help-38

[near vessel]

do the same thing for equation 2

[hybrid bluff]

Ohhhh that makes sense

[hybrid bluff]

Ok

[near vessel]

choose any point, sub the values of x, y, and m, then solve for b

and finally write the equation

[hybrid bluff]

Thankk you soo muchh

[near vessel]

got it?

[hybrid bluff]

I got the two equation and from there I can figure out if there is a solution or no

[hybrid bluff]

Yes

[near vessel]

yep

good job

[hybrid bluff]

I asked so many people none of them could help me. Thank youuu‼️🙏

[near vessel]

no problem

[hybrid bluff]

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[turbid gazelle]

what is the locus of z^2 = |z|^2 + 4

[supple saffron]

[turbid gazelle]

<@&286206848099549185>

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[deft storm]

Have you been helped @turbid gazelle ?

[strange ice]

so imput x+yi as z and expand?

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[steady turtle]

Hi, may I know is the answer for my highlighted parts correct?

For the complement of B, is there any?

And what is the meaning of [a.]?

The a here not refer to a.?

I thought just like this

[hallow spruce]

It refers to the subsets listed in part a

[steady turtle]

And also if B={a,b,c,d,e}, is there’s a complement? Since it contains all the elements in it, I don’t know how to determine

I’m taking this as minor, and because I didn’t really learn proving and logic, modern algebra is quite hard for me

[hallow spruce]

I assume you're referring to this example 8 at the top

[steady turtle]

Yes

[hallow spruce]

Well by the definition of the complement, B' contains every element in X that isn't in B

What elements are those?

[steady turtle]

Not for me🥲 the lecturer speaks too fast and assuming the students already know, skipped many parts but I’m not from math school

[steady turtle]

But X and B contains the same element, same number of elements

[hallow spruce]

Ok, so what elements are in X that aren't in B?

[steady turtle]

Empty set?

[hallow spruce]

Well the answer I was looking for was "no elements"

If we want to be precise, the empty set is not in X

But yes B' = empty set

[steady turtle]

Oh, I thought there’s an answer for this. Anyway, I checked the textbook, there’s a sentence write that empty set is a also a subset

[hallow spruce]

It is

[steady turtle]

So basically, no element is empty set? And empty set doesn’t not need to write in between {}?

Yea, it’s my first time to record video and trying to revise this subject myself

[hallow spruce]

A set with no elements is the empty set, yes. And not only do you not need to write it without brackets, you shouldn't

[steady turtle]

Okay thanks, after I revise and find something I’m not clear, I’ll ask again. There’re quite a number of parts which I can’t understand even I go through YouTube

[hallow spruce]

Fair enough, good luck

[steady turtle]

Hi, actually for AxB and BxA, it’s not equal right? Is the order of (a,b) and (b,a) matters in this case?

For me, my logic told me the order matters

[hallow spruce]

It says "ordered pair" so order indeed matters

[steady turtle]

Okay and normally for number of elements, we write modulus A or n(A) better?

[hallow spruce]

Either

#(A) is sometimes used too

|A| is best imo

[steady turtle]

Thank you so much for this moment

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[wraith hinge]

what exactly defines something like "the integral of a matrix"?

[wraith hinge]

because this is pretty nonsensical at first sight

[wooden plover]

they're differentiating and integrating component by component

what they plan to do with that idk

but that's definitely what they intend to say

@wraith hinge

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[meager widget]

What does it mean by factorise over irrational field? Can someone help

[meager widget]

And is the answer to what i means by factorise over complex field

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<@&286206848099549185>

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[idle pewter]

I found it 11

[idle pewter]

Found it 11

If you guys don't help me I will make an Rick n Morty season 7 episode 4

[near vessel]

how you got 11

show your work

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[hybrid wharf]

how it become the second line?

[sly pewter]

-(-f(x)ln(e^x))

ln(a/b) = ln(a) - ln(b)

then simplify

[hybrid wharf]

ohhh thankss

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[frozen nest]

How to do this?

[frozen nest]

I have no idea how to start

[sly pewter]

Is it a Taylor series thing?

[dapper swift]

Consider the Taylor series when x = 1

[dapper swift]

Yes

[frozen nest]

What should I do first?

[dapper swift]

Well you'll notice that the Taylor series for e^x is not the right one to use

[vague citrus]

think and make uese of compedendo and dividendo

[dapper swift]

The numerator has an even Taylor series and the denominator has an odd one

[frozen nest]

Yes

[dapper swift]

Which functions can you think of are related to e^x in this way?

[dapper swift]

Wtf no

[frozen nest]

Up one?

[dapper swift]

Yeah whichever ones come to your mind

If you're really stuck, ||observe that e^x + e^(-x) is even and hence there are no terms with odd exponents||

[rough goblet]

south ABCD

[vague citrus]

I mean why not?

[dapper swift]

Lmao wdym

[rough goblet]

always b clearing doubts

-austin, sometime

[dapper swift]

Ohhhhhhh

[rough goblet]

also idk what compedendo and dividendo is

[dapper swift]

It's a competition trick

If a/b = c/d, then (a + kb)/(a - kb) = c/d

That's one of the rules

[vague citrus]

nothing too complicated tho

[rough goblet]

what?

[dapper swift]

Ah wait I see what you mean

[solid kiln]

ren (what's lemonsaurus?)

[vague citrus]

The property I were to use is
if a/b=c/d then (a+b)/(a-b)=(c+d)/(c-d)

[rough goblet]

this?

[dapper swift]

Well it would be a good way to check the statement but unfortunately it's not a proof

[sly pewter]

What is the name of this

[rough goblet]

that... doesn't seem very right?

[dapper swift]

Componendo and dividendo

[sly pewter]

Ohk

[rough goblet]

i still think taylor series are best FYI

[vague citrus]

there is a basic proof for that

[rough goblet]

probably cosh and sinh

yep, cosh and sinh

[vague citrus]

The next step would be use of taylor series

[rough goblet]

no need

[frozen nest]

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[dapper swift]

Yes exactly

Long story short, it's (cosh 1 - 1)/(sinh 1 - 1)

[trail ingot]

[dapper swift]

The assumption is a/b = c/d

[trail ingot]

i was more just surprised this is an actual thing

[sweet blaze]

Next is friendo

Lol

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[wraith hinge]

im stumped on this

[shrewd anvil]

Do u know ur graphs? @wraith hinge

[wraith hinge]

i know the x and y axis and all that

[dapper swift]

Technically the graph could be anything

But actually just assume it's a quadratic in the form x = ay^2 + by + c

Then substitute each point (x, y) in

To get 3 linear equations in a, b, c which you can then solve simultaneously

[frozen plover]

yeah this question really isn't fair

it's some sort of -sqrt(x) if i were to guess

[dapper swift]

Also I think (3, 2) is actually the vertex (the x-increase is 1 then 3)

That will make it much easier

[frozen plover]

doing that would give you both halfs though

[dapper swift]

Yeah you would need to choose the lower half

[frozen plover]

probably gonna end up both b and c being 0 anyways

or just b 0

[dapper swift]

Yeah basically assume it's in the form $y = 2 - a \sqrt{x - 3}$

[solid kiln]

south

[dapper swift]

Then it's much easier

[wraith hinge]

i figured it out tysm

idk how to close

[dapper swift]

.close

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[dapper swift]

No worries

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[flat wing]

can someone help me oon this question please

[dapper swift]

!status

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[flat wing]

no idea

i under stand of nth roots of complex numbers work

but when it comes to these questions

im lost

[dapper swift]

Try drawing the triangle out

[trim grotto]

Ok, so your first step would probably be to draw out everything the question tells you on your own argand diagram and just seeing visually where B could be first

[dapper swift]

There are two possibilities for B cause you can either go clockwise from A, or go anticlockwise

[flat wing]

Something like this?

[dapper swift]

Yeah

Also a reminder that multiplying by a complex number with |z| = 1 is a rotation

You need to find the number such that when you multiply it by a point, it rotates it by 60 degrees

[flat wing]

?

Wdym

Like the argument?

I know the argument for ones of the angles

Is -pi/4

Or no?

[trim grotto]

Yes, that's true

[flat wing]

Does that help with the question? Or no. Since I’m still not seeing what I need to do

[trim grotto]

It does help, as since the triangle is equilateral you know the arguments of your possibilities for B need to have arguments of -pi/4 +- pi/3

[flat wing]

Oh ok

But why +-pi/3?

[trim grotto]

Because internal angles in an equilateral triangle are all 60 degrees, or pi/3 radians

[flat wing]

Ohhhhhh

Ok

🤦

Finally got the answer

Thank you for the help

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[remote narwhal]

Hi, I just want to make sure I setup the problem correctly.

[remote narwhal]

For part b

Also, is the 2 supposed to be out front?

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[feral grail]

how would I find the missing variable here

[feral grail]

I got stuck

anyone can help please?

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[zinc kiln]

Lecture material leaves something to be desired: AC Circuit Pictured. Nodal analysis equation with complex numbers: both polar coordinates and rectangular coordinates. Solved for V phasor but I couldn't replicate it by hand or with calculator.

[broken belfry]

This is your solution?

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[zinc kiln]

This is lecture material where the professor skipped over the hard part: rearranging for the V phasor without messing up the complex numbers somehow

It seems simple at first. There is only one unknown! Shouldn't I be able to punch it all into a graphing calculator?

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[left aspen]

What does the symbol in the middle mean?

[left aspen]

its not < btw

[frozen sluice]

angle?

[left aspen]

I think so too but I'm not quite sure how that works here

[left aspen]

Like how would I calculate anything with that expression

its apparently in polar cordinates

[frozen sluice]

well I assume you just plug things in, is there a question that you're supposed to answer?

[left aspen]

hm, not really, I guess

[left aspen]

do you know how/when this formula is used?

its an electricity formula

[frozen sluice]

well I can't really tell you without more context but I'm assuming the sqrt(2) is magnitude

[left aspen]

I see

hmmm

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[burnt spade]

hello, how can i solve this integral?
$$
5050 \frac{\int_0^1\left(x^{50}-1\right)^{100} d x}{\int_0^1\left(x^{50}-1\right)^{101} d x}
$$

[solid kiln]

Renoir

[burnt spade]

here's what ive done so far

$$
\begin{aligned}
& \text { Let } n=\int_0^1\left(x^{50}-1\right)^{100} \
& \Longrightarrow 5050\left(\frac{\int^1\left(x^{50}-1\right)^{100} d x}{\int_0^5\left(x^{50}\right)\left(x^{50}-1\right)^{100} d x-\int_0^1\left(x^{50}-1\right)^{100} d x}\right)=\frac{n}{\int_0^1 x^{50}\left(x^{30}-1\right)^{100}-n} \
& \int_6^1\left(x^{50}\right)\left(x^{50}-1\right)^{100} d x=x^{50} n-\int n \cdot 50 x^{49} d x \
&
\end{aligned}
$$

[solid kiln]

Renoir

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[night fox]

The question was:-
1)Determine how many of the smallest 2024 hewitt numbers are divisible by 216?
2)Either prove or disprove that there exist two hewitt numbers whose sum is equal to 9x2^k for some positive integer k

[neon dirge]

hm could you briefly provide a definition for them

[night fox]

The sum of three consecutive numbers cubes like 1^3+2^3+3^3

[lusty delta]

well this is interesting

[lusty delta]

my advice to the other person who asked was “try to express the hewitt numbers algebraically”

have you done this?

[night fox]

I indeed have done it to no avail as it requires us to check the first 36 even numbers in the equation by hand

[lusty delta]

then can u show what u did

[night fox]

So n^2+2 can never be 72 and hence we know that numbers 72-2025 will have 28 numbers divisible by 216. Which requires us to check the first 72 numbers

But since odd numbers won’t satisfy this condition anyway we only check even numbers.

Also for these questions we had about 20 minutes so there must be an obvious solution

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<@&286206848099549185> ?

[night fox]

<@&286206848099549185>

[velvet vector]

hi

[night fox]

Um I mentioned the question above

[night fox]

<@&286206848099549185>

[supple saffron]

hi

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[night fox]

This one

<@&286206848099549185>

[night fox]

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[civic bolt]

I forgot how HL works so i took a guess, is this right?

[gritty harness]

HL is when hypotenuse (H) and one of the 2 legs (L) are equal on both triangles

Which is true for this question so yeah

[gritty harness]

Only applies on right angled triangle

It's basically SAS

[civic bolt]

wheres the other side?

I thought they have to be consecitive

[gritty harness]

AB and ZY both are marked by 1 cut

That means they're equal

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[short trellis]

numbers 1/2 , 1/6 , 1/12 is a arethmatic sequence (they are not necessarily nexto to each other) what is the biggest possible difference that the sequence can have

[tawdry apex]

1  1   1
-, -, --,
2  6  12


It kinda looks like a sequence where 2 previous numbers are being multiplied

[shrewd ridge]

this is something with gcd

like suppose it's 4, 39, 7004

then it must be 35

[tawdry apex]

not 6966?

the biggest difference in this sequence

[short trellis]

well thees numbers can be not the next numers

[shrewd ridge]

this is a random selection of a proper sequence where all differences are the same

[short trellis]

so 1/2 could be n1 and 1/6 can be n23

[short trellis]

yes

so how do i find it

[shrewd ridge]

the biggest difference would make 2 numbers next to each other

[short trellis]

wdym?

[shrewd ridge]

so like maybe it goes 1/2, 1/6 ... 1/12
then the difference is −1/3, but you can't get from 1/6 to 1/12 by doing −1/3

but the other way wortks

1/2 ... 1/6, 1/12
the difference is 1/12, and you can get from 1/2 to 1/6 by doing −1/12

generally you just find gcd of 2 given differences

for fractions 1/n it's the same as lcm of denominators

somehow

[short trellis]

wel thoes can also go like 1/2...1/6...1/12

[shrewd ridge]

yes

it's possible that you can't put them next to each orther, just in this case you can

and if you can, this must be largest difference

cuz it can't be even larger than that

[short trellis]

so what is it?

[shrewd ridge]

−1/12

because that's the difference between 1/6 , 1/12

it can't be larger, in absolute value

and it doesn't contradict 1/2 being in the sequence

[short trellis]

what if none o them are next to each other

the its not solvable?

[shrewd ridge]

whoops

yeah, biggest would mean smallest absolute value

but that's not solvable i guess

this is always solvable

if you;re looking for biggest difference

but here it could be interpreted as looking for smallest absolute value

because we see it's negative

maybe the original question doesn;t have this problem

anyway, if you can;t put them next to each other, it's still solvable

for example 4, 7, 12 they can't be next to each other, but difference of 1 is possible, that's the largest

it's gcd(3,5)

[short trellis]

so what is the biggest diffrance in my problem

[shrewd ridge]

there's no limit to the "biggest" difference

because the difference is negative

if we assume we can rearrange the numbers
1/12, 1/6, 1/2
then 1/12 works

if the question allows rearranging, there's no problem

[short trellis]

is it infinyt then?

[shrewd ridge]

0 rather

like −1 / 17348927348927348972389 is larger that − 1 / 748239

smaller absolute value gives larger number, with the largest limit of 0

[short trellis]

ok thanks

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[wraith hinge]

Find the standard equation of circles with a radius of 2 that are tangent to both the vertical line ( x = -1 ) and the horizontal line ( y = 3 ).

any ideas to solve it ?

[frosty slate]

Its pretty easy

Usually the x^2 + y^2 = r^2 is the standard form

But u can modify it

To this form

Where

[wraith hinge]

yeah

[frosty slate]

when a and b shift the center of the circle off each x and y axis

See how as b = 1, the center gets offset by the y axist by 1 unit

@wraith hinge

U should use this form instead

[frosty slate]

Then just adjust a and b such that it satified this

[wraith hinge]

i know

[frosty slate]

👍

[wraith hinge]

but how to find the center ?

of the circle

[frosty slate]

U have the radius

And the fact that the circle tangent to verticle and horizontal line

which mean...?

Well, I expect u should figure it out

[wraith hinge]

center is r distance from both points ?

[frosty slate]

U getting there

Yes

[wraith hinge]

so it will be
( -1 + 2 , 3 + 2 )
?

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[sick bramble]

need help understanding how the numerator and denominator both suddenly got (x-1)

and what happened to the 1

[dapper swift]

1 = (x - 1)/(x - 1)

[dapper swift]

So the first step was that they substituted f(x) = (x + 1)/(x - 1) into the inner part of f(f(x))

So you have f((x + 1)/(x - 1))

Replace x with (x + 1)/(x - 1) wherever x appears

[sick bramble]

so with the numerator it was (x+1/x-1) + (x-1/x-1)

oohh

u see

i*

what about the x-1 denominators? what did they do to them?

@dapper swift

[dapper swift]

So they just multiplied by (x - 1)/(x - 1)

That's how you know they cancel out

[sick bramble]

i see, thanks!

[dapper swift]

Npnp

[sick bramble]

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[proper shore]

Find the shortest distance from the point with coordinates ( (3; 1; 7) ) to the point, the coordinates of which are positive and satisfy the inequality

[
\frac{3x}{y} + \frac{y}{21z} + \frac{7z}{x} \geq 3(1 - (y + 7z)^4)^{1/4}.
]

[solid kiln]

kitten

[proper shore]

@dapper swift

@lime yew

[dapper swift]

Please don't ping random helpers

And also only ping helpers after 15 minutes

[proper shore]

Ok

[dapper swift]

Also you've asked the same question three times by now

This question is hard

You could ask in #competition-math where it won't get buried when the help channels closed

But I can't guarantee you will get an answer

[proper shore]

Ok

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[old lance]

seems like a minimization problem

like you have to find the minimum of a function but I dont know how you would get that function from the info given

you'd have to find the minimum of the distance function between the point (3, 1, 7) and the inequality given, But I haven't taken multivar calc yet so I can't help much

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.close

.close

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[sour talon]

is there an explanation

[marsh forum]

taylor expansions

,w series expansion of ln(1+x)

[solid kiln]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[marsh forum]

there you go

[marsh forum]

you still here?

[sour talon]

i understood but ln(1 + u) = u for the first derivative only

[marsh forum]

yes, but this is the taylor series

[sour talon]

yeah yeah but it still an approximation

[marsh forum]

it is, yeah

[sour talon]

yeah thanks for help !!

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[balmy spear]

Was trying to figure out a way in which I could get you to the solution a bit more slowly but you will either know that rule or you wont

If you didnt before, now you do

[past mountain]

Oh sweet thank you

[safe apex]

👍

your welcome 😂

[past mountain]

Hold on how would I plug this in?

[balmy spear]

Well your angle is angle 1 in the picture

X is the big arc, Y is the small arc

Pick the numbers appropriately and bung it in ye olde calculator

[past mountain]

Oh right on thank you!

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[slender pasture]

need help with this factoring and solving quadratics problem im stumped

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[past mountain]

I can help with that

So when doing these type of questions

A quadratic question like that will result in (2x - 3) (x - 1)

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[past mountain]

Can someone fact check me? I think it's 20.9

[potent coral]

If in cm^2 then you're right

[rapid dock]

whats the question

[past mountain]

'Find the area of the blue sector"

[rapid dock]

alright

well the radius is given

means the diameter is 8

of the full circle

and if the angle is 150

then the other angle is like

210

[past mountain]

Oh crap

You're right

[rapid dock]

and circumference is pi times diameter

the circumference is about 21 or smt

[rapid dock]

and youre wrong about..?

[past mountain]

Never mind sorry

I thought the wrong thing for a sec

[rapid dock]

how did you get 20.8

or 20.9*

@past mountain

[past mountain]

Yea sorry, I was getting water

[past mountain]

A first I thought the 4 cm was 4 cm^2

[rapid dock]

oh alr

[potent coral]

What

The formula is .5r^2 theta

[rapid dock]

but whyd you think that

[potent coral]

How's it not 20.9

If cm is used for radius

[rapid dock]

did i say it wasnt

i was asking how he got it

the answer

to see his work

[potent coral]

Oh

[rapid dock]

;/

[past mountain]

Sorry I closed my tab that had my work :(

[rapid dock]

oh dan

dam*

[past mountain]

So It's not 20.9

[rapid dock]

it is

[past mountain]

Oh

[potent coral]

It's not if you're to use m^2

Depends on the unit

[rapid dock]

well yeah but it doesnt say ^2

thats why i asked how he got the answer

if he thought it was 4cm ^2

[past mountain]

On the topic of this

This would't happen to be 4.6

[potent coral]

True

[past mountain]

Yesssssss, thank you!

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[frosty glacier]

Given $(ax)^{\log a} = (bx)^{\log b}$, prove that $x = \frac{1}{ab}$

[frosty glacier]

I tried solving for x but it didn't quite work out...

[viral merlin]

a^log b no?

[solid kiln]

Sb

[frosty glacier]

Oh sorry

[viral merlin]

you can just take log on both side 3rd step

that should do the trick

[frosty glacier]

I do not get it, what am I supposed to do?

[viral merlin]

a = b
log a = log b
on third step
did it not work?

dont tell me i tunnel visioned this again , lemme try on paper

[frosty glacier]

You don't know if a=b...

[viral merlin]

no no not the variables

i meant to take it as an example

i was reffering that take log on lhs and rhs

[frosty glacier]

oh ok]

[viral merlin]

yeah and that seems to work

try

[golden sparrow]

@frosty glacier how much is a^(loga)?

[frosty glacier]

$\log (x^{\log(\frac{a}{b})})= \log(\frac{b^{\log b}}{a^{\log a}})$

Like this?

[frosty glacier]

wdym

[solid kiln]

Sb

[viral merlin]

yepo

as we know log x^alpha = alpha log x

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[gray rapids]

any luck?

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[tulip valve]

is this correct

[lunar galleon]

it says less than 49

[tulip valve]

oh so im assuming B?

[lunar galleon]

year

yeah

[tulip valve]

appreciate uuuuu thanks

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[pastel linden]

i know i have to use the chain and quotient rule to get started but idk how to apply the quotient rule with the square root and simplify with the chain rule

[ionic pendant]

you may want to start by simplifying the expression (log rules) before taking the derivative

[pastel linden]

i tried that but got to the wrong answer

i can show my work

gimmie a sec\

idk if i cancelled out correctly

[rugged latch]

u changed the 4x-2 into 4x+2 line 3

[pastel linden]

oh

ohh

so its 36 at the top instead of -10?

and now i just use the chain rule?

i looked it up but idk how the derivative of that turned into the right side

welp guess im cooked

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[lofty stone]

Can someone help please with confirming an answer

[lofty stone]

Is this e?

[split chasm]

can you show your work

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[lofty stone]

I dont have any work i j did it mentally

[split chasm]

do proper work on paper

that'll give you a clearer idea of the situation and whether what you're doing is valid / mm akes sense

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[late beacon]

I just don’t know how to do it

<@&286206848099549185>

<@&286206848099549185>

[wraith hinge]

Hey

[wraith hinge]

First you need to figure out the bounds of integration

@late beacon

[late beacon]

Oh thanks

That’s so helpful

Please don’t do it step by step and walk me through it over the course of an hour

I don’t learn that way

Pls just show me how to do it

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[brittle pier]

could someone help me wrap my head around mathematical induction?

[brittle pier]

im stuck on proving k+1

This is where im at. Now, i dnt know what to do after this step after replacing

[neon dirge]

your last estimate is insufficient

3*3^(K+1) is good

Because you can turn it into 3^(K+2)

but for the first part you want to get (K+1) * 2^(K+1)

to fulfill the induction

so firstly you can take a 2 away from the four:

4K * 2^K = 2K * 2^(K+1)

and then use an estimate:

≥ (K+1) * 2^(K+1)

and then you have it in desired form :]

And the proof is done

[brittle pier]

could you possible break it down a bit more

when you mean take 2 away from the four

is that on the left hand side where i highlighted yellow?

[neon dirge]

Well you had 4K*2^K + 4*3^(K+1) right

[brittle pier]

yes

[neon dirge]

and you want to get
... ≥ (K+1)*2^(K+1) + 3^(K+2)
at the end right?

[brittle pier]

right to match what i had in P(n=k+1)

[neon dirge]

except with K+1, exactly

so the steps would be:

4K * 2^K + 4 * 3^(K+1) >
4K * 2^K + 3 * 3^(K+1) =
4K * 2^K + 3^(K+2) =
2K * 2^(K+1) + 3^(K+2) ≥
(K+1) * 2^(K+1) + 3^(K+2)

sry for typing like a sloth, am on phone

[brittle pier]

no worries at all, give me a sec to read it

in that second step, where does the 3 come from?

4K * 2^K + this -> 3 * 3^(K+1) =

how do we go from 4 -> 3

4K * 2^K + -> 4 * 3^(K+1) >
4K * 2^K + -> 3 * 3^(K+1) =

[neon dirge]

Well we estimate

4>3

you can estimate wherever you like

[brittle pier]

oh so really, i can put whatever i want as long as less than above?

omg

[neon dirge]

yop

that's the core principle of estimation

you can estimate whatever you want

BUT

you need to end up where you want to be

so if you throw away too much dough there may not be enough anymore to bake the cake

what a shitty analogy

[brittle pier]

💀

[neon dirge]

It's more like, in many proofs in analysis

You constantly need estimates like here

And you are allowed to estimate as you wish

But often you are very restricted in how much you can estimate

[brittle pier]

i see

[neon dirge]

which becomes kind of an artform

to be very delicate with the estimates

[brittle pier]

on the last 2 steps, how do you

2K * 2^(K+1) + 3^(K+2) ≥
(K+1) * 2^(K+1) + 3^(K+2)

[neon dirge]

We know that K≥3 right

[brittle pier]

right

[neon dirge]

which means 2K will always be greater than K+1

because if K≥3

then:

K+K ≥ K+3

[brittle pier]

oh wait, i think i see now lol

[neon dirge]

≥ K+1

[brittle pier]

i got confused but you were making the inequalities

i see i see

[brittle pier]

so

Then from here, you just pick the best one?

[brittle pier]

where did you come up with k+k >= k+3?

[neon dirge]

K≥3

:D

Add K

[brittle pier]

wow it really be that easy for you

i feel like im learning alien language

[brittle pier]

thank u sm !!

.close

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[neon dirge]

it's the language of perfection

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[cedar coral]

Hello, im trying to find the inflection points and concavity of relativistic mass
I already got the second derivative, but idk it seems wrong cuz desmos says there are no inflection points

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[tribal jetty]

what happened to c^4

mm

i dont think the simplification is correct

where are the denominators

[cedar coral]

my thought process was that if the denos were to be 0 it would be undefined
so like the only parts of interest are the numerators

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[cyan oxide]

u have to combine them into one denominator

$\frac{a}{b} + \frac{c}{d} = 0$

[solid kiln]

CowCat

[cyan oxide]

$\frac{a}{b} = -\frac{c}{d}$

[solid kiln]

CowCat

[cyan oxide]

$ad = -cb$

[solid kiln]

CowCat

[cyan oxide]

not juts $a + c = 0$

[solid kiln]

CowCat

[cedar coral]

ohhh okay gets tysm

did i do first division wrong

cuz like
-m0 and c^4 ... are supposed to be together

or nah

@cyan oxide

[cyan oxide]

both ways will end up at
1 + 2v^2/c^2 = 0

2v^2 = -c^

which is false, so the second derivate can not be zero

[cyan oxide]

and that matches which the graph

whats left is to find concavity

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[wraith hinge]

Pls help. 1. Idk where to start

[stone wagon]

for 28, think about the ways you can write out the formula in terms of (x+y1) * (x+y2) for y1*y2 = -12

[wraith hinge]

Formula? Which formula?

[stone wagon]

x^2 +bx -12 =

0

[wraith hinge]

Yeah

[lament plaza]

(x+4)(x-3)
where b= 1
so roots are -4 nd 3

[wraith hinge]

I can just set b to 1?

[dapper swift]

You can just sub each answer choice in, so if the answer were A then 3^2 + 3b - 12 = 0

Or 3b - 3 = 0, lmao

[wraith hinge]

Huh? How am I supposed to subtract 3b from 3? 3b is a variable right?

[dapper swift]

Well 3b - 3 = 0 means b = 1, and b is a positive integer

We're just substituting x = 3 in

[lament plaza]

i mean
i did by observation

say roots are x1 and x2

so x1.x2 = -12

x1 + x2 = -b

[wraith hinge]

How am I supposed to know if b is 1 😭 is it because it’s an integer?

[lament plaza]

i think the way the question is framed is asking you to observe

since it says

what could be the solution

[wraith hinge]

…..

Is there a way to find the solutions in another way?

Or is that the only way?

I’m supposed to figure out what b is equal to right?

And find the roots after

[lament plaza]

if you wanna show workings

you make cases

[wraith hinge]

?

[lament plaza]

first lets say
one root is 3

so you put x = 3

nd form an equation in b

then put x =4

[wraith hinge]

So u’re just suggesting substitutions?

[lament plaza]

yes

there's no other way

[wraith hinge]

If I do 3 then the equation is indeed true

[lament plaza]

the data would be insuuficient to deduce the value of b

[lament plaza]

then you plug in x = 4

b would be -1

which does NOT satisfy the condition for b be a positive integer

[wraith hinge]

Yea

So substitute is the only solution? Really?

[lament plaza]

yes

i dont really recall any other way

nor there should be any to my knowledge

[wraith hinge]

How about for number 25?

.

[lament plaza]

k must be 3

just expand

det = 0

since we have exactly one solution

[wraith hinge]

I still don’t get it

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[blazing geode]

heeyyyyyyyyyyy

[blazing geode]

hey guys

heyyyyyyyyy

[wraith hinge]

Someone help here please: #help-29 .

[blazing geode]

only matrices that possess same numbers of rows and columns are applicable of matrix multiplication.

Why I get a cross

[golden sparrow]

Because what you said is not true.

[blazing geode]

matrix multiplication works on every matrices regardless of their rows and columns

[golden sparrow]

That is also false. You should study this topic.

[dapper swift]

Write the number of rows and columns each matrix has, in that order

Then the middle two numbers must match to be able to multiply them

The outer two numbers are your dimensions of the new matrix

[blazing geode]

I see, thank you so much

[dapper swift]

No worries

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@blazing geode Has your question been resolved?

[blazing geode]

I'm wondering that if there's any intuitive way to think about matrix operation.

[supple copper]

Perhaps you should watch 3blue1brown’s series on linear algebra

It provides a fantastic geometric intuition for what matrices are doing

[blazing geode]

I see

[blazing geode]

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[shell basalt]

quick question

[shell basalt]

a perfect square always has a discriminant = 0... Why?

because a discriminant = 0 means only 1 solution, but perfect squares have two?

my question arises from this, where the solution is setting the discriminant = 0

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[delicate lance]

how do I derive this?

[wraith arch]

now this time we can rewrite this as $(tan(x^3))^2$

[solid kiln]

y0shi

[wraith arch]

and then we just have to apply the chain rule like two more times after the power rule

[delicate lance]

ohhhhh

is there other ways to do it bc i was seeing if i could find the derivative of tan^2x

i dont wanna know other ways ngl

[burnt spade]

chain rule

[burnt spade]

chain rule

[wraith arch]

well $tan^2(x)+1=sec^2(x)$

[solid kiln]

y0shi

[wraith arch]

if you really want

[delicate lance]

ohh ok

[wraith arch]

but it’s unnecessary

[delicate lance]

yea i dont wanna know that way

is this correct?

[wraith arch]

yep

[delicate lance]

tyty

[wraith arch]

yw!

[split chasm]

missing ) or extra (

[delicate lance]

oh ok

guys is this last step necessary?

this is what i did

[dapper swift]

How do you get $(\cos x)(x^3 - 2x + 4)$

[solid kiln]

south

[delicate lance]

chain rule

so derivative of sinx is cosx and then plug the x^3 - 2x + 4 into the x and then the derivatiev of that

[wraith arch]

yeah that’s right

maybe put an extra parenthesis around cosine

just in case

[delicate lance]

oh ok

[dapper swift]

Yes, so you shouldn't have cos x

You have cos(x^3 - 2x + 4): the inner function stays the same

[delicate lance]

ohohok

[dapper swift]

One version of the chain rule is d/dx f(g(x) = f'(g(x)) * g'(x)

[wraith arch]

oh wait

i didn’t even see that

[dapper swift]

Lol

[wraith arch]

good catch

[delicate lance]

guys what abou this one

this is what i did

[wraith arch]

that’s right

[wraith arch]

yep

[delicate lance]

thx so much

[wraith arch]

yw!

[delicate lance]

does this need to be simplified?

my teacher did this but i dont understand how he got the - sign

[split chasm]

there shouldn't be () around the cos in the initial expression

- sign comes from derivative of cos

[delicate lance]

ohh

did they do cos inverse of pi over 3 ?

this is a diff question

[split chasm]

no

no inverse here

just cos

pi/3 is a special angle

the cos of that is simply 1/2

[delicate lance]

oh ok

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[broken pond]

Do I find the area of the rectangular pizza and than find the square root of the area for the square side?

[outer rivet]

That sounds sensible

[broken pond]

378 x 168 = 63504

square root of 63504???

[dapper swift]

Yeah you should use a calculator clearly

[broken pond]

okay, the square root of that is 252

[dapper swift]

Yes

[broken pond]

so how much longer 378 is to 252???

[dapper swift]

Yes

[broken pond]

252 divided by 378 = 0.66666...

so 67%?

[dapper swift]

Wrong way around

Cause 378 is longer than 252, so your ratio should be more than 1

[broken pond]

oh yeah sorry, 378 divided by 252 is 1.5

150%

[dapper swift]

Nearly

Cause 100% is just the same as the original

A 100% increase is actually 100% + 100% = 200% of the original

[broken pond]

so it's 200%???

[dapper swift]

No 200% was just an example

[broken pond]

oh okay, thank you for your help

[dapper swift]

So it's 50%

a 50% increase

[broken pond]

yep, thanks for the clarification

[dapper swift]

Npnp

[broken pond]

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[flat wing]

how would i answer b

[flat wing]

have no idea what im suppsed to do

the answer for a

[frosty slate]

You have the answer already, what do u expect 💀

[flat wing]

for a yea

but there no point of looking at the answer for b

if it doesnt explain it well

and if i dont get it

[frosty slate]

Honestly, there isnt anything much to explain from the answer

I think maybe you need to pratice more?

[flat wing]

maybe

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[young arch]

.close

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[proper shore]

Find the shortest distance from the point with coordinates ( (3; 1; 7) ) to the point, the coordinates of which are positive and satisfy the inequality

[
\frac{3x}{y} + \frac{y}{21z} + \frac{7z}{x} \geq 3(1 - (y + 7z)^4)^{1/4}.
]

[solid kiln]

kitten

[proper shore]

@lime yew

[solid kiln]

Pratham_Shetty

[proper shore]

But I don't know this

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[wraith hinge]

Do you notice the terms x/y, y/z, z/x very nicely cancel out once multiplied?
And x,y,z are positive

Which inequality you can use for that?

[solid kiln]

icebal²

[wraith hinge]

Hint ^

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[open patio]

Two regular six-sided dice are randomly rolled once. Whats the chance of getting one 4 and one 5?

[open patio]

i get that it is 2/6 on the first but then idk

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[flat wing]

what does it mean by limting velocity?

[flat wing]

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[digital hull]

If $A = \begin{pmatrix} 1 & 2 \ 0 & 1 \end{pmatrix}$, find an invertible matrix $P$ such that $P^{-1}AP = \begin{pmatrix} 3 & 0 \ 0 & -1 \end{pmatrix}$.

[solid kiln]

Gasper34

[digital hull]

I have already tried to find it, but I forgot that P must be invertible

I tried doing PP^-1AP = PX (X is the result) so AP = PX

Is this ok or should I try another way?

[ionic pendant]

do you know about diagonalization?

[digital hull]

Nothing was ever said about that topic in class, if there is another way, it would be the best, if there is no other way, I listen to you

[ionic pendant]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/07%3A_Spectral_Theory/7.02%3A_Diagonalization

you will need to know how to find eigenvectors

[digital hull]

Give me a minute please

Thank you

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[flat wing]

why is x= 0 when t = 0 should it not be 10k?

[flat wing]

since initially it was 10k?

[ionic pendant]

what was the question?

[flat wing]

mb forgot that

here

[ionic pendant]

so x is the amount of contaminant

the reservoir starts out uncontaminated

[flat wing]

oh ok ty

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[icy nova]

is this simplification correct

idk how log2(z) * (ln2) simplified into lnz

[golden sparrow]

Use the properties of logarithms

[outer widget]

you can right log2(z) as ln(z)/ln(2)

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[wraith hinge]

Is this supposed to be a proof? I'm so lost....

[velvet bolt]

have a read then come back

https://en.wikipedia.org/wiki/Monotone_convergence_theorem

[wraith hinge]

Thank you.

[velvet bolt]

no worries

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