help-39

[snow latch]

I have no idea what to do here. I thought at first to use W = Vq and use E = hf to connect them but the answer is so different and I have no idea what I'm doing.

maybe if I use Ek = 1/2mv^2 but its not entirely right

[sharp vigil]

note that the de Broglie wavelength of a particle with momentum $p$ is [ \lambda = \frac hp ]

[jolly parrot]

cloud

[snow latch]

yep

λ = h/mv

I did Vq = 1/2mv^2 to try to get a equation in terms of v because the final equation isn't in terms of v

,tex $ \sqrt{\frac{2Vq}{m}}

[jolly parrot]

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

[snow latch]

and then i can sub that in to λ = h/mv

BUT

the answers say that its actually

,tex $ λ = \frac{h}{\sqrt{2qVm}}

[jolly parrot]

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

[snow latch]

so idk how they got m to not be m^-1

[sharp vigil]

,, \frac{m}{\sqrt m} = \sqrt m

[jolly parrot]

cloud

[snow latch]

but where did the m come from

top m

[sharp vigil]

from mv

[snow latch]

, tex $ m \sqrt{m}

[jolly parrot]

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

[snow latch]

h/mv so it will be m x sqrt 2Vq/m

does it not effect the other variables in the equation?

[sharp vigil]

what do you mean by affect the other variables?

[snow latch]

like its not exactly m x rt m

,tex $ \frac{h}{m\sqrt{\frac{2Vq}{m}}}

[jolly parrot]

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

[sharp vigil]

by exponent rules we can split the square root into a product of two square roots

or alternatively write $m = \sqrt{m^2}$ and combine the two

[jolly parrot]

cloud

[snow latch]

ok yes true

but does that mean that m / rt m = m x sqrt 1/m

[sharp vigil]

yes, because sqrt 1 = 1

[snow latch]

,tex $ \frac{m}{\sqrt{m}} = m \sqrt{\frac{1}{m}}

[jolly parrot]

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

[snow latch]

im stupid

thanks

we can just blame the fact that i'm sick for that oversight 🤦‍♀️

cloud, thank you, you're always the one to help me out

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[dreamy wind]

How do i do this?

this is my work so far idk if im headed in the right direction

[sterile tusk]

you can rewrite $\int\frac{\sec^3(\theta)}{\tan^6(\theta)}d\theta$ as $\int\frac{\cos^6(\theta)}{\cos^3(\theta)\sin^6(\theta)}d\theta$

[jolly parrot]

y0shi

[dreamy wind]

oh shoot fax i didnt think abt that

[dreamy wind]

ty so much

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[naive hemlock]

.reopen

Yo

[naive hemlock]

.close

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[fossil drum]

I'm not sure on what they are asking here

[fossil drum]

I evaluated the double integral

but what do they mean by upper and lower bound?

is that like a max and min thing?

[sterile tusk]

like the bounds of integration

[fossil drum]

theres only two inputs though

wouldnt there need to be 4

[sterile tusk]

thats true

huh

have you done a question like this before?

[fossil drum]

Heres a better picture

[tropic saddle]

I think they mean number <= value of integral <= number

but not sure

[fossil drum]

No I havent, just regular double integrals over rectangular regions is what ive done

[sterile tusk]

i see

[fossil drum]

https://myopenmaths3.s3.amazonaws.com/ufiles/706699/Solution_for_Students_29.pdf

Here's a written example

but I didnt understand it

and its hard to read

[burnt dust]

It's this yeah

They are doing a very crude estimate

All they are doing is that the value of integral is bounded between the area of the domain times max of integrand on the domain and times min of the integrand on the domain

@fossil drum

[fossil drum]

oh I see

I think im a little confused on something though, how come the max of the integrand function times the area of the domain is greater than the integral value?

oh wait

I see

its basically taking that max point and forming a cube? @burnt dust

like if we were to integrate on that constant over a rectuangluar region it'd create a cube

[burnt dust]

Not a cube, but yeah

[fossil drum]

oh ok

[burnt dust]

It is a rectangular prism

[fossil drum]

oh yeah

[burnt dust]

I think the better way to think about is

That the integral is the same as the average value of integrand integrated over the same region

And the average is between the min and the max

[fossil drum]

hmm ok I see

thank you for the help

[burnt dust]

Np

[fossil drum]

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[quaint sluice]

can someone double check for me pls

[spare lark]

If derivative goes from + to - its a max not a min

And if it goes from - to + its a min

So min would be x=-2,4

[quaint sluice]

yeah i had it in the wrong order

[spare lark]

So what about max ?

[quaint sluice]

x=0

is anything else on there wrong?

[spare lark]

Checkin

All good next

[spare lark]

No

[quaint sluice]

alright thx

.close

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[vast berry]

When B goes down 1 meter does A go up 1 meter (diagonally)

[pearl pond]

@vast berry Has your question been resolved?

[upbeat glade]

Show the entire question.

@vast berry

[vast berry]

ok

[pearl pond]

@vast berry Has your question been resolved?

[final hinge]

Yes since the string is inextensible

[vast berry]

Thx lmao common sense failedme

[pearl pond]

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[gilded hollow]

Show that a square always gives the remainder 0 or 1 when dividing with 4

[gilded hollow]

I'm having a real hard time proving stuff in discrete mathematics..

$a^2 \equiv 0 \mod 4 \
a^2 \equiv 1 \mod 4$

[jolly parrot]

Merineth

[gilded hollow]

Would this be a good start?

If i start inputting values into a such as 1, 2, 3 and 4 i can clearly see that a^2 mod 4 always gives a remainder of 0 or 1 but i don't think that's how they want me to prove it..

Ideas?

[cursive wraith]

it is how they want you to prove it

a is either 0,1,2 or 3 mod 4

so a^2 mod 4 is either 0^2, 1^2, 2^2 or 3^2

otherwise you can do a smaller case division even/odd :
if a = 2k, then a^2 = ...

if a = 2k+1, then a^2 = ...

[gilded hollow]

Hmm

I'm not sure i understand that

[cursive wraith]

which one you don't understand

[gilded hollow]

1 mod 4 = 1, 4 mod 4 = 0, 9 mod 4 = 1... and so on

Isn't this proof enough?

Showing the pattern

[cursive wraith]

no

you have to be rigorous in a proof

[gilded hollow]

I don't know what rigerous means

[cursive wraith]

it's not because a sequence goes 1,0,1,0 for the first 4 terms that it's gonna keep doing that

you have to clearly show the pattern goes on forever

[gilded hollow]

Ah i see

[cursive wraith]

without checking every number individually

[gilded hollow]

So we start by stating that since it's mod 4 we can have a remainder of 0,1,2 or 3

[cursive wraith]

yep

[gilded hollow]

and

we have to prove why it's NOT gonna be 2 or 3?

Or prove that it HAS to be 0 or 1

[cursive wraith]

well you've reduced down to 4 possibilities

so you can just check them 1 by 1

what happens if a mod 4 = 0?

etc

[gilded hollow]

well in the case of a mod 4 = 0 it has to be a power of 4, no?

[cursive wraith]

uh

12

is 12 a power of 4?

i think you used the wrong word

[gilded hollow]

It was more of a guess rather than using the wrong word

I'm guessing 90% of the time in discrete math

[cursive wraith]

a mod 4 = 0

just means 4 | a

[gilded hollow]

shouldn't it be

[cursive wraith]

a is a multiple of 4

[dapper kraken]

cant you do a=2k or 2k+1

then expand

[gilded hollow]

a^2 mod 4 = 0

[cursive wraith]

talked about it already

[dapper kraken]

oh sry didny see

[cursive wraith]

yes, if a = 0 mod 4 then a^2 = 0 mod 4

so first case done

[gilded hollow]

what

not so fast

[cursive wraith]

?

[gilded hollow]

a^2 mod 4 = 0

I need to solve this equation, no?

[cursive wraith]

no don't go the other way around please

start from what a mod 4 is

and THEN compute a^2 mod 4

[gilded hollow]

ok. uh a mod 4 = 0 only happens at 4, 8, 12... i.e a multiple of 4?

[cursive wraith]

yes

[gilded hollow]

a mod 4 = 1 only happens when it's 1 away from the multiple of 4, so it should be: 3,5,7,9...

i think?

[cursive wraith]

3?

[gilded hollow]

No sorry not 3

[cursive wraith]

it's not 1 'away from'

it's 1 'more than'

[gilded hollow]

Oh

Right

in that case

a mod 4 = 1 only happens when it's 1 more than the multple of 4: 5, 9, 13...

[cursive wraith]

1,5,9,13,17...

[gilded hollow]

Yeah sorry, i'm trying to do it as fast as i can

a mod 4 = 2 happens at: 2, 6, 10, 14...

a mod 4 = 3 happens at: 3, 7, 11, 15...

and then we are back at the multiple of 4 again, i.e 0

a mod 4 = 0 happens at: 4, 8, 12...
a mod 4 = 1happens at: 1, 5, 9,1 3...
a mod 4 = 2happens at: 2, 6, 10, 14...
a mod 4 = 3happens at: 3 ,7, 11, 15...

So i have these 4 cases on a

But the per defenition of a square, a square only gets, 1, 4, 9, 16...

But then again, that isn't rigorous

https://tenor.com/view/cat-cats-orange-cat-orange-cats-cat-weird-gif-17010424884748616697

<@&286206848099549185>

[last moth]

Hey! Do you follow what the previous person said about every number either having remainder 0, 1, 2, or 3 when you divide by 4?

That's actually very similar to the q I helped you on previously haha

[gilded hollow]

Hehe c:

[gilded hollow]

Well we did conclude this

[last moth]

Okay so that means we can write the number either as 4k+0, 4k+1, 4k+2, or 4k+3

Are you good with that

It's either 0, 1, 2, or 3 more than a multiple of 4

[gilded hollow]

Hmmm where does the 4k come from?

[last moth]

Think about what we do when we divide something with remainder

We split the number into one part which is divisible by 4 and one part which isn't

For example: 13 divided by 4 is 3 with a remainder of 1

So 13 = 4(3) + 1

[gilded hollow]

Yes, that would be using euclides iirc

[last moth]

Yeah it's one of the steps in Euclid's algorithm, but the idea of division with remainder is really important in number theory as a whole too

[last moth]

OK so we have four cases, let's just square them and see what happens

What's 4k, squared?

[gilded hollow]

16k^2

But the problem is i dont really get the 4k

a = 4k
This i understand since it mentioned it in the book for the definition of evenly dividing

a | b -> b = ak

[last moth]

Do you understand why 13 = 4(3) + 1

[gilded hollow]

Yes

[last moth]

Okay so

The k in that example is 3, and the remainder is 1

[gilded hollow]

a = 4k + 0 a = 4k + 1 a = 4k + 2 a = 4k + 3

[last moth]

In other words, the quotient is 3, and the remainder is 1

If you divide any number n by 4, you can rewrite the number as n = 4 * (the quotient) + (the remainder)

That's the definition of division with remainder, basically

[gilded hollow]

Okay, that makes sense

[last moth]

Cool, it's a very important idea that will apply to very many problems

[gilded hollow]

But is there a reason to why we square them?

[last moth]

The problem asks you to

[gilded hollow]

Isn't it squared from the start?

[last moth]

We're saying a is either 4k, 4k+1, 4k+2, or 4k+3

We haven't calculated a^2 yet

[gilded hollow]

a^2 = 4k + 0
a^2 = 4k + 1
a^2 = 4k + 2
a^2 = 4k + 3

[last moth]

The problem wants you to look at a^2

[gilded hollow]

oh

In that case

a^2 = 16k^2

[last moth]

yup

and then what's that, mod 4?

[gilded hollow]

Uhm, i'm not sure how i apply mod 4 to an equation

[last moth]

We want to divide 16k^2 by 4 with remainder

What do you get when you divide 16k^2 by 4

[gilded hollow]

4k^2?

[last moth]

Yup

[last moth]

So can you write that in this form

16k^2 = 4 * (the quotient) + (the remainder)

[gilded hollow]

I'm not sure i follow

We start with the case when the remainder is 0, yes?

a = 4k + 0

So we take and square it since it's supposed to be a square?

a^2 = 16k^2 + 0^2

and since 16k^2 is divisible by 4 with a remainder of 0, it's 0?

[last moth]

Yes

[last moth]

I was just asking you to practice writing it in this form

Because the concepts of quotient and remainder are very important

i.e. fill in the things that should go in (the quotient) and (the remainder)

[gilded hollow]

hmm ok

a = 4k + 1

a^2 = 16k+1

[last moth]

Not quite

[gilded hollow]

oh

(4k+1)^2

[last moth]

Yeah

And then how do you expand that out

[gilded hollow]

16k^2 + 8k + 1

[last moth]

Good!

And then what do you get when you divide that by 4

[gilded hollow]

4k^2 + 2k + 1/4

[last moth]

Yeah, so the 1/4 part is the part that we can't divide

Which gives us a remainder of 1

[gilded hollow]

oh i see

[last moth]

You can write this as: a^2 = 4 * (4k^2 + 2k) + 1

That clearly shows that the remainder is 1 when you divide by 4

[gilded hollow]

i see

$a^2 = (4k+2)^2 \iff a^2 = 16k^2+16k+4$

[jolly parrot]

Merineth

[last moth]

Yup

[gilded hollow]

And this is divisible by 4 which results in a remainder of 0

[last moth]

Yup!

[gilded hollow]

and lastly:

$a^2 = (4k+3)^2 \iff a^2 = 16k^2 + 24k + 9$

[jolly parrot]

Merineth

[gilded hollow]

$9 = 4 \cdot 2 + 1$

[jolly parrot]

Merineth

[gilded hollow]

results in remainder of 1

So now we have proven all the cases for mod 4 where the remainder for a = 0,1,2 and 3

[last moth]

Yup! :)

[gilded hollow]

It wasn't that hard, however i'm not really able to solve it unless i'm given information such as to write it in the form a = 4k + remainder

[last moth]

Btw as the helpers said earlier, it's also possible to do 2k, 2k+1 instead of 4k, 4k+1, 4k+2, 4k+3

But I think that's less obvious to see

Since there's no 2 in the question

Dividing by 4 is the most obvious thing to do

[last moth]

Well, quotients and remainders are one of the concepts you learned, right?

You mentioned learning Euclid's algorithm for example

It's really really important to know that you can divide any number n by any number d and write the result as n = d * (the quotient) + (the remainder)

In fact we did that in the previous problem too!

Now that you know how useful this trick is, you can try applying it to other problems going forward

[gilded hollow]

I hope so :S

Since i get stuck at all of them lmao xD

I'm never able to start solving a problem since i don't know what to write

Based on what you mentioned earlier with hthe n = d * (the quotient) + (the remainder)

$x^2+x+2 \equiv 0 \mod 4$

[jolly parrot]

Merineth

[gilded hollow]

Since mod 4 results in a remainder of 0,1,2 or 3

It should be able to be written as

$x^2+x+2 = 4k + a$ where a is 0, 1, 2 or 3?

[jolly parrot]

Merineth

[last moth]

Is this the thing you wanna prove

[gilded hollow]

Yes

[last moth]

Maybe start with writing x = 4k + a instead

Similar to what we did with this problem

we wrote a = 4k + (something), instead of a^2 = 4k + (something)

So that we could then square it

[gilded hollow]

Oh right makes sense

what about the +2?

Can it be ignored for now?

[last moth]

You'll write x = 4k + a

And then you'll want to calculate what x^2 + x + 2 is

and then divide it by 4

That's generally how you might proceed about this problem

[last moth]

Er, I don't think this statement is true though? If x = 0, you get 2 on the left-hand side but 0 on the right-hand side

Is there some condition on x?

[gilded hollow]

it just says "solve the squared congruent..."

[last moth]

Oh

That means you don't want to prove it's true then, you want to find the values of x so that it's true

This same procedure should work though

[gilded hollow]

Oh

[last moth]

Write x = 4k + a, then calculate what x^2 + x + 2

Then check if it's divisible by 4

[gilded hollow]

hmm

$x(x+1)+2 \equiv 0 \mod 4$

[jolly parrot]

Merineth

[gilded hollow]

This wont help?

[last moth]

There's nothing wrong with it

Try writing x = 4k + a and calculating out though

[gilded hollow]

ok

I'm not sure if this reasoning is correct but:

$x \equiv 0 \mod 4$ means that x has to be a multiple of 4 for the remainder to be 0, no?

[jolly parrot]

Merineth

[last moth]

Yes

That's correct

[gilded hollow]

Doesn't that mean that we want to find $x^2+x+2 = 4$

[jolly parrot]

Merineth

[gilded hollow]

hmm

[last moth]

No, it can be equal to any multiple of 4

It'll be simpler to do the method I said though

[gilded hollow]

I'll do your method but i'm just checking x^2+x+2 = 4^x should be able to find my solution?

Nvm i'll just do it like you said

[last moth]

No

4^x is not multiples of 4

That's powers of 4

If you wanted multiples of 4, that would be 4n

[gilded hollow]

oh right

$x = 4k+a$

[jolly parrot]

Merineth

[gilded hollow]

I don't know how to calculate this, we have 3 missing variables

or do i replace a with 0,1,2, or 3

[last moth]

Calculate x^2 + x + 2

[last moth]

You will at the end

[gilded hollow]

$x^2+x+2 = 4k + a$ like so?

[jolly parrot]

Merineth

[last moth]

No

x = 4k + a

Plug that formula for x into x^2 + x + 2

[gilded hollow]

oh

$(4k+a)^2 + (4k+a) + 2$

[jolly parrot]

Merineth

[last moth]

Yup

And then expand

[gilded hollow]

$16k^2+8ka+a^2 +4k+a+2$

[jolly parrot]

Merineth

[gilded hollow]

$16k^2+8ka+4k+a^2+a+2$

[jolly parrot]

Merineth

[last moth]

Good!

And now you can plug in different values of a

to see which ones will result in the whole thing being divisible by 4

[gilded hollow]

$16k^2 + 8k0 + 4k+ 0^2 + 0 +2 \implies 16k^2+4k+2$

[jolly parrot]

Merineth

[gilded hollow]

This results in a remainder of 2

when a = 0

[last moth]

Yup, so that's not divisible by 4

What about when a=1

[gilded hollow]

$16k^2+8k+4k+4$ is div by 4 when a = 1

[jolly parrot]

Merineth

[last moth]

good!

[gilded hollow]

$16k^2 + 16k+4k+8$ div by 4 remainder 0 when a = 2

[jolly parrot]

Merineth

[last moth]

Yup!

[gilded hollow]

and finally

$16k^2+24k+4k+14$ is not div by 4 since $14 = 4*3 + 2$ remainder 2

[jolly parrot]

Merineth

[last moth]

Wonderful!

Good job, I think you're getting a lot better at this :)

So what would be your final answer

As to what values x can be

[gilded hollow]

Hmmm

Well we concluded that the only time that $x^2+x+2 \equiv 0 \mod 4$ is true when our remainder of a = 1 or 2

[jolly parrot]

Merineth

[last moth]

yes! perfect

[gilded hollow]

I'm not entirely sure what to make of it tbh since a isn't x

answer is x = 4k+1 or x = 4k+2?

[last moth]

yes

[gilded hollow]

Solving the algebraic expressions isn't hard

I just can't write them

You went from "since mod 4 is remainder of 0,1,2, or 3 thus x = 4k + a where a = 0,1,2,3 thus (4k+a)^2+4k+a+2 mod 4 thus insert a = 0,1,2,3, or 4 thus finding x"

That is a lot of knowledge to know in order to solve something as complex as this

[last moth]

It's the same method that we used to solve the last problem

You'll get the hang of it with practice

[gilded hollow]

hehe there aren't any more practice problems :) next is Fermats little theorem

[last moth]

Ah, sounds fun, good luck! The rest of the material will probably build off of this anyways

So you'll get more practice anyways

[gilded hollow]

I hope so

tysm

.close

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[gilded hollow]

Use Fermats little Theorem to calculate $2^{3011} \mod 31$

[jolly parrot]

Merineth

[gilded hollow]

Could someone help me clarify a few things

Fermats states that :

$2^{30} \equiv 1 \mod 31$

[jolly parrot]

Merineth

[gilded hollow]

So we can rewrite it as

$2^{30} \cdot 2^{2981}$ ?

[jolly parrot]

Merineth

[gilded hollow]

or do i figure out how many 2^30 fits into 2^3011?

[sharp quest]

you don't need to figure out exactly how many, but that's the idea yeah.
What is a multiple of 30 that's less than 3011?

[gilded hollow]

100

[sharp quest]

what's 100?

[gilded hollow]

Hmm well

[random ermine]

i think she means 30*100=3000

[gilded hollow]

yes.

does that mean we have

[sharp quest]

then yes, you have 2^3000 * 2^11

[gilded hollow]

And since we know 2^3000 is congruent to 1?

we get 1 * 2^11?

[sharp quest]

yep

[gilded hollow]

so we reduce it down to $2^{11} \mod 31$

[jolly parrot]

Merineth

[random ermine]

yes

[gilded hollow]

Do i have to do it manually now? calculate 2^11

[random ermine]

you could it's not that bad

or note that 2^5 = 32 is 1 mod 31

[gilded hollow]

oh that's neat

didn't see that

so we have 2^1 left

remainder 2?

[random ermine]

think so

[gilded hollow]

,w 2^3011 mod 31

[jolly parrot]

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[gilded hollow]

[random ermine]

great

[gilded hollow]

I def need more practice but they literally only had one example in the book lol

on the exam we got

$2019^{2019^{3}} \mod 11$

[random ermine]

u could also have used 2^5 = 1 mod 31 at the beginning

[jolly parrot]

Merineth

[gilded hollow]

I assume i split it up into

$2019^{2019} \cdot 2019^{2019} \cdot 2019^{2019} \mod 11$

[jolly parrot]

Merineth

[gilded hollow]

and i know that

[random ermine]

no

[gilded hollow]

no?

[random ermine]

that's 2019^(3*2019)

[gilded hollow]

oh right

[random ermine]

you want 2019^(2019*2019*2019)

ig first what's 2019 mod 11?

,w 2019 mod 11

[jolly parrot]

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[random ermine]

,w factorize 2019

[jolly parrot]

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[random ermine]

can you reduce the problem any further

[gilded hollow]

dont do it for me

I'm still at the start

$2019^{10} \equiv 1 \mod 11$

[jolly parrot]

Merineth

[gilded hollow]

So how many times does 10 fit into 2019

201 + 9 times

[random ermine]

this is good

[gilded hollow]

Couldn't i technically rewrite it with that in mind such that:

$2019^{9^3} = 2019^{729}$

[random ermine]

9^3 = 729

[gilded hollow]

fuck

[jolly parrot]

Merineth

[gilded hollow]

$2019^72 * 2019^9$

[jolly parrot]

Merineth

[gilded hollow]

$2019^9$

[jolly parrot]

Merineth

[random ermine]

yep

[gilded hollow]

6^9 ?

[random ermine]

yep

[gilded hollow]

$6^9 \mod 11$

[jolly parrot]

Merineth

[gilded hollow]

Hmm lemme think

now i'm not sure how to simplify further

it's still a very large number

[random ermine]

we started with a much larger number

this is the easy part

[gilded hollow]

We need to write 6^9 in a way where mod 11 is congruent to 1?

[random ermine]

not necesssarily

you can simplify it any way

eg use 6^2 = 36 = 3 mod 11

[gilded hollow]

Yeah that was what i meant

almost

wait so you want to write it as

$6^26^26^26^26^1$

[jolly parrot]

Merineth

[gilded hollow]

but since 6^2 is 36

and 36 mod 11 is : 36 = 11 * 3 + 3

it becomes

$33336$ ?

[jolly parrot]

Merineth

[random ermine]

mhm

[gilded hollow]

3^4 * 6 = 81 * 6 = 486

$486 \mod 11$

[jolly parrot]

Merineth

[gilded hollow]

486 = 11 * 44 + 2

so remainder 2?

[random ermine]

yep

good job

[gilded hollow]

,w 2019^2019^3 mod 11

[jolly parrot]

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[gilded hollow]

<3

[gilded hollow]

Thank you

That wasn't too hard tbh

A lot of simplification and substitution

[random ermine]

yeah just keep reducing

[gilded hollow]

wtf is this tho

[random ermine]

just a slightly different approach

[gilded hollow]

"Since [2019/11] = 6 and 11 are primes,
...."

since when is 6 a prime?

[random ermine]

11prime

for flt

a can be anything

[gilded hollow]

i'll ignore that solution xD

thanks tho!

.close

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[jovial hare]

any mind helping me solve this im relearning this work and struggling to remember

[jovial hare]

piecewise is the only thing really

[scenic aurora]

splitting the case maybe?

like |x| = -x, x <0

but if x > 0, |x| = x

so you have 2 cases

[jovial hare]

yeah i know i just dont dont know the notation to write a piecewise

[scenic aurora]

another way is to square it

[scenic aurora]

like... this?
$$|x| =
\begin{cases}
x & \text{if } x \geq 0 \
-x & \text{if } x < 0
\end{cases}$$

[jolly parrot]

nonstationary nickname

[vital estuary]

honestly, i would solve it by splitting into x^2-2=2x+6 and -(x^2-2)=2x+6, then verify solutions by plugging them back into the original equation

[jovial hare]

yes !

[jovial hare]

yeah and for this case what would the restrictions be?

[vital estuary]

no restrictions

[jovial hare]

then what would go in for the piecewise?

[vital estuary]

nothing

when you plug in to the original equation, you will remove any solutions which do not work there

[jovial hare]

but even you added restrictions i get its an example but i have never seen a piecewise which doesnt have them\

[vital estuary]

by plugging them back into the original equation, you dont have to worry about restrictions

basically you get all the good solutions and solutions known as extraneous solutions which you then remove by verifying

[jovial hare]

ahh alright then ty

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[eager yoke]

can somebody confirm this? so essentially
first derivative

• related to gradient
  when dy/dx = 0, this is stationary
  second derivative
• related to nature of stationary point
  d2y/dx2 > 0, minimum
  d2y/dx2 < 0, maximum

thank you!

[spare lark]

First derivative give us variations, maximum and minimum, second derivative is for inflexion point and concavity up/down

[eager yoke]

am I able to ask questions about this please?

[spare lark]

Sure

[eager yoke]

are you able to give me a very simple example using the first derivative for maximum and minimum please?

[spare lark]

f(x) = x^2

So first derivative is f'(x) = 2x

[eager yoke]

yes

I understand so far

[spare lark]

And you look when does 2x is positive for f to be increasing and where negative for f to be decreasing

Like its negative on (-inf,0] and positive on [0,+inf)

[eager yoke]

I see

[spare lark]

What about 2x = 0 then ?

Where f'(x)=0 ifw

We have a minimum

Because f'=0 at x=0 and change her sign at x=0

[eager yoke]

ty

.solved

[pearl pond]

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[little silo]

hi

[little silo]

who knows cramer rule good

[pearl pond]

@little silo Has your question been resolved?

[pearl pond]

@little silo Has your question been resolved?

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[pearl pond]

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[dense sedge]

can anyone help me with this?

[bitter meadow]

8

[dense sedge]

how?

[bitter meadow]

96÷64=1.5 12÷1.5=8

[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.

[dense sedge]

do i do the same here?

[bitter meadow]

Yeah

[dense sedge]

so 30 divided by 9

and divided by 6?

[bitter meadow]

Other way

[dense sedge]

?

[cloud shard]

30/J = 9/6

[bitter meadow]

9÷6 answer ÷ 30

[dense sedge]

i get it now

TYsm

[bitter meadow]

Cheers

[cloud shard]

good job

[pearl pond]

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[dense sedge]

i'm stuck agin with this

[dense sedge]

can anyne help

[unique granite]

I think you just scale it

15/3 h/7

find h

[dense sedge]

u have to find the h mm not to scale it

[unique granite]

try 35

[dense sedge]

u sure?

is right

[unique granite]

you are scaling it

because

15/3 to h/7 is a multiplication in size kinda

[bitter meadow]

I got 34.5

[unique granite]

so you solve for h

i just did 15 times 3/7

to find the multiplier

[dense sedge]

what?

[unique granite]

to find h you have to see how they got 7

if they are similar shapes imagine the second shape is just 2 times bigger

so how would 3 scale to 7

you would find that by dividing 7/3

[dense sedge]

k

[unique granite]

therfore doing the same thing with 15

[bitter meadow]

I did 7÷3=2.3
2.3×15=34.5

[unique granite]

thats because you rounded up to a single digit

if you put the fraction in its more exact

or did 2.333333 repeating

its a repeating decimal so the answer sways

[bitter meadow]

Yeah

I know what you mean

[dense sedge]

so do i do the same with this one?

[unique granite]

yea

[dense sedge]

so 16 divided by 14 answer divided by 42

[unique granite]

no

[dense sedge]

how?

[unique granite]

m/16=42/15

14*

solve for m

opposite of dividing is multiplication

[dense sedge]

42/14 = 3

3 times 16 = m

[unique granite]

sorry I worded that wrong

they equal eachother so your finding how 16 got to 14

[dense sedge]

so 16/14

[unique granite]

so 14/16

[dense sedge]

= 0.875

and what now?

[unique granite]

you were right 16/14

sorry

[dense sedge]

all good

[unique granite]

but then you multiply 42 by 16/14

[dense sedge]

so 42 times 1.14285714

[unique granite]

no

sorry im completley turned around

restart

[dense sedge]

k

[unique granite]

m/16=42/14

[dense sedge]

yes

[unique granite]

16 multiplied by something equals 14

[dense sedge]

k

[unique granite]

14/16= that something

so 16*14/16 equals 14

therefore

42*14/16 equals m

[dense sedge]

but the answer will be 34.875

[unique granite]

how many decimals did you type in

[dense sedge]

do u think the answer will be 48?

[unique granite]

yes

[cosmic charm]

m/42 = 16/14

16/14 is 8/7

42 is 7*6

[light helm]

boing your manipulation is incorrect

[cosmic charm]

the rest you smart fellas can figure it out

[dense sedge]

so we do divided by and than times?

so this one is 9 divided by 6 times 8?

[cosmic charm]

make the ratio first

then decide times or divide

[dense sedge]

idk what ratio

can u explain

[light helm]

as long as you don't mess up the order of operations, do multiplication in an order most convenient to you

[cosmic charm]

the sides of two similar shapes are in ratios

[dense sedge]

yes

[cosmic charm]

so in that shape, ||r/8 is equal to 9/6||

[light helm]

$\frac ?? = \frac ??$

[jolly parrot]

ℝαμΩℕωⅤ

[wheat saddle]

hi anybody Spanish?

[light helm]

note that there are multiple valid ways to set up the ratio

[cloud shard]

me

[dense sedge]

i got my answer thank alot

[wheat saddle]

Bro me ayudas con una operación que no entiendo Xd?

[cloud shard]

si, mándame un Dm

[pearl pond]

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[winged lance]

can someone explain how did it get to r=1 and r=5

[naive dirge]

Do the mathematical operation

It is not that hard as it seems

[winged lance]

cant

i tried

[naive dirge]

What

You failed Mickey Mouse

[winged lance]

i tried bro

[naive dirge]

You will pay the price

[winged lance]

ik its easy

i did my steps

and i find nothing wrong

look

wait sending pic

[winged lance]

@naive dirge come solve it

its 2r^2+7r-1 anyways

i mistakenly wrote wrongly above

but still cant solve it

[naive dirge]

What

You failed Mickey again

And you dare to text me

[winged lance]

😂

help me then

[naive dirge]

Okay

[winged lance]

hmm

i got it

.close

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[haughty hamlet]

i dont understand b

[pearl pond]

@haughty hamlet Has your question been resolved?

[naive hemlock]

See

@haughty hamlet

In 2nd one you have to solve the diff equat

Find y=f(x) then plot it for x from 0 to 1

And then tell why your approximated y is greater than the true y

Approximated y means by Euler method

Welcome np

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[haughty hamlet]

Nice name

.close

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[jolly parrot]

zzzz

[pearl pond]

@minor ferry Has your question been resolved?

[pearl pond]

@minor ferry Has your question been resolved?

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@minor ferry Has your question been resolved?

[floral rune]

Yess

Ight

Do differentiation in part a

Soo then

Ummm wait a sec

I have to solve that

[pearl pond]

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[short scaffold]

For this expression how do i know which integer values of x and y satisfy

[short scaffold]

y^4-x^4+3y^3+y^2+x^2y^2-y^2x^2-2y^3x-8y^2x-4yx-x^3+3x^2-2yx^3+y+x

$\ y^4-x^4+3y^3+y^2+x^2y^2-y^2x^2-2y^3x-8y^2x-4yx-x^3+3x^2-2yx^3+y+x=0$

[jolly parrot]

AbdulkareemSyria

[short scaffold]

is there a way to know which positive integer values for x and y satisfy

the above

i also know it is equivalent to

$\left(y+1\right)\left(y+2\right)\left(y-x\right)^{2}=\left(y+x\right)\left(y+x-1\right)\left(x+1\right)^{2}$

[jolly parrot]

AbdulkareemSyria

[short scaffold]

<@&286206848099549185>

[pearl pond]

@short scaffold Has your question been resolved?

[short scaffold]

<@&286206848099549185>

[tacit wolf]

I don't find and good solution but i think i will try using wolfram to find it if u want

[pearl pond]

@short scaffold Has your question been resolved?

[short scaffold]

that would be good ig

[pearl pond]

@short scaffold Has your question been resolved?

[pearl pond]

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[midnight haven]

can anyone explain Euler's method and it's improvement on solving first order differential equations?

[pearl pond]

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[mint hatch]

i have quite a few problems to solve but please help with this one first

[dusty pilot]

Use Pythagoras

Well nvm actually

[mint hatch]

im lowkey cooked please help my dads finna beat my ass

[dusty pilot]

Wtf

[rose parrot]

Stewart can help

[mint hatch]

whose that

[rose parrot]

Stewart's theorem

[mint hatch]

sorry i have no idea what that is is there any chance at all u could solve it for me

[dusty pilot]

I had never heard of this

[spare lark]

Never heard of this too

[dusty pilot]

Ig its not too hard to prove

U use like the cosine formula

For the 2 adjacent angles

[spare lark]

Very casual theorem ig

[mint hatch]

whats the answer tho

[naive hemlock]

Fr

Apply it

And get the answer

[pearl pond]

@mint hatch Has your question been resolved?

[mint hatch]

what about this one

[thick dagger]

use the cosine rule

or wtv that it is called

[spare lark]

The second one is just pythagore

[pearl pond]

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[pearl pond]

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[pale sundial]

Given a right triangle ABC, right-angled at A, with median AD. Draw DH//AC and DK//AB(H=AB, K=AC). Prove that quadrilateral ADHK is a rectangle

[pale sundial]

i dont know where to start

Given a right triangle ABC, right-angled at A, with median AD. Draw DH//AC and DK//AB(H=AB, K=AC). Prove that quadrilateral ADHK is a rectangle

.close

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[torn willow]

can someone help me with this quesiton pls

[torn willow]

,rotate

[torn willow]

the 2nd one

[cunning veldt]

so for the first vector you draw a vector thats 6 units horizontally away from the origin, and 0 units vertically away from the origin
for the second vector you draw a vector thats -2 units away from the previous vector horizontally, and -3 units from the previous vector vertically

you keep repeating that until you finish the last one

that you should give you a path of interconnected vectors

tldr; for each vector, draw it starting from the tip of the previous one

[torn willow]

but why is there an addition symbol

[cunning veldt]

because thats what vector addition represents

[torn willow]

so do we add?

[cunning veldt]

when you add 2 vectors youre drawing vector one, then drawing vector 2 where its origin lies at the tip of vector one

[cunning veldt]

in this case it wants the path so no
If it just wanted the final destination then you wouldve added everything in the top and everything in the bottom

[torn willow]

wait lemme try

wait the terminal point of the previous vector is the initial point of the vector

right?

,rotate

[torn willow]

That looks right to me

Oh ye I just checked the answer

And it’s correct

[cunning veldt]

yes

[torn willow]

Ty

[cunning veldt]

wow thats nice they made you draw a boat

[cunning veldt]

Youre welcome!

[torn willow]

Oh and one more thing

To represent vectors do you use angled brackets, big brackets or square brackets

[cunning veldt]

typically you express them like in the book here with normal brackets
But in a more advanced (university) math class (or maybe even earlier im not sure)
You might express them as <A , B , C> horizontally instead of vertically

but id advise you to just stick with your cirriculums representation which in this case is the normal brackets

[torn willow]

oh

ok

tysm

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[midnight haven]

what why how

[fickle schooner]

Recall the factor theorem

[midnight haven]

each one?

like

i want to solve the question without the multiple choice

can i do that?

um

[jolly parrot]

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

[fickle schooner]

https://en.m.wikipedia.org/wiki/Factor_theorem

First paragraph of that page

[midnight haven]

what

[fickle schooner]

Seems pretty clear but here we go

(x - a) is a factor of some function f(x) if and only f(a) = 0

[midnight haven]

f(-1)=0

[fickle schooner]

Yes

What would (x - a) be then

[midnight haven]

how should i find the function of f(x) tho

wait

so f(x)=n/x-a

[fickle schooner]

huh

[midnight haven]

so its b?

[fickle schooner]

Bo

*no

a is -1 like you said

What is (x - a) then?

[midnight haven]

oh -- = +

its c

[fickle schooner]

Mhm

Yh

[midnight haven]

but like

where did u get x-a from

is it a rule

[fickle schooner]

You can prove it

[midnight haven]

alright thanks

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[midnight haven]

what do they mean by a root of this function

[red garden]

The roots of a function are where f(x) = 0

so where it crosses x axis

[midnight haven]

oh

so b?

yup b

thanks

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[dense iris]

Why can I not move the half outside the integral here?

I wanted to rewrite it as 1/x with the half outside

however, I do not get it in the term lnk

but 1/2 lnk then

ohh

fuckkkkk

they want me to rewrite it wiht the power rule

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[little dawn]

is this right?

[pearl pond]

@little dawn Has your question been resolved?

[little dawn]

<@&286206848099549185>

breathes..

it’s noted

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[near flame]

How would i start solving this??

[thick dagger]

use the ratio
of "similar" triangles

cuz they are parralel u can conclude that CED is similar to BEA

from then use the ratio

[near flame]

Ohhh i kinda understand thank you

[thick dagger]

"kinda" 🙄 🤔 ?

[near flame]

Enough to try get an answer

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[storm thicket]

How would one approach this

[frank goblet]

maybe rewrite sin in terms of exponentials

or some substitution maybe

[storm thicket]

Tried it

Didnt work

[storm thicket]

Such as

[frank goblet]

dunno yet, im trying things out

[storm thicket]

Okie

Hm sin as an exponential might actually work

[frank goblet]

damn this is a difficult one

the answer can be checked, but the way there is usually just an approximation sadly

[frank goblet]

oh good idea splitting it

why 4i tho?

[storm thicket]

Yeah mistake saw it now

[frank goblet]

shouldn't it be 1/2i in the front

[storm thicket]

Basic addition is the worst part in calculus

[frank goblet]

yeah haha

i have an idea

the original integral will be real right

so when splitting the integral

can't we rewrite e^ix to cos?

or is that what you did

[storm thicket]

Ok this is correct

[frank goblet]

where did the 2 go?

[storm thicket]

When timing by 2i on denominator it cancels with the 2 of the -a to a property

And 1/i=-i