help-17

[vast shale]

For 31st?

[bleak tundra]

yuh

[vast shale]

You should first try to visualize this figure and breakdown into its smaller parts

[bleak tundra]

do i find surface area of it first?

like the cylinder only

[vast shale]

The surface area is the part on which you can touch or paint

[vast shale]

Right?

[bleak tundra]

umm its like the base

so like

the circles on top and bottom

on the cylinder

[vast shale]

Yes , first find surface area of the cylinder including its base, cruvy side and upper part left

[bleak tundra]

uhm

alr gimme a minute

or two

LMFAo

[vast shale]

Take your time

Be aware that some portion upper base of cylinder is nullified by base of Cone

You cannot paint or touch that area anymore

[bleak tundra]

so i cant find the surface area without the cone being part of it?

[vast shale]

What do you mean?

👀

[bleak tundra]

cuz rn im solving for surface area of the cylinder. It being SA=2pi10times6+2pi10squared

so without the cone being part of it.

[vast shale]

First tell me the word equation of SA of cylinder according to figure

SA of Cylinderical portion = Curved Part + Base + Upper Base - Base of Cone

Do u get this point?

[bleak tundra]

oh

ohh i get it now

can i do them seperately and then add them up

?

wait

nvm

how am i supposed to find area of the curved part

base times height?

or wut

[vast shale]

How?

[bleak tundra]

idk

never done this

[vast shale]

Its a basic formula?

[bleak tundra]

idek

[vast shale]

You must have that formula somewhere written in your textbook

Its a basic formula

[bleak tundra]

uhmmmmmmmmmmm

lemme checkk

[vast shale]

It will give u volume

[bleak tundra]

no i dont have anythign like it on my packet

cuz we aint use textbooks LOL

just a packet

[vast shale]

Hmm

[bleak tundra]

it sucks

[vast shale]

Curved surface is

[bleak tundra]

i mean i got formulas

[vast shale]

2pi × R × H

Circumference × height

If u apply logic too it , it will make sense

[bleak tundra]

so i do this? and then circumference times height

or is that two options

[vast shale]

No

Its a same thing

I think your basics arent clear with these type of topics

Circumference of a circle is 2pi × r

[bleak tundra]

ohhh

[vast shale]

I am going back to work now

[bleak tundra]

wuat

[vast shale]

Cya and Gl

[bleak tundra]

byee

oh yeee im so dumb

i forgot what formula for circumference was

LOL

i got it

wait on this do i have to find surface area without the base of the cone?

surface area being the curved surface, upper base and bottom base?

@ancient panther

<@&286206848099549185>

mb

spelt it wrong

nvm

.close

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[celest isle]

who can explain me from 0 and or give me good videos to watch

[regal bane]

You really just want videos on graphing exponential functions

If you can graph g(x), you're done

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[gritty kayak]

Stuck on 2e and 2f

[gritty kayak]

How are the answers (1 0) and (2 4)

I got (0 -2) and (2 12)

Pretend these brackets are vectors

[flat whale]

What's 2(x+1) +3?

[gritty kayak]

Good point

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[vast shale]

I’m having trouble with question d

This is my working out

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[vast shale]

I’m having trouble with question d

[vast shale]

This is my working out

[autumn sleet]

d. y"+5y+4y'=0, y=e^(-2x)sinx

Need to verify that function is solution of DE

[vast shale]

Did I do something wrong?

[autumn sleet]

Yes , in the very last step

3y-8y=-5y

You calculated it as -4y

Otherwise all correct

[vast shale]

Ahhh yessss I got it

Lmao I can’t count😂

Thank you

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[glacial scroll]

a^2+b^2=1 and x^2+y^2=2 what can be the lowest value of ax+by?

[keen umbra]

nothing else?

just these informations?

[glacial scroll]

Yes

[keen umbra]

ok

is this olympiad maths

[glacial scroll]

Nah

[keen umbra]

oh ok

one sec

[keen umbra]

ok got it

so our first step is to add both of the equations

can try?

[glacial scroll]

Yes

[keen umbra]

so what do you get?

[glacial scroll]

a^2+b^2+x^2+y^2=3

[keen umbra]

yes

so now we try to construct ax+by

you know how?

[glacial scroll]

No

[keen umbra]

use (a+b)^2 =a^2 +b^2-2ab

[glacial scroll]

Ok

[keen umbra]

any clue?

[glacial scroll]

Yes

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[bitter copper]

Help im being dumb with Latex

[bitter copper]

Wait

$\x^2$

[twin meteor]

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

[bitter copper]

What

$x^2$

[twin meteor]

Nacros

[bitter copper]

What

[sly sierra]

fyi, there is #latex-testing

[bitter copper]

It’s fine I sorted it out, I was testing it but I kept getting 2x but then I accidentally missed out the slash when I was tryna show my error and it worked

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[blissful saffron]

nobody knows without more context

[autumn sleet]

again, more context will help us understand

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[vast shale]

i'm stuck on question 2f

[vast shale]

i'm getting two different answers

[autumn sleet]

Can you show your work?

[vast shale]

sure

the answer in the book was the second one

[autumn sleet]

first can be converted into second answer by using the equation in the question

[vast shale]

oh mb

thanks so much

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[unborn wadi]

Why does the taylor series in regards to Eulers method show the error? I am at a loss. Please help me.

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[unkempt steppe]

Ik it shoupd be simple but idk why is it like that

[unkempt steppe]

From photomath

If we do the multiply by 1/49 so we can add shouldnt the 100 be 4900

[calm minnow]

so do you remember how to factor @unkempt steppe ?

[unkempt steppe]

Factor what

The only factor is between 51/49 times b²

[calm minnow]

the $\frac{51}{49}b^2 + b^2$

[twin meteor]

jpsz

[calm minnow]

You can factor out a b^2 right?

[unkempt steppe]

So to factor it I would need to 51b²/49+49b²/49

[calm minnow]

You could do that. By factor I mean this:

$b^2(\frac{51}{49} + 1)$

[twin meteor]

jpsz

[calm minnow]

But you get the same thing by doing this

[unkempt steppe]

Oohhh

Now I got it

It looks like O just forgot

Lol

Thx

.close

[calm minnow]

np

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[void relic]

If I put everything to the power of 10 to get rid of log 10

[void relic]

and one of the sides is like 2+6

would it be 10^2 + 10^6

or 10^2 multiplied by 10^6

so if it was like $log10^x = 2 + 6$ would it go to $x = 10^2 + 10^6 or 10^2*10^6$

[vast shale]

uh

[twin meteor]

Aorliei

[vast shale]

2+6=8

whats the base of log?

[void relic]

nevermind I checked on my calculator you multiply them bothj

log base 10 sorry

I get it now anyways but thanks

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[silent saffron]

Hello. Please help

[urban token]

what can i help with here

how far have you got?

[silent saffron]

The center is 8,4 correct?

[urban token]

almost

[cobalt dock]

8,-4

[silent saffron]

Oh yes forgot the negative sign

This is how far I got, I move the center to 8,-4

[urban token]

okay

[cobalt dock]

does that show as correct?

I can't see the radius clearly

but remember, the radius is r^2 = 4 -> r = 2

so from the center to the edge of the circle should be two 'squares' on the grid

idk I can't see the image clearly

[silent saffron]

Got it thank you

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[silent saffron]

hi. please help

[paper depot]

do you know how to find the area of a circle?

[silent saffron]

no

[paper depot]

did you think to look it up before posting here?

[silent saffron]

?

what do u mean

[paper depot]

did you look in your notes, textbook or Google (as applicable) for "area of a circle" or "area of a circular sector"?

[silent saffron]

no notes or google

[paper depot]

no google? are you legally forbidden from using search engines?

[silent saffron]

area of circle = pi * r^2

[paper depot]

ok, see, great.

now for this sector.

the angle of this sector is 144°.

can you tell what fraction of the circle's area is taken up by the sector?

recall that a full circle is 360°.

[silent saffron]

144/360 = 2/5

[paper depot]

great

are you able to continue from here?

[silent saffron]

area of circle is 78.54 if you round it

right?

[paper depot]

you should not round

Express your answer as a fraction times pi

[silent saffron]

25pi

right?

[paper depot]

that is the area of the circle, yes

[silent saffron]

so now what do i do?

@paper depot

my laptop is shutting off in 5 minut

so i have to do this quick

[paper depot]

i said:

can you tell what fraction of the circle's area is taken up by the sector?
you replied:
2/5

[silent saffron]

thats the answer?

[paper depot]

no thats not the answer but its what you need to use to get the answer

the area of the circle is 25pi

your sector is 2/5 of that

whats 2/5 of 25pi

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[silent saffron]

.reopen

[vocal sleet]

✅

[silent saffron]

A circle centered at A(2,5) goes through the point B(5,4). What is the slope of the tangent line to the circle at point B?

@reef grove

Please help me with this

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[wooden latch]

not sure how these arent correct?

[worthy citrus]

Wants the equation of the line not a number

[wooden latch]

so how would I put that together?

[opal obsidian]

x=-3?

[wooden latch]

so then it woukd be y=0 for the second one

?

[calm minnow]

no, how do you find the horizontal asymptote?

[wooden latch]

its the blue dotted line lmaoao

[calm minnow]

based on the equation tho, how do you find it?

you know x = -3 because the bottom would be zero and undefined.

How do you know the horizontal asymptote based on the equation

[wooden latch]

u have to getthe numerator and the denominator to be equal

so it would be leading coefficient of numerator

over the leading of denom

so -7/1

[calm minnow]

there ya go.

which is -7

[wooden latch]

OH

im an idiot

[calm minnow]

No, you are not. You are smart. You just figured out the problem

[wooden latch]

❤️ xd

tysm

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[sleek flame]

question, if I solve a linear system, x=-29, y=6 what would x be in modulo 7?

[vast shale]

-29 = ? mod 7

add an appropriate amount of 7s (or do the division)

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[naive heron]

Let n be the least positive integer greater than 1000 for which:
gcd(63, (n+120))=21 and gcd((n+63),120)=60.
What is n?

source: 2020 AMC 10A #24

[naive heron]

idk where to start

[vast shale]

use crt to lift mod lcm(21,60)

[naive heron]

whats that

[vast shale]

chinese remainder theorem https://en.wikipedia.org/wiki/Chinese_remainder_theorem

[naive heron]

oh ok

but why 21 and 60

its asking about 63 and n+120

and n+63 and 120

[vast shale]

crt gives us all possible n

[naive heron]

wait

[vast shale]

the first gcd gives n+120 = 0 mod 21

and the second gives n+63 = 0 mod 60

then crt and check

[naive heron]

yes

oops looks like i actually dont know what crt is

i was thinking euclidean alg

[vast shale]

the proof of crt uses euclidean alg

there's probably a way to do it with euclidean alg only

[calm minnow]

Heh, I found it...

Not sure if I did it the algebraic way.

Basically I asked: what is a number greater than 1000 that has a gcd with 120 that is 60

[vast shale]

(aops has some solutions https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24)

[calm minnow]

Wow, that first problem is exactly the same. Nice find!

Ok, not quite. They want the sum of the digits. But pretty close

[naive heron]

lemme look

[vast shale]

aops pretty much has all comp maths questions

[naive heron]

yeah im studying for comp math lol

[calm minnow]

Haha, all the pieces are coming together

[naive heron]

so im literally supposed to apply the euclidean alg

[calm minnow]

Good to know

[naive heron]

except its kinda hard to understand

i should prolly turn off my music

the solution is exceptionally hard to understand but i think i got it

thanks!

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[gritty meteor]

Can someone help me with finding out CED if I put my calculation into the calculator there is a wrong answer coming out. Did I set it up wrong or something ?

[gritty meteor]

the answer should be CED = 50.48 but if i type it in either i get an error or 89.64

@strange crater could you help

@hard atlas

someone pls help

[hard atlas]

dont ping specific people

[gritty meteor]

you helped others and seem like a person who could help thats why i pinged you

[hard atlas]

and you're pinging me for a second time when I just told you not to ping

[gritty meteor]

just replying 💀

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@gritty meteor Has your question been resolved?

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@gritty meteor Has your question been resolved?

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@gritty meteor Has your question been resolved?

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[bleak tundra]

i need help

[urban token]

With?

[bleak tundra]

number 32

[bleak tundra]

i dont know hwo to do it at all

please help

whast the formula for this

...

but nah i want to understand it

and master it

when its asking about volume and surface area as well

nvm its taking too long

ima go review my stuff

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[vast shale]

571149.82

[vast shale]

ft sq

base area is 646x646

lateral area = 1/2(646)(√ 226829) x 4(triagnles)

total 646x646 + 1/2(646)(√226829)

= 571149.82

@bleak tundra

hope this helps

i can provide a diagram in dms as well i found online

also this isnt my problem

im just helping

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[rugged vortex]

I wouldn't open a help channel just to do this

[vast shale]

how else was i supposed to reach him

and i didnt open this

originally

whatever

.close

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[rugged vortex]

.once it's closed, just leave them be

[solar moat]

Hello

[solar moat]

Can anyone help me with my problem please?

[rugged vortex]

Just post it.

[solar moat]

Integrate by Cauchy's formula.

I'm very confused about how to solve this

[flat whale]

What's C

[solar moat]

It's just how you write the equation

It isn't used in solution

We are to use Cauchy's formula for this

[flat whale]

Definitely not

[flat whale]

It absolutely is

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.reopen

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✅

[solar moat]

How

[flat whale]

https://brilliant.org/wiki/cauchy-integral-formula/

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[mellow rampart]

i thought this would work

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[onyx meadow]

Help

[onyx meadow]

I know the area of each tile will be 36 inches^2

Then I multiplied that by the number of tiles which is 1080*36=38880 inches^2

That would be the whole area

Is that correct so far?

[strange crater]

it's easier if you change the squares' area to sq ft.

[onyx meadow]

0.25ft^2

For one cube and then multiply by 1080 giving 270 ft^2

Giving B as the answer

Perfect

Thanks for always helping!!

[strange crater]

yw

[onyx meadow]

I’m gonna send another question

I always get confused with these symbols

[onyx meadow]

But I know it is a plugging in problem

[strange crater]

you had the first part right

[onyx meadow]

Yeah confused with that second pet

Part*

[strange crater]

try it

[onyx meadow]

And how to write it

Ok

[strange crater]

you'll have to plug in "-t" for "y" for the second expression

[onyx meadow]

Is that correct

[strange crater]

not quite. the 3rd term should be written (-t)^2, just to be clear

and the final term shouldn't have a square for both letters

[onyx meadow]

Hmm

Why would it be -t^2

For third

[strange crater]

because you're replacing "y" with "-t", so the substitution must occur before you square it

y^2 --> (-t)^2

or, y^2 = y * y --> -t * -t

[onyx meadow]

Yes I got that

But isn’t that what I put

I didn’t put it in parentheses

[strange crater]

having the parentheses just makes it clear what you meant. it was unclear to me without them whether you meant "-(t^2)" or "(-t)^2"

[onyx meadow]

Got it

And third term is +ts^2

I’ll rewrite it

[strange crater]

yes, and then combine terms

[onyx meadow]

These are the answer choices

[strange crater]

are you sure?

[onyx meadow]

B is 2st^2

[strange crater]

ok that's good then

[onyx meadow]

I got this

I don’t know where I went wrong

[strange crater]

be careful with the signs of everything

[onyx meadow]

I got it

Yeah gotta be really careful

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[vast shale]

I need help with this question

[lime gorge]

The probability of it landing on heads doesn’t change

Nor does the probability of landing on tails change

It just so happened that they got 5 tails and 4 heads

Think about a regular quarter

If u flip it 99 times and get 88 heads and 11 tails

What’s the probability of getting a heads on the next flip

Still 0.5

Doesn’t change

[vast shale]

ahh i see

that makes sense

very nice example btw

this question tricked me

[lime gorge]

Yep that was its intention Xd

[vast shale]

lol

thx anyways

take care

[lime gorge]

Np

[vast shale]

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[soft geode]

Hi

[soft geode]

for this type of equation how do you know if the graph is supposed to be like

how do h know which one

also why is there two lines if the graph isn’t supposed to surpass the asymptotes?

[timber orchid]

You can test values

See if for a positive x, you get a pos or neg y

[pallid zenith]

also, look for a mismatch between the signs of the leading coefficient

[pallid zenith]

one is horizontal, one is vertical

[soft geode]

Can you elaborate

I’m confused

[timber orchid]

For your f(x), plug in say x = 1

Does your f(1) give you a positive number?

[timber orchid]

If yes, then red

If not, then blue

[soft geode]

Is there a way to look at the equation and know if it’s left or right

Wait

Oh

[pallid zenith]

using this method dont you have to identify where the horizontal asymptote is, first

or hope its along y=0

[soft geode]

I have both the asymptotes

this is the problem I am working with

[pallid zenith]

to me its easiest to identify places where the signs are gonna match

but it looks like you identified which one it is already?

whats your question specifically on that question

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@vast shale Has your question been resolved?

[floral pike]

ok well the period is [0,12)

so you have

5.63cos(2pi/12 x) + 6.85 = 10

5.63cos(pi/6 x) = 3.15

cos(pi/6 x) = 0.55950266429840142095914742451155

blech

now, arccosine both sides

and solve for x

a scientific probably won't return all 3 answers

I mispoke

I mean it won't get you all 3 2 roots in (0,12] or [0,12) (shit vision or something)

depending on how you define the domain

I'm actually struggling to remember how to get all 3 rn

let me think a little longer

I'm going to try to use euler's formula something

OH

right

I just need the period

ok, we left off here:

5.63cos(pi/6 x) = 3.15

cos(pi/6 x) = 3.15/5.63

arccosine both sides

pi/6 x = arccos(3.15/5.63)

this value (RHS) = 0.9770106949 rad

which is about 56 degrees

this is in Q1

we have the same cosine value at pi - arccos(3.15/5/63)

(we are just subtracting this same reference angle from 180 degrees, in the unit circle. You have seen this before)

here's the crucial step then

pi/6 x = arccos(3.15/5.63) + 2pi*k
pi/6 x = pi - arccos(3.15/5.63) + 2pi*k

solve each of those for x where k is a counter (0,1,2,...) such that your result is still less than 2pi

if you wait a few more min I'll write it out

oh...and I messed up the quadrant I needed

it's ok it will be correct on paper

no idea why I thought there were 3 intersections in (0,12]

there's just 2

@vast shale

also, sorry "theta" is actually arccos(3.15/5/65) (the whole thing)

but I indicated in the picture that it was just the inside of the inverse function

by mistake

I make a lot of mistakes ._.

but I get there

...and that means you will too 🙂

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[violet fox]

hi

[twin meteor]

JoeTheLazy1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

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@violet fox Has your question been resolved?

[jolly epoch]

can you provide more info?

[violet fox]

sigma is standard deviation and sigma squared is variance

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[violet fox]

.close

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[solid sage]

I’m quite stuck on the first part

[solid sage]

I tried using half angle but I couldn’t do much since there was plus minus

[trim walrus]

Ye so

You should he aware that 2sin(A)sin(B) = cos(A-B)-cos(A+B)

Are you??

[solid sage]

Yes

[trim walrus]

Now assume

A-B = alpha (a)
A+B = beta(b)

Add then and you'll get

a+b=2A

Subtract them and you'll get

b-a=2B

Hope this is not confusing!!

[solid sage]

It’s good!!

[trim walrus]

And substitute in

[solid sage]

I’ll try it

🙂

Give me like couple minutes

[trim walrus]

Ohk

[solid sage]

Yep worked

Thank you 🙏

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[empty frigate]

a and b are just variables, in this case they're representing the coordinates of the centre of the circle

so if you have a circle of radius 2 with centre (3,1), that would be a = 3, b = 1, r = 2, so any (x,y) satisfying (x-3)^2 + (y-1)^2 = 2^2 would be on that circle

[trim walrus]

Yep, like in the circumference

Not actually inside

But the area is not given

[empty frigate]

well there would be lots of them

any point that's on the circle

[empty frigate]

for instance here's this circle plotted on a graph

the line goes through every (x,y) that satisfies that equation

for instance (1,1)

yep

[trim walrus]

Isn't the Point D just (5/2,-5/2)

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[abstract minnow]

hi

[abstract minnow]

here , the answer why 3/8 ? i got 3/16

[main spade]

Why is there no question?

[abstract minnow]

this a duplicate integral for the area

Like that

[vast shale]

Okay did u draw ur region of integration

[main spade]

That iterated integral does not give the area of the region T.

[north ruin]

i think they mean the integral over the area not actually the area

so the integral is correct

[abstract minnow]

they mean that the x and y runs on that what was given

[vast shale]

.

Because you can make this simpler on you

[main spade]

Note: You would call this the double integral of the function f(x, y) over the region T (after you use the theorem to express it as a iterated integral.)

[abstract minnow]

first i solved the internal then the other and ended up with : 3/8* ( -1/cos^(pi/3) - -1/cos(0) )

and thats 3/8*(-1/2 + 1)

[main spade]

I get 3/8 evaluating the iterated integral.

Did you get to a point where you had

$\frac{3}{8} \int_{0}^{\frac{\pi}{3}} \frac{\sin x}{\cos^2 x} \dd{x}$?

[twin meteor]

stabulo

[abstract minnow]

yes

i failed a -1 somewhere

[main spade]

Use the substitution u = cos(x) but this causes the lower bound to be a greater value than the upper bound so you must flip and include a negative after that procedure.

[abstract minnow]

okay , sorry that was it

i got it

3/8

[main spade]

Nice. 🙂

[abstract minnow]

i got sin(x)*y^-2 part wrong

[main spade]

Ah.

[abstract minnow]

forget to divide by -2

thanks !

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[crimson pumice]

can someone calculate this differential equation?

[crimson pumice]

I know how to do this

but the answer I get is different to the one in the book

this is differential equations about logistical equations

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@crimson pumice Has your question been resolved?

[wraith venture]

,w solve 9000 = 10000 / (1+199e^(kt)) for t

[twin meteor]

Results provided by WolframAlpha

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Requested by themateo713#5303

[outer warren]

show work

(as well as the full question)

[crimson pumice]

give me a second

original question

its like 4 parts

[outer warren]

is the red your work or is it given

[crimson pumice]

given

Heres my working out for that original question

,rotate

[crimson pumice]

this is their answer

I have no clue how they got 1/1291

@outer warren

[outer warren]

probably a typo

the approximation of 46
uses 1791

,w 10log(1/1791)/(log(49/249))

[twin meteor]

Results provided by WolframAlpha

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Requested by ramonov#0921

[outer warren]

,w 10log(1/1291)/(log(49/249))

[twin meteor]

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[crimson pumice]

ah I see

so I actually did it right then?

thanks for calculating it, I was feeling I was going insane

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[bitter copper]

$\int_{a}^{b}\frac{(x+3)^2}{x(x+1)^2}$

[bitter copper]

Help

[twin meteor]

Nacros

[bitter copper]

Ah wait I can split it up

[brittle minnow]

just divide doe

remember that x=0 and -1 and points of discontinuities

[bitter copper]

Yeah I just took partial fractions, integrated quite nicely, pulled together the ln functions and it worked out nicely to pi(ln(2^17/3^8) - 2/3)

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[glad pendant]

How to do? I tried everything

[paper depot]

,rccw

[paper depot]

multiple issues youve got there

confusion between decimals and percentages, confusion between number of correct answers and percentage of correct answers, confusion between the probability that a variable equals 0.1 and the number 0.1 itself...

[glad pendant]

The answer written is not what I’m concerned about, It’s a old test from somebody else

[glad pendant]

Wow would I get the right answer?

How*

[paper depot]

be a bit more organized than the other person lmao

yeah so ok

let X be the random variable for the number of right answers by the student

X ~ Bin(20, 0.8)

do you understand this notation?

[glad pendant]

I do, I understand

I just don’t know how to input the probability in the binomial probability equation

[glad pendant]

I just don’t know the trials

How would I input the 10 and 20%

[paper depot]

well, how many questions is 10% of 20?

and how many is 20% of 20?

[glad pendant]

1 and 2

I tried that already

What would the probability be?

[glad pendant]

Would the probability be 4/5 or 1/5

[paper depot]

no, 10% of 20 is not 1.

[glad pendant]

No, 10% of 10 is 1

[paper depot]

and the probability of getting a question right according to the problem is 4/5.

[paper depot]

sure is, but the test doesn't have ten questions, it has twenty.

[glad pendant]

Ohhh I see what ur saying shit okay

How’s that?

It gives us really really small numbers like 0.0000000000013

[glad pendant]

Help!!!

[paper depot]

really small numbers make sense here actually

with a 20-question, a student who gets questions right about 80% of the time should be expected to get around 16 questions right out of the 20

so getting between 2 and 4 should be rare.

[glad pendant]

Ohhh okok

I mean it shouldn’t be that rare though like the decimals are really long

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[hidden niche]

$y=t^{ln(t)}$

[hidden niche]

Bad timing xd

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[solemn storm]

Can someone verify this proof (proof following question)

[twin meteor]

GoldenPhoenix

[worthy citrus]

looks fine

[solemn storm]

B) Given that $\psi(I)=I, \
\psi(xx^{-1})=\psi(x)\psi(x^{-1}) \
\implies \psi(I)=\psi(x)\psi(x^{-1}) \
\implies I=\psi(x)\psi(x^{-1}) \
\implies \psi(x)^{-1}I=\psi(x)^{-1})\psi(x)\psi(x^{-1}) \
\implies \psi(x)^{-1}=\psi(x^{-1}) \
\qed$

[twin meteor]

GoldenPhoenix

[worthy citrus]

also fine

[solemn storm]

cool, thought so, just worried I would be missing some weird little thing

[worthy citrus]

yeah there isnt really much room to miss anything in questions like these

[solemn storm]

I'm sure, but when you don't have many tools to start with and you're not well practiced with the tools you do have, you wanna make sure you're in good form to start with

prevents trouble down the line

[worthy citrus]

yeah of course, looks good

[solemn storm]

ty! now to do the other ones since I feel I got a good grasp. also I'm just realizing I should have been using phi instead of psi, but meh, consistency satisfies context.

!close

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[undone pier]

can someone conceptually explain to me why sin cos and tan are both used for trig functions but also hyperbolic functions?

[undone pier]

I'm completely familiar with how sin cos and tan relate to the unit circle and their applications in trig

but seeing sinH, cosH, and tanH sort of confuse me

[wheat wolf]

I don't know what u are saying are meaning like tan x or tan "theta"

[river minnow]

Cosine and sine are defined as x and y coordinates of a point on the unit circle, right?

[hard atlas]

sinh is a different function to sin. they are not the same. the name was just chosen this way cause they have very similar properties

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@undone pier Has your question been resolved?

[undone pier]

ah, so they have nothing to do with one another ?

thank you

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[hard atlas]

.reopen

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✅

[hard atlas]

they do

do you know complex numbers?@undone pier

[undone pier]

to a very basic level yes

[river minnow]

Can you answer this btw

[hard atlas]

there are identities like sin(ix)= i sinh(x) which relate them

[undone pier]

yes this is what I was taught in terms of just cosine and sine

[hard atlas]

which are the reason why they have very similar properties

just from a complex perspective

[river minnow]

Similarly cosh and sinh are defined are x and y coordinates of a point on the curve x^2 - y^2 = 0

[hard atlas]

but from the real perspective you dont see that

[undone pier]

x^2 - y^2 = 0

is this a function or a relation ? what would you call this?

[river minnow]

= 1*

I meant x^2 - y^2 = 1

[hard atlas]

compared to x^2+y^2=1 for the unit circle

[river minnow]

Functions are relations

[hard atlas]

you notice that if you replace y by iy you get the other one back

[undone pier]

yes thats true

[hard atlas]

which is essentially this

[undone pier]

okay

thank you for this, i feel like i understand them conceptually now

have a good day

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[hidden niche]

5 ways to call the derivative ?

[hidden niche]

I have 3

1: The instantaneous rate of change

2: The derivative

3: The speed of a particle in a determined point in time

4: ?

5: ?

🥺

[karmic imp]

Why do you need 5?

[thin vale]

maybe you could call it the difference quotient evaluated at a point?

[hidden niche]

That's the activity

[hidden niche]

Yeah good one

I need 1 more, maybe a name that other sciences use

Outside of math

[strange crater]

the slope of the tangent line at a given point

[hidden niche]

That

[vast shale]

Inverse of the integral?

[hidden niche]

❤️❤️🙏👍🛐🛐🥺

Yes

Thank you

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[quiet bloom]

What is the answer of this? i am confused

[paper depot]

do you know your exponent laws?

[quiet bloom]

yes

[paper depot]

also dont ask "what is the answer?" bc thats a surefire way to get people telling you we don't give out answers

(because we don't)

[paper depot]

ok, can you name a few?

[quiet bloom]

if the base is same the exponent get subtracted when divison

and added when multiplication

[paper depot]

not the best wording but it'll do

so can you apply that here?

[quiet bloom]

yes

5 + -3

which is 7^2

but then theres 7^2 / 7^2 and i am not sure about that

i think it is 1

[paper depot]

what's a number divided by itself

[paper depot]

you think it is 1, but you have your doubts?

[quiet bloom]

idk

[vast shale]

When the numerator and denominator are the same, the answer will always be 1 (for nonzero numbers)

[paper depot]

is something stopping you from definitively saying "yes, 7^2 / 7^2 = 1"? yes/no @quiet bloom

[quiet bloom]

no

[paper depot]

ok then problem solved?

[quiet bloom]

yes

[paper depot]

cool

[quiet bloom]

thank you both

[paper depot]

anything else you want to ask?

[quiet bloom]

nope

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[sly coral]

can someone help me do this

ik how to do inverse functions

but

i dont get this one

[azure rover]

basically invert the sides, the lower part goes on top and the upper part goes on bottom

[sly coral]

what

[karmic imp]

It's not that inverse

[karmic imp]

Did you try anything?

[sly coral]

yh

but its like

stupid

[karmic imp]

What did you try

[sly coral]

so

i heard

[karmic imp]

Can you show your work

[sly coral]

u need to

do

y=f(x)

so

y=2x/x-5

so

i got

x-5(y)/2=2x

[karmic imp]

Please use parentheses when typing fractions in text form

[sly coral]

its just

y =2x/x-5

[empty frigate]

is that $\frac{2x}{x}-5$ or $\frac{2x}{x-5}$?

[twin meteor]

bee [it/its]

[karmic imp]

See

That's why

Because order of operations, reads the first way

[sly coral]

ohhhhhhhhhhhhhhhhhhhhhhhhhh

[karmic imp]

While you meant the second

[sly coral]

the latter

[empty frigate]

so 2x/(x-5)

[sly coral]

yes

:/

[vocal sleet]

@sly coral Has your question been resolved?

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[pulsar scroll]

I was solving a differential equation for simple harmonic motion and this appeared in the mark scheme

[pulsar scroll]

why is that the amplitude

for y, where y is the displacement at time t from the equlibrium position

I dont understand how they got from the second line to the third line basically

[zinc sequoia]

this is light damping?

oh wait no its forced

oh so basically youre findinf the amplitude of th ocmbined wave

2 sinusoidal waves of the same period can be combined into one

youre aware of this?

[pulsar scroll]

I got complez roots so is it not underdamping?

[zinc sequoia]

no its not being damped complex roots dosent mean its not under daming

its not under dmaping since its homogenous

either way your aware

[pulsar scroll]

is this to do with the form Rsin(something + something) I forgot exactly what

[zinc sequoia]

yeh

exactly

[pulsar scroll]

ah

[zinc sequoia]

R(sin (t+ alpha)

for some value of alpha

and the ampliutude is the value of R

[pulsar scroll]

what is alpha?

[zinc sequoia]

its an angle

[pulsar scroll]

also would it be wt instead of just t

[zinc sequoia]

are u alevel?

[pulsar scroll]

yh

[zinc sequoia]

uk?

[pulsar scroll]

yeah

[zinc sequoia]

haha same bro

[pulsar scroll]

oh nice

[zinc sequoia]

no just t

[pulsar scroll]

what board u doing?

[zinc sequoia]

same perios

edexcel

period

not wt

[pulsar scroll]

oh ok

[zinc sequoia]

this is a pure 2 concept the r(sin(t+alpha

[pulsar scroll]

yh I think it's single even

u do further I'm assuming?

[zinc sequoia]

yeh