help-13

[fleet remnant]

shouldnt it be k^2 -k

i see it now

for some reason i thought that k was gonna be left

but instead 2 is left

[vagrant elbow]

nvm, you said you got it

😭

[fleet remnant]

for some reason i thought 2k / k is k hahaha

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[sweet flicker]

I have two radiuses. One is smaller than the other. Both create surfaces, which depend on their square times pi.
This means there should be a quadratic relation between the surfaces.
Instead of calculating the surface by both radii, I should be able to write something like this:

A_2 = A_1 ^ x

How do I find this x

[sweet flicker]

And in the same swing

if I write X²/Y

can I also write X/Y^-2

?

[iron cedar]

What exactly is the goal here, are we trying to get a ratio of these surfaces or something else, could you clarify?

The relation between the area of the surface and the radius is not linear, sure. But what does "having a quadratic relation" between surfaces mean?

I'm trying to understand your thought process.

[iron cedar]

For instance, if we know one radius is double the other, we can say that the ratio of their corresponding surface areas are the square of this ratio. This would make one 4 times larger than the other. But that does not really imply A_2 = (A_1)^x .

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[brisk nova]

.reopen

[brisk nova]

help please

[silver oxide]

rearrange

so that the sqrt is on the otherside on its own

[brisk nova]

ight

[silver oxide]

then square both sides

[brisk nova]

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[sweet flicker]

It's part of a whole more complex equation and I needed that difference of "change" over surface to be a potency for convenience.

Solved it by normalizing all values, getting X as log b (y)

where y is my entire equation and b is the normalized shenanigans

(I had an experiment, and later realized values were all equally wrong, but depending on a potency of one surface that is different)

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[crimson sedge]

(a) is a real, positive number. Break it down into two parts (a=b+c) so that its product (bc) has the largest possible value. (It's a really basic related rate problem, I know you have to find the maxima, but I'm not sure where to start anymore)

[silver oxide]

hmm

idk how to prove it but i know the values for b and c

[agile wyvern]

am gm inequality says

[crimson sedge]

I mean I know theres subtitution with (bc)

like since a=b+c than c=a-b and (bc) = (b(a-b))

[vernal minnow]

correct

[crimson sedge]

you than have to derive but I'm not sure where I'm going

[vernal minnow]

derivative of b(a-b) with respect to b

is what you're looking to do next

because at this point b is your only "variable". a is "constant" for purposes of this calculation, so you treat it like you would any other constant, say 3 or -1 or pi

and you just substituted out c so that's no longer part of your equation

[crimson sedge]

I found a/2

[vernal minnow]

that seems right

[crimson sedge]

I see

[vernal minnow]

so it's maximized where b=a/2

what's c?

also formally you'd have to check the 2nd derivative is <0 to make sure you're at a maximum and not a minimum

[crimson sedge]

c is half of a?

[vernal minnow]

yeah so they add up to a

[crimson sedge]

ahh I see

thank you

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[versed plaza]

i tried to solve this but i am getting my answer other way

i don't know how to use the hint and how the author wants me to solve

i tried to solve it by doing sum of N terms on both sides of the inequality |zn| <= an

then using triangle inequality and then taking lim N -> infinity

and since the real series of an converges it implied that the series for zn converges too

but idk how to solve it using cauchy criterion

this is from the book fourier analysis by stein

i tried to find solution online but there was none

[gritty viper]

,w Cauchy criterion

[wraith dagger]

Results provided by WolframAlpha

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Requested by a_mog

[gritty viper]

It doesn't seem to me like you've described a valid alternative solution

[gritty viper]

in any case, you should really be able to get started with this

You shouldn't be stuck right at the beginning

[normal isle]

Is that the math solver thingy?

[gritty viper]

Yeah

[normal isle]

Ok

I have seen some people use it

[gritty viper]

So what's S_j-S_i in this case

[versed plaza]

i actually know what the cauchy criterion is but i am not getting how to use it to prove it

even if i use the given definition of series convergence with the series for an i can't get forward

[gritty viper]

!show

[cedar kiln]

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

[versed plaza]

is the argument sound?

[cedar kiln]

@versed plaza Has your question been resolved?

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@versed plaza Has your question been resolved?

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@versed plaza Has your question been resolved?

[azure swift]

I dont think this is valid

youve shown that S_n is bounded

this does not imply convergence

this why the hint is also the better route imo

it sorta encodes the same sort of idea youre going for, but youre not using all the relevant information

[azure swift]

(imagine for example if S_n was alteranting between signs)

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[uncut phoenix]

can someone help me understand this

[uncut phoenix]

how do they know y^2 <= z <= 2x
and not 2x <= z <= y^2
and other limits such as x<=3 and not x>=3
the entire calculation is spread everywhere and haywire and I want to understand how to get it in order

when i visualize them using graphs its extremely easy to understand but i wont have desmos in an exam

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[uncut phoenix]

<@&286206848099549185>

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@uncut phoenix Has your question been resolved?

[bold kettle]

I found this one tricky to visualize also. I typically do these by looking at projections onto the xy, yz, or xz planes, but those aren't too helpful here at first. Fundamentally, we should know that z=2x and x=3 form planes and that z=y^2 forms a "trough" with a parabola for the sides. Let's take x=3 and draw what the yz plane should look like. You have the parabola y=z^2 and a horizontal line z=6 (because z=2x). The region inside these two curves is bounded below by the parabola and above by the horizontal line formed by z=2x. This can justify that the bounds for z are y^2<=z<=2x.

Now the problem bounds x. If we integrated x first, we'd simply say that x goes from the plane z=2x to 3, but we're integrating with z first, so let's substitute z=y^2 and say that y^2=2x, so x=(y^2)/2 becomes our lower bound.

For y, we have to notice that z ranges from 0 to 6. From the parabola, we get 6=y^2, so y=+/-sqrt(6).

[uncut phoenix]

ah i see

since y is being integrated last, it needs to range from a constant to another constant

so its like taking a huge cube, chopping it in different directions and shapes

you can achieve a shape by chopping in different orders

[bold kettle]

Exactly, yeah

[uncut phoenix]

hmm

[bold kettle]

Your final bounds should be constants since you're finding a volume

[uncut phoenix]

so i'd have to use mostly intuition for questions like these

visualizing the shape of the region

[bold kettle]

Yeah. I've done it a fair bit but I was having trouble with this one also

Just gotta keep practicing and the intuition comes

[uncut phoenix]

i see

thanks!!

ill just solve more problems ig

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[fierce laurel]

Where did I go wrong in this integration

the answer is 3pi/2

i got 4

[humble karma]

Something went wrong here :

You probably didn't apply the power reduction formula properly. You should be getting the integral of a constant at some point, hence the pi in the answer.

[fierce laurel]

correct

its 1/2 not 2/3

and there should be a negative sign in front of the first part of reduction formula when using sine

although wouldnt rlly matter here

thanks for your help like 3 days in a row lol

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[fierce laurel]

.reopen

ok uhhh

[cedar kiln]

✅

[fierce laurel]

i switched it to 1/2 and now my answer is 3

@humble karma thoughts, my good sir? 😭

[humble karma]

This is still wrong

It should be constant

[fierce laurel]

oh

it should be a 1 right

because 2-2 = 0 and anything to the power of 0...

oof

yeah i did n-1 instead of n-2

awesome, thanks. got the right answer!

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[sand patio]

in class whenever my professor would do cos(pi) it would = -1? but every time i do it in my calculator its 0.99849715

[mighty shuttle]

cos(π)=-1, yes

[sand patio]

oh yeah the unit circle

[mighty shuttle]

did you set it to radians ?

[sand patio]

nvm

yeah completely forgot for a sec

[dire thorn]

🤨

[sand patio]

yea lol

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[gentle pelican]

hi

[gentle pelican]

can someone help me this problem 😭

how can i explain this

@everyone

sorry for the ping but i rly need help on this

[thick heath]

I don't know if it works for you but you can use arccosine(sqrt(3)/4)), which gives you the angle and you can verify if it is within the interval and with that value use tangent to see if the value is positive or negative.

[bleak lark]

do you know that thetha is either in the third or fourth quadrant

and knows that the cossine is positive (First and fourth quadrant)

so theta is on fourth quadrant

and tan in fourth quadrant is negative

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[quick sand]

<@&286206848099549185>

[quick sand]

Question number 6

[ancient wind]

Rs Agrawal?

[quick sand]

Can anyone help me to understand the approach to solve this question

[quick sand]

Nope

[ancient wind]

Aight

[quick sand]

Can you help me

[ancient wind]

Nope

[crimson sedge]

is it just question 6 that you need help on?

[quick sand]

Yess

[crimson sedge]

could you first describe what you understand about the question?

[quick sand]

I want to understand the approach to solve this question

X1,X2,X3,.....X100 be positive integer

[crimson sedge]

right

[quick sand]

Xi +Xi+1=k for all i

[quick sand]

It mean that if I put any value in i then it will come out as k right?

@crimson sedge

[crimson sedge]

k is a constant

so

[quick sand]

K is constant yeah

We have X10= 1

Now we have to find X1

[crimson sedge]

yeah the value of X1

[quick sand]

But how?

[crimson sedge]

so do you think that X1 would have a lesser value of 1 or tell me why you don't think so?

[quick sand]

No I dont think that X1 have lesser value than 1 cause it is a positive integer

[crimson sedge]

I meaning that do you think that X1 would have a lesser value than X10 because X10 equals to one?

[crimson sedge]

ohh nvm I read it wrong lol

okay moving forward

[quick sand]

So X1 < 1 right

@crimson sedge

[bleak lark]

I think that must be it

[quick sand]

Yeah I was also think the same that option b is correct and you proved it

Thank you

[crimson sedge]

yes I meant that

[quick sand]

Thank you buddy for your help

[crimson sedge]

you're welcome

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[heavy crypt]

Which of the following statements is valid for any real number x < -1?

[heavy crypt]

i tried all this random crap

if u can read it

[atomic tendon]

Sure seems like C

[heavy crypt]

it is but why

[delicate elm]

sqrt((x+1)^2) = -(x+1)

[atomic tendon]

in the first one, A, those 2 things are identitical for x < -1

[heavy crypt]

yh

[silver oxide]

the square root of a square is just the absolute value.

[atomic tendon]

for B, since x is square alone, it will always get an additional +1 before you take the root, so it will always be positive and bigger than the the one of the left (since on the left, you're essentially losing 1

D is obvious

so C it is

[atomic tendon]

also this, which i forgot

[heavy crypt]

ok but i dont want to use the absolute value so i worked around it just squaring a lot

[opaque root]

If u expand inside sqrt of C u can realize 2x is negative

[atomic tendon]

then my logic is right

for A, those 2 things are obviously the same

[silver oxide]

x must be less than -1

question states x is a real number less than -1

[heavy crypt]

i tried to work c out tho but you'd get 0 < 0 so what does that even mean

[silver oxide]

c. is just the triangle inequality sqrt[(x+1)^2] = |x+1| which by the triangle inequality, will be less than |x|+1

[heavy crypt]

ok idk i kinda prefer my method and i forgot to put -x-1>=0 for A

and not just >

[silver oxide]

nothing wrong with ur method, just time consuming

[heavy crypt]

yh

[silver oxide]

for a multiple choice question like this, counter examples would work best

[heavy crypt]

do i just plug in values then

-2 for example

[silver oxide]

yep

[heavy crypt]

okok

im so used to writing everything out and not having multiple choice haa

[uncut mason]

x1+x2+x3+x4+...+x10 = 5k. x2+x3+x4+...+x9 = 4k. this means that x1+x10 = k. therefore x1 = k -1

correct me if i'm mistaken

[silver oxide]

well yeah, if u spot the triangle inequality then u instantly know its the answer, but like I said you'd have to spot it

[heavy crypt]

idk the triangle inequality

whatd you do

[delicate elm]

doesnt this just work

[silver oxide]

yes it does

[delicate elm]

and sqrt(x^2) would be -x

[atomic tendon]

You don't need even to think about it that hard, very clear to see that for A, you're essentially working with the same number (you're always going -1)

for B, super clear that the right side will be bigger since you're immediately squaring instead of making it smaller by +1 like the left hand side

D is beyond obvious why its wrong, since the left hand side will be a positive number

So C it is

[heavy crypt]

.close

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[heavy crypt]

i dont get this

where does this come from

okk

ig im still kinda confused bc some of these statements r true for x = -1

but like A for example doesnt that also just work for any x < -1

what did i do wrong for c tho

the domain is wrong should also be =

but other than that the actual calculations

.reopen

[cedar kiln]

✅

[atomic tendon]

just plug -3

in all of these

[heavy crypt]

actually

scrap everything ive done

for c if you'd do

actual calculations

then sqrt(x²) +1 >= 0

but that wouldnt work

like i get working w absolute values but what if u decide not to

then im confused

[atomic tendon]

are you trying to prove this

or what

[heavy crypt]

?

[atomic tendon]

like what are you trying to get

you know that C is the correct answer

you even know why

[heavy crypt]

can i not try multiple options?

[atomic tendon]

ok l o l

[heavy crypt]

its not that deep

yh

but for the right side youd do this basically sqrt(x²) +1 >= 0

but u cant solve that

but u can square it if u say x>=0

but that wouldnt work w the given

so is that y u cant

omg i cant type anymore

okok

thx for sticking around and helping

i mainly got confused bc normally we'd just say x>=0

for that part

but if the given is stating otherwise it probably would work even for other values

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[versed mural]

i really dont understand why i cannot do (x**+0)(x+**0)(x-1)

[bleak lark]

you can

[floral arrow]

Context?

[versed mural]

here ^

[floral arrow]

I don't see any x-0 or x+0 there

[tribal kite]

you can

[versed mural]

is (x-0) more common?

[floral arrow]

In any case, x+0 = x-0 = x, I don't know what you're trying to achieve

[bleak lark]

It doesn't matter if you are subtracting or adding 0

[tribal kite]

no there's literally no difference

[versed mural]

thanks!

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[odd plover]

can someone explain

[odd plover]

I can show what i got

but i dont know how

[tribal kite]

abc=40, 2ab+2bc+2ac=100

let a=2

you have 2 equations and 2 unknowns you can solve it with simultaneous eqns

[odd plover]

ok but how do u do the simultaneous

ive never done it with a value like abc = 40 cause

they're all multiplies

[tribal kite]

2bc=40

20/b=c

substitute this

4b+2bc+4c=100, 20/b=c

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[fierce laurel]

Where am I going wrong in my integration?

[fierce laurel]

The bounds of integration are correct, I checked the answer key.

[fierce laurel]

this guy again 😭

[fierce laurel]

you right

let me compute this now

thank ya G 🫡 appreciate it

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[tulip crow]

do i multiply both sides by -1/-1 ?

[lyric narwhal]

!nosols

[cedar kiln]

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.

[turbid ether]

mb mb

[gritty viper]

Factor both sides

[tulip crow]

i dont know how to factor the denominator

[turbid ether]

how would you factor (a^2 - b^2)

[tulip crow]

for the numerator?

x(x+3)(x-3)

[turbid ether]

for this

[tulip crow]

(a+b)(a-b) ?

[neat raven]

what's the question?

[tulip crow]

up

[turbid ether]

well first factor out the gce from the numerator

[tulip crow]

i did

[turbid ether]

then see how the bracket on the numerator and denominator are the same except one of them is x^2 - 9 and the other is 9 - x^2

lets say we have (x-1)/(1-x)

[tulip crow]

9-x^2 is diff from x^2-9

[turbid ether]

yeah im just saying theyre flipped around

do you know how you can cancel these out?

[tulip crow]

multiply by -1?

[turbid ether]

yeah

[tulip crow]

its answer choice 1

[turbid ether]

yeah

[tulip crow]

what if i was doing long divsion

[turbid ether]

wdym

[tulip crow]

the long polyonmial divsion

[turbid ether]

not sure

[tulip crow]

.close

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[neat raven]

calculate tan25 deg without a calculator.

[neat raven]

you can use tan60, tan90, tan0 and tan 45, and tan30.

[long arrow]

well, I don't think is a way to combine them to get tan(25)

if you don't need an exact value then I'd recommend

[crimson sedge]

Use taylor expansion?

[long arrow]

doing 180 deg = pi --> 25 deg = 25/180pi

and then plug 3.14 (or more accurate approximation) and calculate

[neat raven]

you just converted them in radians, didn't you?

[long arrow]

yes but it gives an opportunity to get an approximate value

[neat raven]

oh, i'll see.

[crimson sedge]

That's just the angle in radians?

[long arrow]

then Taylor or using tables of trig values

or some approximation for such an angle

[crimson sedge]

Taylor is just too chefs kiss

[neat raven]

So there ain't any direct way to solve. Okay.
👍

[long arrow]

By the way, it's an exact question you got?

or it's a part of some problem

[neat raven]

Oh.
it was.

If the base of a triangle is 25m, and angle of deppresion = 25 deg, find height.

it's approx 14m.

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[neat raven]

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[bitter widget]

1+5-59531967/465*64-6431+4655164/4642351255=????

[mortal hemlock]

,w 1+5-59531967/465*64-6431+4655164/4642351255

[wraith dagger]

Results provided by WolframAlpha

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Requested by flappie.

[neat raven]

have you encountered this in a question?

[runic garnet]

@bitter widget if you troll again…I will not hesitate to ban you…don’t think u can mess with me…I am the server king….dont mess around here…

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[untold shadow]

can some one explain how k=2 column is being filled ?

[untold shadow]

i have been tring to understand cross entropy, and my professor told me to understand this problem

i am confused what the table in the answer denotes and how entries 2nd column is decided

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[untold shadow]

<@&286206848099549185>

[carmine bronze]

The problem states that for k other than k=1 and k=n, the selected candidates is random. There is no rhyme or reason other than perhaps Rank 1 is chosen three times, Rank 2 is chosen two times, and Rank 3 is chosen one time.

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[shut wadi]

idk how to find x

the scale factor is 1.5

these are the similar triangles seperated

<@&286206848099549185> does nobody know how to solve this

this cannot possibly be an Einstein level problem that nobody can solve

[simple pebble]

x=10

probably

[shut wadi]

I just figured that out

but

through guessing

how do I actually figure it out through an euation or formula or something

[simple pebble]

I was getting food whenever this help channel was opened

the ratio of 5 to 3 should be the same as x to 6

since the denominator is doubled, double the numerator too

5/3 = 10/6

[shut wadi]

ok

[zealous fable]

just use ratio of similar triangle

$\frac{6+3}{6} = \frac{5+x}{x}$

[wraith dagger]

NKH

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[south terrace]

for this question, when I am solving for sin(x)=1/3 do I keep my calculator in deg or rad mode?

and at the end do I include the solutions for sin(x)=1/3 as well or only sin(x)=-1/2

[loud patrol]

its says 0 to 2pi so that means radians

[loud patrol]

they are both solutions so you should include both

[south terrace]

oh I see, should i include the solutions for sin(x)=1/3 in the conclusion?

i checked online, some include it some don't so i'm really unsure

[loud patrol]

yeah

its just like how if you have (y-2)(y+1) = 0

y = 2 and y = -1 are both solutions

[south terrace]

alright, i have another question too

so for this one as i am working on the inequality should it become x=1.28 or be x>1.28

[loud patrol]

its good to draw a unit circle for these

(pi-2)/4 is basically 1/4

in the first quadrant

cosine starts at 1

and then gets smaller and smaller going to zero

so an angle bigger than 1.28

will give a smaller cosine value

however

if you go far enough around

cosine will decrease again

like looking at this

which part of the circle do you want

[south terrace]

i'm not sure

if the final answer is an interval or what

i'm so sick of these questions

i don't know what to do

its the small things that are tripping me up

[loud patrol]

so you have 4cos(x)+2 > pi

which means you want larger cosine values

which you get on the right of those points

[south terrace]

this was my conclusion

in the beginning it was cos(x)>

and now its cos(x)=

[loud patrol]

its not cos(x) =

it should be cos(x) > (pi-2)/4

[south terrace]

oh alright, i will change it

so i should change this too

cos(x)>pi-2/4 is 0 for 1.28 and 5.00

[loud patrol]

i would just say cos(x) > (pi-2)/4 on [0, 1.28) U (5, 2pi]

[south terrace]

this was my conclusion

what should i remove?

[loud patrol]

change the first equals to greater than

[south terrace]

yes

anything else?

[loud patrol]

no looks good

[south terrace]

thank you so much

i appreciate it

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[spice grail]

Is x = 6? If so, why? I was told that it was because the distance between -1 and 5 is 6, but I don't know what the relationship between the value of x and that affirmation is.

[glad fog]

are you familiar with pythagoras' theorem?

[spice grail]

Yeah.

Oh wait bruh.

[glad fog]

you can split that triangle into a smaller right angle triangle

you know the value of h

you can also get the length of the new triangle via the length from the base from A to the foot of h

[spice grail]

No I don't. I know its in x = 2. That's it.

[glad fog]

A-B is a straight flat line

h goes from the y coordinate of the bottom line to the y coordinate of C

you can get h that way

[spice grail]

I don't know the y coordinate of C.

That's a b in the image, not a 6, just in case you thought it was.

[glad fog]

oh thats b

[spice grail]

Yeah.

[glad fog]

if you don't have the value of c you can't determine the value of x based on just the coordinates of A and B

[spice grail]

Then how can I know what the height of that triangle is?

[glad fog]

you can't, since thats b, and thats apparently not given(?)

can you show the original question?

[spice grail]

Sure. it's in spanish, so let me translate it for you:

[glad fog]

ill probably be able to figure it out

math is universal 😉

[glad fog]

ah

you have more valuable information

its an equilateral triangle

that means the length of each side is the same

that means x must be 6

and CB also must be 6

you can use pythagoras' theorem still to calculate the value of (a, b)

[spice grail]

Why is x 6, that's what I don't get, and that was my original question.

[glad fog]

thats a definition of an equilateral triangle

all sides same length

[spice grail]

Oh lol I just had a stroke.

Yeah nevermind I get it.

[glad fog]

👍

[spice grail]

Thanks~

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[regal finch]

hi can you tell me how to rationalise a number

[subtle raven]

can you give an example of what you mean?

[regal finch]

like to rationalise a number

[subtle raven]

aeiou

that's not an example

[regal finch]

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[crimson sedge]

Could someone check if this is correct?

[cedar kiln]

@crimson sedge Has your question been resolved?

[ancient lodge]

,w true or false -8(4x+3)^(-2)+10(5x+1)(4x+3)^(-1)=2(25x+1)/((4x+3)(4x+4))

[wraith dagger]

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Requested by civil_service_pigeon

[ancient lodge]

ngl I have no idea what you did here

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[humble karma]

You want $\frac{n^2}{n^2 + p^2 n }$ to be more or less constant for large n

[wraith dagger]

Azyrashacorki

[humble karma]

If you think of it in the "limit" sense, you want the power in the numerator to be at most that of the denominator ( a bit like when you do a limit with infinity / infinity )

So p^2 * n has to be O(n^2)

And thus p^2 has to be O(n)

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[limber snow]

Is this good enough for proof of correctness? Im not rlly good at these...

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@limber snow Has your question been resolved?

[limber snow]

<@&286206848099549185>

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[crimson sedge]

helloo!

[crimson sedge]

i cant understand how these work

can someone please explain it to me

thank youu

[summer otter]

what u dont understand

[crimson sedge]

im not sure how they get to that conclusio

conclusion

[clear umbra]

look up the double angle identities

[crimson sedge]

yes i just learnt them

[summer otter]

thenuse them

[crimson sedge]

OH WAIT

broooooo how did i not see it

its literally the identity backwards

omfg

sorry its been like 3 hours of math 💀

[summer otter]

lol

[crimson sedge]

i started when my ipad was on 2% and now its charged to 84%

thank oyuu

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[timid cloak]

I am supposed to decide which quantity is bigger A or B. However, I feel the question should be answered as "Cannot be determined" but the answer scheme begs to differ. Stating that the quantities are equal.

[undone halo]

"the least possible length in inches of the longest piece"

do you get what that means

[timid cloak]

I don't think I do cuz to me it could be 37 and the other 2 pieces could be 2 and 1

[undone halo]

i dont know if im reading it properly

[timid cloak]

But that logic doesn't seem to pass

[undone halo]

but it says 40 inches

not 30

[timid cloak]

Sorry 37, 2 1

[undone halo]

basically

whats the smallest number that the longest length can be

while still being the longest length

ex. 37, 2, 1 is valid but 36, 3, 1 is also valid

and 36 is smaller than 37

[timid cloak]

Ahh, I think I got it now

[fleet quarry]

hi

[undone halo]

actually

because different integer lengths

36, 2, 2 isnt valid

36, 3, 1 is

[fleet quarry]

just think which 3 numbers in row add together and makes 40

[fair geyser]

the idea is that you want to divide it in equal pieces, so it would be 13 13 14 with the highest 14, but you can't do 13 13 14 because they have to be different

so 12 13 15 is the exact way they are looking for

[timid cloak]

Ah so 12,13,15

[fleet quarry]

get the group that has the biggest result when u multiple them

[timid cloak]

This hurts my brain, doesn't feels like mathematics, just big intelligence checks which I am drastically failing lol

[fair geyser]

i agree

mathematics is not about this

[fleet quarry]

the longest length shouldnt change if its asking for the minimum length of it

[timid cloak]

What's the quickest way to arrive at this?

40/3 = 13.33 and them maniuplate the numbers

[undone halo]

if u want to have the least value for the maximum length

youd have to make it so that the other 2 ribbons would be as high as possible without being over the other ribbon length

[fleet quarry]

it isnt so difficult

[fair geyser]

i don't know

[fleet quarry]

the minimum length should be 1t

15

[timid cloak]

Yup I got it now

Thank you, all

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[knotty harness]

How do you give a linear combination of P1 and P2 which equals P3

[digital cliff]

you can solve aP1+bP2=P3

[knotty harness]

But how do u do that

[digital cliff]

well the first row would give a+2b=-3 for example

second gives b=7

its simple from there

[knotty harness]

b=7?

Isnt the second row

P2 + 7P3 = 6

[digital cliff]

no?

the 6 isnt even in the first three columns

which are what youre looking at

[knotty harness]

Its an augmented matrix?

Each row equals to

4, 6 and -3

[digital cliff]

no

youre literally just solving aP1+bP2=P3, it has nothing to do with the 4th column

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[neon fjord]

how u do dis🙁

[pulsar lily]

what have you tried?

[neon fjord]

i just did 180-84 but uhm the answer key said 76 im js confused on how it led to that

[pulsar lily]

for these problems we can ignore the multiple choice completely for sanity preservation

[neon fjord]

okok so what do we do for this q 😭

[pulsar lily]

it's a puzzle

you can either try stuff, experiment first

or you can review what you've already learned

[neon fjord]

alright ill try

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[icy mirage]

how do you simplify this

[blazing zephyr]

maybe you could make the single x interms of log_5(something)

[onyx idol]

Bring x down and log5 right?

[subtle harbor]

,tex .exp rules

[wraith dagger]

🫎 Moosey 🫎

[subtle harbor]

,tex .log rules

[wraith dagger]

🫎 Moosey 🫎

[subtle harbor]

well, there's one not listed, namely the fact $a^{\log_{a}(x)}=x$

[wraith dagger]

🫎 Moosey 🫎

[icy mirage]

can you give me a hint on which rules to use

[subtle harbor]

well what rule involves a sum in the expontent

[icy mirage]

the product rule of exponents

[subtle harbor]

mhm

so how else can you write 5^(x+log_5(x))

[icy mirage]

oh wait do i do 5^x + 5^log5x

[subtle harbor]

no...

that's not what the rule says

[subtle harbor]

look closely at this rule

[icy mirage]

isnt a = 5 and x=x and y=log5(x)?

oh 5^x * 5^log5x

oh i got the answer ty

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[simple torrent]

This isn't really a problem, why does AM-GM get tighter if you have less numbers? For example, with the sequence
1 + 2 + 1/x + x, applying the inequality would give 1 + 2 + 1/x + x ≥ 4 * fourthroot(2)
However, if you combine the 1 and 2, applying the inequality would give 3 + 1/x + x ≥ 3 * cbrt(3)

[simple torrent]

And plugging those two equations into the calculator, the second one would get smaller

If anyone could just point me to a resource that explains this I'd appreciate it

[gritty viper]

Really depends on what the numbers are

If you have 1 and 100 it isn't very tight

If you keep adding 100s it'll get tighter

[simple torrent]

Wait sorry can you explain what you meant again

I didn't really understand

[gritty viper]

AM-GM doesn't particularly get tighter or looser as you add numbers

It can do either depending on the numbers you add, as my example shows

[simple torrent]

Huh ok

[crimson delta]

well the second one is smaller which means that the first one is the better bound

[simple torrent]

Well

I'm currently trying to find the minimum value of a constant + a sequence of variables in fractions that multiply to 1

1/x and x is just the simplified version but if I applied the one on top I would just straight up be wrong

So I was kinda confused

[crimson delta]

well then you probably made a mistake

[simple torrent]

Yeah

[crimson delta]

you would of course get an even better bound for 1+1+1+1/x+x

[simple torrent]

Wait I'm so confused rn lmao

The sum of reciprocals is at least 2

So 2 + 3 ≥ 5

[crimson delta]

yes

[simple torrent]

So were both of my top ones wrong

[crimson delta]

no

the first one says (thing) >= 4.7, the second one says (thing) >= 4.3

those are both true

they just arent as accurate as they could be

which would be (thing) >= 5

[simple torrent]

OHHHH

= x just means it could be greater than x, but it'll actually be higher; we just don't know until we apply the better bound

But then how would you know you've already reached the best bound and have gotten the true minimum?

Since if I just took 4.7, even though (expression) >= 4.7 would be correct, my answer to the question (minimum value of expression) would be wrong

[crimson delta]

well yes thats the question

in general you dont

[wicked stag]

3+x+1/x minimum you need ?

[crimson delta]

well no thats obvious

I mean for other more complicated expressions

where you cant just differentiate and stuff

[wicked stag]

This is -∞

[crimson delta]

positive x

[agile wyvern]

am-gm has minimum when a=b=c=d...

so if you can set all your variable expressions equal then you can find your minimum

[crimson delta]

but often you cant have that

so you know that when you apply am-gm you lose some efficiency

[simple torrent]

So uh

Should I just expect to get some "find the minimum/maximum value" problems wrong

[crimson delta]

well presumably the questions you are asked are doable. so no

I was speaking more generally

[simple torrent]

Oh

Do you know what this phenomenon is called

[crimson delta]

which one

[crimson delta]

like this and what was mentioned earlier says, am-gm is more accurate if the numbers are closer together to each other. 1,2,1/x,x are "closer together" thant 3,1/x,x

[simple torrent]

Well, what you said below, how am-gm is more accurage if the numbers are closer together

Or wouldn't it be best to just leave integers out completely? Like just calculate the variables and add the ints at the end

I'm just gonna take a look at what the solution says lmao

Do you mind DMs to ask you more about this later, I think I'll just close this channel for now

[crimson delta]

it basically follows from the proof

but maybe a bit too technical for me to explain it well

maybe the n=2 case

you probably have seen the proof 0<=(x-y)^2=x^2-2xy+y^2=x^2+2xy+y^2-4xy=(x+y)^2-4xy and then rearranging, yes?

[simple torrent]

Yea

[crimson delta]

you'll notice that only at the first step you have an inequality

only at 0<=(x-y)^2

everything else is equalities

so the only "inaccuracies" you can get are from that first step

and that first step 0<=(x-y)^2 is more accurate the closer x and y are together

[simple torrent]

Oh ok

Ty

.close

[cedar kiln]

Channel closed

Closed by @simple torrent

Use .reopen if this was a mistake.

[cedar kiln]

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[autumn tusk]

i've got up to finding x = 0 and x = sqrt...

i have
x (sqrt(27-x^2-y^2) + 3x) = 0
x= sqrt((27-y^2) / 2)

but i'm stuck here

how did they found those other values? sqrt(27/2) and 3 ?

[cedar kiln]

@autumn tusk Has your question been resolved?

[cedar kiln]

@autumn tusk Has your question been resolved?

[autumn tusk]

<@&286206848099549185>

[autumn tusk]

ping me please if someone wants to help

[cedar kiln]

@autumn tusk Has your question been resolved?

[autumn tusk]

bad day

[azure swift]

Alternatively, have you tried using something like Lagrange multipliers?

[autumn tusk]

no i dont know what it is

[azure swift]

I see

[autumn tusk]

i think those exercises are meant to solidify what i've just learnt.

[azure swift]

Hm what about parametrizing the sphere

[autumn tusk]

but even reading the solution i don't get it

[azure swift]

then you'd reduce down to two variables

Oh nvm they pretty much do that

[autumn tusk]

i don't know, i feel you're taking me to another route here XD

i need to unstuck from whre i am stuck.

[azure swift]

Well, reading each step. Is there one in particular where you get lost in such case?

[autumn tusk]

yes, what i wrote above, have you read that? I'm stuck

[azure swift]

Yeah but you said you dont even understand the solution

So that cant only be it

[autumn tusk]

i've got so far

then i don't understand what they do to find those other numbers.

the part i've highlighted

[azure swift]

Alright, could you summarize the idea of the solution up to that point in such case?

(while this may seem tiresome it actually helps)

[autumn tusk]

yea

i 've stup the formula with 3 variables xxx+yyy+zzz

i need to find min and max of this.
i take advantage of the constraint that the sum of the squares is =27, solve for one variable and substitute that in the formula i need to find the min and max, so i have a surface and 2 variables

so now i can find the min and max of that surface, find the partial derivatives , equal to 0 to find the extrema

and i've only found the origin and 2 other formulas in terms of the other variables

[azure swift]

not extrema but critical points, i.e eventual extremas

[autumn tusk]

sorry, yes.

then i tought, boundaries.

0<= x, y, z <= 27

but since i have 2 vars i express z interms of x and y

so i get
0<= x^2 +y^2 <=27

part of me wants to find the value of the 2 partials along the boundaries now

[azure swift]

?

[azure swift]

How?

[autumn tusk]

sorry <= sqrt27

[azure swift]

I mean no? so youre saying x,y,z lie in [0,sqrt(27)]^3 ?

[autumn tusk]

i dont know what you did there

[azure swift]

what does this mean?

[azure swift]

this is me trying to parse your inequality

[0,sqrt(27)]^3 = [0, sqrt(27)]x[0, sqrt(27)]x[0, sqrt(27)]

i.e a cube with sides sqrt(27)

[autumn tusk]

i mean this

[azure swift]

yeah so not this

[autumn tusk]

this follows

[azure swift]

huh?

[autumn tusk]

0<=x^2 <= 27

right?

each number squared can go from zero to 27

[azure swift]

It depends, the reason it follows is not really the way in which you think so far as i understand you.

Try to picture the condtion instead

We have that x^2 + y^2 +z^2 = 27

so a sphere with radi sqrt(27)

try to imagine that

[autumn tusk]

ok

[azure swift]

but since x,y,z are non-negative

imagine that we only look at the first octant

do you see it?

[autumn tusk]

yes

[azure swift]

Okay, so the key here is that they let z be a function of x and y

so the domain of this function has to be x^2 + y^2 <= sqrt(27) with x,y non-negative

do you see why?

[autumn tusk]

oh ok now yea

it's a quarter of a circle

radius sqrt 27

[azure swift]

Yes!

[autumn tusk]

the surface covers all that area

[azure swift]

Can you picture it in your head?

Yes!

[autumn tusk]

ye

[azure swift]

Good, was that your intial argument with this ?

Or has this now been cleared up

[autumn tusk]

mmm. no i did this without thinking geometrically.

but i get to the same result i think

i just remove one variable in the end

[azure swift]

While some may say that pictures are non-rigoures, i think not; its a hidden secret. Use it to get a feel for the problem

its really useful

[autumn tusk]

but that one is not the function i need to find the min and max anyway

that gives me the bounds

[azure swift]

Hm im not following

[autumn tusk]

i need to find min and max of the sum of the cubes

[azure swift]

Oh, yes you mean the one, right yeah

well you see the condition has to be used somehow

and this is how we do it

now it is sort of baked in when we substitute z in the sum of cubes

[autumn tusk]

but ok now i have the bounds alright.

[azure swift]

Right yes

[autumn tusk]

and i also have some functions where i have the min and max.

which i don't really know what to do with them.

x = sqrt(27-x^2 - y^2 )
y = sqrt(27-x^2 - y^2 )

this comes from the partials. I get a zero at the origin and a zero at those points....

what do i do with those now?

[azure swift]

Zero with these?

[autumn tusk]

no, with the partials

this is what's left

[azure swift]

So let's try squaring these

[autumn tusk]

so i have zero if x and y are zero

when i manipulate the expr i get x(bla bla bla x)

so the funciton is zero if (bla bla bla x) is zero

(bla bla bla x) is sqrt(27-x^2 - y^2 )

(simplified)

[azure swift]

Okay, lets stop for a moment

are you lost at how to find the other critical points?

[azure swift]

Do you follow this last bit, which talks about 4 cases?

[autumn tusk]

not really.

i mean i understand that those other 2 expressions should be a point

but .... ?

this i mean
x= sqrt(27-x^2 - y^2 )
y= sqrt(27-x^2 - y^2 )

thats a ... curve?

[azure swift]

Yeah

Lets start with the first one

When y = 0

Thats one of the cases

[autumn tusk]

and why?

because we look at the bounds?

[azure swift]

Oh so you didnt follow this last bit?

[autumn tusk]

yea no, a bit lost there

[azure swift]

Well once we have the partials, we need to solve a system of two equations

[autumn tusk]

i get the same expressions

[azure swift]

Right, when they vanish.

So

let me just post them heere again

When are they zero?

Whats maybe, the trival solution here?

[autumn tusk]

when
x and y are 0

[azure swift]

Yes!

[autumn tusk]

and what iv'e posted above

[azure swift]

So thats one possibility

Right yes!

[autumn tusk]

x= sqrt(27-x^2 - y^2 )
y= sqrt(27-x^2 - y^2 )

[azure swift]

This requires some maturity about how logic is used imo

So we have an or statement and an and

the or arises from the fact that there are multiple solutions

while the and is due to the simultaneity of the equations, i.e they have to occur at the same time

[autumn tusk]

m hm....

[azure swift]

So say when you solve a quadratic

in general you might have two roots

that's an or statement for the roots

both roots dont have to be true at the same time, whatever that means

[autumn tusk]

mmm ok, real roots ok

[azure swift]

Just an example if it helped

while system of equations they usally talk about a solution pair (x,y)

so x and y have to agree at the same time

[autumn tusk]

ye

[azure swift]

so the solution for x and y is a stricter notion

[autumn tusk]

so i tought

let me intercept you here

[azure swift]

So in this case we have a combination of that, yes

[autumn tusk]

i want to know where those 2 curves are the same, that would be another critical point

so i put one into the other basically and result is 3

[azure swift]

Yes and you can see this algebraically aswell

[azure swift]

It's not really different from any other system of equations youve done

but well uh non linear

[autumn tusk]

well i've got the point 3,3

[azure swift]

So to reiterate, you have multiple solutions (that's the or) and you have a solution pair x,y (thats the and)

[autumn tusk]

ok

[azure swift]

so going back to this

We can consider the case when y=0, since its one of our solutions

[autumn tusk]

oh

[azure swift]

Now what happens with your x=sqrt.. equation?

[autumn tusk]

ok undderstood, sqrt(27/2) the y goes to 0

[azure swift]

Yes

and the very same happens if x = 0 for the other equation

[autumn tusk]

ok let me see if i understand correctly what we did here

so i have various isolated solutions, and i want to find when they are all true

so if the solution is just a number, nothing to do.
but if it's a function i can find where they intersect (that would be our 3,3 coords)

and mmm i don't know how to visualize the other part

[azure swift]

it wont be a function, its a curve/equation to be pedantic of me

[autumn tusk]

ok, thanks, right

[azure swift]

Hm, the other part being?

[autumn tusk]

the other part i understand logically, i need to put into a system.
y=0 and x=sqrt()

[azure swift]

Well let me jus stop you if thats okay; let me try something

it might help

[azure swift]

So okay, you have these two equations

they arriesed from a system of eqautions