help-27

[pure vessel]

That's how u do it

Thanks a lot

I have a query as well

[pastel pasture]

No worries

Sure

[pure vessel]

I have moved to grade 11 from 10th last month...In 10th I was a 100/100 scorer in Math but 11th Maths seems tough

Is it normal..And how do I gain 100/100 in my 11th and 12th as well pls tell

[pastel pasture]

That depends on you tbh

[pure vessel]

Any tips ?

[pastel pasture]

Dont hesitate to ask questions to anyone, even if they feel dumb

[pure vessel]

Ok

And any recommendations about Mathematics books for class 11&12

Is Cengage good ?

[pastel pasture]

I'd recommend asking your teachers about that

Or fellow students

[pure vessel]

Oh ok

Thanks a Lot

👍

[pastel pasture]

You can close the channel if you're done

[devout snow]

@pure vessel Has your question been resolved?

[pure vessel]

Oh ok

[devout snow]

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[formal sierra]

Need help on how to not make this 0/0

[woven radish]

MathIsAlwaysRight

[polar chasm]

now try to break it into product of 2 fractions

[woven radish]

MathIsAlwaysRight

[formal sierra]

where did the factored out x go?

[polar chasm]

sin(x)/x

Can you reply to the step you didnt understand to?

[formal sierra]

here. did the x from x(3x+2) go to the denominator of sin(x)?

[woven radish]

MathIsAlwaysRight

MathIsAlwaysRight

[polar chasm]

applying the rule above

it stayed in the denominator of sinx, yeah

[formal sierra]

this is a big help, thank you!

[devout snow]

@formal sierra Has your question been resolved?

[devout snow]

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[novel bay]

How did it get to this answer

[pseudo basin]

,rccw

[pseudo basin]

do you know in general how scientific notation works?

[devout snow]

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[polar trout]

I thought you would do tan^-1(9.7/5.1)
This would give you the bottom left angle and would be the same as x due to alternating angles, what have I done wrong here?

[rare mantle]

I don’t think that is wrong?

[polar trout]

Then maybe it’s the calculator

[polar trout]

My answer I got was less then 90, so that wouldn’t work, right?

[rare mantle]

why not?

the angle is clearly less than 90 degrees

[polar trout]

Waittt

[rare mantle]

what was your answer btw

[polar trout]

I’m dumb

[rare mantle]

,w tan^-1(9.7/5.1) * (180/pi)

[woven radish]

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[rare mantle]

should be this

[polar trout]

62.265

Oh

Thanks

[rare mantle]

yeah

[polar trout]

It was because I didn’t do it to 1dp 💀

Didn’t think to check that, that’s why I was told I was wrong

[rare mantle]

ah lmao

[polar trout]

.close

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[mellow trout]

whats the correct answer? (if i answer wrong it makes me go through 40 minutes of forced video watching)

[mellow trout]

nvm .close

.close

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[spice saffron]

Hey! Can someone help me solve this

[lyric hornet]

what is "breadth"? is it width?

[spice saffron]

Correct

[lyric hornet]

is O at the origin?

[polar chasm]

althought it's not mentioned, I'd just assume it is. Otherwise it would be unsolvable

[lyric hornet]

well you really can't assume

but yea the x and y axes are given

by the figure

so I'd imagine

@spice saffron $\overline{AO}+\overline{OB}=\overline{AB}$ by given and $\overline{AO}=\overline{OB}$ by definition of midpoint can you do the rest?

[woven radish]

XxMrFancyu2xX

[spice saffron]

uh?

im in 9th grade?

[lyric hornet]

oh-

I thought it was a Geometry question

ok well since O is a midpoint we know the distance from point A to point O is equal to the distance between point B and point O, you agree?

[spice saffron]

Okay

But do we know AO?

Please I have an exam tmrw its 12 AM right now

[lyric hornet]

alright

I'll help you

but first I need to know how much you before I blow ur brain up like I did above 😅

[spice saffron]

Right

So we know uh

Points

(x,y)

x axis is abcissa and y axis is ordinate

[lyric hornet]

ok that's something, do you understand points?

[spice saffron]

As far as my knowledge goes yes

[lyric hornet]

ok not a bad start fs

so the distance between point A and point B is 6cm right?

[spice saffron]

Correct

[lyric hornet]

alright

and point O is the midpoint right?

[spice saffron]

mhm

[lyric hornet]

point O is on the y-axis right?

well it's on the origin

but therefore also the y-axis, yes?

[spice saffron]

Right

[lyric hornet]

so then the y-axis cut the rectangle in two, right?

[spice saffron]

w-what okay

But I definitely wont realize that in the exam

[lyric hornet]

you seem unsure, what doesn't make sense?

[spice saffron]

uh

This is confusing

Anyways

Yes it does

[lyric hornet]

point O is on the y-axis right? because it cuts the side in half, the y-axis must cut the rectangle in half

[spice saffron]

right

[lyric hornet]

very bad rectangle but you get the idea?

[spice saffron]

right

[lyric hornet]

so that origin is at the point (0,0), right?

[spice saffron]

Right

[lyric hornet]

and each of those segments must be 3 from the origin in the x direction

[spice saffron]

rightt

[lyric hornet]

because the length of the total rectangle is 6

[spice saffron]

but thats 3m right

[lyric hornet]

3m, 3cm, 3mi

[spice saffron]

but how does that prove that 1cm = 1 unit of x/y axis

[lyric hornet]

whatever your unit of measure it given

[lyric hornet]

you're not trying to prove that

you're trying to find the coordinates of the corners

hence we have that

[spice saffron]

interesting

but isnt the horizontal axis the x axis

and shouldnt the firs tpoint (0,-3) be (-3,0)

[lyric hornet]

yk

i am not awake

good catch

see you're more prepared than you think 😂

but yes (-3,0) and (3,0) are your coordinates

but because the height is 3 we have found all three coordinates

@spice saffron ok i gtg now if this helped a little if any 🥴

[devout snow]

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[devout snow]

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[devout snow]

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[native tartan]

how do i use the bot to show integration

[stone stump]

,w int x^2 dx

[proud perch]

$\int$

[woven radish]

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Mr. Gamer

[stone stump]

do you mean stuff like this?

if the first, go to the website wolframalpha

if the second, its called latex

[proud perch]

$\int_{a}^{b} f(x) \ dx = F(b) - F(a)$

[woven radish]

Mr. Gamer

[devout snow]

@native tartan Has your question been resolved?

[devout snow]

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[ancient dust]

find the real parameter m and solve the quadratic equation 2x^2 - (1 - 2m)x + m -1 = 0 so that its roots so they complete this : 1/x1 + 1/x2 = 3/4

[ancient dust]

how do i find m?

[thin apex]

Start with the quadratic formula?

[ancient dust]

so a = 2 b=? c =-1

[thin apex]

a = 2, b = -(1-2m), c = m-1

[ancient dust]

ok thank you will try now

.close

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[wind crag]

how would i start problem 8? I have no idea what to do.

[midnight echo]

Read the hint

If you put two of them on top of each other you’ll end up with a cylinder…

[wind crag]

what would i do after i have the cylinder

[midnight echo]

What would the volume of that cylinder be?

Compared to the volume of a truncated cylinder? (The one in the question)

[wind crag]

864 pi

[midnight echo]

^

[wind crag]

sorry i don’t really get it

[midnight echo]

This is wrong btw, think about how you can construct a single cylinder from two of those shapes

[wind crag]

ohh is it 180 pi

[midnight echo]

No

[wind crag]

what.

the height would be 20 right

if i combined 2

[midnight echo]

Yes

[wind crag]

what did i do wrong here?

[midnight echo]

Why 3?

Wait I’m dumb

Yeah you’re right

[wind crag]

because the radius is 3

[midnight echo]

But that’s the volume of two of them right?

[wind crag]

yeah

[midnight echo]

So you have to…

[wind crag]

divided by 2?

[midnight echo]

Yeah

[wind crag]

okay ty!

.close

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[fallen elk]

can i get some help with quesiton c?

[graceful cosmos]

Plug the point into one of your functions, and solve for the unknown

[fallen elk]

ok let me try that

like this?

[proud perch]

hint: the functions should be the same up to a constant factor

the leading coefficient

[fallen elk]

wym

[graceful cosmos]

You're missing any kind of unknown

[fallen elk]

oh

[graceful cosmos]

Since you wanted a family of solutions for part b, you'd need something like an a in there

[fallen elk]

yeah

i used k

[graceful cosmos]

Ah I see it

[fallen elk]

so then i just solve?

[graceful cosmos]

Then yeah! That looks pretty good

[fallen elk]

ok ill give it a shot

incorrect

i got -4

they got -2

and they have 2x

[devout snow]

Show your work, and if possible, explain where you are stuck.

[main gull]

I see how they got it

-4 = 2 * -2

They then took 2 and multiplied it to (x + 1/2) to get rid of the fraction

@fallen elk

[fallen elk]

can you show me a visual picture

[main gull]

When you write out your equation, you have y = -4(x + 1/2)(x - 2)(x - 6), correct?

[fallen elk]

yeah

[main gull]

Now the solution didn't want to have a fraction, the 1/2

So it split -4 to be -2 * 2

[fallen elk]

then it became

2x + 1

[main gull]

So you have y = -2 * 2 * (x + 1/2)(x - 2)(x - 6)

Multiply 2 into the x + 1/2

And you get 2x + 1

[fallen elk]

got it

thank you

.close

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[fallen elk]

can i get some help with 21 A

[wicked turtle]

consider what the factors of such a function must look like

[fallen elk]

i really just dont understand what simpkfiied form means

[wicked turtle]

i don't know what they mean by that either, presumably either the factored form or the ax^3 + bx^2 + cx + d form

[fallen elk]

its ax^3 form i just checked the answer

[wicked turtle]

ok

[fallen elk]

so what do i do from there

[wicked turtle]

start by writing down the factored form, then you can just multiply it out to get the other form

[wicked turtle]

no, the roots are not 1 and +-sqrt(6), they are 1 plus or minus sqrt(6)

[fallen elk]

oh sorry

[wicked turtle]

also i suggest instead of writing one factor with +-, write it as two separate factors, one + and one -

yea, like that

ok that's correct

now to get it into the other form, you just multiply everything out

[fallen elk]

ok

[wicked turtle]

probably easiest to multiply the two rightmost factors first

but any order will work

[fallen elk]

so this is what i have so far

[wicked turtle]

suggestion for a shortcut for the product of the last two:
$(x-1 + \sqrt{6})(x-1 - \sqrt{6}) = (x-1)^2 - 6$

[woven radish]

Bungo

[wicked turtle]

where i used the fact that $(a+b)(a-b) = a^2 - b^2$

[woven radish]

Bungo

[fallen elk]

right

i got to gto

.close

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[main relic]

bit unrealistic but sure

[fathom thicket]

https://tenor.com/view/pokemon-ash-hat-turned-around-throw-gif-8995789

[main relic]

we can make small steps

so what help do you want

that doesnt seem very >13 but whatever

what are you confused on

multiplication is like, adding something multiple times

if i say something like, 5x7

im adding 5, seven times

5+5+5+5+5+5+5

yes

do you play games

so its like

Apple x4

four apples right

multiplciation is the same

5 x7

seven 5s

get it?

[fathom thicket]

just so you know

there is a 13 age minimum policy we have here

on discord in general

pretty sure grade 5 is like 11 year old max

[main relic]

lets just not be a snitch

at least, snitch after i explain this to them

yeah

theres 6 1s

1 x6

[silk hill]

Are you sure you should be on discord?

[fathom thicket]

i heavily doubt the 10 year old kid is gonna remember this within the next 3 hours

[main relic]

then there are 500 489s

heres a hint

500 is 1000/2

so you can do 489*1000/2

which is wayyyy more easy

and if you see numbers like

999, 101, 299

example 7x999

we can say 7x1000-7x1

7000-7 is way easier than just calculating 7x999

[frosty cradle]

<@&268886789983436800>

[main relic]

ok im done explaining you guys can do whatever you want with this 10 year old now

wait

how are we gonna close this channel

.close

[devout snow]

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[main relic]

ok

[devout snow]

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[sleek estuary]

why is there a 4, what i remembered was a(x-h)^2 but that was a long time ago can anyone refresh me why there is a 4?

[grand siren]

i think we probably don't have enough context here to answer this...?

although have you considered that maybe a = 4p

[sleek estuary]

i am studying calc 2

3D

[naive estuary]

and...?

[frosty cradle]

I think p is the distance from the vertex to the focus or something

[grand siren]

yeah that's not really context
i still don't know what any of these letters mean

[frosty cradle]

yes, see here: https://math.stackexchange.com/a/2392684/993372

[sleek estuary]

before traces

[long fog]

correct'

[devout snow]

@sleek estuary Has your question been resolved?

[devout snow]

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[restive river]

Anyone need help

[devout snow]

Please don't occupy multiple help channels.

[supple knot]

Not how this works

.close

[devout snow]

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[restive river]

Anyone need help

[ebon ice]

yes

[restive river]

OK send the problem

[pseudo basin]

@restive river this isnt how this works

[pseudo basin]

@ebon ice open your own channel

[fathom thicket]

lmao

[ebon ice]

ik the first one is wrong

[fathom thicket]

just go and browse

all the help channels

that are literally being used

[ebon ice]

wha

[fathom thicket]

not you

[ebon ice]

they literally asked if i needed help?

[pseudo basin]

@restive river you dont go asking if anybody needs help. the people that need help are the entire Math help (OCCUPIED) category. you go into one of the channels and help the person there.

keep in mind !nosols of course.

[restive river]

oh

[fathom thicket]

you should still

make your own

help channel

for it

!help

[devout snow]

Please read #❓how-to-get-help

[ebon ice]

okay..

[naive estuary]

hmmm

[restive river]

Ugh Idk what to do here

[naive estuary]

@restive river seriously?

[restive river]

I accidentally asked someone for help

and

uh

did something wrong

I just want to help.

[pseudo basin]

and weve told you how to do that

[restive river]

Ok fine

I'mma help someone

BYE

[devout snow]

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[hollow ether]

Does part 1 of the theorem hold true vice versa? Like if bn converge if and only if an converges?

[stone stump]

thats what if and only if means

<=>

[devout snow]

@hollow ether Has your question been resolved?

[devout snow]

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[quasi wasp]

I need help with some simultaneous equations regarding unit vectors

[restive river]

substitute a and b into pa+qb

[quasi wasp]

LLike this?

@restive river

[restive river]

yes

now equate coefficients of each component

[quasi wasp]

What do you mean by that sorry?

[restive river]

i and j form a basis

eg if pi + qj = 5i + 6j, then p=5, q=6

[quasi wasp]

Like this? @restive river

[restive river]

p and q are different

you should keep them seperate

so 3pi -5qi becomes (3p-5q)i

then equate 3p-5q with 30

do the same for j

[quasi wasp]

Figured it out now, thank you

.close

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[pastel pasture]

$$y = |x|$$
$$\ln y = \ln |x|$$
$$\dv{y}{x} \cdot \frac{1}{y} = \frac{1}{x}$$
$$\dv{y}{x} = \frac{|x|}{x}$$

[woven radish]

NEONPerseus

[pastel pasture]

Would this be a valid derivative for the absolute value function

It comes with the bonus of being undefined at 0!

[pseudo basin]

wdym

it is perfectly defined at x=1

[pastel pasture]

Since the absolute value isn't differentiable at 0

Oh

Nvm

I get the joke

[pseudo basin]

but also i think this is kind of overcomplicated

though correct it is

[pastel pasture]

Hmm

What would a not so complicated derivative be

As a function

[stone stump]

just do it piecewise

[pastel pasture]

Isn't that clunky

[stone stump]

ehh

[pastel pasture]

$\dv{x}|x| = \sgn x$

[woven radish]

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

[pastel pasture]

No native latex support for the sign function

Noob

Anyways, thanks

.close

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[sage ember]

n = 1 and L=1, This is a fourier coefficent, all I get is 0.

[sage ember]

I literally have no clue, it says (in swedish) to note that the integration from 0 → L is the same as the integration for 0 → L/2 times 2. After that you are suppose to use partial integrations for 0→L/2. Still get c = 0 or c = c.

After som GPT chatting, he gave me an idea of how to solve this. The trouble is with f(x) in the integral, what is it really? If we take L/2 then we know that f(x)=x exactly, i think :P Which gave me the result c = 0.31828. If this is correct you can close this help thingy or correct me completly. I will be away for a moment.

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[proper violet]

A bucket of mass M is lowered from above using a rope subjected to a tension of modulus T. The acceleration of the bucket has modulus a and points downwards. Which of the following relations is correct?
(A) Mg+T+Ma=0
(B) T=-M(g+a)
(C) T=M(g-a)
(D) T=-M(g-a)
(E) T=M(d+a)

[small quartz]

What do you think

\sum F = Ma

So....

@proper violet

Mg downwards, T upwards

Mg - T = Ma

T = M(g-a)

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[daring violet]

if someone could check this!

[silk hill]

Ok, could you tell me, what is $\frac{n!}{(n-1)!}$

[woven radish]

Try_hard

[fringe flint]

it will N?

[daring violet]

1/n+1?

[fringe flint]

no

it will be n

[fringe flint]

this equations value

equation's*

[silk hill]

The answer will be n

now what about $\frac{(2n-2)!}{(2n)!}$

[woven radish]

Try_hard

[daring violet]

Oh sorry I did plus instead of minus

I understand

I got the answer, thank you!

.close

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[short iron]

[(a^{-4} +b)^{-3}]

[woven radish]

dopediscorduser

[short iron]

How would you go about simplying this into a single fraction?

[\frac{1}{(\frac{1}{a^{4}} +b)^{3}}]

[woven radish]

dopediscorduser

[short iron]

I assume this is the first step?

[supple knot]

it already is a single fraction?

[short iron]

Sorry simplified to a single fraction with no negative exponents

The book is saying

[\frac{a^{12}}{(a^4 b + 1)^3}]

[woven radish]

dopediscorduser

[short iron]

I'm not sure how they got that answer though

[winter patrol]

multiply numerator and denominator by (a^4)^3

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[young crane]

Is this how the direction of the forces will go?

Or the other way round

[surreal frigate]

The positive ones repel each other and both attract the negative

and the negative attracts the positives

[young crane]

So is it correct?

[young crane]

Or does the arrow go the other way around

[surreal frigate]

both ways

[young crane]

The way I use it doesn't matter?

[surreal frigate]

is any of them in place, like it should not move?

[young crane]

Huh?

You see those blue arrows I drew right?

[surreal frigate]

yeah

[young crane]

I was given a question like this a long time ago

Basically this

And what I did was draw the arrow the same way I have

Here

But I now became hesitant since it's just a negative charge in the middle

I was wondering if it would look like this

[surreal frigate]

It's both at the same time

(-) ----> <---- (+)

[young crane]

what about for this

[surreal frigate]

still the same

[young crane]

So why was I told to just draw one arrow

[surreal frigate]

maybe the exercise says that one of the points is static

[young crane]

But does this affect the overall solving of the question

Using parallelogram law of forces of vectors

[surreal frigate]

depends on what the question is

[young crane]

sqrt(f1^2 + f2^2)

the question is to find the resultant force acting on the -16 charge

[surreal frigate]

then yeah you drew the arrows right

the first ones

[young crane]

this?

[surreal frigate]

yeah

[young crane]

Okay so why did I draw it correctly?

How do I know the direction

[surreal frigate]

cause you're examining the forces acting on -16

[young crane]

i see

ohhhhhhhhhhhh

so if it was +20C

it would be the other way

[surreal frigate]

yeah

no

wait

(-) ---<--- (+)

yeah

[young crane]

great

thanks!

Can I leave this open for a bit

Just in case

[surreal frigate]

sure yeah

[young crane]

I can use parallogram law of vectors riht @surreal frigate

parallelogram*

[surreal frigate]

yeah that's how I would do it

[young crane]

I got 4.82x10^11

[devout snow]

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[devout snow]

@young crane Has your question been resolved?

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[karmic hound]

Yo yo

[karmic hound]

I already change the formula to 1/3 B X h

But what do I do for pi?

Do I divide by 4?

Or put the pi decimal

Done

192π= 1/3 πr2 x 4?

/π both sides?

Im not sure

144= r2

The get square root?

I got 12

But like I keep putting the equation in and I’m not GETTIKG IT AHHHHH

Angry lol

How

Magic

Amazing thank you

Now

7 more problems to go

And a quiz in the morning. 😟

How@does this look

It doesn’t feel right

Im so failing this quiz

Crap

[devout snow]

@karmic hound Has your question been resolved?

[karmic hound]

Yo

So

100π S.A

How do I find the volume

Of a sphere)

[devout snow]

@karmic hound Has your question been resolved?

[meager hollow]

looks good to me

[meager hollow]

theres a formula

[meager hollow]

where did you get 100 pi from

[karmic hound]

It said the Surfae area was 100 pi

[devout snow]

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[lone crane]

what the fuck is this

is it integration by parts??? u sub??? i tried everything and nothing is working and i want death

[patent marsh]

well tell death "not today"

give me a moment to look it over

[lone crane]

thank you my god

[upper schooner]

What did you try for the IBP?

If you don’t mind sharing

[patent marsh]

yes you are correct that you'll need to do IPB

[lone crane]

okok

do i just have to do it twice

[upper schooner]

The start looks fine to me(!)

[lone crane]

but then once i get the integral, how do i even find a value

💔💔💔

[patent marsh]

correct!

[upper schooner]

Do it again on the integral you have

[lone crane]

take the limit to infinity?

okok

[patent marsh]

you'll need to use a limit

yep

see? you got this

[lone crane]

omg 😭😭😭

[patent marsh]

you don't need us at all

[lone crane]

okok let me try

and thennn

i will come back if i have questions

[patent marsh]

sounds good

[lone crane]

can i take the 1/3 out of the integral?

okie ‼️

[upper schooner]

Constants you can pull out of any integral

[lone crane]

okie okie perfect

so i can also pull out the 2?!

[patent marsh]

also once you're done solving this have you ever heard of the IBP tabular method? It can make IBP less verbose

but we'll try it after you're done w/ the problem

[lone crane]

i have not

am i being silly or doing good

[patent marsh]

cool we'll go over it afterwrads

[lone crane]

i feel like that’s wrong

and idek how id take the limit of that now

LOL

💔💔💔💔

[upper schooner]

Yea don’t make any substitutions for it

If that’s what you tried to do

[lone crane]

i didn’t do another usub or integrate by parts again

oh

I FORGOT AN X

[upper schooner]

Just apply IBP another time on this - you seem to have divided by 2x instead in yours

[upper schooner]

,rccw

[lone crane]

oh wait

why’d i put du in there

[upper schooner]

[lone crane]

i am so silly

I SEE

I UNDERSTAND NOW

okok

i try again

[lone crane]

yes

i was silly

and i out du back in

[naive estuary]

should be 2/3 int x e^(-3x) dx

[lone crane]

i am struggling

HOW THE FUCK LIMIT 😭😭😭

dead

[naive estuary]

x^2 e^-x when x -> infinity is 0

[lone crane]

oh u take the limit of the original thingy??

but then what’s the point of integrating 😔😔😔

it wants the answer as a reduced fraction

[naive estuary]

also the last one should be e^-3x

[upper schooner]

You take the limit of what you get on integrating (the integration limits)

[naive estuary]

not e^3x

[lone crane]

yah that’s what i thought

[upper schooner]

IBP gets you those terms you’re working with

[lone crane]

but idk how i would take the limit of that ☹️

[lone crane]

ah yes i fixed

[upper schooner]

Then you work out the integral and get the limit from that

[upper schooner]

See e.g. this, “exponentials beat powers”

[lone crane]

i’m jus a little confused on how like

[naive estuary]

well...you have F(x) now, limit F(b) - F(a), as b -> infinity...

[lone crane]

like how would i do this and get a fractional answer

hold on

let me try

[naive estuary]

answer is 2/27

[floral citrus]

@

[lone crane]

damn

how’d u do that

[floral citrus]

are you @naive estuary to help me with a matice promblem?

[lone crane]

i like am trying but i getting 0 💔

[naive estuary]

whats your F(a)

limit F(b) is 0

[lone crane]

i’m just using infinity i guess

[naive estuary]

huh?

[lone crane]

like f(infinity) - f(0)

[naive estuary]

I mean...you "can"

[lone crane]

what would i do instead?

[floral citrus]

can someone help me with 3#

[lone crane]

bro

[naive estuary]

you know F(infinity) is 0, just find F(0).....

[lone crane]

oh

duh

[naive estuary]

Sry I hate matrix

[lone crane]

Holy fuck

I GOT IT

THANK I KIND SOUL

FOR UR

PATIENCE

[naive estuary]

oh btw, they may deduct points for writing like this...

[lone crane]

YES ik it’s just like for sake of little brainpower rn

[naive estuary]

lol

[lone crane]

ty again 🫶

.close

[devout snow]

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[floral citrus]

@naive estuary who can help em with matrix?

[naive estuary]

idk...just wait

[devout snow]

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[rich dagger]

Help.

[rich dagger]

How do you know when a triangle is right-angled by following these elements:

Triangle a) 8 cm, 7 cm and 5 cm
Triangle b) 16 cm, 8 cm and 10 cm
Triangle c) 10,5 cm, 7 cm and 9 cm

[cinder hazel]

If I have a triangle with sides 3cm, 4cm and x, can you find x?

right angled triangle

i should add

[rich dagger]

Yes

[cinder hazel]

how do you do this?

[rich dagger]

we can use the Pythagorean theorem

a^2 + b^3 = c^2

[cinder hazel]

okay

and Pythagoras Theorem only applies to right angled triangles

What is the value of x in my question above?

[pseudo basin]

b^2

[rich dagger]

Ohhh

[rich dagger]

How

[rich dagger]

Yesyes

[hidden dragon]

What's c, 10.5cm, 7cm and 9cm?

Why did you write 4 sides in c

10, 5, 7 , 9

[rich dagger]

What?

[rich dagger]

5^2 + 7^2 = 8^2

<@&286206848099549185>

[silk hill]

No, this triangle will not be right-angled in this case

[restive river]

First check, if it's a valid triangle by checking that the sum ofany two sides is larger than the third and then follow the pythagoreus theorem

[rich dagger]

I know.

[rich dagger]

Yes, but none of them are right then? Have you seen them?

[restive river]

wait

There is one. Try again

[rich dagger]

Is it triangle b)?

[restive river]

check c once again

[rich dagger]

Triangle c) 10,5 cm, 7 cm and 9 cm

This one?

[restive river]

no c

the sides are 10.5,7 and 9

[rich dagger]

Yes, should I check it?

So it should be right?

[restive river]

check it

[rich dagger]

10,5 is the baseline or width

[rich dagger]

^

[restive river]

the hypotenuse is the longest side

[rich dagger]

Yes

7^2 + 9^2 = 10,5^2

[rich dagger]

That give us 11,4

when we draw it

[restive river]

ohh sorry, my mistake. None of them is right

Had a calculation mistake, my bad

[rich dagger]

so they are all wrong?

[restive river]

yep

[rich dagger]

So it's a trick question

[restive river]

u can answer that none of them is right as they don't follow the pythagoreas rule

[devout snow]

@rich dagger Has your question been resolved?

[devout snow]

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[winged wagon]

Is that right?

[winged wagon]

I feel like the 4 shouldnt be negative

But idk

<@&286206848099549185>

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@winged wagon Has your question been resolved?

[winged wagon]

<@&286206848099549185> its been 15 mns now

[rare sluice]

please repost the question more clearly

[winged wagon]

No i just wanna know if the math is right after i found cos t and sin s

The -4 sqrt21/ 25 - -12/25

When adding the fractions

And subtracting them

Is that part done right?

[devout snow]

@winged wagon Has your question been resolved?

[devout snow]

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[desert matrix]

I learned in limit notation that when x approaches a- it is approaching the value of a without ever getting close to it.

[desert matrix]

How would we write x approaching and being equal to a but not superior to and visa versa?

I'll send an example

[proud perch]

like a left hand and a right hand limit?

[desert matrix]

Yes

[proud perch]

yea

you can write

$\lim_{x \to 0^+} f(x)$

[woven radish]

Mr. Gamer

[proud perch]

this gives the limit from the right

[desert matrix]

So you can use the same notation?

[proud perch]

yea

[desert matrix]

Even if x is equal to a?

[proud perch]

just remember the plus and minus signs for a left and right hand limit

[desert matrix]

Even if x is equal to a?

[proud perch]

the limit doesn't care about the value at a

that is irrelevant

[desert matrix]

That can't be right.

Am I misinterpreting the workbook?

[proud perch]

f(a) has NOTHING to do with the limit of f(x) as x approaches a

lol

what this is saying is that the limit of a function as x goes to a exists if and only if the left limit equals the right limit

[desert matrix]

I'm still confused.

<@&286206848099549185>

[zenith jacinth]

Cest a propos de quoi ? @desert matrix

[desert matrix]

Les limites.

[zenith jacinth]

Oui qu'est ce qu'elles ont ?

[desert matrix]

Donc je fais les règles de la limite à gauche et à droite.

Je voulais savoir...

[zenith jacinth]

Oui, si une limite existe, on a la limite à gauche qui est égale à la limite à droite

c'est une équivalence

[desert matrix]

Ouais c'est ça qui me gosse.

Comment est-ce qu'une limite peut exister?

Comment peuvent-ils être équivalent?

[zenith jacinth]

La limite à gauche de f en a est la même que la limite à droite de f en a, donc lim f = f(a)

Et donc si lim f = f(a), les limites à gauche et à droite existent, ce sont deux propositions équivalentes

[desert matrix]

Ok. Je comprends la logique qui suit.

Mais je ne comprends pas comment f(a) de gauche peut être égale à f(a) de droite.

Visuellement.

[zenith jacinth]

C'est une histoire de limite

la valeur d'une fonction en un point approche une certaine valeur

fais un dessin

[zenith jacinth]

oui, la fonction approche une valeur en 0

[desert matrix]

Woahh

Slow down.

Quoi?

De où il sort le zéro?

[zenith jacinth]

C'est pas 0 ?, je sais pas tu n'as pas gradué ton repère

[desert matrix]

C'est un point indéfinie.

[zenith jacinth]

définie, je vois a en bas

[desert matrix]

Donc comment peut-il avoir un point d'intersection.

[zenith jacinth]

t'entends quoi par point d'intersection ?

[desert matrix]

Ben c'est pas ça que ça veut dire quand f=f(a)?

Qu'ils se rencontrent à un point?

[zenith jacinth]

c'est qui "il"

[desert matrix]

Les deux fonctions?

[zenith jacinth]

Pourquoi, y a deux fonctions ?

c'est quoi ta fonction au juste ?

[desert matrix]

Att j'ai mal élaboré ma confusion.

Ok ok on recommence.

[zenith jacinth]

[desert matrix]

Ok ya une fonction.

[zenith jacinth]

Oui

[desert matrix]

De un côté x se tend vers a de la gauche sans s'y rendre

Et de l'autre côté x se rend vers a de la droite sans s'y rendre.

[zenith jacinth]

??

[desert matrix]

Mais ma questions c'est comment ces deux limites peuvent-ils être égal?

[zenith jacinth]

Parce qu'ils tendent vers la même valeur ?

[desert matrix]

Mais ils ne se rendent jamais à cette valeur.

[zenith jacinth]

c'est ce que "limite" veut dire

[desert matrix]

Ou au moins de ce que je peux comprendre.

J'écoute.

[zenith jacinth]

limite ca veut pas dire que la fonction est égale à cette valeur, à moins que la fonction soit continue à ce point

exemple : lim 1/x quand x tend vers l'infini

1/x s'approche beaucoup de 0 mais sans jamais être égal

[desert matrix]

Ça tend vers 0?

[zenith jacinth]

Oui

[desert matrix]

Ok donc la limite en d'autres mots c'est quoi?

[zenith jacinth]

Je viens de le dire, c'est la valeur qu'approche une fonction en un point

sans forcement l'atteindre

[desert matrix]

Ohhh

Ok ok

J'ai une autre question.

[zenith jacinth]

[desert matrix]

I can't help being stupid.

[zenith jacinth]

[desert matrix]

Ok ok

[zenith jacinth]

[desert matrix]

Si x est égalle à ou plus grand que 2 comment est-ce qu'on l'écrirait avec la notation d'une limite?

*La partie en haut c'est ce qu'on est dit de faire.

En bas c'est ce que j'ai patenté.

[zenith jacinth]

$\lim_{x \to 2^{+}}$

[woven radish]

Herels

[desert matrix]

Ah c'est bon j'ai compris.

Désolé pour toutes ces questions.

Mais saches que tu m'a vraiment aidé.

[zenith jacinth]

No problemo

[desert matrix]

Merci bcp.

Tu parles l'anglais?

[zenith jacinth]

C'est pas ma langue de base mais je me débrouille

[desert matrix]

Kk cool

.close

[devout snow]

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[mint imp]

Ok so I need help calculating inverse kinematics for a robot arm Im building
I'm going to include several pictures and an explanation so ill let you know when im done

[mint imp]

My arm has 4 degrees of freedom, two sliders are attached to the linear rail and the central column and the end effector rotate.

I included the video to show how the arm moves.

Currently, Ive solved the forward kinematics but am having trouble going the other way

I represented the joints with several vectors and did vector addition to add them together

Each endpoint will be a matrix of [X, Y, Z, theta] where theta is the direction the end effector is pointing

I need to solve the equations for Y1, Y2, thetaC and thetaE where thetaC is the angle of the central column and thetaE is the rotation of the end effector

[lusty sapphire]

cool

[mint imp]

I am willing to solve the inverse kinematics a different way but this is the easiest way I could think of, if there is a better way lmk

ok im done, sorry for the long explanation

[lusty sapphire]

I'm at work rn, but I may be able to look later

Unless someone else can help you right now

[mint imp]

ok, thank you

I'm not expecting an answer right away, its not exactly straighforward

to me anyways

[devout snow]

@mint imp Has your question been resolved?

[devout snow]

@mint imp Has your question been resolved?

[devout snow]

@mint imp Has your question been resolved?

[devout snow]

@mint imp Has your question been resolved?

[devout snow]

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[scenic surge]

so Im done one part of the question

but im stuck for the second half

x_1 * x_2 = c/a

[graceful cosmos]

Instead of dividing one of the factors, say "by zero product property, this equals 0"

Which is a lot more rigid

Otherwise, wow that's slick. I didn't know you could get this without actually solving for x1 and x2

[scenic surge]

wait wait what

oh u mean divin by x_1 - x_2 ?

[graceful cosmos]

Btw you can solve for x1 and x2, which is an easy method to solve part 2

[scenic surge]

alr lemme try

[graceful cosmos]

Probably not the best way to do it, though. Let me think about that.

[craggy escarp]

I have a method which is simple and gets you both the sum of the roots and their product

Equate ax²+bx+c with its factorization a(x-x1)(x-x2)

Since x1 and x2 are roots right?

Compare x term of left and right side of the equation to find the sum

Compare the constant term of left and right side of the equation to find the product

[scenic surge]

$c = a(x_1 - x_2)(x_1 + x_2)$ ?

[woven radish]

Calc II Victim

[craggy escarp]

Nuuu, look

ax²+bx+c = a(x-x1)(x-x2)
ax²+bx+c = (ax-a.x1)(x-x2)
ax²+bx+c = ax²-a.x.x2-a.x.x1+a.x1.x2
ax²+bx+c = ax² - a(x1+x2)x + a.x1.x2

The coefficient of x on the left is: b

The coefficient of x on the right is: -a(x1+x2)

They must be equal!

b = -a(x1+x2)

So: x1+x2 = -b/a

[scenic surge]

HOLY SHIT

OMG

i get it

and for c

c = a.x1.x2

divide both sides by a

to get

x1.x2 = c/a

brahhh I woulda neevr thought of that

[craggy escarp]

Hahahaha!

Now you know!

What about the other question, what have you done?

The x1 ² + x2 ²

[scenic surge]

other question?

wait which one

oh shit

i didnt even read dat

[craggy escarp]

Okayy

Try, and then I will show you how to do it

[scenic surge]

would this be valid?

[craggy escarp]

Exactly what I did!

[scenic surge]

nice

oh shit

theres another part to it

so x_1 + x_2 = 47 and x_1 * x_2 = -59

[craggy escarp]

Should be easy for you!