help-28

[minor crater]

notice that $\sqrt{a^2+b^2} \in \bR$

[glossy valve]

artemetra

[minor crater]

there's no imaginary part

right?

[hot lark]

Yes

[minor crater]

awesome

going back to this

that means that $a-bi \in \bR$ too

[glossy valve]

artemetra

[minor crater]

right?

[hot lark]

So b is 0

[minor crater]

yes

exactly

[hot lark]

Ok

[minor crater]

substitute b = 0 into this

[hot lark]

A=A

[minor crater]

yep

that means a can be anything

[hot lark]

Ok?

[minor crater]

so b=0, a any real number

[hot lark]

Oh

[minor crater]

so it must be a flat line

[hot lark]

Infinite on both sides?

[minor crater]

yes

[hot lark]

Are you 100% sure about this

[minor crater]

yes

[hot lark]

Ok thx

[minor crater]

100%

[minor crater]

np

oh

actually

[minor crater]

no

it's only the non-negative reals

sorry

$a \geq 0$

[glossy valve]

artemetra

[minor crater]

my bad

[hot lark]

?

[minor crater]

$\sqrt{(-2)^2} = 2$

[glossy valve]

artemetra

[minor crater]

≠-2

[hot lark]

Why

We are in C

[minor crater]

a is real

[hot lark]

Ok

I guess that

If we had like

Sqrt z = conj z

Or some sht like that

We would have had a line on both sides

Right

[minor crater]

eh

if we had $\pm |z|$ then yeah sure lol

[glossy valve]

artemetra

[hot lark]

No not that

Sqrt z

[minor crater]

that would be a completely different problem

[hot lark]

Nah look

[minor crater]

$\sqrt{z} = \overline{z}$

[glossy valve]

artemetra

[minor crater]

is that what you are talking about

[hot lark]

Yea

[minor crater]

yes, that would be a completely different problem

[hot lark]

Sure but look

2 values are = Conj z

[minor crater]

,rccw

[minor crater]

mm

yeah

[hot lark]

1 distance only tho

[minor crater]

wait

[hot lark]

Well

If I put

Sqrt conjz

[minor crater]

is conj z -theta

[hot lark]

Thats infinite distances

[hot lark]

Yes

[minor crater]

yes it is

got it

[minor crater]

well, feel free to solve for that

,w sqrt(z) = conj(z)

[glossy valve]

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Requested by arte_m_etra

[hot lark]

?

Oh yea

But thats because the modules are different

So we get 1 distance

Right?

Can u help me make it so that its not 2 results

But 2 lines

On opposite sides

@minor crater

[minor crater]

as in 2 axes?

[hot lark]

What?

,w sqrt(z) = z + i

[glossy valve]

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Requested by DeFra#2727

[hot lark]

,w sqrt(z) = z

[glossy valve]

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Requested by DeFra#2727

[hot lark]

,w sqrt(z) = conj(z)

[glossy valve]

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Requested by DeFra#2727

[hot lark]

,w |z| = sqrt(z)

[glossy valve]

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Requested by DeFra#2727

[hot lark]

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[tired oak]

What is $\sum_{n=0}^\infty\frac{n^k}{n!}$?

[glossy valve]

jsidind810

[tired oak]

$k\in\mathbb{N}$

[glossy valve]

jsidind810

[plush egret]

what is

[tired oak]

what does it equal

does it converge

[plush egret]

try ratio test

[tired oak]

but i need a value

im not interested in testing

[plush egret]

,w sum[n^k/n!, {n,0,Infinity}]

[glossy valve]

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Requested by janniku

[plush egret]

hmm

okay, sure, this makes sense i think

$\sum \frac{x^k}{k!} = e^x$

[glossy valve]

jan Niku

[plush egret]

so if x=1 ...

@tired oak

[tired oak]

what

[plush egret]

im giving you the answer

[tired oak]

then sum 1/k! = e

[plush egret]

yea

[tired oak]

thats not what i asked

[plush egret]

take k=0 in your sum

[tired oak]

but k needs to be able to vary

i need an expression for $\sum_n n^k/n!$ in terms of $k$

[glossy valve]

jsidind810

[plush egret]

this is pretty suggestive

[tired oak]

not really

hint mepls

[gritty rose]

The table's values are growing

Every term is at least e

[tired oak]

no shit but whats the coefficient

[gritty rose]

Chill your attitude

[tired oak]

n!-1 or what

no

[gritty rose]

[tired oak]

no to this

i meant

not you

whats the coefficient on the e

[gritty rose]

That's a different question from this

[tired oak]

how

,w sum[n^k/n!, {n,0,Infinity}]

[glossy valve]

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[tired oak]

ffs

[tired oak]

do you understand what i need

[gritty rose]

Oh yes I see your question now

[tired oak]

Clearly $\sum_{n=0}^\infty \frac{n^k}{n!}= e\cdot A(k)$

What is $A(k)$

[glossy valve]

jsidind810

jsidind810

[tired oak]

do you understand

what im asking

@gritty rose

[gritty rose]

Yes

[tired oak]

do you know the answer

[gritty rose]

Yes

[tired oak]

What is it please riemann

what is A(k)

[serene hazel]

!nosols

[full forum]

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.

[tired oak]

This is not homework

[serene hazel]

we don't give out answers

[tired oak]

This is not homework

Riemann

can you please tell me the answer

for gods sake

[glacial pasture]

[tired oak]

riemann

where did you go

[glacial pasture]

likely tired of your attitude

[tired oak]

What attitude did i have

Ive been trying to get help all day in the « help channels » and i havent got anything

Nobody helps

People here dont understand that sometimes the best help is just the answer

Then you reverse engineer it

[unkempt thorn]

its not

you have to learn the process

[tired oak]

Yes it is

They are equal

[unkempt thorn]

¯_(ツ)_/¯

you aren't going to get the answer

its agains server policy

[tired oak]

Well he hasnt even given me a hint

[tired oak]

Something to do with taylor series?

can you give a hint? something? that isnt just saying you know the answer?

[gritty rose]

Nope, I just know the answer and not any hints

[tired oak]

Why not? Why are you marked as Helpful?

That’s not very helpful

[gritty rose]

Blocked

[tired oak]

Lol

[gritty rose]

!volunteers

[full forum]

Helpers are just people volunteering their time to help you. Be polite and patient.

[tired oak]

help

You haven’t helped yet

[full forum]

@tired oak Has your question been resolved?

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[frank snow]

amc mock

@mystic spruce

<@&286206848099549185>

[astral belfry]

if you find the probabilty of eventually landing at 2, 3, 4 by hand, a pattern should emerge

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[desert sonnet]

how do i get from one to another

[pulsar stream]

i would say how can we get rid of this nasty 1+4e^(-x)?

what can you do with fractions ?

[desert sonnet]

i dont get either of these posts...

[pulsar stream]

do you know how to simplify this?

[desert sonnet]

yeah you change it to put the 5 on top and then its 5/4 right

[pulsar stream]

it's not 5/4.. 🙂

try multiply the top and bottom by 4 (you can do this with fractions)

[desert sonnet]

damn what...

oh its 4/5

[pulsar stream]

exactly

so the first image i sent, it looks kinda similar to the second image right?

[desert sonnet]

ok wait gimme a min ima try to do this on my lonesome now

[pulsar stream]

can you identify what you did in the second example and apply it to the first?

aight

[desert sonnet]

how would i do -e^(-x) * 7e^(-x)

is that just -7e^(-x)

nvm got it

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[torn jolt]

What number theory concepts would i use to solve this problem?

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[full forum]

Please don't occupy multiple help channels.

[wicked frost]

Looking at the solution here and I'm a bit confused. Where did the 1/x-3 go?

[devout valley]

Setting $F(x) = \int_0^x e^{t^2} \dd t$, you have $\lim_{x\to 3} \frac{F(x^2 - 9)}{x - 3}$

[glossy valve]

@devout valley

[devout valley]

For which they used l'opital's, working out derivative of the top and bottom

[wicked frost]

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[mental blade]

Can someone please explain how to solve this equation?

[hasty wing]

Do you know about substitution?

[mental blade]

yes

[hasty wing]

Try to eliminate A or B from one equation using the other via substituition

[mental blade]

Can you please give more help?

[hasty wing]

Solve for A or B using the first equation

what do you get?

[mental blade]

Ok thanks

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[boreal sinew]

Need help to solve this

[boreal sinew]

I do know that root cube I = - I and root cube -i = I

But I don't know the steps to get this value

So pls someone assist me to this

[muted flare]

wdym solve

You have an expression that's true

[boreal sinew]

Nah I meant how to prove this one

My bad

[muted flare]

$\sqrt[3]{-i} = -\sqrt[3]{i}$

[glossy valve]

jewels!

[muted flare]

Is it just not that

[boreal sinew]

Yeah

[muted flare]

uh so what's the problem

[boreal sinew]

I just need to find their individual value

Like what's value of root cube I

[muted flare]

write $i$ as $e^{i\pi/2}$ I suppose

[glossy valve]

jewels!

[muted flare]

And work in the polar form

[boreal sinew]

Wait let me show you

Something

@muted flare

Can you prove that expression by using this method?

I just don't know how to start

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[fading knoll]

The last part where it says loci satisfying 3+4i

isn’t it all of them ?

<@&286206848099549185>

Is anyone there

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@fading knoll Has your question been resolved?

[formal scarab]

have you tried plugging in 3+4i into each?

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[gloomy lance]

I have a question about the ways I can solve this. I know i can just use the squeeze theorem, but can i just plug in 0 in x², but not in 1/x, and since anything times 0 is 0, the limit would be 0. Is it allowed to plug in the value in just one place? Or is that wrong and it's just a coincidence that it worked for this case?

[scarlet herald]

You can plug in 0 here in both places because cos is a bounded function which goes between -1 and 1

A bounded function times 0 is 0

[scarlet herald]

It being a bounded function also gives you the ability to use the squeeze theorem

Well not really

But makes it easy to do so

[gloomy lance]

I know, what I want to know is if I can just plug in the 0 in one place and not the other to evaluate a limit.
For example, limit of x*1/x as x approaches 0
If I plug in 0 just in the first x, and use the argument that "anything times 0 is 0" then the limit would be 0. Is that correct?

[scarlet herald]

Its not

As you can see that the limit of x*1/x is 1

Because the x's cancel

[brave blaze]

you cant multiply 0 with something that isnt defined

[gloomy lance]

I see. Thank you

Nice pfp by the way lol

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[brave blaze]

oh a Bernkastel fan

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[thorny seal]

Find the sum of the exponents of the prime factor of the squad roots of the largest perfect square that divides 14

[thorny seal]

<@&286206848099549185>

[full forum]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[thorny seal]

Pls can som1 answer

[torn jolt]

!show

[full forum]

Show your work, and if possible, explain where you are stuck.

[thorny seal]

I don't know where to start

@torn jolt

If u ain't gonna answer why stop my 15 min countdown

<@&286206848099549185>

[torn jolt]

i was bg

theres legions formula u might need to look at

it’s related to exponents

[thorny seal]

@torn jolt I think found the answer is 9 pls check for me

[torn jolt]

show your work pls

i don’t understand neither the questions last part

no its not u just added 7+2

i think i found the answer

14 = 7*2

largest prime here is 7 so let 7^2 = 49

but 49 its not divisible by 14

so therefore no solution

[thorny seal]

Waiit

I said 14!

Soortty

14! I meant

Now check pls

[torn jolt]

wer are doing so wrong

its not 14!

where u got 14!?

[thorny seal]

@torn jolt

No I meant the question isn't 14 it is 14!

@torn jolt

[torn jolt]

ohhhh

ok now i can find

so its the same method

legions formula

u need to find prime factorial of 14!

[thorny seal]

Yah I did it

[torn jolt]

give me

[thorny seal]

It's 2^11×3^5×5^2×7^2×11×13

[torn jolt]

take the square root of it

[thorny seal]

But it won't give me a square

[torn jolt]

so what u think

[thorny seal]

We bring put the largest square

[torn jolt]

keep in mind that it should be smaller so u can do -1 from exponents that is odd

[thorny seal]

Which is 2^10×3^4×5^2×7^2

And root of this is simply 10080

Which has its prime factorization as 2^5×3^2×5×7

So we add 5 and 2 and 1 and 1

Which is9

Am i right ?

[torn jolt]

its the sum of exponents

[thorny seal]

Yah

[torn jolt]

wait

u should get 18

[thorny seal]

Why the said of the root

[torn jolt]

read this

[thorny seal]

It's the prime factorization of the root of that number

Not the number

[torn jolt]

yes but i took -1 from the odd exponents

wait

[thorny seal]

Yah but u r supposed to root all the exponents first

[torn jolt]

oh yes ur right

as question says

[thorny seal]

Yahh

Tnx

Lemme send u another 1

[torn jolt]

so 5+2+0*4 = 7

[thorny seal]

Nah

If u divide all the individual exponents by two then u add

Pls help me with number 22

[torn jolt]

is that system of equations

seems like

u can try to solve them

from given expression

its 12

E

[thorny seal]

Plss teach me @torn jolt

[torn jolt]

the clever methodis look at the 9n and 6n

does something ring a bell?

i have to go

sorry

[thorny seal]

<@&286206848099549185> can som1 explain this

[full forum]

@thorny seal Has your question been resolved?

[thorny seal]

<@&286206848099549185> pLS CAN someone explain that solving

[warm cairn]

I I was u

I would have individually checked for every option

It will hardly take a min

[full forum]

@thorny seal Has your question been resolved?

[thorny seal]

@warm cairn what if its is a theory question I need to know how to solve it

[full forum]

@thorny seal Has your question been resolved?

[thorny seal]

<@&286206848099549185> can anyone show me a free ebook to le1rn about everything on prime numbers

Bro I don't think it's free

[thorny seal]

<@&286206848099549185>

No 22

.close

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[thorny seal]

.reopen

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✅

[thorny seal]

<@&286206848099549185>

[fiery glacier]

Hi

What do you need help on

[thorny seal]

Pls no 22

Of that pic

[thorny seal]

No 22

[thorny seal]

Bro pls don't leave

[fiery glacier]

Ok

[thorny seal]

No 22 plss

[fiery glacier]

Factor

If it has 2 factors and it is prime

Then on of the two factors needs to be positive or negative one

[thorny seal]

Wdym I dont understand

Ohh

U mean factors like 1 and itself

Ping when u r back

[thorny seal]

Bro

[fiery glacier]

I was back lol

The first one is the product of n minus 5 and n minus 1

[thorny seal]

Ok

[fiery glacier]

So n minus 5 or n minus 1 is plus minus 1

[thorny seal]

Yah

OK go on

[fiery glacier]

O wait I factored wrong

[thorny seal]

Ohh ok

[fiery glacier]

2n plus 3 is prime

So 12 doesn't work

And -1

[thorny seal]

Why

[fiery glacier]

And then use eleimination

12 times 2 plus 3 is 27

Which isn't prime

[thorny seal]

What if it is theory question

What do I do

[thorny seal]

Bro

[fiery glacier]

Bc

Test the valies

[thorny seal]

Bro am saying what if I don't have options

Like multiple choices

What do I do

[fiery glacier]

not sure you can do it without options

bc none of them factor

[thorny seal]

I really don't understand what u mean by there r no factors

[thorny seal]

Amma ask someone else

.close

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[thorny seal]

.reopen

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✅

[thorny seal]

No 22

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[fiery glacier]

bro i think you are overcomplicating it

the best way is to eliminate the options with 12

then eliminate the options with -1

bc 2n+3 isn't prime for 12 and -1

meaning the only answer choice left is 7

and then you can verify it

[thorny seal]

I'd understand

I dont

Why r we doing all this

Bc a dude told me the answer was 12

[fiery glacier]

huh?

2n+3 is equal to 27 = 3^3 which isn't prime

[thorny seal]

Ok

[thorny seal]

<@&286206848099549185>

nine husband and wife couples wanted to take photos in 3 rows. row 1 has 4 people, row 2 has 6 people, row 3 has 8 people. they agreed that every pair of husband and wife must be in the same row and every two people next to each other must be husband and wife or the same sex. define many different possible positions

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[thorny seal]

.reopen

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✅

[thorny seal]

<@&286206848099549185>

nine husband and wife couples wanted to take photos in 3 rows. row 1 has 4 people, row 2 has 6 people, row 3 has 8 people. they agreed that every pair of husband and wife must be in the same row and every two people next to each other must be husband and wife or the same sex. define many different possible positions

<@&286206848099549185>

[gritty rose]

chill your pings

[thorny seal]

Bro pls can u answ

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[fiery copper]

Its been 4 hours, RIP

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[lament horizon]

@pseudo frigate no one else is on and this question is hella hard

[pseudo frigate]

1. What question
2. Please don't ping random users
3. 

[lament horizon]

3(5x-10) =50 solve for x

[gritty rose]

A few helpers are on. Be patient

[cloud meteor]

oh no

is this a joke-

[lament horizon]

no im just autistic and in 7th grade

[cloud meteor]

okay so

first you distribute the 3

what does it become?

[lament horizon]

to be honest idk

[deft tulip]

7th grade but undergraduate tag 🤔

[lament horizon]

why do i have that

wait holup

how the hell do i have undergrad

i dont even know what that us

is

[deft tulip]

so multiplying the number through the parenthesis...here is another example:
$7(2+5x) = 14 + 35x$

[glossy valve]

Soosh

[deft tulip]

that's similar to your problem, now how would you do yours?

[lament horizon]

oh so multiply both sides

so it becomes 30-15x

[cloud meteor]

15x-30

[lament horizon]

oh backwards

sry

[cloud meteor]

yeah

now add 30 on both sides

ADD

[lament horizon]

45x-60

[cloud meteor]

no

15x - 30 = 50
15x - 30 + 30 = 50 + 30
15x + 0 = 80

15x = 80

[lament horizon]

ok

thx

[cloud meteor]

np

[lament horizon]

so x= 5.3333

.close

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[cloud meteor]

NOO

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[reef forum]

am i doing this right? and what next

[bitter beacon]

parallel lines have equal slopes

[reef forum]

o

oh wait so m=1/6?

thank u

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[mortal berry]

Hello my question I don’t understand

Find -2u - 4v and |-2u - 4v| If u = 2i - 3j and v = <1,-2>

This is a new topic for me and my teacher sent videos that don’t pertain to this and I don’t understand what I need to do. This is an online class and just need help understanding what I need to do. This is also just a practice problem.

[mortal berry]

<@&286206848099549185>

So after some thinking I think I figured it out would this be right

$-2u = -4i+6j$ as $U$ is $2i-3j$

So you would do

$-2(2i-3j)$

To get $-4i+6j$

[glossy valve]

SlickCyph

[mortal berry]

The rest of the question
$-4v = -4i+8j$ as $V$ is $<1,-2>$
So you would do
$-4(1i-2j)$
To get $-4i+8j$

[glossy valve]

SlickCyph

[mortal berry]

Then you add the i and j

$(-4i-4i) + (6j+8j)$

Which you get

$-8i+14j$

Would this be correct?

[glossy valve]

SlickCyph

[mortal berry]

<@&286206848099549185>

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[tall meadow]

<@&286206848099549185>

[pastel nacelle]

a- x=1 and x= -4
b- for the x axis (1/2;0) for the Y axis (0;-1)
c- the domain of f (x ∈ R/x≠0) range of f : [-2;3 ]

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[tall meadow]

But I think the domain is [-3,3]
And Range [-2,3]
?

And how do i get f(-1)

[pastel nacelle]

I might be wrong about the domain as for f(-1) its 0 because the function intercepts with the y axis in (0;-1) which is (x;f(x)) therefore the value of f(-1) is 0

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@tall meadow Has your question been resolved?

[tall meadow]

. close

[gritty rose]

.close

[tall meadow]

#.close

[gritty rose]

.close

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[finite epoch]

i really need help with this graph

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[finite epoch]

i don’t know where to begin

[grim skiff]

That red line is the tangent at point A of the curve, and it wants you to find the gradient at point A so you can find the slope of that line

[finite epoch]

im proper failing maths and i really don’t understand how to do that

[grim skiff]

Do you know the slope formula?

[finite epoch]

nope💔

my teachers set us homework on this and we’ve never covered it haha

[grim skiff]

So you're saying, something like this doesn't look familar to you?

[finite epoch]

i’ve never seen it ever 😭

he has a thing for doing homework on things we’ve never done

[grim skiff]

https://www.youtube.com/watch?v=2p-NBIKpeXQ&ab_channel=Mathcation
Something like this may help

[finite epoch]

thank youuu i’ll check it out

[grim skiff]

The process for your question is similar, you need to identify two points along the red line, then you can apply that same process

[finite epoch]

i’ve been doing what the videos saying but it said incorrect, i was confused with rise and run but i know change in y and x

let me give it another go

nope

wait i put it in wrong oops my bad

one more try

I GOT IT

THANK YOU

[grim skiff]

Show what you are doing

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[ashen minnow]

In the factory, a new type of component was developed that is 60 kg lighter than the previous one. As a result, it is possible to manufacture 2 more components from 840 kg of metal than from 900 kg of metal before. Find the mass of the new type of component.

appreciate any pointers on this

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@ashen minnow Has your question been resolved?

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@ashen minnow Has your question been resolved?

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@ashen minnow Has your question been resolved?

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@ashen minnow Has your question been resolved?

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@ashen minnow Has your question been resolved?

[shy hamlet]

ok so if I understand the problem correctly, you have 2 types of component, new and old, and have to find the kilograms per unit of the new one.

First step guide:
What I did to help organize it is a table that compares the information from both. First you have the total amount of kilograms used (900 in old /840 in new) then you have the amount produced, lets say you produced n of the old one, how much did you produce of the new one? hint:||n+2||.

Second step guide:
Now, what you need to find is the kilograms per unit of component, lets call that "density", now you derive a formula for "density" that uses the data given above and relate the density of the new component and the density of the old component (hint:|| using the fact that old density is 60 more than new density||

After that you have enough information and equations to solve it. remember to check that your solution makes sense and solves the given problem

hope this helps! c:

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[torn jolt]

I don't know how to rotate

Anyway my problem is on first line of second page

I can't understand how it directly got there

Plz someone

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[torn jolt]

<@&286206848099549185>

[viral jasper]

,rotate

[viral jasper]

But it only does the most recent picture,so that doesn't help

[viral jasper]

We need more context. Where is the equation (i) it talks about?

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@torn jolt Has your question been resolved?

[torn jolt]

It's in page 1

@viral jasper

Thanks bro plz help me sorry for late reply

[torn jolt]

X²-y² = a it's in page 1

[viral jasper]

Oh I see. It was hiding

[torn jolt]

Yeah

[viral jasper]

Solve for y^2 in (i) and put it in (ii). Then solve for x^2

[torn jolt]

Ok I will test that wait

@viral jasper bro I got no where

[viral jasper]

Oh I see the pain point. Put solution for y^2 in (iii). I think your book meant to say (iii) instead of (ii)

[torn jolt]

Oh yeah

I should have noticed that

Thanks

@viral jasper

[viral jasper]

Np

[torn jolt]

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[tepid cairn]

help

[tepid cairn]

i have exam

and i need to study

for da exam

[iron cradle]

topic?

[tepid cairn]

and da question is hard

[tepid cairn]

thats the quetion in the book
simplifier A=√32+3√18-5√8

if someone knows how to do that please explain it to me

bruh

[iron cradle]

do you know some properties of square roots that could help you simplify this expression?

[tepid cairn]

√32+3√18-5√8
=√16+√16+3x√6-5√8

that doesnet makes any sense

[iron cradle]

not exactly, we can write √32 = √(2 x16) = √2 * √16 = 4√2 for example

[tepid cairn]

and the rest of it?

[iron cradle]

maybe try yourself first a method similar to how i simplified sqrt(32)

[tepid cairn]

ok a sec

[iron cradle]

cool

[tepid cairn]

it wont effect anything

they wil stay the same

like

√32=√16*√16

uh

why u did √(2*16)

[iron cradle]

this is not how square roots work

note that some number x = (√x)^2

[tepid cairn]

wait i will just picture the exersice

laptop's camera is

i cant do 4 √16?

[iron cradle]

no it doesnt work like that

[tepid cairn]

A=√32+3√18-5√8
=2√16+3√18-5√8

that doesent change the result

right?

[iron cradle]

but √32 is not 2√16

√32 = √(2 x 16) = √2 x √16 = 4√2

[tepid cairn]

oh hell nah

i got 1/20 in the controle exam

now in synthese am going to get 0

[tepid cairn]

i dont have to change the √32 to make the exersice

i can just count it right/

?

[iron cradle]

to simplify the expression, you will have to

ill solve this question, and then you can attempt it for another one

[tepid cairn]

ok

[tepid cairn]

u will give me a exercise who looks like it?

or i will find a exercise myself

[iron cradle]

A=√32+3√18-5√8 = √(16 x 2) + 3(√(9x2)) - 5(√(4 x 2)) = 4√2 + 3(3√2) -5(2√2) = 4√2 + 9√2 - 10√2 = 3√2

notice how i get everything in terms of √2

[tepid cairn]

i will write it

and try to apply it with another exercise who looks like it

[iron cradle]

cool good luck

[ruby magnet]

Please have we maths 55 a here ?

[iron cradle]

harvard maths 55?

[tepid cairn]

what i need to do if a question says

devlopper

alo

@iron cradle where u gone

[iron cradle]

dw im here

[tepid cairn]

in the next quetion they said

devlopper (1-√3)^2

[iron cradle]

you mean expand?

[tepid cairn]

our maths french

idk what is it in english

i dont what is it even in french

bc am arabic

[iron cradle]

ggs

[tepid cairn]

ggs wha

[iron cradle]

i would assume expand

do you know the binomial theorem?

[tepid cairn]

am going to translate it

english and french are not my language but we use french in our math idk why

only prep peuple have it with the arab language

idk what is a binomial theorem? but i translated it

wait

look

[iron cradle]

do you know how to expand (a-b)^2

[tepid cairn]

(1-√3)^2=1^2-√3^2=1-3=-2 is that right?

[tepid cairn]

yes the 2 be multiply for both

[iron cradle]

not exactly, it is that (a-b)^2 = (a-b)(a-b) = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2

[tepid cairn]

aaa

i remember a rule or idk what is it in english

(a*b)^2=a^2xaxbxb^2 or idk if it is like that

u can close the ticket or idk what is it

i need to go

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[last locust]

does the laplace transform seem correct?

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[vapid locust]

This is what I have so far but idk what the last step should be

realized .heic prob wont render so this is the work lol

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[fluid zodiac]

Could someone help me with this?

[torn jolt]

alright

first thing i would do

which is probably slow

is get an idea of what amount of musicians i could possibly have for each situation

if you have a graphing calculator u can just set up some equations

so try to set up an equation for the 1st part

[split swift]

you can also use the chinese remainder theorem

[fluid zodiac]

Thanks

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[dawn echo]

how do u solve cot(-pi + pi/4)

[quaint prawn]

Do you know the unit circle

[dawn echo]

yes but if its on an axis how do u solve it

[quaint prawn]

!original

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

[quaint prawn]

What is -pi + 3pi/4

In terms of pi

[dawn echo]

-pi/4

[quaint prawn]

Yes

And what is cotangent in terms of sine and cosine

[dawn echo]

cos/sin?

[quaint prawn]

Okay

Do you know cosine and sine of -pi/4?

Using the unit circle

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[dawn echo]

-1/root2

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[wintry mantle]

I want to evaluate a limit, similar to that of the definition of e. I know the result by plugging it in wolfram alpha, however I don't understand the steps of how to get to the result. I looked up Taylor series, and Maclaurin series, but I'm not familiar with these and I did not understand how to do the series expansion.

• Why is the result e^{a/b}, how can I derive it?
• Why does the constant c in the denominator vanish?

[patent sierra]

its about taking log both sides

The power 1/x comes down so

We can do the taylor expansion of log(1+x) as x and ignore the rest cause x tends to zero so the number just get more smaller

do that and youll see how

[glacial pasture]

saw this a while ago, could use it here

[patent sierra]

yeah this one's like a formula for it

[wintry mantle]

okay, thanks, I will try to do the log transform!

how can I get the first term of the Taylor series at x=0? When I try to plug in x=0, and do f(0), I get a division by zero -- I think I'm misunderstanding

[patent sierra]

(1/x ) • log(1+f(x)) = f(x)/x

in the first example the f(x) is x so the numerator and denominator x cancels

[main ginkgo]

you can do this without taylor expansion though

[patent sierra]

your supposed to cancel them

[main ginkgo]

@wintry mantle you could solve this with l'hopital instead of taylor series

[wintry mantle]

What would be the simpler approach to solve my second equation? 🙂

[main ginkgo]

$\ln\left(y\right)=lim_{\left(x\to0\right)}\frac{\ln\left(1+\frac{xa}{b-xc}\right)}{x}=\frac{0}{0}$

[glossy valve]

Combustion

[wintry mantle]

so I would take the limit above and below

and then differentiate both of those?

[main ginkgo]

[wintry mantle]

alright I think I get the correct answer now, L'Hopital makes it much simpler. I never took any classes for this, so it's new territory for me. Thanks for the help

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[elder cape]

Multivariable calc help needed

[red blade]

okay

bestie

pls ask

[elder cape]

gradient P(s,t) = (6000e^t - 4000s, 6000se^t)

"find all points at which the gradient is perpendicular to the boundary" simply means this:

[red blade]

yes ofc

okay so what is the gradient of g

and also the gradient of P

no like calculation wise

[elder cape]

partial derivaitve of s and t

[red blade]

okay

[red blade]

grad P(s,t) = (6000e^t - 4000s, 6000se^t)

so that's given as this

[elder cape]

that i'm not too sure...

idk what g is 😭

This was the orginal P:

from this we get the gradient of P to be:

grad P(s,t) = (6000e^t - 4000s, 6000se^t)

[red blade]

aight bro

[red blade]

imma help you out here

so that gradient looks good

now set $f_s = 0$ and $f_t = 0$

[glossy valve]

nosqldb

[red blade]

for the inside of the boundary g

[elder cape]

wdym?

Why r we doing that?

[red blade]

it's not just the boundary we need to worry about

[elder cape]

yeah so I have to find the gradient of this right?

[red blade]

or wait nvm we don't acc have to do that

bc it's just asking for "find all points at which the gradient is perpendicular to the boundary"

[red blade]

yeah

so that would be like

<2s, e^t> right

[elder cape]

what has be confused tho is

its an inequality...

[red blade]

g(s,t) = 11/3

[red blade]

but we only care about the boundary

not the inside

[elder cape]

oh ok

so yeah the gradient is just

(2s, e^t) after taking our partial derivatives yeah

[red blade]

okay

then just solve

$(6000e^t - 4000s, 6000se^t) = \lambda (2s, e^t)$

and then lmk

[glossy valve]

nosqldb

[red blade]

aight pookie bear

it's up to you now

[elder cape]

ok I will try to solve this and get back to ya

[red blade]

https://tenor.com/view/its-the-way-you-act-prayag-gif-5561611490017780398

@elder cape let me know

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[elder cape]

don't we also have a third equation?

[elder cape]

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[visual cape]

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[torn jolt]

Why a function needs to be bijective in order for it's inverse to be defined

[torn jolt]

<@&286206848099549185>

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[azure grove]

assume otherwise

say its not bijective

which means its either not one one or not onto

[torn jolt]

yes

Or many one and into

[azure grove]

if its not one one, then f(a) = f(b) = c for some a ≠ b
then if g is the inverse function, you'll have g(c) = a = b
which is not a function

[torn jolt]

okay

And why it needs to be onto

[azure grove]

and if its not onto, then there must exist some b in the range which is not the image of any element in the domain

which means g(b) does not exist

so g can't be completely defined

[torn jolt]

Wdym

Can you express this mathematically

[azure grove]

i did...

[azure grove]

this aint no biology 😂

[torn jolt]

Okau

Okay

Explain with help of f(x) = sinx

[azure grove]

say the domain is [-pi/2, pi/2] and range is [-2, 2]

then this is not a bijective function, do you see why?

[torn jolt]

range can't be [-2,2]

[torn jolt]

it's bijective if [-pi/2,pi/2] > [-1,1]

[azure grove]

yeah, but we need a function that's not onto

so i just included the extra part

[torn jolt]

Okay

[azure grove]

but now sin inverse is not defined on [-2, 2]

so function has to be onto

[torn jolt]

Yeah

Right

Thanks

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[azure grove]

Welcome 🙂

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[pulsar sand]

Any Idea on how to calculate the summation of nth fibonacci number over 5^n?

[elfin stream]

'over 5^n'

[pulsar sand]

I tried using Binet's Formula but it seemed I got stucked and it looks like a no good path, is there a better way?

[elfin stream]

Oh you meant nth Fibonacci number divided by 5^n

[pulsar sand]

Yah

[fast peak]

hmm binet sounds somewhat promising tho. where did you get stuck?

[pulsar sand]

Idk what to do with that 5^n part

[fast peak]

combine it with the other something^n

[pulsar sand]

Hmmmm

This seems really hard

[elfin stream]

I would recommend showing your work

[brittle steeple]

I would suggest reviewing exponent laws and seeing if any of them apply to this situation

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[unreal hull]

wrong solution

[hollow turtle]

Draw a diagram, and figure out the area that you are trying to find

Your answer should be a number, not an expression

[uncut nebula]

yea drawing the area is the easiest way to see

[glacial pasture]

have to find your intersections

[unreal hull]

oh forgot to set them together

x^2 = 4 is -2, 2

and then i do definite integrals from there

gotcha

[glacial pasture]

mhm

well

what integrals

[unreal hull]

huh

[glacial pasture]

ig it works just to do one here, thats not always the case though

[wise wyvern]

It should be 4x - x^3/3

[analog tundra]

How

Line cuts parabola

In two points

[glacial pasture]

since (-2)^2=(2)^2

[analog tundra]

Ok wait communication skill issue my end

[hollow turtle]

Honestly @unreal hull , this is so much easier to think about if you draw a diagram

[hollow turtle]

You have made a sign error in your working out, and your integral is also the wrong way around

[hot herald]

its alright if you take absval at the end
apart from chaining equations that i dislike/don't recommend, you had another sign screw up

[unreal hull]

Oh so it should be -2 on top?

[hollow turtle]

The limits are the right way around, but the integrand should be $4-x^2$

[glossy valve]

HazelOak

[unreal hull]

Oh

[unreal hull]

Why?

How do I know for the future which order they should go in

[hollow turtle]

Because 4 is at the top of the region you are trying to find, and x^2 is at the bottom

If you did it the other way around, you'd get the same answer with a negative sign in front

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[fast marlin]

Can I conclude anything of a and b

All I know is that they’re less than 1 right?

[queen crater]

Apart from a^2+b^2 = 1?

[fast marlin]

Yeh

[queen crater]

Not much else to conclude without context

a+b > 1 I guess

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[torn jolt]

Why is

[torn jolt]

A^0= 1

Definition

[umbral dome]

A³ = 1×A×A×A (3 times)
A² = 1×A×A (2 times)
A¹ = 1×A (1 time)
A⁰ = 1 (0 times)

[torn jolt]

Dang

In general

[torn jolt]

Makes sense fr

[torn jolt]

Why is 1A = A and also in 1/1=1 <— like here, one in deno is ignored, why

[torn jolt]

even my dad understood this explenation

[umbral dome]

if you multiply or divide anything by 1 it must stay the same.

2×3 = 2+2+2 (3 times)
2×2 = 2+2 (2 times)
2×1 = 2 (1 time)

division is the inverse of multiplication so a similar logic applies

[torn jolt]

if anything i would have asked why 0A=0

but thats just the definition of 0 element in a field

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