help-17

[eternal glacier]

for this question can I just use RREF on the right matrix to find basis?

or still a few more steps involved? D = PAP^-1 required?

[topaz bolt]

You can solve AX = -4X i think

[eternal glacier]

ummm

how did you get that formula

[topaz bolt]

-4 is assumed to be an eigenvalue

[eternal glacier]

so this will not work?

[topaz bolt]

Whats that? Mind un-abbreviating lol

[eternal glacier]

RREF on this

[topaz bolt]

Ohhh like gaussian elimination?

[eternal glacier]

just by looking at this, column 2 and 4 are probably nuked

yes

so i should be left with only 2 columns for the basis

[topaz bolt]

I mean if youd wanna get the determinant its gotta be 0 since -4 is said to be an eigenvalue right

[eternal glacier]

<@&286206848099549185> anybody know if RREF will work on this to find the basis?

or if I can only solve with the formula AX = -4X

[eternal glacier]

yeah this determinant is toast, 2nd column makes that obvious

so no inverse to this square matrix

but dunno if that really helps at all for the basis, i just need a set of 2 vectors in R^4 to find basis

[topaz bolt]

You dont have to calculate the determinant tho

[eternal glacier]

I know lol

[eternal glacier]

not sure why it was brought up

[topaz bolt]

The matrix looks nice for a system of 4 equations to solve

Since some lines have 0s you can get a relation between two variables reducing the dimensionality of the eigenspace

[eternal glacier]

i'm just asking if RREF on the second matrix can be used to solve, or if AX = -4X is the only way to solve

to find the basis

[topaz bolt]

I mean through rref you can get the determinant but its already known to be 0?

[eternal glacier]

huh?

RREF just finds the pivots

to solve for each variable

[topaz bolt]

On a matrix with no variables?

[eternal glacier]

???

you do realize this is referring to x1, x2, x3, x4, right?

each row is an equation with those variables

if you have not heard of RREF before maybe AX = -4X is not the correct formula to be using here

[topaz bolt]

So were doing the same as solving AX=-4X

Just that you wrote (A+4)X = 0

And are trying to find the kernel of A+4

Or no?

[eternal glacier]

sorry, just to double check..
have you taken Linear Algebra as a course before?

RREF was the taught during the first week, we learned about it

[topaz bolt]

Yes

[eternal glacier]

alright just making sure

bc if you are talking about kernel maybe you are advanced with it?

we haven't learned about that yet, but we are first year

[topaz bolt]

I guess language barrier?

Since i studied in the french system

[eternal glacier]

yea, sorry my bad, maybe it's called something else

RREF is a different acronym for you?

[topaz bolt]

It looks like gaussian elimination/gaussian pivot

[eternal glacier]

well RREF would be Guasse-Jordan elimination specifically

REF is just regular Guassian elimination

RREF requires each pivot to have 0 both above and below

REF just requires 0 below each pivot

and an extra step of backward substitution to solve

i prefer RREF because it's very clear, only one solution to RREF

REF can potentially have unlimited possible solutions, and extra work with the backwards substitution

so I am hoping that with RREF I can finally solve for the basis lol, it possible

[topaz bolt]

Ok i see what you mean

[eternal glacier]

but this feels like it could be a more complicated question, when it's referring to Eigenspace

but maybe not though, that's why I am checking

One sec, I will finish the work and post

to show what I mean

[topaz bolt]

Alright

[eternal glacier]

would this be the correct answer?

for the basis

set of vectors

based upon the free variables x3 and x4

hmmm, I tried and it's not correct, dang..

[topaz bolt]

Indont have a paper on me rn im outside but let me check lol

@eternal glacier for the two first rows you can sort of deduce for an eigenvector X = ( x y z w) you have x = y and w = -z

So the eigen basis would be i guess (1 1 0 0), (0 0 1 -1)

[topaz bolt]

Not sure since there still two other rows

[topaz bolt]

Actually oops x = z

And z = -w

So y is free

And maybe the eigenbasis is (1 0 1 -1) and (0 1 0 0)

[eternal glacier]

x = z?

[topaz bolt]

Yea so 1 dimension reduced

[topaz bolt]

Less degrees of liberty

[eternal glacier]

oops

I said x2 = -x4

but it should be x3 = -x4

[topaz bolt]

Ah

[eternal glacier]

so maybe this is correct?

[topaz bolt]

I guess so

[eternal glacier]

mine is closer to yours, but (-1) scalar for x4, that matters, right?

[topaz bolt]

No its ok

[eternal glacier]

oh

[topaz bolt]

Since x3 = -x4

[eternal glacier]

OK I will give this a try

[topaz bolt]

Same as x4 = -x3

[eternal glacier]

so there are many possible solutions for this question

like infinitely many actually

[topaz bolt]

Im not sure if infinite lol

[eternal glacier]

if it's just based on scalar, i think so

like i could swap all of the 1's for 43434's and it would still be correct I think?

technically it equates to the same thing just using different basis'

[topaz bolt]

Yea it depends on what "axis" you choose

[topaz bolt]

Different basis

But same eigenspace

[eternal glacier]

right

bad wording, fixed now

OK i will give this a try

fingers crossed lol

tyvm 🙂

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[topaz bolt]

No problem

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[round bough]

Help please

<@&286206848099549185> sorry it’s cause my deadline is in 5 minutes

I know how to do the question

Ik whag a prism is

Just confusing cause it tells me I’m wrong

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[restive compass]

Im having trouble understanding this

[restive compass]

I dont get like how you determine y_1 from y_2

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[marble tinsel]

idk what the equation is

[frozen bobcat]

they just want you to rewrite it as an equality.

[marble tinsel]

aight one se

se

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[buoyant turret]

i got the answers but unable to justify how to use inequality signs

answer is D) all of the above

but how may i conclude the signs given in the options

pls help if some one can..

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[torn spoke]

When multiplying something by a variable expression which might be 0, do we have to consider 2 cases: when the expression =0 and when it isn't every time we do such a multiplication?

[torn spoke]

Extraneous and missing roots from multiplication and division dumbfound me a bit

[tight sonnet]

just give an example

[torn spoke]

Say we have 1/(x+1) = x + 2 and we try to multiply by (x+1)

Do we have to separately consider cases when the multiplier is 0 and when it isn't?

Do we have to do it when it is (x+2) for example?

[tight sonnet]

i mean multiplier cant be 0 in this case

beacuse its in denominator

[torn spoke]

So

Is there any chance we will get extra roots or miss some roots when we multiply this by (x+1)?

[tight sonnet]

as we said x+1 cannot be equal to 0

so x cannot be equal to -1

if after multiplication youd somewhat got x=-1 as a sol youd need to disgard it

beacuse in the orginal x-1 was in denominator

[torn spoke]

What if we multiply by (x+1)^2 instead?

I'm genuinely confused by when we get new / lose old roots

[frozen bobcat]

if you totally cancel a factor, you probably lost a root.
if you introduce a new factor, you probably introduced a new root.

like x = x(x-1)
if you divide by x, you lost a root.

[torn spoke]

An interesting opinion

What's the logic behind it?

[frozen bobcat]

x = x(x-1) is a quadratic equation. dividing by x turns it into 1 = x-1, a linear equation.
Quadratics have 0, 1, or 2 roots, linear equations have 0,or 1 roots. So it's possible to lose a root (in this case, we lose x=0)

[unborn coral]

To add on to that, anytime you divide by any expression, you're implicitly saying that that expression is non-zero (e.g. you can only divide by x above if you assume x is not 0) so thats why you lose a root because you threw a solution away by dividing

the best way to avoid is not to divide by expressions but rather to gather and factor

so e.g. x = x(x-1) becomes x(x-1) - x = 0 or x(x-1-1) = 0 or x(x-2) = 0 and then you get both sollutions

[torn spoke]

@frozen bobcat why do you say probably? Are there specific requirements for it to be the case or is it random?

[frozen bobcat]

depends on what you divide by
like the other guy says, by dividing you're assuming that expression isn't 0

so if x=0 is a solution, and you divide by x, you've excluded that solution

[torn spoke]

@unborn coral so the exact reason for losing the root is assuming that x isn't some value while it is? Or are there cases when this statement is incorrect?

[unborn coral]

Anytime you divide by an expression you're assuming that x isn't some value but its possible it actually was a solution of the original, that's all! often when we divide "by cancellation" you are exactly getting rid of a solution

[torn spoke]

So as I understand. Whatever solutions we have, we can make a multiplication chain of them on LHS (eg x=1,2 => (x-1)(x-2)) while having 0 on RHS. Thus if we divided by one of these values we lost one of the roots. Am I undeniably correct?

I think I am

So we shouldn't divide by any value unless it is known to be not 0. Correct?

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[manic basin]

I have a question about infinite series

[manic basin]

Can someone help?

[iron parrot]

what is your question?

[manic basin]

I need to tell whether this converges or diverges

which i got to be divergent

but

if n=2 or 4 then wouldn't it be undefined?

[copper crypt]

lol true

Good point

[heavy yoke]

the series should have started at a high enough n that it wouldn't happen, since it doesn't really make sense to add undefined tetms

[manic basin]

so is this a mistake?

[heavy yoke]

i would say so. since the intention is to test how the series behaves with large n, i would assume they mean to start with n=5 or something similar where there aren't any undefined terms

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[compact vale]

How can I solve this?

[real cloud]

average * how much it counts added together

[compact vale]

Nvm I think I got it

I multiply each number by what percentage they count for and then add and round?

i got 83

[real cloud]

sounds good

[compact vale]

thanks

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[near relic]

there was a shape similar to this in a book that said this shape was the domain of a complex function. It also said the domain of a complex function has to be an open subset of complex numbers..

What exactly makes this shape "Open"?

[near relic]

I think I'm taking open and closed titles very literal but I could really use guidance here 🙏

Cuz if for it to be open every point has a defined point all around it on a tiny scale

what happens at the border

wait can I ping now

I think

<@&286206848099549185>

[kindred barn]

what u need ? help wit?

[near relic]

I'm trying to understand what exatcly the difference between an open domain of the subset of complex numbers for a complex function is but I'm not understanding a thing

[near relic]

This figure is meant to represent the open domain subset of C for a complex function

but I don't quite understand what makes it "Open"

[kindred barn]

like the [ and ( things in a domain?

[near relic]

THe figure

[kindred barn]

Shape is not open the domain is open and closed

like open and closed means it includes the end points and closed means it doesnt include those

[near relic]

but if you make the little epsilon ball on one of the endpoints some of the points won't be in the domain no?

OH

[kindred barn]

what?

[near relic]

so the epsilon ball will yield a point in the domain for any point in the shape besides the border that's what you mean?

if it's an open domain

[kindred barn]

Yes

[near relic]

so why does it have to be an open subset?

Why can't the domain be closed

[dull bear]

What's the context to that claim?

(and assuming the boundary isn't included, yes, pick any point and you can form a little ball about it that would fully be in that blue region)

[near relic]

Up to now I've understood that the domain of a complex function must be a closed subset of complex numbers, is this wrong?

[dull bear]

I mean, unless there's a property they're talking about, a complex function could have a open/closed/both/neither domain

e.g. the identity function

[near relic]

Man I thought I was starting to understand

[dull bear]

Do you have a screenshot/picture of the claim they're making that the domain must be open?

[near relic]

it wasn't stated exactly it's just what I've been understanding

so then if the domain can be closed too

what would the blue region look like if it was a closed domain

[dull bear]

The boundary would be included as well

Open would have the boundary completely excluded, closed would have it all included

[near relic]

what's the point to that then

like why include the boundaries

[dull bear]

Properties that depend on being closed, I guess(?)

[near relic]

i see

I think I understand, so the domain itself won't have an effect on the shape itself

rather the domain type

Ah!

Well thanks a lot for the help

I'll close the channel now for other people. Have a good day / night

.close

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[empty python]

How did they get the values 1/4 and 2

[frozen bobcat]

$\ln(b) - \ln(a) = \int_a^b\frac1x\ dx$ combined with log rules. That's how they're finding values.

[twin meteor]

Zybikron

[empty python]

Please just tell me how they got the values 1/4 and 2 in this context

Or don’t help

[wicked shard]

Wym

They just told u

[empty python]

In this context please

Not a general

[frozen bobcat]

$\ln(2) + \ln(4) = \ln(2) - (-\ln(4)) = \ln(2) - \ln(4^{-1}) = $\ln(2) - \ln(1/4))$

[empty python]

Thing

[twin meteor]

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

[empty python]

Why would they randomly start subtracting

[frozen bobcat]

to use this

because the whole set of problems you posted is just using Fundamental Theorem of Calculus, so they're setting up everything to get back to ln(b) - ln(a) for some values a and b.

[empty python]

This isn’t making any sense why would the b value be 2 then ? It would’ve been 4

[frozen bobcat]

that's one way to do it. There isn't a single correct answer here.

[empty python]

Can u explain that way

Why would b be 2 and not 4

[frozen bobcat]

b is the upper bound of the integral, and should be the larger value in ln(b) - ln(a)
You can use b=4, but it changes everything else.

[empty python]

Exactly my point

[wicked shard]

What

[frozen bobcat]

$\ln(4) + \ln(2) = \ln(4) - (-\ln(2)) = \ln(4) - \ln(2^{-1}) = \ln(4) - \ln(1/2)$

[twin meteor]

Zybikron

[empty python]

Why would they then do 2 - 4

I’m trying to understand their way please

[empty python]

If u aren’t helping don’t say irrelevant stuff

[frozen bobcat]

they don't. They do ln(2) - ln(1/4)

[frozen bobcat]

to use Fundamental Theorem of Calculus, which relates the area under the curve to a difference
in this case ln(b) minus ln(a)

[empty python]

Can u explain the part where I am saying why

Like why are they doing that

And In that order

[frozen bobcat]

ln( ) + ln( ) isn't relatable to an integral as it is written. They need to rewrite it as ln(b) - ln(a)

[empty python]

Why r they doing it in that order

[frozen bobcat]

i am explaining why they change it to minus.

[wicked shard]

They want to

[frozen bobcat]

because they can. You can do it an any order.

[empty python]

Please stop

[wicked shard]

???

Its the right answer

Its because they want to

You can do it the other way

Nobodys stopping you

[empty python]

Ok so how do I generate more values I want to get the same graph

Of In 8

[wicked shard]

Any two numbers that multiply to get 8

[frozen bobcat]

if you're looking for a fundamental reason that the answer book has that specific answer, there isn't one. You can rewrite that area in a dozen different ways.

[wicked shard]

Theres 2 per pair

[empty python]

2 x 1/4 is not 8

[wicked shard]

2*4 is

You can generate 2 integrals from this

[empty python]

Ok can someone please just tell me the general way I can use to generate as many values I want to get the same graph of In 8

[wicked shard]

Find 2 factors of 8

They can have decimals, doesnt matter

[novel iris]

A property of logs is that -ln(a) = ln(1/a)

In this case we have ln(1/4) thus its the same as -ln4

[empty python]

Ok

So

[wicked shard]

1/a to b and 1/b to a are both valid integrals

That work

[wicked shard]

I told you

[wicked shard]

.

[frozen bobcat]

8 = ab
ln(8) = ln(ba) = ln(b) + ln(a) = ln(b) - ln(1/a)
pick any two positive values of a and b.

[empty python]

Ok so for 1 x 8 it would be 1 and 1/8

[novel iris]

yep

[wicked shard]

Ye

[empty python]

Jeez man thanks sorry if I had some attitude

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[wicked shard]

Its fine

[novel iris]

no worries

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[empty python]

This is very helpful as well thank you

[empty python]

Oh no

.stop

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[dusk bear]

Help

😭

I dont rlly get trig identities

Like

Ik them

But i dont get the problem

[buoyant notch]

How do you get a denominator of sinxcosx

[dusk bear]

Idk 😭

[buoyant notch]

You have two fractions

You multiply to get a common denominator right

[dusk bear]

Yeah

OH

OHHH

[buoyant notch]

[dusk bear]

SO THE FIRST BOX IS COS^2

[buoyant notch]

Yes

[dusk bear]

What happens to the sin+cos/cos+sin

Like the left one next to the first box

How did they get rid of it

Wait

[buoyant notch]

?

[dusk bear]

Im dumb

OKOK

I MISUNDERSTOOD IT

[buoyant notch]

[dusk bear]

SO THE SECOND BOX WOULD BE

Would it just be cos sin

[buoyant notch]

Yes

[dusk bear]

Woah

THANKYOU

[buoyant notch]

Np

[dusk bear]

I FINALLY FINISHED THIS ASSIGNMENT WAHOOOOOOO

:DDD

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[terse plume]

b)
"To remain at the same elevation, where we can walk at the location/point
(2, 0, 1)? Justify appropriately."
I already now the gradient from f function

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[bright parcel]

how do i go on about solving
(y^2)+2y=2x+1 from (-1,-1) to (7,3)

[urban edge]

context?

[bright parcel]

sorry

Find the length of the given curves

@urban edge

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[bright parcel]

Find the length of the given curves
(y^2)+2y=2x+1 from (-1,-1) to (7,3)

help pls

[normal heath]

You could solve it for y=... and do as the previous questions, or solve for x=... and use basically have x and y swapped

[bright parcel]

idk how to do it all

can you guide me?

[normal heath]

Have you done the previous questions?

[bright parcel]

no

i just want to know how to do this

do we find the derivative of the function first?

[normal heath]

Yes correct, the length of the curve is given by this integral

[twin meteor]

π=√g

[bright parcel]

first

we do
dy/dx (y^2) + dy/dx 2y = 2

now we can factor the dy/dx

which can be dy/dx (2y+2)

and now im stuck

[normal heath]

That won't work as you'll have y in the expression

[bright parcel]

wait

[normal heath]

You first want to solve for y=...

Solve a quadratic equation

[bright parcel]

y'=2/2y+2

[normal heath]

Yes, but the problem here is y is in the equation for y', meaning you can't really do the integral w.r.t x

[normal heath]

But there is an easier way: you can solve for x=... instead, and swap x and y in the equation for arclength

[bright parcel]

y is in the equation for y' , i dont get this part

[normal heath]

y' = something with y

[bright parcel]

yes

what is wrong with that

[normal heath]

Well y' is used in this integral

[bright parcel]

isnt the equation for arc length

[twin meteor]

π=√g

[bright parcel]

yeah

so basically

[bright parcel]

y' is this

so we just plug that there

[normal heath]

That would give $\int_a^b \sqrt{1 + \qty(\frac{1}{y} + 2)^2} ,\dd x$

[twin meteor]

π=√g

[normal heath]

How do you do this integral without knowing what y is?

[bright parcel]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh

im stupidddddd

we need x there

instead of y

[normal heath]

yeah we need only x's in there

[bright parcel]

quick question

what is this for

[normal heath]

Well that's exactly the same equation

[bright parcel]

is there any difference?

[normal heath]

Technically no, but it's easier to see that you can also find the arclength for a function like ( x = ) f(y) = something with y

[bright parcel]

so we basically need to find x'?

[normal heath]

YEah exactly

You could also find y', but it's just longer and more algebra

[bright parcel]

ok so how do i find x'

[normal heath]

Well what is x written in terms of y?

[twin meteor]

π=√g

[bright parcel]

idk

[normal heath]

How to change this into x = ...?

[bright parcel]

oh

x=((y^2)+2y-1)/2

[normal heath]

Yep very nice

Now what is x' (basically the same as calculating y' but the x and y are swapped)

[twin meteor]

π=√g

[bright parcel]

x'=2y+2

[normal heath]

Not exactly

[bright parcel]

wait

idk

[normal heath]

what is the derivative of y^2?

[bright parcel]

2y?

[normal heath]

yeah

[bright parcel]

ok

[normal heath]

but here you divide by two

Probably forgot that

[bright parcel]

how do i do that then

[normal heath]

You divide by 2...

[bright parcel]

oh

y

[twin meteor]

π=√g

[bright parcel]

yeah

true

[normal heath]

Now just put that into this equation and integrate

[bright parcel]

ok

[normal heath]

$L = \int_a^b \sqrt{1 + (y+1)^2} ,\dd y$

[twin meteor]

π=√g

[normal heath]

Any clues on what a and b are?

[bright parcel]

yeah just give me a second im writing it down

on my notes

[bright parcel]

is it a problem to do dx instead of dy in the end?

[normal heath]

wdym?

[bright parcel]

you have dy in the end of the integral

[normal heath]

yes, dy here is to show which variable we are integrating with respect to

I could write like like $L = \int_a^b \sqrt{1 + (a +1)^2} , \dd a$

[twin meteor]

π=√g

[normal heath]

The curve looks like this

We want the length between (-1, -1) and (7, 3)

Between what values does y vary?

like y goes from what value to what value

wait

Maybe

Yeah yeah ofc

It's just to show here y varies between -1 and 3

Going from the first point to the second point

Yeah sry you dont need the graph at all, it can just be helpful to visualise

a and b

which are -1 and 3

They simply come from the boundary points (-1, -1) and (7, 3)

(-1, -1) and (7, 3)

(it's in bold)

Lol just @ me if you want

dw ill wait

Ill play skribbl

Yeah good question

It's cuz we wrote it as x= y^2/2 + ... or basically f(y) = y^2/2 + ...

So the variable here is y, which is why we look at the y component of the points

nw

[normal heath]

if it were y = x^2 + ... or f(x) = x^2 + ... then you would be taking the x component of the points

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[harsh thunder]

Trying to prove this formula for calculating some permutations in an excercise by induction, so far I verified this is true for n=8 so now I've got to prove that this formula implies that it's also true for n+1 and I'm done:
I think that if we substitute n+1 to n we can decompose the first part as ((n-2)!(n-3)!)/((n-7)!4(n-8)!) but I'm a bit stuck with how to proceed from here

[harsh thunder]

like this if S(n) implies S(n+1) than the formula is true I just need this final step but can't quite get it

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[empty python]

Where did the In 2 come from, in the pictures is the question highlighted and my work which I got half of it

[golden night]

integral of 1/x is ln x

[spark storm]

since when was the integral of x^-1 = 0

[empty python]

Ok but how is it used in this context

[spark storm]

in line 4

you wrote the integral of x^-1 as 0

[empty python]

Yes cuz -1 plus 1 is 0 then the reciprocal of 0 is 0 so 0 times 1 which is 0

[spark storm]

wait no no no

reverse power rule does NOT apply on x^-1

look what happens

you get 1/0

which is undefined

[golden night]

reciprocal of 0 is undefined

[empty python]

Wait hold on

I’m confused

[empty python]

How

Pls explain

[spark storm]

well

ur dividing by 0

[empty python]

Then how am I supposed to do this question

[spark storm]

the integral of x^-1 is special

it's lnx

[empty python]

Is that something I should memorize

[spark storm]

think back to your differential calculus days

differentiating lnx gave you 1/x

[empty python]

Bruh I’m 17

I’m learning it rn

[spark storm]

wait but you should've learnt differentiation before integration right

[empty python]

Yes I know this

[empty python]

Yes

[spark storm]

yep

so it's the inverse

of that

[empty python]

Ohhhh

I seee

Let me try again

It didn’t work

@spark storm

@golden night

[spark storm]

no

lnx isn't equal to 1/x

[empty python]

How

[spark storm]

nvm oyu didnt

but

isnt ur answer correct

thats correct

???

ln1 is 0

[empty python]

The answer is 11/8 plus In 2

[spark storm]

yh thats what you have

the ln1 is 0 so you can ignore it

[empty python]

How

[spark storm]

e raised to what power gives you 1?

[empty python]

Oh ok

Thank you

[spark storm]

always ask yourself

that question

[empty python]

Lol

I should

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[pastel cloak]

Yo real quick, how would you be able to deduce from this information that p(t) = 120

[pastel cloak]

Natural law of growth: P'(t) = k*p(t)
Solution to law of growth = p(t) = P(0)e^(kt)

[heady ibex]

the population starts at 60

so when it has doubled it is 120

which means you want to find t for which p(t) = 120

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[pastel cloak]

Ah I see thank you

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[plain walrus]

question 35, i just need help, ik where to start

[plain walrus]

i just dk where to proceed lol

[subtle helm]

does that say cot(0)?

[plain walrus]

cot Q

[subtle helm]

oh

use a sum/difference of angles identity for tangent

and the fact that Q = (P + Q) - P

[plain walrus]

yeah i did that i just dont know where to proceed next lol

[subtle helm]

what did you get

[plain walrus]

this is what i have so far

[subtle helm]

if you do it this way, you can just multiply by 1-2tanq on both sides and combine like terms

[subtle helm]

this way just gives you an expression with no unknowns right away

[plain walrus]

and the fraction will cancel out correct?

[subtle helm]

the denominator gets cancelled yes

[plain walrus]

did i do this right? then i bring tan q on one side and subtract the two?

[subtle helm]

you distributed wrong on the lhs

[plain walrus]

oh

oops mb

[plain walrus]

okay fixed, now i flip since its cot and do 15/ 10?

[subtle helm]

yes

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[plain walrus]

Alright, got the answer, thank you so much!

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[strange gull]

Why is h here 0.25? i thought h = b-a/n which is 1/5

[strange gull]

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[ionic dagger]

i got the area correct but perimeter wrong

[candid nymph]

Well yeah you forgot a pi

[potent anchor]

perimeter of the larger semicircle's "circle" part is 1/2 * 2 * r * pi = 2pi
perimeter of smaller semicircle is 1/2 * 2 * r * pi = 1pi

• 2 from exposed larger semicircle portion

[ionic dagger]

so add those all?

[potent anchor]

yeah i think its 3pi + 2

[ionic dagger]

okay its correct thanks!

Also, I don't know where to start here

[potent anchor]

break the problem up into smaller problems

[ionic dagger]

ok

i got the area

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[radiant monolith]

Im trying to learn the process on how to get the solution to this. I can graph it as it's a Standard form and have found the intercepts but how do i interpret the graph or solve to get the fraction solution?

[copper crypt]

Does this part make sense to you

[radiant monolith]

That's the part im confused on. Like where did they get those fractions? Like sure i can probably chuck every A-D equation into the calculator and see if they satisfy the equation but i feel that's slow and there must be a better faster way to solve it. I look at the graph and maybe there must be a way to read it to get the solution without having to manually type every single problem into the calculator

[radiant monolith]

i may be brain fogged right now but how does that convert over to x=-1 2/5 and y=1 3/5?

[copper crypt]

well x could be anything between -2 and -1 and y could be anything between 1 and 2

but just look at the options

[radiant monolith]

Wait hold up i think i just figured it out. x is basically equal to -1.4 and y equal to 1.6. The intersecting point is in the middle of one and two.

[copper crypt]

x value of the intersection point is between -1 and -2, the y value of the intersection point is between 1 and 2

[radiant monolith]

yeah you right thats what i meant. Thank you. i just needed to stare at the problem and look at the graph a bit closer

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[forest rock]

How is part b possible? i also put it through desmos and the values were different as wellthe 2nd picture is what the answers were but i'm confused about how it worked it out

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[smoky heart]

Hey guys,

I'm trying to get myself to understand the combination of:

A store sells 4 varieties of donuts. Vanilla and Chocolate are two of the varieties sold. How many ways are there to select 14 donuts so that at least 5 Vanilla donuts and at most 4 Chocolate donuts are selected?

My initial thought is C(12,3) + C(11,3) + C(10,3) + C(9,3) + C(8,3).

[regal bane]

Can you detail your initial thought a bit more?

Oh wait, I think I see what you did

Nope, I take it back. Why 12?

[smoky heart]

Yeah, I really have trouble figuring out what technique to use. I ended up reading about the "Stars and Bars" theorm and using that.

But tbh I'm pretty lost

[regal bane]

Oh yep that's why 12

So okay, you took the 5 vanilla you have to take, then split into cases

Take no chocolate? Then you have 3 choices that can repeat and you have to take 9 items

[smoky heart]

Yeah instead of using stars and bars is there a better more fundamental approach?

I haven't really learned that yet

And found it from just trying to get help online

[regal bane]

This is a very good question for stars and bars lol

[smoky heart]

Oh lol

Then I'll do stars and bar s

I eman it never hurts to learn

mean*

Could you walk me through the stars and bars then?

So initially we have 14 that we want to select from.

[regal bane]

We take 5 vanilla. No choices are made here. We still need to take 9 more. We split into cases:

• How many ways if we agree to 0 chocolate?
• How many if we agree to 1 chocolate?
• etc.

It looks like that's how you did it

[smoky heart]

Yeah I was kidna just following the online post

Where I basically held one constant

and then went down

[regal bane]

But there's 11C2 ways to take 9 donuts, given you can choose from 3 types

As it's 2 bars and 9 stars

[smoky heart]

just watched a video on stars and bars again

[smoky heart]

okay gotcha, so it looks like using stars and bars it's

9 + 3 C 3
8 + 3 C 3
7 + 3 C 3
6 + 3 C 3
5 + 3 C 3

and adding these up should give the final soln

it's decrementing b/c i'm holding 5 constant for the vanillas and then chocolates is being subtracted form 4

[regal bane]

There's 11C2 ways to choose 9 donuts from 3 types

[smoky heart]

or is my bars wrong in that i only need bars between chocolate and vanilla

and don't care about the randoms

[smoky heart]

oh i see so i should have 2 bars then

[regal bane]

For discussion's sake, I'll call the other two types "mint" and "strawberry"

So for the case where there's no chocolates being chosen:
We are choosing 9 donuts between vanilla, mint, and strawberry. Note that we've already decided chocolate is off the table

That's 9 stars, 2 bars.

11C2

[smoky heart]

oh i see

i think my misunderstanding is the use of the bars

so the bars are to divide/control a variable to a specific number?

[regal bane]

In this case I'm using it to divide groups

[smoky heart]

sorry i'm drawing this out on paper

i see!!!

[regal bane]

You can imagine the stars are literally donuts and the bars are separations between them

[smoky heart]

okay so
xxxxx | |
V C M S

like that?

yeah

[regal bane]

We've already decided to take 5 vanilla even before we started with stars and bars, so they won't be involved

The stars and bars are meant to count the next 9 we take

[smoky heart]

oh

so the next nine should be this then:

|| xxxxxxxxx
C M S

[regal bane]

Yeah that's a good case! It would signify that we just took 9 strawberry

[smoky heart]

i see

but it could also be some amount of mint and strawberry

[regal bane]

Our total would be 5 vanilla, 9 strawberry. 14 donuts

Another case is
xxx|xxx|xxx
signifying 3 of each

[smoky heart]

right

[regal bane]

That's V M S, not C M S. This is the case where there's 0 chocolate being taken

[smoky heart]

okay gotcha

[regal bane]

We can step into the next case, where we agree to take 1 chocolate. Then, we choose 8 donuts from the other 3 types

There's 10C2 ways to do this.

[smoky heart]

this is assumingw e've taken out the 5 vanilla right?

[regal bane]

11C2 + 10C2 + 9C2 + 8C2 + 7C2 is what I'm finally getting at

Yep. That was before doing anything

[smoky heart]

gotcha

[smoky heart]

oh i see, do I need to reinclude the 5 that we pulled out at the start

[regal bane]

We're counting the number of choices we have to make.

There's no choice involved in the 5 donuts at the start

So we've already included it, so to speak

[smoky heart]

gotcha

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[lunar kite]

i'm confused with these two formulas. well i know that the first one i can find std dev by finding the square root after. but it's the right hand side i'm confused why they're different

[finite mist]

Hope this help

[lunar kite]

"well i know that the first one i can find std dev by finding the square root after. but it's the right hand side i'm confused why they're different"

[finite mist]

Which part

[lunar kite]

for the second one, shouldn't it just be this?

i found two different formulas so i'm confused

here, why are these two different?

[lunar kite]

they are the same?

also what are the | bars for after Σ?

[finite mist]

probably absolute

[lunar kite]

so they are the same formula?

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[lunar kite]

<@&286206848099549185> if someone can please confirm for me

or @finite mist if you're still there

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@lunar kite Has your question been resolved?

[ripe bison]

Type the question please.

[lunar kite]

Are these two formulas the same?

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@lunar kite Has your question been resolved?

[worthy bear]

Yes

It’s the same

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[vocal sleet]

Please don't occupy multiple help channels.

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[gentle umbra]

did i do this wrong

[gentle umbra]

i took the derivative, plugged in the xyz respective values, and put it into the form x+x't

in x for example x=2 and x' is 2/sqrt7

[spiral inlet]

x=2, but you plugged in 2 for t

x is 2, but t is not 2

[gentle umbra]

so i leave the entire derivative as the slope?

[spiral inlet]

No, not quite

Look at the original parametric equations

[gentle umbra]

oh

[spiral inlet]

What value of t would correspond to the point (2, ln4, 1) ?

[gentle umbra]

oh

so i set it equal to 2 and solve for t

and then plug t in?

[spiral inlet]

That works, but all three coordinates need to have the same t

[gentle umbra]

oh so it’s just 1

cause z=t

[spiral inlet]

Yeah, t=1 at that point

[gentle umbra]

alright thanks i’ll try again

[spiral inlet]

and you can do a quick check to make sure that it works for x and y as well, which it does

yeah no problem 👍

[vocal sleet]

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[median echo]

hello, how do i multiply higher level logarithmic functions with non linear logarithmic functions composed within exponential function along with a rational function added to the side. my teacher said i multiply and then i add but i dont understand?

[thin vale]

let me ask my sir @sly sierra

zEIC

[sly sierra]

zEIC

[median echo]

?

is that mnemonic

[nimble jungle]

zEIC

[lime gorge]

stop asking questions, just feel the math and you'll understand

[thin vale]

xEIR

[sly sierra]

Explaining "zEIC" to Albert Einstein would require breaking down the concept in a manner understandable to someone with his level of expertise in physics.

Assuming "zEIC" refers to the "Electron-Ion Collider" (EIC), I'd explain it as a cutting-edge particle accelerator designed to collide high-energy electrons with heavy ions (such as gold or lead nuclei) to study the fundamental structure of matter. I'd emphasize its role in exploring the strong force that binds protons and neutrons within atomic nuclei, potentially unraveling mysteries of nuclear physics and the behavior of quarks and gluons. I'd relate it to Einstein's own contributions to understanding the fundamental nature of matter and energy, highlighting how the EIC builds upon and extends the legacy of scientific inquiry that he helped pioneer.

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[odd matrix]

can someone explain this last step

[odd matrix]

how do they know its concave down and how do they know the value is max

HELP!

[rare gate]

second derivative is negative so its concave down

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[mystic forge]

hello, I was struggled on finding a function of order n, for every positive integer n. can someone give some hint? here the order refers to fofofo...f(x) equals x, n is the number of composition

[mystic forge]

like this

[regal bane]

Consider functions of the form:
f(x) = (ax + b) / (cx + d)

It's surprisingly helpful to know linear algebra in this case, if you do

[mystic forge]

my linear algebra is not very good

[regal bane]

All good, still very do-able

[mystic forge]

I'll give it a try now

thx

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[restive crag]

I missed math this Monday and now I have a quiz tomorrow and I feel like my math brain fell asleep so I need help remembering how to do this. We did it a week ago but I haven’t practiced.

I’m attaching the problem I’m working on along with the steps, but it isn’t clicking for me for what I should actually do for step a

I’m confused on how to figure out the end points and critical points on f’(x) <= 0 for all [a, b]
I don’t really understand what this is telling me or what to do with this information

[dull bear]

can you say what type of function f is on the interval [a, b], if it has its first derivative non-positive on that interval?

[restive crag]

Oops sorry didn’t mean to send that omgosh

Uhhhh, I’m thinking

I want to say exponential but I don’t think that’s right, next I wanna say is quadratic but idk how i would know

[dull bear]

it happens, don't worry

[dull bear]

Welllll not necessarily either of those, I'm looking for a simpler description of a function whose first derivative is always non-positive (sorry also mistyped originally, I meant non-positive, trust me to want to say "negative" and then change my mind to "non-positive", but to mess it up )

[restive crag]

Oh haha ur good, so it looks like the function is decreasing with the highest point being 0 on this interval

Is that the answer you’re looking for?

[dull bear]

"the function is decreasing" is exactly what I wanted, yes

[nobody saw that ]

Or at least non-increasing, so basically from x = a to x = b, you're "going down" (or at least, you'll never go back up)

So you know at least one point where you have an absolute maximum, and one point where there's an absolute minimum

[restive crag]

Sorry someone came up and was talking to me, I understand the absolute max being 0, but not the absolute min

Cuz couldn’t it just go all the way to negative infinity?

[dull bear]

It wouldn't: you have a closed inverval [a, b], so the function must be bounded

However, there's nothing to say that the absolute maximum must be zero, nor what it actually is, afaik?

[restive crag]

How would I find what the function is, so that I can know what the interval is?

[dull bear]

Well you don't need to know what the interval or the function exactly is, you can state facts in general based on "a" and "b"

[restive crag]

Oh

So more generic answers than actual numbers?

[dull bear]

There are some "simple" examples of what f could be, and a few more "flavoursome" examples too

[dull bear]

Yep exactly

[restive crag]

Okay omgosh bless. So the absolute max would be B and min would be A?

Oops hehe

[dull bear]

Other way aroud

[restive crag]

Okay!

[dull bear]

Hehehe I think I've infected you now

[restive crag]

it happens

[dull bear]

But yea, the absolute maximum must occur at x = a [at least], and the absolute minimum must occur at x = b [at least]

[restive crag]

And there wouldn’t be any critical points since they didn’t tell us in the problem

[dull bear]

Cause from there you're "going down" and all

[dull bear]

There may be, they say that f'(x) <= 0, so there may be a point where f'(x) = 0, but you know that said point would only be local at best because of the decreasing-ness

[restive crag]

Okayyyyyy, I think i understand

[dull bear]

So, like, if I said that f(x) = -x^3 and on the interval [-3, 3], then you know the minimum is -27 and that's at x = 3, and the maximum is 27 at x = -3, but also that you have a critical point at x = 0, cause f'(x) = -3x^2

[you can also see that said f' is non-positive too!]

But of course at x = 0, you don't have a maximum or minimum

[restive crag]

Okay I have a clarifying question, with the interval being [A,B] wouldn’t the smallest number be listed first so that A would be the smallest/min?

Ik you said it was the other way around but my brain is just stuck on that, so that B would be the max

[restive crag]

I guess I don’t fully get this

[dull bear]

As in the domain is [a, b], the smallest number is listed first sure, but those are the inputs to the function

So the smallest x value is x = a, the largest x value is x = b, but nothing is said about what f(a) and f(b) would be

Of course, the image of f would now look like [f(b), f(a)] because f(b) is the smallest and f(a) the largest

[restive crag]

Okayyyyyy

[dull bear]

So they're asking you where the absolute maximum and absolute minimums are, they aren't asking what their values are (though of course you could find them in terms of the variables you have)

[dull bear]

Does that clarify your question?

[restive crag]

Gotcha yes tyyy

also, another question, if I have multiple math questions even if they’re different, should I close the channel and open a new one?

Also tysm for your help!!

[dull bear]

Depends, how different are they? Which ones were you thinking of asking?

Some people stay in the same one, sometimes they open new ones each time, it all depends really

[restive crag]

Well for question b ig, so it’s not too different haha

My brain is still sleeby

[dull bear]

In that case, you can stay in this one if you want (and are happy with me? )

[restive crag]

Yes! Omgosh tysm

So I rewrote the problem, and uhh

[dull bear]

[restive crag]

Then…. To find the end points, it would just be A and B

But for the max/min

[dull bear]

Well, for now, let's think critically (sorry that one was terrible, just had to )

So you basically told me that you know that g'(x) < 0 means that g is [now strictly] decreasing for those values of x where it's true

[restive crag]

Yeah that makes sense

[dull bear]

I'm sure you can tell me what g'(x) > 0 means, and what c being a critical number means?

[restive crag]

It means, that as the function is decreasing, at c it is undefined? Cuz a critical point is when the function equals 0 or is UND

[dull bear]

Oh yea sometimes people do that undefined thing

But it's more that the function's derivative is either zero or undefined

If it were zero then we would be happy

[restive crag]

Yea that’s what I mean

[dull bear]

If not, then

[restive crag]

😭

[dull bear]

For now, let's assume it's zero

What would that tell you about the point x = c? You have g'(c) = 0, and then g'(x) > 0 for x < c, and g'(x) < 0 for x > c

[dull bear]

[actually we might be more fine, cause they tell you the functions are continuous ]

[restive crag]

That…. Would tell me that… uhhh, C is……………….

I feel like it changes, like, the function isn’t linear and has a curve

[dull bear]

Well there could be a "curve" sure, and the function would certainly not be linear [throughout its domain]

[restive crag]

Wait would it be the first, where we find it not on the entire domain but an interval?

Between x<c and x>c? @dull bear

[dull bear]

We could do that we're dealing with the function on an interval after all

Remember that c is the only critical number on the interval [a, b], by the fact they said that any critical numbers mentioned are all that exist

[dull bear]

Also for the first part, I was looking for "g is [strictly] increasing"

So you have that g is continuous, and that when x < c, g is increasing, and when x < c, then g is decreasing...

So if you're increasing as you approach c, and decreasing as you leave c, then what are you at c?

[dull bear]

[restive crag]

We are at the highest max at c? @dull bear

[dull bear]

!rats

[vocal sleet]

[restive crag]

I love rats

[dull bear]

Yep, that's it x = c is the absolute maximum, because you're gonna be less than g(c) either side

[dull bear]

Heheh figured

[restive crag]

What would the absolute min be? Woukd it be a?

[dull bear]

Aha! Very good question

Wellllll, I mean, cause here we don't have the function g or the interval...

[restive crag]

so, therefore we don't have enough information to tell us

[dull bear]

Correct

[restive crag]

yayayayaya

[dull bear]

You could e.g. have -|x| on the interval [-2, 3], for which the minimum is at x = 3, or on [-4, 1], for which the minimum is at x = -4

With the critical point being x = 0 (the absolute value function is not differentiable at x = 0, but is differentiable everywhere else, and is continuous everywhere it's defined)

[restive crag]

wait where did all the numbers come from

[dull bear]

I made them up

[restive crag]

oh

[dull bear]

You can choose examples that prove your point

[restive crag]

that makes so much more sense

[dull bear]

The fact that I came up with those two examples allows you to see why you can't determine where the minimum would be, cause even the same function, but with different intervals, have them in different places (one has them at "a", the other at "b")

You could even make an example where the minimum is not at either a or b

[restive crag]

ohhh!! i get what you're saying

[dull bear]

So for (b), we have the maximum definitely occurring at x = c, because g is continuous (the fact it is matters!) and that you're increasing to the left, decreasing to the right
but we have no idea where the minimum would be, there are at least two different examples of where it could be

[restive crag]

gotcha! thanks for taking the time to explain it to me!

[dull bear]

A pleasure to on to the next one?

[restive crag]

yes pleasssee

I rewrote it just so we can see the problem easily

This is like the first one where we are on an interval [a,b] but unlike the first one, the function is looking silly and confusing. We also have the second derivative

[dull bear]

Yeeaaa this one is... "interesting"

It's doable, but kind of naughty

[restive crag]

ofc it is ahhh

the only time we use a second derivative is in the second paragraph in the summary

when we are finding global extremes over the entire domain

[dull bear]

the second derivative is useful in a couple of ways here

Well, the fact that we have a second derivative means we have a first derivative of h

[restive crag]

[dull bear]

Now, the fact we have a first derivative, can you say anything about h from that?

[restive crag]

it means... that, well, at least for the second derivative, the function is decreasing and never reaches 0

[dull bear]

Well, the first derivative is always decreasing sure, but "never reaches 0"?

[restive crag]

yeaaaaah because it only has < 0.... right?

[dull bear]

Well, erm, the second derivative is never zero sure, but the fact the second derivative is strictly negative doesn't mean the first derivative must never be zero, and in fact, if you let me put my Chamillionaire on for a moment...

[dull bear]

can you see what I'm trying to hint at with the reaction here?

[restive crag]

yesss... i get that just because the second derivative isn't negative doesn't mean the same for the first...

absolute full power mode is happening rn

the first derivative is increasing?

[dull bear]

Well, it won't always be

🎶 they see me Rolle-ing, they hating 🎶

[restive crag]

wait i don't get it

[dull bear]

alright maybe you might not have heard of it called that way

[restive crag]

no i haven't ahhh

[dull bear]

But Rolles theorem (also see Mean value theorem) tells you that if you have a differentiable* function h on an interval [a,b], and you have h(a) = h(b), there must be at least one point c in the interval (a,b) where h'(c) = 0

[restive crag]

o h

i've never heard of those

[dull bear]

interesting they haven't told you those then, cause that's quite important for arguing about the function

[restive crag]

he hasn't taught us a lot

[dull bear]

hmmm, how dare he not

[restive crag]

for reaaaal

[dull bear]

Anyways, you can trust me on that one, as you see, I never make mistakes

[the mean value theorem states that if you have a differentiable* function h on the interval [a,b], then there's some point c in (a, b) such that h'(c) = (h(b) - h(a))/(b - a)]

But besides, anyway, the fact is, we have [at least] one point c in between a and b where h'(c) = 0, and then we have that h''(c) < 0, so what can we say about the point x = c?

[restive crag]

would c be a critical point, and it would be the absolute... minimum?

[dull bear]

It would be a critical point yep, but it would be a maximum - the only one though, because we know that f' is always strictly decreasing, so that's the only point where it's zero