help-36

[tiny nova]

(w+√2)(w-√2)

[spare folio]

okay

you have still w

xD

[tiny nova]

So we can essentially take the I out of the root

Well it equals i

So set each factor equal to it

[spare folio]

how

[tiny nova]

Wait

Oh no

This question is making me annoyed

Hehehehehehe

[spare folio]

...

[tiny nova]

I made a silly mistake

Sorry

[spare folio]

no problem

[tiny nova]

Hmmmmm

Can you factor w^2-2-i

Let me see

Nope

Can't factor

Oof

[spare folio]

...

[tiny nova]

Did the synthetic division get anywhere

[spare folio]

what is this actually lol

im not used to hear that maybe

[tiny nova]

It's like a different way to divide polynomials

You take the coefficients

And set them in a certain way

It's hard to explain without drawing it

[spare folio]

okay

[tiny nova]

It's essentially the opposite of polynomial long division

But it still divides

Instead of subtracting you add

[spare folio]

maybe i dont understand it clearly right now but looks like

uh

you know

oh wait

wait wait wait

ahaha

okay go on

[tiny nova]

With synthetic division?

[spare folio]

but i think i found smthng u know

[spare folio]

yes

no

[tiny nova]

Well you take the coefficients

And the coefficients of the other polynomial

[spare folio]

i mean this doesn't work for us i think

[tiny nova]

Well yeah

[spare folio]

let's write

2+i

[tiny nova]

I think one polynomial has to be monomial

[spare folio]

just 2+i

[tiny nova]

What about the root

[spare folio]

what is the height

of 2+i

1 right?

in the complex

plane

[tiny nova]

Well how would you combine unreal and real numbers

I guess we could believe you

[spare folio]

sorry

imaginary line

and real line

this is the angle

between us

[tiny nova]

Ohhhh

Okay

[spare folio]

e^itheta thing

you know

[tiny nova]

I though you meant only in one direction

Yeah

[spare folio]

when it is π rad

[tiny nova]

Have seen that before

[spare folio]

it is just -1

[tiny nova]

Okay

[spare folio]

because it is π rad just left side

and with length of 1

of course

|-1| = 1

[tiny nova]

Yeah

[spare folio]

so let's just try for 2+i

[tiny nova]

Okay

[spare folio]

height is 1

horizontal length is 2

tan(theta) = 1/2

[tiny nova]

Yeah

[spare folio]

theta = arctan(1/2)

[tiny nova]

Yeah

Well now we can try QSFR method

[spare folio]

26.56°

[tiny nova]

Okay

That works

[spare folio]

and square of it

[tiny nova]

Wait

Why do we square it

[spare folio]

root

sorry

sqrt of it

[tiny nova]

Yeah

[spare folio]

√(2+i(

)

[tiny nova]

Okay

5.153

I think

[spare folio]

but it has length

also

[tiny nova]

Okay

[spare folio]

wait what is e^i26.56

write in radian

[tiny nova]

Can I use a calculator

[spare folio]

must do this

[tiny nova]

Heehe

[spare folio]

okay it's not

necessary

just dont know about it

but know about it

ahaha

[tiny nova]

Okay

[spare folio]

I'm talking about the representation of number

you dont know what is e^i17427

but

you know

ahaha

okay so

cos(theta)×r must be 2

[tiny nova]

Well if you are referring to it being x+iy

[spare folio]

yes

[tiny nova]

Okay

[spare folio]

cos value is

0.89

approx

[tiny nova]

Wait

Don't you do sin

[spare folio]

it doesnt mater

[tiny nova]

Huh

What do you mean

[spare folio]

it will come after this

just one is enough

[tiny nova]

Cosine does adjacent over hypo

But the adjacent is already 2

[spare folio]

okay i mean cos(arctan(1/2)) ≈ 0.89

and this must be

[tiny nova]

Okay

[spare folio]

2

with length

[tiny nova]

Okay

[spare folio]

of the original number

[tiny nova]

Well that is the horizontal value

[spare folio]

yed

yes

so just

2/length = 0.89

[tiny nova]

Okay

[spare folio]

length/2 = 1/cos

length = 2/cos

[tiny nova]

Okay

[spare folio]

≈ 2.23

[tiny nova]

Okay

[spare folio]

when you multiply 2.23 by cos(arctan(1/2))

you will have 2

[tiny nova]

So

[spare folio]

and

when you multiply by 2.23 by sin(arctan1/2)

you will have 1

it must be like that

let's check if you want

[tiny nova]

Okay

But did we even get the answer

Okay

[spare folio]

so

we know that

the number is

2.23×e^i26.56

[tiny nova]

Okay

[spare folio]

but it's just 2+i

[tiny nova]

Squareroot that

[spare folio]

its square root will be

√(2+i)

[tiny nova]

Yeah

[spare folio]

but let's write 26.56 in radians first

because it wont work like that

[tiny nova]

Yeah

[spare folio]

xD

≈ 0.463

[tiny nova]

Okay

[spare folio]

so take half

[tiny nova]

Now we take the root

Wait what

[spare folio]

lets say 232

root is 1/2

XD

[tiny nova]

Okay

[spare folio]

so we have

[tiny nova]

Ohhhhh

Okay

[spare folio]

bottom part is square root part

[tiny nova]

You applied it to the exponent

[spare folio]

yes

[tiny nova]

That's why it is half

[spare folio]

yes

[tiny nova]

That makes more sense

Okay

[spare folio]

so we have x+iy form now

[tiny nova]

But what about the 2.23

[spare folio]

but when multiply this

by i

we will have ix-y

and we will just plus i then

and we will have (x+1)i-y

you know

[tiny nova]

Yeah

[spare folio]

okay just learn about

what is x and y

and find it

[tiny nova]

Hmmmmm

[spare folio]

it is a factor which has a share in x and y values

[tiny nova]

Let me think

[spare folio]

it has an important mission actually xD

or just

write it as

cos and sin

and let us be more mathematical

we did so many

approx

ahaha

[tiny nova]

Hehehe

[spare folio]

this is not natural

xD

[tiny nova]

How would we represent the em^i*0.232 as sin and cos

[spare folio]

like this

e^ix = cosx+isinx

r×e^ix = rcosx+risinx

[tiny nova]

Oh

Yeah

[spare folio]

and

[tiny nova]

You said that before

[spare folio]

when multiply by i

it is ircosx-rsinx

[tiny nova]

We get -1

[spare folio]

so -rsinx is real part

of it

[spare folio]

it is for x = π

[tiny nova]

Yeah

[spare folio]

because cosx = cosπ = -1

sinx = sinπ = 0

[tiny nova]

Okay

[spare folio]

cosx+isinx = -1+i0

= -1

it is a special

[tiny nova]

Okay

[spare folio]

situation for it

[tiny nova]

Yeeee

[spare folio]

we have 0.232 rn

not π

[tiny nova]

Oh

[spare folio]

not 3.1415

[tiny nova]

Oof

Well now would we calculate sin(0.232)

And cos

[spare folio]

yes

but we have one more step

adding i

because it was

i+i√(2+i)

we found what √(2+i) is

[tiny nova]

Okay

[spare folio]

we found what i√(2+i) is

[tiny nova]

Yeah

[spare folio]

but not i+i√(2+i) yet

so add one i

to it

[tiny nova]

Okay

[spare folio]

this will add to the imaginary part btw

so +1 for the imaginary part

increase the coefficient of i by 1

so this is the number

[tiny nova]

I+I(2.23 * cos(0.232) + 2.23 * sin(0.232) +1

Hmmmmm

[spare folio]

this is wrong actually

[tiny nova]

The one I wrote?

[spare folio]

we add i to the expression with cos and sin

[tiny nova]

Okay

[spare folio]

yes wrong representation i think

[tiny nova]

I+I(2.23 * cos(0.232) + 2.23 * sin(0.232)+1)

That?

[spare folio]

no

there is just i for cos

not cos and sin

[tiny nova]

But isn't it I * √2+i

The whole thing

[spare folio]

i+i(2.23cos(0.232))-2.23sin(0.232)

and sinus's sign is negative

[tiny nova]

Okay

[spare folio]

because we multiplied it by i

before

or wait

okay i understand

what you did

um

yes it is true but there is something missing

[tiny nova]

Is it the I in front of cos

[spare folio]

sin must have a factor of i in front of it

[tiny nova]

Oh I was close

I+I(2.23 * cos(0.232) - 2.23 * isin(0.232)+1)

[spare folio]

Now i will write without the numbers 0.232 and 2.23

we have something like r*e^ix

as √(2+i)

so it's equal to this: r*(cosx+i*sinx)

which is rcosx+risinx

[tiny nova]

Yeah

[spare folio]

and we will multiply it by i

it will be

ircosx-(minus)rsinx

[tiny nova]

Yeah

[spare folio]

minus is important because there is

[tiny nova]

I

[spare folio]

i²

you know

ircosx-rsinx

and +i

[tiny nova]

The i^2 makes it -1

Yeah

[spare folio]

yeah

i+ircosx-rsinx

just plug in

r = 2.23

x = 0.232

and wait

there is one more

to look beautiful xD

i+ircosx = (rcosx+1)i

so write it as (rcosx+1)i-rsinx

plug in now

and rcosx+1 is your y

-rsinx is your x

from now on

[tiny nova]

Oh yeah

[spare folio]

like x+iy

[tiny nova]

Because if is x+iy

Yeah

[spare folio]

yes

i will check the result

[tiny nova]

This is slowly coming together

[spare folio]

like i+i√(2+i) versus

this one

xD

they're not the same because we did some approximations

[tiny nova]

0.512

I Think

[spare folio]

let's check what is -rsinx

so let's check what is -2.23rsin(0.232)

[tiny nova]

Yeah

[spare folio]

and let's see if they are close

with

-0.34...

nope they are not

ahahaha

[tiny nova]

I got 0.512 for it

Hehehe

[spare folio]

how

[tiny nova]

Calculatoring

Hehehehehe

[spare folio]

bro..

it was all in vain

[tiny nova]

Hehehehehe

[spare folio]

ahahah

okay not actually

[tiny nova]

Oof

[spare folio]

but

we probably made a mistake somewhere because it looks pretty accurate

[tiny nova]

Well we got in x+iy form

[spare folio]

But i feel a little more comfortable rn

ahaha

[tiny nova]

Yeah

[spare folio]

i have a way at least

you know

you can try this too

i think i need to go now

and

see you later

[tiny nova]

Okay

I think I need to go too

[spare folio]

good

[final saddle]

@crude axle Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[eager breach]

I have to solve this riddle:
You let yourself be lulled by the magic of the place and stumble against an unexpected object.
Furious, you get up and inspect the object of the crime: an alarm clock.
You love the ticking sound that accompanies the regular movement of the hands and you study it deeply.
Unbelievable, it is poorly built!!!
If the "hour" hand goes around in 12 "hours", the turn takes 56 "minutes" for the "minute" hand and 55 real seconds for the "seconds" hand.
If you think it is better to repair it, find a good address, report this alarm clock and answer 0
If you think it's fun to know how many real seconds it takes for the three hands to be aligned again (after the alarm clock reads "midnight"), answer by giving the number of real seconds.

[eager breach]

I did this :

But i think it's false because i didn't learn how to solve it

<@&286206848099549185>

[tranquil pine]

can you send the question

[eager breach]

How can i translate this riddle by an equation or by anything else so i can solve it ?

<@&286206848099549185>

[tranquil pine]

let me see

do you need th4

answer or the eqaution

?

[eager breach]

well i prefer to have the equation so i can try to solve it

[tranquil pine]

so

I am not sure if I am right but

the riddle says that it takes 56 minutes for the minute hand to make a full circle and the seconds hand take 55 seconds

right?

[eager breach]

yes

[tranquil pine]

so it is asking after how many hours will the three hands be aligned agains?

[eager breach]

exactly

no how many seconds sorry

[tranquil pine]

oh ok

then how many minutes in a hour

[eager breach]

60

[tranquil pine]

in reality 60, according to the clock 56

[eager breach]

yes

[tranquil pine]

how many seconds in a minute

60 in reality, 55 according to the clock

how many hours for the hour hand to make a circle

[eager breach]

12

[tranquil pine]

exactly

so the total seconds will be

12 x 56 x 55 seconds

[eager breach]

ahhh okay that'why i am wrong bc i did 126060

[tranquil pine]

that is in normal clocks

as the clock is broken

we will have to use the time given to us

[eager breach]

okayyy

so i just need to change that in the equation ?

[tranquil pine]

yes

[eager breach]

alr ty

Hag1

.close

[final saddle]

Channel closed

Closed by @eager breach

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

Please don't occupy multiple help channels.

@glad phoenix Has your question been resolved?

[woven sandal]

Please don't occupy multiple help channels

[final saddle]

@glad phoenix Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[devout hazel]

If im rolling 5 dice- how many ways can i get a total of 11?

[devout hazel]

I know im suuposed to use generating functions

[cold gorge]

Have you tried anything?

[devout hazel]

but i dont know how to find the coefficient

i know i need to find the coefficient of x^11

or

i can just find the coeff of x^6 after factoring out the x^5

<@&286206848099549185>

[mint relic]

ok

can u make me understand it in a good way pls

[mint relic]

.

[devout hazel]

i need to count the ways to find a total of 11 from 5 die

i know the approach should be genrating functions

by finding the coefficient of x^11 in (x+x^2+x^3+x^4+x^5+x^6)^5

but i dont know how to go further

[mint relic]

ok

[tiny nova]

What about permutations

[devout hazel]

my prof said to use generating functions

[tiny nova]

Oh okay

I don't know what those are

[devout hazel]

ok thanks

should i ping helpers again?

idk

[mint relic]

wait just a moment

[devout hazel]

i really need some help

ok

[mint relic]

no

You're on the right track! Let's break down how to find the coefficient of x^11 in the expansion of (x + x^2 + x^3 + x^4 + x^5 + x^6)^5 and relate it to the number of ways to get a total of 11 on 5 dice.

Generating Function: Each term (x, x^2, ..., x^6) in the expression represents rolling a specific value on a single die. Raising the entire expression to the power of 5 signifies rolling 5 dice (each term contributes its value to the total).

Coefficient of x^11: We need the coefficient of x^11 because each x in the expansion represents the value rolled on a die. To get a sum of 11, we need 11 x's.

Combinations: The coefficient tells us how many ways we can arrange the x's (dice rolls) to get a specific sum (11).

Here's how to find the coefficient:

Each term in the expansion (x + x^2 + ... + x^6) contributes at most one x to the power of 11.
So, we need to find combinations of these terms where the sum of their powers equals 11.
For example, (x^5 * x^3 * x^3) or (x^6 * x^2 * x^2 * x) are valid combinations as they all contribute x^11.
However, directly listing all possible combinations can be tedious. Here's a shortcut:

Use the Binomial Theorem to expand (x + x^2 + ... + x^6)^5.
This will give you a sum of terms where each term is a product of 5 factors chosen from (x, x^2, ..., x^6).
The coefficient of x^11 will include all the terms where the product of the powers of x in those 5 factors equals 11.
Finding all such combinations is mathematically challenging. There are more advanced techniques to solve this, but for this specific case, you can search online for pre-calculated coefficients of multinomial expansions.

Knowing the coefficient of x^11 will tell you exactly how many ways you can roll 5 dice to get a sum of 11.

[devout hazel]

bruh

i can use chat gpt too

i need help undersstanding it cuz i have to explain my proof

and it game me 2 answers- 210 and 250 lol

but thatas not the point- i dont understand it

[mint relic]

i am not using gpt

[devout hazel]

ok - i know why im using generating functions - i dont know how to find the coefficient

<@&286206848099549185>

[final saddle]

@devout hazel Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[trail dove]

Is there a quicker way to do this other than just testing each zero individually, or is that the only option?

[craggy river]

trial and error

get one root then divide it with the whoe term you will get the other two roots

[trail dove]

So thats a no then xP
alright, just wanted to be sure.

.close

[final saddle]

Channel closed

Closed by @trail dove

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tranquil pine]

Just looking at taking these rational functions and identifying their graphs and for the first time I'm seeing 0's identified by the numerator? I thought you could only identify the denominator to have 0's to solve for the vertical asymptote...

[final saddle]

@tranquil pine Has your question been resolved?

[zenith pollen]

the numerator being 0 gives you x-intercepts (y=0), which helps drawing too

[tranquil pine]

@zenith pollenwow that seems really obvious now, noted and thank you so much. 🙏

.close

[final saddle]

Channel closed

Closed by @unkempt dune

Use .reopen if this was a mistake.

[glacial torrent]

Sorry fkr the late response but what do you do with 3x

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[karmic kindle]

So it’s 180 degrees= 3x

[karmic kindle]

Divide by three

So x=60 degrees

Or pi/3

[vital furnace]

?

[final saddle]

@karmic kindle Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[subtle knot]

Ive been stuck on this problem for a while now pls help

[marble agate]

this is just geometric sequence

a_1=2187 and a_4=1536

and you need to find a_10

[subtle knot]

huh

whats an a

[marble agate]

a sequence 'a_n'

geometric in this instance

[subtle knot]

thanks

.close

[final saddle]

Channel closed

Closed by @subtle knot

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[teal robin]

Would I be able to receive help on this problem?

[tranquil pine]

sure

where are you stuck

[teal robin]

Well I am thinking that I have to put the probabilities for each one into an equation

But not for sure

Like one equation for one-litre cap and one equation for 2-litre cap

[tranquil pine]

do that

[final saddle]

@teal robin Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tranquil pine]

What just happened here

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[wintry cedar]

help please

[wintry cedar]

I need assistance

@loud sundial save me please

[final saddle]

Please do not ping individual helpers unprompted.

[wintry cedar]

I think i've got #1-2, #6 and #8

#1 should be (-3.pi)

#2 should be (-r,theta)

#6 should be (2,pi/3)

#8 should be (r,theta)

I'm lost on everything else

[final saddle]

@wintry cedar Has your question been resolved?

[wintry cedar]

.close

[final saddle]

Channel closed

Closed by @wintry cedar

Use .reopen if this was a mistake.

[wintry cedar]

.reopen

[final saddle]

✅

[wintry cedar]

please dawf

need help

[gilded anchor]

I think it’s safe to assume that the two with 3s match

[wintry cedar]

how would you do #3?

[gilded anchor]

With the R, theta

My best guess so far

Although I could be completely wrong

[wintry cedar]

thank you.

.close

[final saddle]

Channel closed

Closed by @wintry cedar

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[limber isle]

this is sorta a guess

i think its right

since

1/2^4 is 2

but

cojecture

barely educated guess

[static fractal]

we have an even root of 2x, so we require 2x>=0, so x>=0

[limber isle]

even root..?

[static fractal]

as in 4th root

for even number roots you need a non-negative input if we're working with real numbers

then note that the root is also non-negative

but we're then multiplying by -4, so the left side is non-positive whenever the root is defined, while the right side is positive

and the root is defined whenever x>=0

[limber isle]

ohh

wait so

why is x>0 if 2x>0

[static fractal]

divide both sides by 2

[limber isle]

and you mean greater than or equal to right

[static fractal]

yes

[limber isle]

2x/2 is c

[static fractal]

that's why I wrote 2x>=2

[limber isle]

x

0/2 is like

[static fractal]

0

[limber isle]

0

yeah

oh

ok that was less hard than i thought

thank you!!

.close

[final saddle]

Channel closed

Closed by @limber isle

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[scarlet bridge]

Can someone check my work?

[scarlet bridge]

Please and thank u

[novel rover]

2nd page, how did you get $\int \pi sinx dx = \pi$?

[soft zealot]

prime

[scarlet bridge]

Dont u take out the constant, so that it will be equal to pi times the integral of sin of X?

Cause pi isn't dependent on X right?

so thats why we take it out?

[novel rover]

yeah

but you just left it as pi 😭

[scarlet bridge]

it isnt dependent on X so it will stay as pi right?

if not, what should it be?

Lets say Pi of x sin of X DX then u can take out the constant, but Pi isnt dependent on X so it will stay as Pi. So if we take it out, that will be equal to Pi times the integral of Sin of X

[final saddle]

@scarlet bridge Has your question been resolved?

[final saddle]

@scarlet bridge Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[waxen jay]

I wanted to solve this vectors but i keep getting it wrong

[waxen jay]

I have two vectors <0,575*2> and the other vector is given in an angle and magnitude, i was thinking to add them and get the magnitude

when I convert both vectors to component form I get $<0,1150>$ and $<575\cos(15),575\cos(15)>$

[soft zealot]

guy

[waxen jay]

this is my work, but the answer should be 1712

and i dont understand why

<@&286206848099549185>

[stone ermine]

I think you added the "1150" to the wrong one

[waxen jay]

how so?

[stone ermine]

displacemtn in x: 1150 + 575 cos(15)

displacement in y: 575cos(15)

[waxen jay]

wait lemme just make sure that i understand

do you agree these are the two vectors we're working with?

[stone ermine]

no

[waxen jay]

ookay im not sure then lol

what are the two vectors we are working with

[stone ermine]

I have a different axis: I called the upword direction x. but you called it y

if I change it to your cords

I have <0,1150> and <575sin(15) , 575cos(15)>

[waxen jay]

hm, how did you get sine in your x position of the vector?

[stone ermine]

sin is in the y position when the triangle is in the normal orientation but in this case the tryangle is rotated (look at picture) and in this case sin = opposite/long end

[waxen jay]

oh the triangle is rotated i see

so the total length vector in component form would be <575sin(15) , 575cos(15)+1150> right?

@stone ermine

[stone ermine]

I believe so

[waxen jay]

ohhh my god

okay youre right lmao

thank you 🙏

.close

[final saddle]

Channel closed

Closed by @waxen jay

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[edgy jacinth]

If z = 2-3i is a solution to the equation 2z^3 + bz^2 + 30z-13 = 0 where b ∈ R , then find b and the three solutions. Any help would be much appreciated thank you

[green ibex]

recall that complex solutions come in conjugate pairs

[edgy jacinth]

so z = 2+3i would also be a solution?

[green ibex]

yes

[edgy jacinth]

then do you make (z-2i-3)(z-2i+3)(z-x)?

[green ibex]

with a constant at the front

you could do that yes

[edgy jacinth]

Then equate both of them yea?

[green ibex]

sure

that would work

[edgy jacinth]

is there a better way of doing it?

[green ibex]

well you could plug in (2-3i) into it, and that would allow you to solve for b immediately

[edgy jacinth]

interesting

I got 7 for B does this look right?

[green ibex]

+4za

not -4za

[edgy jacinth]

cheers the rest of it good?

[green ibex]

yeah

[edgy jacinth]

Now how do i find the third solution?

[green ibex]

(2z-a) will be your extra factor

so just set that to 0

[edgy jacinth]

Ahhh cheers

thank you

[final saddle]

@edgy jacinth Has your question been resolved?

Channel closed

Closed by @edgy jacinth

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tranquil pine]

This is the easiest shit ever but I forgot how to slove it

[vast summit]

m²+4n²=(m²-4mn+4n²)+4mn

[tranquil pine]

How R you able to put that +4mn?

[vast summit]

I subtracted 4mn then I added back the 4mn

[tranquil pine]

Aren't you changing the thing

[vast summit]

the -4mn inside the bracket can cancel with the +4mn outside

[tranquil pine]

Yea ik 1st grade math, I'm saying how is that not changing the equation

Nvm

I gotit

.close

[final saddle]

Channel closed

Closed by @runic mountain

Use .reopen if this was a mistake.

[tranquil pine]

oh mb I read this wrong

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[silver maple]

a) y = ??

[ivory vessel]

what have you tried

[silver maple]

b) x = ??

[dusty jay]

i just wanna know how to do this type of question

[ivory vessel]

well, find the equation of the line

maybe it would help if you sketch it out

[silver maple]

I have given you a hints for both questions

[magic rampart]

gradient of the x axis is 0

[dusty jay]

u actual rat

cocomelon go away

thanks guys

i got it

.close

EW

[ivory vessel]

type it again

[dusty jay]

/close

.close

[ivory vessel]

editiny doesnt work

[final saddle]

Channel closed

Closed by @dusty jay

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[warm python]

The region between two concentric spheres of raidus a and b contains a charge denisty if $\rho (r)= \frac{\Omega}{r}$, and a point charge q is placed at the centre, find $\Omega$ such that the electric field between a and b is constant

[soft zealot]

Why am. I here

[warm python]

I started by finding the charged enclosed at a radial distance r from the centre
that would be $\int_a^r\frac {\Omega}{r} 4 \pi r^2dr$
not sure how to proceed as omega will ultimately depend on r

[soft zealot]

Why am. I here

[warm python]

I could use guass law

but what would I set E to b?

*be

[signal vector]

In the region between the spheres with radius a and b you want the field to be constant?

[warm python]

yes

it must be 0, right?

so $ E(4 \pi r^2)= $\int_a^r\frac {\Omega}{r \epsilon_0} 4 \pi r^2dr+q$

[soft zealot]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[signal vector]

Yes that gives you the field at a distance r

[warm python]

yeah, but how do I ensure it's constant?

Lebinez rule?

nah

that wouldn't work

[signal vector]

Can't you just integrate it?

[warm python]

I could but between what limits?

[signal vector]

I don't think it's possible to place a charge such that the field is constant

[warm python]

a and b

I need to find omega

not q

[signal vector]

Omega would be a function of R in that case

[signal vector]

With the limits you have rn

a to R

Since for every r you want the field to be constant

[warm python]

hmm, ok

[signal vector]

And yes probably use leibniz rule

Idk this just seems sketchy if omega isn't a constant you can't integrate it

[warm python]

it's a constant according to my book though

[signal vector]

Wait omega is a constant?

How is that possible

[warm python]

I did find this answer online

but it makes no sense to me

[signal vector]

E isn't independent of r if Q-2πAa^2=0 though

oh yes it is

The r^2s cancel

[warm python]

oh shit, yeah just realised

thanks

.close

[final saddle]

Channel closed

Closed by @warm python

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[alpine jay]

how do I do part d? the answer to part c is 18m

[alpine jay]

idk the number of bounces until it reaches rest

[inland kettle]

original height times (0.8)^number of bounces = peak height per bounce

[alpine jay]

i already know that

[inland kettle]

was answering this

at rest, peak height = 0

[alpine jay]

yes

[inland kettle]

technically

[alpine jay]

but there is no value of n for which peak height =0

[inland kettle]

i know

that's because we're ignoring friction and energy loss and thermal stuff

[alpine jay]

so how do i do it

yeah

[inland kettle]

limits? idk

[tranquil pine]

It will get to zero after an infinite number of bounces

[alpine jay]

how would i do part d without calculating number of bounces before rest

[tranquil pine]

So you need limits

[alpine jay]

yep

[alpine jay]

i hate limits

is there a way to do it without it

idk how to do it with limits anyways

[tranquil pine]

And to calculate the total time you'll need to calculate a series of inifite terms

[alpine jay]

ok

[tranquil pine]

(sum of infinite terms)

[alpine jay]

for part c i did that

[inland kettle]

again. infinite series.

yk the total distance

[soft zealot]

ren

[alpine jay]

yep

what do I do from there tho

s=dist on each bounce

[inland kettle]

just input distance??

the total distance covered = d

[alpine jay]

no?

[inland kettle]

yes

for this case

set whatever the total dist is to "d"

then just solve for t

look at what it says after part c.

[alpine jay]

so ur saying root(18/5) is the time

[inland kettle]

what?

what's the total distance that the ball travels b4 cming to rest?

[alpine jay]

18m

[inland kettle]

so yes sqrt 3.6

[alpine jay]

but the answer is 11.3s in ms

[inland kettle]

i meant distance as in answer to c

[alpine jay]

answer to c is 18m

[inland kettle]

hm

u sure??

[alpine jay]

yes

[tranquil pine]

I found 22m total distance travelled

[alpine jay]

it is 18m

i think u did 10*2+2

but it should be 10*2-2

[tranquil pine]

Why minus 2? Aren't you considering the first bounce?

[alpine jay]

tell me what the height of the 1st bounce is

ig ill ask for help later

.close

[final saddle]

Channel closed

Closed by @alpine jay

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[past bronze]

can someone help with this

[past bronze]

no idea on how to start

[marble agate]

ok so im not sure if its correct solution but using the fact that f''(x)>0, f'(x) increases i calclated (f(3)-f(1))/(3-1) and (f(5)-f(3))/(5-3)

i got 0.6 and 0.9

so avarage increase on interval [1,3] is 0.6 and avarage increase on interval [3,5] is 0.9

as the increases always increases

[past bronze]

i did not understand your appraoch

[marble agate]

the f'(3) needs to be between 0.6 and 0.9

[past bronze]

also why did you use average increage

[past bronze]

correct

0.7

[marble agate]

f''(x)>0 so f'(x) is increasing right?

[past bronze]

i do not know why but yes makes sense

[marble agate]

we were going thru this before 😭

but ok lets agree about that

[past bronze]

no like how can i prove that?

[marble agate]

derivative of f represents the increase of f

if the derivative is positive the increase is positive

[past bronze]

k so if f'(x) decreases then f(x) decreases and f''(x) decreases?

[marble agate]

no 😭

[past bronze]

why not

sorry if i am botherig you

and self studying and need help

[marble agate]

f'(x) can decrease but as long its positive f(x) increases

its like basics of calculus

[past bronze]

man

lets just complete with the question

i do not understand that

😔

[marble agate]

lets agree that is true for now

[past bronze]

ok

[marble agate]

by evaluating this: (f(3)-f(1))/(3-1) i can calculate avarage increase of f in interval [1,3]

[past bronze]

yes

[marble agate]

so the closer to x=3 the increase is bigger beacuse we took the avarage and the increase itself increases because f''(x)>0

so f'(3) needs to be bigger than 0.6

[past bronze]

so you imply from that that instantenous velocity is always > average velocity

[marble agate]

in this case

beacause f'(x) increases

for all x

and interval [1,3] is right below 3
instantenous velocity is > average velocity

for interval [3,5] is right above 3

so instantenous velocity is < average velocity

[past bronze]

what

i think vice versa

[marble agate]

😭

[past bronze]

bruh

[marble agate]

i edited

[past bronze]

i suck

[marble agate]

ok now its good

[past bronze]

this is wrong tho

instantenous velocity is > average velocity

[marble agate]

for interval lower than 3

in this case

and instantenous velocity is < average velocity for interval higher than 3

its not always

not even always in this case but in intervals

if the interval was [2,4] for example i could tell nothing about it

[past bronze]

i understood how you solved it

and how you got the answer

but its just that i do not understand what am doing

idk if that makes sense

[marble agate]

i get you dont understand logic behind it?

or am i missing what are you trying to say?

[final saddle]

@past bronze Has your question been resolved?

Channel closed

Closed by @past bronze

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[wild plank]

2x²-(2k÷1)x÷2=0

[tiny gorge]

what is the ÷1 doing for us?

it's so hard to read, maybe it should be +1?

[wild plank]

Take it up with my teacher

[marble agate]

it should be "+" i think

[tiny gorge]

yea ÷1 makes no sense unless they're trying to be obscure

[wild plank]

It's not

[outer yacht]

its a +

[wild plank]

It's not

[outer yacht]

can't be

ok then its inconsequential

[wild plank]

I don't make the question

[outer yacht]

it's just a 2k then

[tiny gorge]

just ignore the ÷1 then

because it has no effect

which is why it makes no sense to write it

[marble agate]

so its 2x^2-kx=0

[outer yacht]

2x² - 2kx + 2 = 0?

[wild plank]

No

÷2

At the end

[outer yacht]

uhm

[marble agate]

am i right @wild plank ?

[outer yacht]

okay

[tiny gorge]

i think they're both supposed to be + but what do i know

[outer yacht]

x² - kx = 0 then...

[marble agate]

me too

[wild plank]

2x²-(2k÷1)x÷2=0

[marble agate]

2k/1=2k

[outer yacht]

2x² - kx = 0...

[marble agate]

so 2kx/2=kx

[outer yacht]

still no solution..?

one root has to be 0 from observation alone

[final tangle]

what makes you so sure those aren't supposed to be "+"

[outer yacht]

0 x 4 = 0

[wild plank]

It's my class work

[outer yacht]

so k might as well = 0

its a plus man

[final tangle]

that's not justifcation

[wild plank]

That's what I got as well

It might be a typing error

[outer yacht]

which is what its supposed to be

[wild plank]

But on paper it's divide

[final tangle]

did someone explicitly tell you those are division symbols?

[outer yacht]

let's say its a plus to make life easier

[final tangle]

or are you making an assumption

[outer yacht]

or harder actually

[wild plank]

No, but it's a divide clearly on my paper

[wild plank]

Blud it's on my paper

[final tangle]

well the paper is missing a lot of pixels

[marble agate]

so k=0 that it

[outer yacht]

2x² - (2k + 1) + 2 = 0

this should be if its not a divide

[tranquil pine]

hey yall

[outer yacht]

last time i checked your spm or o level syllabus they never use division symbols anymore

[marble agate]

yeah if thats photo of your paper it missing a lost of ink lol

[outer yacht]

i'm assuming spm

[wild plank]

It's a scan

[tranquil pine]

what are we doing?

[outer yacht]

a poor scan*

[wild plank]

It is

[outer yacht]

discriminants

[outer yacht]

oh come on its a plus

or else answer is straight up 0

spm will never make it that easy

[tranquil pine]

oh ok

b2-4ac?

[outer yacht]

maybe

[tranquil pine]

wait let me get my board

[wild plank]

My teacher taught us sum of roots and product of roots so it might be related

[tranquil pine]

it is

sum of roots= -b/a

product of roots=c/a

[wild plank]

-b/a and c/a

Yep

[tranquil pine]

yeah

[outer yacht]

(x - a)(x - 4a) = x² + (2k -1)x + 2

4a = 2

[tranquil pine]

b2-4ac=0, real and equal roots

[outer yacht]

(x - 0.5)(x - 2)
therefore 2k - 1 = -2.5
k is -1.75

[tranquil pine]

b2-4ac>0, 2 distinct roots

[outer yacht]

by pure observation thats my working

[tranquil pine]

b2-4ac<0, no real roots

[outer yacht]

its not discriminants

[tranquil pine]

then what

[outer yacht]

wait i made a mistake

[wild plank]

It's SOR and POR

[outer yacht]

2(x - a)(x - 4a) = 2x² + (2k - 1)x + 2

[tranquil pine]

oh ok

[outer yacht]

compare

[wild plank]

Assume 1 root is alpha and the other is 4 alpha

[tranquil pine]

what do you want me to do

[outer yacht]

yep

[wild plank]

Find possible value of K

[tranquil pine]

ok ok can you type the question all at once

[outer yacht]

wait no

let me retry this

[tranquil pine]

send the questio please

[wild plank]

My febble attempt

I gave up on the last line

[outer yacht]

(x - a)(x - 4a) = x² + ((k - 1)/2)x + 1
4a = 1
a = 0.25
-2.5 = k - 1
k = -1.5