help-27

[lilac portal]

Yeah, but AM GM inequality literally comes from geometry

I've found a great picture that demonstrates it

So sqrt(xy) should be positive

[signal crag]

is there any way to justify this algebraically

[lilac portal]

There is

[signal crag]

could you explain why the square root is always positive algrebraically

[lilac portal]

Yeah I can, but u have to do proof by contradiction

First assume that it's the other way around

[signal crag]

so if i want to prove this in an exam, would i need to draw up a shape to prove that the square root is always positive

[lilac portal]

I mean, u can. But if u want to do it algebraically, u need to assume always the sqrt(xy) will be always positive. There's a proof on why, but it is too complicated for you rn

[signal crag]

so if i assume the square root is always positive, i should not theoretically lose any marks

right?

[lilac portal]

Because in order for the sqaure root to become positve, u need two positives or two negatives

I mean, u can prove it using proof by contradiction

It works better in this case

[signal crag]

if i have to all that for this, then i might as well prove it using a^2+b^2 >= 2ab

[lilac portal]

OOOOH I get what ur saying rn, ur saying something like this right??

[signal crag]

replacing a with sqrt(a) and b with sqrt(b)

[signal crag]

yeah

[lilac portal]

AAAAH

No, just assume that x and y are positive.

We only use +- if we don't know if the input of square root is positive or negative

since we assume that x and y are both positive, then no need to do that

and ur going to square it anyways

that eliminates the negatives

And we generally dont use negatives for x and y on am gm inequality. There is a separate proof for that. Here's what I found that can be useful https://www.vedantu.com/question-answer/can-a-geometric-mean-be-negative-class-10-maths-cbse-601b71bdea1e82352687be77

[signal crag]

ah

so its just a general rule

that means i can assume that the square root will be positive

[lilac portal]

Yeah

[signal crag]

nice

thanks so much

[lilac portal]

And at the last part of ur proof

If u really want to prove it u can assume two proofs for that one

(a+b)^2 >= sqrt4ab

[signal crag]

why sqrt(4ab)

[signal crag]

i was just wondering why the a+b is also not getting rooted

[lilac portal]

Ah i'm not really good at typing stuff here

Don't really know how to use latex

[signal crag]

ah no worries

[lilac portal]

But u can also use this

First is draw a rectangle with side a and b

[signal crag]

cool

[lilac portal]

Make 4 of this

Arrange it this way

The side length of the inner square is this

So
4 of the rectangle (ab) = (a+b)^2 - (a-b)^2

If we erase the (a-b)^2, we have 4xy <= (a+b)^2

Since (a+b)^2 is the whole square where it is made up of 4 rectangles ab plus some small square

We erased that small square so it is now bigger

Do what u did here but at the last proof

[lilac portal]

then this

[devout snow]

@signal crag Has your question been resolved?

[signal crag]

I see

That’s pretty cool

Thanks for all your help

I really appreciate it

.close

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[onyx parrot]

I don't understand this

I mean

I don't understand substitution method

!help

[devout snow]

Please read #❓how-to-get-help

[lilac portal]

U just replace x with given

x = 2y +4, so replace the x in 2x with 2y +4

[onyx parrot]

so its: " 2(2y +4) - 4y - 8 = 0 "?

[lilac portal]

In this case, there might be ordered pairs in the given

Are there ordered pairs in the multiple choices?

[onyx parrot]

(X,Y)

[lilac portal]

There we go

[lilac portal]

so from here

solve for y

if it is 0 = 0, then there are infinite ways to solve system of equations

If there is y = something, then only 1 solution

[onyx parrot]

ah alr

[lilac portal]

if it is not equal, then there are no solutions

[lilac portal]

since here it is 0 = 0

Then just replace all of those ordered pairs with their respective variable

Ordered pairs are always written like this
(x,y)

so just replace the x's and y's

[onyx parrot]

ah ok thank you

can you suggest a youtube video for this topic?

[lilac portal]

U can try khanacademy or just search in youtube

[onyx parrot]

ah ok

[lilac portal]

but what I did here so it is much easier is use either elimination or just add the two equations together

[onyx parrot]

elimination method?

[lilac portal]

Better to just add the two equations together because we always get 0 = 0 if we eliminate it

[onyx parrot]

but how do I know if I should add them or subtract them?

is there a way to tell?

[lilac portal]

U want to make the coefficient of the variable equals to zero

[lilac portal]

Here he used addition here because -y +y = 0

U can either use subtraction or addition, as long as the coefficient of x or y = 0

[lilac portal]

But since it's 0=0, then it means that the equation is just the same

so u can use either equation to substitute the numbers in the ordered pairs

[lilac portal]

either sub it to the x and y of the first equation or the second equation, it would work

[lilac portal]

Use the x's and y's here to verify if they're a part of the solution

Remember, ordered pairs are written like this (x,y)

Since we're trying to find out the ordered pairs that does not satisfy the solution, then we need to find a pair where if we substitute it, it doesn't spit out the corresponding x or y

[onyx parrot]

When/How will i know if I'll be using subtraction then?

[onyx parrot]

as in substitute it to the equation?

[lilac portal]

Yes

[lilac portal]

You can actually choose either to add or subtract them

[onyx parrot]

ohh

[lilac portal]

here we can use subtraction, lemme show you

[onyx parrot]

ok

[lilac portal]

Can't really write it neater at word

[onyx parrot]

why did we multiply it by -1?

oh wait

[onyx parrot]

is it because of: 3x " -y " = 5

[lilac portal]

here is a better explanation

[lilac portal]

but yes

You don't actually need to subtract it or add it

Both works

[onyx parrot]

oh

[lilac portal]

U can choose

[onyx parrot]

wait what method did you use here

elimination? substitution?

[lilac portal]

It's called elimination if u do that

Since ur eliminating the other variable

By making the coefficient into 0

Also known as Gauss method

Can also work for more than 2 variables

There can be x + y + z + f + g

[onyx parrot]

do we need to multiply the equation by it's Y?

[lilac portal]

and it will work

[lilac portal]

because it's -y just like what u said earlier

[lilac portal]

this one

[onyx parrot]

i think im triggering you or something, sorry im having a hard time to understand 🫠

[onyx parrot]

we have to turn it into positive?

[lilac portal]

Yeah, for subtraction

[onyx parrot]

ohhh

[lilac portal]

SInce ur confused on why here they just add it

[onyx parrot]

so we multiply an equation to it's y if it's y is a negative, so we can subtract?

[lilac portal]

they add it since it is much easier

[lilac portal]

As long as we can eliminate the y's or the x's

[onyx parrot]

how do we know if we can eliminate it?

if the coefficient is 0?

[lilac portal]

Here they add it since -y + y = 0

Here I multiplied it first by -1 and then subtracted it

both will spew the same answer

5x = 10

x = 10/5 or 2

[lilac portal]

That is earlier i said this

[onyx parrot]

i think i understand this one better

[lilac portal]

This

[lilac portal]

and this

[lilac portal]

Yeah can't really properly put it together since I used word there

So don't worry if ur confused if u need to add it or subtract it

Both works

[onyx parrot]

so either way, if i subtract and add, i get the same answer right?

[lilac portal]

Just make sure that u eliminate the variable properly and solve the others properly

[lilac portal]

yeah

Just pick which is easier

[onyx parrot]

ahh alright

hold on im gonna read through everything so i can understand

ok thank you so much i understand it

i will take note of everything here

.close

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[tiny rain]

hi can someone help me with this? i wrote out the proper definition but i cant seem to make sense of it

.close

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[tranquil wing]

Hi! I have a math problem I just cant figure out. So we have a deck of playing cards however, there has been 4 cards added to it. The added cards are king of spades, queen of spades, jack of spades and 10 of spades. The question is how the probability of getting two pairs when taking 5 cards from the deck.

[tranquil wing]

I just can't figure out how to start so a hint or a push in the right direction is greatly appreciated 🙂

I know how to get the probability when using a regular 52 card deck but when the 4 spade cards are added, I just cant figure out how to add them to the equation

When doing it for a normal deck, you take (13 choose 1) * (4 choose 2) * (4 choose 2) * 44 to get the number of combinations of two pairs. When doing it for the deck of 56, should I use like (17 choose 2)?

<@&286206848099549185>

[devout snow]

@tranquil wing Has your question been resolved?

[tranquil wing]

<@&286206848099549185>

[devout snow]

@tranquil wing Has your question been resolved?

[tranquil wing]

<@&286206848099549185>

[lilac portal]

This is how I did it @tranquil wing

I did a Probability tree diagram

And Imagine all scenarios that can happen (already written at the picture)
and then multiplied it to each other

I just translated the top part to factorial notations at the bottom

[tranquil wing]

Didn't know it would become so complicated 😄 Thanks for the help!

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[devout snow]

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[sand escarp]

| |x+1| - |x-3| | = |x|
Please help me solve this example. Actually I have two of them, but I hope I will make the second one myself. It looks a bit like this

[coral tulip]

Want no. Of solns or value of x?

[sand escarp]

Apparently, "X"

[coral tulip]

I tried using

Graph

x=-4 and x=4 are 2 solns

There are 2 more solns

Wait

I think i solved it

[sand escarp]

I have a example, but i cannot do anyway

bad quality on discord.. uhm

[coral tulip]

Can u draw graph of $y = ||x+1|-|x-3||$?

[woven radish]

Anagh

[coral tulip]

@sand escarp

I solved it

[sand escarp]

Hm.. I don't know, but i think It shouldn’t be decided through the graph. On the example I threw above there it’s like no

[coral tulip]

Well using graph its easy to find solns

You can see the intervals in which the solns are so u can easily open mod and then find x

,rotate

[sand escarp]

it's correct graph?

[coral tulip]

Well we don't need a graphing calc

We can draw graph using transformations

[coral tulip]

Yes

Now if you plot y=|x|

Intersection of both graphs are the solns

[sand escarp]

I’m not very good at this, but I think I understand a little bit about you sayed

[coral tulip]

I think it can be solved algebraically but it will be lengthy

[sand escarp]

I don’t know if I drew it correct. But it was like this

[coral tulip]

Little bit incorrect

y=|x| at x = 4; y=4

See your line

[sand escarp]

Oh, yeah

i got it

Maybe i must to draw normall graph to find answers, because i little confuse

[coral tulip]

Noice one

[coral tulip]

,rotate

So x=-4 and x=4 are 2 solns

[coral tulip]

Can you see ?@sand escarp

The other 2 solns are between 0 and 2

[sand escarp]

Places where they touch?

[coral tulip]

Now see the sign of |x+1| and |x-3| between 0 and 2 and remove the mod sign

[coral tulip]

Yes intersection points

[sand escarp]

between 0 and 1 is 3th solns

and between 1 and 2 (close to 2) 4th solns?

i correct understand?

[coral tulip]

See no. 2 and no.3

They are between 0 and 2

[sand escarp]

For example, the third solution is 0.5
I take 0.5 and do the following calculations, but without modulo signs?
(0.5+1)-(0.5-3)=0.5?

[sand escarp]

the second on your graph

[coral tulip]

It is not 0.5

2/3 is the soln

[sand escarp]

2/3 is second solution

[coral tulip]

do you know how to open mod?

[coral tulip]

Yes the second one marked on my graph

[coral tulip]

?

[sand escarp]

Hmm yeah, one second

if [x] < 0 to [x] = -x

[coral tulip]

WHAT ABOUT |x+1|?

I need to get a sleep now

Can you dm me tomorrow or someone else will explain you

[sand escarp]

|x+1|= 0
x + 1 = 0
x = -1 ?

[sand escarp]

Uh. ok

[devout snow]

@sand escarp Has your question been resolved?

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[vocal marsh]

Question

[vocal marsh]

when solving $x^2 = a$ for some $a > 0$ and solving for $x$

[woven radish]

stacheg0d

[vocal marsh]

This is talking about when I’m supposed to put a

In front of a radical or answer

Is “a” in this text speaking about the final answer or the initial problem? Because I’d imagine it’d work if it was on the initial problem

I’ve had problems where a was negative and it needed the plus/minus

<@&286206848099549185>

[winter brook]

a is just a, the a is the same a that u start with

the point of having a>0 is for you to not have a sqrt of a negative number

[woven radish]

Jukelyn

[vocal marsh]

But what happens if having a sqrt of a negative number is okay

[winter brook]

then you would have imaginary numbers

so u'd just write it in terms of i

[vocal marsh]

Yeah, which I’m on

[winter brook]

but here a>0 so u don't have any

[vocal marsh]

But can a < 0 ?

[winter brook]

sure

[vocal marsh]

For me to put the plus minus

[winter brook]

but ur problem said for it to be a>0

[vocal marsh]

Yeah yeah you’re right

[winter brook]

the plus minus is related to taking the sqrt of x^2

[vocal marsh]

That was actually an example from someone here last night

So basically whenever I’m solving for an x^2 I’m going to use the plus minus?

Because there would be two answers for x ?

[winter brook]

exactly

but the imaginary part comes in to play when whatever inside the radical is negative

[vocal marsh]

And if it broke down into an irrational radical I’d still have the plus minus assuming I was solving for an x^2 ?

[winter brook]

for example: $$x^2=-4$$ has the solutions that x=2i and x=-2i

[woven radish]

Jukelyn

[winter brook]

wym

[vocal marsh]

Like x= 2i plus minus radical 7

[winter brook]

sure if that's an answer then it's perfectly fine

but the +- in this case is kinda the same since imaginary roots come in pairs, aka congugates

[vocal marsh]

And if I wasn’t solving for x I wouldn’t need the plus minus

[vocal marsh]

Yeah that’s one of the things that makes me question when etc

[winter brook]

it's not a quesiton about when tbh, u always have the +-, it's just that it's kinda an overlap of definition

[vocal marsh]

AWESOME

Yeah that part really tripped me

The thought of overlap made me think I wouldn’t need it

[winter brook]

essentially

[vocal marsh]

For some strange reason

[winter brook]

no no, it's even more reason to need it

it's serving dual purpose essentially

[vocal marsh]

Yeah wow, very glad that got cleared up

[winter brook]

yuhr

[vocal marsh]

Big help! I appreciate it big time!

.close

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[mild basin]

Is this proper syntax for the derivative of a composite function?

I’ve never seen it written this way.. Newton syntax?

[shut spear]

pretty sure the ' should be between f and (

but I could be wrong

[mellow panther]

yes

[mild basin]

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[little copper]

why wouldn't the (a_1b_1 + a_2b_2 + \ldots + a_nb_n) be squared?

[woven radish]

nablasleep

[little copper]

since we're finding the discriminant of the quadratic equation of f(x), wouldn't b2 - 4ac be (4\left(a_1b_1 + a_2b_2 + \ldots + a_nb_n\right) ^2- 4\left(a_1^2 + a_2^2 + \ldots + a_n^2\right)\left(b_1^2 + b_2^2 + \ldots b_n^2 \right))

[woven radish]

nablasleep

[devout snow]

@little copper Has your question been resolved?

[devout snow]

@little copper Has your question been resolved?

[devout snow]

@little copper Has your question been resolved?

[little copper]

fine

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[lethal osprey]

could someone explain to me

how this rule works

[gleaming socket]

what is the question?

[lethal osprey]

im not sure how they went from the one on the left hand side to the one on the right hand side

[gleaming socket]

factor out 5 in the denom

[hot steeple]

you take 1/ on both sides

and then take the 1/25 to the other side

and then divide everything by 5

[lethal osprey]

ah

ok i got it now

thank you

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[devout snow]

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[smoky gyro]

Can someone properly explain to me how to rationalise the denominator? I don’t fully understand it. Help would be appreciated

[exotic stump]

get the root out of the denominator by multiplying by 1

i.e. sqrt(2)/sqrt(2) for example

= 1

[smoky gyro]

Wdym by multiplying 1? Like sqrt(2)/sqrt(2) • 1?

[weak cove]

You are allowed to multiply by sqrt(2)/sqrt(2) because it is the same thing as multiplying by 1

essentially it changes nothing

you wouldn't just be allowed to multiply by a random number, like 5, as it would change the value of your fraction

but you are allowed to multiply by 1, or by 5/5, or by sqrt2/sqrt2

as they are all equivalently, doing nothing to your fraction

what they do end up doing though is rationalizing your denominator

[smoky gyro]

I thought sqrt(2)/sqrt(2) became 1 because you can cancel out the numerator and denominator

[weak cove]

You can

[exotic stump]

it is 1

[weak cove]

and it does

[exotic stump]

that's why you can multiply by that

it doesn't change the value

[weak cove]

but the method is to ignore the allowed cancellation, in order to rationalize the denominator

for example

,, \frac{5}{\sqrt{7}}

[woven radish]

AustinU

[weak cove]

If we want to get rid of the sqrt(7)

in the denominator

,, \frac{5}{\sqrt{7}}*\frac{\sqrt{7}}{\sqrt{7}]

[woven radish]

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

[smoky gyro]

Uh

[exotic stump]

i believe in you

$\frac{5}{\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}}$

[woven radish]

a disappointing son

[weak cove]

Thank you

lol

You are allowed to multiply by that , because it is 1. Like you said earlier water beam, it could just cancel and be multiplication by 1

but instead of cancelling them, and essentially doing nothing for yourself

multiply across, like you do with fractions, and you will see that it transforms into something without a radical in the denominator

[smoky gyro]

Oh this right?

[weak cove]

^ and because you didn't do anything besides technically multiplying by 1, they are equivalent forms of the same number

Yes exactly right

[smoky gyro]

Alright I’ll try do this

@weak cove I think I did it wrong

Possibly

idk

[pseudo basin]

no, this is correct

[smoky gyro]

oh okay good

And if I have something like 3/3 + sqrt(2)

Does the process change

[devout snow]

@smoky gyro Has your question been resolved?

[weak cove]

well, if you have (3+sqrt(2)) on the bottom, you have to multiply by something called the 'conjugate' which basically just means swap the sign from positive to negative or vice versa. so you would multiply by (3-sqrt(2)) / (3-sqrt(2))

try it out and you will see why it works

@smoky gyro

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[desert anchor]

hii so my friend sent me this question and im not sure how to solve it

[devout snow]

@desert anchor Has your question been resolved?

[desert anchor]

<@&286206848099549185> please help 😕

[devout snow]

@desert anchor Has your question been resolved?

[strange meteor]

Ya what’s up

[desert anchor]

this question

i cant solve it

[gaunt helm]

hint: break down the entire figure into only 5 triangles, take note which angles can be subtracted using the central pentagon and which angles are counted twice

[desert anchor]

thank you!!

.close

[devout snow]

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[devout snow]

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[pallid blade]

Why is this wrong?

i used integration by parts

integrade dx and differentiate the Square root

then i just added a +7 to the integrated term to make calculation simpler

then i used "absorption"

[pallid blade]

maybe my mistake is here

[fiery pivot]

Partial Integration?

[pallid blade]

yea

[fiery pivot]

I think the second line supose to be

$\dots - \int {\frac{x}{\sqrt{7-5x^2}} dx}$

[woven radish]

Welf

[fiery pivot]

then solve it by substitution

[pallid blade]

how

[keen marsh]

Oo im getting integrals rn

at school

[pallid blade]

u forgot the *5x

because chain rule

[pallid blade]

.

and 1/2

[fiery pivot]

suppose
$t=7-5x^2$ \
this and that, then
$\int { \left( -\frac1{10} \right) t^{\frac{-1}2} dt}$

[woven radish]

Welf

[devout snow]

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[ocean thorn]

Does anyone know where I can go from here?

[ocean thorn]

How can I simplify this?

[devout snow]

@ocean thorn Has your question been resolved?

[devout snow]

@ocean thorn Has your question been resolved?

[devout snow]

@ocean thorn Has your question been resolved?

[ocean thorn]

Can someone please help

[nova idol]

what

depends what f is

what is C3

[ocean thorn]

Those are just constants

[nova idol]

whats C1

[ocean thorn]

Those are all constants except f

[nova idol]

i dont see how you can get that equation, can you show the entire question

[ocean thorn]

Wdym

[nova idol]

what is the problem

[ocean thorn]

I wanted to find a way to find the solutions for f

[nova idol]

what

do you mean isolate f?

[ocean thorn]

Yeah but it already is isolated, I wanted to find the solutions for it algebraically

[nova idol]

f is not isolated

you have a cubic function there

not easy to solve

might not be factorable

what is the actual problem you need to solve

maybe you took a wrong turn

[ocean thorn]

I used Planck's law

[nova idol]

maybe you messed up there

i dont understand what you want

there may not be a simple equation for f

[ocean thorn]

I can't help you understand if you don't make a attempt to try to understand. I wanted to find a way to find the solutions for f.

[nova idol]

Algebraically?

[ocean thorn]

Yes

[nova idol]

use the cubic formula

so you have f + f^3(C2C1) = C3 + C1

you can find a, b, c, d for the cubic equation

a = 1

[devout snow]

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[devout snow]

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[magic barn]

for this qn how do i form an equation for the material is expanding?

[devout snow]

@magic barn Has your question been resolved?

[restive river]

Hmmm... Does that mean it expands horizontally at a speed of 0.2 cm/s or 0.2 cm/s on both left and right?

I'll assume it's the first one

[magic barn]

should be both left and right

[devout snow]

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[fast helm]

i dont get the du sub part

i get dx=e^-1x du

dk how to sub it in

[restive river]

yo

[restive river]

so you know u = e^x, du = e^x dx , dx = du/e^x

try to rewrite everything in terms of u

like e^x = u

so what is e^(2x)

[fast helm]

u^2?

[restive river]

yes

so you know dx = du/e^x

can you sub that in for dx

in the integral

[fast helm]

ok

got it

.close

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[rugged jewel]

How would you solve this?

[restive river]

Weird... (1-x)/3=(y+2)/4 is already the equation for a line without the need for lambda

Oh wait

Lambda specifies a point on the line

[rugged jewel]

I already have something,
A1: p1(1,-2,0) and direction vector <-3,4,lambda>

[restive river]

https://www.desmos.com/calculator/09rcrkmq4l

[rugged jewel]

But i dont know for A2

[restive river]

Actually I just realised this is wrong

https://www.desmos.com/calculator/kzbptkdxne

[restive river]

To be honest with you I was following but then you mentionned direction vectors and my brain went nope

WTF is that angled bracket notation and why do I see it everywhere

[rugged jewel]

Vector notation

Basically the same as ()

But to show its a vector, and not a point

its <>

[restive river]

Okay but why does your vector have 3 values?

[rugged jewel]

Tbf, i dont know

its just in the book lmo

[restive river]

Okay honestly I am more confused than you

Fancy looking 2 = 2?

[rugged jewel]

Z=2

Book also says, for A2: point(0,1/2,2) and direcetion vector <1,2,0>

But,

I dont know how they got those values

[restive river]

The way you write Z is as outrageous as how I write 1, uppercase "i", and lowercase "L" the same

[rugged jewel]

I could send you a pic, but its not in english

[restive river]

Ooooooh

Wait

Okay my brain just did 360

Alright so I entered this question thinking I could solve it and I'm starting to think maybe I just wasn't able to read the question properly to notice it is much more complicated than expected

[rugged jewel]

I'm just going to send you a picture

And translate

[restive river]

Yeah, just as I thought

[rugged jewel]

Only thing that is given, are the first 2 equations

[restive river]

I only thought I could solve it because I didn't realise I couldn't fully parse the question

Now that I see what the problem actually is it is clear I shouldn't have entered this help channel in the first place

[rugged jewel]

☹️

[restive river]

I recommend you ping helpers while replying to the original message, and if no one answers open a new help channel

This rarely happens...

[rugged jewel]

<@&286206848099549185>

Thx anyway Labyrinth!

[devout snow]

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[devout snow]

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[blissful trellis]

how do I find a side of a right triangle if i only have the hypotenuse and the angle?

[wary ermine]

u can use rignometry

trignometry*

[blissful trellis]

whats the formula

[wary ermine]

which side

is it the bottom or the left/right

[blissful trellis]

bottom

[wary ermine]

so find cos(that angle)

then u know the ratio of the bottom to the hypotenuse

wait sry

cos(that angle)

[wary ermine]

yea basically

[devout snow]

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[bitter patrol]

9=0.95 and 2=0.30, log_10(9/2)

[bitter patrol]

do u just divide 0.95 by 0.30

anyone?

[lilac portal]

wait so 9 and 2 were used not as numbers here but as a placeholder/variable?

[bitter patrol]

yea

[lilac portal]

Oh yea so just divide it definitely and if ur not comfortable dividing decimals, try to translate it into fractions

[bitter patrol]

and for log_3 how do you find the value

[lilac portal]

either use a calculator or rewrite the log function back to exponential function

[bitter patrol]

what would that be

[lilac portal]

Like this

[bitter patrol]

oh alr

[lilac portal]

But I think u can just use a calculator on this one

or maybe u will use law of logs

since there is a division

[bitter patrol]

thnals

ima try it out

[lilac portal]

yeah

gl

[bitter patrol]

👍 ]

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[last plank]

Hey what's up, how can i subtract 3x from both sides?

[last plank]

15x + 10y = 150
Reduced the equation by dividing by 5
3x + 2y = 30
Now how can i subtract 3x from both sides??

[eternal marsh]

do you want something like 2y = 30 - 3x ?

[last plank]

Can 2y = -3x + 30 count as correct?

[eternal marsh]

yes

[last plank]

Yeah then

How do you subtract x for both sides btw

This one gets me confused

[eternal marsh]

like how do I get to 2y = -3x + 30

[last plank]

Yeah

[eternal marsh]

so we substract 3x in both sides like 3x -3x + 2y = -3x +30

and 3x-3x = 0

so we can say that 2y = -3x + 30

[last plank]

I'm very sorry, I didn't quite get that, hope your patient enough

[eternal marsh]

np what don't you understand ?

[last plank]

The 3x-3x part

[eternal marsh]

so you want to substitute x from both sides right

[last plank]

Yeah thats what i wanted to do

[eternal marsh]

so you can do (3x + 2y) - 3x = (30) - 3x I put bracket to show you that's our previous equation and we substract in both sides 3x

do you understand this part

[last plank]

Hmmm, i tried to process this one, but sorry, this one was quite confusing

[eternal marsh]

what do you don't understand ? I'm just substracting both sides, you can do that

[last plank]

Wait, do we just simply just take the, 3x and write two of them down like 3x - 3x = 0 and thats it

[eternal marsh]

we take each member of the equation and subtract 3x

so in the left member which is 2y +3x we substract 3x so we have 2y +3x -3x = 2y because 3x-3x = 0 and on the right member of the equation which is 30 we substract also 3x because if we substract 3x on a side we need to substract it on the other to keep the equality, so we have our -3x +30. Finally we get 2y = -3x + 30

[last plank]

I almost forgot to tell but this is application of linear equations in two variables just a reminder

[eternal marsh]

?

[last plank]

Application of linear equations in two variables

[eternal marsh]

yes and

[last plank]

Our current lesson

[last plank]

Ah

Gotcha chief

Thanks very much

Appreciate it!

[eternal marsh]

np good luck 🙂

[last plank]

Thanks 😊

[devout snow]

@last plank Has your question been resolved?

[devout snow]

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[gray agate]

how do we solve this

i am stuck completly rn

[devout snow]

@gray agate Has your question been resolved?

[polar chasm]

Integration by parts?

[polar chasm]

What do you get after integration by parts

[devout snow]

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[jade sundial]

Hi im currently confused with this:
How did he get from the left to the right term?

[supple knot]

let m = n+1 and factor out m

[jade sundial]

I dont understand, can you explain please?

[earnest stirrup]

give the value (n+1) a new name

can be m

or just call it apples

so u have n+2 apples on the right, right? 🙂

how many are there on the left?

[vast igloo]

did u get the ans?

[jade sundial]

No Im still confused but I think I have to think about it

[vast igloo]

let me see

[vast igloo]

can u send the complete quesion?

is the objective is to find the value of n?

[earnest stirrup]

no, they just transformed the expression

[jade sundial]

I think I got it

I can post it, mom

[lusty sapphire]

n(n+1) is just n times (n+1). That is, you add (n+1) a total of n times.

Also, 2(n+1) is adding n+1 two times

So you add (n+1), first n times, then 2 times. So, you add (n+1), in total, n+2 times

[jade sundial]

I hate maths....
But thanks you all, helped me a lot

[devout snow]

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[azure swan]

What is the question asking?

So there are 3 different flags, call them ABC

What specifically does "any number of them may be used mean"

[weak cove]

1 at a time, 2 at a time, 3 at a time

Think of it as 3 questions, and add up all the possible signals

[azure swan]

Meaning 3 spaces are available?

[weak cove]

how many signals can be made (1 flag at a time) if you have three flags +

[azure swan]

In which ABC can be placed

In any order

And abc can repeat?

Still not sure

[weak cove]

No, you only have three flags

[azure swan]

I am not sure what you mean

[weak cove]

you can put up all 3 at once (in a certain order)

or you can put up only 2

or you can put up only 1

and wants to know how many different signals you can make in total

[azure swan]

Is this a permutation?

[weak cove]

using any amount of the flags

[azure swan]

Assume, I am a moron, and then explain

[autumn sandal]

Can anyone spot my mistake ? Π/4 and -1.326rad aren’t answers

[weak cove]

Basically, you have three flags like you said, A B and C. It wants to know how many signals you can make, and it is giving you the information that you do not have to use all three flags (although you could)

[weak cove]

this channel is occupied use a different one

#❓how-to-get-help

So you can think of it as three questions, and add up all the posibilities

If I asked you,

you have three flags, if you raise one at a time, how many signals can you make?

what would you answer?

[azure swan]

Well, 3, at a time I do 1 signal

[weak cove]

how many total signals can you make though

raising only 1 flag at a time

its not a trick question I promise

[azure swan]

What exactly do you mean when you say "at a time"

I think I am slipping at that

[weak cove]

Like you couldn't raise both flag A and B at once

you must only raise one flag

[azure swan]

It's mutually exclusive

[weak cove]

I'm not sure if I would say that, but maybe

[azure swan]

Exclusive*??

Just exclusive alone

[weak cove]

Sure

It is like I am asking you

you have three colors, if you can only use one color at a time, how many different colors can you draw with

and the answer is clearly 3

[azure swan]

Yes

[weak cove]

but then what if you could use two colors at a time?

and what if you could use three?

add up all the results and it will be your answer

and replace colors with flags/signals

[azure swan]

Lemme stare it it for a sec

[weak cove]

It is a permutation

basically the calculation you do is 3 choose 1 + 3 choose 2 + 3 choose 3

logically it is the same as the example I just gave you

[grizzled yew]

Isn’t the answer 2^3-1

[crimson tiger]

that's if the order doesnt matter

here, the order does matter I guess

[weak cove]

Yes, because it is a different signal if say flag blue is on the top or on the bottom

[grizzled yew]

Oh

[azure swan]

Okay so, if I were to word it,

I have 3 colors, Red Blue Green

And I cannot use the same color again if I have used it (the pencil color gets vaporized)

I have 3 colors to choose from

So i use 3 colors, and then two, and then 1

[weak cove]

and this is because you only have one of each flag

[azure swan]

So it's 3 factorial?

[weak cove]

no

I'll give you some examples and then let you figure it out

If the flags are Red Blue and Green

here are some signals I can make

1. Red Blue Green

2. Red Green Blue

3. Green Red Blue

4. Blue

5. Blue Red

6. Green

7. Green Blue Red

etc....

maybe now you get the point

[azure swan]

Order matters, and I can use any amount from those 3

Something ticks

[weak cove]

Order matters, you can use 3, or 2, or 1

so find out how many different ways you can do 3

and how many different ways you can do 2

and 1

and then add them all together

[azure swan]

3!+2!+1!

[weak cove]

No

[azure swan]

I am embarrassed

[weak cove]

Don't be embarrassed

[azure swan]

Lemme try again

[weak cove]

^ The way you should be computing each

is 3 choose 3, 3 choose 2, and 3 choose 1

not just plain factorials

[azure swan]

Ohh fk

So I have 3 flags, I can make any signal, and order matters. And I always have 3 flags, doing 3!+2!+1! Suggests that my flags are getting reduced along with the number of possible outcomes I can do with them (because ofc, there are less flags).

And doing 3!, Suggests that I find number of possible outcomes for 3 flags which can go into 3 spaces. But this isn't correct because, I am not taking into account the other times when I can have two spaces for 3 flags and so on.
So

3! Says #of pos outcomes for 3 spaces that can hold flags.
3!/1! Says #pos outcomes for 2 spaces.
3!/2! And so on

The answer is be

B

I think I missed the "any number of them may be used part"

[azure swan]

Does this sound right?

?

[weak cove]

the answer is b, good job.

that does sound good

[azure swan]

.close

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[tight estuary]

I dont undetstand this step

[exotic stump]

common denominator

[tight estuary]

Oh so were first trying to subtract the Numerator and because they don't have the same denominator we multiply by the most "close" to common denominator... and because what ever we do to the top we do to the bottom we also multiply the Common denominator to the bottom.
am i right?

if 1/5 was 1/7 would we multiply the bottom by 7(h+5)?

.close

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[orchid verge]

do you have any tips for algebraic word problems, they trip me up bad. In this recent module of mixture and combined rates word problems, I messed up on the quantities alot. What is your tips for overcoming this

[tight estuary]

Practice

lots of practice

[stone stump]

write out explicitly what quantities are known and which ones aren't

[vast rain]

well before any of that

I find most people don't actually try to parse out the plain English meaning of the words before trying to run numbers through whatever

a lot of those word problems tell a story, and you should listen to understand what they're asking

[devout snow]

@orchid verge Has your question been resolved?

[devout snow]

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[red helm]

Is anyone familiar with phase diagrams for a system of linear differential equations ? I’m having trouble with understanding what the axes represent ?

[devout snow]

@red helm Has your question been resolved?

[devout snow]

@red helm Has your question been resolved?

[red helm]

.close

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[kind sonnet]

There is this trigonometric identity
Sin(2x) = 2sinxcosx
Can you extend this to any coefficient a?
=> Sin(ax) = asinxcosx?

[grizzled yew]

No

[kind sonnet]

You sure? Then how would you solve
Lim x-> 0+ f(x)
Where f(x) = sin(qx)/x

Without using the derivative rule / hospital rule or whatever

[devout snow]

@kind sonnet Has your question been resolved?

[wicked turtle]

try writing $$\frac{\sin(qx)}{x} = q\frac{\sin(qx)}{qx}$$

[woven radish]

Bungo

[kind sonnet]

But what will this achieve

[wicked turtle]

probably you know how to compute $$\lim_{x \to 0}\frac{\sin(qx)}{qx}$$

[woven radish]

Bungo

[kind sonnet]

I believe that would be 1 yes

[wicked turtle]

ok then

[kind sonnet]

Ohh

[wicked turtle]

what can you say about the limit of the original fraction

[kind sonnet]

It's q?

[wicked turtle]

yep

[kind sonnet]

Damn that was so simple

[wicked turtle]

yea, no trig identities needed

[kind sonnet]

Thanks

[devout snow]

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[wooden wraith]

You want to find dy/dx, right?

Multiply both sides by your denominator

and basically you can just solve the equation for dy/dx in terms of x and y

Lol

Shouldn't be too much different than what you just did

As long as you know the derivatives of those trig functions

Yep

right idea but the second term should be negative

because you subtracted dy/dx*3xsec^2(y)

well basically yeah but you're swapping some + and - signs still

In the last step the 3 in the numerator became negative, and the one in the denominator became positive

Honestly it seems like you understand this pretty well

the signs was just a minor mistake

Yep that's basically it

also, taking the derivative lmao

[wooden wraith]

hm

oh

The xy part, you need to use the product rule as you do the chain rule

$\frac{d}{dx} (xy)^\frac{1}{2} = \frac{1}{2}(xy)^{-\frac{1}{2}} \cdot \frac{d}{dx}[xy]$

[woven radish]

tatpoj

[wooden wraith]

yeah, exactly

[devout snow]

@dusty pelican Has your question been resolved?

[devout snow]

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[balmy bloom]

How would I do question 4

[nova idol]

which part

[balmy bloom]

4a

I tried simplifying but it was different than the book answer

[nova idol]

all its asking is b(n+1) - b(n)

i would not think you need to simplify

since its a cubic

[balmy bloom]

I don’t understand

[nova idol]

but if you do need to, there are a lot of ways you could mess up doing the algebra

what did you write

[balmy bloom]

I tried this

It was wrong

[nova idol]

thats just b(n)

reread the question

[balmy bloom]

So I need to add a 1

For the value of a day

[nova idol]

no

b(n+1) is not the same as b(n) + 1

[balmy bloom]

So what I wrote was b(n)

[nova idol]

yes

[balmy bloom]

And how would I change that to b(n+1)

[nova idol]

do you know what b(n) means

[balmy bloom]

Yea it’s = to (2n+1)^3

[nova idol]

so what is b(n+1)

[balmy bloom]

You add another (2n+1)^3?

[nova idol]

okay look

whatever is inside b()

is the input to the function

imagine f(x)

if you add one to x

say d = x + 1

then f(x+1) = f(d)

[balmy bloom]

How would that translate into the function

[nova idol]

what is a function

[balmy bloom]

The equation

[nova idol]

If we change the input

How does the equation change

[balmy bloom]

We put n+1 instead of n

?

[nova idol]

Yes

[balmy bloom]

And how do we add that

[nova idol]

Wdym how

[balmy bloom]

Do we do 3n

Instead of 2n

[nova idol]

No

It’d be 2(n+1)

[balmy bloom]

oh and then I multiply

[nova idol]

Yes

[balmy bloom]

And then I continue normally?

[nova idol]

Yes

[balmy bloom]

do I simplify then

[nova idol]

Yes

[balmy bloom]

and then the question is asking the increase between day n and day n+1

[nova idol]

Yes

[balmy bloom]

So what I did was say n

Day*

[nova idol]

Yes

[balmy bloom]

And I do say n+1 and see the difference

day*

[nova idol]

Yes

[balmy bloom]

how would I see the difference

Do I just subtract normally

[nova idol]

Yes

[balmy bloom]

Do I put each equation in a bracket to subtract

Or just normal subtracting without brackets or does that not matter

[nova idol]

Doesn’t matter

[balmy bloom]

Okay can I try doing the question and come back or will the room close

Or do I just open a new room if I’m stuck

[nova idol]

You have around 20 min

[balmy bloom]

Okay Ill only need a few minutes

At most

Thank you I’ll be back if I mess up

[nova idol]

No problem

[balmy bloom]

I’m not the best at math so I kinda struggle with this stuff

[nova idol]

That’s alright

We all do at some point

[balmy bloom]

Thanks for help

I got the question correct

Thank you for the help

I might open another room later on if I’m stuck on another problem

[nova idol]

Yay good job

[devout snow]

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[ionic river]

would this be 2^20

[mellow panther]

yes

[ionic river]

ok

[devout snow]

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[weak cove]

Could anyone familiarize me with the method of variation of parameters?

<@&286206848099549185>

[analog trellis]

What would you like to know?

[weak cove]

Well it wasn't covered in lecture, so I am wondering if it is simple enough that an explanation here could get me going

if you think it might be too much for here I can go find a video

but I don't know what the method is

Those questions are for the context of how I will be using it

b looks like I could do using undetermined coefficients instead though?

[analog trellis]

You can use either method sometimes. Variation of Parameters has one crucial advantage though. That is that it can be used when the right side of the DE has a infinite numbers of linearly independent derivatives (undetermined coefficients can't be used for this).

Are you wanting to learn about the entire topic?

Or just see a few examples?

[weak cove]

How would the right side of the equation have infinite amounts of linearly independent derivatives? I guess whichever you are willing to do

Learning more would be better, but I respect your time

[analog trellis]

For (a), tan(x) when you differentiate it just gets bigger and bigger.

[weak cove]

oh, you mean as you keep differentiating got it

I thought you meant for a single derivative at first

[analog trellis]

Yeah. The method of Undetermined coefficients can only be used when it has a finite number of them.

[weak cove]

that's good to know

[analog trellis]

It might be worth giving this a slow read:

[weak cove]

Are these from an online textbook that I might be able to pull up to see better?

[analog trellis]

In this last screenshot it does mention that the method of Undetermined Coefficients will usually be easier.

Well the quality will be the same. It's from a book called Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard and these screenshots start at page 233.

[weak cove]

Just on discord the pictures are small even when I click them is all

thank you, I will read that

[analog trellis]

I'm not sure how to fix that sadly. 😢

[weak cove]

No worries I found the book already