help-28

[slate hemlock]

Can you see why? @worthy bridge

[worthy bridge]

yeah

[slate hemlock]

So you have both height and base now

[worthy bridge]

ok

ty

[slate hemlock]

Yw

[dull escarp]

Hi how to multiply decimal

[worthy bridge]

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[slate hemlock]

Open a new channel

[dull escarp]

how

[slate hemlock]

Go in the math help (AVAILABLE) section

help-18

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[atomic vortex]

how would I simplify something like this any further

[atomic vortex]

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

Just so that its to find the derivative of it

I was thinkin of simplifyin it if possible

some guy did -sin^2(x)/x

idk how

All I could do was 1/x(-csc^2(x))

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[atomic vortex]

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[bright trellis]

Is basic physics problems helpable here?

[bright trellis]

There is an object(mass isnt known)
This object is pushed by some Force and it traveled 0.4m

When they added 0.2kg on the object with the same force it went only 0.2m

I need to find Mass of the object

[wide sundial]

Something’s weird about this

A forced applied on an object means acceleration

[bright trellis]

Yes

[wide sundial]

It doesn’t move distance

[bright trellis]

Object was still

Then some force pushed it

And it went 0.4m

And in the same time with the same force it but with added 0.2kg it went only 0.2m

[wise wyvern]

I'm going to have to ask, when you say it went 0.4m and 0.2m, it's safe to assume we're talking same time?

[wide sundial]

That force has to be applied over some timeframe for the object to move

[bright trellis]

Yes

[wise wyvern]

Okay. That makes more sense.

Then x is proportional to acceleration.

Acceleration is inversely proportional to mass.

Use that.

[wide sundial]

Ah that is a lot better way to ask what I was trying to ask haha

[dawn glacier]

Help

[wise wyvern]

#❓how-to-get-help

[bright trellis]

Yes

Then im gonna = them right?

[wise wyvern]

equate what

[bright trellis]

F=ma
F=(m+0.2)a

[wise wyvern]

use a different variable for the second a

It's different.

Then you equate yes.

[bright trellis]

Think I got it

Thanks

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[half remnant]

how are you supposed to work out these pronumerals?

[half remnant]

ik that there are 2 angles equal to 90

do you have to prove TOP is an equilateral triangle or smth

[simple totem]

is there more information than this?

do we know OPA is straight

heck even then alpha could be anything < 90 if you make TA long enough

[half remnant]

im assuming that OPA is straight, O is the centre and TA is a tangent

ive figured out that a = 2b

[simple totem]

that looks about right, but im still not sure if you can solve for all angles without some more info

[half remnant]

damn alr

[simple totem]

is anything else given to you?

[half remnant]

nope

[simple totem]

what were you asked to do

[half remnant]

i honestly thought u just had to prove TOP is equilateral

[simple totem]

well it doesn't have to be

[half remnant]

yeah

[simple totem]

i think there's a missing dash on one of these edges to show they're equal to the radius (probably the chord)

in which case TOP is indeed equilateral and this is very solvable

[half remnant]

yeah if TP had a dash this would be solveable

alas, it does not

I will skip this question for now. Thanks for the help though

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[simple totem]

OK

oh

.reopen

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✅

[simple totem]

@half remnant PA does have a dash i didn't even see it

i thought it was gamma'

[half remnant]

oh u meant that

[simple totem]

lmao

nono i was talking about TP but this works too

[half remnant]

alralr

[simple totem]

so OT = OP right

[half remnant]

mhm

[simple totem]

and TA^2 = (2OT)^2 - OT^2

TA^2 + OT^2 = OA^2 = (2OP)^2

[half remnant]

isnt it OA*PA

[simple totem]

instead of what

[half remnant]

instead of (2OT)^2 - OT^2

[simple totem]

is that an identity

i was just using pythagoras

[half remnant]

its like this thing

[half remnant]

ohhh

[half remnant]

yeah i get this

[simple totem]

this is an x, 2x, xroot3 triangle right?

[half remnant]

yes

[simple totem]

so the angles should be trivial. are you allowed trigonometry?

[half remnant]

yes

ohh OTA = 30

[simple totem]

then sin(gamma) = x/2x = 1/2 -> gamma=30

[half remnant]

coz -sin(30

yeahh

ok this is starting to click

and A is 60

since the identity sqrt3/2

yes yes yes i get this

thank you so much 🙏🙏🙏🙏🙏🙏🙏

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[dense raft]

Can someone explain how this works? They moved out the V, but im confused how.

[torn jolt]

They first multiplied both sides by 2

To get 2s = (v_0+v)t

[dense raft]

right

[torn jolt]

Then multiplied both sides by 1/t

to get 2s/t = v_0+v

Then they subtracted both sides of v_0

To get v = (2s)/t - v_0

[dense raft]

im just thinking im not gone 😛

[torn jolt]

[dense raft]

ok got it thank you very much. so the 2 being inside the () doesnt matter?

as in you can normally multiply both sides with it

[torn jolt]

I mean it was improper of me to multiply both sides by 2 first

But you would preferably multiply both sides by 1/t first

[dense raft]

right, doing that would remove the ()

and you would reach the same conclusion, sweet 😄

thakn you, much appreciated.

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[torn jolt]

Yeppie

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[latent cobalt]

How far are you on this problem?

[glacial swift]

Sorry so far im stuck i think i got a rough idea of what to do

[latent cobalt]

Ok, and what are you thinking so far/where are you stuck?

[glacial swift]

Its a related rates problem i think so i have 2 sin pi = 2

I got no idea if this is right tho

[glacial swift]

Am i slightly on track?

[latent cobalt]

That's not what I got

[glacial swift]

Oh 😭 could u explain

[latent cobalt]

So what did you get when you took the derivative of y cos(x) = 2x?

[glacial swift]

Dy/d tsin(x) =2 dx/dt right?

[latent cobalt]

I think you forgot the y on your left hand side

You should be using product rule here

[glacial swift]

Oh ya

So its just Dy/dt y sin(x) =2 dx/dt right?

Then i plug in the values?

[latent cobalt]

I don't think so, I think your product rule should give you something more complicated

[glacial swift]

Sorry im kinda blanking

[latent cobalt]

So for product rule, you have f(x)g'(x)+f'(x)f(x) right?

[glacial swift]

Yep

But where would i apply it?

[latent cobalt]

well on the LHS, you have y cos(x) right?

[glacial swift]

Yep

[latent cobalt]

So, f(x) = y, and g(x) = cos(x). What are your f'(x) and g'(x)?

[glacial swift]

F would be y' and g would be sin x

[latent cobalt]

g' would not be sin(x)

[glacial swift]

-sin?

[latent cobalt]

Since you are taking the derivative with respect to t, you have to use the chain rule on it

And yes it also needs to be negative

Do you know the chain rule?

[glacial swift]

Yeah but isnt the derivative of x just 1?

Oh wait

[latent cobalt]

If we were taking the derivative with respect to x, yes, but we are taking it wrt to t

[glacial swift]

-sin(x) dx/dt?

[latent cobalt]

Yep!

[glacial swift]

So something like this?!

[latent cobalt]

I'm not quite sure what you are showing me

[glacial swift]

Oh is this the answer?

Like after i got the dx/dt i just plug in the values

[latent cobalt]

Not quite, you've made another error somewhere, but we are definitely on the right track. What did you get for the derivative of the equation?

[glacial swift]

The derivative i got y-sin(x)+dy/dt cos(x) dx/dt

[latent cobalt]

Very close, but not quite

Could I see a picture of you work?

[glacial swift]

Yeah

[latent cobalt]

Ok, so it should be f'(x)g(x)+f(x)g'(x), you have an extra addition in there

Also the dy\dt cos(x) should not be negative

[glacial swift]

Okay ty

Idk why i put that there

[latent cobalt]

Do you understand what we have gone over so far?

[glacial swift]

Yep!

[latent cobalt]

Cool, so you can keep solving it and let me know if you get stuck/what answer you get and we can go from there?

[glacial swift]

Okay!

I think i made a mistake on it

[latent cobalt]

Yes it isn't y-sin(x) it is y*(-sin(x)). Does that make sense?

[glacial swift]

Ohhh yeah

So it comes out to -2pi -2= dx/dt

[latent cobalt]

No

Do you wnat to redo your work and send me another pic?

[glacial swift]

Yes please

[latent cobalt]

We are getting close here, and your doing great lol

[glacial swift]

Ok i really appreciate ur help

[latent cobalt]

How is it going?

[glacial swift]

Im working on it rn

Ngl im kinda stuck but heres what i got so far

[latent cobalt]

I think we are running into an issue with the RHS of you equation. What is the derivative of 2x in this case?

[glacial swift]

2x is just 2?

[latent cobalt]

If we were taking it with respect to x, but remember, we are taking the derivative with respect to t

[glacial swift]

So its 2 dx/dt?

[latent cobalt]

Yep!

I think right now you might be able to get it right, because what you had before that was very close to being right

[glacial swift]

So the 0 cancels out both y and dx/dt then i divide the 2 out on both sides and get -1= dx/dt?

[latent cobalt]

YES!!!!!

[glacial swift]

WOOOHOOO

UR SO SMART

[latent cobalt]

Haha, I'm majoring in math at college, so I have the advantage of spending all my time doing it

[glacial swift]

Yeah i make way too many mistakes when it comes to math

[latent cobalt]

Haha, I feel that. TBH I'm actually just procrastinating my HW by helping other people tonight

My stuff was to hard so I decided to take a break

[glacial swift]

You mind helping with one more problem?

Dont feel pressured to either

[latent cobalt]

I wish I could, but it's 4 AM where I live and I have to go to bed. But if you want you can dm me and I'll help you with it tomorrow

[glacial swift]

Ok tyy! Goodnight

[latent cobalt]

No problem. Goodnight!

[glacial swift]

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[next basalt]

I need some help with logarithms.

40 = 10^x -0,2 * 10^x

[cold quarry]

@next basalt 10^x is common on the right side of equation so factor it out

[next basalt]

so i move 10^x to the left side of =

?

[cold quarry]

no

hold on

my phone is uploading picture

[next basalt]

Hmm

[cold quarry]

ya

[next basalt]

I don't understand

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[jovial dagger]

i need help with calulating a bore stonke and pistion zise for a 8 culinder engine

[jovial dagger]

1 pistion has 93.75cc on 1 cylinder but how big dose the pistion need to be and how long dose the stoke need to be to make that work

(need) 750cc spit on 8 cylinders

what is the bore stroke and pistion size

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[jovial dagger]

how many times do i need to devide 3622 to get to 750

16,15129874042647 bore

19,67145359410917 stoke

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@barren escarp Has your question been resolved?

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[void arch]

When they say 2 consecutive S’s

Is that essentially SSS and more

[frosty geyser]

seems like it, SS to SSSS

[void arch]

Why is it SS?

Isn’t that just one consecutive S?

[frosty geyser]

English isnt my main language so Im not sure if consecutive includes the first element

It could also mean SSS and SSSS

From examples I find online it says that {1,2,3} are 3 consecutive numbers

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[next abyss]

I solved for the linear combination but it was wrong

[ivory cairn]

you were given the coordinate vector of x, and the basis.
x is the linear combination of of the basis vectors with the coefficients of the coordinate vector

What you did was the process to find the coordinate vector

you should do the product of the basis matrix with the coordinate vector to find the original x vector.

B*[x]_B

[next abyss]

the inverse of B times <4,1>?

nvm

[next abyss]

I'm confused by what this means

[ivory cairn]

put the vectors of B into a matrix.
Multiply by [x]_B

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@next abyss Has your question been resolved?

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[hushed cairn]

prove that . if n is a perfect square then 2n cannot be a perfect square ?

[cold quarry]

if n is a perfect square then sqrt(n) gives a integer. but sqrt(2n) = sqrt(2)*sqrt(n) doesnt give a integer since its multiplied by sqrt(2)

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[hushed cairn]

oh so when sqrt(2) times n then it dont equal perfect square

[rare dock]

a positive integer is a perfect square iff all the primes in its prime factorization have even exponents

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[twin wolf]

For 5a) how would we do that

[twin wolf]

The origin is just x = 0 and y = 0

So it would be numbers above than 5?

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[wise wyvern]

If you're talking (a) then, not quite.

You don't want the distance to be greater than 5. Nor do you want it to be less than 5. You want it to be exactly 5. From origin.

so if you have your point (a,b) the distance of that point from origin should be 5. How'd you do that? Well, there's distance formula.

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[faint scarab]

During foil, can i say -L * -L = L^2 or is it -L^2?

[faint scarab]

it affects the trinomial i am factoring

[eager violet]

A negative times a negative is a positive

[faint scarab]

Okay, so i CAN give it a positive coeffecient?

[eager violet]

Wdym

Why wouldn't you be able to have a positive coefficient

[faint scarab]

Just a thing ig i never learned. Although the answer to any L would be positive in that place i didnt know if it has a coeffecient of 1, or if "it has a -1 coeffecient but when you plug in L it will go away"

[eager violet]

L has a positive coefficient but that doesn't mean L is positive

Idk what exactly you're saying

[faint scarab]

Okay so, im finding an eigenvalue. I have the equation like (-x -4) (-x -3). Dont know if the proper foil should result in x^2 + 3x + 4x + 12

[eager violet]

It should, yes

[faint scarab]

tyty

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Please don't occupy multiple help channels.

[fathom arrow]

i have one try left

why 4 is wrong?

the critical points i have are 3 and 1

i tried those

f(-1) = 12
f(1) = 8
f(3) = 4
f(6) = 58

it should be 4?

oh -1 is actually -12

that's the one?

[rain quail]

someone tell me what is science notation

[short latch]

open a new channel

[short latch]

yup!

[short latch]

#❓how-to-get-help

[fathom arrow]

thank u

[short latch]

definitely -1

[fathom arrow]

yeah i put it in and it is right

[fathom arrow]

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[rare zenith]

How to find the gradient of $\frac{z}{\sqrt{x^2+y^2+z^2}}$

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Please don't occupy multiple help channels.

[glossy valve]

todadqa

[rare zenith]

I can see that the gradient of $\abs{x}^2 is (2x, 2y, 2z)$

[glossy valve]

todadqa

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[spiral bronze]

how can i integrate this to get the answer

[spiral bronze]

i get the first 1/5v^5 but dont know what to do to get the rest

[torn jolt]

convert the q into indices format ; do each of them seperately

so for -3/2 sqrt(v^5) ==

-3/2 v^{5/2}

[spiral bronze]

ive never had to use indeces format and i have no clue how to apply it to this im bad with fractions

[torn jolt]

.

[spiral bronze]

right

[torn jolt]

now apply the rule that u know

ignore -3/2 for now

[spiral bronze]

same for 6 v^{1/8}

[torn jolt]

suppose its just v^{5/2} ; integrate it and then multyiply the ans by -3/2 after all done and then simplify

yes

[spiral bronze]

hold on where did the /2 come from here, is it just assumed because its sqrt its ^1/2 so place the top number over 2

forget that question i got that

[torn jolt]

k

[spiral bronze]

so now ive got

[torn jolt]

some weird formatting at the end i sppose

[spiral bronze]

im supposed to have gotten very similar to this format though

whats this

[torn jolt]

the final answer

[spiral bronze]

my book and an integral calculator both get different answers than yours for some reason

[torn jolt]

i doubt

show ur book ans

[spiral bronze]

i have shown it

[torn jolt]

this is the book answer??

bru

[spiral bronze]

correct

[torn jolt]

i trust tht book ans

but uhh

thats weird

yes

book is correct @spiral bronze

so did u get it?

[spiral bronze]

nope hes pulling the denominators out of thin air

[torn jolt]

no

book myt be rong

wat is the answer that u r getting

is more important

[spiral bronze]

I get everything he does until the last fraction

[torn jolt]

YE

the last fract of book is rong

this is wat i get wen i do nd ye

16/3 x^9/8 shud be right

book last fract is rong

they prob did a misprint etc

.

@spiral bronze

[spiral bronze]

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[torn jolt]

I am not sure how to solve it with this formula

[simple totem]

the limit they gave you gives you the slope of the tangent at x=a

so if you want the slope of the tangent at x=2, you'd want to find the limit $\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$

[glossy valve]

maximo

[torn jolt]

okay and I can do that by inputting a very small number for h?

[simple totem]

i mean kind of but that's not in the spirit of the question

[torn jolt]

how do I solve it that way then?

[simple totem]

compute f(2+h) and f(2), and do some algebra until you can plug in h = 0 and get a number

have you done limits before?

[torn jolt]

I tried and couldnt get it

let me try again

[torn jolt]

This look better?

[simple totem]

looks good

[torn jolt]

perfect thanks

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[cursive badger]

hi guys quick question

what happens with the 2

after sin

do i times the bracket

or flip it to other side

[forest widget]

not a fan of the notation at all, but it's equivalent to

$2 \sin (2(\frac{x}{4}-\frac{\pi}{3}))$

[glossy valve]

IV

[cursive badger]

so times the bracket

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[cursive badger]

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[tired perch]

Hey guys, could someone please try and explain the definition of a direct sum of vector spaces ?

[tired perch]

Especially in a dimension > 2

[fast peak]

well for starters, what is the definition and what do you not get about it

[tired perch]

I can seem to find a convenient definition in english 🙁

E1,…Ep are "under" vector spaces of E

[fast peak]

"subspaces"

[tired perch]

Subspaces thank you

The regular sum is the first equation

The direct sum is the second one

I’m having trouble visualising a direct sum in a vector space

[fast peak]

well can you visualise the regular sum?

[tired perch]

For example these 3 vectors aren’t a direct sum yet they fill out the whole space right ? I’m so confused

[tired perch]

Good point, i’m not too sure to be honest

[fast peak]

if anything, the span of those vectors is the whole space

[tired perch]

I guess you just sum up each vector and end up with a point, a line or a apace ?

[fast peak]

not the vectors themselves

but we are talking about adding subspaces, not adding vectors

[tired perch]

Yes that’s what i meant mb, the direct sum is looking at the sum of the span of each vector

But the definition is for all vectors of the subspaces

[fast peak]

lets try to add two lines "visually" and then figure out how this corresponds to the addition of certain vectors in those subspaces given by the lines

[tired perch]

In this particular combination of vectors, the definition says that the 3 subspaces engendered by these 3 vectors is not a direct sum right ?

[fast peak]

you can imagine taking one of the lines and sliding it along the second one

https://www.desmos.com/calculator/k8bqqxnijy

something like this

[tired perch]

The sum would give us 1 unique line right ?

Which corresponds to the sum of every combination of vectors on each line yes ?

[fast peak]

every vector you can reach by sliding along is in the sum of those two lines

https://i.imgur.com/BvRtbq1.png

[tired perch]

I follow 👍

[fast peak]

clearly v and w are linearly independent so if you take all possible combinations of them you reach the whole space

which you can imagine as sliding the blue line until it intersects with the orange one at a_1*w and then going along the blue line to a_2*v

[tired perch]

Yes 👍

[fast peak]

and we see that just with the two lines we already reach the whole space, we don't need a third one

https://i.imgur.com/tecqLZu.png

we could imagine taking a third line and sliding it along the blue one while the blue one slides along the orange one

[tired perch]

The bottom blue line right ?

[fast peak]

https://www.desmos.com/calculator/fvmfx0biqd

something like this

although hmm no that isn't that nice

urgh I don't really wanna think about the relevant functions I need

[tired perch]

Yeah never mind i’m out of time need to get back at it

[fast peak]

ok

[tired perch]

Thx for trying to help bro, appreciate you @fast peak ❤️

[fast peak]

anyway the point is, direct sums are just those where we don't have "unnecessary lines"

like the purple one here

[tired perch]

Oh shit

Okkkk i think i get it haha

[fast peak]

they are still like the other sums, just "more efficient"

and if you wanna think about it in terms of linear independence, the basis vectors of the relevant subspaces are all linearly independent

which gets you an obvious choice for a basis for the sum

[tired perch]

That’s what was fucking with me

I was wondering if they had to be linearly independent

[fast peak]

but here if we wanted to build a basis out of the basis vectors for the orange, blue and purple line, we know that we need two of them but there is no obvious choice which two of the three we should choose. while before when we just had the two lines, we just take the two basis vectors and are done

yes try proving that this is the case from the definition

it follows pretty fast

[tired perch]

What do you mean ? You just take all the basis vectors and create a basis vector for the sum right ?

[fast peak]

yes you combine all the bases of the subspaces and get a basis for the sum

while with a regular sum you can't do that, you might end up with too many vectors

[tired perch]

Gotcha makes way more sense

Thx bro love you

❤️

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[stray cloak]

Hello, I've been stuck in this question: $\phi: G \rightarrow G$ bijective, $\phi(ab)=\phi(a)b, \forall a, b \in G$. Must $\phi$ be an isomorphism?

Well, it isn't because then it necessitates that $\phi(b) = b$ which isn't always the case. But i can't really seem to give a convincing proof for it.

[glossy valve]

soda

[full forum]

@stray cloak Has your question been resolved?

[stable plover]

what is \bar{G}?

[stray cloak]

Oh sorry. They’re both groups

[stable plover]

it's just an arbitrary group?

well it has to contain G

[stray cloak]

yep they're just arbitrary groups

wait no, I was wrong

there's no G bar

rewrite: $\phi: G \rightarrow G$ bijective, $\phi(ab)=\phi(a)b, \forall a, b \in G$. Must $\phi$ be an isomorphism?

[glossy valve]

soda

[stable plover]

you just need to check that $\phi$ is a homomorphism

you're given the property that $\phi(ab) = \phi(a)b$ for all $a$ and $b$ in $G$, so you can deduce (as you have done) that $\phi(b) = b$ for all $b$ in $G$

then the homomorphism property follows by combining the two equations

[glossy valve]

tushar

[stray cloak]

I see, so since $\phi = \phi^{-1}$, we got 2-way homomorphism = isomorphism

[glossy valve]

soda

[stray cloak]

actually i'm not sure about $\phi = \phi^{-1}$...

[glossy valve]

soda

[stable plover]

bijection + homomorphism = isomorphism

it is the case that the inverse of a bijection is a bijection, the inverse of a homomorphism a homomorphism, the inverse of an isomorphism an isomorphism, but i don't think that's related to this problem

you're given that it's a bijection

and bijection <=> invertible

but you know nothing about the behavior of the inverse function

[stable plover]

homomorphism from G to G' and homomorphism from G' to G does not imply isomorphic

not sure if that's what you're saying

[stray cloak]

yea, that's what i was wondering about

thank you!

[stable plover]

e.g. you can have trivial homomorphisms between any two groups

[stray cloak]

i'll review the chapter on homomorphism now. thank u!

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[coral rapids]

I'm tryna solve for x

What should I do to end up with the circled equation

[sharp vine]

rewrite it as quadratic equation in terms of x

[trail oasis]

Yes developp the square

[coral rapids]

Is that the only way

Sounds like a lot of work g

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[sleek brook]

is this correct?

[sharp flame]

Yup

[torn jolt]

Yes

[sleek brook]

and this?

[sharp flame]

Yes

[sleek brook]

so both answers are correct?

[sharp flame]

There's also a third correct answer 😂

[karmic elbow]

they are equivalent

[sharp flame]

But yes both are correct

[sleek brook]

oh alright

Thanks

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[summer ibex]

how do i solve this

its a bonus on my old test im reviewing it

and just one thing to is R(x+2)
is R like a polynomial or a varialble

[sharp flame]

If R(0) = 0 then there is no constant term

So c = 0

[summer ibex]

yea i figured that

but than i got R =0

but than that means R(x) = 2x

[sharp flame]

The roots of the equation are then -b/a and 0

[summer ibex]

but in question a it says prove R(2) = 0 and with my anwser it will be 4

?

wait a=0 too right

[sharp flame]

If a = 0 it won't be a quadratic

And then 0 would be the only solution

[summer ibex]

cause we dont have a x² just a x

[sharp flame]

But you know that 2 is also a solution

[summer ibex]

idk what a quadratic is

[sharp flame]

You can just construct the equation I mean

[summer ibex]

i know like the ax³ + bx² +cx +d =0 thing on the calc only

[sharp flame]

wut

What grade math is this

[summer ibex]

10

i know how it looks but i mean we didnt take it yet

[sharp flame]

Uh

A quadratic is any degree 2 equation

ax^2 + bx + c = 0 is the general form

[summer ibex]

im not usa so its not like that man

[summer ibex]

yea

but so is R(x+2) a polynomial

[sharp flame]

Yeah it is

[summer ibex]

ohh

[sharp flame]

Put x+ 2 in place of x

For R(x)

[summer ibex]

okay

what grade are you in btw

[sharp flame]

11

[summer ibex]

nice

wait

doesnt that create an infinate pardox.

like you can keep replacing R(x+2)

in R(x)

i think i almost got it

[sharp flame]

Nope

It simply means that the polynomial evaluated at x

Gives the same value if evaluated at x + 2 and you add 2x

[summer ibex]

i got 2a=-b

and c=0

[sharp flame]

Sorry I don't have paper on me right now

[summer ibex]

its fine

[sharp flame]

How did you get that relation between a and b

[summer ibex]

im doing it

ok so

wait

i did something wrong one sec

i forgot the +2x

doesnt matter its still correct

okay so

i did the R(x)=R(x+2)+2x

than i made R(0)=0

[stable plover]

what are you working on? are you trying to show that R(2) = 0?

[summer ibex]

so therefore in the first one
R(x)=ax²+bx+c
R(0)=0+0+c

so c=0

[summer ibex]

yea im tryna solve the bonus

[stable plover]

substitute x = 0 into this equation

[summer ibex]

than the new R(x) we do a bunch of things and get

[stable plover]

that'll give you R(2) = 0

[summer ibex]

wait ill just get it for u

[sharp flame]

ax^2 + bx = a(x + 2)^2 + b(x + 2) + 2x

[summer ibex]

what

[sharp flame]

You did this right?

[summer ibex]

wait but its inside the paranthesis

[stable plover]

you can solve a) by simply substituting x = 2 and x = -2 into R(x) = R(x+2) + 2x

[karmic elbow]

I suggest you do this question in order of a) then b)

[summer ibex]

wait so R(x+2) is not a polynomial

[sharp flame]

It is

[karmic elbow]

R(x+2) is a polynomial

[summer ibex]

ok

[sharp flame]

What you did is right

[stable plover]

this

part a) will help you solve part b)

[summer ibex]

ok than how can u make it R(0)=R(0+2)+2x0

u cant go inside the polynomial right

[stable plover]

you're done

[summer ibex]

ok

[karmic elbow]

^

[stable plover]

R(0) = 0

[summer ibex]

?

[stable plover]

so simplify the terms on each side

[summer ibex]

yes

[stable plover]

R(0) = R(2)

[summer ibex]

^

[stable plover]

and you're given R(0) = 0

[stable plover]

i don't follow what this means, but you're simply asked to show R(2) = 0

[summer ibex]

ok so what u are saying works but what i mean is
if P(x)=P(x+2)+whatever
if you subsitute x for lets say 3
when u do P(3)=P(x+2)<-- this paranthesis i should not be able to go in and say x=3

[karmic elbow]

Why wouldn't you be able to

P is a function. P(x) is a function of x. You can manipulate the parameters however you'd like

[summer ibex]

ok i think i get it

i found out 2a=-b

but how can i find a

[karmic elbow]

Have you found the answers to a) yet?

[summer ibex]

yes

like he said i just do R(2) and R(-2) and ill get the anwsers

[karmic elbow]

I feel like Neon might've misled you a bit on this specific question. There's no need to find the vertex of the parabola, there's a much cleaner solution

[summer ibex]

?

[karmic elbow]

Given that you know

1. c = 0
2. When x is 0, R(0) = 0
3. When x is 2, R(2) = 0
4. When x is -2, R(-2) = -4

You can solve this through factored form algebra

[summer ibex]

i got it i think

YES

also whats a factored form stuff

[karmic elbow]

Standard form of a quadratic equation can be translated into a factored form, if you know the roots.

Roots of a polynomial equation is basically which values of x = 0 for R(x).
We know that R(0) = 0, and R(2) = 0, meaning that when x - 2 = 0, and x = 0

You can put this in the form of a*(x-0)(x-2) = 0, for whatever the value of 'a' is.
This is simplified to ax(x-2) = 0. We can solve for a if we have a point besides the roots, such as R(-2) = -4
Plug this in, a(-2)(-2 - 2) = -4, and solve for a. You then have a. If you plug the value of a back into the factored form, you can multiply out the brackets to get the standard form ax^2 + bx + c

[summer ibex]

i can start my workout now

[summer ibex]

bro are u albert eistein number 2 holy shit thats so complex

man i thought i was smart XD

all i did was that R(2) and R(0)

[karmic elbow]

It's probably simpler and more intuitive to solve this way in the long term with future equations, rather than trying to find the vertex and convert from vertex form to standard

[summer ibex]

R(0) told me 2a=-b

and R(2) told me 16a+3b=-4

but i dont know the quadratic thing btw

i use mr calculator

[karmic elbow]

Ah, system of equations, that also works given the formula. But will probably be slower in the long term with larger numbers

[summer ibex]

u know that 0 divided by zero is x

or anything

i did the equation of the calculation of the quadratic equation on my own

the triange of truth

[stable plover]

system of equations was the solution i saw

R(0) = 0 => c = 0
R(2) = 0 => 4a + 2b = 0
R(-2) = -4 => 4a - 2b = -4

then pretty simply elimination

[karmic elbow]

both are intuitive in this situation

[summer ibex]

bro what im getting b=1 when i do R(-2)

[stable plover]

correct

[stable plover]

this system yields a = -1/2, b = 1

[summer ibex]

wait

oh right i made a tiny calculation mistake and made it
16a+3b=-4
instead of
16a+4b=-4

one last thing in question 1 the teacher said its impossible but like if X=0 and b=-1 shouldnt it be possible

[stable plover]

what is impossible?

part 1?

[summer ibex]

yes

[stable plover]

a polynomial being identical to 0 means that the polynomial is equal to 0 for all inputs x

[summer ibex]

huh

doesn't it mean equal 0

[stable plover]

yes

for all x

[summer ibex]

oh so x cant be 0

let me tell u what she told us was the reason

[stable plover]

f(0) = b + 1, so it's clearly not identically zero whenever b is not -1

if b = -1, then f(1) = a + 3, which is not zero unless a = -3

[stable plover]

exactly

that's the intuitive explanation

mine is analytical

polynomial being identically zero means all the coefficients for each power of x are zero

this is equivalent to being equal to 0 for all inputs x

[summer ibex]

ohhhh

i get it

not just x=0

x = anything

[stable plover]

yes

[summer ibex]

cuz than it would be p(0)=0

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[ember tartan]

So i'm not getting that basis for row space at all and i have no clue what i'm doing wrong

basis should be 1 -1 3 right???

[forest widget]

if you look closely the answer key is very close to what you have, just that r2 is added to r1 again

i guess the text wanted rref

[ember tartan]

i see now 😭

thank you

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[tropic steppe]

I'm doing the euclidean algorithm

[tropic steppe]

gcd(5,21)=1, they are asking me to find the s and t, of s5+t21=1 by doing the extended euclidean algorithm

i did so, i even used the online calculator, but when plotting in the numbers, supposed s=1, t=-4

i get this:

that's weird, it should be 1=1 right?

[dusky island]

switch t and s

[tropic steppe]

okay

ayo tf

thanks for the help, i don't know how i didn't see that.

how can that be?

did i do it oppositely?

because there it says s is 1, and t is -4, how can it be oppositely

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[vagrant linden]

hey why isnt this convergent? doesnt diverge/converge test if the area of a curve is infinite or finite?
wouldnt you say -cos(infinity) if finite because it cancels its own area every time it goes below the axis, and vice versa?

[glass crystal]

its undefined

it approaches no value

since it oscillates

[vagrant linden]

what if it was about the x axis by like 5

would it still be undefined

[glass crystal]

wdym?

like cos(x)+5

[vagrant linden]

5 units

ye

[glass crystal]

yeah its still undefined

it has no limit

[vagrant linden]

so it just has to approach a value

[glass crystal]

when x goes to infinity

[vagrant linden]

if it doesnt its undefined

ok that makes sense

[glass crystal]

thats the defintionof a limit

[vagrant linden]

ohhhh

that makes sense, i thought we were just measuring area

[glass crystal]

well idk

but cos(inf)

seems like a limit

or maybe you were doing

integral of sin(x)

from 0 to infinity

[vagrant linden]

that was the og prob

[glass crystal]

or something like this

yeah it has no value

its not defined

[vagrant linden]

ok cool, that makes a lot of sense if we are measuring limits

ty

[glass crystal]

well

when you have a infinite symbol

in a integral

that means we are taking a limit

the limit of the integral from a to y of f(x) dx

when y goes to inffinity

[vagrant linden]

we arent measuring area under the curve?

[glass crystal]

well an integral is an area under a curve

[vagrant linden]

ye

[glass crystal]

but we are taking the limit of the ares

as you consider the function further and further

[vagrant linden]

why

its an integral not a limit

[glass crystal]

because thats how we deal with infinity

when you have

an infinity symbol

theres a limit somewhere

in this case

its a limit of integrals

but still a limit

[vagrant linden]

but why do we shift from an itegral to a limit when we use infinity

[glass crystal]

we dont shift

thats the definition

integral from a to infinity

by definition

is the lim of integral from a to b

when b goes to infinity

[vagrant linden]

oh i remember

yeah that tracks

[glass crystal]

so here

if you calculate the integral of sin

from 0 to y

you get

cos(y)-1

and this has no limit

when y goes to infinity

so your integral is undefined

[vagrant linden]

alright that makes sense because really we are measuing the limit

dope

is there anything else i should know?

[glass crystal]

idk

about this not necessarely

[vagrant linden]

life advice

whats the most important minecraft tip

etc

[glass crystal]

dont write the infinity symbol

if you are not familiar with what you are doing

unless its lim( ...) =inf

like cos(inf) simply does not make sense

its like dividing by 0

[vagrant linden]

ok, i will try to apply that to every area of my life as that sounds like life advice

[glass crystal]

be on point with your water bucket tricks

[vagrant linden]

appriciate the wise words they will not go unotes

^

i will go waterbucket into infinity and get 0/0 in minecraft and life

appriciate the advice on the real tho, thanks for explaining this

have a good one lol

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[torn jolt]

I think the topic of this is " estamation of parameters"?

Can someone help me this

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[torn jolt]

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

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[torn jolt]

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[paper kelp]

is there a short way to solve this question or do i have to manually guess

[smoky wing]

m^2 + 2017 = n^2

[hollow herald]

and then write: (n^2 - m^2) = 2017

Factorize the LHS and break RHS to factors to compare

[woven snow]

[hollow herald]

I knew 2017 was prime =_=

[woven snow]

i didn’t

[paper kelp]

I have to manually guess then?

[woven snow]

wdym?

[hollow herald]

(⁠╯⁠°⁠□⁠°⁠）⁠╯⁠︵⁠ ⁠┻⁠━⁠┻

[paper kelp]

like guess values of m

[hollow herald]

n^2 - m^2 = 2017

[torn jolt]

When you just like, ignore what they said

Completely so

[paper kelp]

ye but 2017 cannot factor

[hollow herald]

(n+m)(n-m) = 2017 • 1

n + m + n - m = 2017 + 1

[woven snow]

•

[paper kelp]

2n=2018

[woven snow]

how do u get such symbols

[smoky wing]

•

It’s default if you’re on iPhone

[hollow herald]

next you can assign n + m = 2017 or 1 whichever

but m is positive so

:/

[paper kelp]

so n = 0 and m =1? as n+m is 1 you say?

[woven snow]

hm?

[paper kelp]

but it doesn't work no?

[woven snow]

it is one of case

which is not satisfied here

[hollow herald]

here we FIXED the only possible value for n don't you think?

[woven snow]

in naturals

[paper kelp]

so n = 1009

[woven snow]

yes

[paper kelp]

m^2+2017 = 1009^2

[woven snow]

,calc 2018/2

[glossy valve]

Result:

1009

[woven snow]

yes

[hollow herald]

now my guy will solve 1009^2 - 2017 ._.

and take sqrt

[woven snow]

or well n-m=1

[paper kelp]

so that's the only possible thing left so answer is 1

[hollow herald]

yessir but did you get what we did?

[hollow herald]

[hollow herald]

[hollow herald]

[paper kelp]

you put m^2+2017 as n^2 then you shift n^2 to the left and 2017 to the right so it becomes n^2-m^2=2017 then you did n+m+n-m = 2018 and then n is 1009 and that is the only possible value so then as there is only one possible value of n, m can only be one thing as well

and m is positive so the answer can't be 2

[hollow herald]

(⁠ヘ⁠･⁠_⁠･⁠)⁠ヘ⁠┳⁠━⁠┳ just to be sure.. you understood why:

1. n + m + n - m = 2018?

2. why n = 1009 => m has only 2 possible values?

right?

[paper kelp]

yes

[hollow herald]

perfect.. Good luck

[paper kelp]

m has 2 possible values because m could be a negative number or positive number and question says m has to be positive so rule out negative so answer is one

[paper kelp]

thank you

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[paper kelp]

I have no clue how to solve this

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[crude yarrow]

i tried 2 different ways to approximate this sum but the precision was not good enough to be able to mark the correct ans (which, testing with wolfram, is D)

the approximations were decent but the problem asks for even more precision 😕

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[left crown]

Help

How do i solve this

[stable plover]

$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4}\sqrt{2}$

[glossy valve]

tushar

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@left crown Has your question been resolved?

[wanton osprey]

Still need help with this one?

@left crown

You have 5sqrt2 minus 3sqrt8

You need to find the highest perfect square that is a factor of 8

This would be 4

What times 4 gives you 8? 2

[left crown]

Is my answer wrong?

It's "Adding and Subtracting Raducal Expressions"

[wanton osprey]

So we now write it as 5sqrt2 minus 3sqrt4sqrt2

Yes I know

Now square root of 4 can be simplified to 2

5sqrt2 - 3x2sqrt 2

[wanton osprey]

5sqrt2-6sqrt2

Yes

[left crown]

Like that bro?

[wanton osprey]

U did it right

[left crown]

Ohh okk so i misplaced 1 number

Hahaha thanks

[wanton osprey]

Mhm XD

[left crown]

Thank you bro

How about this one

[wanton osprey]

Perfect

That’s the answer

[left crown]

I followed this example

[left crown]

Thanks

For the help brooo

Now i can go to sleep

Thanks

[wanton osprey]

LOL yes u need rest

Any time, no worries

[left crown]

Ye yeah byeee

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[neon mulch]

how is E(Y)=5?

[neon mulch]

I found E(X)=13/6 and squared that to find E(Y) but it's wrong

[onyx glen]

yeah because $E(X^2) \neq [E(X)]^2$ lol

[glossy valve]

Ann

[neon mulch]

what's the correct formula?

[mighty ridge]

there's no direct formula, think about the law of Y

and calc the expected value of Y