help-28

[broken sage]

ooh thx!!

[torn jolt]

np

come with more questions help me get helpful

[broken sage]

will do when i get stuck at problems

<3.

[torn jolt]

aight close it for now

[broken sage]

i got one question though, if i multiply 1 row with the values of another row, does the value not change?

[torn jolt]

yes

multiply R1 by R3?

value of determinant changes in general

few cases where it remains the same

[broken sage]

so how does this work?

[torn jolt]

cases like all row elements are same , making it euivalent to multiplying with a scalar

[broken sage]

??

[torn jolt]

for this mb i thought you were asking other question

[broken sage]

?

[torn jolt]

let me latex

[broken sage]

um alr

[torn jolt]

eh what's the error

[
\begin{vmatrix}
1 & 1 & 1\
A^2 & B^2 & C^2\
BC & CA & AB
\end{vmatrix}=0
]

[
\text{Multiply } C_1, C_2, C_3 \text{ by } A, B, C \text{ respectively}
]

[
\Rightarrow
ABC
\begin{vmatrix}
A & B & C\
A^3 & B^3 & C^3\
ABC & ABC & ABC
\end{vmatrix}=0
]

[
\text{Factor } ABC \text{ from the third row}
]

[
\Rightarrow
A^2B^2C^2
\begin{vmatrix}
A & B & C\
A^3 & B^3 & C^3\
1 & 1 & 1
\end{vmatrix}=0
]

[
\text{Interchange rows}
]

[
\Rightarrow
A^2B^2C^2
\begin{vmatrix}
1 & 1 & 1\
A & B & C\
A^3 & B^3 & C^3
\end{vmatrix}=0
]

[

A^2B^2C^2(A-B)(B-C)(C-A)(A+B+C)=0
]

[
\Rightarrow A+B+C=0
]

[
(x-a)+(x-b)+(x-c)=0
]

[
\boxed{x=\dfrac{a+b+c}{3}}
]

[glossy valve]

firestepper

[broken sage]

ok, you didnt put comma so i misunderstood that you said to multiply one row by another

[torn jolt]

[broken sage]

thanxies!

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[torn jolt]

need help with this

so for the part a, its simple P(z) = (z-b)Q(z) + r > 0

for part b

we know that for P(x) divide by (x-b) with b > 0 is the upper bound if there is no negative signs on the Q(x) and r(x)

which is true in this case

so that means if we take any number greater than b

P(x) > 0

if P(x) = (x-a)Q(x) + r, where Q(x) has alternating values then
say Q(x) has n degree then
then if all the coefficient of Q(x) are alternating then there are two ways
first coefficient is negative then
P(x) < 0
first coefficient is positive then
P(x) > 0

right?

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[stuck fiber]

Would I lose marks if I wrote this as my answer?

Instead of this

Like would I lose marks for leaving out the x1 x2 x3 vector

[queen crater]

If you wrote x = ..., I'm sure you wouldn't

If not, depends on your teacher I guess

[white badge]

if you merely wrote the numbers I think you might have an issue with the rubric, but if you at least wrote x = [numbers], should be fine-ish?

best to just ask your teacher.

[stuck fiber]

Alright thanks!

❤️

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[west river]

Help me gng how do i solve derivative of a complex function when it is on power of 2

[delicate torrent]

,rccw

[white badge]

is that a 1 in the outer exponent?

[trail barn]

that's probably a prime for derivative

[west river]

This may help with understanding my question i dont speak english well

[neon basin]

ja govorim malo srpski

[white badge]

,rccw

[west river]

Pomazi kume

[west river]

Ako je broj u zagradi na kvadrat i onda sve to na prim sta se onda desi

[neon basin]

pa ovo je formula za odvod

$(x^n)' = n \cdot x^{n-1}$

[glossy valve]

USS-Enterprise

[west river]

E to

[neon basin]

I ovo je "chain rule" neznam kako se zove

[west river]

Hvala sve najbolje

[neon basin]

nema problema

tamo ti prvo treba odvod x^2, ali tvoj x je (x-2), tako da imaš 2*(x-2)^1, i zatim puta odvod (x-2), znači 1. Dakle 2*(x-2)*(x-2)' = 2*(x-2)*1 = 2*(x-2)

[onyx glen]

this is wrong

it should be 3x^2

[onyx glen]

odvod == derivative?

[full forum]

@west river Has your question been resolved?

[neon basin]

Yes

[onyx glen]

does puta just mean times

[neon basin]

Yeah

At least I think derivative is 'odvod'. I speak Slovene not Serbian, but I assume it's the same for derivative

Might be 'izvod'

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[cobalt atlas]

Hey guys I'm stuck right now

[cobalt atlas]

no calc

[void magnet]

How is sin(90°-80°) = sin(10°)?

Wait

You wrote sin(80°) = sin(10°)

[turbid badge]

do you know identities for complementary angles?

[cobalt atlas]

no

[void magnet]

How is sin(80°) = sin(90°- 80°)

[turbid badge]

sin(90°-x)=cosx
cos(90°-x)=sinx

[nova briar]

sin^2 is an even function

[cobalt atlas]

I was following along in the book and the book said to take 90 - theta

[cobalt atlas]

so It should be cos instead now ?

[turbid badge]

yes

[void magnet]

ik but read again what i wrote

[turbid badge]

also what sheesh is saying, sin80° =/= sin(90°-80°)

[nova briar]

Oh yeah, that's wrong

[cobalt atlas]

dude I am so confused what do I do with the square root

sin(90-x)=cosx

[onyx glen]

where are you seeing a root

[cobalt atlas]

sin^2(80)

[onyx glen]

that's just a square, not a root...

[cobalt atlas]

oh mb

[nova briar]

why don't you plug x=10 to this, and use an identity?

[cobalt atlas]

cos(80)

[nova briar]

what is cos(x) if x is 10?

[cobalt atlas]

cos(10)

I can't use a calc

[nova briar]

So can you continue from the beginning knowing this, and see where it leads you?

[cobalt atlas]

got it

I just dropped down the square it turned into 1 + 9 is 10

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[still current]

how cnai transform g(x) = ln 3(5-x)-4 into f(x) = log_c(b)(x-h)

[still current]

g(x) = log_e (3(5-x)) - 4

e^y = (3(5-x)) - 4

[delicate torrent]

$g(x) = \ln(3(5-x)) - 4$ => $f(x) = \log_c(b)(x - h)$

[glossy valve]

1 divided by 0 equals Infinity

[still current]

yeah

[delicate torrent]

hold on

the -4 was in ln or not?

[still current]

nah its just -4 at the end

[delicate torrent]

then how do you have this?

[still current]

i did ln = log_e

then e^y left side

[delicate torrent]

yeah

but since the -4 was outside the ln

then you're wrong

that only happens when the -4 is inside the ln

[still current]

not gonna lie i dont know if its inside or utside

but this is the equation

g(x) = ln 3(5 - x) - 4

[white badge]

do you not have an image of the original equation?

[white badge]

gods, this is another badly-written question...

[still current]

e^y = 3(5-x) - 4 i meant

[frank elm]

Is 3 the base of log?

[still current]

yeah

[delicate torrent]

e is the base?

[still current]

e is the base

log_e (3(5-x)) - 4

[frank elm]

U r contradicting urself

[frank elm]

Okay

[still current]

what now?

[white badge]

it would probably help to convert that 4 into a ln as well so that you can do something with the two lns together.

[still current]

i thought about that but i didnt know how

[white badge]

$\log_b(b) = 1$ for any base $b$, so if you want 4...

[glossy valve]

Nicole

[still current]

u gotta do + 3 then

[white badge]

well that'd be shifting the problem to the + 3.

[white badge]

think about how you can get a 4 from the equation I mentioned in here.

[still current]

wait should i erase what i wrote earlier

[still current]

log_16(2)

[white badge]

well that is wrong for two reasons.

log_16(2) = 1/4 being one of them, and the other is that in my equation I had log_b(b) and no numbers.

consider just log_b(b) = 1. how can you get a 4 on the right without addition?

[still current]

sorry im back

u can multiply by 4

@white badge

[white badge]

there we go.

so what do you get?

[still current]

wait we multiplied by 4 the right side

[white badge]

well I assume you know you need to multiply both sides of an equation by the same number, right...?

[still current]

4(e^y) = 4(3(5-x) -4)

[white badge]

I'm pretty sure you'd end up needing division here.

subtraction rule.

but I'm about to head off to bed, so I will defer to future helpers.

[still current]

alright thanks

.close

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[still current]

how can i solve the equation 7^x = 8^(x-5)

[void nova]

Do you have access to logarithms?

[dry arch]

try taking the log of both sides

[still current]

ok

log (7^x) = log (8^(x+5))

[dry arch]

now you can use log(a^b) = b log(a)

[still current]

xlog (7) = (x-5)log (8)

[void nova]

xlog(7) = **(x - 5)**log(8)

[still current]

then what

[dry arch]

collect like terms

[still current]

log 8 / log 10

[dry arch]

put together the terms with x

first step: distribute log 8

but first, i think your - changed to +

so you should fix that

[still current]

how do i distribute the log 8

[dry arch]

(x-5) log 8 = x log 8 - 5 log 8

[still current]

ohh

ok ok

[void nova]

log(8) and log(7) are normal numbers...

[still current]

so now its log (8/5)

if im not mistaken

[void nova]

"It" refering to what?

[still current]

the substraction

wait nvm

[dry arch]

no, you should collect like terms

so move the x log 8 to the other side of the equation where the x log 7 is

[still current]

x log 7 - x log 8

[dry arch]

yeah and factor out x

we are solving a linear equation in x now

like ax+b = 0

[still current]

x (log 7 - log 8)

[dry arch]

just one more step

|| divide by || || log 7 - log 8 ||

[still current]

x = (-5 log 8)/(log 7 - log 8)

[dry arch]

close

yeah

looks good

[still current]

oh alright tysm

[dry arch]

you're welcome

[still current]

ima do 4 more of these

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[sharp vine]

It looks like basic statistics, i.e. mean, median and mode

https://www.youtube.com/watch?v=A1mQ9kD-i9I

This one should be helpful

Obviously just to get familiar with the topic. I see there that your teacher also solved some text exercises, but they're based on the basic knowledge

I hope you can understand it on your own then, good luck

However if you get stuck, just come back and ask a particular question

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[full forum]

Please don't occupy multiple help channels.

[nimble ridge]

<@&268886789983436800> need full ban probs

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[civic dust]

I am working on this problem, and Im trying to figure out why I am getting different answer to a and I am trying to understand what b wants from me

[civic dust]

Here is the working

And appearantly the answer is wrong, it should be

[clear root]

You've already replaced x and y in the second line

so what you need is y = ...

[civic dust]

Yes that's how you find the inverse of function

[last gazelle]

I think the algebra mistake is here

[civic dust]

Switching the places temporarily

[last gazelle]

you skipped some steps in your head

assuming that that symbol is a y??

[civic dust]

Yes

It's x and y simple function

[clear root]

yes but you then didn't find y = f^{-1}(x) you had x = some function of y and then some arbitrary algebra to get that result

[civic dust]

What am I doing wrong

[clear root]

1+5/y = x is not the same as
y = 1+5/x

[civic dust]

Yeah that's the issue, I cant figure it out myself

Where the mistake is

[void nova]

There

[clear root]

how'd you go to the next line from there

[civic dust]

I’m stuck in this step

[clear root]

expand

and then factor all the terms with y

[civic dust]

But this will give me additionall work

[void nova]

(Also please please please do not write your y like that, it's completely unreadable)

[clear root]

and a correct answer

what's your priority?

I don't think this is supposed to be here

[civic dust]

Oh yes I need LOGAN PAUL TO HELP ME

I NEED HIS FUCKING COINS!!!!!!!!!!!!!!!!!1

[void nova]

<@&268886789983436800>

Here we go again...

[civic dust]

I think Logan paul himself wanted to help me

With 5th grade algebra

Am I brain dead at this point?

[clear root]

what's x(y+5)

[civic dust]

But I expanded this

And it gave me the same result

[clear root]

yeah I'm asking

What would you get if you expand it

[civic dust]

x*y+5=y

[clear root]

that's wrong

[civic dust]

Oh shit right

Yeah

[clear root]

why do you think they've written (y+5) and not xy+5 if they're the same

you have to distribute x to both of those terms

[civic dust]

Yeah my mistake

Thank you

Im still getting wrong answer

That's why I hate studying math

[clear root]

show work

[civic dust]

Wait I am redoing it since it was a mess

Stuck same

Again

[flint hull]

dont do the last two steps

also its kinda hard to keep track i would recommend using y instead of f(x), and you can switch it back to f at the end when you write your answer

[civic dust]

I switch it in the beginning and the switch it back

The first 2 steps just signalize I switched the variables

[lost topaz]

so what are you trying to do here again?

[civic dust]

Find the inverse of a function f

And im getting beaten by algebra from 5th grade

[lost topaz]

go to that step,

[civic dust]

I am

[lost topaz]

now i want you to put all y's to the left

[civic dust]

I can't

[lost topaz]

you can

swap signs

[civic dust]

If I divide the whole equation by y then 5x will be divided by y

[lost topaz]

i dont ask you to multiplu

[civic dust]

and it stays

[lost topaz]

or divide

[civic dust]

ok?

[lost topaz]

just make it so everything that has y is on the same side

which gives you?

[civic dust]

5 = y - xy

Wait what

Wait

[lost topaz]

5x

[civic dust]

5x yes

[lost topaz]

now you see it?

5x = y - xy

[civic dust]

Nope

[lost topaz]

factor the y out

whats left

[civic dust]

5x/y = -x

[lost topaz]

no, factor the y out like (xy + 5y) = y(x + 5)

from the right side

5x = y - xy

[civic dust]

oh my god thank you

I literally can't believe

Im so bad at algebra

[lost topaz]

its ok man, we all have our goods and bads

just practice

practioce beats talent if talent doesnt practice

[civic dust]

Yeah but the issue is that I will be writing my college exam in 2 years and I am getting beaten up by math from 5th grade

I know

[lost topaz]

you can do it

[civic dust]

I hope

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[spice grail]

can someone help me integrate this using trig sub

[prime pier]

good so far. $\sec(\theta) = \frac{1}{\cos(\theta)},$ so you can use the chain rule to differentiate, then simplify

[glossy valve]

ηασιβ ♥

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[fallow ridge]

I understand the solution but wouldn't directly prime factorizing 123420 number be more efficent?

[rare dock]

what would that mean

[fallow ridge]

instead of going through each time and checking is the lefted number is divisble by which

[fallow ridge]

ah what do you mean

[rare dock]

describe to me how you would factor it ‘directly’

[fallow ridge]

because it would be more faster

[rare dock]

what would be faster

describe it

[buoyant ravine]

but what would be your algorithm?

[fallow ridge]

it would be more efficent and faster to get it's all prime factors

[buoyant ravine]

what is "it"

[rare dock]

how are you going to get the prime factors

[fallow ridge]

By prime factorization

[buoyant ravine]

this is prime factorization though

[spiral vigil]

are you suggesting to skip the step with the 10s?

[rare dock]

[fallow ridge]

hmmm

[fallow ridge]

Yeah well

but

[fallow ridge]

That was smart i think?

[buoyant ravine]

so like if i were to be like "prime factorize 123420", how would you do it?

[fallow ridge]

i would do it by creating the table. hmmm Ok i understand. He did the same thing but just went through each step

Alr

Thanks for help

everyone

[buoyant ravine]

np :)

[rare dock]

[fallow ridge]

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[rare dock]

bro avoided providing an algorithm to compare to like the plague

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[grim wigeon]

Does anyone know how to get permissions to send messages in the combinatorial structures channel?

[lime trellis]

get the undergraduate role

in #info

[grim wigeon]

Appreciate it lil bro

[warped frost]

anything else?

[grim wigeon]

Yeah

[warped frost]

sure, ask ahead and I'll pin it for ya

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[little nexus]

napıonuz la zekı ınsanlar

[little nexus]

yanı kısmen

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[warm mango]

May someone assist me with this ?

[full forum]

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

[warm mango]

a mix of 1 and 2

[knotty grail]

m<1 is the obtuse angle here?

should be yeah

[warm mango]

Yes

[warm mango]

It is obtuse yeah

yo ?

[knotty grail]

are you aware of exterior angle property?

[warm mango]

Yes I'm pretty sure

Sorry for late responses I didn't see your message

[crude granite]

it is very simple problem

[knotty grail]

use that in this triangle

observe how 1 is an exterior angle to the triangle in this image

[crude granite]

angle 1 is 100

[warm mango]

180 - 50 - 50 =80?

[vocal flax]

But you gotta help him
how to solve tho

[knotty grail]

thats the missing angle inside the triangle

[warm mango]

Kinda the best I got

[crude granite]

so answer is H and A

[knotty grail]

1 is outside the triangle

[warm mango]

180 - 80 = 100 or sum?

[warm mango]

Erm

[crude granite]

understand?

[vocal flax]

Can’t give answers like that 😭

[crude granite]

if you don't understand, I can explain

[warm mango]

Yes and no ish

[crude granite]

what do you mean?

[crude granite]

don't send msg to me

[knotty grail]

notice how 80 deg and angle 1 are supplementary here

you involuntarily used angle sum property to find the missing third angle inside the triangle

[warm mango]

Alright uhh

[crude granite]

so understand everything?

[warm mango]

I would say I've got a good understanding of it

[crude granite]

good

and anymore problem?

[warm mango]

I have a few more but I'll go in another slot unless you wanna continue this one

[crude granite]

no continue

I can help you.

tell me your problem.

[knotty grail]

no need
continue here

[warm mango]

uhhh alrighty

I think I got the 3. but could we go over both?

[crude granite]

angle FOH + angle HOG = 180 degree

but angle FOH is 105 degree

so can you solve the problem?

[warm mango]

I'll do my best one second

[crude granite]

ok

so answer is ?

[warm mango]

My answer is 75°

[crude granite]

good

right

[warm mango]

Alright

[crude granite]

how old are you?

[warm mango]

15 says it in my bio thingy

[crude granite]

are you american?

[warm mango]

I think so 😱

Actually no I'm from Bangladash

😱😱

[crude granite]

ok

and next problem

[warm mango]

yh I'm american

Alright I'll show a photo

[crude granite]

are you boy?

[warm mango]

Male yes

[crude granite]

so is this a problem?

[warm mango]

Yh honestly I don't think so

Aren't the answers like right there ?

[crude granite]

what?

it is really simple problem.

[warm mango]

Oh because like they're vertical angles, they're the same or no

[knotty grail]

vertically opposite yes

[warm mango]

ah

[knotty grail]

nvm vertical angles it is

seems many call it vertical

[crude granite]

any other problems?

[warm mango]

Yes

[crude granite]

tell

[crude granite]

so don't you solve this problem?

[warm mango]

Not yet that's why I asked

[crude granite]

the property of isosceles triangle, yz = zw and xz is vertical as yw and then xyw is isosceles triangle

so answer is?

[crude granite]

hey you there?

[warm mango]

Yes sorry I was working on it but my answer for the first one is

XW = 7.5

And for the second I got

BF = 4.2

[crude granite]

yeah you are great

[warm mango]

Alright thank you I've just got one more page left I need help with if you don't mind helping

[crude granite]

ok

[crude granite]

so the biggest angle's opposite side is the largest

[warm mango]

Alright

[crude granite]

so solved?

[warm mango]

I'm pretty clueless here

[crude granite]

so?

what can I help you?

[warm mango]

nvm I think I got it

Alright I'm done for tonight thanks for the help from both the guy from earlier and you

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[spiral hamlet]

This has an error. No?

[devout valley]

,rccw

[devout valley]

Yeah if they mean the second year

[spiral hamlet]

I was thinking that since the initial was 20, the t should be 2 because it implies after the second year

[delicate torrent]

how about letting the increase ratio be $r$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

then $20r^2 = 25$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

with the condition that $r > 0$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

and solve for $r$

[glossy valve]

1 divided by 0 equals Infinity

[devout valley]

Regardless, I understand OP is asking about the options being given (and gather the impression they know how to do the question otherwise), as none of the options given are correct

[delicate torrent]

poor OP\

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[frank vapor]

Uhm

[frank vapor]

I need help again

I cant solve number 3 idk what im doing wrong

[zinc yacht]

Factor out 2/3 from the last given equation and see what's left

[frank vapor]

Wdym

[zinc yacht]

OP=2/3(OA+(1/2) OC)
(obviously all of these are vectors )

[frank vapor]

Im finna cry

I don’t get it

..

[zinc yacht]

I am just taking 2/3 common from right hand side

[frank vapor]

I have my midterm exam tomorrow

I don’t understand linear combination vectors

And order in R sys of inequalities

[frank vapor]

Can u

Show me on like

A paper

A note

Anything

[zinc yacht]

Hold on

[frank vapor]

Okay

[zinc yacht]

The RHS should seem familiar then

Do you recognise the RHS term?

[frank vapor]

No.

[zinc yacht]

Could you do the earlier sub parts of the question?

[frank vapor]

U mean

Number one and 2

I did them

And I found difficulties in number 2

I did number one easily

[zinc yacht]

Did you get to the answer of 2?

[frank vapor]

Oh wait

wtf

P is given

I need to get collinear points

…

Idk wtf im doing

I thought I should get p

[frank vapor]

Yes

I can take

OP and PQ

Get coordinated of it

And do determinant

And prove it 0

To prove collinear

[zinc yacht]

I mean there's a much shorter way

[frank vapor]

What is it

[outer nova]

AB = kAC => A, B, C are collinear
in this case OP = (2/3)OQ

[frank vapor]

Oh

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[patent vapor]

How do i graph with e

[white badge]

if you replace that e with another number, say, 4, can you graph it?

[patent vapor]

Wait i can just put this into a calculator

[white badge]

or that, yeah, I suppose.

[patent vapor]

Ok thank you

.close

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[spring oak]

∑(n=1 → 128) (-1)

[minor crater]

[minor crater]

do you have a question to ask

don't troll in help channels, please

[white badge]

how is this diagram related to your first message...?

[full forum]

@spring oak Has your question been resolved?

[delicate torrent]

!xy

[full forum]

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

[delicate torrent]

gửi câu đi còn triệu hồi đại ca

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[patent vapor]

1-4 need some help

@wary condor

[remote gale]

!status

[full forum]

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

[patent vapor]

1

Somebody help

[outer nova]

take a few random x's and see if the result grows or shrinks ig

[remote gale]

Do you know what do the mean?

[void nova]

$y = a^x$ is INCREASING if $a > 1$

[remote gale]

Xd

[patent vapor]

English please

[remote gale]

Yea yea do you know what is the question?

[glossy valve]

Alberto Z.

[void nova]

We are writing in English!

[patent vapor]

What if its 1

[void nova]

It's a constant

[patent vapor]

If its less its decreasing

[remote gale]

Its its constant, not its a constant

[remote gale]

Yea

[patent vapor]

Ok thank you

Quiz time

[remote gale]

Actually, no, only if its bigger than 0

[void nova]

$1^x = 1$ for all values of $x$

[glossy valve]

Alberto Z.

[patent vapor]

What does that mean

[remote gale]

Which part you dont understand

[patent vapor]

When is it decrease or increase

[remote gale]

There is the expression $a^x$

[glossy valve]

Roy

[patent vapor]

Please faster

[remote gale]

Lets first pretend x > 1

[patent vapor]

I have 1 minute

[remote gale]

If so, then if a > 1

Then its increasing

[patent vapor]

Ok thanks

[remote gale]

A<1 decreasing

A=1: constant

A=0 also constant

a<0 is not defined like this

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[cedar sapphire]

Help with this number 1 as I wasn’t there for it. The slope and secant line equation

[remote gale]

,rccw

[torn jolt]

what do you understand by secant line of a function

[full forum]

@cedar sapphire Has your question been resolved?

[hot herald]

pretty much the same as queston 2

first get f(-1) and f(3)
to get your two points

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[cedar sapphire]

It’s opposite of cos

[onyx glen]

secant function in trig ≠ secant line of a function

[hot herald]

opposite also a bad word

[cedar sapphire]

Did I correct some of it so far sorry it’s messy

[sonic stratus]

Try to take the photo such that we can see the original f(x) and any other context to the problem

[cedar sapphire]

Number 1 by the way

[torn jolt]

this is what you call secant

[hot herald]

that info gives you one of the points you want
now calculate f(3)

[torn jolt]

secant line to a function/curve is the line which cuts the it at two or more points

yes as omeganato says you need f(3) and f(-1)

then you can find the slope

[sonic stratus]

@cedar sapphire your first problem is finding the two points of interest.

aka, ⁨-1,f(-1)⁩ and ⁨3,f(3)⁩

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[ivory grail]

Hello! Im having an issue with computer science about asymptotic analysis, especially in time complexity. I am kind of confused about the reason why substitution method especially for big O notation. Here is my question

Given that the machine independent time is: T(n) = 2T(n/2) + n

we educated guess that T(n) = O(n log n)

I understand how to deduce this utilizing mathematical induction step by step, so I am just trying to grasp the concept on reason why one uses the assumption of T(m) where m < n, aka given some value of c > 0 and n_0 > 0, prove that T(n) <= c(n log n) where m < n.

let m = n / 2 (so that it always satisfies m < n)

I thought the implication should be like:
T(n) ==> T(n / 2)

and not T(n / 2) ==> T(n)?

[ivory grail]

for context I assumed we are proving that it is true still for a smaller amount of input, instead of a bigger one? Im just a bit confused

[full forum]

@ivory grail Has your question been resolved?

[ivory grail]

I was also researching a bit, especially the big omega notation, Ω, it seems it uses the assumption of T(m) where m < n as well for induction(except it shall follow the definition of Ω instead of big O), is my initial idea wrong about the logic on how they prove it?

namely, this is my thought process:

1. u first prove base case P(n_0) for some c_0, such that c >= c_0 > 0

2. then u utilize this fact and assume and instantiate T(m), m < n so that it is true, and prove T(n) from this. this is the part Im unsure, why does it work that when T(m) = O(g(n)) where m < n ==> T(n) = O(g(n))

proves that T(n) = O(g(n)) for all n > n_0?

[full forum]

@ivory grail Has your question been resolved?

[full forum]

@ivory grail Has your question been resolved?

[full forum]

@ivory grail Has your question been resolved?

[ivory grail]

alright, after studying for a tad bit longer, I kind of understand this now, writing it here for better understanding:

1. we start by proving P(n_0)

2. we then prove the implication (k < g ∧ P(k)) --> P(g)

and to understand this, I try to substitute k for any value g / b where b > 1, then it must be true that g / b < g, hence:

P(g / b) --> P(b * g)

and since b can be any arbitrary value greater than 1, for any initial value of k that satisfies P(k), we can always find a value b > 1 such that g = bk (since g > k), which further proves for any number larger than any given k that is true, all input sizes are also proven as well

so hence P(n) is true for all n > n_0

closing this naw :P

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[robust slate]

In the future, please use the ,rotate command.

,rotate

[robust slate]

Your use of $a_n=ar^{n-1}$ could be relevant. You have the fourth and sixth terms, so consider setting $n=4$ and $n=6$.

[glossy valve]

Civil Service Pigeon

[robust slate]

Further hint: ||Divide the equations||

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[slate knot]

sorry abt that

[slate knot]

ty

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[robust slate]

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[grim remnant]

I was in class and my professor gave this equation.

I am stuck on part B. My professor didn't show how to get the answer: $855. However, I understand the previous steps

[rare dock]

pretty handwriting

[remote gale]

yea

[delicate torrent]

your professor really wants you to use a variable for this?

[grim remnant]

Yes 😭

[delicate torrent]

he's crazy

[grim remnant]

I am unsure how he got $855 as the answer

[delicate torrent]

agreed

[hidden belfry]

ITS THIS GUYYYYYYY

[remote gale]

bruh

[hidden belfry]

Also why do u need a varaible for this problem?

[delicate torrent]

Oh my god

[remote gale]

xD

[hidden belfry]

ye

we all here

so is the problem resolved?

[remote gale]

prob

[bold summit]

They're typing. So I don't think it's solved

[grim remnant]

That's a very good question. He was using the 1 step method for percent, algebra

[warped frost]

I don't think anyone addressed this question

[remote gale]

xD

[grim remnant]

Its not solved yet. I don't know how my professor got $855 as the answer for part B

[remote gale]

uhuh

[hidden belfry]

oh

whatd u get?

also btw is that a 1525 or 1625?

[grim remnant]

It's 1525, i accidentally wrote it wrong

[remote gale]

ok listen carefully

the truck is 1525

the gas is 500

you have 3000-1525-500=975 left

[hidden belfry]

no u gotta account for boxes

[remote gale]

boxes are 1.5 per box

stop

lemme talk

so 1.5 per box

[hidden belfry]

oh k

[remote gale]

with 80 boxes its 80 * 1.5 = 120

another 120 from the 975 is 975-120=855 left.

@grim remnant

[delicate torrent]

Ay he's cookin

[grim remnant]

Oh? Holy shit. For real?

Okay this makes sense

[delicate torrent]

-# ||calc guy can help a primary school question but can't help an algebra question||

[bold summit]

Can it also be done as 3,000-2,145? 🧐. Wouldn't that get the same answer too or is that an incorrect way?

[remote gale]

yea

[remote gale]

-# 🔪

[grim remnant]

Well thank you. I will remember this for my exam trmw

[delicate torrent]

Same way tho

[grim remnant]

.closed

How I do this?

[delicate torrent]

Remove the d

[grim remnant]

Oh 😭

[delicate torrent]

!done

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

[grim remnant]

.close

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[river relic]

why is it 33000-33000(1.04)^n/1-1.04 instead of just 33000(1.04)^n/1-1.04

[tacit siren]

$\sum_{i = 0}^{n-1} ar^i = a \left(\frac{1 - r^n}{1 - r}\right)$

[glossy valve]

phoenixperson

[tacit siren]

@river relic

if you want I can help you derive the formula

it's p ez

[full forum]

@river relic Has your question been resolved?

[river relic]

I know the formula

I just dont know where you get another 33000

[tacit siren]

well if you use the formula, you see that you distribute out the $a = 33000$ onto 1 and $r^n$ in the numerator when you multiply

[glossy valve]

phoenixperson

[river relic]

Ohhhh

I didnt see that you multiply

I get it now

Ty

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[agile lion]

how do you go about finding the inverse of this

[onyx glen]

same as with any other function

[knotty grail]

replace f(x) with x and vice versa
now find an expression for this f(x)

$x = \frac{-e^{f(x)}+1}{e^{-f(x)}+1}$

[agile lion]

oh ok

so wait

lemme try smth

[onyx glen]

easier if you write y instead of f(x)

[knotty grail]

okay since this looks kind of intimidating maybe first compress the huge mess by $e^{f(x)} = t$

[onyx glen]

unless you bloated it on purpose

[knotty grail]

true

[agile lion]

would i need to use natural log for the e's

[knotty grail]

$x = \frac{-e^{y}+1}{e^{-y}+1}$

[knotty grail]

i mean after solving for t yes

[agile lion]

eulers number and the exponential is whats throwing me off

[knotty grail]

taking $e^y=t$ gives $x = \frac{-t+1}{\frac{1}{t}+1}$

[agile lion]

i know this sounds stupid but wheres t coming from

or are we just assigning a new variable

[knotty grail]

new variable for simplification purposes yes

[agile lion]

ah ok

[knotty grail]

seems to give a quadratic in t

[agile lion]

so we'd get 2 answers yes?

[slate lily]

No, one answer

[agile lion]

im still working it out

[slate lily]

Since there are exponentials involved, most likely you will have to take logs

Since e^y must be positive, we generally look for the branch that satisfies the original function's range.

[agile lion]

do the 1's cancel out

can i just ignore them

[slate lily]

You can't make that assumption

[agile lion]

i give up

.close

[full forum]

Channel closed

Closed by @agile lion

Use .reopen if this was a mistake.

[slate lily]

what why?

Don't give up so easily

[agile lion]

ive been stuck with this question since yesterday

i tried going to a tutor but they didnt help one bit

[slate lily]

Would you find it easier if I worked it out and posted the workings here?

[agile lion]

i feel so stupid for not getting it

[slate lily]

Then we can go through it together if you don't understand some parts?

[agile lion]

sure

because ive tried alot of things, went through my textbook, looked online found nothing

[slate lily]

Ok, give me a minute to write it out

[agile lion]

my biggest weakness is with doing stuff with rational/irrational functions not sure why

[slate lily]

First, Multiply the numerator and denominator by $e^x$ to eliminate the negative exponent:$$f(x) = \frac{-e^x + 1}{e^{-x} + 1} \cdot \frac{e^x}{e^x} $$$$ f(x) = \frac{e^x(1 - e^x)}{1 + e^x}$$

Next, to find any inverse we swap the x and y. So: $$x = \frac{e^y(1 - e^y)}{1 + e^y}$$

Now, we isolate the $e^y$ terms. We set $u = e^y$ to make the algebra easier to see:
Substitute $u$:$x = \frac{u(1 - u)}{1 + u}$

Clear the fraction: $$x(1 + u) = u - u^2x + xu = u - u^2$$

Rearrange into a quadratic form in terms of $u$: $$u^2 + (x - 1)u + x = 0$$

Now we solve for $u$ using the quadratic formula:$$u = \frac{-(x-1) \pm \sqrt{(x-1)^2 - 4(1)(x)}}{2(1)} $$$$ u = \frac{1-x \pm \sqrt{x^2 - 2x + 1 - 4x}}{2} $$$$u = \frac{1-x \pm \sqrt{x^2 - 6x + 1}}{2}$$

Recall that $u = e^y$. To solve for $y$, we take the logarithm of both sides. $$y = \ln\left( \frac{1-x \pm \sqrt{x^2 - 6x + 1}}{2} \right)$$ Hence, the inverse is $$f^{-1}(x) = \ln\left( \frac{1-x + \sqrt{x^2 - 6x + 1}}{2} \right)$$

[glossy valve]

Ajay

[slate lily]

Have a look through and ask me anything

[agile lion]

mk gimme a moment

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[opal wolf]

Why is (t_m) and (u_m ) monotone

[queen crater]

(t_m) is increasing, (u_m) is decreasing

[split hatch]

t_m is increasing because if A ⊃ B then inf A ≤ inf B

[opal wolf]

This sequence is oscillatory , so won't the infimum will have same value for all m and similarly supremum

[queen crater]

Not necessarily, no

Plus, as I said in your previous channel, increasing and decreasing include staying constant

[full forum]

@opal wolf Has your question been resolved?

Channel closed

Closed by @opal wolf

Use .reopen if this was a mistake.

[opal wolf]

Since the sequence is monotone...

[full forum]

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[topaz kelp]

ok so im practicing geometry proofs right

[topaz kelp]

and its a pretty classic one

[paper nebula]

why why are you here

[topaz kelp]

so what im trying to do is label the intersection of the two external angle bisectors as D

i need help bro

[paper nebula]

you literally do aops rbo 😭😭😭

anyways

[topaz kelp]

and then

prove that AD is the angle bisector of angle A

[topaz kelp]

bro im so geeked in geometry

[brittle steeple]

This proof is kinda random and hard to guess at

[rare arch]

but not in geometry dash ???? bro this game is so cool

[brittle steeple]

Do you know how to show the interior angle bisectors are concurrent?

[paper nebula]

<CBD = 90 - <ABC/2

[topaz kelp]

uh

not really

[paper nebula]

<BCD = 90 - <ACB/2

[topaz kelp]

yeah ive labelled all of that already

[paper nebula]

solve for <BDC

js go angle hunting gng

[topaz kelp]

wait

but

how does getting BDC help me

[paper nebula]

ok nvm im dumb

but go angle hunting

<ABD = ?

[topaz kelp]

90 + <ABC/2

right?

[paper nebula]

yeah

same for <ACD

[brittle steeple]

I'd give some names

[paper nebula]

yeah

why named D as the intersection of angle bisectors

[brittle steeple]

A = 2α, B = 2β, C = 2γ

[topaz kelp]

ok

[paper nebula]

uh since im too lazy to type greek

i'll just call A = 2a, B = 2b, C = 2c

so <ABD = 90 + b, <ACD = 90 + c

wait

<ADB and <ADC?

[brittle steeple]

Do they sell mathematicians keyboards

You'd need like a Greek key that functions like a shift key

idk where some letters like theta/phi/psi/etc would go but maybe if I spoke greek I'd know

[topaz kelp]

øk

[paper nebula]

wait maybe label <ADB = x and <ADC = y

and then show that <BAD = <CAD

by finding an algebraic equation

[topaz kelp]

lowk dont think theres enough information

[paper nebula]

nah

then why would they ask ts

rbo 😭😭😭

[topaz kelp]

no like

i mean

not enough information to solve for the angles algebraically

[paper nebula]

law of sines?

[topaz kelp]

theres probably a geometric way

[paper nebula]

lowk

wait

[brittle steeple]

well

[paper nebula]

draw angle bisectors of <ABC and <ACB

those are concurrent

[topaz kelp]

yeah

[paper nebula]

and then maybe right triangles?

[brittle steeple]

you can "solve" for D right

you have angle side angle

[topaz kelp]

i mean

i can like

kind of do that?

i drew the altitudes

from D to BC

and AB and AC

[brittle steeple]

There's a dumb sketch I have

[brittle steeple]

Let's consider our point to definitely be the angle bisector of the exterior angles

Draw a circle centered at our point tangent to BC

can we show the circle is tangent to the other two sides?

[topaz kelp]

i mean

isnt that like

the definition of the excenter

so

yeah

we can

[topaz kelp]

yeah

thats basically what i did

[paper nebula]

blo how do you prove it's tangent 😭😭😭

wait im dumb

definition 🤦‍♂️🤦‍♂️🤦‍♂️

[topaz kelp]

???

ok

so

basically

uh

lets say

we draw perpendiculars from D to BC, AB, and AC

let the points they meet at be E, F and G respectively

tehn

by AAA similarity

because of the external angle bisectors

FBD is similar to BDE

but since they also share a hypotenuse

theyre congruent

so DF = DE

[paper nebula]

thats the excircle triangles

right

[topaz kelp]

and by the same logic

[paper nebula]

lololol

[topaz kelp]

DG = DE

and so

those perpendiculars are all the same length

and so

[brittle steeple]

Oh yeah this is a nice way to make the circle thing tangible

This should go through then

[topaz kelp]

they're the radii of a circle centered at D

[topaz kelp]

and the circle is tangent to AB AC and BC

ok

now

how do i get that its the bisector of <A

[paper nebula]

FBD, EBD, and ECD, GCD are all congruent by HL

or wait

[brittle steeple]

Same triangles this time wrt to A instead of B and C

like

same kind of triangles when you just look at the bisected angle and circle and ignore the irrelevant stuff

[topaz kelp]

soo...

AFD and AGD?

which triangles?

[paper nebula]

AFD is cong to AGD lolol

by HL

[topaz kelp]

bruhhhh

i forgot they share a side 💔

[paper nebula]

and then the result follows

[topaz kelp]

i was only looking at the right angles

and

DF and DG

and i was like

"hmm thats only an angle and a side not enough to prove congruency"

ok

ty for the help❤️

[brittle steeple]

It's not hard to imagine how the proof for 3 interior angle bisectors occurs now probably

[paper nebula]

the place where they intersect is known as the incenter

but the proof is trivial by incenter proof 😭😭😭

[brittle steeple]

they just intersect at the incenter (center of inscribed circle), though since proving a priori you can inscribe a circle in any 3 lines is hard, you probably make the proof go through using the same perpendicular line thing

[brittle steeple]

Well you have to know the incenter exists, which is the same problem we were having here

[paper nebula]

ig

but you could also "construct" it first