help-10

[drifting wraith]

we both just made up a model

[timid silo]

Who is "they". Please show a picture from the task!

If I have 255 Green Colors, 255 Red Colors, and 255 Blue Colors - what is the amount of possible colors I can create? Or at least, what would be the equation to find out?

This looks like a made up task from "King TK" himself. I have never seen a task written like this.

[obtuse pebble]

@random egret Has your question been resolved?

[obtuse pebble]

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[haughty drift]

Hi uh Im pretty confused with this

[haughty drift]

Why I cant do direct cross multiplication in the first one?

[proven zephyr]

Number 16?

which number?

[obtuse pebble]

@haughty drift Has your question been resolved?

[haughty drift]

17

[obtuse pebble]

@haughty drift Has your question been resolved?

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[modest crypt]

Operations on the square roots

[modest crypt]

how do I solve this

[obtuse pebble]

@modest crypt Has your question been resolved?

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[opal solar]

https://www.bbc.co.uk/bitesize/guides/zpbcjty/revision/2

[obtuse pebble]

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[soft burrow]

$(0,2) \cup [3,4]$

[warm shale]

m_g

[soft burrow]

by simplifying we get the set {1,3,4} right?

[kindred cedar]

Only if you do it with natural numbers.

[soft burrow]

how is it with real numbers?

[kindred cedar]

You can't really simplify it further.

[soft burrow]

it's an interval

so real numbers

[kindred cedar]

Two intervalls

[soft burrow]

the minimum is 1 right?

with real numbers

[kindred cedar]

Yes

[final thunder]

Doesn’t it just have to be bigger than 0

[soft burrow]

yeah but how do I define it

as a number

[final thunder]

Can’t really define it

[soft burrow]

so there is no min

[final thunder]

Probably if not too sure

[soft burrow]

i'm not sure too

anybody knows?

[final thunder]

Ah

I got it

There’s no minimum because

Let’s say m is our minimum

But we can do (0+m)/2 which is halfway between 0 and m

This leads to a contradiction

[soft burrow]

ok, i think this is enough as explanation to write

the supremum is 4 ?

[obtuse pebble]

@soft burrow Has your question been resolved?

[obtuse pebble]

@soft burrow Has your question been resolved?

[obtuse pebble]

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[elfin zenith]

please help, I'm really stuck

[final thunder]

I mean sinx = cosx only when x=pi/4 in the interval (0,pi/2)

[elfin zenith]

were the angles*, translation mistake, sorry

[elfin zenith]

yes, this is the problem,

I tried to use wolfram alpha but the answers were not useful

[final thunder]

,w solve sinx=cosx

[warm shale]

Results provided by WolframAlpha

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

Requested by Pure#9389

[final thunder]

Yeah…

[final thunder]

So you got 45 and 45 angles

Lmao

Wth

[elfin zenith]

so, is there are any other solutions?

[final thunder]

Don’t think so

Not possible for a triangle to have these ratios probably

Ah await

Acute triangle

[elfin zenith]

yeah, a triangle with 3 acute angles

[final thunder]

Nvm I thought it meaant only 1 was acute

[elfin zenith]

is this even possible?

yeah, this question has a mistake I guess

[final thunder]

Yeah that’s not possible lol

[elfin zenith]

my teacher wants the answer by Tuesday

[final thunder]

[elfin zenith]

well, can I change the question?

I want to make it: proof that this cannot happen

[obtuse pebble]

@elfin zenith Has your question been resolved?

[obtuse pebble]

@elfin zenith Has your question been resolved?

[obtuse pebble]

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[timid silo]

$$
|{X:X \in P(B),|X|=6}|=28 \text{ find |B|}
$$
$$
\text{ let a1,a2... be elements of P(B), suppose we are constructing P(B), first step: include a1 or not, second step: include a2 or not, etc...}
$$
$$
\text{chossing 6 times to include the item yields 28 sets thus}
$$
$$
|B|=C\binom{2^n}{6}=28 \text{ whatever it is}
$$

[warm shale]

Zermelon

[timid silo]

is my proof rigorous ?

,ask cosec 56

[warm shale]

Results provided by WolframAlpha

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Requested by Kousei Arima#0408

[timid silo]

?

.close

[obtuse pebble]

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[obtuse pebble]

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[uncut robin]

$\f:A\rightarrow B \text{ is surjective}
\g:B\rightarrow C \text{ is not injective}
\(g\circ f): A \rightarrow C
\\text{Show that } g\circ f \text{ is not injective}$

[uncut robin]

$\\text{I can write } B \text{ as } f(A) \text{ so my functions are}
\f:A\rightarrow f(A) \text{ and } g:f(A)\rightarrow C$

[warm shale]

Thunder7

Thunder7

[obtuse pebble]

@uncut robin Has your question been resolved?

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[uncut robin]

$\f:A\rightarrow B \text{ is surjective}
\g:B\rightarrow C \text{ is not injective}
\(g\circ f): A \rightarrow C
\\text{Show that } g\circ f \text{ is not injective}
\\text{I can write } B \text{ as } f(A) \text{ so my functions are}
\f:A\rightarrow f(A) \text{ and } g:f(A)\rightarrow C$

[warm shale]

Thunder7

[timid silo]

since g is not injective, then g(x)=g(y) with x≠y. Then consider the preimages of x and y

[obtuse pebble]

@uncut robin Has your question been resolved?

[uncut robin]

wdym

[timid silo]

its the definition of injectivity

[uncut robin]

yeah I get that

but after that?

[timid silo]

since f is surjective, you can find an element z such that f(z)=x

then use those two facts to construct a counter example to injectivity

[uncut robin]

$\\text{So } \exists x_1, x_2|g(x_1)=g(x_2) \implies x_1 \neq x_2
\\text{But } x_1 \text{ and } x_2 \text{ can be written as } f(z_1)=x_1 \text{; }f(z_2)=x_2
\\exists f(z_1), f(z_2)|(g\circ f)(z_1)=(g\circ f)(z_2) \implies f(z_1) \neq f(z_2)$

[warm shale]

Thunder7

[timid silo]

yup, now show that z_1≠z_2

[uncut robin]

how so?

[timid silo]

f is a function

what does this tell us if z_1=z_2?

[uncut robin]

f(z_1)=f(z_2)

do I have to use the inverse?

[timid silo]

nah you dont need to use the inverse

just use the contrapositive of that statemenet

[uncut robin]

what statement

[timid silo]

z_1=z_2 => f(z_1)=f(z_2)

[uncut robin]

but z_1 != z_2

[timid silo]

yeah exactly, thats what we want to show

[uncut robin]

so if z_1=z_2 then f(z_1)=f(z_2)
so if z_1!=z_2 then f(z_1)!=f(z_2)?

I'm lost

[timid silo]

no, we have that if f(z_1)≠f(z_2), then z_1≠z_2

[timid silo]

adding that to the end of here completes the proof

[uncut robin]

but how do you know is it always prop1 = prop2 then prop3 = prop4
prop3 != prop4 then prop1 != prop2

[timid silo]

they are equivalent, its a theorem from logic (you've probably seen it in a proofs course before) https://en.wikipedia.org/wiki/Contraposition

[uncut robin]

ohhh ok

thanks a lot

.close

[obtuse pebble]

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[long ermine]

Hi I would like to know how to get this V + 5/2 - 5/2 in the solution here:

[long ermine]

this would be the given homogeneous coefficient 🙂

[long ermine]

i really dont know how did my prof get v + 5/2 - 5/2 in that solution

<@&286206848099549185>

.close

[obtuse pebble]

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[long ermine]

.close

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[timid silo]

hi

[timid silo]

how do i rationalise this

[high lily]

multiply numberator and denominator by conjugate of denominator

[timid silo]

is that 5+rt3

i always mess up somewhere tho bc my answer is wrong

[high lily]

that is the conjugate yes

[timid silo]

yeah

but i think i multiplied it wrong

i got 20rt5 + 4rt8/22

[high lily]

yeh, you multiplied wrong

where's rt(8) coming from

[timid silo]

4rt5 x rt3

[high lily]

doesn't tell me how you're getting 8 under the root

[timid silo]

i’ll show my working

[high lily]

$\sqrt{a} \times \sqrt{b} \red{\neq} \sqrt{a+b}$

[warm shale]

ℝamonov

[timid silo]

what does it equal

[high lily]

you added the 3 and 5 instead of multiplying...

[timid silo]

ohhhh

wb this

[kindred moss]

9 and 2

3sqrt(2)

go from there

Instead of root18

To be clear

[obtuse pebble]

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[obtuse pebble]

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[safe imp]

Find x (indicies and surds)

[safe imp]

this is the question im stuck on, im not sure where to start with it

[untold flax]

√5 = 5^1/2

get 25^(x-1) in base 5 as well

and then apply indices laws

[safe imp]

so like (5^2)^x-1 ?

[untold flax]

ye

but there's something mroe you can do to that

[safe imp]

what else is it i can do i cant see it

[untold flax]

bottom one 👀

[safe imp]

ohhhhhh so 5^2x-2

[untold flax]

yep :D

[safe imp]

yay i eventually got it tysm

[untold flax]

👍

[obtuse pebble]

@safe imp Has your question been resolved?

[obtuse pebble]

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[scarlet oriole]

https://media.discordapp.net/attachments/903486480075354182/1030227213800460408/Screen_Shot_2022-10-13_at_14.15.04.png

[scarlet oriole]

Apparently it is not

But i'm not sure why

[zenith raft]

is longer like... in curve length?

how about if f is constant?

[solar trellis]

You should find a counterexample

[scarlet oriole]

Ok

Here is the answer sheet

But I am confused

I don't know why it is 2sqrt(1+f') instead of sqrt(1+f')

I want to think this is wrong

Since it is a weird question

[obtuse pebble]

@scarlet oriole Has your question been resolved?

[obtuse pebble]

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[scarlet oriole]

.reopen

[obtuse pebble]

✅

[scarlet oriole]

@ruby fulcrum

[obtuse pebble]

@scarlet oriole Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[sullen imp]

The top one😭

I solve for Y and input it in but i keep getting one of the wrong answers

[granite lark]

so what did you find y as?

[sullen imp]

-4 and 0

Its supposed to be -4 and 2

So im prob doijg the math wrong

WAIT

[granite lark]

hang on let me try work it out here

[sullen imp]

I meant solve for x

I solved for Y on the second eqtn and then plugged it into the first

To find X

[granite lark]

you solved for y on the 2nd equation?

[sullen imp]

Yeah

Y= 12/x

[granite lark]

as in xy = 12?

ohh

[sullen imp]

Then put thay into the first

[granite lark]

okay

so what does that give you when you sub y into the first equation

[sullen imp]

-4 and 0

For x

But it should be -4 and 2

Did u get sum else

[granite lark]

I mean just as the equation

[sullen imp]

3x - 12/x = -16

[granite lark]

so 3x-24/x = -6

[sullen imp]

Wait

Whyd u get 24

OMG

I didnt see the two

[granite lark]

yh

[sullen imp]

In front of y

Omfg

[granite lark]

also 6 not 16

[sullen imp]

Omg

Sorry

Im really tired

Thank u sm

[granite lark]

np lol

[sullen imp]

.close

[obtuse pebble]

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[fiery vapor]

does anyone know how to do this?

[uneven palm]

find f'(x) first

and then plug in 11

[fiery vapor]

ok ill try that

idk how

[uneven palm]

have you learned how to find the derivative of something like this?

more specifically, quotient rule?

[fiery vapor]

this is an example that he gave us

do i solve it like this?

[uneven palm]

yea that's an example of quotient rule

but do you understand what to do from it?

[real viper]

hey guys i'm new

[fiery vapor]

not entirely

do i plug in the 11 after i solve for f'(x)?

[uneven palm]

yeah

but do you know how to solve for f'(x)?

[fiery vapor]

i think so

ill just follow how he does it

ill let you know

[uneven palm]

[real viper]

i have a problem too

in mathematics

maybe u guys gonna laugh abt it

bc might be very easy

[uneven palm]

hey this channel is occupied, you should use one of the available ones

[real viper]

right

yo and how this chats work

like someone come to help u or smth

[fiery vapor]

this is what i got so far

idk if im on the right track

[fiery vapor]

is this where i plug in 11?

[uneven palm]

yep

[uneven palm]

looks right to me

[fiery vapor]

so that goes in the first box?

[uneven palm]

yeah

and make sure to simplify your derivative for the second box

[fiery vapor]

did i simplify this right?

[uneven palm]

in the last step

5x - 40 - 5x = -40

[fiery vapor]

oh yeah oops

wait so its like the same thing

[uneven palm]

but otherwise it seems right to me

[fiery vapor]

i got it right

thank you very much

W teacher

how do i end this channel thing

[uneven palm]

no problem lol

it’s .close

[fiery vapor]

.close

[obtuse pebble]

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[timid silo]

is this correct?

looks good to me

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

thanks

[obtuse pebble]

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[late mango]

I need someone to check my answer and see if i did anything wrong

[mental plaza]

aye aye

wassup

hm

ur first one

man

[late mango]

Aye

[mental plaza]

It was going so well

then idk what happened

[late mango]

I just need checking on what i did right and wrong

[late mango]

Because i remember my instructor telling me about if the denominator is greater than the numerator then you add or subtract

[mental plaza]

what’s going on there

[late mango]

Its just how my instructor showed me how to do it

[mental plaza]

u got the answer

1/x

but idk what

u did with it

[late mango]

1/x to -4/4

[mental plaza]

so u got $(x^{-4})^{\frac14}$

[warm shale]

Springsskateboard

[mental plaza]

right

then u got

[late mango]

Basically

[mental plaza]

$x^{\frac{-4}{4}}$

[warm shale]

Springsskateboard

[mental plaza]

yea?

what’s -4/4

[late mango]

Power

[mental plaza]

what’s the result

evaluate

-4/4

calculate

[late mango]

So i got 1/x and x is to the -4/4th power

[mental plaza]

nope

if you want to do it like that

the negative

become positive

b ur first

but first*

[mental plaza]

answer this

[late mango]

So its just 1

[mental plaza]

what’s -4 divided by 4

[late mango]

Hmm

[mental plaza]

-4 divided 4 is -1

[late mango]

-1?

[mental plaza]

yeye

[late mango]

Yea

[mental plaza]

nice

so u get

$x^{-1}$

[warm shale]

Springsskateboard

[mental plaza]

which is

$\frac1x$

[warm shale]

Springsskateboard

[late mango]

I see

But then whats the point of having x to the -4/4th power hold up

[mental plaza]

wdym

it’s just the exponent

the power

[late mango]

Lemmie show an equation me and my teacher went over that he said was right

[mental plaza]

yeye

sure

alr I checked

the rest

of ur work

and the second one

super close

third one

is right

[late mango]

Ignore the notes

[mental plaza]

wait nvm

third one isn’t right

it’s wrong too

[late mango]

Damn

[mental plaza]

alr lemme see

[late mango]

He said it was correct so

I kept doing the method he taught me

[mental plaza]

what u did

in ur notes

is perfect

what u did

in ur work

isn’t

[late mango]

How

[mental plaza]

I see ur mistake

common mistake

repeated

throughout

ur work

for example

when u have

$x^{-\frac12}$

[warm shale]

Springsskateboard

[mental plaza]

it becomes

$\frac{1}{x^{\frac12}}$

[warm shale]

Springsskateboard

[pine reef]

Yo

[mental plaza]

see the exponent

it becomes positive

the sign flips

[late mango]

Yea

[mental plaza]

yea

[late mango]

So im just flipping

[mental plaza]

u didn’t do that

[pine reef]

I still cant find out how to get a help channel

[mental plaza]

alr try

#help-22

[pine reef]

Ok

[mental plaza]

u can see question 2

in ur work

u didn’t flip

ur final answer has -2

[late mango]

And it becomes positive again correct

After

[mental plaza]

yep exactly

when u take reciprocal

[late mango]

So for number 2 instead of what i got what would it have been

[mental plaza]

$x^{-\frac25} = \frac{1}{x^{\frac25}}$

[warm shale]

Springsskateboard

[mental plaza]

so it’s $\frac{1}{\sqrt[5]{x^2}}$

[warm shale]

Springsskateboard

[late mango]

So basically my whole issue is i didnt turn that negative number positive

[mental plaza]

yep

for the first one

that wasn’t the prob

idk what u did

for the first one

because u got 1/x

but idk what u did with it

[late mango]

So i got it but didnt get it fully

Because of the -4/4

[mental plaza]

ye I think

u got confused

[late mango]

So do i reanswer it as -1/x =1/x

[mental plaza]

what

why is - = +

[late mango]

Because -4/4 is -1 so then it turns into 1 no?

Or does it stay negative

[mental plaza]

yes

turns into +1 when u take reciprocal

1/

[late mango]

Hmm

[mental plaza]

for the third

u didn’t need

to take

1/

[late mango]

How come

[mental plaza]

u didn’t have negative power

not needed

u already had

$x^{\frac43}$

[warm shale]

Springsskateboard

[mental plaza]

which is

what

[late mango]

1 something

[mental plaza]

no need

[late mango]

1 1/3

[mental plaza]

in terms of

roots

what does

the 3

in the denominator

tell u

if u wanna

express

it root form

is it square root

cube root

fourth root

[late mango]

So instead of what i got what would it have been

[mental plaza]

u tell me

[mental plaza]

it’s this

but

if u wanna express it

in root form

what would it be

it’s like

how u did

in ur part 3

2*

[late mango]

I got lost what

[mental plaza]

alr so

if I

give u

$x^{\frac12}$

[warm shale]

Springsskateboard

[mental plaza]

the 2

in the denominator

tells me

it’s a square root

ye?

so

I can write it

as

$\sqrt{x}$

[warm shale]

Springsskateboard

[late mango]

Which is 1

[mental plaza]

what

[late mango]

Because x by itself is 1

[mental plaza]

?no

wdym

[late mango]

And if you square x its 1/2

[mental plaza]

the power is 1 yes

[late mango]

Just to dumb it down for myself yea

[mental plaza]

so ye

for ur qn

$x^{\frac43}$

[warm shale]

Springsskateboard

[mental plaza]

what does

the 3 day

say

[late mango]

Lemmie rethink

[mental plaza]

alr alr

sure bro

take ur time

[late mango]

Yea im kinda struggling on this

Oh

That the numerator is greater than the denominator so it wouldnt need to be subtracted? Or sum like that

I remember my teacher telling me that and it clicked when i realized this

@mental plaza

So 1 wouldnt need to be over it

[mental plaza]

wassup

sorry I was

helping someone

else

alr

lemme read what u wrote

[mental plaza]

what

no

u heard wrong

[late mango]

What

Damn im being lied to

[mental plaza]

even when numerator > denom

you don’t have

to take 1:

1/

if it’s positive

[late mango]

You dont have to have 1 over at all then

[mental plaza]

unless

it’s negative

then yes

[late mango]

Unless its like my first question

[mental plaza]

you have to

[late mango]

Yea

Okay

If its negative then its 1 over

[mental plaza]

ye

so it becomes positive

[late mango]

And if its not then its written like x to the 4/3 power

[mental plaza]

yea

[late mango]

Alright that makes sense

[mental plaza]

nice

[late mango]

So in conclusion question 1 is -1/x

Question 2 is 1/x to 2/5

Question 3 is x to the 4/3

[mental plaza]

2 and 3

yea

1

no

where

is the negative

coming from

tho

[late mango]

Oh right becomes positive

My bad

But i got one more question

What if at the beginning

Both powers are the same number

On x

Do they just cancel out and become nothing

Just x?

@mental plaza

[mental plaza]

aye aye

wassup

[mental plaza]

wdym

like

$x^2 \cdot x^2$?

[warm shale]

Springsskateboard

[late mango]

Yea

[mental plaza]

no

u add the powers

use exponent rule

[late mango]

If both numer and denom are the same number after what happens

[mental plaza]

ah numerator and denom

so like

$\frac{x^2}{x^2}$

[warm shale]

Springsskateboard

[mental plaza]

yes

they cancel

and become 1

[late mango]

Because I was doing this self made eqution and uh

Lemmie just show

[mental plaza]

self made

o-O

[late mango]

Im just writing random numbers

And checking if im doing it rifht with new note

Notes

[mental plaza]

alr

so far

aye

aye

ye *

ur right

u have

0/5

in the power

right

so what is 0/5

tell me

what’s the value

[late mango]

Just 0

[mental plaza]

nice

so u have

$x^0$

[warm shale]

Springsskateboard

[mental plaza]

anything

ANYTHING

to the power 0

is?

[late mango]

0

[mental plaza]

anything except 0*

[mental plaza]

no

it’s 1

[late mango]

1

[mental plaza]

use ur calculator

I’ll see

[late mango]

I dont use a calculator thay often

[mental plaza]

ull*

but ye

[late mango]

Also that is what i did ask

[mental plaza]

anything

to the 0 power

is 1

[late mango]

If it jist becomes 1

[mental plaza]

yea

[late mango]

Alright thats cool

Yea i think thats everything i needed to know

[mental plaza]

nice

[late mango]

.close

[obtuse pebble]

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[hazy raft]

Coordinates of vertices of HYPERBOLa PLPLSPSLSLSPLSLSLSPSLS

[obtuse pebble]

@hazy raft Has your question been resolved?

[obtuse pebble]

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[mental mortar]

How would i start number 4

[misty tree]

Gauss’ formula is n(n+1)/2

To find the sum of the even numbers in a set, take the highest even divided by 2. Add up the numbers 1 to this new number, and multiply that by 2

So like adding up the evens 1-28, it’s 14•15/2•2=210 (sorry I multiplied wrong)

This works because you can just factor out a 2

Since 2n is always even, you don’t need to worry about subtracting 1 to make it even

It should be 2n(2n+1)

You know what

I think I did it wrong again

I’m gonna get a sheet of paper

[mental mortar]

Ok

[misty tree]

It’s n/2(n/2+1)

For b, we know that Gauss’s formula is n(n+1)/2

Substitute 2n+1

(2n+1)(2n+2)/2-(n/2)(n/2+1)

Fml

Since it is 2n, the right side should be n(n+1)

[mental mortar]

Wait

Why sub in 2n plus one

[misty tree]

I need to rethink my whole explanation right quick

[wild swallow]

,rccw

[warm shale]

Couldn't find an attached image in the last 10 messages.

Oct 14, 2022, 2:53 AM

[wild swallow]

ugh

,rccw

[violet sentinel]

wait, isn't the sum of just the odd integers just n^2 and the sum of even integers n(n+1)?

[misty tree]

Yeah I messed up

[violet sentinel]

ah I see! No worries

[obtuse pebble]

@mental mortar Has your question been resolved?

[misty tree]

No, it would be (x+1)^2

@violet sentinel

It is true that adding up odds gives squares, but it needs to be switched up

For example, 7+5+3+1=16

2x+1=7, so x=3

4^2=16, so (x+1)^2 is the sum of the numbers 1 to 2n+1

[obtuse pebble]

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[weak drum]

I’m pretty sure I did this wrong but I’m not sure where I messed up

[random depot]

lul

use ibp

[weak drum]

i’m p sure this is solvable without IBP

[random depot]

Then just simplify your way down

[weak drum]

it doesn’t look like you treated u(t) as the heaviside step function?

i might be misunderstanding

[random depot]

u = tau -1 , dv = e^-2tau

[weak drum]

still confused
how does this cover u(tau + 1)?

[weak drum]

like if you have this u and dv pair
what’s happening to the heaviside function?

oh, i see that you put everything inside the heaviside function in the last part
how does that work?

the integral doesn’t evaluate to a step function either btw

[obtuse pebble]

@weak drum Has your question been resolved?

[weak drum]

nvm my answer was correct (except for the -2t in the exponent for the t>=1 case
that should be 2t)

[obtuse pebble]

@weak drum Has your question been resolved?

[obtuse pebble]

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[worldly compass]

cup is also an angle

[worldly compass]

i have no work unfortunately i have no idea where to start or how to even get the answer

[bold salmon]

qcm as in q centimeters ?

Quite confusing question

[worldly compass]

it’s 9cm

[bold salmon]

Oh it’s a 9 my bad

[worldly compass]

no it’s fine

[bold salmon]

Square triangle or random tho ? Is that all the information given ?

[worldly compass]

all of the numbers have cm value

that’s all the information given

[bold salmon]

You could try using the cos rule but I’m not sure it would be the shortest way of doing it

[worldly compass]

i’ll try doing that

thanks

[obtuse pebble]

@worldly compass Has your question been resolved?

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[cunning spruce]

The bisector BD of the triangle ABC divides the side AC into segments AD = 6 cm and DC = 8 cm. Find the length of the bisector if the angle DBC = 120 degrees

[royal basin]

microsoft paint has a text tool. might help your diagram a bit better

first suggestion that comes to mind is to do some angle-chasing bullshit with the law of sines...

have you made any progress so far otherwise?

[cunning spruce]

Not really, I've tried using the ratio of adjacent sides to the bisector, but from that we only have a small ratio that BC = 8k and AB = 6k. Also, using the formula for the length of the bisector, find it with some values ​​and others to equate. Sorry for the inaccuracy of the translation, I use google translator. Now I'll throw off the photo from the notebook.

[royal basin]

what language are you translating from?

[cunning spruce]

from ukrainian

[royal basin]

right

can you send the original? i can probably figure out enough of it to understand what the problem is about

[cunning spruce]

original of task?

or my message

[royal basin]

oh, you're translating your messages through google translator... sorry

ok but definitely send the photo from the notebook you mentioned

[cunning spruce]

it will took about 5-8 minutes

I rewrite a little, because it was so mess

[obtuse pebble]

@cunning spruce Has your question been resolved?

[obtuse pebble]

@cunning spruce Has your question been resolved?

[obtuse pebble]

@cunning spruce Has your question been resolved?

[obtuse pebble]

@cunning spruce Has your question been resolved?

[cunning spruce]

<@&286206848099549185>

[pallid canyon]

from 28k^2 = 28 + 14 b, find b in terms of k, and then put the value of b back into one of the previous equations 64k^2 = ... or 36k^2 = ...

You'll get a quadratic equation with only k that you can solve

[obtuse pebble]

@cunning spruce Has your question been resolved?

[obtuse pebble]

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[earnest shell]

If the first urn has 2 blue balls and 8 red balls, the second urn has 6 blue balls and 4 red balls, and the third urn has 8 blue balls and 2 red balls. What is the probability of drawing 2 blue balls?

i solved this question

[earnest shell]

and i got 0.472

i did BBR + BRB + RBB =0.472

is it correct ?

[civic zealot]

how many balls are being drawn? and are they each being drawn randomly from (possibly) different urns? are they drawing it with replacement or without?

[earnest shell]

i think we have to draw from each urn one ball

so in total 3 balls

[civic zealot]

that looks right then

[earnest shell]

aight thank you so much !

.close

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[frank night]

hello

[frank night]

can someone explain if i have to prove ax^2+bx+c>=0 then why delta<=0

[rigid lintel]

delta being?

[frank night]

b^2-4ac

[rigid lintel]

ahh well you know what the value of delta means right

[frank night]

yes

[rigid lintel]

so if ax^2+bx+c>=0 what does that mean for the function

like if you draw it what is the lowest possible point

[frank night]

0

[rigid lintel]

exactly

[frank night]

oh delta is the minimum value..

[rigid lintel]

no

wait is it

[frank night]

it's actually -delta/4a

.close

[obtuse pebble]

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[dry flint]

hi

[timid silo]

for the first one, can you think of a linear combination that equals 0?

[dry flint]

they all have to equal 0

oh wait no

they just all equal each other

[timid silo]

right, if we denote the coefficients as x, we would just need to set x1 = x2 = x3 = 1 for example

[dry flint]

YEA

yea*

does that mean its dependendt?

[timid silo]

yes one counterexample is enough

[dry flint]

ahhh i see

what abt the second one

so i get v1 = -v2

v2 = -v3

v3 = -v1

can i just say the only soltuion is 0 lol or do i have to provei t

[timid silo]

wait i gotta think a bit about that one

[dry flint]

wait i got it

the only way x = -x is 0

.close

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[fierce egret]

I have this question "How many intersection points do you get if a line and a circle intersect?". Is this the answer "A line and a circle can interact in one of three ways: they may intersect in two points, they may intersect in one point, or they may not intersect at all." or do I have to solve something?

[ivory peak]

Yes that is accurate

[fierce egret]

Okay, thank you!

[ivory peak]

A line can intercept through two coordinates of a circle, one (a tangent), or none

[fierce egret]

Thanks a lot

[ivory peak]

np

[fierce egret]

I also have this question

What formula can I use for this question "In 3D space (XYZ), find the angle between an XY plane and the line passing through the points (0,0,0) and (1,1,1)."?

[ivory peak]

idk sorry

[fierce egret]

Fair enough

Thanks again

.close

[obtuse pebble]

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[ivory peak]

nvm

I misread

apologies

[dusk mauve]

solve the system of equations

x+y=60

x+16=y

where x is games lost and y is games won

so answer is c

[ivory peak]

answer is D

[dusk mauve]

sorry ans is D we were solving for y..

[unborn brook]

idk how to get that answer

[fierce egret]

Was this meant for me?

[dusk mauve]

no

[unborn brook]

.lose

.close

[obtuse pebble]

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[drowsy girder]

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b + c$$

[warm shale]

Pluton

[drowsy girder]

Prove

If $abc = 1$

[warm shale]

Pluton

[drowsy girder]

And they are all positive numbers

Seemed easy at first look but idk how to use abc = 1 to make it homougene or however you spell it.

[final thunder]

Seems like a job for AM-GM

[drowsy girder]

Yea

But still i dont see any obvious answer for this

I mean any obvious way to use AM-GM on this

[final thunder]

I’ll try something one sec

[wooden cipher]

yeah this does look like am-gm

maybe its titus

or maybe im just rusty

[drowsy girder]

I mean it must be am-gm its not cauchy-schwarz i never in my whole life used qm-am and i dont believe its am-hm

[final thunder]

I got it

Consider this $\frac{a}{\sqrt[3]{abc}} + \frac{b}{\sqrt[3]{abc}}+ \frac{c}{\sqrt[3]{abc}}$

Great

[warm shale]

Pure

[final thunder]

@drowsy girder

[wooden cipher]

oh damn

[drowsy girder]

Ye thats a + b + c

[final thunder]

Express it in such a way that’ll allow you to use AM-GM

[drowsy girder]

$$\frac{a + b + c}{\sqrt[3]{abc}} \geq 3$$

This prob not it

[warm shale]

Pluton

[drowsy girder]

I mean ik its not it

[final thunder]

No like from here

[drowsy girder]

Thats should ideally be RHS right

[final thunder]

Write this as 3 fractions still

[drowsy girder]

So ig $$\sqrt[3]{\frac{a^3}{abc}} + \sqrt[3]{\frac{b^3}{abc}} + \sqrt[3]{\frac{c^3}{abc}}$$

[warm shale]

Pluton

[drowsy girder]

Should be rhs

[final thunder]

Yes

And there’s a clever way to write this

So that we can exploit AM-GM

[drowsy girder]

Oh $$\sqrt[3]{\frac{a^2}{bc} \cdot 1} + ....$$

Ig

[final thunder]

It’s a bit more than that you basically want to multiply by 1

(hidden 1)

[warm shale]

Pluton

[final thunder]

[drowsy girder]

Oh abc?

[final thunder]

More like $$\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}} + ....$$

[warm shale]

Pure

[final thunder]

So from here i multiply it by b/b

[drowsy girder]

Oh and you get that
$$\frac{\frac{a}{b} + \frac{a}{b} + \frac{b}{c}}{3} \geq a$$

[warm shale]

Pluton

[drowsy girder]

Which when paired with others prob gives the original inequality

[final thunder]

Yeah

So

Doing the same for the other fractions yields, $$\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}} + \sqrt[3]{\frac{b}{c} \cdot \frac{b}{c} \cdot \frac{c}{a}} + \sqrt[3]{\frac{c}{a} \cdot \frac{c}{a} \cdot \frac{a}{b}}$$

[warm shale]

Pure

[final thunder]

Applying AM-GM here basically and you’re done

[drowsy girder]

Yeah thanks alot.

[final thunder]

[drowsy girder]

Have a nice day.

.close

[obtuse pebble]

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[grave dove]

I know what brackets and parenthesis denote, but what's weird is that a similar problem doesn't use them for a correct answer (that I know is certainly correct). It's a closed circle, which is defined, but do you not use brackets in this context?

[stark nebula]

You would use brackets here since it’s less than or EQUAL to -3

[grave dove]

Yep that's right. Thanks for the reminder lol. Silly me

[stark nebula]

np

[grave dove]

.close

[obtuse pebble]

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[junior hull]

im not sure if this is exactly the right channel for this but I was wondering if someone could help me understand finding the derivative of an inverse function?

[junior hull]

like for starters, how does it equal to that? isnt what they are saying on the left the same as the right so why is the right on 1/m when the left side says it is equal to m?

[obtuse pebble]

@junior hull Has your question been resolved?

[warm shale]

hR1487

[junior hull]

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[obtuse pebble]

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[obtuse pebble]

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[surreal grotto]

How do you write, $(\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n.....}}}}})$ in terms of series?

[timid silo]

$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n.....}}}}}$

[stark nebula]

Bots dead

[surreal grotto]

I don't know why the bot isn't working but i think you got the idea

[timid silo]

weird

a series? not a recursive sequence or something?

[surreal grotto]

No a series

[obtuse pebble]

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[winter geyser]

What does the word horizontally indicate here?

[winter geyser]

Like, I understand west of north, it's a triangle in the 2nd quadrant

But 4580m horizontally really trips me up

Confusion

[sour jasper]

You’re working in 3 dimensions, right? So horizontally means that if the summit were actually at ground level, then it’d be 4580 meters away from you. Put another way, the distance from the camp and the projection of the summit point onto the floor is 4580 meters.

So the hypotenuse of the triangle in the second quadrant is 4580 meters

[obtuse pebble]

@winter geyser Has your question been resolved?

[winter geyser]

Oh

I get what you're saying, conceptually

But still can't understand how the word horizontally fits there

Maybe there's a different meaning of the word i don't understand

Why not just say the diagonal distance from the summit to the camp is 4580 or something

rip

[sour jasper]

Well the diagonal distance could also be interpreted as the direct length of your displacement vector

Like you attached a zip line from the summit to the camp

The word horizontal makes sense to me because this distance happens to be the distance of the displacement component parallel to the horizon

Like imagine you standing to the side, facing both the camp and the mountain upon which the summit rests. From this viewpoint, pretty clearly you can see why 4580 meters is called a “horizontal” distance

[winter geyser]

Oooh

That makes more sense

I'm pretty new to the 3D plane

But I'm starting to understand what I'm actually doing here

[winter geyser]

And I just now actually understood what you said here

It's not a zip line from top to bottom, it's just if you looked at the area from above and treated it as a 2D plane for x and y

It's the hypotenuse for just the x and y, ignoring the z

In that case, does this seem correct to you?

[sour jasper]

I haven’t checked any of the arithmetic but you’ve plugged all the right numbers into the right functions

So looks good to me

[winter geyser]

Sounds good

Thank you so much!

[sour jasper]

No problem!

[winter geyser]

.close

[obtuse pebble]

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