help-28

[tacit siren]

And d as another

[wild charm]

I'm doing trade math and desperately need help with it no bot or a.i. can't even answer the problems.. even using a calculator is difficult. If I can get through the final quiz I wont ever have to do it again..

[tacit siren]

rip

Open ur own help channel pls and thank

!help

[full forum]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[deep hull]

ai cant even do simple arithmetic

ive seen it fail at 7*5

thought it was 42

[tacit siren]

<@&286206848099549185> anybody wanna give this a try? Rules are only used provided theorems, solve for d in 5 applications of those theorems. Best we can do is 6

[deep hull]

the equation for solving x and y are 180-2x + 180-2y + 58 = 180, right?

[tacit siren]

Yeah

Well solving x+y

Can't solve for x and y and d individually with just 2 equations

[deep hull]

once you solve x+y what happens

[tacit siren]

You plug it into x+ y+d =180 and solve for d

[deep hull]

ohh i see

that makes sense

dang still no one

[white flicker]

d=32

[deep hull]

ope

how did you get to that

[white flicker]

wha

[deep hull]

what angle theorems did you use

[white flicker]

wait

didn't you said that you had the answer already

[deep hull]

yeah but it uses 6 angle theorems

we need to use 5

[white flicker]

wait

6 angle theorems

holy

you mean you used the term of 6 angles or you mean 6 angle term

[deep hull]

6 angle theorems

these

[white flicker]

BRUUUUUUUUUUUUUUUUUUUUUUUUUUUH

I only used 3 of them

[deep hull]

uh how

[white flicker]

I used a+b+c=180
ITT

[deep hull]

like what did you do

[white flicker]

Wait

ok

i did it wrong at the begining

let me fix it

its not 32

[deep hull]

pretty sure d is 61

i mean i guess i can put that in and get 1/6 marks for the question

better than nothing

[deep hull]

lol nobody answered to this

[tacit siren]

rip

[deep hull]

is it against the rules to use it again

after a certain time period

say like 30 minutes

[tacit siren]

Hmm not sure actually

I've never opened a help channel so I've never actually read the rules

Idk even if it's against the rules I don't think anybody will care spesh if u wait

[deep hull]

i mean this is all it says

[white flicker]

no its 79

[deep hull]

i think pinging twice after 30 minutes isnt abusing

[deep hull]

why

[white flicker]

I did so many calculations

its gotta be 79

[tacit siren]

I haven't actually calculated it I just know how to calculate it

[white flicker]

58+3d+x+y=180

[tacit siren]

Where does 3d come from

[white flicker]

418-2x-2y=180

sorry i mean 360

i did it wrong agian

lol

i just keep forgetting to watch what i marked

[tacit siren]

hmm how are those angles x + d and y+d if u don't mind me asking

[white flicker]

breh

x+d+y=180

[tacit siren]

oh rip

[white flicker]

huh?

wdym

[tacit siren]

Either way he can't use quadrilateral theorem

No I mean as in rip for me

[deep hull]

yeah just the ones on the page lol

[white flicker]

nooooooooooooooooooooooooo😡

[tacit siren]

Hmm

[white flicker]

ok

i lose this

you win

[tacit siren]

Maybe they'll let you prove it adds up to 360 with sum of angles of triangle theorem tho

[white flicker]

you don't have to though

[deep hull]

its kinda annoying when i search for answers to geometry problems and the internet says to use trigonometry and im like "im not supposed to know that"

[deep hull]

not this one

just in general

ive had it happen many times

[tacit siren]

ah shit I think this is still 6 steps if you count the proof of sum of angles is quadrilateral is 360 as 2 applications of SATT

[white flicker]

axiom doesn't needs to be proved

[tacit siren]

oh

[white flicker]

its already proved

quadrilateral is 360

for sure

[tacit siren]

Sum of angles of quadrilateral is 360 isn't an axiom/theorem

That he can use

He has to prove it from sum of triangles probs

[deep hull]

something like that just makes me think that ur overthinking

the answer is meant to be really simple

[tacit siren]

But it is the final boss too :p

[white flicker]

no

bro

have you learned the diagonal

one triangle is 180

[deep hull]

the format is listing the angle theorems used in finding the angle

[white flicker]

right?

[tacit siren]

Mhm

[white flicker]

you cut the shape into two triangle

[tacit siren]

Yeah

[white flicker]

which it means the tottal is 360

[tacit siren]

That's how I'm saying you prove the total is 360

[white flicker]

breh

[tacit siren]

But u gotta use sum of angles is 180 on 2 triangles

[white flicker]

My teacher said it doesn't needs to be proved

[tacit siren]

Yeah but his rules do

[deep hull]

this is how the answers are meant to look

i mean i could email my teacher to see if i can use that

[tacit siren]

yeah

[deep hull]

uhh i mean i guess ill do that now

feels a little unsatisfying tho

[white flicker]

aight

good luck

[deep hull]

like theres something im not seeing

[white flicker]

[deep hull]

i guess ill close this for now

.close

[full forum]

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[silver forge]

I'm faced with quadratic functions rn and I've forgotten how to begin solving them. I don't understand the tutorials online for some reason. Can someone help in explaining the steps? I have the answer(it said 1.48), but I want to understand how to arrive there.

[glacial pasture]

so we're starting off with -3.75(t-0.2)^2+6.15=0 right?

[silver forge]

yes

[glacial pasture]

whats your instinct on what to do first if we are trying to isolate t

[silver forge]

distribute ^2 then divide?

[glacial pasture]

when you say distribute ^2 what do you mean?

[silver forge]

t^2 ?

[glacial pasture]

$(t-0.2)^2 \neq t^2-0.2^2$ if thats what you mean

[glossy valve]

AℤØ

[silver forge]

yes

[glacial pasture]

yeah you cant do that

exponents dont work that way

[silver forge]

Ohh

[glacial pasture]

the first thing id do is move 6.15 to the other side

[silver forge]

-3.75(t-0.2)^2=-6.15 ?

[glacial pasture]

yeah

what might you do next?

[silver forge]

Oh round off both sides?

[glacial pasture]

nope

[silver forge]

I mean

mb

square root 💀

[glacial pasture]

not yet

[silver forge]

Oh shoot hang on

[glacial pasture]

this will be the step after this one though

[silver forge]

Do you multiply both sides by -1?😭

[glacial pasture]

i mean, yeah i suppose you could

you need to divide both sides by -3.75

[silver forge]

AHHH I SEE

√(t-0.2)^2 = √1.64
?

[glacial pasture]

yup

[silver forge]

Okay I think I got it tyvm😭

.close

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[trail fulcrum]

I'm supposed to be able to solve for moments of mass and inertia, as well as centre of mass, in spherical and cylindrical coordinates

[trail fulcrum]

But I don't see anything in my textbook about it

Are the equations the same?

[full forum]

@trail fulcrum Has your question been resolved?

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[frank grove]

hi

[frank grove]

is this new theorem I made valid?

its called

The Half Theorem

$k(2) = (k(1/2)*4)$

[glossy valve]

Agent78

[rare dock]

[torn jolt]

i think this

may not be true always

[jolly terrace]

https://tenor.com/view/senator-palpatine-anakin-skywalker-phantom-menace-gif-10607636

[torn jolt]

ago crazy tho

go

[frank grove]

yay

I have my new theorem!

i made

[torn jolt]

[frank grove]

i mean

I think it works every time

though

[torn jolt]

wait i did it wrong

did i?

is it like

recirpocal

or

what

[frank grove]

bro

$k(2) = (k(1/2)*4)$

[glossy valve]

Agent78

[frank grove]

plug k with a number

[torn jolt]

i thought u meant

function

wise

haha

[frank grove]

k

try that

[torn jolt]

i mean

i dont need to try it

to know its true

i think

i was confused cuz ithought u meant k(2) as in k(x), x=2

[frank grove]

lol

k is replaced by a number

fyui

[torn jolt]

i know

[torn jolt]

^

[frank grove]

so does that work?

[torn jolt]

i mean

i think

[frank grove]

if so

[torn jolt]

its just reflexive property

k(2)=k(1/2)*4 k(2)

[frank grove]

oh mon

[torn jolt]

any number x is equal to itself x

[frank grove]

uh

not to thids

$k(2) = (k(1/2)*4)$

[glossy valve]

Agent78

[torn jolt]

k(2) = x

k(1/2)*4=k(2)

x=x

[frank grove]

omg fine

i'll find another theorem

[torn jolt]

good luck brother

[frank grove]

heres another one!

I found

[torn jolt]

go ahead

lets see

[frank grove]

The Tesseric Theorem

$a^4 + b^3 + c^3 + d^2 + e$

ta da1

[torn jolt]

whats the conclusion

[frank grove]

wait I did it wrong

$ax^4 + bx^3 + cx^2 + dx + e$

[torn jolt]

conclusion?

[glossy valve]

Agent78

[frank grove]

to solve a tesseract

a 4D cube!

.close

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[worldly wind]

i setup the equation 10.9^2 + b^2 = 13^2 to find segment pb (this is question 27 btw). i got some weird long decimal and we never get those kinds of numbers on tests and other homework so i think i did something wrong.

[full forum]

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• Show your work, and if possible, explain where you are stuck.
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[torn jolt]

I'm confused about $a_n = 2$

[glossy valve]

What should I do

[torn jolt]

shouldn't it be $\frac{n}{n+1} a_n = 2$

[glossy valve]

What should I do

[torn jolt]

wait

[short bridge]

Where is $a_{n} = 2$ ?

[glossy valve]

Enemagneto

[lucid pendant]

Yes, and the fraction tends to 1 as you take the limit

So you can replace
$$\lim_{n \to \infty} \frac{n}{n+1}a_n$$
by
$$\lim_{n \to \infty} a_n$$

[torn jolt]

$$\frac{n}{n+1}a_n - \frac{2n-1}{n+1}$$

[glossy valve]

What should I do

Yuese

[torn jolt]

hmmmm

[lucid pendant]

$$\lim_{n \to \infty} \frac{n}{n+1}a_n = \left( \lim_{n \to \infty} \frac{n}{n+1} \right)\left(\lim_{n \to \infty} a_n \right) =\lim_{n \to \infty} a_n $$

[glossy valve]

Yuese

[lucid pendant]

Does this make it more clear? I believe the first equality is only valid provided that $(a_n)_n$ is convergent, so I suppose it's not super rigorous

[glossy valve]

Yuese

[torn jolt]

Yes it makes sense

TY

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[muted flare]

$\int_{-1}^1 \log (\sqrt{1 + x} + \sqrt{1 - x}) \dd{x}$

[glossy valve]

jewels!

[muted flare]

I'll be back in a bit, I tried subbing x = cos 2u and x = sin 2u but didn't end up anywhere with it

[full forum]

@muted flare Has your question been resolved?

[muted flare]

.close

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[versed swan]

P(E ∩ F) = P(E) + P(F) − P(E ∪ F) ≥ P(E) + P(F) − 1 , what this is trying to tell
can you just guide me step by step trough this
Image
are there any assumptions, involved here

[versed swan]

can you just guide me step by step trough this
Image
are there any assumptions, involved here

[onyx glen]

none

P(E ∪ F) ≤ 1

therefore -P(E ∪ F) ≥ -1

[versed swan]

hmm

in proof where RHS disapper after 2nd line

@onyx glen sry for pinging can yu guide ahead?

*further

[onyx glen]

the 1st line proves the left half of the goal inequality

the 2nd line begins work on the right half

[versed swan]

this is trying to tell P( E Intersection F)=min(P(e),p(f)?

[onyx glen]

≤

not =

[versed swan]

yes <=

hmm ok

TY

but\

wait

in image why thers = sign then?

[onyx glen]

because they say P(E) + P(F) - max(P(E),P(F)) = min(P(E),P(F))

[versed swan]

can yu give me exampl, its confusin for me please

[onyx glen]

P(E) = 0.8, P(F) = 0.4

max(P(E), P(F)) = 0.8

P(E) + P(F) - max(P(E),P(F)) = 0.4 + 0.8 - 0.8 = 0.4

[versed swan]

yes

how this is true?

last doubt

@onyx glen

[onyx glen]

$P(E \cup F) \geq \max(P(E), P(F))$

[glossy valve]

Ann

[onyx glen]

probably proved earlier

[versed swan]

yes

thnx.

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[sturdy summit]

Lets say an equation 4a(4-1) would the answer be 4a + 3 or 12a?

[fair oracle]

4a(4-1) is not an equation

[sturdy summit]

What would be the answer

[pseudo cape]

It's an expression, and simplified, it's equal to 12a

[sturdy summit]

Thank you very much

Answer I was hoping for

[fair oracle]

We do the paranthesis first, thats the way to avoid ambiguity with different order of operations.

[pseudo cape]

You can either use the distributive property or first simplify the expression in the brackets

[sturdy summit]

I was just confused wether it was gonna be a 4a + 3 or 12a

[sturdy summit]

If we distribute its just gonna be the same answer anyways

[pseudo cape]

Yes, 16a - 4a = 12a

[fair oracle]

How did you get 4a+3?

[sturdy summit]

I was dumb thanks for the help

[sturdy summit]

I forgot the PEMDAS rule

[fair oracle]

Don’t worry about it, the order of operations should be evident from the context

tbh PEMDAS is added to get same answers in elementary math questions

[sturdy summit]

I mean it is yeah

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[stoic parcel]

log₄(x^3) + log₂(√x) = 8

[stoic parcel]

how do i make the base same?

[torn jolt]

Do you know the change of base rule?

[stoic parcel]

yeah

log b A = log c A/ log c B

[torn jolt]

Use it to change that log4 into log2

[stoic parcel]

yes

idk how to do it

[torn jolt]

What would a and b be?

[stoic parcel]

what do i do from here

im soooo confused

@torn jolt

please help

[torn jolt]

Why'd you change the second one? The base was already 2

Also since when is log2(4) equal to 16??

It seems you calculated 2^4 instead of log2(4)

Ask yourself: 2^what=4?

[stoic parcel]

2

[torn jolt]

Yes. Now please do this again, but only apply the change of base formula on the left term (the right one already has a base of 2)

[stoic parcel]

@torn jolt

helppp

[torn jolt]

I would definitely not hire you as a photographer

$\frac32\log_2x+\frac12\log_2x=8$

[glossy valve]

Labyrinth

[torn jolt]

@stoic parcel This is what you currently have, right?

So now you just need to factor out the log2(x)

And the rest is pretty obvious

[stoic parcel]

yea

so i get 3/4 log 2 x

[torn jolt]

Uuuuh no you don't?

Since when does 3/2+1/2=3/4?

[stoic parcel]

doesnt law of logarithm apply here?

[torn jolt]

???

What "law of logarithm"?

And why would it apply to the addition of 2 fractions?

[stoic parcel]

log A + log B = LOG (AB)

[torn jolt]

How does that apply to 3/2+1/2?

Oh wait

Right

log(3/2)+log(1/2)=log(3/4)

[stoic parcel]

YEHA

[torn jolt]

Except you'll notice 3/2 and 1/2 are out of the logarithms

So this rule does not apply

You just have 3/2+1/2

[stoic parcel]

ohh so it doesnt apply when theres a base?

[torn jolt]

???

???????

Logarithms always have a base

[stoic parcel]

how do we know if smth is out of logarithms

im so confused

[torn jolt]

You see how there's parentheses?

Like, log(2)?

[stoic parcel]

yeah

[torn jolt]

Well, 2 is inside the parentheses here

If you wrote 3log(2) you'd have 3 outside and 2 inside

It's that simple

[stoic parcel]

ohh

[torn jolt]

If you factor out the log2(x) in the equation you have you get (3/2+1/2)log2(x)=8

As you can see there's not a single logarithm wrapped around either of the fractions

All you have is 3/2+1/2

[stoic parcel]

ohh okay

i can do from here

thanks a lot

[torn jolt]

@stoic parcel Remember to plug your solution back into the original equation to check that it's all correct. Also, don't forget to .close

[stoic parcel]

okay

thanksss

.close

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[lucid sierra]

uhmm

[lucid sierra]

do i did this right

or there is more to simplify

[nimble current]

the expression at the top is kinda blurry, could u send another pic but zoomed in on that part

[lucid sierra]

you mean number eight?

im asking about number 9 but sure ill send it too

[nimble current]

oh

[nimble current]

note that $288=2^5\cdot9$

[glossy valve]

lpieleanu

[lucid sierra]

oh wait

and what should i do with that

[nimble current]

then, we have $$\log_2(288)=\log_2(2^5\cdot9)=\log_2(2^5)+\log_2(9).$$ Try, to continue simplifying the expression you got from there

[glossy valve]

lpieleanu

[lucid sierra]

ohh i see but how about the divide by two part

[nimble current]

What did you get for the numerator?

[lucid sierra]

log 2^2 (288^1) = 1/2 log2(288) right?

[nimble current]

no, currently, using the progress above, the expression is $$\frac{\log_2(2^5)+\log_2(9)}{2} - \frac{1}{6}.$$

[glossy valve]

lpieleanu

[lucid sierra]

ohh i see thanks

so (5+log2(9))/2 - 1/6

[nimble current]

yep, and this can be simplified just a little bit more

[lucid sierra]

cant find it

no clue

[nimble current]

first, express it as $$\frac{5}{2}+\frac{\log_2(9)}{2}-\frac{1}{6},$$ then combine the $\frac{5}{2}$ and $-\frac{1}{6}$ and use the fact that $\log_m(n^p)=p\log_m(n)$

[glossy valve]

lpieleanu

[lucid sierra]

ohh i see now thanks

soo (3log2 (3))/2 +7/3

[nimble current]

$9=3^2\neq3^3$

[glossy valve]

lpieleanu

[lucid sierra]

BEAUTIFUL THANKS

[nimble current]

what do you get as your answer?

[lucid sierra]

log2 (3) + 7/3

[nimble current]

👍

[lucid sierra]

you help so muchh ty

i might ask more could i ping you again?

[nimble current]

you're welcome

[nimble current]

sure

[full forum]

@lucid sierra Has your question been resolved?

[lucid sierra]

not yrt

,rotate

[lucid sierra]

this correct? @nimble current

[nimble current]

yep, very nice job!

[lucid sierra]

THANKSS YES

now final question ig

how abt this one?

any simplification?

should i make log2 (32)/log2 (10)

@nimble current sorry

[nimble current]

you made a tiny error, $\log_3(3^{-2})$ is $-2\log_3(3),$ not $\frac{\log_3(3)}{-2}.$

[glossy valve]

lpieleanu

[lucid sierra]

oh sorry

[lucid sierra]

and after that? its all complete?

log(32)/-2

[nimble current]

yep ✅

[lucid sierra]

thanks now my homeworks done i love you

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[glossy valve]

Joseph.P

[onyx glen]

$(x-16)P(2x)=16(x-1)P(x)$

[glossy valve]

Ann

[onyx glen]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[onyx glen]

hold on

P(x) = (x-2)(x-4)(x-8)(x-16)Q(x) ?

how did you get this?

fair cop

alright

ok so you got that Q must satisfy Q(x) = Q(2x) for all real x

what polynomials do you know of that do that

[glossy valve]

Joseph.P

[onyx glen]

ok fine except for those

i mean it doesnt matter actuaally

but still

what polynomials do you know of that do that?

[tribal oxide]

Ye constant polynomial

[tribal oxide]

Constant polynomials

[onyx glen]

great

then you're done aren't you

[tribal oxide]

No because I need to prove Q is constant

[onyx glen]

you need to prove that if Q satisfies (∀x ∈ R \ {1,2,4,8,16})[Q(x) = Q(2x)] then Q is constant.

consider the polynomial R(x) = Q(x) - Q(3).

well, yes, R does satisfy R(x)=R(2x) as well.

what else can you tell me about R?

and you can prove that?

[tribal oxide]

No

I’m dumb

[tribal oxide]

It would implies that Q is constant but that’s is just an assumption

[onyx glen]

R has 3 as a root

due to its property of R(x) = R(2x) it also has 6 as a root

and 12, and 24, and so on.....

[tribal oxide]

So R is constant and equal to 0 because it has an infinity of roots ?

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[untold shale]

Hi, i need help to find the domain of a function

[untold shale]

Basically this is the function

i know that the numerator should be ≥ 0

while the denominator should be different from 0

in the first photo i try do solve that but i'm stuck and idk how to continue

<@&286206848099549185>

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[sinful yew]

SIR
HOW I CAN GET PYQ CHAPTERWISE

[narrow helm]

Calm down with the caps, and this is a public server, so dont start with 'sir'.
For your question, we dont know what PYQ's you're asking for.
Also try #book-recommendations or search online (there are many online sources that provide PYQ's)

[sinful yew]

i want them for free

pdf

[narrow helm]

yeah they are available online.

[sinful yew]

give me link

i am so confused

[narrow helm]

wdym give link.

idk what exam you are talking about? be specific.

[sinful yew]

I am talking about board class 12 isc

[narrow helm]

also, this is not a question involving maths

so I am closing the channel, ask your other questions (not math) in #discussion

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[torn jolt]

Need help to prove Q2b with induction

[torn jolt]

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[torn jolt]

How does this shake out any differently to achieve the logical equivalence its asking for

[torn jolt]

i know you can prove the question using a truth table, but can you also do it using properties of logical equivalence?

I saw someone do something wild with distributive law but that in my book seems to only work if you have three statements p q and r

but theres only 2 here

also i dont know how it would work when two of the statments are negated even if theyre in the right general form

of two or statements being and'd

or two and statements being or'd

When i tried left hand side it seemed even more helpless

i just checked with a truth table

the way that guy applied the distributive property is incorrect as they were not logically equivalent

the work above is mine just to show the question

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

well i proved it with a truth table

and like the problem says those are logically equivalent

but still cant figureout how to do it with properties

<@&286206848099549185>

[lavish condor]

You can apply the distributive law here

The law as listed in your book doesn't have to have different variables as it's inputs

It can be terms

[torn jolt]

Hmm at the very least the way that person did it was incorrect

hmm

ill try some stuff out

[lavish condor]

Let me write the law as this

[glossy valve]

Watching Forms

[lavish condor]

You can choose

[glossy valve]

Watching Forms

Watching Forms

[lavish condor]

That help?

[torn jolt]

so i can just rename it like normal math

and use the properties

yeha

yeah

ill try that out

[lavish condor]

Yes, the laws of logic are written in such a way that you can put in any logical term, not just variables

[torn jolt]

thank you

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[tepid root]

I'm not really understanding

The second part of the problem

https://cdn.discordapp.com/attachments/830088413419798538/1153020999122550944/446847607025696798.png

is it just this ?

[rocky vale]

They're looking for the volume of air you'd have to add

To inflate the balloon from radius r to radius r+1

[left bone]

hint: how much air does a balloon of radius r+1 have, and how much air is already in the balloon

[rocky vale]

This is the total volume of the r+1 balloon, not the amount you'd have to add to the already partially inflated radius r balloon

[tepid root]

hmm

https://cdn.discordapp.com/attachments/830088413419798538/1153023871994314792/446847607025696798.png

adding more volume is like inflating

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@tepid root Has your question been resolved?

[tepid root]

I think i got it

https://cdn.discordapp.com/attachments/830088413419798538/1153030006805643414/446847607025696798.png

epic latex fail

but its something like this

x is the ammount I add

to make r^3 go to (r+1)^3

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[tepid root]

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[vagrant flame]

How do you solve this?

[serene hazel]

show the full question

[serene hazel]

are you familiar with exp rules?

thats what the question asking

[torn jolt]

can you take out the common thing from both numerator and denominator, cancel it, and the solve with 5 to the powers of like 1,2,3,4

[vagrant flame]

you multiply by 5^-2004?

[torn jolt]

if i told you , i had 3^69 , could i have written it as 3^68 times 3??

similarly can i do something likewise here?

[cyan wedge]

You can if you do it on the top and bottom. That's a valid way to do it

[vagrant flame]

alr

[cyan wedge]

Not conventional, but it's good

[vagrant flame]

i think that is the way my teacher taught us

how about if the expoents wouldve have been negative

[cyan wedge]

Good teacher then. Never did it that way tbh

[cyan wedge]

just multiply by the opposite, so the positive

[vagrant flame]

how did you learn it?

[serene hazel]

,tex .exp rules

[glossy valve]

Akira 🍉

[serene hazel]

look up for negative exp rule

[cyan wedge]

using what Eragon was talking about so (5^2008-5^2006)/(5^2007+5^2004) becomes 5^2004(5^4-5^2) / 5^2004(5^3+1) = (5^4-5^2)/(5^3+1)

[torn jolt]

true

its the same thing

but in different ways

[vagrant flame]

thanks everyone

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[austere oracle]

How do you solve this?

[tiny bridge]

Can you say anything about the number of roots of f(x)?
Are you familiar with complex conjugation?

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[humble solar]

this is in chinese but all the info given is:
AH perpendicular to BC at H
m angle C is 35 degrees
AB+BH=HC
the question is:
What is the measure of B?

[humble solar]

I know that ABH is congruent with ADH but dk how to solve this

[short bridge]

Are you familiar with concepts of similar triangles?

Oh wait. That might not work.

Anyway, have you studied trigonometry?

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[amber pewter]

trying to figure out how to do these on a BA II using the finance features if anyone knows how

[full forum]

Please don't occupy multiple help channels.

[umbral swallow]

all i know is the compound formula lmao

should solve ur problem

[amber pewter]

lol not that simple but thanks

<@&286206848099549185>

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[hot mango]

confused about this problem

as when you find out what n is supposed to be in these, there is typicly one variable

however the 4th derivitive of this has a fex x's in it makeing it two variables

so im not sure what to do

this is the 4th derivative

<@&286206848099549185>

...

<@&286206848099549185>

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[hot mango]

wow

.close

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[sharp portal]

I have no idea what they're asking. Wording wise, this wasn't covered in class:

[proud delta]

<9

[sharp portal]

?

[proud delta]

thats the answer

[proud delta]

hmm

[sharp portal]

That still doesn't help me understand what they're asking me to do

[proud delta]

ah

[brave blaze]

basically if you have :
-a < x < a, it can be rewritten as : |x| < a

thats it

[proud delta]

it is |x-5| < 9

[sharp portal]

Ah

[proud delta]

yeah thats what i was doing now

[sharp portal]

Was greater than, which made sense since all the numbers are larger than 9

[proud delta]

havent looked closely to the inequality up there

yes, indeed

[sharp portal]

Thank you

[proud delta]

no problem

[sharp portal]

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[winter sphinx]

How do I cancel out the bottom term? I know it’s possible but I get confused by the negative sign

[shrewd hamlet]

mutliply the top and bottom by -1

then see if u can do some manipulation

[winter sphinx]

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[magic bronze]

State minimum degree of this function

[severe linden]

!show

[full forum]

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

[magic bronze]

4 turning points so I got 5 since it's 1 more than there is turning points but it's wrong

[severe linden]

How many zeros do you see?

Arguing with turning points seems more difficult.

[magic bronze]

4

[severe linden]

yes but some zeros are not singular zeros

for example the zero at 2 looks like a twofold zero

it looks like a "parabola zero"

Do you see any other zeros that are are not normal zeros?

,wolf plot x²

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[magic bronze]

I'm confused we never learned that

[severe linden]

I mean this is degree 2 but has only one zero.

[severe linden]

that surprises me

I think turning points are not reliable.

[severe linden]

How many turning points do you see here?

i see 0 but it's degree 2.

[magic bronze]

1

[severe linden]

no

turning point means bending in one direction and then bending in the other direction

there is no such point here, it all bends only in one direction

How have you introduced turning points? Did you use derivatives?

[magic bronze]

This is power functions

[severe linden]

yes?

How did your lecturer tell you to identify turning points?

[magic bronze]

Polynomial functions have a series of hills and valleys also known as turning points

We also learned about local max/mins points with turning points

[severe linden]

Are you perhaps confusing local minima and maxima with turning points? I see in english turning points apparently just refer to minima and maxima. Sorry for the confusion.

[severe linden]

This function does have 1 local minimum.

[severe linden]

In any case neither of this will work here.

We need to count zeros (porbalby eaisest) or otherwise account for threefold zero.

,wolf plot x³

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[severe linden]

you see this has no maxima or minima and just 1 turning point

however it's degree 3.

It's called a threefold zero and it's whats happenign in your function at -3

if you want I can explain the way to count zeros (it's easy). Let me know if you need any more help.

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[torn jolt]

<@&286206848099549185>

[soft axle]

where is the question ⁉️

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[light sonnet]

Please do not open a message with the helper ping

Post your question

[torn jolt]

ok my fualt

[torn jolt]

I dont eeven understand it

[soft axle]

when x=0 what is y?

[torn jolt]

0 right?

[soft axle]

when y is 0 what is x?

[torn jolt]

0

[soft axle]

wrong

[torn jolt]

hold on let me get my notebook to right down some things

[light sonnet]

https://www.youtube.com/watch?v=ruTcNElXdzQ&ab_channel=MrsALovesMath

[torn jolt]

ok im back

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[torn jolt]

heko

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@torn jolt Has your question been resolved?

[slate pasture]

Do you know slope formula?

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[tacit rapids]

im confused why c) is a polynomial?

[proper hawk]

what

that's the most common one

,w graph x^2

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[tacit rapids]

it's something like y= x^2 right

yeah

but like

x^2 is not a polynomial

it's only 1

[proper hawk]

????

[tacit rapids]

not more than 2

[proper hawk]

you are describing a linear functio

n

[opal pewter]

Monomials r a subset of polynomials

[tacit rapids]

so monomials are also polynomials?

[proper hawk]

yes

[opal pewter]

Lol

It is how it is

[tacit rapids]

ok

so y = 1

1 is a polynomia?

iiuc

[opal pewter]

Yep

[proper hawk]

yes

[tacit rapids]

alright i get it now

thank you so much guys

🙂

[opal pewter]

Look for the general format of polynomials

[opal pewter]

Uhh

[tacit rapids]

ooooh since constants are raised to the power of 1

they go into the

[opal pewter]

Not THAT one

[tacit rapids]

linear (degree 1) form

right?

[tacit rapids]

no?

[tacit rapids]

isn't it the same thing?

[opal pewter]

That's not the general format

They show the general form in special cases

[proper hawk]

cute shark

[tacit rapids]

but linear(degree1) form is enough to prove constants are also polynomials?

[opal pewter]

Not realy
In linear polynomials defination
We assume a≠0

[tacit rapids]

ah i see

ah i see

no sharp turns

so y = |x| is not a polynomial but y = x is a polynomial

https://math.stackexchange.com/questions/2342385/why-is-absolute-value-function-not-a-polynomial

oh also

this explains why

y = |x| not differentiable

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[lucid sierra]

log (54) if log (3) = a and log (5) = b

[lucid sierra]

i dont get where 5 come from

[uneven sequoia]

hey can someone help me

[narrow helm]

!help

[lucid sierra]

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[narrow helm]

wtf

[full forum]

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[random canopy]

Hi, I am trying to prove the triangle inequality and I am stuck. I did the first case of a,b >= 0, but I need to do the other cases and I am getting a bit confused, and not really sure what to do

[random canopy]

I just want to finish this so I can sleep please

[torn jolt]

constants are degree form^0

[random canopy]

<@&286206848099549185>

[full forum]

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

[red blade]

Okay

So what have you done so far

[random canopy]

I believe I need 2 more cases

[red blade]

Did you make this format btw

[random canopy]

We are required to use this format

[red blade]

Oh I hate it

Okay anyway

[random canopy]

Yeah I do too

lmfao

[red blade]

But with the individual case work

[red blade]

Keep it organized imo

A >= 0 b >= 0

A >= 0 b < 0

A < 0 b >= 0

A < 0 b < 0

[random canopy]

Yeah thats what I was thinking

[red blade]

You got it from here?

[random canopy]

its more like my brain is fried from doing proofs for 17 hours and i have no clue whats going on kinda thing yk

[red blade]

Go to sleep

[random canopy]

its like i think and then i dont

[red blade]

Wake up early

You’ll be fine tmr

[random canopy]

i have classes tmrw

i wanna finish this off then sleep

its my last thing anyways

[random canopy]

wouldnt this

[random canopy]

and this

both be practically the same?

@red blade would this be right

[red blade]

Thinks this works

[random canopy]

i dont know how to do the other two cases though

[torn jolt]

idk why would you write >= in the last statement as |a|+|b| = |a+b| when a,b < 0

[random canopy]

oh

Is this fine then?

@red blade what do i do from here?

i don't get it

wouldnt you have more subcases or wtv?

idk

i dont get it

@red blade @torn jolt Could I get some help on this one?

[red blade]

What’s up

[random canopy]

I don't get how to prove case 3

[red blade]

Case 3 is?

[random canopy]

|a| + |b| = b-a

[random canopy]

this

[red blade]

No like

Is a positive

B positive etc

[random canopy]

a is negative b is positive

[red blade]

Uh make another sub case

Where b + a is positive and b + a is negative

[random canopy]

i dont get it

[red blade]

Do it and you will understand

[random canopy]

so i make another case?

[red blade]

Yes

Im going to sleep rn

But I’ll check in with you in the morning

[random canopy]

one more thing

[red blade]

Oh

What’s up

I have a minute or so

[random canopy]

do i really need a case of where b < 0, and a>=0

since I am going to prove b>=0 and a<0

and they are both the same

[random canopy]

ye its short

[random canopy]

do you also need to do one for where b+a is = 0?

[red blade]

Oh just include 0 in one of them

[random canopy]

wdym?

so would i need 2 or 3 sub cases

i think ill just not go to my morning lecture

i already am ahead in that class anyways

.close

[full forum]

Channel closed

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• Be polite and have a nice day!

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[full forum]

@distant grove Has your question been resolved?

[full forum]

@distant grove Has your question been resolved?

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

[tough ore]

hi!does anyone know how can i find the origins of certain mathematical formulas or theorems?for example,I was wondering how the trigonometric form of complex numbers was discovered and it came to be?

[tough ore]

pls ping me if u answer

[blissful nova]

like the first time?

historically?

[torn jolt]

this was a good watch https://www.youtube.com/watch?v=5PcpBw5Hbwo

[blissful nova]

@tough ore

[full forum]

@tough ore Has your question been resolved?

[full forum]

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• Be polite and have a nice day!

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

[digital chasm]

why am i still here after 15 minutes ..

[light sonnet]

The lack of parentheses

[digital chasm]

huh

[light sonnet]

It's (70 * 30)/(7 * 3)

[digital chasm]

yeah thats what i put

[light sonnet]

No you did not

Lack of parentheses

[digital chasm]

but i didn't add any parentheses on the other one

[light sonnet]

If you don't use parentheses in the denominator, (70 * 30)/7 * 3 then by order of operations that is read as $\frac{70 * 30}{7} * 3$

[glossy valve]

dldh06

[light sonnet]

So it does 70 * 30 * 3 then divided by 7

[digital chasm]

ohh

but I don't understand why

[light sonnet]

Because 7 * 3 is grouped in the denominator

(70 * 30)/7 * 3 then by order of operations that is read as $\frac{70 * 30}{7} * 3$

[glossy valve]

dldh06

[light sonnet]

Because PEMDAS

You do Multiplication and Division from left to right

So 70 * 30 = 2100

2100/7 = 300

300 * 3 = 900

But (70 * 30)/(7 * 3) you do parenthesess first

So 2100/21

= 100

[digital chasm]

right

[digital chasm]

whats pedmas?

[light sonnet]

Are you more familiar with BODMAS?