help-23

[thin bridge]

its basically reading what you see

lets take something simpler

3 raised to** what power** gives 3^2

[split anvil]

do I have to do something with the ln? cause I don't really understand logrithms very well at all

[thin bridge]

what is the power of 3 in 3^2

no

you're still overthinking

[split anvil]

9?

[thin bridge]

you don't need to understand logs for this

no

not what i'm asking

i don't care about the simplified numerical value of 3^2

[split anvil]

forgive me, I am a baby at math

3*3?

[thin bridge]

not what i'm asking

$3^{\text{what value}} = 3^2$

[flat frigate]

ℝαμΩℕωⅤ

[split anvil]

2?

[thin bridge]

yes

i'm literally just asking for the expression of the power

its the same principle for what you have

your h(x) is
e^[whatever this is]

[split anvil]

Ok so how does that apply then? is there some way to simplify ln(4)cos(x^6)?

[thin bridge]

you could simplify the ln(4) a bit if you wanted

but that's what i was asking for

that's your h(x)

[split anvil]

oh right yeah ok. I didnt think the ln(4) was a part of it though

[thin bridge]

well it is

ln(4) is part of the power

[split anvil]

so I have this then?

[thin bridge]

yes

[split anvil]

Ok so now I have to figure out the derivative of that

not sure what to do with the ln(4)... I assume I have to convert ln(4) into something and cos(x^6) into -sin(x^6) then finally find the derivative of x^6 to plug in

[thin bridge]

constant multiple rule

ln(4) is just a number

[split anvil]

oh its a constant? ok then I just drop it right. its 0

so just this?

[thin bridge]

no

you don't erase because its a constant

derviative of 5x isn't 1

$\dv{x}(k\cdot p(x)) = k\dv{x}p(x)$

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

you can factor out the constant, and take the derivative of the main part for less tedious work

[split anvil]

not sure I understand. What part of this is wrong?

[thin bridge]

missing ()

you want to product of ln(4) and -sin(x^6)
what you have there represents a difference

[split anvil]

of right, because converting cos to -sin... ummm ok

[thin bridge]

$h'(x) = \ln(4) \cdot (-\sin(s(x))\cdot s'(x)$

[flat frigate]

ℝαμΩℕωⅤ

[split anvil]

ok right, in my head that's what it was but I needed to write it that way for sure

ok so finally I just get the derivative for x^6 which is easy enough, 6x^5

[thin bridge]

yes

[split anvil]

sorry, had a text. Ok so then I have to basically compact it all back down again right?

[thin bridge]

put all your components together and simplify what you can

[split anvil]

ok so here is what I got, but I am sure I will need to simplify something

[thin bridge]

your forgot the power of 4 in the first part

e^ln(4) in the second part can be simplified back down to 4

and then bring all the constants to the front

[split anvil]

oh right ok... hmm lemme give it a try

Umm ok I think this is close. I really struggle with simplifying things correctly

[thin bridge]

ln(4) is a constant, you should bring that to the front

and you can also apply power law for log to simplfy that a little

[split anvil]

ok one sec, I had if different let me see if I can get it

I can simplify the log? umm hmmm how is that?

So WolframAlpha gave this, so I think I am getting close

I'm not sure how they got -15 at the beginning though and not sure how to get
4^cos(x^6) to become 2^2cos(x^6)+1

omg i submitted what I had and it worked!! I have been stuck on this like all freaking day! Thank you!!

[safe radish]

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[distant hinge]

someone tell me how is this wrong i doubled checked and tried it in another form and still wrong

[distant hinge]

calc 1^

[gusty trench]

firstly you're missing a dt

if you have a function y(t), the differential is given by dy = y'(t) dt

[distant hinge]

ohh ok

thx

[gusty trench]

i think your derivative looks correct

[distant hinge]

i frogot abt it

[gusty trench]

so just adding the dt should do it

[distant hinge]

yeah just did and its correct

thx

[safe radish]

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[halcyon turret]

Hey calc 2 is really kicking my butt over the summer and I have this test due in 30 minutes so if anyone wants to help a bit it would be greatly appreciated

[gusty trench]

it unfortunately is against #rules to get help for a quiz/test

[halcyon turret]

can we just pretend its homework for a bit

[gusty trench]

💀 i'm not willing to but if you can convince some other helper, go for it lol

[halcyon turret]

anyone please

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[woven axle]

What am I doing wrong?
400=-10x&2+180x-330
-10x^2+180x-730
x=180 root 32,400-4(10)(730)/20
x=180 Root 32,400-29,200/20
x=180 Root 3200/20
x=180+=56.57/20
x=9+-2.83
x=11.83 and 6.17

[dusty fiber]

you can just calculate the soccer ball's profit

x is given for it

[woven axle]

7.5 right

[dusty fiber]

yes

then subtract that from 400 to see how much profit the footballs need to make

[woven axle]

mannn

Alright

ill try and get back to you

alright so football needs to make 167.5 but would I plug in that vlue for its x too right

[dusty fiber]

no

167.5 would be the profit

youre trying to get x

[woven axle]

ohhh my lord

My fault

so its y

[dusty fiber]

yes

[woven axle]

im doing smt wrong

can I see what you did

[dusty fiber]

,calc (-6)7.5^2+1007.5-180

[flat frigate]

Result:

232.5

[dusty fiber]

i got 232.5

i definitely for sure did this earlier

did not just do this right now

[woven axle]

I got that earlier too for the soccer ball

[dusty fiber]

,calc 400-232.5

[flat frigate]

Result:

167.5

[woven axle]

yep

But I would like to see your x=-b+-Root b^2-4ac/2a for the football

[dusty fiber]

hang on let me write this down one sec

i got an answer but it seems wrong

but i think its actually right

what answer did you get?

[woven axle]

-22.5+-20.67

I think I'm wrong as well

idk it doesn't feel right

[dusty fiber]

that cant be right

[woven axle]

wait I have a question

[dusty fiber]

that means the price for the football would have to be negative

[woven axle]

since its 4ac for that one part right

and a was -4

do I put in -4 or just 4

I think I'm getting a brainfart rn'

[dusty fiber]

-4

[woven axle]

whatchu get

[dusty fiber]

and c should come out negative

[woven axle]

uhhh

where'd you go

https://tenor.com/view/huh-verne-over-the-hedge-turtle-disbelief-gif-24744867

[dusty fiber]

how did you get that answer?

what work did you do

[woven axle]

oh mannn

Give me like 2 min to write this out

[dusty fiber]

you can write it on a piece of paper and take a picture

or on a drawing program and take a screenshot

[woven axle]

x=-180Root 32,400 (16)(-317.5)/-8
x=-180Root 32,400-5080/-8
-180Root27,320/-8
x=-180+-165.287628/-8
x=22.5+-20.6609535

[dusty fiber]

what are the commas

as in the number 32400?

[woven axle]

yep

[dusty fiber]

b² is 6400..

[woven axle]

wait

[dusty fiber]

use the equation for the footballs

not the two added together

[woven axle]

brooo

I was using 180

I read 180 not +80

oh nahhhhhhhhhhhhh

ill get back to you...

10+-4.54147553

[dusty fiber]

thats closer to what i got

how did you get that root on the right

[woven axle]

wat

I didn't even give you my steps whatchu mean...

oh wait

I dont understand what you mean

[dusty fiber]

i dont think thats right

what did you do to get there?

[woven axle]

mannnnn

alright Ill write it here

x=-80Root 6400-5080/-8
x=-80Root1320/-8
x=-80+-36.3318642/-8
10+-4.55

so the answers are 5.5 and 14.5

[dusty fiber]

that looks right to me actually

i found an error in my work from you showing me yours

both of those answers are correct

they should be at least

[woven axle]

damn man you had me panicking

I thought I was gonna have to write another

Whatchu get though

[dusty fiber]

i got the wrong answer bc of a mistake i made

im checking your answers right now

yes they are right

[woven axle]

https://tenor.com/view/spongebob-squarepants-horray-finally-about-time-gif-5909754397376002885

We spent an hour man

Alright have a nice day man

.close

[safe radish]

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[woven axle]

.reopen

[safe radish]

✅

[woven axle]

Wait actually what do you think the exact words I need are

This is an open ended question but I need key words and idk what I am missing

I need you back crepas

[safe radish]

@woven axle Has your question been resolved?

[woven axle]

nvm I got it

.close

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[lean otter]

#help-17 message

The answer given is true

@long oracle @tardy mango

$x_i = 0$ doesn't act as a counter example

[flat frigate]

username30001

[lean otter]

The condition is on $a_i$

[flat frigate]

username30001

[lean otter]

To find a counter example we have to find a $a_i$ which satisfy the given condition but still not bounded.

[flat frigate]

username30001

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@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[crisp mural]

ive been trying to understand what axiom of choice intuitively means, not just the 'sets' definition of it because well then it's hard to imagine what's the controversy about it

but i'm not really qualified enough to actually understand the 'why & how' of this depth yet

so could someone give me an analogy, or a metaphor as to what and why makes this axiom so interesting?

one thing i found everywhere is that it's needed when there are infinitely many sets; each of which have non distinguishable elements and they're all non empty; so apparently the axiom of choice gurantees a way you can still select one element from each set;
Why do we need the axiom of choice to select elements from an infinite array of identical elements? Can't we randomly select one from each?

[fathom adder]

The only thing i remember is that it could lead to error if badly used

The answer deserve a book ngl

[crisp mural]

well probably any axiom used badly would result in an error cuz well

[fathom adder]

Yeah but this one is a destroyer xd

[crisp mural]

xD i see

[crisp mural]

there's this explanation almost everywhere and no reason for it, like it doesn't really make any sense to me as to why you need an axiom for this

and they'd proceed to say that this axioms leads to other things like banach tarski and would mention a few implications too fancy for me to understand

instead of giving some clean and clear reason, so i don rlly know

[kind tinsel]

you can’t randomly select one from each because the description of such a choice would be too long for first-order logic to handle

[crisp mural]

how-

[kind tinsel]

like the axiom says “yes we can pick one at random”

like your intuition says the axiom of choice should be true

[crisp mural]

if you have an infinitely big bag of identical balls, each one has just as much probability of being here, or being there; why can you not select one randomly?

it's the same for finite bags as well, or atleast it should be

oh we don't need the axiom for selection of identical objects from a finite number of sets?

is that so?

[kind tinsel]

the technical description involves the fact that proofs have to be finite length, so how are you going to just tell me “pick one at random”? you have to actually describe a method for picking the ball using ONLY axioms that are in ZF, and saying “pick randomly” is equivalent to the axiom of choice

you don’t have an axiom of probability

[crisp mural]

ohhh i see

[kind tinsel]

but indeed it is VERY intuitive to say “pick at random”, which is why the majority of mathematicians don’t give that much thought to AC

[crisp mural]

so basically; we're dealing with an event such that no 'actions' are defined yet

and a unique action is the same as an axiom

so the 'action' of picking randomly, is a unique action that we refer to as the axiom of choice, atleast in this case?

[kind tinsel]

actions are usually either equivalent to theorems about existence of objects, or axioms about existence of objects

in this case, “picking something from each set” is precisely the same as the existence of a function from the index set to the union of those sets

[crisp mural]

indeed

[kind tinsel]

and if you can’t establish the existence of such a function, then the “action” isn’t valid. and there are “models” of set theory where the axiom of choice is false (that’s what it means for choice to be independent of ZF)

some sets don’t need the axiom of choice to have such functions by the way

for example you can always choose a basis of finite dimensional vector spaces

no matter what

also if you have a “chain” of sets A_0 \subset A_1 … \subset A_kappa for some ordinal kappa you can also have a choice function

and there are more interesting examples

but off the top of my head these are two that are easy to prove you don’t need AC for

[crisp mural]

I see, can I say that this scenario is different than the one I originally used above, since you don't need to define a choosing basis, as you can prove one already exists naturally?

one more thing, why or why not do we 'need' the AC for selection of identical objects from a finite number of non empty sets?

and when to know that AC is a must to be able to meaningfully sort things?

[kind tinsel]

that’s about how proofs work, mathematicians require a proof of a fact (like existence of a function) to be finite in length, there is no “do this process and this process and this process…” if the thing in the dots is infinite. finite dimensional vector spaces only use “…” because the thing in the dots is finite, and one could, given enough time, write down the entire process for choosing a basis for a 1000000000 dimensional vector space

but in the infinite case, the … will literally be an infinite proof if one tried to write it out in some cases

in the case of stratified sets, which is unique in the sense that it looks like you’re choosing infinite things, one may just say “for all i in the index set, let f(i)=a_0”

[crisp mural]

ohhhhh right this makes much more sense now

[kind tinsel]

and that is a valid finite length proof

using AC allows us to give finite length proofs to things that we believe we should do as “infinite mathematicians”

[crisp mural]

yesyes am somewhat able to wrap my head around it, thanks a lot! I tried watching so many youtube videos and articles to find one that could somehow help me line my intuition with it and i found none

[kind tinsel]

yeah youtube videos get this one very wrong unfortunately

probably better to open a proper set theory book

probably roitman

[crisp mural]

hmmm actually i would probably be unable to understand because of how frequently the notations and everything are used as an almost substitute to language in them

[kind tinsel]

yeah but you can get used to the language

it’s precise

and that’s what you want in math

[crisp mural]

well that's truee

yupp

[kind tinsel]

and you can always come back to the discord to ask clarifying stuff

like remember that you’re not alone

[crisp mural]

^-^

thanks a lot again! Have a good day <3

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[woven axle]

quick check
2-(3x)^2
2-(-21)^2
2-(441)
2+(-441)
-439?

[woven axle]

is that the correct way to do it

[icy lance]

seems alright

[woven axle]

Thanks man

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[edgy vine]

i need help with e)

[faint seal]

The question’s language is pretty self explanatory: show that phi=0 is a root of the equilibrium temperature equation, no matter what T is

[edgy vine]

yeah, how?

[faint seal]

Just input phi=0 into the equation

[edgy vine]

and then x(0)=0

[faint seal]

Yes

[edgy vine]

is that litterally all?

[faint seal]

Yes

[edgy vine]

cos i did do that just thought it was too easy

[faint seal]

You get 0=0, which is true for all values of T

Hence phi=0 is an equilibrium value for all T

[edgy vine]

huh, do i then have to explain why thats an equilibrium

[faint seal]

It is an equilibrium because it satisfies the equation

[edgy vine]

ok

thanks

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[glass field]

how.

[final halo]

change of base formula

you have the right hand side

[glass field]

ah

thanks

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[coral shoal]

If f:[0,4] --> R, f(x) = 4x-x^2 find the approximate value of L(σ,f) and U(σ,f) using four equal parts

did i put the numbers correctly?

i feel like i did a mistake since x belongs to [0,4]

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[warm bramble]

It's a logical based question that I'm not able to understand very well

[warm bramble]

<@&286206848099549185>

Here's the solution for this too just help me understand why all the remaining games were assumed to be draw

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[brazen sleet]

hello

[empty gyro]

Hi. Did you have a question?

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@brazen sleet Has your question been resolved?

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[craggy parcel]

need help on number 1

[limber shore]

use the gauss thing

just multiply by a number and plus/minus the other one

option 2: throw everything in the calculator if ur allowed

[craggy parcel]

then how do i find the dimension for that space

[limber shore]

no idea for the second question..

dont even understand what it is asking

💀

[craggy parcel]

bro the story of my life with this damn course

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@craggy parcel Has your question been resolved?

[proper bramble]

the dimension is how many free variables you have

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[exotic dome]

Ill help

[exotic dome]

nah i cant do that

[main mural]

wut?

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[grand quiver]

This is the question. Per the given information, I've established f(0)=10= a/(1+b) and f'(0)=2 but I'm not really sure where to go from there

[grand quiver]

and I'm not even sure I'm right with the givens

[short jackal]

You need to compute f'(t)

Then f'(0)=2

[short jackal]

Plug in t=0 into that for f'(0)

[grand quiver]

finding f'(t) would be through implicit differentiation, right?

[short jackal]

Chain rule

Do it however you can

[grand quiver]

okay

thanks

.close

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[short jackal]

Np

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[dusky geyser]

I;m sorry I already asked for help and closed the previous thread on this but I still can't seem to wrap my head around why I'm getting a different answer for 6 2/3 - 12 3/4 without converting to improper fractions.

[sonic sierra]

-7 11/12 is not same as -7 + 11/12

-7 11/12 = -(7 11/12) = -(7 + 11/12)

Which is -7 -11/12

[safe radish]

@dusky geyser Has your question been resolved?

[dusky geyser]

Oh I actually see it now

Thank you!

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[proven rose]

how did the 2 in the denominator in the 2nd ss get there?

[proven rose]

<@&286206848099549185>

[white umbra]

please wait 15 min before pinging helpers

[proven rose]

oh ok sorry

[white umbra]

just multiply the numerator and denominator both by 2 tho

to get from screenshot 1 to screenshot 2

[proven rose]

sory ive been up for a while i think im missing something, but if i multiply the whole thing by two, wouldnt the top have a 6x^2

uh i did not understand

[flat frigate]

faiyrose

[proven rose]

ah

oh i see now

tysm

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[summer comet]

Hello im having trouble understanding this can anyone help me?

[safe radish]

@summer comet Has your question been resolved?

[safe radish]

@summer comet Has your question been resolved?

[white umbra]

SInce this is a conceptual physics thing, I would try the physics server instead

discord.gg/physics

[summer comet]

Thanks my friend

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[open cradle]

Hello, mathematicians. i want to ask this

So, i know which ones are linear. (g) for example isnt linear and it is quadratic.

My question being, what expression is (f) called?

[quiet plume]

It's also quadratic. The degree of xy is 2, even though they aren't the same variable.

[quiet juniper]

not a mathematician so cant answer srry

[open cradle]

I've heard of the term two-variable linear expression. Does it apply in this scenario or not?

[quiet plume]

No. Two variable linear expressions look like $Ax + By$

[flat frigate]

Azyrashacorki

[quiet plume]

Essentially, you add the degrees of variables multiplied together in a given term. If for every term, the highest is 1, then it is linear.

[open cradle]

My last question, for (d), 4x + y - 8, is it a linear expression or not?

[quiet plume]

It is.

[open cradle]

Very well. Then it is solved. Thank you Mr. Azyrashacorki.

[quiet plume]

No worries 🙂

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[woven axle]

Can someone teach me?

[woven axle]

Is the rectangle referring to the green part only, I really dont know where to beig

begin

[pure reef]

I think so

seeing the diagram

and the answer

[woven axle]

Can you walk me through?

nvm

.close

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[lime dust]

What did you try?

[scenic holly]

Also I had another try before that one

[turbid adder]

Try just changing 1-cos2x into 2sin^2x at the beggining and go from there

[scenic holly]

I've done that

[scenic holly]

.

[turbid adder]

Yes but you also did a bunch of unneccessary steps on the wat

way*

Dont split it into different fractions

[lime dust]

It is a 2 step limit

[scenic holly]

Don't really get what you mean

[lime dust]

2sin^2x/(xsinx)

Sin^2x = (sinx)(sinx)

[scenic holly]

where would you get 2sin^2(x)?

[lime dust]

1-cos(2x)=2sin^2x

[scenic holly]

How?

[lime dust]

Do you know the identity of (1-cos(2x))/2?

[scenic holly]

No information about the identity on the internet

And I was not given of it, therefore, the exercise could be solved without the identity

I wanted to know what I did wrong

[lime dust]

The double angle

[scenic holly]

Because the actual answer is 2

[lime dust]

If you are not allowed to use identities, are you allowed use l’hopital?

[scenic holly]

Is that about derivatives?

[lime dust]

Yes

[scenic holly]

Not yet

[lime dust]

Ok one last question

Do you know what is cos(2x)?

[scenic holly]

Can't I solve the exercise with fractions?

[scenic holly]

cos^2(x) - sin^2(x)

[lime dust]

Ok and cos^2x?

Use pythagorean identity

[scenic holly]

1 - sin^2(x)

Oh boy

1 - 2sin^2(x)

[lime dust]

Exactly

Now u subtract 1 from both sides

Multiply by -1 both sides

[scenic holly]

That's pretty useful, thank you very much

[lime dust]

And divide by 2

Well no need to divide by 2

Because the original was 1-cos(2x)

[scenic holly]

But still, is there any other way to solve the example without using the identity

Because I was using "lim_(x->0) (1-cos(x) / x) = 0"

[lime dust]

You cant say that it is 0 because u dont know the limit of 1/sinx

From the left it tends to -inf and from the right to +inf

So you are not able to calculate this limit like that

[scenic holly]

And what if I do something like that?

[scenic holly]

That's technically correct

[lime dust]

In the second line u are changing the limit

U multiplied by sin^2x the numerator

[scenic holly]

My bad

[lime dust]

If the first step is wrong then the rest makes no sense

Cause u are analyzing something different

[safe radish]

@scenic holly Has your question been resolved?

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[steep magnet]

Why is it not just zero? Theres (x+3) in the numerator so no matter what it just goes to 0 right?

wait I just saw

.close

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[scenic holly]

What's the webpage?

[steep magnet]

Omptest.org. Its really good cause every question is repeatable, the numbers change automatically. But you have to pay 60 euros for it 🥲

[scenic holly]

It's a hole, isn't it (x != -3)?

[steep magnet]

Yeye just noticed haha

[scenic holly]

Oh boy, is it just for universities or?

[steep magnet]

Thats what its made for but anyone can use it

[safe radish]

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[humble lance]

Hi guys can someone explain how to find AB

[safe radish]

@humble lance Has your question been resolved?

[humble lance]

<@&286206848099549185>

[cold aurora]

Haven't you already found out AB? Do you not understand that, or is that someone else's solution?

[safe radish]

@humble lance Has your question been resolved?

[humble lance]

I did find out but I don’t understand how

Like i did backwards that’s why I can get the answer

<@&286206848099549185>

[obtuse jungle]

@humble lance ask from help🗣🔜✔🗣🗣https:/@@t.co/O0YLd549uJ

[humble lance]

.close

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[timid turtle]

You know those lines on race tracks to ensure everyone runs the same distance by making sure people start further ahead if their towards the outside? How do they get that line?

[scenic saddle]

measure distance from the centre of the track

for simplicity imagine running in a circle around a point you are really close to vs really far away

[timid turtle]

But how do they get the exact difference

Like the exact line

[scenic saddle]

for the circular track example: measure distance of the runner from the centre, figure out the circumference, and adjust the start of the track so everyone runs the same distance by finding the arc you need to "cut off"

[timid turtle]

Thanks 🙂

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[viscid grail]

Suppose Ax=b is a compatible linear system with n variables and x_0 from R^n is one solution. Then x∈R^n is a solution <=> x=x_0+v where v is a solution of the homogenous system Ax=0.

[viscid grail]

just a quick question

x_0 is a vector or a scalar?

[arctic raven]

it's a vector

it lives in R^n

[viscid grail]

and v is a vector or a scalar?

[arctic raven]

v is necessarily a vector by context

in saying Ax=b or Ax=0 where A is a matrix, it's inferred that x, b, and 0 are each vectors

v being a solution to a matrix equation like this also implies v is a vector; in this case it satisfies Av=0

[viscid grail]

alr

ty

.close

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[thorny mirage]

I need help

[thorny mirage]

I don’t know where to start

[arctic raven]

what information do you learn from the premises?

[eager dew]

okay, from this text you can dissect quite a lot

i am not familiar with the english terminology of things, so bear with me lol

[arctic raven]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[eager dew]

the text says that it is a parabola which is opened downwards

oh ... sorry.

[arctic raven]

no problem, just doing that preemptively lol

since you were about to mention terminology I thought it'd be relevant

[eager dew]

now i dont know how much i can help without accidentally blurting out too much lol

[arctic raven]

yeah lol

[eager dew]

i can visualise the text a little bit tho, right? would that count as just guiding someone onto the right path

[arctic raven]

I'm not an authority so you can do as you like honestly

[eager dew]

ok then lol

this is an example of a downwards opening parabola, try making a function for this

@thorny mirage

[thorny mirage]

-x^2

My bad I was afk

[eager dew]

no problems, but exactly

or in other words -1 * x^2,

[thorny mirage]

Yes

K has to be less than 0 tho

[eager dew]

it says more than

[thorny mirage]

So does that mean that the vertex is below 0

Mb I worded it badly

[eager dew]

it says that k is greater than 0, meaning the vertex also has an y value above 0

[thorny mirage]

Mb looked at it wrong

[eager dew]

can you read out the vertext x - value out of this question?

[thorny mirage]

0

[eager dew]

sorry, i meant from this problem

or in this case, the h value

[thorny mirage]

-6/2a

[eager dew]

but you dont have a, right? there is also another solution

[sonic sierra]

Hello

f(1)= f(5)

Try using this

[eager dew]

you know how a parabola is symmetrical in nature

[thorny mirage]

yeah

Damn what other way

Ohhh

[eager dew]

you seeing it?

[thorny mirage]

H+k/2

Is that what you want

[eager dew]

not really

[thorny mirage]

Actually no

[eager dew]

a vertex is always the 'mid' point of a parabola

[thorny mirage]

Do you like want the solutions /2?

[eager dew]

you have been given these two points,

and you need to find the middle between them,

how would you do that?

[thorny mirage]

A+b/2

[eager dew]

sorry,its just hard for me to try to help without blurting out the answer

[eager dew]

exactly

you have been given both a and b in the question

[thorny mirage]

No you’re good

[eager dew]

just as bear bear mentioned, it said f(1) = f(5),

[thorny mirage]

Oh 1 and 5

[eager dew]

exactly! :)

[thorny mirage]

6/2

[eager dew]

and since you know that the vertex is always in the middle of parabola, you know that the vertext has to be at exactly that x value

which as you have mentioned, is 6/2 or 3

im assuming that the constants a and b have to be whole numbers

you mentioned that a has to be negative, right? what is the smallest negative whole number you can think of?

[thorny mirage]

-1

[eager dew]

exactly, so now you have a = -1
your function becomes
f(x) = -1x² + 6x + b

your condition is that f(3) has to be above 0

or
f(3) > 0

Since you already have the x value, you can just put it in the new equation and solve for b

[thorny mirage]

Where did you get f(3) from

[eager dew]

thats where the vertex must be

[thorny mirage]

Oh shoot

[eager dew]

3 is the x value of the vertex

[eager dew]

because it is also the mid point

[thorny mirage]

Plug in 3 for X

[eager dew]

yes

[thorny mirage]

And then just solve for b

[eager dew]

exactly

try that and tell me how it goes

[thorny mirage]

I don’t have a paper with me so I’m doing mental math so Tia taking me awhile

F(3)=45+b

But then idk how youre supposed to solve that now

[eager dew]

ok, so our other condition was f(3) >0

[thorny mirage]

So are we able to plug in anything greater 45?

[eager dew]

yes, kinda

[thorny mirage]

Ok let’s do -44

[eager dew]

i have just realised you solved it wrong

[thorny mirage]

Ok

[eager dew]

try it again

[eager dew]

specifically this step right here

[thorny mirage]

-45+b

And then idrk where to do from there

[eager dew]

thats also not right...

the function is
f(x) = -1* x² + 6 * x + b

=-1* 3² + 6 * 3 +b
= -9  + 18 + b
= 9 + b

im just solving it right now for you because this isnt part of the original problem lol

[thorny mirage]

Oh shoot why did I multiply 6 by 9

[eager dew]

thats exactly what i thought you did lol

but anyways your condition is that f(3) (ie 9 + b) has to be bigger than 0

you can either use it using an inequation

or by solving for the border where it becomes negative/positive

[thorny mirage]

So 9+b>0

[eager dew]

yeah

that works!

what do you get for b then?

or what conditions do you get for b?

[thorny mirage]

B>-9

[eager dew]

exactly!

[eager dew]

what does the answer have to be then?

[thorny mirage]

-8

Solutions above -9

[eager dew]

that is right but i meant to the orignal question lol, what choice has to be right?

[eager dew]

. this one

[thorny mirage]

II only

Is that wrong or

[eager dew]

..

[thorny mirage]

Both

Both

[eager dew]

yeah lol
and remember, we were assuming that both a and b had to be whole numbers

if they werent only a would have been true

[thorny mirage]

Oh thank you bro

Life saver

[eager dew]

here is the function where b = -9 (doesnt fulfill the b>-9 condition which is why k is also 0 and not >0)

and here it is with b > -9, fulfills the condition k>0

[eager dew]

no problem lol

btw, a and b HAD to be whole numbers

[thorny mirage]

Is that what they mean by integers

[eager dew]

if a was anything other than -1 (the smallest integer) the condition f(1) = f(5) wouldnt have been fulfilled

[eager dew]

yeah

whole numbers are integers

[thorny mirage]

Why did it have to be the smallest integers available

[eager dew]

i mixed up whole numbers and integers throughtout the whole explanation

[thorny mirage]

Why did it have to be -1 again?

Sorry o just have to understand it

[eager dew]

here

we have said that the vertext has to be at the mid point between 1 and 5 right? which was 3

[thorny mirage]

Oh if you plug in -1

[eager dew]

and since -6/2a also gives us the vertix x value, both of them must be equal

[thorny mirage]

You get 3

[eager dew]

exactly! only -1 fulfills the equation

[thorny mirage]

Okay okay thank you

I usually want to like

Understand the thought process

So this helped me a lot!

Thank you

[eager dew]

no problem :)

hope i havent have complicated the explanation too much

[thorny mirage]

It’s in the SAT

so like solving this is going to take awhile

But with this

Cut down my time

[eager dew]

oh its in the sat? isnt that the college admission test type thing?

[thorny mirage]

Yeah

[eager dew]

oh that sounds fun. well, lots of luck to you! do your best on the sat

[thorny mirage]

Thank you!

[safe radish]

@thorny mirage Has your question been resolved?

[safe radish]

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[wispy cliff]

Hello - I'm trying to find the volume of these functions rotated about the X axis:

x=sqrt(y), x=y/4.

First, I convert them both to y= functions, giving me y=x^2 and y=4x. I then find the bounds of integration by setting these two equal to each other. I end up getting 0 and 4. I both graph theese functions on my calculator and create a table to see which is the "outer" and "inner". My outer is 4x, my inner is x^2, so my equation for the volume is:

[wispy cliff]

I've completed the problem but my teacher says my solution - 2048/15 * pi is incorrect.

Is that solution actually correct? If not, does anyone have any idea where I went wrong?

[lime dust]

U got negative volume?

[wispy cliff]

oh no that was just a hyphen

[lime dust]

Do u have the original problem?

[wispy cliff]

let me type it up

I'm using the disk method, the one with the

equation

[desert swallow]

I got the exact same answer

[wispy cliff]

Right? It's confusing.

[desert swallow]

Do you know what the right answer is?

[wispy cliff]

No, even according to mathway my answer's correct.

but mathway's really unreliable

[desert swallow]

But did mathway give an answer at least?

[wispy cliff]

Yeah, the same one we're getting.

[desert swallow]

Oh, I misread correct as incorrect

[safe radish]

@wispy cliff Has your question been resolved?

[safe radish]

@wispy cliff Has your question been resolved?

[safe radish]

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[halcyon light]

the 1st 2 inequalities I got

but

for the third one

can't I just say y<12

and y has to be greater than the point of intersectiojn

and also my original answer had

x is greater than or equal to -5

and that x is less than 1

idk why I can't define the inequalities that way

[knotty anchor]

Generally these questions have many different answers that are defined in the markers mark scheme and not the mark scheme you have

[halcyon light]

oh yh tbf it says this

but it doesnt say im gonna be awarded a mark

also spurious means fake apparently

[fringe dirge]

if y < 12 arent there some regions outside the quadratic still shaded

[halcyon light]

or bogus

[knotty anchor]

No in the examiners mark scheme there are many

[halcyon light]

right

[halcyon light]

wait what

oh yeah

i see what u mean

true

[halcyon light]

dyu think this inequality w

is worth writing down

cause that was part of my initial answer

[fringe dirge]

this is what you've defined

[halcyon light]

yeahh i get u

[fringe dirge]

well if its in your working beforehand it doesnt matter really, as long as the three inequalities at the end are clearly shown to be your three answers

[halcyon light]

but u don't think they would award me any marks?

ugh ok

I'll jus get the equation of the 2 lines and curve next time

thanks

[fringe dirge]

i mean seeing as its only a 2 mark question I dont think theres enough to hand out anymore

[halcyon light]

yh true

[fringe dirge]

is this an a-level question by any chance

[halcyon light]

yeahhhhh

year 1 xd

[fringe dirge]

yea the answer sheets and things looked familiar

i assume the previous parts got you to find the equations

so really the final part is just testing you know how inequalities work

[halcyon light]

.close

[safe radish]

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[halcyon light]

.reopen

[safe radish]

✅

[halcyon light]

I'm stuck here

on b

I got the deravative equal to -1/12

I do not know what to do now

whole graph btw

[fringe dirge]

so what does -1/12 represent

[halcyon light]

the gradient function

[fringe dirge]

well yes but what does that mean in this case

[halcyon light]

the steepness of l?

[fringe dirge]

is it the gradient of l or the gradient of C

[halcyon light]

C

[fringe dirge]

yes and what relationship does l have with C

[halcyon light]

tangent to C at the point B

[fringe dirge]

oops I meant the other relationship it has

[halcyon light]

normal to C at a

[fringe dirge]

yes so we have the gradient of C at A, cant we work out the gradient of L at A then

[halcyon light]

wait isn't that what I worked out tho

cause I took the negative reciprocal

[fringe dirge]

oh lol didnt realise you had done that

so we know the gradient of L is -1/12

can we form an equation for L then

[halcyon light]

thing is tho

idk the y coordinate

or the x coordinate

I'm supposed to use

[fringe dirge]

well youre given a coordinate that is definitely on the line

[halcyon light]

yeah even if i use the -7/2 tho

how am i meant to work out the y coordinate

when theres a k constant

in the y = equation

[fringe dirge]

oh yes my bad, you know the gradient of l is the same as at B dont you

[halcyon light]

yeah

[fringe dirge]

and you are trying to find an equation for the x coordinate at B

so cant you create an equation with your -1/12

[halcyon light]

oh shit am I meant to write it like

y = -1/12x

[fringe dirge]

this is using the deriviative of C, which youve already found

[halcyon light]

so y = 12x

[fringe dirge]

$\frac{\text{d}y}{\text{d}x}=\frac{6}{7}x^{2}+\frac{2}{7}x-\frac{5}{2}$

this is the derivative of C you got?

[flat frigate]

noah

[halcyon light]

i just realised I wrote +5/2 instead of minus

but yeah

[fringe dirge]

so you want to find the x coordinate at which the gradient of C is -1/12

or whatever the gradient of l is

[halcyon light]

yeah i messed up

gradient of c is 7

then negative reciprocal is -1/7

[fringe dirge]

so now this with -1/7

[halcyon light]

yeah I got it now

thanks

.close

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[limber jewel]

Can someone help me through this slowly

[limber jewel]

I know the basics of change of order of double integrals

[safe radish]

@limber jewel Has your question been resolved?

[limber jewel]

I'm stuck at finding the new x limits according to me it should range from 0 to y^2/4a but my class notes say it ranges from y to y^2/4a

From the graph i say the new x should be 0 to the y curve

[cerulean coyote]

[0 \leq x \leq 4a] and [x \leq y \leq 2\sqrt{ax}] so, just assume equality for bounds, then you get $x=0, x= 4a, y=x, y=2\sqrt{ax}$. x is less than $y^2/4a$, and so is y=x, so y is the lower bound. or else, you integrate for all values of x from 0 to $y^2/4a$.

[flat frigate]

fish

[cerulean coyote]

essentially, the left bound for x values increases as y increases, so the lowest bound of x being 0 always is not correct

[limber jewel]

How is x less than y on the line?..i understand that y increases as x increases, so does that mean if i integrate from 0 to y^2/4a i would get some area that is not inside the region..?

[cerulean coyote]

yeah. Essentially the bounds of x should oftentimes be in terms of y. so $x_1(y)\leq x \leq x_2(y)$.

[flat frigate]

fish

[limber jewel]

Oh i finally get it

Thank you very much i get the rest of the problem

[cerulean coyote]

you're welcome

[limber jewel]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[languid forge]

does anyone have a tip/trick to help remember the product to sums and sum to product formulas?

[fathom adder]

Can you post it ?

The formulas

[languid forge]

these ones sorry

[fathom adder]

Oh i dont have for this one sorry

[languid forge]

allgg!! im sure someone else wouldhave a trick 😭

[fathom adder]

I have for others formula but thoses i just learn them like a barbarian

[short jackal]

This is a more advanced and time consuming trick

If you forget an identity

It can be rederived using the complex exponential definition of sin and cos

[cold aurora]

Okay. That's definitely an overkill.

You can derive these easily from simple addition of angles identities.

As long as you know, Sin(A+B) and Cos(A+B), you can derive all of these.

[short jackal]

I'm assuming you know no trig identities

[languid forge]

yeahh i suppose i could do that

but it would be time consuming esp in test situation

hmm ig ill just have to keep repeating it to myself lol

[short jackal]

Yes, this would be the backup just in case

[languid forge]

ok! good idea then

its just hard because no matterhow much i practice these type of questions it doesnt stick to my head so im worried iwont be able to recognise when to use it when i need to 😭

[safe radish]

@languid forge Has your question been resolved?

[lean otter]

im doing the same type of problems rn. if you do a bunch of problems they start to stick. your also practicing for your test at the same time.

[languid forge]

oh yeahh ig i sjoild just keep practicing then!

thanks

[safe radish]

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[elder kernel]

Is this correct

[cold aurora]

No.

[elder kernel]

.close

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[safe parrot]

In how many ways can the word TUNNELBANESTATION be divide into 4 different parts such that
a) each part has at least one letter, example : TUN-NEL-B-ANESTATION

[safe parrot]

Am i thinking correct that i first have to assign 4 letters to my 4 parts (or boxes) ?

And since i have 17 letters and 4 parts it should be. 17P4 ?

That would be the total amount of combinations that exist for when we choose the first 4 letters into the 4 parts?

[covert yoke]

@safe parrot stars and bars. Instead of thinking about it in terms of parts, think about the spaces between letters either having a separation or not. From this perspective it's reasonably easy to see that you get 16 choose 3 ways.

[safe parrot]

how did you know it's stars and bars ?

[covert yoke]

It's a common combinatorics technique

TUNNELBANESTATION becomes

T.U.N.N.E.L.B.A.N.E.S.T.A.T.I.O.N

Focus on the dots some of them (3) can become -s instead.

T.U.N-N.E.L-B-A.N.E.S.T.A.T.I.O.N

Any selection of 3 .s to turn to dashes will be a valid way of separating the word into 4 parts

And vice versa

[rotund wolf]

do the letters have to be further arranged while in the group?

cause that makes the problem significantly harder

[safe parrot]

But how did you determine that it's solved with stars and bars?

I cant seem to understand the difference when to choose permutation, combinations, stars and bars, boxprinciple, multinomial, binomial theorem

[covert yoke]

So stars and bars are useful for when you need to split something up without really rearranging it

[safe parrot]

I was under the assumption that stars and bars is i want to split something where each element is not distinct. Meaning there is no difference between them

Do we not care which order they are in?

T | U | N | N E L B A N E S T A T I O N So this could be one solution?

[covert yoke]

Well, if you can rearrange the parts, then you would need some other approach, possibly starting with stars and bars

But perhaps you are misinterpreting the book or your lecturer?

[safe parrot]

if i use stars and bars approach here

i get 20C17

[covert yoke]

How do you figure that?

[safe parrot]

$\binom{n+k-1}{n} = \binom{17+4-1}{17} = \binom{20}{17}$

[flat frigate]

Merineth

[covert yoke]

So stars and bars, as I generally conceive it is as follows

I have n stars in a line and I want to insert m bars between them

Such that at least one star is in each group

*|***|***|***

And example with n = 10 and m = 3

You can see by inserting m bars you get m+1 parts

So when we ask for parts we have to subtract one from the parts to get the bars

Now you have n-1 places to put m bars, this is (n-1) C m

[safe parrot]

I'm not sure how to interpret stars and bars anymore since everyone keeps saying different things

[covert yoke]

Ok, you explain it how you learned it originally

[safe parrot]

I was told that i have a box 4 boxes where i want to place 17 items inside them. How many different combinations can i get out of it. And that is under the assumption that each "star" are identical

What confuses me even more here is that the word TUNNELBANESTATION since stars and bars goes under the assumption that each "star" are identical. But we can clearly see that the word has different unique letters...

[covert yoke]

Same thing actually, except instead of "bars" you're dealing with "parts" so your "bars" are one fewer

So because we cannot rearrange the letters in the word they are functionally identical

In other words it doesn't matter what the letters are

[safe parrot]

Mhm i see

i cheated and looked at the answer and they solve it like this :

[covert yoke]

You can see that if the string instead read HAVEANICEDAYKEVIN it wouldn't change the answer

[safe parrot]

$x_1+x_2+x_3+x_4 = 17$ \
where : $x_1 \ge 1, x_2 \ge 1, x_3 \ge 1, x_4 \ge 1$

[flat frigate]

Merineth

[covert yoke]

I have no idea what they are trying to do

[safe parrot]

That's apparantly the solution on how to solve it

[covert yoke]

That's not a full solution

[safe parrot]

no the answer is what you gave but the solution is this

This is exactly the reason why I’m so confused. Combinatorics never has a clear way to solve the problems, they are always new and always introduce something different which I can’t figure out

[covert yoke]

Well, the good news is there are only a handful of tricks

And you'll figure them out pretty quick

[safe parrot]

I'm truly truly desperate at this point

i'm like 30 problems in now and i haven't been able to solve a single one on my own

The tricks seems to be endless

how did you get 16C3?

[covert yoke]

From stars and bars. You have 17 letters, that means there are 16 places a dash could go. Next, there are 4 parts, and so there are 3 dashes separating the parts

[safe parrot]

wtf

I was told n = the amount of stars, k = how many boxes

I have 4 boxes

17 letters

[covert yoke]

In that formulation the answer is (n-1)C(k-1)

The two formulations are equivalent.

[safe parrot]

stars and bars formula is $\binom{n+k-1}{k}$

[covert yoke]

Are you certain about that?

[safe parrot]

sorry over k

[flat frigate]

Merineth

[safe parrot]

that's the formula for stars and bars

n is 17 k is 4

20C4

but 20C4 has the possibilities where there are no stars between bars

[covert yoke]

You might be using the version where a bin can be empty

[safe parrot]

,rotate

[safe parrot]

If i want to make sure i have at least one star between bars

n would be reduced by 4

which means i have 13 balls to place inside the boxes

13+4-1 = 16

16C4

I dont know how you or the book get 3

when we have 4 boxes

[covert yoke]

The point is we aren't really considering the boxes themselves.

We're considering the borders between the boxes

[safe parrot]

the borders between the boxes?

[covert yoke]

That's the "bars" part of stars and bars

[safe parrot]

But we have 4 boxes?

this is one possibility

so k = 4?

[covert yoke]

So I don't like claiming this when I cannot read Swedish, but it's possible that you may have misread or misinterpreted the book.

At least, the book's presentation of the formula differs from English Wikipedia's in the case of possibly empty boxes.

[safe parrot]

"The number of combinations with repetitions of size k among n given objects is equal to ..." where the "..." is the formula

Our number of combinations are 4? Since we want to have them in 4 different parts

and we are given 17-4 letters to work with, n = 13

I fail to see where k = 3, since we are dealing with 4 boxes / 4 parts/ 4 arrangements of the letters