help-39

[pearl pond]

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[burnt lodge]

I am a little confused on how I am supposed to do this question

[uneven cypress]

Do you know that for some function F whose derivative is f, $\int_a^b f(x) dx = F(b) - F(a)$ @burnt lodge

[jolly parrot]

Adhi

[burnt lodge]

yes

[uneven cypress]

Try using this result

[burnt lodge]

thanks, I think I got it

.close

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[slate matrix]

can anyone tell why the case where radius of S is 12 is rejected?
I'm getting 2 answers
48 when we take radius of S as 8
and 72 when radius of S as 12
But the tests solution only has 48 as answer. Is 72 also correct or am I missing something?
Even after making graph on desmos everything seems correct for 72 but no one was talking about this so I'm confused.

[frank violet]

Can you send the solution here?

I mean yah there should be two cases for circle S

[tardy reef]

you are right, power of point A is 4, so that its tangent is 2 units, and thus center can be (8,8) as well as (12,12)

[pearl pond]

@slate matrix Has your question been resolved?

[pearl pond]

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[crimson nebula]

This is my attempt for Q3.

However I could not show why u is an upper bound of s

[random ermine]

what's x

[crimson nebula]

Where is x?

I think u r talking about s

[random ermine]

ur talking about x

read ur question

[crimson nebula]

Sorry , it should be s

[random ermine]

you literally showed u is an upper of S

you wrote it down

[crimson nebula]

In the second half , i showed that no number less than u can be an upper bound

But i doubt whether I showed correctly that u is an upper bound of s

I wrote that line in the proof , but not sure there's enough reasons to justify it

[random ermine]

yes it's correct

fix s

u + 1/n >= s for all n

[crimson nebula]

does u beinf inf (u+S') implies that u is an upper bound of S..?

[random ermine]

so s is a lower bound of u+S' = {u + 1/n: n in N}

[crimson nebula]

Yup that's the idea

[random ermine]

so s <= inf(u+S') = u

[crimson nebula]

Okhay . Thanks

Rest all is fine ?

[random ermine]

yes

[crimson nebula]

Ok ty

.close

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[spiral coyote]

For the condition x ^ 2 + y ^ 2 - 4x + 3 = 0 Find:

(i) maximum and minimum value of x ^ 2 + y ^ 2

(ii) maximum value of x ^ 2 + y ^ 2 + 2y

[spiral coyote]

Idk why what i tried first for the second one is wrong

[dapper kraken]

consider doing this graphically

[spiral coyote]

Why can't i use the range of the first part to solve the second partTT

[toxic lichen]

what does this mean exactly

i -> [1,9]?

[spiral coyote]

I reused the range from the first part

[spiral coyote]

Oh the numbering

[hard crystal]

What do you mean by that?

[toxic lichen]

ok so you found that min(x^2+y^2) = 1 and max(x^2+y^2) = 9 in part (i)?

[spiral coyote]

Yes

[toxic lichen]

what this gets you is that the range of x^2+y^2+2y is contained in [-1, 11]

but you do not get that either of these bounds are sharp i.e. achieved for some value of theta

you get that x^2 + y^2 + 2y ≥ -1 for all eligible (x,y) -- that much is correct -- but you don't know that -1 even can be achieved.

same story with 11.

[spiral coyote]

,w 5-sqrt(20)

[jolly parrot]

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Requested by ._ch3rry_.7

[spiral coyote]

,w 5+sqrt(20)

[jolly parrot]

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Requested by ._ch3rry_.7

[spiral coyote]

I dont get it :>

[toxic lichen]

the minimum of the sum of two functions is not always equal to the sum of their individual minima.

i can give you a simple example where your kind of logic fails, if you want.

[spiral coyote]

That's what i cant wrap my head around

[spiral coyote]

Yes please

[toxic lichen]

ok

let's consider two functions on the interval [0, 5]:

f(x) = x^2
g(x) = -6x

can you fill in the blanks here?

The minimum value of f(x) for x ∈ [0,5] is ____, achieved at x = ____.
The minimum value of g(x) for x ∈ [0,5] is ____, achieved at x = ____.

these should both be quite simple. don't overthink it.

and don't try to jump ahead.

[spiral coyote]

Okay

0,0;0,0

?

[toxic lichen]

are you sure about all of these

[spiral coyote]

Oh wait no

Its -6

[toxic lichen]

also copy down the full thing please

with blanks filled

[spiral coyote]

can you fill in the blanks here?

The minimum value of f(x) for x ∈ [0,5] is 0, achieved at x =0.
The minimum value of g(x) for x ∈ [0,5] is -30, achieved at x = 5.

[toxic lichen]

ok, correct.

notably, these x-values are the only places where each function reaches its respective minimum, yeah?

[spiral coyote]

Yes

[toxic lichen]

now, your logic would have us believe that the minimum value of f(x)+g(x) would be 0 + (-30) = -30... but at what x would we achieve it?

[spiral coyote]

Idk

Its not in the domain?

[toxic lichen]

incorrect

the correct answer would be we wouldn't

because f and g don't achieve their minimum at the same x

if they did achieve their minima at the same x, then that shared x would also be the minimum point for f+g -- but it's not.

in fact the true minimum of f(x)+g(x) = x^2 - 6x is -9 achieved at x = 3. which is equal to neither of the two functions' minimum x-value.

[spiral coyote]

Wouldn't it be $3\pm i \sqrt{21}$

[jolly parrot]

ch3rry

[spiral coyote]

Okay so

[toxic lichen]

no

complex numbers and ordering don't mix

i will advise against trying to open that can of worms

[spiral coyote]

My answer could've been correct if both functions achieved their minima at the same theta?

[toxic lichen]

yes.

and the same for maximum.

btw, when you're finding minimum and maximum values, complex numbers should not be appearing in the picture

[spiral coyote]

Oh thts bcz you asked when x²-6x=-30

[toxic lichen]

i didn't ask you to literally solve that equation

[spiral coyote]

At what x isnt solve for x 😭?

[toxic lichen]

the correct train of thought in optimization terms would be:

in order for f+g to be -30, we would need to have both f(x)=0 [which happens at x=0] AND g(x)=-30 [which happens at x=5].
but we cannot have both. therefore there isn't any such x.

[spiral coyote]

True

Ah okay i think i get it

Same applies to maxima too right?

[toxic lichen]

yes

[spiral coyote]

Right so what i wrote just isnt possible because theta can only hold one value at a time

Thnx♡

.solved

[pearl pond]

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[sharp smelt]

So what I've come to so far is $U(f)≥L(f,P)$ ,where $P$ is some arbitarry partion.

[jolly parrot]

wai

[sharp smelt]

Notation

[naive zinc]

You done

U(f) is an upper bound

Any upper bound >=sup

[sharp smelt]

oh right

🤦

Thanks!

.close

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[little scroll]

Hi, why is the answer to this x < -7 , +♾️ > and not x < -7 , 172 >

[toxic lichen]

0x < 172 (a.k.a. just 0<172) is true always, for any x; it's not the same as x < 172

@little scroll

[little scroll]

Can 0 < -4 be true @toxic lichen

[autumn fossil]

Well, is 0 < -4?

[little scroll]

No, so it only works on positive numbers

[autumn fossil]

It's never true. 0 < -4 cannot be true. Is there some extra context to it? Is it part of some larger inequality? If so, can u send the full question?

[autumn moat]

Can you re write the question pls?

[toxic fractal]

!original

[pearl pond]

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.

[drifting vale]

The answer to this is that x is any real number

[little scroll]

Yas that

[drifting vale]

Ye, so what

[little scroll]

Please wait i will transleate

[toxic fractal]

post the original first, translate later

[autumn moat]

What the question

[little scroll]

The solution to the first inequality is every real number x, so the solution to the second inequality x \geq -7 is also the solution to the system.

This is answer

So if the answer is any real number it can be even negative number

[drifting vale]

Yes, it can be a negative number

<@&268886789983436800>

[little scroll]

Only when 0 > - ♾️?

[drifting vale]

Wdym

The inequality is true for any number x

[little scroll]

But can it be 0 < - 5 is this also answer or is it that it doesnt have answer

[warped violet]

I don't think you grasp the meaning of inequalities.

[little scroll]

I agree

[warped violet]

0 < -5 literally says "0 is less than -5"

That is never true.

[autumn moat]

Fs

[drifting vale]

Your inequality does not say that

[bronze heath]

no real solution, wait maybe no solution

[little scroll]

I get that but how 0x < 172 can be any real number

[warped violet]

Because 0x is equal to 0

Anything multiplied by 0 is 0

Therefore you get 0 < 172

Which is true, 0 is less than 172

Any x you use, you end up with 0 < 172

So it's true for all x

[drifting vale]

But the inequality does not imply 0 < -5

[little scroll]

So x > 4 cannot be any real number?

[autumn moat]

Use the formula b² -4ac

[warped violet]

Well pick x = 3

You get 3 > 4

Is 3 greater than 4?

[drifting vale]

It's anything to the right of 4 on the real number line

[little scroll]

Okay i think i am starting to get picture

[warped violet]

0x < 172 is just a special case

Something like x > 4 is just that. x must be greater than 4

[autumn moat]

???

[toxic lichen]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[bronze heath]

inequality -> equality

[little scroll]

I get this now, thank you very much❤️

[autumn moat]

I know

I was tryna solve his question

[drifting vale]

It's not the question tho

They wanted to solve 6

You're trying to solve 1

Oh wait

Ye, I see now. Sorry

[little scroll]

Thank you i solved first but i was stuck on what 0x meant

[autumn moat]

Ah alr

[toxic lichen]

ok but your handwriting could use a lot of improvement

your x does not look like x

[autumn moat]

Yeah 🤣

[warped violet]

Oh I thought it was n

[drifting vale]

Me too

[autumn moat]

Xd

[little scroll]

Same🤣

[autumn moat]

My handwriting sucks

[warped violet]

Okay just to be sure

What's your solution now

[warped violet]

Well

-7 is included

Because x is greater or equal to -7

[drifting vale]

Use the forearm a bit, not just the fingers @autumn moat

[little scroll]

You are right i forgot to use [

[warped violet]

But also my suggestion

Don't just leave 0x < 172

Write something like 0 < 172

And $x \in \mbb{R}$

[jolly parrot]

USS-Enterprise

[warped violet]

Or at least the last part

And then you find the intersection of that and x ≥ -7

[little scroll]

Like this?

[autumn moat]

Why do you draw the lines btw

[little scroll]

My friend taught me to solve it like that

[bronze heath]

better write +infinity instead of R

[toxic lichen]

what is this R meant to be

[little scroll]

I think i didnt understand where should i write this @toxic lichen

[toxic lichen]

if you want to write +infty then write +infty.

that R refers to the entire real line.

[drifting vale]

First inequality says "x is any real number"

Second inequality says "x is greater than or equal to -7"

So x is any real number greater than or equal to -7

It's the intersection

[-7, +inf)

So what Ann said applies only to the first inequality

It's the first statement

[little scroll]

I corrected my answer

Thank you guys you are real help❤️

And please give this thread to someone in need

[bronze heath]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

@little scroll Has your question been resolved?

[little scroll]

.close

[pearl pond]

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[digital hinge]

i need help on this exercise please
i don't know how to solve it

[digital hinge]

i've tried to use the classical formula of the limit but i can't seem to go anywhere useful and my exam is tmr

<@&286206848099549185>

[pearl pond]

@digital hinge Has your question been resolved?

[glossy trench]

What do you know about finding directional derivatives? Is the limit definition the only way you've seen how to do them?

the easiest way to do this is taking the dot product of the gradient and a unit vector in the direction

[digital hinge]

okok i think i got it, but i didn't know i was supposed to do that,
thank you!

[pearl pond]

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[digital hinge]

.close

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[spiral coyote]

A variable circle w passes through the fixed-point A(p, q) and is tangent to the x-axis. Prove that the point diametrically opposite to A lies on (x – p)² = 4qy

[compact ridge]

have you tried drawing it out?

[spiral coyote]

"Is tangent tk the x axis" means tht x axis is a tangent right

As in it touches x axis at one point

[compact ridge]

mark A, the point on the x-axis, and the diametrically opposite point and then it's the right angle inscribed by the diameter

[toxic lichen]

yes

that is what "tangent to the x-axis" means

[spiral coyote]

The eqn of circle would be x²+y²-(p+h)x-(k+q)y +hp+kq=0

[bronze heath]

h and k is?

[spiral coyote]

Point diametrically opp to A

.close

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[frank violet]

that was fast

[spiral coyote]

I got it lmao

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[prime hill]

Someone pls help me with my proj hahahaha I did the design I jst need help with all every math parts

[ivory basin]

Well what do you need help with

[prime hill]

Math - Rounded Park
Front Page

• Make sure to have enough space for the strip of colored paper and make sure that the strip is cut cleanly (be generous with the strip but not too much)
  Second Page
• The actual Park
• Overlay the grid to the park and place some landmarks on the x- and y-axis
• Maximize the space and fill it of possible
• Limit the landmarks (5-7 only)
  Third Page and onwards
• Isolate each landmark, make sure to zoom in/out to maximize the space
• EACH page is reserved for ONLY one landmark
• The circular landmark will be placed on the left side of the page while the right side is reserved for the FINAL (this means no solutions) equation in general and standard form
  2nd to the last page
• This is where your table is located (only one table)
• Area/Landmarks, Center, Radius, Equation in Standard form, and Equation in general form
• Maximize space
  Last Page
• This is your solution page (can be 1-2 pages)
• The ONLY solution needed is for the general form (no need for standard)

So this was the instruction

and here are some examples from my classmates

my problem is here is the design that Ive made

I need help with all every math partss which is needed

oh

<@&286206848099549185>

[sudden jungle]

can you tell me what grade you are on

maybe I can help you

ah well

so like what math part?

@prime hill

bruh what

[prime hill]

Grade 10

[sudden jungle]

oh dang

welps I can still try

[prime hill]

hahaah

[prime hill]

See the examples from my classmates and instructions like the graph and grid thingy an yes the every math needed i need help I jst made the design an idk what to do next

[sudden jungle]

ok lemme read the instructions again

well let's start with the coordinate plan then

so first we need to figure out the area of the play area

[prime hill]

Oki

Btw is this good?

That’s what I made

[sudden jungle]

I mean you need coordinates

it has requirements

[prime hill]

Yepp thats why I need

And why imm telling

[sudden jungle]

ohhh

k

so first you do the main lawn then

it has the center at (0, 0) which u do

the radius is 30 meters

which means the left most point should be (-30, 0)

and the right most should be (30, 0)

[prime hill]

Where did u get those

30 meters

[sudden jungle]

in the instructions

💀

did I read it more careful than u

it's in the google docs part

[prime hill]

Wait where-

[sudden jungle]

under coordinate plan(cartesian grid)

[prime hill]

I cant see ot

[sudden jungle]

oh wait that's example shoot

[prime hill]

Oh..

Yeah

I was gonna say it

Thats an example

[sudden jungle]

well you can decide the dimensions by yourself right?

[prime hill]

U have to base it from my design and thats my problem that’s why im here hahahaha

[sudden jungle]

*?

[sudden jungle]

hmm you say this is project

ok

[prime hill]

Yes

Exactly

🥲

And im cramming sht

[sudden jungle]

well I guess you can decide where the center of the pond is?

[prime hill]

Type shi im sobbing

[sudden jungle]

dang

[prime hill]

Huh

Crammer godz

[sudden jungle]

base on the requirements

[prime hill]

Im so dumb with maths

[sudden jungle]

you are supposed to use most of the space

for your grid

[prime hill]

Yep

[sudden jungle]

Second Page
The actual Park
Overlay the grid to the park and place some landmarks on the x- and y-axis
Maximize the space and fill it of possible

ok then

[prime hill]

Did i did it wrong

[sudden jungle]

you have examples, which is great

[prime hill]

Yepp

[sudden jungle]

no you did it right

so u can also suppose that the center for the lawn is (0, 0)

[prime hill]

I jst dont understand the math parts okay like with my own

[sudden jungle]

and the radius is 20

[prime hill]

U mean main

Main lawn

WAIT

[sudden jungle]

yeah

[prime hill]

MAIN LAWN IS THE WHOLE CIRCLE?

[sudden jungle]

yes

[prime hill]

Can we like label the parts first

[sudden jungle]

so you want to label the coordinates for the centers?

k

[prime hill]

I have
fountain
Playground
Parking lot
Coffee shop
Fish pond

[prime hill]

Yes

all maths

Wait fountain is main lawn?

[sudden jungle]

to make your life easier

hang on

no the fountain isn't

you just happened to make them have the same center

the fountain and the main lawn

and to make your life easier

u could literally get rid off the foundtain tbh

*fountain

[prime hill]

wut

Ahahahha

[sudden jungle]

see this example

[prime hill]

Ye

[sudden jungle]

he/she made the landmarks' centers on the x and y coordinates

so you can do that too

[prime hill]

What do u meann omg

[sudden jungle]

ok so now we are only talking about the examples for now

[prime hill]

Ye

[sudden jungle]

u see that for the playground thingy

that its center is on the x-axis

[prime hill]

Ye

[sudden jungle]

maybe it's a bit off

[prime hill]

ughhh I hate this

Im so tired

Its 1:15am

[sudden jungle]

holy

do you have school

[prime hill]

I dont wanna restart

[sudden jungle]

ok

[prime hill]

Nope its sunday

[sudden jungle]

I'll help u base on your thing

[prime hill]

Oki

🙂

[sudden jungle]

lemme put it here again

how about start with the fountain

so like it's center is (0, 0)

its

so then you can set its radius as 5

[prime hill]

Wait

What is radius hahaha

[sudden jungle]

BRUH

[prime hill]

hahahahahaahha

[sudden jungle]

ARE YOU ACTUALLY 10th grade

[prime hill]

Yes

REFRESH PLS

[sudden jungle]

what's your grade in your math

[prime hill]

secret

Im not passing maths bruh

[sudden jungle]

the radius is half of its diameter

[prime hill]

👽

[sudden jungle]

diameter is the longest line in the circle

[prime hill]

YES

IM RIGHT

[sudden jungle]

k

[prime hill]

im jst not certain okay

😭😭

[sudden jungle]

bruh I'm 8th grade 💀

[prime hill]

OH

MY GOSH

[sudden jungle]

well I'm studying honors geometry

[prime hill]

We didn’t study circles in g8

[sudden jungle]

*taking

[prime hill]

Ohhh

That’s why

[sudden jungle]

u know the area formula for circle?

[prime hill]

Okay

[prime hill]

I forgot

[sudden jungle]

......

[prime hill]

Hahahahahahaha

Sorry bro

[sudden jungle]

A = pi * r^2

a stands for area

r is radius

[prime hill]

yeaa

What now

[sudden jungle]

so do what i say maybe

the rightmost point for the fountain

make it (5, 0)

and leftmost is (-5, 0)

[pearl pond]

@prime hill Has your question been resolved?

[pearl pond]

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[lost steppe]

,rotate

[lost steppe]

Hi could I have some help on how to start w this pls

[toxic lichen]

what if the -2^n wasnt there

[lost steppe]

Then it would just converge to 3

Yh idk where to go w that

[honest spear]

I think what ann was hinting at is that as n gets larger the 3^n is the dominating term. 2^n doesnt grow fast enough and its "as if" 3^n is the only term there

theres probably some neat algebraic trick to show it converges to 3 as well

oh yeah

[lost steppe]

But as an analysis q, idk how to show that it converges to 3

[honest spear]

start with $$3^n -2^n$$

[jolly parrot]

Nyxzore

[honest spear]

and factor out the 3^n

thats my hint

[lost steppe]

Hm Alr

Oh if I show that (1 - (2/3)^n)^1/n converges to 1

Then it’s done

[honest spear]

fr

and thats geometric so its a done deal

[lost steppe]

Or I could use Bernoulli’s?

Inequality

Wait not it’s a negative

[honest spear]

u dont want to just do this by inspection?

geometric goes to 0

you get $1^0$

[jolly parrot]

Nyxzore

[south inlet]

I mean, Bernoulli’s does work if the part in the bracket is greater than -1, so it can be negative

[cinder flower]

,av eugene_krabs_has_cake

[jolly parrot]

eugene_krabs_has_cake's Avatar

Click here to view the image.

[honest spear]

could say find ln of the limit and eventually just get 0 and you just have e^0=1

[crude pewter]

$log(3^n-2^n)=nlog3+log(1-(\frac{2}{3})^n)\sim nlog3- (\frac{2}{3})^n$ So \$\lim (3^n-2^n)^\frac{1}{n}=e^{\lim \frac{log(3^n-2^n)}{n}}= e^{\lim \frac{nlog3- (\frac{2}{3})^n}{n}=e^{log3}=3$

[honest spear]

exactly

[south inlet]

Not again 😭😭

[honest spear]

can we call the latex police?

💀

[dire vortex]

Could you use sandwich theorem too? $(1-\frac{2}{3})^{1/n} \leq (1-(\frac{2}{3})^{n})^{1/n} < 1$

[jolly parrot]

Ollie

[honest spear]

many ways to skin a cat fr

[jolly parrot]

vs
Compile Error! Click the reaction for more information.
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[pearl pond]

@lost steppe Has your question been resolved?

[crude pewter]

Clearly $e^{\lim \frac{-(2/3)^n}{n}}=1$

[jolly parrot]

vs

[pearl pond]

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[empty reef]

How can I write my final answer in the form circled above

[empty reef]

(Preferably using the logarithms method and not indices)

[prisma kernel]

well you can write log(6) as log(2*3)

[empty reef]

https://tenor.com/view/speed-ishowspeed-reaction-reactions-meme-gif-8141100727479891642

Oh my God

[prisma kernel]

you saw it?

[empty reef]

Wait yes and what then

[prisma kernel]

break it into $\frac{log(3) + log(2)}{log(3)}$ - 1

[jolly parrot]

doctorstrangejr

[prisma kernel]

now separate the fraction

!nosols please

[pearl pond]

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.

[polar river]

Let him answer it bro

[empty reef]

Okay let me do that

[prisma kernel]

sure

[empty reef]

I’m sorry I haven’t touched this topic in ages so I don’t recall those lil things

[prisma kernel]

no problem mate, even I didn't spot it for a few seconds

[empty reef]

Okay tysm I got it now

[prisma kernel]

Alright that's nice, what did you get?

[empty reef]

Log3(2)

[prisma kernel]

yes that's correct

[empty reef]

Thanks 🙏 imma close this now

.close

[pearl pond]

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[pearl pond]

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[abstract pagoda]

Someone help me to prove lagrange theorem

[drifting vale]

Which one 💀

[polar river]

💀

[toxic lichen]

lagrange was quite a prolific guy and has multiple theorems named after him.

which one do you mean

op offline 💀

[polar river]

I think he mean all 💀

[warped violet]

They did use singular form

So guess not

[pearl pond]

@abstract pagoda Has your question been resolved?

[abstract pagoda]

Alll

[drifting vale]

So you want like 30% of calculus proven

[abstract pagoda]

Well in the group theory one

[drifting vale]

What do you not understand exactly

In the proof

[abstract pagoda]

I just want proof

The equality of the cosets

Of group G

[toxic lichen]

so you want the group theory one.

[abstract pagoda]

Yes sis

[toxic lichen]

https://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)?wprov=sfla1

i gtg but at least we managed to get you to say what you ac want

[abstract pagoda]

Its enough to help me to understand the proof of
|giH|=|H|
for all i in {1,2,…,|G|}

Uhh

[pearl pond]

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[sly mulch]

Hi i have a test in like 5 days and i cant fail it or else i have to repeat the year💔 soo im stuck at monomials and polynomials (we didnt have the teacher for quite a while in middle school so like i never did those and now im in high school and if i dont know these things they wont let me pass💔 ) i tried doing some of these exercizes but they never turn out the way the book says they should.

[robust oar]

,rccw

[robust oar]

can you show work?

[sly mulch]

Yes sure one second

[polar river]

Is the right side the answers?

[pastel umbra]

It is

[sly mulch]

yes

i did n.188 and n.193 but i didnt finish them bc they were totally wrong(the numbers of the fractions were way too big) ill take a photo of the others i did rn

[polar river]

Try answering some small ones first before going some big ones

5 days is so big you can do it

[sly mulch]

Okay ill try doing the first one(which is n.186) and show the work🙏🏼

[sly mulch]

i hope so😭

[polar river]

Do it step by step first before going all out to avoid confusion

[sly mulch]

im stuck at a term

if i have (100a2) : a4 does the monomial become negative? like the result is -100 a2?

[polar river]

Show work

[cinder flower]

beautiful handwriting

[sly mulch]

Hahahah thanks i think its kinda messy

[cinder flower]

😭 it’s not

[polar river]

It should be (10a²)²

Then what's next

Look back at you're solution

[sly mulch]

so then its 100a4 :-a4

ohh

[polar river]

Yeah

Wait

[sly mulch]

yes?

[polar river]

Why's there's -?

100a⁴:a⁴ is the same as 100a⁴/a⁴

[sly mulch]

so i write 100

[polar river]

Yes

[sly mulch]

Okayokay

now its 2a6-a6+a6+100

[polar river]

Look carefully at you're solution don't rush

[sly mulch]

Okokk

[sly mulch]

wait but is this wrong

because the result is 102a6

[polar river]

Nope

[polar river]

Look carefull at (-a)⁶

For example if i have (-1)² what it will be?

[sly mulch]

+1

?

[polar river]

Yes

How about (-1)⁴?

[sly mulch]

+1

[polar river]

Yep

Then (-1)⁶?

[sly mulch]

+1

[polar river]

What about this?

Remember a is a variable

[sly mulch]

+1a?

[polar river]

Each individual variable always has 1

[polar river]

Correct

Apply PEMDAS or BODMAS

Exponent first before parentheses or bracket

[sly mulch]

but when i do 2a6-1a6+1a6 its still 2a6

[sly mulch]

sorry what is that

[polar river]

Look again at you're problem

[polar river]

You have -(-a)⁶ right?

Don't forget the negative outside

It will become -(+a⁶)

Then -a⁶

Did you get it?

[sly mulch]

wait but doesnt it become + when there are two - ?

[sly mulch]

like instead of -(+a6) its +a6

[polar river]

Exponents apply before the outer minus sign

It's an order of operations

-(-a)⁶ isn't the same as -(-a⁶) if we have -(-a⁶) first then we can apply directly the sign so it will become + but the exponent is in the outside that's why we multiply the exponent to the inside

-(-a)² is the same as -[(-a)(-a)]

[sly mulch]

OH WAIT

i think i get it

[polar river]

-(-a)⁶ is the same as -[(-a)(-a)(-a)(-a)(-a)(-a)]

Do you get what I'm saying?

[sly mulch]

Yesyes

ill rewrite it one second cuz its creating me confusion

[polar river]

We must solve the inside of the parentheses first before the outside

[red halo]

It's basically the order that you have to solve the equation in

👍

[polar river]

Remember the PEMDAS rule. Parentheses first, Exponent, Multiplication, Division, Addition, then lastly Subtraction

[sly mulch]

i think i did it

when the coefficient is 0 we remove also the letters right

[polar river]

Yep

[sly mulch]

I solved it🥹

thank you

i’ll do the one after it

187

[polar river]

Show answer after you solve it

[sly mulch]

Okay

[polar river]

Dont rush, take your time answering it

[sly mulch]

I finished it,is it correct?

[polar river]

Try medium ones

[sly mulch]

Okk

[polar river]

Before you try answering it, arrange it first, change the : into fraction form to avoid confusion

[sly mulch]

what do u think is medium? because in the page they’re all described as easy level

[sly mulch]

Alright

[polar river]

Yeah they look easy if you arrange them correctly

[sly mulch]

I’ll try doing the n.200

[polar river]

Ok

[sly mulch]

Ok i finished it but i think i made a mistake

the book says the result is xy4 but i got xy3

[polar river]

Wai

Did you distribute the exponents to the variables?

[polar river]

You forgot to multiply the outside exponents to the variables inside

[sly mulch]

In which term?

[polar river]

All of them

[sly mulch]

Oh😭

i’ll redo it

[polar river]

Look carefully

[sly mulch]

i got ittt

idk why i get so distracted while doing math i keep forgetting about lots of things

[polar river]

Do you get it now?

You can just get rid of the 1

[sly mulch]

Yes,but there is one that i repeated 2 times yet i cant do it. Is it okay if you help me with it or do you have to go?

[sly mulch]

oh alright

[polar river]

Ok

[sly mulch]

okok its the n.201

[polar river]

Show

[polar river]

What part are u stuck

Don't divide first the 12/16 and 27/8

[sly mulch]

Huh why

[polar river]

Solve first the 27/8 -3 and so on

Because there's a negative sign that connects to it

Wait

[sly mulch]

yes but then its connected to the other multiplication

like

idk what order i have to follow

[polar river]

Yesyes you're correct

That was on me my bad

Use pemdas rule

[sly mulch]

No problem

[sly mulch]

uh so i do the multiplication first

12/1•1/16

?

[polar river]

Yes

[sly mulch]

Okok i got 12/16 is it correct

wait

i have a doubt

the multiplication was 12•1/16 but i turned 12 into a fraction by making it 12/1,was that correct or should i have done 12•1 and 12•16?

[polar river]

12/1 is the same as 12

[sly mulch]

Ohh okay okay

[sly mulch]

so the answer to this is correct

because i wrote

12/16

[polar river]

What's your answer

[polar river]

Yes

[sly mulch]

Okay okay

when i went to divide 12/16 and 27/8 i think i made a mistake

[polar river]

Check carefully

[sly mulch]

Okok

i divided the letters

and i got b2 c4

now i have to multiply the numbers

[polar river]

Correct

Yep

[sly mulch]

Do i simplify 16 with 8 or do i leave it like that?

[polar river]

Simplify

So it will be simple

[sly mulch]

Okok

[fiery willow]

What is this about

[polar river]

This

[sly mulch]

12/2•1/27 i got 12/54

[polar river]

Simplify it more

[sly mulch]

4/18?

i divided them both by 3

[polar river]

More

[sly mulch]

2/9?

[polar river]

Yep

[sly mulch]

wait so i can simplify in two steps using two different numbers?

my teacher told me we cant😭😭

[polar river]

You can just use 6 in that so it will not be 2 steps

Find the biggest number to simplify so it will just one step

[sly mulch]

ohh i didnt think of that

Right

alright now i have 2/9b2 c4-3c•[b4c:(-6b)2

[polar river]

Yes

[sly mulch]

sorry if it looks confusing i dont have the powers option in my keyboard

[polar river]

Proceed to the right side [b⁴c:(-6b)²]

[polar river]

It's fine, I'm just using my phone too

[sly mulch]

Okok

[sly mulch]

[b4c:(36b2)]?

[polar river]

It's not 36b²

Wait

[sly mulch]

huh but why

6x6=36

[polar river]

It's correct mb

[sly mulch]

Oh npnp

ok now this is what i find confusing

[polar river]

Where

[sly mulch]

to divide this,do i turn b4c into 1b4c?

so that i do 1b4c:36b2

or do i js divide the letters and keep the number 36

[polar river]

Just divide the letters no need to put 1

[sly mulch]

Okok

[polar river]

Keep the 36

[sly mulch]

36b2c

?

[polar river]

Nope

It's still b²c:36

You only divide the letters

[sly mulch]

wait what

[polar river]

This $\frac{b^4c}{36b^2}$ to this $\frac{b^2c}{36}$

[jolly parrot]

sice19

[sly mulch]

and how do i do this

[polar river]

Remember: is a fraction

[sly mulch]

1:36?

[polar river]

Yep it's the same

[sly mulch]

yeah that was what confused me

[polar river]

That's why i said you don't need to put one because every variable has a 1

[sly mulch]

here

[sly mulch]

Oh okok

[polar river]

Every variables that has no numbers has an imaginary 1

Like a, b,c theres 1 in each of them

1a,1b,1c

[sly mulch]

Alright ill keep that in mind

so now how do i do b2c:36 because when i do it on the calculator the number is decimal😭

like i did 1:36

[polar river]

Don't do decimal

Just keep it on 1:36

We can just use fraction

[sly mulch]

Ok i do 1/36 b2c and leave it like that

[polar river]

Yes

[sly mulch]

Now i have to multiply it by 3

[polar river]

Same as 2/9

Yes

[sly mulch]

i simplify

36 with 3

so it becomes 12

[polar river]

Yes

Yes

[sly mulch]

i got 1/12 b2c4

[polar river]

And no

[sly mulch]

😭

what did i mess up

[polar river]

The letters are above

[polar river]

Here

You didn't put it in the bottom

1and 36 are seperate

Remember what's beside 1

[sly mulch]

so what do i do

[polar river]

Let's go back

You have this right? b⁴c:36b²

[sly mulch]

Yes

[polar river]

Then you did is divide the letters

b²c:36 right?

[sly mulch]

Oh yesyes

[polar river]

After that you multiply the -3c³?

So it will become -3c³(b²c:36) right?

Then -3b²c⁴:36

[sly mulch]

yup

wait what

[polar river]

Then you see what i did in the letters?

[sly mulch]

OHH

yes u added the c

[polar river]

I multiplied it

Yes the letter exponents will be added if i multiply it

[sly mulch]

but wait

[polar river]

Then it will become -b²c⁴:12

Simplification

[sly mulch]

shouldnt we divide the letters and multiply the numbers since it isnt a multiplication?

wait

oops nvm

it is a multiplication

😭😭

[polar river]

All I did is step by step so that you can understand

[polar river]

It's multiplication because it has parentheses

[sly mulch]

wait so

can u repeat one second

we have

3c3•1/36b2 c

[polar river]

The letters b²c are beside of 1

[sly mulch]

okay okay

[polar river]

Wait

Let's go back

[sly mulch]

Yes?

Okok

[polar river]

First we have $-3c^3[b^4c:36b^2]$ right?

[sly mulch]

this is what i wrote so far

[jolly parrot]

sice19

[sly mulch]

Yes

[polar river]

This is correct

[sly mulch]

Alright thanks

So then

Wait

now what do i do

[polar river]

Try to solve that using what you've wrriten

[sly mulch]

Okok

wait i’ll write it on digital really quick

[polar river]

Ok

[polar river]

Yes

Remember before b⁴c:36b²?

[sly mulch]

Yes

[polar river]

We only just divide the letters so it becomes b²c:36

[fiery willow]

Yes

[polar river]

It never goes down

[sly mulch]

Okok

so now what remains is

2/9 b2 c4- 1/12 b2 c4?

[polar river]

Yes

[sly mulch]

Ok ill do it now

do i simplify 12 with 2or leave it like that

[polar river]

Just minus 2/9-1/12

Leave it like that

[sly mulch]

Okok

33/108 b2 c4

[polar river]

How did you get 33/108?

[sly mulch]

WAIT SHIT

i added them

instead of substracting

😭

its 15/108 b2 c4

[polar river]

Simplify it

[sly mulch]

i simplified them both by 3 and it turned out 5/36 b2 c4

[polar river]

Yep

[sly mulch]

Its correct

Thank you so much🙏🏼

[polar river]

You're welcome

Practice more

[sly mulch]

Alright

i’ll do the others alone and if i have doubts i’ll ask (at the very end) i dont really wanna force you to help me because it has also been 3 hours 😭 if you dont feel like keep going thats okay

[polar river]

Bruh what

It's 3hrs already??

[sly mulch]

yes

we started at 22:27

if im not mistaken

[polar river]

I never noticed

[sly mulch]

oh wait shit

23:27

[polar river]

Well, math is fun

It's 8am here

[sly mulch]

Yeahh when u understand it hahaha

[polar river]

Doing math since yesterday

[sly mulch]

i forgot about timezones

[sly mulch]

Hahahahahaha

[polar river]

Peace out ✌️

[sly mulch]

wait what does that mean sorry

im not good at english

😭

[polar river]

Just goodbye

[sly mulch]

Ohh

goodbyee

[polar river]

Byee

[pearl pond]

@sly mulch Has your question been resolved?

[pearl pond]

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[exotic scaffold]

for the last part would there not be a + C for the general term

[heavy onyx]

you may add a +C

[unborn abyss]

and should!

[pearl pond]

@exotic scaffold Has your question been resolved?

[rotund nymph]

the whole thing is messy
there is no exponent by the -1

there would be a +c yes

[exotic scaffold]

ok thanks got it

yea i thought so too just wanted to make sure

.close

[pearl pond]

Channel closed

Closed by @exotic scaffold

Use .reopen if this was a mistake.

[pearl pond]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[urban jacinth]

hi guys what does question 3 mean?

[urban jacinth]

(3) i'm confused ab

[compact ridge]

have you found the partial derivatives?

[urban jacinth]

yes; $\nabla f(x,y) = \left(\frac{1}{2\sqrt{x(y+1)}}, \frac{-\sqrt{x}}{2(y+1)^{3/2}}\right)$

[jolly parrot]

Moonful

[urban jacinth]

i'm not even sure what (3) is asking tho

[compact ridge]

right, so the values where y + 1 = 0 and where x/(y + 1) < 0 are not in the domain of A

but then there are values apart from these ones

such that either the x- or the y-partial derivative is undefined

[urban jacinth]

so x=0

[compact ridge]

exactly

[urban jacinth]

$\frac{\partial f}{\partial x}(0,y)$ is undefined

[jolly parrot]

Moonful

[urban jacinth]

how do i show that

[compact ridge]

cause sqrt(x) is in the denominator

so that's it

[urban jacinth]

is that a proof tho?

my lecturer's answers look like this

he used first principles - and i don't understand whats going on 😭

i mean of course i can read the working out

but i don't get why he's doing it

[compact ridge]

I feel that's unnecessary

[urban jacinth]

the problem is like

the unit i'm doing is very rigorous

[heavy onyx]

<@&268886789983436800>

[urban jacinth]

so i feel like he'd want a proof besides "x is in the denominator"

[compact ridge]

but your prof concludes that $\lim_{t \to 0} \sqrt{\frac{1}{t (y + 1)}$ doesn't exist

[jolly parrot]

south
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[gray otter]

yeah, just show that the limit does not exist

[compact ridge]

I mean it is more rigorous in that they are taking the derivative on the path (0, y) + t (1, 0)

and yes you do need to take the limit

[urban jacinth]

oh so we're just saying "there is an infinite gradient on the y axis which isn't a real number, and so the parial derivative at x=0 does not exist in the x direction (so not all partial derivatives exist on x=0)

and the second bit he did is redundant?

[gray otter]

the partial derivative exists if the limit / differential quotient exists, so yeah

[urban jacinth]

ah so just to clarify

infinity doesnt count as "existing" when it comes to gradients?

bet thank you very much

both ofyou

.close

[pearl pond]

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Closed by @urban jacinth

Use .reopen if this was a mistake.

[pearl pond]

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[wary vault]

Hi there. Im trying to prove that for p,n in N, if a_pn, a_(pn-1) , ... , a_(pn-(p-1)) converge to the same number, l, then a_n converges to that number too. I was too lazy to define n_0 to be the "maximum" number thats convenient for these inequalities to hold, but i think it is pretty obvious that there exists one. If my assumption isnt wrong, is the rest of the proof okay?

[wary vault]

also, in the last inequalities i skipped the part where we have, for instance: |a_pk -l|<ε and then implying its |a_n -l|< ε

shit i made a small mistake defining n_1 hold on

[rustic gate]

[wary vault]

hi snow

[rustic gate]

i think this is fine

[wary vault]

okayy thank you

.solved

[pearl pond]

Channel closed

Closed by @wary vault

Use .reopen if this was a mistake.

[rustic gate]

could probably be worded better

but the idea is right

[wary vault]

true im bad at words

[pearl pond]

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[urban jacinth]

hey everyone! could someone help me out with (2) please?

[urban jacinth]

here's what i have

$\lim_{x \to \infty} \frac{x+mx}{x^2 + (mx)^2 + x(mx)} = \lim_{x \to \infty} \frac{(1+m)x}{x^2(1+m^2 + m)} = 0$

[jolly parrot]

Moonful

[urban jacinth]

then i go to polar

$x = r\cos(t), \ y = r\sin(t)$

[jolly parrot]

Moonful

[urban jacinth]

and we say

$\lim_{r \to \infty} \frac{r(\cos t + \sin t)}{r^2 (1 + \sin t \cos t)}$

[jolly parrot]

Moonful

[urban jacinth]

what happens now?

what do i do now haha

[compact ridge]

since you already know the limit is 0, cause the idea is r/r^2 = 1/r

you just need to bound (cos t + sin t) / (1 + sin t cos t) for all t

[urban jacinth]

perfect

my real question is why do we bound it?

/why do we show that (cost + sint)/(1+sintcost) is bounded?

[compact ridge]

cause you can easily find the limit of something like 2r/r^2

[urban jacinth]

oh i mean yes but why is it important to show that (cost + sint)/(1+sintcost) is bounded? like what would happen if it wasn't bounded?