help-26

[feral cargo]

I need help for calculating area

[feral cargo]

Here is a picture

[acoustic tangle]

The area is already given though

[feral cargo]

I mean calculating x and y

The values

[acoustic tangle]

Right, can you notice any properties that x and y should satisfy?

[feral cargo]

Never mind I know

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[pulsar grove]

<@&286206848099549185> If i get x(t) = something and y(t) = something. y'(x) is calculated by y'(t)/x'(t). Is there anyway for me to calculate y(x) ?

[neon iron]

yes

[pulsar grove]

could you tell me how please

[neon iron]

can you please send the question first, it might get a little hard to explain like this

[pulsar grove]

how about this one

normally i will just solve for t from x(t)

then substitute t by x into y(t)

and ill get y(x)

but with some equations x(t) is too complicated its pretty hard to solve t respect to x

[neon iron]

just find x'(t) and y'(t), now you know y'(x). then you can integrate y'(x) to find y(x)

this is the proper method of doing these kinda questions

[pulsar grove]

find intergrate of y'(x) isn't better than solve t respect to x

so there's no easier way ?

[neon iron]

you will have to substitute the value of t when you're done calculating y'(x)

only then you'll be able to integrate

[neon iron]

i guess not

[pulsar grove]

thank you so much

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[neon iron]

your welcome

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[junior linden]

how to solve those

[neon iron]

use the definition of e

[junior linden]

idk what that is

[junior linden]

okayyy ty

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[polar drum]

Let $M$ be a non-empty subset of $\mathbb{R}$. Proove that $\sup{-x\ | \ x\in M}=-\inf(M)$.\\
Proof. Let $a=-\inf(M)$ then for $\forall x\in M$ holds that $\inf(M) \leq x$. When multiplying this equation by $-1$ we get $-x \leq -\inf(M) = a$, therefore $a$ is upper bound of ${-x \ | \ x\in M}$ . \
Let there be some $b<a, \ b\in\mathbb{R}$ such that $\forall x\in M:\ -x\leq b$. By multiplying both sides by $-1$ we get $-b\leq x$, implying that $-b$ is lower bound of $M$. Since $\inf(M)$ is the greatest lower bound of $M$, $-b\leq\inf(M)=-a$ and thats a contradiction with $b<a$, thus we prooved that $a$ is the least upper bound of ${-x \ | \ x\in M}$.

[thorny flame]

Slowaq

[polar drum]

is my proof correct?

<@&286206848099549185>

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[polar drum]

<@&286206848099549185>

pls help

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[jolly agate]

Hello! I have come this far

How do I now determine what X and y should be ?

Do i need to only look at the (1-xy) part or do I need to consider the whole function ?

So in order for 1-xy = 0, Do i solve for x = 1/y and put it in the other equation as X or how do i proceed ?

Here is a better pic

(Firs is partial derivative of x and the other is partial derivative of y)

Alr no problem I found the critical points I think, if y = 0 then x = 0

If y =/= 0, y = y, x = 1/y

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[quaint mesa]

How'd I prove that the polynomial p defined by
P(x)=x^3-kx^2+2kx-4k has atmost one real root for k>0

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[jolly agate]

Hello i would need some help to get the coefficient of the hessian

I just know that If I would just do the formula right over it would be very long, just wondering if there is any tricks in these types of equations ?

can we keep the factorization ?

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[turbid cedar]

.reopen

i need help with understanding geometry

[turbid cedar]

especially bearings

<@&286206848099549185>

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[quasi prairie]

can someone help me with this please?

[quasi prairie]

<@&286206848099549185>

[lean mist]

hi

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[quasi prairie]

hi

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[green latch]

I have tried solving this question by this method

But it isn't matching with the options

Why?

The actual solution is like this but I don't really care about the solution

[green latch]

I want to know why this method is wrong

[sweet shard]

You need to write the last term of the sum

[green latch]

It will be 1/2 cuz every term is 1/2 there right ?

[prisma coral]

no

also its not convergent so u cant do it infinitely

it must terminate at n

[neon iron]

yup i agree

[green latch]

Yeahhhhhhh

I forgot that it's convergent I can't use this method

[sweet shard]

You can use this method to solve for Sn

[green latch]

But why the last term isn't 1/2 but instead it's anything else , because when I was solving this , I substracted 3 terms and they all were 1/2

[sweet shard]

You just need to do this

[green latch]

Ogey let me show you wait

Like this ?

[sweet shard]

The last term's numerator is not n-1

Would be easier to find the exact denominator first

Find the nth term of just the denominator.

[green latch]

Since the denominator is a GP

I can write the last term in the terms of 2 to the power of some exponent ?

Is that what you're referring to?

?

[sweet shard]

Yes

2^n or something similar

So what's the numerator

[green latch]

Is it 2^(n)-1

Divided by 2^n

[sweet shard]

Right you should be able to find Sn/2 now and then double it to get Sn

[green latch]

Ogey thanks I have solved it 🫂🫂🫂

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[shy cedar]

how to continue?

[shy cedar]

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[molten crystal]

f= x^3+y^3+z^3
x^2+y^2+z^2 = 27
x,y,z > 0
I need to parametrize the boundary curve on f given the contraint condition.

the reasoning is
i can rewrite in terms of 2 variables
f= x^3+y^3 + sqrt(27-x^2+y^2)^3

i can parametrize the boundary of the variables of the new f in x and y
x= sqrt(27) sin(t)
y= sqrt(27) cos(t)

plugging in f becomes

[molten crystal]

is this correct? z part becomes zero

since i have to take derivative of this, doesnt really matters, cause the fisrt sqrt27 i have there is just a scalar.
but the solution writes this

why only 27?

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[timid junco]

@molten crystal is this the question from yesterday?

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@molten crystal Has your question been resolved?

[molten crystal]

yes but not same question

you didnt answered me to my question yesterday, you've just solved the problem without reading my question 😄

@timid junco

[timid junco]

oh

sorry

i mean was it right?

what was the question?

[molten crystal]

unfortunately it's bed time now, i need to sleep

no worries, my new question is this one i've just wrote

[timid junco]

ahh

ok well gn

[molten crystal]

if you wantto give it a shot, ill read tomorrow

[timid junco]

i cant help with derivatives

[molten crystal]

my problem here is not so much derivatives

i have a different number than in the solution.

[timid junco]

yes I cant help with integrals either

i havent learned that stuff yet

sorry

[molten crystal]

didnt say anything about integrals, but ok no worries

[timid junco]

alright well gn

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[neon iron]

Hello, I am trying to prove by induction that $(a+b)^n = \sum_{k=0}^n{n\choose k}a^kb^{n-k}$, I am at the inductive step and have arrived at this:

$\sum_{k=0}^n{n\choose k}a^{k+1}b^{n-k}+\sum_{k=0}^n{n\choose k}a^kb^{n+1-k}$, I am not sure where to go from here though to relate it to get a sum up to n+1

[thorny flame]

Ware

[neon iron]

I feel like it has to do something with the bounds or something because I've seen that you can do this somehow

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[lost bloom]

anyone able to help me with my algebra

[lost bloom]

im stuck

[neon iron]

Did you know that slope =Change in y coordinate /Change in X coordinate

This will help you in finding y

[lost bloom]

ok

thank you

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.reopen

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[neon iron]

Yo

[lost bloom]

what if by chance slope isnt defined

[neon iron]

Slope is not defined only when change in X coordinate is 0

[lost bloom]

ok

thank you very much

[neon iron]

So you will have change in X coordinate=0

[lost bloom]

ah

thank you

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[tidal garnet]

Hi, I get the induction part all the way to the n = k+1 part. By the transitive property of inequality, if If a < b and b < c , then a < c. so when we want to prove that 20+(k+1)log(base2)k less than or equal to (k+1)^2, then doesnt the logic follow that anything less than 20+k+1logbase2k is ALSO less than (k+1)^2? How come in this answer scheme it writes 20+(k+1)logbase2(2k) is greater than 20+(k+1) could u please explain this to me?

[neon iron]

Dope pfp 👌

[tidal garnet]

thx

[woeful drift]

How come in this answer scheme it writes 20+(k+1)logbase2(2k) is greater than 20+(k+1) could u please explain this to me?
where does it say this?

[woeful drift]

it writes 20+(k+1)logbase2(2k) is greater than 20+(k+1)
they never write this

[tidal garnet]

what

oh sorry i meant 20+k+1(logbase2(k+1)

[woeful drift]

now we're talking

[tidal garnet]

they wrote that the 2k part is greater then that

soz

[woeful drift]

they just rewrote the log bit using log(ab) = loga + logb

[tidal garnet]

yes i understand hwo they did it. but the question asks to prove 20+klog2k<k^2.

for a<b, if b<c then a<c

as 20+(k+1)log2(2k) is greater than thelog K+1 part, then it could also be greater than (K+1)^2

we are trying to prove that its less than (K+1)^2

[woeful drift]

they are making use of transitivity

the thing on the left is what you start with (let's call it a) and by replacing k+1 with 2k they've created something bigger (let's call it b)

then by using the assumption above they create something even bigger (c and then d). So they've shown b<c thus a<c as desired

[tidal garnet]

we wanna prove 20+(K+1)log2(K+1) < (K+1)^2.
shouldnt we approach this by finding someting less than (K+1)log2(K+1) because we can guarantee it will be < (K+1)^2 as well?

in the answer they write 20+(K+1)log2(2k) greater than (K+1)log2(K+1). then this could also be greater than (K+1)^2 for all we know

which is not what we are trying to prove

[woeful drift]

because we can guarantee it will be < (K+1)^2 as well?
you can't guarantee this

[tidal garnet]

how do we know that 20+(K+1)log2(2k) is greater than (K+1)log2(K+1) but NOT greater than (K+1)^2

sorry for the jargon i was referring to this link to help https://k12.libretexts.org/Bookshelves/Mathematics/Analysis/07%3A_Sequences_Series_and_Mathematical_Induction/7.03%3A_Mathematical_Induction/7.3.03%3A_Induction_and_Inequalities

[woeful drift]

we know it's not greater than (k+1)^2 because the last line of inequalities shows it isn't

if you're asking how we know that ahead of time, that's not so easy to answer

[tidal garnet]

so i kinda get the logic but can u explain in simpler terms, like maybe an example

[woeful drift]

i'm not quite sure what kind of example you're looking for

i also just had a big dinner so i'm operating at like 50% brain power lol

[tidal garnet]

oh ok

why in the induction proofs they say b>c

[woeful drift]

where exactly do they say this?

[tidal garnet]

brb

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[tawny swift]

what am I doing wrong

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[sweet shard]

Show your work and someone can review it

[lusty goblet]

Put random values of n & t, so the answer can be verified

[tribal lance]

What would a simple one be?

[lusty goblet]

Real numbers

[tribal lance]

1 and 1?

but final answer doesn't have a t

since it's simplified...?

[lusty goblet]

Put which can be easily calculated

[lusty goblet]

It provides that it doesn't depends on t

Try to eliminate t

[tribal lance]

We did eliminate t because I'm getting the simplified answer to be 10 - 9.9^n

but if we use t = 1, n = 1

we get 1.9 for orginal expression

but simplified isn't 1.9

[lusty goblet]

Put other real numbers & check, may u not get magnitude equal but if values are close to each other then the answer is right

[tribal lance]

okay I changed my answer to this:

10 - 9.1 (0.9)^n

that gives the magnitude very close

but not the same

[sweet shard]

Are you just guessing formulas?

[tribal lance]

no

let me show my work

[sweet shard]

Yes you should always do that

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[frosty iron]

not sure where to start

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[frosty iron]

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[neon iron]

factorizing might help

did you try that?

[frosty iron]

factorizing what?

[neon iron]

111

for first case

222 for second

so on

[frosty iron]

would that even work

[neon iron]

let me check

3*37

guess not :P

[neon iron]

yo

@frosty iron

playing roblox really does help in getting ideas 😎

[neon iron]

for second equation

write b+3 as b+2+1

then multiply

and then substitute equation 1 into 2

then take ratio of equation 1 and 2

you will get b+2

from that you will have b

same for all other variables :))

write down what i told and you will understand

oh shit

i suck

sorry!

it doesn't work

[frosty iron]

oh):

[frosty iron]

real

[neon iron]

sorry again i thought i had it

[frosty iron]

what if

Let P(x) = (a + x)( b + x + 1)( c + x + 2)(d + x + 3)

and find some relation for P(5) - somethign = number

so P(x) = x^4 + kx^3 + lx^2 + ox + p

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[brazen sandal]

area between curves

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[spring wolf]

I would be integrating $\sqrt[3]{2x} - \frac{x}{2}$ from 0 to 4 in this case. Is there a particular reason you integrate from -4 to 0 also, and add it?

[thorny flame]

eththorn

[brazen sandal]

i thought the integral would be -4 to 4

[spring wolf]

usually you have to find the intersection between the two curves and integrate the area they bound for these questions

[brazen sandal]

instructions for this question is just sketch enclosed by the given curves and find its area

and when i put both cbrt 2x = x/2 i would get + or - 4 so i thought we integrate from there

the text book answer for this question is 4

[spring wolf]

yeah that makes sense actually. I used Wolfram and it gave 0 and 4 as the only solutions, but that is incorrect. -4 to 0 is also a solution as you've identified correctly. Your mistake is in the algebra

because the integral from 0 to 4 for the area is 2, and we can double it by symmetry to arrive at 4. I'll have a close read of the integration steps.

[brazen sandal]

this is what i arrive to, is this not correct?

[spring wolf]

okay so the integral of the cube root of 2x is not (3/4)(2x)^4/3. The two is raised to 1/3, not to 4/3.

because $\sqrt[3]{2x} = \sqrt[3]{2} \cdot \sqrt[3]{x}$

[thorny flame]

eththorn

[brazen sandal]

how come shouldnt the integral be adding 1 to 1/3? making it 4/3?

[spring wolf]

because it's a coefficient term. like $\int 2x dx = x^2 + c$ not $\int 2x dx = \frac{1}{2} \cdot 2^2 \cdot x^2 + C$

[thorny flame]

eththorn

[spring wolf]

using the properties of integrals, you can break apart the integral and take the term outside the integration sign, if you visualise that better.

[brazen sandal]

oh okay so only x would be x^4/3 and 2 is stil cuberoot

[spring wolf]

yeah

[brazen sandal]

i think i got it thanks

.clsoe

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[fair dagger]

I have an expression with 4 variables, all of them are randomly chosen real numbers between 0 and 1
How do I find the expected value of the expression?

[wooden osprey]

!xy

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[fair dagger]

No context, I have an expression, I want to find its expected value, I do not know how, because all I can find on the internet doesn't cover what I need

[static viper]

Wouldn't it just be the average?

Nvm

[fair dagger]

Yea, and how do I calculate that?

[static viper]

Do you need a proof? Or just an answer

[fair dagger]

A proof would be nice

[static viper]

Honestly, isn't every sum except 0 ans 4 equally likely? @fair dagger

[fair dagger]

wdym

[static viper]

If you can choose any real number

There would be infinite cases for each sum no?

[fair dagger]

Yes

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[fair dagger]

<@&286206848099549185>

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[grim prairie]

am i making the correct step?

[warm monolith]

Why did the number stay the same after you squared it?

[grim prairie]

which one?

[grim prairie]

equation after (5)?

[warm monolith]

oh nvm i get what you did now

[grim prairie]

cannot solve the sysmem of equs

but the process should be right hmmm

<@&286206848099549185>

[warm monolith]

the absolute value is wrong, it should be the square root of that

|z²|=sqrt(a²+b²)=sqrt(841)

=29

[grim prairie]

29^2 =841

|z|^2 = a^2 + b^2

|z^2| = |z|^2 = a^2+b^2

oh ok i made a mistake

=29 is right

[warm monolith]

proceed from that and you should get the right answer

[grim prairie]

a=+-2

[warm monolith]

you forgot to tkae the root of 4

yeah

[warm monolith]

wait

a²=b²+21 not b²-21

[ashen trout]

You cooould eyeball it

[grim prairie]

so much silly mistake haha sry

[grim prairie]

thank you!

[warm monolith]

yw

[grim prairie]

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[junior laurel]

A rectangular prism has a height of $h$ units, and its base is a square with side length $x$ units. The volume of the prism is 128 cubic units. Find the minimum value of the expression $x^2 + 4h$.

[thorny flame]

ZiXO

[mortal steeple]

Do you know how to calculate the volume of a prism?

[junior laurel]

yes

base area * height

and the base here is a square with side x

so the volums = x^2 * h

[mortal steeple]

Wait is a rectangular prism

Basically a 3d rectangle?

With 2 sides equal?

I was imagining a pyramid lmao

[mortal steeple]

Can you use calculus?

If not what about AM GM

[junior laurel]

yes for sure

i can do both

go with AM GM ?

i prefer calculus tho

[mortal steeple]

Ok AM GM is super simple

Tell me what the eqn we get from volume

And tell me what we need the min of

[junior laurel]

ok we need a = b for the minimum of the equation x^2+4h

to use AM GM we can do $\frac{x^2 + \frac{512}{x^2}}{2} \geq \sqrt{x^2 \cdot \frac{512}{x^2}}$

[thorny flame]

ZiXO

[junior laurel]

meaning : ${x^2 + \frac{512}{x^2}} \geq 2 \sqrt{512}$.

[thorny flame]

ZiXO

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[neon iron]

The stability of a system/transformation is defined as if x(t) is a bounded input, we have y(t) be a bounded output.

So if we have y(t) = x(t-2) + x(2-t)

how do i go about doing that

[worthy storm]

triangle inequality?

[neon iron]

how exactly

ok assume the sum is bounded by M

then x(t-2) <= M and x(2-t) <= M

no wait

this is wrong

[flint stump]

Start by restricting x. Thats the premise

[neon iron]

operations on the time won't affect the bound

implying that

x(t) < M, x(2-t) < M, x(t-2) < M

[flint stump]

x(t-2) < 2 ? How?

[neon iron]

typo

[flint stump]

ah ok

[neon iron]

but anyways

absolute value function is not a linear function

so that's where triangle inequality comes in i guess

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[zenith field]

this is technically physics i think

but i think its worth a shot trying

[mortal steeple]

Delta T means variation in T right?

Like the ± part?

[zenith field]

yes

[mortal steeple]

I guess there is a formula for when ± is in the denominator I don't remember however

[topaz sinew]

@zenith field Has your question been resolved?

[dreamy ember]

@zenith field just calculate it with the bigger value and then again with the smaller value

[zenith field]

does that work?

[dreamy ember]

yeah

you can use uncertainty formula if u want

but thats basically how it works

2pi/5.801 = 1.08312106657
2pi/5.811 = 1.08125715147

(1.08312106657 - 1.08125715147 )/2 = 0.00093195755
(1.08312106657 + 1.08125715147 )/2 = 1.08218910902

so your answer is 1.08218910902+-0.00093195755

https://uncertaintycalculator.com/

you can use websites like this

[drowsy mesa]

4/612/9+(-3/8)(-12/9

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[clear plinth]

Ques 9

I solved it halfway

Tho it might not be correct

[keen matrix]

for int 3y^2e^(y^3) don't use IBP, recognize that d/dy(y^3)=3y^2 and use u-sub, everything else seems right

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[elder glacier]

can i get help?

[elder glacier]

determine the limit for lim x->infinity f(x) and lim x->-infinity f(x) if they exist

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[mild furnace]

i need help about a game

[mild furnace]

is it possible to create a square with the given figures?

[gentle night]

Well yeah with the 2 triangles there

[mild furnace]

they are 2 triangles, 4 rectangle trapezoids and 2 other figures

[mild furnace]

using all of them

[cosmic silo]

can u send me a screenshot, dont have the app to view it

[mild furnace]

a mom i'll convert it

[mild furnace]

thanks

[cosmic silo]

ohh

[gentle night]

It is a hard question though

[mild furnace]

you essentially have to recompose them to create a square

[gentle night]

Yeah what I think is that would be more related to trial and error rather than mathematics

[cosmic silo]

exactly

[cosmic silo]

wouldve been better if those were present physically

[mild furnace]

i don't know about that. Maybe you cannot. But you have to demonstrate why

[craggy haven]

one thing math can tell you is how big the square has to be

based on the total area

[mild furnace]

right

but i don't have the measurements

[gentle night]

Then you can try with getting combinations that gives you the side length of the square

Maybe it would be faster to do that

[craggy haven]

can you use something else as a measuring tool?

maybe fold a dollar bill in half

[mild furnace]

i have only the pic sorry not the physical stuff

the area of the trapezoid is 1/2 x (b_1 + b_2) x h?

[craggy haven]

can you open it in an image editor and use a ruler there?

[mild furnace]

ok

[craggy haven]

,tex .plane geom

[thorny flame]

hayley is not british

[mild furnace]

yes

the trapeizium rectangle is: b=5.7. a=4.5. h=1.75

the triangles are: b=5; h=3.85

and the other 2 figures are: 2x4.2 x 3.7x1.85

the area i get is 70

70.03 to be precise

so 8.36 has to be the side of the total square

now?

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[elder glacier]

can someone check my work

[elder glacier]

determine the limit

i did the first part of the task

i wanna know if it's correct or not

[gentle night]

It is correct

[elder glacier]

i did some tiny modifications

is it still correct

just making sure

[gentle night]

Still correct

[elder glacier]

how about if i do -infinity instead of infinity

[gentle night]

Well what do you think

[elder glacier]

e^1 * e^infinity

[gentle night]

So

Complete

[gentle night]

Correct
But you just need to specify that this way the limit doesn't exist

[elder glacier]

Wdym?

[gentle night]

Like write thus followed by the first part of the question before the first arrow
DNE

[elder glacier]

like here?

[gentle night]

No I mean after that solution

[elder glacier]

= infinity
DNE

?

im not sure exactly what you mean

[gentle night]

Like this

$\text{Thus} \lim_{x\to-\infty}(e^{-x+1}) DNE$

[elder glacier]

so i write that separately under my work?

[thorny flame]

Sherif Player

[gentle night]

Yes

[elder glacier]

okay thanks

.close

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[daring oak]

if i have an equation in the form of ( (x ± 0.5) - (y ± 0.5) ) / t, is that equivalent to (x-y)/t ± 1/t?

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Please don't occupy multiple help channels.

[daring oak]

.close

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[vestal sigil]

yes it is

[daring oak]

thank you so much

[vestal sigil]

i arrived late lol

[daring oak]

no worries, i still appreciate it!

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[barren tapir]

Hey there

having trouble finding this

I get a cubic but even with the hint its hella hard to solve

speicifcally the parts on B btw

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@barren tapir Has your question been resolved?

[barren tapir]

<@&286206848099549185>

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[sharp dew]

I found trouble to approximate the value of (83)^0.4 / (2.6)^4

[weak osprey]

Need help understanding how to break down and solve these questions and the answers

[topaz sinew]

@sharp dew Has your question been resolved?

[hoary burrow]

For q1 identify the operation made on the lhs

[weak osprey]

What’s lhs

[hoary burrow]

And see if the same operation is made on the rhs

[hoary burrow]

Left hand side

Rhs is right hand side

For example:
x=1
x-1=0

What operation did I do on both the sides?

[weak osprey]

Equals

[hoary burrow]

No

[weak osprey]

X

[hoary burrow]

operation means like minus plus divide multiply

[hoary burrow]

How about if we rewrite this as:
x=1
x-1=1-1

What operation did I do on both the sides

[weak osprey]

Subtract

[hoary burrow]

Yes

I minus one on both sides

Doing this maintains the equality of the equation

What if I did:
x=1
x-2=1-1

Is the equality of the statement maintained?

[weak osprey]

Yes

[hoary burrow]

No

On the lhs

We minus two

On the rhs

We minus one

Which are not the same

So equality is not maintained

What about
x=1
2*x=1*3

Is equality maintained?

[weak osprey]

Yes because it has 1

[hoary burrow]

Are you sure

Look on the left hand side of the equal sign

What do I multiply x by?

[weak osprey]

2

[hoary burrow]

Then on the right hand side of the equal sign

What do I multiply 1 by

[weak osprey]

3

[hoary burrow]

So is equality maintained?

[weak osprey]

No

[hoary burrow]

Yep

Alright now look at question 1 again

If there's ever a point where mal or Tyler's solution does not maintain equality

Then their solution is wrong

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[weak osprey]

Tyler

[hoary burrow]

Type
.reopen

If you still want help

[weak osprey]

.reopen

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[hoary burrow]

Oh nvm

[hoary burrow]

.close

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[weak osprey]

Tyler

[hoary burrow]

Yes

How about mal

Could they have both made mistakes

[weak osprey]

Maybe they both get rid of the 2

[hoary burrow]

What operation does mal do on both sides during the second step

[weak osprey]

Subtract

[hoary burrow]

How about on the rhs?

Is -11-1=-10?

[velvet aurora]

yes

[weak osprey]

Yes

[hoary burrow]

Welp I tried

[weak osprey]

Oh

[hoary burrow]

Revise your negative numbers frfr

[weak osprey]

I missed a year of school I got dumb

[leaden jungle]

same

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[turbid linden]

how do you factor 8x^3 - 27 without knowing the formula

[sweet shard]

Learn the formula first

,w factor x^3-y^3

[thorny flame]

Results provided by WolframAlpha

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

Requested by rie.mann

[turbid linden]

so no way without the formula?

[sweet shard]

No better way

[turbid linden]

well there's a way then

[topaz sinew]

@turbid linden Has your question been resolved?

[unreal tinsel]

if you were to factor it, you would be using the formula, directly or indirectly

[turbid linden]

well still i awnna know how to do it

without knowing that

[unreal tinsel]

it's like asking for how to walk without learning how to walk

[turbid linden]

wdym

[unreal tinsel]

if you want to know how to factor it you need to learn the formula to factor it

are you trying to derive the formula?

[turbid linden]

not really

im pretty sure those formulas are only there to make things faster

[unreal tinsel]

8x³ - 27

2³x³ - 3³

(2x)³ - 3³

(2x)³ + (2x)²×3 + (2x)×3² - (2x)²×3 - (2x)×3² - 3³

(2x)((2x)² + (2x)×3 + 3²) - 3((2x)² + (2x)×3 + 3²)

((2x)² + (2x)×3 + 3²)(2x - 3)

this is just deriving the formula but using specific values instead of a and b

step 4 just added and subtracted the same values

step 5 used those values to help with the factoring

[turbid linden]

was it a matter of trial and error?

[topaz sinew]

@turbid linden Has your question been resolved?

[topaz sinew]

@turbid linden Has your question been resolved?

[night imp]

No , they derived the actual formula by thinking what terms can be added to factorise
As Riemann said it's best to remember those formula

[turbid linden]

Well I really don't want to memorize formulas as there's lots of them and theyre limited on what they cover

.close

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[loud sail]

Hello everyone I need help with a question, number C, it's from a mock exam I already gave and got back graded, I don't even know how to begin with d.

[neon iron]

did you calculate question a and b

[loud sail]

Oh, yeah I did

Well actually I messed up at a) and needed help to later fix it but I did b) right

I don't know how those two play into thear question though

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[crude kelp]

Hello, good morning, anyone help me understand the Weierstrass Function and its relation in the time domain and frequency domain, I don't have any clue about it but I need to understand it for my curiosity, I'm serious about it and am a speedy learner, could someone tell me sources to consume and how to visualize using Mathematica or any other tool!

[crude kelp]

when we talk about applying the Weierstrass function in certain contexts, such as signal processing, wave propagation
In signal processing, the time domain -as a signal evolving over time- and transformed it into the frequency domain via a fourier or laplace transform
This is what I mainly thinking of and trying to visualize but I'm missing the core pieces of information on how to do it exactly and how would I modify the context of using it, if u have some materials or sources I can rely on to find some understanding on how to use it I would appreciate it.
what I'm trying to do is I'm a trader in the foreign exchange market who has a weird theory about the market and its fractality and its relation to the WF they share a lot of similarities and i don't mainly believe in coincidences.

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[earnest valley]

hello

[merry hill]

hey

[earnest valley]

i need to factorize this

[merry hill]

think what u can take common (a, b)

[earnest valley]

oh ok

I did it

wait the last one is -3

oh wait

nvm

I think i got it

thanks

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[echo breach]

Does $B{n+1} \Delta B{n} \to \phi$ means limit of $B_{n}$ exists?

[thorny flame]

mindovermatter01

[echo breach]

$B_{n}$ is a set sequence

[thorny flame]

mindovermatter01

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[pearl fog]

im learning induction

so ike i dont think you can just directly put in induction here?

[flint stump]

Looks to me like a telescopic problem

[pearl fog]

yeah i think it is but why is it in the intermediate questions of something about induction

[prisma mesa]

you might be using induction later

i wouldnt worry about it too much

maybe once you find the expression for f(n), if there is some, then you're gonna use induction to prove it works

[neon iron]

can you share the source of this question please?

[pearl fog]

https://artofproblemsolving.com/wiki/index.php/Induction#Intermediate

[glossy girder]

I think this problem is inverted, you are given a result proved using induction and finding its properties, instead of the other way round

So like if u solve this, u get a telescopic form

So the induction question might be this: f(n)/f(n-1) = (n+1)/(n-1), prove using induction that f(n) = that result

[hallow depot]

lol

close though

something like that

[flint stump]

there should be a 2 in numerator right?

[hallow depot]

it could easily be!

[pearl fog]

uhh sorry something came up

hm

isnt it like 2/(m)(m+1)

oh wait you alr said that

so its 2/1997

.close gn

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[elder glacier]

For what value of k is f continuous at 2? What is the range of the function?

[sweet shard]

Use the left and right limit definition of continuous and solve for k

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[maiden cipher]

I have no idea how the options of orthagonal directions were calculated

[maiden cipher]

The example problem just... states they exist, and this is what they are

[maiden cipher]

!occupied

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[maiden cipher]

<@&286206848099549185> ? The example problem I'm studying at the moment lacks instruction on how to find directions orthagonal to the gradient

[sweet shard]

set the dot product with the gradient to zero

[maiden cipher]

Like this?

[sweet shard]

yea you have 2 free variables

the book just arbitrarily picked y=1

typical/conventional thing to do

[maiden cipher]

Wouldn't that imply an infinite number of orthagonal directions? Which, I mean, there are in 3 dimensions, but...

Since we're trying to find where there is no change in z, that cuts us down to 2..

[sweet shard]

yes there is

[maiden cipher]

I guess my point is.. I don't see how we can just arbitrarily pick a value for y

[sweet shard]

just like how there are infinitely many lines perpendicular to y=x

y = -x, y = -x+1, y=-x -1

[sweet shard]

pick a different value for y and see

maybe x=1

find that y value

and verify the dot product is still zero

[sweet shard]

you have two unknowns, but only one equation. so there are infinitely many solutions

[maiden cipher]

Right.. and yet, the book states there are only two possibilities

And incidentally, if we pick x = 1, y comes out to 1/3

Just to be clear, the direction has to be perpendicular (in the XY plane) to the dotted line at that specific point

So I understand there can only be two options.. I just don't know how the book calculated (3, 1) and (-3, -1) were the only two

(And I also don't know how it excluded (-3, -1) from being a valid answer, but.. that was going to be the follow-up question)

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[native orbit]

how do i ask it to teach me functions 0.0

[sweet shard]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:functions

[native orbit]

yu

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[native orbit]

how do i close this lol

@sweet shard help

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[halcyon willow]

basic question why is arcos (-x) = pi - arcos( x) but arcsin (-x) = - arcsin (x)

[halcyon willow]

i mean arcsin is a odd fn so that kinda makes sense

not sure about arccos

[neon iron]

this is true ?

u must prove it ?

@halcyon willow

[halcyon willow]

yeah i was wondering about the proof

or a intuitive explanation works too most likely to adjust ans to its domain

[neon iron]

ok so u can study f(x) = arcos (x) + arcos (-x)

[halcyon willow]

but not sure thats rigorous enough

[neon iron]

idk

[maiden cipher]

Have you graphed it?

[neon iron]

me ?

[maiden cipher]

Think about where sin and cos are positive or negative

[halcyon willow]

sin is positive in first two quadrants and cos has 1st and 4th quadrant

[neon iron]

with this u should have a null derivative

[maiden cipher]

Yep.. and if your cos^-1 value ends up in quadrants 2 or 3, how would you make them positive?

[neon iron]

so a constant value

[halcyon willow]

ah yeah substract pi from it

[maiden cipher]

Bingo

[halcyon willow]

thnx

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[neon iron]

help me out guys

[pearl peak]

find LCM of 30,36 and 40

[neon iron]

how

[light mesa]

Heard of prime factorization?

[pearl peak]

yeah, start by finding the prime factors of each of the three numbers

[neon iron]

i mean HOW yOU KNOW WHAT to taKE HcF ANd LcM

[mortal steeple]

It says the minimum distance so that they all can cover in complete steps

So that implies LCM

It's more of intuition and as you do more questions it will become better

[neon iron]

OK

TY

[mortal steeple]

Kk

!done

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[neon orbit]

What is the formula to figure out this type of question?

[neon orbit]

So if 99% is 198 how can i figure out the whole

[dusk lava]

what number times .99 = 198?

[hollow onyx]

Chaotic question indeed😂

[fickle mulch]

.99x=198

[neon orbit]

So 198 divided by .99 = 200

[wary bison]

wow😭 😂

[neon orbit]

So that’s the way to do it ?

[hollow onyx]

yo help me tho

im doing calculus I

[wary bison]

dayum

[fickle mulch]

ayy same

[hollow onyx]

type shii

[wary bison]

calculus 4 much betta

ans is 200?

[neon orbit]

.close

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[tiny pawn]

how to prove de morgans law

[fiery badge]

How would you approach this?

[tiny pawn]

i already saw some proofs in internet

but it looks like its just using 'or' ' and' operators

cant i prove it without these operators

[fiery badge]

When in doubt use Ven diagrams

[tiny pawn]

i think 'and' 'or' operators cant directly used interchangeably with union and intersections

they are bit differents aren they

[tiny pawn]

how isnt it explicitly same. what should i write on the explanation. if you have eyes you will know it or what

[fiery badge]

Hold on I'm cooking it up

[tiny pawn]

okk

[golden blade]

ja genau

[fiery badge]

Proof by eyesight

[tiny pawn]

ohh wow beautiful

but cant i express it mathematically

[fiery badge]

As for 'and' and 'or' I'm fairly sure they are defined with set operations as a basis

so they can be equivalent

I just messed up in my explanation above

[tiny pawn]

but then what is the point of proving it if i use it inside again.

[fiery badge]

So a set union gives us all elements of both sets A and B, as such we could declare set C that holds all elements contained in either set A OR set B

And a set intersection gives us elements contained only by both sets, so only elements that are contained in both set A AND set B

And then the c is just the NOT operator, since it gives us everything NOT in the set

So you can really translate the initial law as

NOT of (A OR B ) = (NOT of (A)) AND (NOT of (B))

[tiny pawn]

its from internet

from second to third line

[golden blade]

I did it I think

[tiny pawn]

let me see it

[golden blade]

am i allowed

try it

suppose the universal set is omega

[tiny pawn]

every one is allowed

[fiery badge]

nosols though

[thorny flame]

bacc (unhelpful)

[golden blade]

use this property

[fiery badge]

(says I after sending the solution in visual form)

[golden blade]

Logically (A U B)^C means universal set without (A U B)

and then you should derive it

[fiery badge]

Proof by logic

[golden blade]

actually defined like that

[tiny pawn]

what if teacher asks me to prove my logic again

[fiery badge]

"It makes sense, QED" lmao

Just kidding

[golden blade]

huh

[tiny pawn]

isnt it literally what we are trying to prove

[golden blade]

du machst davon gebrauch

[golden blade]

You wanna prove this

I write down it differently and I apply the following property

[tiny pawn]

for me it looks same just used omega isnt it

A compliment is omega without A

[golden blade]

yes

[tiny pawn]

how im gonna prove this one

[golden blade]

du stellst mir fragen

[tiny pawn]

ja

[golden blade]

mengen technisch haben wir x ist aus omega und es gilt x ist nicht aus A und x ist nicht aus B

Also

[thorny flame]

bacc (unhelpful)

[golden blade]

das ist halt das was man festellt, ich wüsste aber nicht wie man das jetzt formal beweisen würde

[tiny pawn]

ich weiss auch nicht.

[golden blade]

wenns x nicht in A oder x nicht in B wäre dann wäre das falsch

es muss beides eben gelten, weil wenn das Element in mindestens einer der beiden Mengen ist, hätten wir ein Problem

denn dann wäre es ja nicht mehr außerhalb

[golden blade]

deswegen folgt das

das könnte man so argumentieren

[tiny pawn]

wir haben noch nicht ueber 'and' und 'or' behandelt.

[golden blade]

huh

dann sag halt dass beides gleichzeitig gelten muss

[tiny pawn]

es sieht aus, als ob wir es unmoeglich es ohne and und or zu beweisen koentten oder?

[golden blade]

wie meinst du and und or

[tiny pawn]

hier hast du direkt and geschreiben oder?

[golden blade]

ja

ich köntte es auch theoretisch denke ich so formulieren

\begin{align*}
\Omega \setminus ( A \cup B ) &= { x \in \Omega : | : x \not \in A \text{ and } x \not \in B } \
&= { x \in \Omega : | : x \not \in A } \cap { x \in \Omega : | : x \not \in B }
\end{align*}

[thorny flame]

bacc (unhelpful)

[golden blade]

wenn ersteres gilt folgt das zweite eben

das erste kommt davon, weil es so eine art festellung ist, die gelten muss

das element darf eben in weder noch A oder B sein

das heißt anders ausgedrückt es darf weder in A noch in B sein

und so folgt es dann würde ich sagen

[tiny pawn]

ja ja ich verstehe. es ist logisch.

[golden blade]

prinzipiell ist die schnittmenge and und vereinigung or gleich, der unterschied ist, die runde schreibweise wird in der mengenlehre benutzt, und die "eckige" in der logik

[golden blade]

das ist halt wie als ob du mich fragst, warum folgt aus x-1=0 dass x=1

ok ja das könnte man durch einsetzen zeigen lol

[tiny pawn]

aber die frage ist von unserem Lehrer nicht von mir

[golden blade]

also was ich meine ist, dass ich das jetzt so grundlegend zerlegt habe, dass ich denke es keines noch tieferen beweises erfordert

[tiny pawn]

okay, danke

[golden blade]

jetzt schaust du enttäuscht 😂

[tiny pawn]

nein, ich bin zufrieden.

.solved

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[simple orchid]

Hello

[simple orchid]

What does 12 a) mean?

.close

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[lethal hornet]

can someone help me with this?

[lethal hornet]

@everyone

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@lethal hornet Has your question been resolved?

[reef fjord]

what is the question?

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[reef fjord]

it may help to draw a picture