt sound familiar tho

[hot lily]

Wait- what exactly do you need help with

[hot lily]

Have you done u-substitution

[limber crag]

yes

i just need help with where i went wrong

but

[hot lily]

Oh wait this is diff

[limber crag]

i think i forgot the 1/2

imma rewrite rqw

[hot lily]

Can you explain this? I feel like you got confused because you mixed up x and u in the same integral

[limber crag]

how i got u^2/4?

[ornate bobcat]

The problem is that you multiplied by 4 instead of getting it out as 1/4

[limber crag]

so i shouldve just taken it out as 1/4

instead of 4

[ornate bobcat]

Yes because it is (1/4)*u^2 in the numerator so when you want to take it out of the integral you take it out as it is

[limber crag]

let me start over a bit iwth it

[limber crag]

ah

but in denom id flip

i always mix those for some reason

But I shill need to apply the 1/2 correct?

still*

[ornate bobcat]

Turn it to a fraction multiplied by the main part then take the fraction out

[limber crag]

Not following

Rather, id need to see what youre meaning visually

wait i think i might have it

[ornate bobcat]

I don't understand what you mean by that
I see that the part of 1/4 is the only problem
If I neglected some non-good-practice order of math then if you corrected that your solution should be correct

[limber crag]

I was getting the 1/2 from finding dx

[ornate bobcat]

Oh right

[ornate bobcat]

Good

[limber crag]

i think i got it now

phew

but

if the 1/4 was in the denom

id bring it out as 4 right?

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[full bronze]

once ive expanded and factored and flipped the second fraction. chat gpt os telling me to cancel two x-11's on the bottom with one x-11 on the top i dont understand why.

[hot lily]

This is why you dont use chat gpt

[full bronze]

Here’s what I’ve done

[hot lily]

Oh wait [this](#help-28 message) is correct

[full bronze]

Yea I k ow but don’t know why

[hot lily]

You cancelled out x - 11 in the numerator and denominator

[full bronze]

Is that incorrect?

[hot lily]

No its good

In general, $\frac aa = 1$

[glossy valve]

King Leo

[full bronze]

Gpt is saying to do this aswell

[slate violet]

,w (x+11)^2 / (x-11) divided by (x^2 - 121)/(11x-121) simplify

[slate violet]

ok now that's not correct

[hot lily]

Thats wrong

[slate violet]

yeah the bot is super inconvenient

[full bronze]

So it is 11(x+11) / x-11 ?

[slate violet]

yep!

[hot lily]

WA or CGPT

[full bronze]

Ok

Thank you!!

[slate violet]

WA

[full bronze]

I was pulling hair out for quite a while on that

[hot lily]

Wa is so amazing :(

,w factor ((x+11)^2 / (x-11)) / ( (x^2 - 121)/(11x-121))

[full bronze]

Sweet

[glossy valve]

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[hot lily]

@slate violet youre just using WA wrong

:(

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[minor jetty]

Hi guys I need a little help with where to start when solving this problem

[slate violet]

try subbing u = sqrt(y)

[minor jetty]

Ohhh okay thank you

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[slate violet]

yes then you can rewrite 2^u as a power of e

npnp

or you probably know the rule already at this point

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[vestal jewel]

How do I solve 8

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@vestal jewel Has your question been resolved?

[full forum]

@vestal jewel Has your question been resolved?

[quaint rune]

use pythagorus

[vestal jewel]

Where are the right triangles

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[little plover]

im not sure if i did d correctly

[brisk obsidian]

I don't think you need integrals for part d.

[little plover]

i'll search up the scoring guidelines

and see

BOMBOCLAT I DID MY INTERVALS WRONG 😭

nvm

they're correct

they js put the interval as B

idk why they did this

[brisk obsidian]

They weren't interested in the change in the volume, the question asked about the change in area of the slice of the volume.

They want to know what the rate of change of the area of this slice is at x=-0.5.

[little plover]

im js gonna go to bed 😭

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[visual smelt]

help

[little plover]

i here wsg

[onyx glen]

is there an image on its way?

[little plover]

this is out of my education ann can help u 😭

i didnt learn this yet

[little plover]

it arrived

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[spice grail]

@onyx glen

is the working for this correct

[void nova]

Awful mistake 🙈

[spice grail]

thanks........

[void nova]

Integral of a fraction is not at all the fraction of the integrals of num and den

[void nova]

You have first to split the fraction, and then integrate

[sick vortex]

life would be pretty easy if that was the case

[void nova]

So you have to integrate 3x²•x^(-1/2) + 2•x^(-1/2)

[void nova]

Indeed lol

[cursive condor]

You know how exponent laws work right

[onyx glen]

!noping

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[spice grail]

sorry

[void nova]

Why? 🤔
Also, don't interrupt the chat

By asking useless questions (also not related to maths)

[torn jolt]

Sorry

[spice grail]

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[wintry cave]

When computint this 4x4 matrix, why am i getting -24 instead of 24?

[wintry cave]

https://www.youtube.com/watch?v=fWzUwrt1Z0s this is the method i used

[honest hill]

if you choose a different row or column frmo the first one, you have to look if you start positive or negative

if you choose an even row or column, you start with a negative, then positive, etc

instead of positive then negative

[grim skiff]

Your sign for the 6 is wrong

You can do what bonk said or apply (-1)^(i + j) for the sign too, where i and j is the row and column number

[wintry cave]

Ohh

Got it!

Thank you!

Chatgpt struggled with this

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[honest hill]

!nogpt

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Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

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[vocal quiver]

Can anyone help me with this

[vocal quiver]

I try to prove by contradiction but I can’t end it

[minor crater]

show your work

[vocal quiver]

I have done to this step and can’t think of any idea on what to do or claim next

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<@&286206848099549185>

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[tame raft]

Just for interest sakes, here I have an equation to calculate the current in a diode. Screenshot 1 and 2 are the same equations (except one of the equations has a -1 at the end) but I get no difference to my answers regardless of whether I use this -1 or not. Is there a particular reason for this?

[tame raft]

Logically if you add a -1 then your answer should be 1 less but this isn't the case because the -1 is inside the bracket. My question is WHY is that the case. Mainly asking just to develop a better understanding of this equation.

[tame raft]

I'm thinking this may be some mathematical reason hence I asked it here

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[torn jolt]

the answers are definitely different but my guess is that I(s) might be extremely small and so -1*I(s) is neglectable(insignificant) and so it dosent really change the answer

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[warm laurel]

yo can I get some help

[warm laurel]

so lets say there's a pump, which pumps up water from a well which has its base at depth h below the ground, and a water level of h/2 (initially)
when calculating the work done to the potential energy of the water by the pump, do we have to consider the displacement of center of gravity of the water in the well, or can you just take mgh where m is the mass of the water in the wel?
you can take that the 'pipe' which pumps water up is located at the bottom of the well and it pumps all of the water up to ground level

if you think of it at a 'deeper' level the argument is whether the pump has to do the full work in bringing the water up an mgh amount, or due to the pressures and all that from the water column inside the well, some part of that work is going to be done by gravity in equalizing the water levels inside the pipe and well. Effectively making the net work done on the potential energy of that water by the pump mghx3/4

[next sail]

off the top of my head, the amount of energy required to lift each successive 'layer' of water increases linearly with depth, so it should be fine to take an arithmetic mean of h/2 and h to get 3h/4

[warm laurel]

yeah

some guys said that you have to consider the full height

and that if you take the difference of final and initial potential energies that's going to be a 'net' work done

and that some part of that work is going to be done by gravity

I argue that any work done by gravity is going to be cancelled out and that the final work done by the pump is mghx3/4 but they still disagree

there was a long debate over this but we couldn't come to a conclusion

[warm laurel]

their argument was that within the pipe, the water gets pumped up 'h' by the pump and the pump does all the work in doing that

I argued that due to the equilibrium of water columns in the pipe and well, there's going to be some part of 'bringing the water up the pipe' done by gravity

which changes with time (as the water level in the well decreases) and that the net result of the work done just by the pump would be the same as taking the difference of final and initial potentials of the water

<@&286206848099549185>

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[fringe turtle]

my tutor in red, my answer in black (i know my answer is wrong because I guessed)

[fringe turtle]

could someone explain my tutors logic?

the minimum distance between the Pi1 and Pi2 was done on the last subquestion (11/sqrt(35)) and Pi1 is -3x-5y-z=5

nvm dawg i think something clicked

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[granite grove]

Hi do these answers make sense for you?

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[leaden gust]

hi i have been having a hard time how to find increasing and decreasing in intervals, can anyone teach me? like this question from scratch

[leaden gust]

<@&286206848099549185>

[austere cove]

!15min

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[leaden gust]

Oh got it

My bad, I’ll wait

[royal charm]

For explanations of common concepts, you're better off deserving YouTube first. Then try it and share your work here.

[leaden gust]

Oki

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[untold yacht]

Help

[untold yacht]

How do I get area of the triangle

[sudden chasm]

find the area of the minor sector
then subtract area of triangle from it

[hot topaz]

Note that they already gave you the radius ! (So you can find the area of the circle, and then remember the whole makes 360 degree, so use that to find the area of the triangle minus the gret sector)

[sudden chasm]

the vertex is 120 so each smaller angle is 60

then apply some trig to find the perpendicular and base of the triangle

[untold yacht]

I figured it out

It’s a 30 60 90 triangle

[sudden chasm]

yup
area of triangle = 0.5×base×perpendicular

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[hollow summit]

where does the e_0 thing come form?

[umbral dome]

,, k_e = \frac{1}{4\pi \epsilon_0}

[glossy valve]

cloud