help-19

[unique meadow]

ah

[verbal lynx]

After the equivalent resistance, the general amount of current can be found,

[unique meadow]

right

thats kvl right?

[verbal lynx]

What?

[unique meadow]

what

we use kvl on each mesh to find it right?

[verbal lynx]

I'm not exactly from an english speaking country

[unique meadow]

oh ok

[verbal lynx]

What is a kvl

[unique meadow]

kirchoffs law

[verbal lynx]

Oooh

[verbal lynx]

If you can, sure!

[unique meadow]

idk how lol

im just asking if thats the step

[verbal lynx]

After the resistance I usually get the votage at the nodes (in the modified pic), which is the same as in the original, and then just Ohm's law my way to calculating everything

[unique meadow]

how do we find the voltages

[verbal lynx]

(About the right chain)
At the highest point of the scheme, we take the voltage (или как там потенциал) to be 10V, so that at the bottom it's 0. Then, since all the current is going through that reisitor at the top, since we know the overall current and its resistance, we can calculate the drop in voltage. Then, we know the difference in voltage between the bottom point and that mesh, so the current through each branch can be found (I = U/R). Then, we get the voltage in that rightmost point, since we know both the current and the resistance between the mesh and the point.
Now that we know all the voltages of meshes (потенциалы, на самом деле, я просто не знаю как это перевести) and resistances, the voltage drops, and so, the currents between the meshes, are easily found.

Is the wire just broken in the right pic?

[odd edge]

@unique meadow Has your question been resolved?

[exotic marsh]

Show that the transposed matrix is equal to the inverse plissss 😔

[unique meadow]

yes it is

[raw herald]

!occupied

[odd edge]

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).

[odd edge]

@unique meadow Has your question been resolved?

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[grim wedge]

^

[sterile blaze]

^.^

[verbal sandal]

@grim wedge what's your question

[grim wedge]

sorry

my phone turned off

i’ll show now

[verbal sandal]

np

[grim wedge]

ok here it says simplify right

my answer is 1/8 x^4

and that’s wrong

but why

[verbal sandal]

1st ques ?

[grim wedge]

yes

[pale atlas]

How are you getting x⁴? Can you show your work

[grim wedge]

ok

[grim wedge]

I mean x^3

[pale atlas]

Still want to see your work , powers are multiplication of fraction and whole number so you will end up with fraction only

[pale atlas]

Oh my bad there is a x outside

I missed that

[grim wedge]

yeah

[pale atlas]

2nd line

[verbal sandal]

how did you turned that 4 into 1/2

[pale atlas]

Also x^-½ to x

[grim wedge]

square root of 4 is 2
2^-1 = 1/2?

[pale atlas]

Where is √4 is given??

[jolly coral]

high level maths

[grim wedge]

the power being -1/2

[pale atlas]

That's only on x not 4

[grim wedge]

ah

lemme redo

[pale atlas]

(4x)^-½

Would be that

Its 4(x)^-½

[verbal sandal]

@grim wedge do you know that the exponents inside the bracket and outside the bracket multiply together ?

[grim wedge]

is the answer 1/8 for the exponent?

[verbal sandal]

no sir

how are you getting that

[grim wedge]

is it

64 for the exponent

x^-1/2

[pale atlas]

@grim wedge
(x^a)^b = x^(ab)

[verbal sandal]

@grim wedge

[grim wedge]

am I wrong

[verbal sandal]

yes

very

[verbal sandal]

@grim wedge try to observe and understand what just happened here

[grim wedge]

bruh so the mark scheme is wrong?

I mean coefficient is 64 whoops

[verbal sandal]

what is 4^3

solution says it all nvm

[verbal sandal]

@grim wedge do you understand now how you were wrong and what's the right way to do this

[odd edge]

@grim wedge Has your question been resolved?

[odd edge]

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[gloomy mason]

(24y-8)+7=29
Given the equation, what is the value of 3y-1?

[gloomy mason]

i reached till here

3y-1=22

[pale atlas]

You want to find y , or 3y-1 from it?

[gloomy mason]

3y-1

[pale atlas]

Ok so think in this way , you have 3y+1, what can you add or subtract in it so that it becomes 3y-1?

[pale atlas]

[gloomy mason]

my bad it was calculation error it was 3y-1=22 only

[pale atlas]

Okkk

[gloomy mason]

now what

[pale atlas]

Well you got 3y-1 = 2

You wanted to find 3y-1 right?

[gloomy mason]

ahhhhhhhhhhhh

no

my bad

[pale atlas]

[gloomy mason]

it was 22

[pale atlas]

Given the equation, what is the value of 3y-1?```

This is your question?

[gloomy mason]

yeah

[pale atlas]

Can you show me your work ?

What have you done

[gloomy mason]

sure

[pale atlas]

[alpine hawk]

Can't we just find the y and substitute in the given equation to find?

[pale atlas]

Yes that's the clearest approach

But their finding is wrong, that's why I asked to show the work

[alpine hawk]

Oh yeah I see

[gloomy mason]

sorry for delay im slow

[pale atlas]

How did 24y-8 changed to 3y-1?

[alpine hawk]

Same

[pale atlas]

You took 8 common I get it, but there should be 8 outside bracket

[gloomy mason]

oh

[pale atlas]

(24y-8) = 8(3y-1)

Like this

[gloomy mason]

but its not

[pale atlas]

I mean you haven't done it, it needs to be done that way

[minor cedar]

I think her answer is correct, she just divided by eight and got the answer she was looking for

[pale atlas]

Then /8 will be on Right side too and +7 will become+7/8

[gloomy mason]

how

[alpine hawk]

Yeah

[gloomy mason]

uh i dont understand

[pale atlas]

You can't simply divide it by a number and make the number vanish 😅

[gloomy mason]

oh

[minor cedar]

Oh, that's true, she didn't divide the other part but she's almost there.

[pale atlas]

If you have

2(x-2) not just (x-2)```

[minor cedar]

Here's the most intuitive way to fit whenever you do something to one side of an equation, you need to do the same thing to the other side

[pale atlas]

No bro, they didn't divide by 7 as well

(a+b)+c , if you are dividing by 2 it will become
(a+b)/2 + c/2

[minor cedar]

Why would you need to divide by 7? She can divide by 8. She needs to divide by 8.

[gloomy mason]

oh

so

[pale atlas]

[gloomy mason]

i have to do this :-
(3y-1) 8 = 22

[pale atlas]

I meant the +7 term on left side

[gloomy mason]

like this u mean ?

[pale atlas]

(3y-1)8+7=22

[gloomy mason]

im not she bro 😭

[pale atlas]

Oh you did subtract

[alpine hawk]

Both are the same bro

[gloomy mason]

no but 7 got canclled

[minor cedar]

I think the answer is twenty two over eight

[gloomy mason]

yeah

[pale atlas]

Yeah mb @minor cedar , they did subtract already 29-7

Sorry

[gloomy mason]

guys

i got it

[minor cedar]

It should be 22 over 8.

[gloomy mason]

yeah

divide by 2

and 11/4

[alpine hawk]

Well what's the answer guys ?

[minor cedar]

Yeah. Eleven over 4.

[gloomy mason]

finally

[minor cedar]

I know I shouldn't have doubted myself.

[gloomy mason]

im gonna fail in my sat i cant even solve this

[alpine hawk]

Just learn

[gloomy mason]

ty all of you

ill come back if i have another doubt

[pale atlas]

@gloomy mason just remember any addition, subtraction, multiplication or division you perform on left side needs to be done on right too

[alpine hawk]

Well I got 66/24 lmao

[pale atlas]

Yeah it's same 11/4

[alpine hawk]

Oh yeah mb

[gloomy mason]

alr

[minor cedar]

The most important thing to remember is if you do something to one side of an equation, you have to do it to the other side too

[gloomy mason]

should i close this because i think ill come again in few minutes with a new doubt

[minor cedar]

And instead of thinking of the manipulation as just moving things from one side to the other, what you're actually doing is adding, subtracting, multiplying and divide.

[pale atlas]

Sure

[minor cedar]

You can keep it open it's fine with me.

[alpine hawk]

I already know that bro

[minor cedar]

We're not talking to you

[gloomy mason]

k

[pale atlas]

You replied to their msg

[alpine hawk]

Oh sorry

[alpine hawk]

Yeah

[minor cedar]

Ohh, I'm so sorry it's my fault

[pale atlas]

We are good

[minor cedar]

I intended to reply to him, but I accidentally must have clicked your profile

[alpine hawk]

Lol np actually

[gloomy mason]

alright guys

we got a new problem here

3(2n-2)-1=9/4
Given the equation, what is the value of
2n-2?

[minor cedar]

just expand it

Then find then then plug into the new equation.

[gloomy mason]

expanding means i should multiply 3 with the bracket

right

[minor cedar]

yes

But only the things inside the bracket.

[gloomy mason]

alr

6n-6=5/2

this is what i am getting

[minor cedar]

I don't know if you didn't transformations that I'm missing, but Let's begin more simply six then - 7 equals nine fourths.

6n-7 equals 9/4

[gloomy mason]

oh no

i dont know this

[minor cedar]

What do you get right after you expand the bracket?

6n-6-1 =9/4

right?

[gloomy mason]

yeah

but i added 1 to the rhs and then divided the frac

[minor cedar]

Let's just do 6 N - 7 equals.9/4

-6-1=-7

That's where that comes from.

[minor cedar]

Do you understand this part

[gloomy mason]

howw

[minor cedar]

see this?

[gloomy mason]

yeah

[minor cedar]

-6-1=-7

[gloomy mason]

but how is 6n-7=9/4 😭

[minor cedar]

-6-1 here becomes -7

[gloomy mason]

alr

6n-7=9/4

alr gotcha

[minor cedar]

Do you need more help finding n

[gloomy mason]

i thought this is some other equation my bad for being silly

n=2/3

but ive to find 2n-2

[minor cedar]

I could be wrong, but I don't think N equals two thirds

[gloomy mason]

wait

[minor cedar]

Take this equation and multiply everything by 4 first

That way you can work with whole numbers.

[peak onyx]

oh wait mb

[minor cedar]

Are you okay?

[odd edge]

@gloomy mason Has your question been resolved?

[odd edge]

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[noble solstice]

Hi! I don't understand how [\int_Y g d(F_* \mu_F) = \int_{\mathbb{R}} g(y)dy] specifically how $d(F_* \mu_F) = dy$, doesnt $F_*\mu_F = \lambda$? ($\lambda$ is the Lebesgue measure)

[noble solstice]

oops i think we both tried to occupy this channel

[gloomy mason]

okayy

after that ?

[steel karma]

Sorry I can’t delete this post

[noble solstice]

there are 2 others though #help-22 , #help-29

[steel karma]

Okk

[noble solstice]

Please ask again in another channel

[clever fjord]

woomy

[noble solstice]

<@&286206848099549185>

[jovial rampart]

@noble solstice hello

[noble solstice]

hi

[jovial rampart]

so what do you want?

[noble solstice]

.

[jovial rampart]

I don't know about the question.

[noble solstice]

all good

[jovial rampart]

can you explain about the problem?

[noble solstice]

I dont understand how the lebesgue measure simply turns into a variable "y",

[jovial rampart]

Oh, that is very difficult for me... I'm sorry...

[noble solstice]

No problem :]

[jovial rampart]

I'm very sorry....

Are you student?

[noble solstice]

yes, could we maybe move to a discussion channel if you want to talk

[jovial rampart]

Okay

[noble solstice]

ping me

[noble solstice]

<@&286206848099549185> :3, here is the video btw, at around 10 minutes https://www.youtube.com/watch?v=q3UgXso-1jw&list=PLBh2i93oe2qvMVqAzsX1Kuv6-4fjazZ8j&index=18

[odd edge]

@noble solstice Has your question been resolved?

[noble solstice]

<@&286206848099549185> somebody 😔

[quasi sparrow]

most people aren't gonna watch a whole video to help. do you have some link or pdf that has the same idea

[noble solstice]

theres no need to watch the whole video

maybe i didnt explain myself well

ill explain better when im back from dinner :)

[quasi sparrow]

yea you need to watch the video since you haven't defined $F_*, \mu_F, X, Y$ yet

[clever fjord]

riemann

[noble solstice]

oh yea true true

[quasi sparrow]

Once you do a little more googling or maybe he mentions it in the video, this is the pushforward measure
https://en.wikipedia.org/wiki/Pushforward_measure

[noble solstice]

yea yea ik its the pushforward measure

thats what the video ts about

i just specifically dont understand the substitution

wait ill explain myswlf better

[noble solstice]

$F$ is a strictly monotonically increasing and continuously differentiable function from a measure space $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ with the measure $\mu_F(A) := \int_A F'(x)dx$, to $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ with another measure which is the pushforward measure of $\mu_f$, i.e. $F_*\mu_F$.

In the video its proven that the pushforward measure is just the lebesgue measure $\lambda$.

By the substitution rule:
$$\int_Y g d(F_*\mu_F) = \int_X g \circ F d\mu_F$$
which when putting everything in, the video finds:
$$\int_{\mathbb{R}} g(y) dy = \int_{\mathbb{R}} g(F(x)) F'(x) dx$$

What i dont understand is how $d(F_\mu_F) = dy$, when $F_\mu_F = \lambda$ not y.

[clever fjord]

woomy

[noble solstice]

<@&286206848099549185> i think i explained myself better here

[odd edge]

@noble solstice Has your question been resolved?

[noble solstice]

.close

[odd edge]

Channel closed

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[noble solstice]

if anyone finds the answer pls dm me

[odd edge]

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[crystal helm]

Hiya, I’m honestly not sure what to do on this one..

[crystal helm]

Do I have to differentiate it twice??

[brittle beacon]

Once is enough what happens when you get the first derivative?

[crystal helm]

That would be equal to the gradient

Ohh and I’d say the gradient is 0?

Since it’s the min and max

Have I differentiated it so far correctly?

[brittle beacon]

(Check what your derivative of v is, it's not exactly the same as the original function, small difference there)

[crystal helm]

And id put those values into the quotient formula?

[crystal helm]

Ah

So it’s -1 now

-e

But the power stays the same

[brittle beacon]

Not quite (remember chain rule maybe ⛓️)

Remember the power of $e$ is $\sqrt{2} x - 1$, after all

[clever fjord]

@brittle beacon

[crystal helm]

So I should differentiate that separately?

I can’t tell if I’ve been stupid but

That’s - 1/2, not very clear

[brittle beacon]

Yeah, it's not very clear, but you have $\sqrt{2} x - 1$ and not $\sqrt{2x} - 1$, the $x$ is not part of the square root

[clever fjord]

@brittle beacon

[crystal helm]

Ohhh

So the x just disappears that’s all

[brittle beacon]

Yep, it differentiates just to sqrt{2}

So your overall derivative of $e^{\sqrt{2}x - 1}$ is just $\sqrt{2}e^{\sqrt{2} x - 1}$

[crystal helm]

And that would go next to the e with the same power

[clever fjord]

@brittle beacon

[crystal helm]

And then I use the quotient formula and put it = to 0?

[brittle beacon]

You could, if you wanted to

Quotient rule is probably a decent choice here if you don't mind the formula (but it's given to you in the booklet anyway )

[crystal helm]

I automatically saw the fraction at the start so

You are free now, should be able to get it done myself

Thank you so much

[brittle beacon]

Awwww you should be all good for sure

[crystal helm]

😃

.close

[odd edge]

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[brave crane]

Help

[brave crane]

Helppp

[quasi sparrow]

the bottom is true regardless of the top

[dense sonnet]

^

[quasi sparrow]

$\sum_{k=1}^{2n+1} f(k) = f(1) + f(2) + \ldots f(2n) + f(2n+1)$

[clever fjord]

riemann

[summer cradle]

and the top is true for every n

[quasi sparrow]

i'm much too lazy to see what [N(N+1)/2]^2 equals when N=2n-1

[summer cradle]

i decided it was very probably 4n^4 - 4n^3 + n^2 and sent that message without actually checking

,wolfram expand ((2n-1)*(2n)/2)^2

[clever fjord]

Results provided by WolframAlpha

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

Requested by slayyla

[summer cradle]

mechanical ram wtf

[quasi sparrow]

mech arm

[brave crane]

Hmmm

Is this difficult or everyone is stupid

[summer cradle]

the question is just asked very weirdly

it's not difficult

the question is easiest without even using the top

[brave crane]

Ah i forgot something

[brave crane]

You must prove it with recurrence

[summer cradle]

recurrence = induction?

[brave crane]

Recurrence solving is by two steps

Initialisation
And after that you suppose that p(n) is true
And you prove that p(n+1) is true
after that p(n) is true

[summer cradle]

sure. we call that induction in english

[brave crane]

Ohh

[brave crane]

Soo prove that

[faint knot]

you here?

[odd edge]

@brave crane Has your question been resolved?

[odd edge]

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[mystic saffron]

what are they asking for

i enter the answer they showed and its still not right

anyone know whats going on

[wanton bison]

well yes, but actually no

do you know the rational root theorem?

[mystic saffron]

do i have to do some synthetic division

[wanton bison]

no

[mystic saffron]

not really

[wanton bison]

well it's synonymous to the rational zero test

so what you are doing basically is based on the coefficients (assuming they are integers) of f you are calculating possible roots of f

[wanton bison]

in their solution you can see they did that

which of course includes the actual roots too

but the thing is they wanted you to do that, apply the theorem to list all potential roots of f, not the actual roots

[mystic saffron]

how is it done

[wanton bison]

here

https://youtu.be/Iaq7z7reznM?si=ej88heiWif3DBjWI

[mystic saffron]

i watched it

[wanton bison]

haha ok?

[mystic saffron]

the ones he worked on

is kinda different from this problem :<

[wanton bison]

how?

[mystic saffron]

or maybe not but i dont get it

[wanton bison]

the principle remains the same, regardless of which degree your polynomial is

let us try it (never done it before)

4 -8 -5 +10 +1 -2

[mystic saffron]

thats why i was asking if i needed to do sythetic division cuz in his vid he was find the zeros and he would use it to find x

[wanton bison]

Now we can work with that

doing what he did

4 -8 -5 +10 +1 -2
bring down 4
4
now you multiply 4 by 1 and add that to -8

[mystic saffron]

how did u get these

[wanton bison]

oh i realized that was something different

[wanton bison]

you are given these from your function

the coefficients

watch example 2

[mystic saffron]

ok

i will be back

[clever fjord]

bacc the sigma😔🤞

[wanton bison]

now if you work them all out you will get all possible potential roots as they did

the numerator are all the coefficients of your polynomial

the denominator is the coefficent of the leading term

that's literally it

[mystic saffron]

what do u mean

[wanton bison]

i set commas right

so your first potential roots are 4/4 and -4/4 (the other ones are the same 4/(-4) and -4/(-4) )

that reduces to +1 or -1

which they also have

now next 8 and 4

and so on

[mystic saffron]

oh ok

tysm

[wanton bison]

np!

[mystic saffron]

<33

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[whole hazel]

Ok yo straight up, can someone do this for me and show the steps pls, I give up

[odd edge]

@whole hazel Has your question been resolved?

[whole hazel]

https://tenor.com/view/helpmepls-kuchnahiaatahelp-i-am-underwater-help-me-pls-uncle-gif-21956179

[deep mason]

hi

how r u

[whole hazel]

Hi m good, I just struggling w/ this one question bro

[deep mason]

ok do this

i'll say all the steps

first get rid of the negative exponent by making a fraction

like blah^-4 = 1/blah^4

tell me what. you get

[whole hazel]

Why does 64 become 64^-4?

[deep mason]

hm no i just mean ont he left side

[whole hazel]

I did this already but I did 1/(3x+5)^4 = 64

[deep mason]

just for (3x + 5)^-4

ok good

[deep mason]

perfect

now

you can flip both sides

[whole hazel]

Make 64 into a fraction n cross multiply, or multiply 64 by denominator to cancel out fraction?

[deep mason]

if a = b, then 1/a = 1/b

(as long as a and b aren't 0, which is true here)

[whole hazel]

Oh so I do 1/64?

[deep mason]

perfect

[whole hazel]

1/8^2?

[deep mason]

(3x+5)^4 = 1/64

and yes 1/8^2 works too

[whole hazel]

Ok now what Ginger da saviour

[deep mason]

okok so now

you can square root both sides

so try that once first

[whole hazel]

Why can I square root both sides?

[deep mason]

you can always do the same thing to both sides of the equation

to get a new equation

bc the two sides must be equal

[whole hazel]

Is that why I made 64 into 8^2?

[deep mason]

that helps yeah

[whole hazel]

To give me admin perms to square root both sides?

[deep mason]

(3x+5)^4 = 1/8^2, then sqrt both sides

[deep mason]

well it just helps out, you didn't have to at all

[whole hazel]

Oh I coulda done it w/ the 1/64 as well, I just skipped a step bascially w/ 1/8

[deep mason]

exactlyy

so sqrt(1/64) = sqrt(1/8^2) = 1/8

and then on the left

sqrt[ (3x+5)^4 ] = (3x+5)^2

does this make sense

[whole hazel]

Yes

[deep mason]

ok

and then we also get a +/-

this always happens when we sqrt both sides of an equation

(x^2 = 9 means x = +/- 3, etc...)

[whole hazel]

Ahh okay I see, so put dat next to 1/8?

[deep mason]

yeah perfect

so (3x+5)^2 = ± 1/8

[whole hazel]

What happened to da 1 over the left?

[deep mason]

that was the earlier step

we went from 1/(3x+5)^4 = 64

to (3x+5)^4 = 1/64

[whole hazel]

How did it disappear tho?

[deep mason]

flipped both sides

cause a = b means 1/a = 1/b

or, equivalently, 1/a = b means a = 1/b

[whole hazel]

Oh shit okay

[deep mason]

we did the "same thing" to both sides of the equation, so that's why it's allowed, in general

whether it's adding some number, or multiplying, or "flipping"

but yeah

[whole hazel]

Square root both sides now?

[deep mason]

so steps so far are:

1/(3x+5)^4 = 64

(3x+5)^4 = 1/64

(3x + 5)^2 = ± 1/8

now basically yes, we do another square root

there's a small bit of a side step with logic here

first

notice how (3x + 5)^2 is a number squared

if you square any real number, it should be positive, right?

[whole hazel]

Yes

[deep mason]

cool. so that means... we should choose the + here, in the +/-

(3x + 5)^2 = + 1/8

[whole hazel]

What scenario would prompt choosing the -?

If x wasn't squared?

[deep mason]

if we didn't have (blah)^2 on the left, we'd probably just have to leave it as +/-, we couldn't choose

if for some reason we knew it had to be negative, we could choose -

(an example would be -blah^2)

[whole hazel]

Ok I got ya

[deep mason]

that's a great question too bc we'll use that right now again, for the next step

[deep mason]

so now, like you said, we square root both sides... again

[whole hazel]

3x+5 = 1/64

[deep mason]

sqrt(1/8) on the right

[whole hazel]

Oh yeah

I got confused

Yeah yeah

[deep mason]

np np cool

so perfect

now here, we don't know about the +/-, so we should just put it back in like usual

in general you should have +/-

choosing + before technically just makes things easier... but it wasn't literally required

it's more true, in general, that there always is a +/- (here choosing + let us avoid complex/imaginary solutions)

[whole hazel]

It helped me understand better, I hear u

[deep mason]

anyway...we need it here

[deep mason]

awesome 👍

[whole hazel]

3x+5 = +- sqrt(1/8)

[deep mason]

perfectt

you've been doing well so i feel like you can probably take it from here

[whole hazel]

Subtract 5?

[deep mason]

right

[whole hazel]

divide by 3

[deep mason]

perfect

then that's it

but then the answers you posted just kinda manipulated the solution to put it in different forms

but yeah

[whole hazel]

Ok I see

Thank u so much bro, ily

Helps a ton frl

[deep mason]

awesome def glad i could help man

[whole hazel]

.solved

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[mortal hinge]

can someone help me pls

[empty spade]

ah damn

[mortal hinge]

o

[empty spade]

mb 😭

[mortal hinge]

it's ok lmao😭😭

[mortal hinge]

i got f'(x) = -1x ^ (-2) but the exponent is supposed to be pos he said so i'm lost

[slow yoke]

use laws of indices

[rich bone]

rewrite as 1 over some positive power of x

[mortal hinge]

he said he won't teach us the proof that the power rule for derivatives can be used for any rational number exponent cause there's not enough time. is there a simple way to get an answer with just the power rules for derivatives?

[mortal hinge]

idrk what he's talking about but is it this ?

[slow yoke]

you use the binomial theorem to prove it if yk what that is

[mortal hinge]

oh i think so

i don't think he wants us to use that tho

o i didn't see this thanks

is the answer -1/x^(2) then ?

okey tyy

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[versed depot]

SA=2πrh+2πr^2

of Cylinder

V = pir^2h

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[fast ivy]

this is like th elast step

So

In purple

M= 10 right

V = 5

G = 9.8

but whats

d

i hate how he didnt plug he just solved it magically );

[mystic saffron]

Distance

[forest sky]

d is the distance it travels along the plane

[fast ivy]

oh

so 4

meters

[mystic saffron]

Ye

[fast ivy]

thnx

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[thin topaz]

The equation of a straight line through the point C(−5, 4), given that its segment intercepted by the lines x + 2y + 1 = 0, x + 2y – 1 = 0 is of length 5.
(1) 3x + 4y – 1 = 0 (2) 7x + 24y – 61 = 0
(3) both (1) and (2) (4) neither (1) nor (2)

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[thin topaz]

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[frigid canopy]

Using the Vector space axioms , I want to prove that $0 \cdot v = 0_{v}$. Where $0_{v} \in V \land w+0_{v}=w ; w \in V$

[clever fjord]

Veni, vidi, perii is not f(wai)

[faint knot]

whats stopping you from looking up a general "0 * x = 0" proof that uses ring axioms

[frigid canopy]

What's a ring

[faint knot]

dont worry about it, just google it

[frigid canopy]

Well, my prof said this can be done using using just the vector space axioms

so I'm trying it

[faint knot]

oh I see

also avoid using \cdot if youre not doing dot product, it can be mistaken as a dot product

[frigid canopy]

what should i use in its place

[faint knot]

nothing to denote scalar multiplication as with xy

[frigid canopy]

👍

This is my attempt

From the additive identity axiom, it follows that $v+0_{v}=v$. Now we attempt to compute $v+0v = v(1+0)=v$. so $v+0_{v} = v+0v$. It thus follows that $0v=0_{v}$

[clever fjord]

Veni, vidi, perii is not f(wai)

[faint knot]

that works for me

try having it be v0 instead of 0v though so that you dont have to say you used commutativity

[frigid canopy]

👍

Thanks!

[faint knot]

np

[forest sky]

scalar multiplication doesn't have commutativity because the scalar & vector are not interchangeable

[frigid canopy]

yah

[forest sky]

so writing the scalars on the left is essentially a a style choice

although i think it's one you should follow

[faint knot]

wouldnt it need to be like an axiom

[forest sky]

well in vector addition we need it to be commutative because we want to be able to interchange two vectors: a + b = b + a
but scalar multiplication takes two different types of objects, a scalar and a vector, so it doesn't make sense to interchange them in the same way. for consistency it's still best to write them on the left

the closest thing is the compatibility with field multiplication [ a(b \vb v) = (ab)\vb v ]

[clever fjord]

cloud

[faint knot]

is v + 0v = v(1 + 0) considered the same as v + 0v = (1 + 0)v if scalar multiplication can deliberately not be defined for vector * scalar?

[forest sky]

it's still clear which one is the scalar and which one is the vector, the one on the left is just more confusing/ambiguous notation

[faint knot]

does this imply that confusing notation is enough to say that the multiplication does not make sense instead of less sense?

[forest sky]

depends on whether the grader is feeling generous with the partial credit

[faint knot]

so it sounds like instead you should swap out v(1 + 0) to (1 + 0)v to be consistent with having the constants on the left
if you leave it as v(1 + 0), you dont have to mention the "axiom" that scalar * vector and vector * scalar are very similarly defined
worst case, you get deducted for doing scalar-vector multiplication the wrong direction

[frigid canopy]

Got it. Thanks

[faint knot]

np

[frigid canopy]

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[supple sandal]

Hey i have a question,
How many route exist if the right steps are maximum two next to each other( you can go up ad many as you want) in a 6x6 grid? You start from the bottom left and the finnish is on the top right.

[noble forge]

https://youtu.be/f9VkiIEMHho?si=3VKrbWMp1dFB3OUA

the question isn't pretty clear to me but this video may help

[supple sandal]

Thanks

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Channel closed

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[oblique nest]

Can anyone help in my problem given above?

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@oblique nest Has your question been resolved?

[oblique nest]

<@&286206848099549185>

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[oblique nest]

<@&286206848099549185>

[oblique nest]

.close

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[steep mantle]

assuming the side of the square is 1, can you find the side length of the big red triangle and the small one?

[fluid tundra]

first find radius of circle

[steep mantle]

sqrt5

[fluid tundra]

tbh i think u should be able to do this

[red surge]

ngl this looks very cartesian bashable

[steep mantle]

are these the same

[mental lotus]

yes. by symmetry and congruence of big red triangles

[pliant latch]

is the middl eline at 45 degree

[steep mantle]

yeah

[fluid tundra]

u don't need this info btw

[steep mantle]

sorry had something to do

.close ok i got it

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[steep mantle]

.reopen

[odd edge]

✅

[steep mantle]

how without that info

[fluid tundra]

so

[steep mantle]

am i stupid

[fluid tundra]

radius is sqrt5

[steep mantle]

bruh fuck i see

[fluid tundra]

so automatically u know the smaller triangle has leg length sqrt5 - 2

& ok u see

[steep mantle]

ok i think its time for me to rest :p

[fluid tundra]

weets

[lyric dust]

lol

[steep mantle]

weet

.close alr thanks

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[buoyant hedge]

<@&286206848099549185>

[limber glade]

!15m

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[limber glade]

so what have you tried

[buoyant hedge]

i tried squaring on both sides

[swift lake]

square?

[limber glade]

there are no sides as of now

[swift lake]

that is a cubic

[limber glade]

this ain't an equation

[buoyant hedge]

yh sry cubing

i kept whole equation = x

[limber glade]

ok then

[buoyant hedge]

and then cubed

[swift lake]

Try to modify $ 5\sqrt{2} + 7 $ in form of $^3$

[clever fjord]

Newt

[swift lake]

it is a good form

[buoyant hedge]

bro i cant understand

how would u modify that

[swift lake]

example :

$5\sqrt{2}+7 = 2\sqrt{2}+3\sqrt{2}+3+4$

[clever fjord]

Newt

[swift lake]

hint :
it was in form of (a+b)^3

[buoyant hedge]

kk

ty

[swift lake]

$(a+b)^3=a^3+3a^2b+3b^2a+a^3$

[clever fjord]

Newt

[swift lake]

just as a reminder

[buoyant hedge]

bro pls elaborate

i'm stuck

.close

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[weary creek]

There are 10 people including Ben and Abby. The 10 people are to be arranged in a line. There need to be exactly 3 people between ben and abby. Calculate the total arrangement possible.

[robust tendon]

,calc 876

[clever fjord]

Result:

336

[robust tendon]

by the condition
There are 8 people
and only 3 can stay at a time

So possible outcomes are
8x7x6

[sterile blaze]

sorry what do u mean only 3 can stay at a time

u can count the number of ways u can arrange 8 people. multiply the number of ways ben and abby can be inserted into a group of 8

[odd edge]

@weary creek Has your question been resolved?

[weary creek]

If I do
8p8 X 2p2 X 6 ??

@sterile blaze

[sterile blaze]

yea

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[tough grove]

Can anyone pls help with this

[narrow crypt]

It’s direct proportionality?

[tough grove]

My teacher just said to identify the constant of k

[narrow crypt]

r and C are the variables

The first one, the constant is 2pi

[tough grove]

How?

My math teacher didn't even lecture this to us btuh

[manic marsh]

constants don't change while variables do

[tough grove]

No solution?

[narrow crypt]

U get how direct proportions can be expressed as

y = kx

[tough grove]

Yup

[narrow crypt]

In this case, y is C

And x is r

Because they are to be substituted with a certain number

So k must be 2pi

Because u can’t substitute 2pi with other numebrs

[tough grove]

Ohhhhh thx so much

[narrow crypt]

Can u do no.8

[tough grove]

K=7?

Is it right I'm Abt to sleep?

[narrow crypt]

Yes

[tough grove]

Yey

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[spare mason]

the question is asking for

angle AVM

no further info was given

[tender epoch]

use pythagoras to find the length of AM

[spare mason]

we dont

have enough info though

[tender epoch]

you do

[spare mason]

AM^2 + VM^2 = 18.6

how do we find VM

[tender epoch]

dont think you need VM

you only need to sides of the traingle and AM would be easier to find

[astral raft]

use pythagoras to find AC

[tender epoch]

by using the fact M is in the middle of the square

therefore its the middle of AC

[astral raft]

divide AC by 2 then use pythagoras to find VM

[spare mason]

i am a little lost

how do we use pythagoras here?

[tender epoch]

it says it a sqaure base

so BC is equal to AB which is 11 cm

[spare mason]

rright

so its a triangle with a base of X and 2 sides that are 11 cm

how do we find the base now

[tender epoch]

what do you mean by base

[spare mason]

this is what we have, correct?

X - amc

i could have

just

[tender epoch]

its a right angle trainge

[spare mason]

written that

[tender epoch]

because its part of the square based pyramid

[spare mason]

holy shit thats so stupid

i got it

[astral raft]

AC=√(11^2 +11^2)

[spare mason]

tttthank you very much

.close

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[deep silo]

Is tjis the right start ?..

[unkempt bluff]

the substitution is correct

let u = sqrt(x) + 1

du = 1/(2sqrt(x)) dx

dx = 2sqrt(x)du

so the integrand becomes ln(u)/sqrtx * 2sqrt(x) du

the sqrtx's cancel out

so it becomes 2ln(u) du

do you know what the antiderivative of ln is?

[deep silo]

Well it’s rather complicated

[unkempt bluff]

not really

you just do integration by parts

that is if you dont already have it memorized

[deep silo]

Oh ye

By parts

Wait so I should not

[unkempt bluff]

(hint: differentiate lnu and integrate 1)

[deep silo]

Isolate c ?

X*

Like i did

[unkempt bluff]

wdym isolate x

[deep silo]

Sqrt(x) = u-1

And u = sqrt(x)+1

[unkempt bluff]

what does that have to do with integrating lnx?

[deep silo]

so i can substitute both sqrt(x)

well im just wondering

[unkempt bluff]

after you get the antiderivative of 2lnu you plug in sqrtx + 1 into all the u's you have

[deep silo]

yes i know

But look at what i did

I have ln(u) / (u-1) * dx

[unkempt bluff]

yes that is techincally correct

but it makes simplifying it to 2lnu very tough so just leave the sqrt(x) as it is and do the dx = 2sqrtx du so that the sqrtx's cancel out

and you end up with 2ln u

[deep silo]

How do u do that ?

Because If i solve for Dx in my du/dx

I get 2*du / x^(-1/2)

right?

[unkempt bluff]

i dont use du/dx at all tbh

i have

u = sqrtx + 1

i differentiate both sides

ok wait i get how to use du/dx now

du/dx = 1/2sqrt(x)

du = 1/2sqrtx dx

dx = 2sqrtx du

see where i got it from now?

[deep silo]

welll

Differentiaging sqrt(x) gives me 1/2 * x^(-1/2)

differentiating

[unkempt bluff]

yes thats the same as 1/2 * 1/sqrtx

which is the same as 1/2sqrtx

i just use a different form

[deep silo]

ohhh

[unkempt bluff]

but you differentiated correctly

[deep silo]

so my dx

is

2sqrt(x) du

[unkempt bluff]

yup

so you have the integrand of

ln(u)/sqrt(x) * 2sqrtx

[deep silo]

im bad with the by parts tho not gonna lie

[unkempt bluff]

aka 2lnu

so thats how you get the integral of 2lnu du

[deep silo]

Ye i understood now! I just made it complicated for myself xd

[unkempt bluff]

do you know the D I method?

it makes ibp 10x easier

[deep silo]

D I ?

No

[unkempt bluff]

its just a much easier way to do integration by parts

[deep silo]

ohhh what is this

[unkempt bluff]

you factorise your integral into 2 things multiplied by each other
You differentiate 1 and integrate the other

then you do uv - integral of duv

its the exact same formula as integration by parts its just an easier wa y to memorize how to set it up

[deep silo]

so 2 * ln(u)

[unkempt bluff]

plus its useful when youre repeating integration by parts for many steps

[unkempt bluff]

yeah

integrate 2 to get 2u

differentiate lnu to get 1/u

[deep silo]

this is the formula we have

[unkempt bluff]

yeah this is the same thing as what i said

[deep silo]

so its just 2 lnu - the integrad of 2 * the derivative of lnu

[unkempt bluff]

like i said theyre all the same ibp formula at the end of the day, just different ways of setting it up

[unkempt bluff]

no remember the F(x) is the ANTIDERIVATIVE of the first part which we said is 2

so its 2u lnu - the integrand of 2 * the derivative of lnu

[deep silo]

Ohhhh

[unkempt bluff]

since 2u is the antiderivative of 2

[deep silo]

So the big F

[unkempt bluff]

capital letter = integral, prime' = derivative

F(x) is the antiderivative of f(x)

[deep silo]

Yeee

Okey

[unkempt bluff]

f'(x) is the derivative of f(x)

[deep silo]

Ohhh

And the ln(u) is kept the same

2u*1/u

[unkempt bluff]

yup

simplifies to 2

so its 2ulnu - 2

wait

[deep silo]

JEEESSUS

So nice

[unkempt bluff]

we did something wrong in the middle

the integral of 2lnu should be 2ulnu - 2u

not 2ulnu - 2

lemme try to find the mistaek

wait yeah i found the mistake

its 2ulnu - THE INTEGRAL OF 2

so 2ulnu - 2u

now thats correct

sorry if i confused you there

[deep silo]

Ye exatly

And then now we should be done and just put in the u again ?

[unkempt bluff]

yup thats correct👍

great job!

[deep silo]

Thank you!! I just had a hard time when I did sqrt(x) = U-1

You could do that to right?

but it would be way more complicated

I made it hard for myself i think

[unkempt bluff]

i think it would lead to the same steps and solotion

since its the same substitution ath eart

sqrtx = u-1 is the exxact same as saying u =sqrtx + 1

[deep silo]

I know but I would have a

[unkempt bluff]

but idk maybe it would make it more complicated

so you should just stick to the simpler u = sqrtx+1

[deep silo]

I would not get rid of the sqrt(x) in the beggining

Since it would be (u-1)

[unkempt bluff]

well you would have
du = 1/2sqrt(x)dx
dx = 2sqrt(x)du
so the integrand would be ln(u)/sqrtx * 2sqrt(x) du so it would simplify to the exact same things

what matters is that you dont rush subsituting u in for all of the x's in the integrand and instead first checking to see whta cancels out

in this case we ended up making the problem a whole lot easier by cancelling out the sqrt(x)

[deep silo]

Yes that’s true

This is way more complicated to do

[unkempt bluff]

well you could still convert the 2sqrt(x) into 2(u-1)

then the (u-1)'s would also cancel out

so it cancels out either way

[deep silo]

Oh okey

That’s true

But it says we can simplify this

How ?!

[unkempt bluff]

factor out 2(sqrtx +1)

it becomes 2(sqrtx + 1) (ln(sqrtx + 1) - 1)

[deep silo]

Oh ye

[unkempt bluff]

idk if they consider that simplified or not

i think its the only way to simplify though

[unkempt bluff]

oh right the C good thing you remembered

[deep silo]

Hahahahha was about to forget it tho🤣

This one is much harder 😢

Would u substitute u = 1+x or U= x^2?

[mystic saffron]

separate in 2

1/x^2 + 1/x

[deep silo]

?

[mystic saffron]

the term in the integral

[deep silo]

Ye

[mystic saffron]

u split it

[deep silo]

Sju

Why

[mystic saffron]

so that u can find primitive of (1+x)/x^2

[deep silo]

How does 1 / x^2 + 1 / x be (1+x)\x^2

We need to expand it and multiply by x?

The 1/x

Maybe 1/x^2 * x

[mystic saffron]

uh just (1+x)/x^2 = 1/x^2 + x/x^2

[deep silo]

Ye

But can we write it directly ?

1/x^2 + 1/x is the same thing ?

[mystic saffron]

bah yes

[deep silo]

ye thats new for me

did not know that

[mystic saffron]

x/x^2 = 1/x

u didnt know ???

[deep silo]

no xd

Oh Ye wait that I know

[mystic saffron]

very weird

[deep silo]

We can take out a x

so 1/x

ohhhhhhhhh

ye sorry my bad

[mystic saffron]

i was cofuse bc u do integrals without knowing to calculate stuff

but ko

ok

[deep silo]

Hahahah im sorry

then we can integrad that

And just put in our 2t and t ?

[mystic saffron]

yea

[deep silo]

cause 1/x^2 = x^-2

which is equal to x^-1/1

if we take the integral

[mystic saffron]

try derivate -1/x

[mystic saffron]

yes good

[deep silo]

now i can put the 2t and t in

[mystic saffron]

but can u make it look better ?

like

[mystic saffron]

the x^-1 with 1/x

ok so

what u do now ?

[deep silo]

Lim as t -> inf