help-36

[tiny gorge]

solve that for r

then use that to work out when the value will reach 5000

(this is assuming annual compounding, maybe continuous compounding makes more sense, in which case use the version involving e)

same general approach either way

[soft zealot]

nox💫

[tiny gorge]

yep

[soft zealot]

nox💫

[final saddle]

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[lucid wind]

please explain

[lucid wind]

the right hand part is very confusing

the intersection of all sets
and their summation
so sum of A1 int A2 (let int be a short form for intersection symbol)
thus
A1+ A1 int A2 + A1 int A2 int A3 + A1 int A2 int A3 int A4 ....

I only understood till this part
after that
the negative power summation threw me off

[vital crag]

the sum of all intersections is overcounting so you subtract it off

write out and expand the statement for n=2 and 3 and you'll see a pattern

[lucid wind]

this is the solution
please tell me which pattern you are talking about

@vital crag

[vital crag]

the solution is too terse and doesn't show n=3

write that out if the base case isn't enough for you

the right hand side is overcounting

subtraction is the correction

https://www.onlinemathlearning.com/image-files/xvenn-diagram.png.pagespeed.ic.qZI8K3AfLE.webp

the intersection of A and B gets counted twice in |A| + |B|

[final saddle]

@lucid wind Has your question been resolved?

[lucid wind]

the |A| represents cardinality of A right?

[vital crag]

yes just like your notes

[lucid wind]

alright
i think i got it now

see
for n = 3
this will be the pattern I will always get @vital crag

[final saddle]

@lucid wind Has your question been resolved?

[vital crag]

the "pattern" should be the thing you're trying to prove.

[final saddle]

@lucid wind Has your question been resolved?

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[split steeple]

What graph has this description? "The function is increasing for all x where it is defined."

[tiny gorge]

many graphs have this description, can you be more specific about what you are looking for?

[split steeple]

I mean I just need like a hint honeslty doesnt have to be any exact one

[tiny gorge]

do any ideas come to mind?

[split steeple]

I feel like this one is simple and just a straight line but im not sure exaclty how it could look like

[tiny gorge]

yea, a straight line can work, can you be more specific about what needs to be true of the line?

[split steeple]

It needs to go to x right?

Like towards + x

[tiny gorge]

not sure what you mean by that

[split steeple]

Towards the right lol

[tiny gorge]

for example, is this an increasing function?

[split steeple]

I guess it could be since the x is increasing but y is decreasing at the same time

I could be wrong though

[tiny gorge]

do you have a definition for "increasing function"? you need to understand the definition if you're going to be able to answer this

[split steeple]

I mean the description I got was "The function is increasing for all x where it is defined.". and honestly the confusing part is where it says "for all x is defined". I mean I think an increasing function is when it goes diagonally up to the left but like I said im not sure so please correct me if im worng

to the right**

ohhh

Increasing functiion

Now i think i get it

Is it when its like "x^2"

[split steeple]

But what could they mean with "for all x is defined" do you think?

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[stark apex]

how would I solve for x?

[tiny gorge]

the left hand side is a product of three things

if it equals zero, then one (or more) of those three things has to be zero

[stark apex]

why did my teacher set the x^2 + 2x + 1 into (x+1)^2

[tiny gorge]

because that way it's easy to see what values of x make it zero

[stark apex]

oh ok

.closr

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[sacred lynx]

How many prime numbers are in k

[vital crag]

do you know what 13! equals ?

[sacred lynx]

13*. 12*. 11 and so on

[vital crag]

yea, factor out a common term from 13! + 2

[sacred lynx]

Common term ?

2?

[vital crag]

like ab + ac = a(b+c)

common term to ab and ac is a

[sacred lynx]

2 right

[vital crag]

yea

[sacred lynx]

Ok so i factor 2

[vital crag]

repeat for +3, +4, ..., +13

[sacred lynx]

Wym?

[vital crag]

repeat means do the same process

can you list out all values of k in this inequality?

[sacred lynx]

Ik what repeat means

Well i could use a calculator but don’t want to do that

[vital crag]

you don't need to write the numbers fully

what's the next number after 13!+2

[sacred lynx]

+3

13!+3

4

5

[vital crag]

.

[sacred lynx]

And so on

Until 13

[vital crag]

so what was confusing about this

[sacred lynx]

So do j find the prime numbers thro those

And thats my answer

[vital crag]

did you factor out a 2 from 13! + 2?

[sacred lynx]

2(13!/2+1)

[vital crag]

is that prime?

[sacred lynx]

Do i not need to calculate to say that?

I got a suspicion it is tho

You here im not sure what to do

Do i factor every number until 13 and tell if there prime or not?

[vital crag]

2 * integer is even

therefore 2 * (13! / 2 + 1) is not prime

[vital crag]

you factored 2 out of 13! + 2, and i explained why it's not prime. repeat for the rest

[sacred lynx]

3 is prime right

[vital crag]

3 * (integer that's bigger than 1) is also composite

[sacred lynx]

So that not it ether

[vital crag]

would probably benefit you if you wrote out all terms instead of "and so on"

[sacred lynx]

So 13! Is. 13 *12 *11 *10 *9 *8 *7 *6 *5 *4 *3 *2 *1

[vital crag]

do you see how 13! + 7 can be factored?

[sacred lynx]

Nope

[vital crag]

rememeber this

[sacred lynx]

Do i factor 7?

[vital crag]

what's a common term between them. yes 7

[vital crag]

you did this for 13!+2, then 13!+7. repeat for the rest between 2 and 13

[sacred lynx]

7(13!/7+1)

Ok

3(13!/3+1)

4(13!/4+1)

5(13!/5+1)

6(13!/6+1)

8(13!/8+1)

9(13!/9+1)

10(13!/10+1)

11(13!/11+1)

12(13!/12+1)

13(13!/13+1)

@vital crag

You here?

I get *2 is even

But i factored all of them now what

Do i find witch one gives me prime numbers

?

[vital crag]

What

[vital crag]

Do you know what prime numbers are

[sacred lynx]

Yes

Numbers that divide by 1 and themselfs

Like 3

[vital crag]

Only by 1 and themselves

4 is divisible by 1 and 4

But 4 is also divisible by 2

[sacred lynx]

So its even

I factored thees all out

But how do ik wich one is prime or not

Can you just pls help me do this problem

I factored all of the now what

How do i teel wich is even and witch is prime

[vital crag]

Is 9 prime?

[vital crag]

I've been helping this whole damn time

!vol

[final saddle]

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

[sacred lynx]

9 is not prime

[sacred lynx]

Ik sorry just want to get the problem done

Sorry

Prime numbers are 2,3,5,7,11,13

[vital crag]

Look at the definition of prime again and look at your list

[vital crag]

.

[sacred lynx]

Yes 1 and themselves

How do ik witch ones divide by 1 and themselfs tho

From my list

[vital crag]

you're not reading this

[sacred lynx]

I am

1 and themselfs

[vital crag]

then you're not understanding it

[sacred lynx]

Ok so the goal is to find prime numbers in my list right?

[vital crag]

do you really doubt that

[sacred lynx]

So the thing im not getting is how to do that

[vital crag]

beucase you're not understanding what prime number means

even though you keep saying it

[sacred lynx]

I mean 3 is prime but *3 is not?

[vital crag]

.

[sacred lynx]

So it wouldn’t count in the prime list

[vital crag]

7 * (any integer > 1) is also composite

because 7 * (13! / 7 + 1) is divisible by 7

since 7 isn't equal to 1 and 7 isn't equal to 13! + 7, 13! + 7 is composite

[sacred lynx]

4 is even

[vital crag]

?????

[sacred lynx]

*4

[vital crag]

are you even reading what i'm saying

[sacred lynx]

I am

[vital crag]

can you read what i say

before moving on

[sacred lynx]

So what there are only composite and even numbers in the list?

Can you tell me one of the members from that list that would be prime

[vital crag]

?????????????

you have the whole list

what part of this did you not undertand

[sacred lynx]

So 7* is prime

Look

[sacred lynx]

I need to find prime numbers that are also k

[vital crag]

????????????????

7 is prime

what does 7* mean

[sacred lynx]

Times by 7

[vital crag]

that's not a number

prime number only applies to integers

[sacred lynx]

So how eould i find if a number is prime if I don’t calculate it from my list

[vital crag]

by looking if it factors

[vital crag]

.

[vital crag]

did you read this?

integer > 1 is like 2,3,4,5,6,...

do you know the word composite?

[sacred lynx]

That divedes by other then 1 and themselfs

[vital crag]

is 7 * (integer > 1) prime or composite?

[sacred lynx]

Composite

[vital crag]

???????\

so why did you say that

[sacred lynx]

But then wich one would be prime from the list

All of them would be composite or even no?

[vital crag]

.

[vital crag]

.

[vital crag]

even implies composite for integers > 2

[sacred lynx]

But i don’t need composites i need primes

How do i find primes

[vital crag]

yes

[vital crag]

can't find them if they're not there

[sacred lynx]

So there are none?

So there are 0 prime numbers in k?

@vital crag

?

Can you tell me so i can sleep

Pls

The answer is 0??

@vital crag

[vital crag]

i already confirmed how many times are you gonna ping me

[sacred lynx]

Ok thanks

.close

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[ruby locust]

hi

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Please don't occupy multiple help channels.

[ruby locust]

is this answer correct?

the third one?

[whole halo]

yes

[vital crag]

#help-5

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[radiant zinc]

i dont get how to do 15 c can i have help please?

[vital crag]

did you do parts a) and b)

[radiant zinc]

yes

[vital crag]

show

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[radiant zinc]

.reopen

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✅

@radiant zinc Has your question been resolved?

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[dapper creek]

help

[final saddle]

Please don't occupy multiple help channels.

[dusky sail]

.close

[dapper creek]

.close

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[ionic hound]

.

Can someone help me find the restrictions for the following

[ionic hound]

The root 2x + root 2x - 2

[mint orbit]

$\sqrt{2x+\sqrt{2x-2}}$?

[soft zealot]

jan Niku

[ionic hound]

Yes

[mint orbit]

what have you tried?

[ionic hound]

I set 2x = 0, so x can't equal one

But then ik that's wrong

[mint orbit]

hmm? why that condition

[ionic hound]

Ik I have to set the entire radical equal to 0

[mint orbit]

start from the inside

lets start here

$\sqrt{2x-2}$

[soft zealot]

jan Niku

[mint orbit]

say this was the only thing; can you tell the restrictions here?

[ionic hound]

X > 0

[mint orbit]

hmm i dont think so

lets say here

$\sqrt{2x}$

[soft zealot]

jan Niku

[mint orbit]

so we'd usually say: the input to a square root cant be negative

[ionic hound]

If we input anything above 0, we should get a defined answer

[mint orbit]

yea

but specifically because

2x becomes positive at x=0

well, nonnegative

and it increases as x goes up

so, let's take that logic to the next step up

$\sqrt{2x-2}$

[soft zealot]

jan Niku

[mint orbit]

where is the inside equal to 0?

if we set $2x-2=0$, what is $x$?

[soft zealot]

jan Niku

[ionic hound]

X = 1

[mint orbit]

yea

and which way has valid x?

so at x=1, we have 2x-2=0

do you get what im asking?

i mean we found where it just started to not be negative

and now we wanna include everything positive

which way do we have to go?

[ionic hound]

X must be greater or equal to 1 then

[mint orbit]

yea, this seems right

we can check by graphing

,w plot 2x-2

[soft zealot]

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Requested by janniku

[mint orbit]

looks about right

okay, so we actually solved one of the pieces of your problem

for $\sqrt{2x-2}$, we have $x\geq 1$

[soft zealot]

jan Niku

[mint orbit]

now lets look at the whole thing

$\sqrt{2x+\sqrt{2x-2}}$

[soft zealot]

jan Niku

[mint orbit]

the trick here is that we have a pretty strong restriction now

x will be at least 1

is it possible for $2x+\sqrt{2x-2}$ to be less than 0 if $x \geq 1$?

[soft zealot]

jan Niku

[ionic hound]

No

[mint orbit]

yea

you can try to solve it

[ionic hound]

It will be above 2 or more

[mint orbit]

yup

so, we don't pick up an additional restriction here

The restriction on the inside root is enough to make sure this one is positive too

and thats it

[ionic hound]

Wow

[mint orbit]

,w plot sqrt( 2x + sqrt(2x-2))

[soft zealot]

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Requested by janniku

[mint orbit]

and it looks like we got it right

[ionic hound]

I was over complicating it too much

[mint orbit]

x >= 1

[mint orbit]

its like a onion

[ionic hound]

Tysm 🙏🙏

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[stark apex]

whats the approach to this?

he multiples by du/dx which is what u=x^2 -1 is?

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[candid sapphire]

i need help with triangles

[candid sapphire]

the traingle rules

translated it's side angle side

side side side

and angle angle

for similar triangles i think

[final saddle]

@candid sapphire Has your question been resolved?

[junior token]

What have you tried?

[candid sapphire]

im jsut overally confused

[junior token]

With the meanings of side angle side and the rest?

[candid sapphire]

yes

im confused how im supposed to know if they're similar

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[hushed musk]

Hi

[hushed musk]

Help

I dont get what im doing wrong

[final tangle]

show work

[hushed musk]

Its rlly messy but i basically just did synthetic division and found -1 to work

[final tangle]

redo the work, so that it's less messy

[tiny gorge]

plug in x=0 to see that yours doesn't have the same value as the original

[hushed musk]

Ok 😭

[final tangle]

can't tell exactly what's wrong without seeing what you did

[hushed musk]

But why

[tiny gorge]

if you multiply your factors you can see that the x^2 term is -5x^2, but it needs to be -3x^2, perhaps that can guide you to what went wrong?

[final tangle]

its a simple method to check whether a solution is wrong

[hushed musk]

Ok so this is my sythenic division

[final tangle]

ok, in that case, your issue is the misapplication of factor theorem

[tiny gorge]

if space is not at a premium, consider writing it out more carefully, use a whole page if you need to

[hushed musk]

Okk

[final tangle]

doing synth div with -1 and obtaining a remainder of zero
indicates that "-1" itself is a zero
and the respective factor is (x - (-1)) = (x+1)

[hushed musk]

Ok so here

[final tangle]

and seeing as they're asking for linear factors specifically
it seems they want you to factorise the x^2-4x+5 over complex as well

[hushed musk]

Would i have to use i?

[final tangle]

yes

[hushed musk]

Ohhhh 😭😭

Ok tyy!

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[stark apex]

what tells us to find the derivative?

[autumn geode]

What does stop mean to you

[stark apex]

0

Idk

[autumn geode]

0 in s(t) or s’(t)

[stark apex]

s(t) ?

[half salmon]

motion's or movement's derivative is velocity. for something to stop it has to have a 0 velocity, something can be at position 0 and still be moving. s`(t)

[stark apex]

oh because velocity is a derivative thats why I need to set the derivate equal to 0?

[half salmon]

yes? I am just wondering if they mean forward motion as a velocity or a position function...

[stark apex]

not sure either

(havent even taken physics)

[autumn geode]

Position probably

[stark apex]

So when im plugging in t = 16 for part b

Am iplugging into a derivative or no

[half salmon]

where are you getting t = 16?

[autumn geode]

Oh wow looks good to me

[stark apex]

Oh ok nice

[autumn geode]

Ask dragon breath tho

[half salmon]

then yes, you would plug it into s(t) to get the distance travelled

[stark apex]

Ive been studying since 4 pm and the common sense is flying by me

[stark apex]

Oh ok

[autumn geode]

Go to sleep!

[stark apex]

Ok i wil right now thx for reminding me

im gonna hit the hay

count some sheep

[autumn geode]

Have a good night you did good work today

[stark apex]

Thanks 😄

.close

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[solid flare]

What can I use for congruency on this, I’m having a brain fart atm

[solid flare]

Can I say pmo is congruent to nmo?

[astral moss]

Yes

[solid flare]

What statement would prove it though?

[final tangle]

are you trying to prove that the triangles are congruent?

[solid flare]

Yes

[final tangle]

then you can't say that <PMO is congruent to <NMO as there's currently no info immediately indicating that

[solid flare]

What can I say?

[final tangle]

identify wat's given to you in the diagram based on the markings

[solid flare]

I’m trying to prove both sides of o are congruent btw

[final tangle]

can you show an uncropped image of the question

[final tangle]

,rotate

[final tangle]

do you know the valid justifications for congruence?

[solid flare]

Somewhat

Hello?

[final tangle]

what are the valid justifications for congruence

[solid flare]

SAS AAS SSS ASA HL

Is that what you mean?

[final tangle]

yes.

consider what you currently have

[solid flare]

I could do HL

[final tangle]

yes

[solid flare]

But then what goes in the last box before congruency, mo is congruent to mo, reflexive prop?

[final tangle]

yes

[solid flare]

Ok

Thank you

This makes more sense now

[final tangle]

3rd line is a bit dodgy though

[solid flare]

What’s wrong with it?

[final tangle]

think they may want you so state that MO is a hypotenuse of both triangles or something

[solid flare]

So switch 5 and 3 sort of?

[final tangle]

don't really need to give <P = <N its own line

[solid flare]

What would go there then?

[final tangle]

that can be combined with line 2

[solid flare]

I can’t just leave it empty tho

[final tangle]

think they may want you so state that MO is a hypotenuse of both triangles or something

[solid flare]

Ok

[final tangle]

ask your teacher to clarify exactly what they want

[solid flare]

Got it, I can do that tomorrow

Thank you for help

.close

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[slate cedar]

Help me out plz. This is crazy~

[whole halo]

how did you get 370

[slate cedar]

I got it by −3x+400=−2.7x+397

[whole halo]

when you solve for x, what did you get

(it shouldnt be 370)

[slate cedar]

I got 10

[whole halo]

now lets say x = 0

and you plug x=0 into -3x + 400

would it really make sense to say you ran 400 meters at the very beginning of the race?

[slate cedar]

no

[whole halo]

you read it as 400 - 3x

at the beginning, youre at 400 meters

then every second you run 3 meters

thats why its 400 - 3 * x

[slate cedar]

oh~

so 400-3*10 then?

because my x is 10

right

[whole halo]

that gets you 370 again

so no

what about the 3 * 10 part of it

what would 3 * 10 mean

[slate cedar]

it would mean 30?

[whole halo]

.

you run for 10 seconds

= you ran for 30 meters

= 30 meters is your answer

[slate cedar]

mean in math terms is equal so I was kinda confused sorry I am dumb

What I mean is mean means equal in math terms making me confused

sorry, and thank you for helpping me out. bye

.close

[final saddle]

Channel closed

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[whole halo]

np

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[mental hamlet]

i don't really agree with this equation

because

we have this

and we can find b using it

right?

and we already know what a is because we have the points (1, -4) and (9, -4), right?

half way between 1 and 9 is 5

and the distance from 5 to either point is 4

4 = a

and we can also find c becuase of the focal points

it's equal to 1

from 5 to either 4 or 6 is 1

so c = 1

right?

and now we can do 16 = b^2 + 1

15 = b^2

oh wait

nvm i agree with the equation

b^2 = 15

LOL

,close

.close

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[gray acorn]

here do i sub

the function in the trig?

then what do i do with the csc and cot

[lyric summit]

change csc into 1/sin and cot = cos/sin

then you see easily, what u sub you have to use

[gray acorn]

already did

thing is

i cant integrate

if its not a single varaible in the trig function right

[lyric summit]

in numerator you got cosine, right?

and in denominator you got

sin^2

[gray acorn]

yes

[lyric summit]

so thikn

cosine

is

a derivative

of sine

hence

u = ?

suich thin that removes cosine

so u = sin

[vapid knot]

u don't need to sub anything

this is a standard anti-derivative

[lyric summit]

if u = sin, then du = cos dt

[vapid knot]

d/dx (cosec x) = -cosec(x)cot(x)

[gray acorn]

yes so i can remove a trig function

[lyric summit]

precisely

[gray acorn]

but the function inside them is still pi t

what do i do with that

[vapid knot]

um u don't need to take substituitions here ig

[lyric summit]

dervative of inside , = a constant

[vapid knot]

this holds i believe

[solar glade]

But u have to divide by pi to hold it

So u better sub first not to make a mistake

[lyric summit]

or use subs twice as has been mentioned above

[inland kettle]

if you don't want to sub, just remember that {f(ax)}' = a*f'(ax)

comes up quite a bit in integrals

[gray acorn]

oh yeah i ahve done this

im left with 1/u

but can i integrate this if the function iinside is pi t and not just t

[solar glade]

Use this

[gray acorn]

isnt this just chain rule?

[inland kettle]

yes it is

[solar glade]

u=pi x

[inland kettle]

but just handy is all

[gray acorn]

ah

[inland kettle]

what have you done so far?

[solar glade]

Can u make csc(u)cot(u)du?

[gray acorn]

yeah this is what i started with

[solar glade]

Then do u= pi t

[gray acorn]

i thnik i ended with

wait

[solar glade]

And u solve the problem in one step

[inland kettle]

@gray acorn where ya at bro
im lost here idk what ur doing rn bc idk ur working

[gray acorn]

i made pi t into

u

[inland kettle]

no just show

[gray acorn]

okkkk

[inland kettle]

$\int_{ \frac{1}{6}}^{\frac{1}{2} }-\frac{4}{\pi}\csc\left(u\right)\cot\left(u\right)du$

[soft zealot]

ren

[inland kettle]

you have this??

[solar glade]

That step should not have been shown

[gray acorn]

yeah and then i made cot into cos/sin

and csc into 1/cos

[steady prairie]

how about you do u=csc(pi t) , du= pi csc(pi t) . - csc(pi t ) cot(pi t)

[inland kettle]

$\frac{4}{\pi}\int_{\frac{1}{6}}^{\frac{1}{2}}-\csc\left(u\right)\cot\left(u\right)du$

[soft zealot]

ren

[inland kettle]

okay

csc = 1/sin

not 1/cos

[solar glade]

The minus also outside

[inland kettle]

nah

[gray acorn]

i was stuck at this

[inland kettle]

dont do that

and it's 4/pi

not 4pi

[gray acorn]

oh rigjt

[inland kettle]

hang on

bro has du = pi dt
and dt = pi du

xD

[gray acorn]

LOL

yeah ignore that

its fixed now

[vapid knot]

dude - csc(u)cot(u) is the derivative of csc(u)

[inland kettle]

!nosols

[final saddle]

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.

[inland kettle]

@vapid knot at least wait

sighs

[vapid knot]

okay mb sorry

[gray acorn]

i thought of that idk why i didnt keep going

its cuz

[inland kettle]

it's fine

[gray acorn]

im used to just u sub the function inside the trig function

[inland kettle]

anyway, syrex, now that you remember that... what is the integral?

[soft zealot]

ren

[gray acorn]

cuz someone told me i cant integrate if the function inside isnt just a single variable

[inland kettle]

?

[gray acorn]

yeah

i had like sin(2x and they said i have to u sub the 2x

and turn it into u

[inland kettle]

not necessarily

[gray acorn]

1 varaible

[inland kettle]

you can if you want

2x is the same variable as x

[gray acorn]

wait

so youre saying i can

[inland kettle]

yes you can do this integral w/o u-subs

[gray acorn]

int(sin(2x) = -cos(2x)

oh wtf

someone told me specifically i cant yesterday

[gray acorn]

wait so this

is possible

[inland kettle]

no

[vapid knot]

no

u need to apply the reverse chain rule

[inland kettle]

remember that {f(ax)}' = a*f'(ax), where a is a constant

[inland kettle]

not necessarily but yes the logic of the reverse chain

@gray acorn are you there fam

[gray acorn]

ahh okk

i havent heard of reverse chain rule yet tbh

[inland kettle]

okay look imma explain rq, one sec

[gray acorn]

ah

[soft zealot]

ren

[inland kettle]

do you understand why this works

[gray acorn]

yeah i do

okay

so

lest say i int

sin(pit)

its gonna be -cos(pit) /pi

[inland kettle]

you mean

$\frac{-\cos(\pi t)}{\pi}$

[soft zealot]

ren

[gray acorn]

yes

[inland kettle]

yes

,w antiderivative sin(pi*t)

[soft zealot]

Results provided by WolframAlpha

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

Requested by justarenegade

[inland kettle]

see

[gray acorn]

ah

[inland kettle]

now anyway, back to your original question

[gray acorn]

okay ty idk why it wasnt taught

[inland kettle]

$\frac{4}{\pi}\int_{\frac{1}{6}}^{\frac{1}{2}}-\csc\left(u\right)\cot\left(u\right)du$

[soft zealot]

ren

[gray acorn]

okay this

[inland kettle]

solve

what's the integral of -csc (u) cot (u) du

[gray acorn]

its just csc

[inland kettle]

csc what

[gray acorn]

man its so dumb

[inland kettle]

you can't say "csc" in an exam
you could mean csc 2, csc pi, csc x, csc pi*u

[gray acorn]

how did i not see it

yeah like

csc(pi t)

[inland kettle]

you mean csc pi*t = csc u?

good

[gray acorn]

yes

[inland kettle]

now evaluate

$\frac{4}{\pi}\left(\csc\left(\frac{\pi}{2}\right)-\csc\left(\frac{\pi}{6}\right)\right)$

[soft zealot]

ren

[gray acorn]

its 4

i believe

[inland kettle]

$\frac{4}{\pi}\left(\frac{1}{\sin\left(\frac{\pi}{2}\right)}-\frac{1}{\sin\left(\frac{\pi}{6}\right)}\right)$

[soft zealot]

ren

[inland kettle]

no

[gray acorn]

oh i forgot about the pi

so -4/pi

[inland kettle]

yes

gj

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[gray acorn]

okay ty!

.close

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[spare glacier]

I have no idea how to start

I just know you have to make a function that makes them a bijection

But i dont know how to derive such functions

[trail mango]

hint: map 0 to c and 1 to d

what’s a simple function that would do that?

[spare glacier]

Uhh

No idea

[trail mango]

think about how a high schooler might do it

[spare glacier]

maybe i can do

c* I {0} (w) + d * I {1} (w)

I is the indicator function

[trail mango]

[spare glacier]

😭

Wait

cx + d?

nah wait nvm

[trail mango]

right line of thinking

but yea not quite

[spare glacier]

c + dx

[trail mango]

not quite still

you’re trying to make a line passing through (0,c) and (1,d) right?

[spare glacier]

yeah

is that a function of the slope or sumn

y - y0 = m(x-x0) or sum

[trail mango]

there are many ways to do it but it’s just a high school algebra exercise

pick whichever one you want

[spare glacier]

Yeah i forgot everything in highschool

[trail mango]

i never remember those formulas

i just put y = mx + b and compute the slope and then use one of the points to determine b

[spare glacier]

can you give me a smol hint for the function

[craggy plank]

yes it is

[trail mango]

this is a massive hint lolz

the slope is d-c

put y = (d-c)x + b

passing though (0,c) means c = 0 + b

[spare glacier]

y = [(d-c)x]/c

[trail mango]

no what

[spare glacier]

aw

I saw a formula for the equation passing throufh two points

[trail mango]

i just put the answer above

[spare glacier]

so y = c + (d-c)x?

ah well yeah u did just put it

[trail mango]

yes

[spare glacier]

Oh no i was right i arrived at y - c = [(d-c)/1]x

I thought that u had to divide both sides by c

damn

im cooked

[spare glacier]

How about number 2

is it just the same?

[trail mango]

no that one is harder

[spare glacier]

damn…

uhh

So how do i start?

[craggy plank]

Good helper

[spare glacier]

agree

[trail mango]

[spare glacier]

Oh i can close it rn if you dont feel like it anymore

you’ve helped me more than enough HAHA

really sorry for the trouble

[trail mango]

it’s a bit difficult to make a bijection between those 2 sets

do you have the schroder bernstein theorem available to use?

[spare glacier]

wtf

damn

aint that real analysis

uhh

[trail mango]

no it’s set theory

[spare glacier]

oh

we havent gone there yet

[trail mango]

gone to what?

[spare glacier]

we havent tackled set theory yet

its in the latter stages of the class

[trail mango]

the problem you posted is about sets but ok lol

[spare glacier]

idk, this is apparently part of counting techniques

[trail mango]

well anyway sounds like the answer to this is no

so i’m going to bed

[spare glacier]

Is it not possible without set theory concepts?

okay

thank you for the time! @trail mango

.close

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[buoyant yarrow]

can someone help me do this using the t formula

[white tiger]

whats the t formula

[static stump]

has the question the specifically said it ?

[buoyant yarrow]

no but i wanna know if its possible to do it using the t formila

it said something about R cos (x-a) or smthn

[soft zealot]

Solomaniac

[white tiger]

oh that

whats the problem in using it

[buoyant yarrow]

@static stump @white tiger yeah ik but when i tried using it i didnt get the correct solution

[white tiger]

!showwork

[final saddle]

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

[buoyant yarrow]

idk what i did wrong

[white tiger]

dont worry just show it and we'll find it eventually

[buoyant yarrow]

well so basically i did 8(1-t^2)/1+t^2 - 15(2t)/1+t^2 = 7(1+t^2)/1+t^2

then i got it to 8(1-t^2) - 15(2t) = 7(1+t^2)

i expanded: 8 - 8t^2 - 30t = 7+7t^2

[static stump]

15t^2 +30t -1 =0

@buoyant yarrow

[buoyant yarrow]

yeah

[static stump]

Great what's the value of t ?

[buoyant yarrow]

0.0328 and -2.0338

when i do arctan 0.0328 i get 1.89 degrees

i multiply by 2 to get 3.76 which is apparently wrong

[static stump]

Did you check -2.0338 ?

[buoyant yarrow]

i got 232.4 for my other solution

[static stump]

Does that match ?

[buoyant yarrow]

nope

its 248

[static stump]

x = 248 ?

[buoyant yarrow]

yea

actually lemme sub my solutions into x

and see if solutions may be wrong

[buoyant yarrow]

@static stump i subbed my solutions i got from arctanning these numbers and i was right

the solutions provided are wrong

[static stump]

Yep. Told you. That's sometimes a possibility. Solutions given in the books sometimes turn out to be wrong.

[buoyant yarrow]

yeah tysm for ur help

[static stump]

Glad to help you.

[buoyant yarrow]

ye

.close

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[tired knot]

Lol I’m doing the same thing as you

I’m at 7G now

Also use the formula

You learn it for a reason

This is like the most basic question they give rn

Which should take u 15 seconds

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[jade fable]

ayy

[jade fable]

"the ratio between A and B" how do I interpret it

A:B or B:A

[wintry kindle]

typically A/B

eg A=9, B=3, then the ratio between A and B would be 3

but you can always explicitly mention it to make it clear

since both 3 and 1/3 are valid ratios between them

[jade fable]

is it wrong to interpret it as B/A

[static stump]

Kinda

Because conventionally, it's A/B

[jade fable]

I see.

[solar glade]

Well, its all about the meaning of what you are saying

The ration BETWEEN A and B cannot be B/A because thats the ratio between B and A and saying that means A/B = B/A which normally is false

[final saddle]

@jade fable Has your question been resolved?

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[true thunder]

Can someone explain me how to calculate the domain of this function?

[fluid bay]

do you know the definition of domain

?

[true thunder]

yea

where the function exists in R

[fluid bay]

yah so

here a function is root

[true thunder]

In this case it would be where
sqrt(x-1) -x +3 >= 0 AND x-1 >= 0

[fluid bay]

can inside of root there can be a -ve value?

[fluid bay]

yeah thats it

[true thunder]

Yea the problem is calulating the result of the first disequation

[fluid bay]

just find the intersection

[fluid bay]

why?

just solve

[true thunder]

I have to make a system containing the equations:
x-1 >= 0 ---> x>=1
sqrt(x-1) -x +3 >= 0

[true thunder]

Ye but the answer I get is [2,5], while the actual solution would be [1,5]

[fluid bay]

show me your work

[true thunder]

I just threw it in the trashcan

[fluid bay]

tf

find a better reason to trick

i am not dum

[true thunder]

What I did was

moving -x+3 to the second member of the disequation

elevating everything to the power of 2

then moving back the second member to the first

[fluid bay]

there will be cases

2 cases

one is x-3>=0

and x-3<0

[true thunder]

how do I contemplate them in the process

[fluid bay]

just make cases

[true thunder]

wdym

[fluid bay]

dont you know what a case is

[true thunder]

no

[fluid bay]

like possibilites

for 2 possibilities

[true thunder]

yea alr but how do I consider the case in the process of calculating the result

[fluid bay]

you solve each case and then find the intersection of their individual results

[true thunder]

how do I solve each case

[fluid bay]

..

solve these

sqrt(x-1) -x +3 >= 0 when case 1

then with case 2

[true thunder]

ok so I have sqrt(x-1) >= x-3

Do I simply replace x-3 with 0

[fluid bay]

no

[true thunder]

I literally don't know what to do

[fluid bay]

while solving this consider x-3>=0

one time

and then

consider x-3<0

[true thunder]

Im not understanding

[fluid bay]

ping helpers

[true thunder]

<@&286206848099549185>

[fluid bay]

i gtg sry

[true thunder]

kk

[stone wagon]

why do you get 2?

[true thunder]

I have sqrt(x-1)-x+3; i then moved -x+3 to the second member and after this elevated to the power of 2 everything

getting the equation x^2 -7x +10 <= 0

[stone wagon]

why <?

[true thunder]

inverted the sign

[stone wagon]

huh

[true thunder]

or else i would have had -x^2

[stone wagon]

aha i see what happened

[stone wagon]

they are not equivalent

[true thunder]

So I should have kept -x^2 ?

[stone wagon]

no that's not the error

the squaring both sides part is changing the equation

[true thunder]

So what do I do

Im not understanding the cases VAIBHAV is mentioning

Also photomath mentions it but I can't understand them

<@&286206848099549185>

.close

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[viral garden]

hello i have a math question and i wanted to know how to solve it so, i am very lost so can you show me step by step please

[wintry kindle]

hm I think a few hints will guide you:

notice that the areas ADF and CFE are the same

because ADEC is a parallelogram

and notice that DEF is a copy of AFC (shape-wise), just with a different size

with these two informations you can calculate the area of AFC

which means you'll have all areas except the area of DBE, which I think you'll get too when you're at that point :) @viral garden

[final saddle]

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[icy wind]

can someone help with 4

i have 0 clue

[soft zealot]

NEON

[icy wind]

yes

[gritty solar]

Alright

Do you know the triangle inequality?

[icy wind]

no

[gritty solar]

Hmm alright

Do you know how to expand $|z_1 + z_2|^2$

[soft zealot]

NEON

[icy wind]

maybe? i don’t think so

[gritty solar]

Okay wait do you intuitively understand

Why those statements are true

[icy wind]

why k is in real givers?

numbers?

[gritty solar]

because it says k is positive

[icy wind]

or the statements in the q

yes i understand

[gritty solar]

k > 0 only makes sense if k is real

[icy wind]

yes

[gritty solar]

alright

[gritty solar]

And this?

[icy wind]

yes i kind of do understand

[gritty solar]

What does |z| = k represent on the complex plane?

[icy wind]

the mod of z or distance to z is k

[gritty solar]

Yes but what shape would it be

[icy wind]

circle

radios k

[gritty solar]

good

yeah

perfect

centered at the origin

[icy wind]

yup

[gritty solar]

Now

The greatest possible real part that any point on this circle could have would be k

Do you agree?

[icy wind]

yes

[gritty solar]

Right good

[icy wind]

wait why not 2k

[gritty solar]

The circle has a radius k

[icy wind]

wait nevermind

yeah ok

[gritty solar]

I want you to visualize the circle

[icy wind]

ok

done

[gritty solar]

Yeah and it should be clear how k is the largest possible real part