help-10

[twin ferry]

wait I did it wrong for sure now

[knotty crow]

answer should be a

[twin ferry]

I just remembered ln a become nothing

not a

fuck

[knotty crow]

after applying rule it becomes 1/(1/x) = x

[twin ferry]

yeah I forgot ln a become nothing

there go my grade

well thank

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[buoyant horizon]

I know how to solve limits without the trig functions but I'm stuck on how to appraoch this

[knotty crow]

can you solve

[warm shale]

Modus

[knotty crow]

?

[buoyant horizon]

negataive infinity?

[knotty crow]

yea

so now

u know what arctan(x) tends to when x goes to -inf?

imagine graph of arctan(x) if you know that stuff

[buoyant horizon]

someting like this right

-pi/2?

[knotty crow]

yep

[buoyant horizon]

oh i see

thanks

caan i ask one mroe uqesiton

I'm aalso stuck on this

I know the vertical asymptotes is when denominator is 0 for the x values

but i'm not sure about the horiztonal ones (maybe when numeraator is 0?)

[knotty crow]

so in this case there are no any

since D = 3^2 - 4(4)(2) < 0

to find H.A. examine limits

[buoyant horizon]

So vertical asymptotes only exists if the D > 0?

[knotty crow]

in +/- inf

[buoyant horizon]

and D >= 0 or D > 0

[knotty crow]

D >= 0 means quadratic has at least one root

so ye

[buoyant horizon]

wait I think I mixed it up

Ok so I know that vertical asymptotes exist when there are roots in the denominator right

And since D<0 there are no roots

so there arae no vertical asymptotes

correct?

[knotty crow]

yes that's true, also important thing is we can't have common factor in the numerator and the denominator

(they would cancel out)

[buoyant horizon]

ok

for the horizontal asymptotes, why do I examine at infinity and -infinity

[knotty crow]

look at the previous example with arctan

lim x to inf gave you an H.A. x = -pi/2

[buoyant horizon]

oh is it only possible to haave 2 horizontal asymptotes?

[knotty crow]

yeah

[buoyant horizon]

I see thanks for you help

appreaacite it

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[humble coyote]

confused on how to do b

[wild swallow]

solve for the kernel of A^T?

[humble coyote]

so A^Tx=0?

[wild swallow]

that'll set x orthogonal to the columns of A

[humble coyote]

ah

thank you

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[timid silo]

how did they get rid of 9x?

[timid silo]

these were the original eq

[daring rock]

9x = 0 (mod 9)

[timid silo]

ohhh right

thank you!!

[daring rock]

no problem 👍

[timid silo]

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[icy cape]

And yes, I did have to draw this...

How do I find C?

Do I just add 52 and 46? (I think not)

Or do I add them and divide by 2?

Someone please help

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[coral marsh]

how to prove this w induction?

[coral marsh]

0 <= m <= n-1

base case would be m,n-1 = 0?

its true for m,n-1 = 0

not sure what to do for inductive step

[obtuse pebble]

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@coral marsh Has your question been resolved?

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[grim crater]

So for my work

The given was sin A = 15/17 in Q2

Sin B = -5/13 in Q3

so i used the pythag which formula is +- sqr 1-(sin)

and I got cos A 5/13

Cos B -4/5

but for some reason None of them are right and I just cant seem to find the answer

I WOULD be willing to go on voice channel for helo

[obtuse pebble]

@grim crater Has your question been resolved?

[grim crater]

no

<@&286206848099549185>

[bold bane]

Both of those cosines are incorrect.

[grim crater]

How are they incorrect

cause im pretty sure

[bold bane]

sin = O/H and cos is A/H which means the denominator will be the same.

[grim crater]

if yyou go cos A = sqr (positive since in Q2) 1-sin A

[bold bane]

That should be sin^2(A).

[grim crater]

I prob forgot that

mb

[bold bane]

You could also use the Pythagorean Theorem to find the Adjacent side.

sin(A) = 15/17 = O/H. Adjacent = sqrt(17^2 - 15^2)

[grim crater]

ok so with that i got Cos A : -8/17

and Cos B -5/13

[bold bane]

cos(B) is not correct.

[grim crater]

how ?

[bold bane]

,w sqrt(13^2 - 5^2)

[warm shale]

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[grim crater]

Ohhhhhh

gotcha

yeah i see i took the oppisite rather then adjacent

i forget sohcahtoa sometimes

ok so my new cos B would be -12/13

[bold bane]

👍

[grim crater]

and For Sin(A+B) i get -140/221

[bold bane]

That's what I get as well.

[grim crater]

i think i got the rest I appreciate it !!

[bold bane]

np

[grim crater]

thank you so much

[bold bane]

yw

[obtuse pebble]

@grim crater Has your question been resolved?

[serene carbon]

transform the function a= h= k=

y=-3x^2+24x-5

ok.

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[stone sequoia]

hi I am unsure of how to do this problem, any help would be appreciated

[high lily]

do you know how to find lcms of expressions?

[stone sequoia]

I thought you multiply the denominators

[high lily]

that'd be a possible common denominator
but its not the in the options (and ideally you'd use the lowest common denominator)

[stone sequoia]

Okay

How do you find the lowest one?

[high lily]

which goes back to my initial question

do you know how to find lcms of expressions?

e.g. would you be able to find the
lowest common multiple (lcm) of 4 and 6

[stone sequoia]

Yes

12

[high lily]

what about something a little more complicated like lcm of
4x and 6x^2

[stone sequoia]

No

[high lily]

what about lcm of x and x^2

[stone sequoia]

No

[high lily]

do you know your exponent laws

[stone sequoia]

Yes

[high lily]

note that x^2 = x * x

[stone sequoia]

Would x be the lowest common denominator in x and x^2

[high lily]

no

what's the lcm of 2 and 4?

[stone sequoia]

4

[high lily]

you didn't say 2 here right?

[stone sequoia]

Yes

[high lily]

you can pretty much apply the same principle as you just did to identify the lcm of x and x^2

[stone sequoia]

Would it be 2x?

[high lily]

no

you said you knew your exponent laws and i also gave you

note that x^2 = x * x

x^2 = x * x, is a already clearly a multiple of x

the lcm of that and x will be x^2

[stone sequoia]

I see

[high lily]

just like how 4 is the lcm of 2 and 4 since 4 is a multiple of 2

[stone sequoia]

Okay

[high lily]

so for the lcm of all your expressions here, you can consider the constants and each variable separately and identify the lcms of
1,6,8
x^2, x^4, x
y^3, y^5, y

[stone sequoia]

The first one is 24

The second is x^4

And the third would be x^5

[high lily]

you mean y^5?

[stone sequoia]

Yes

Sorry

[high lily]

and the lcms of those denominators (the lowest/least common denom) would be the product of those 3 things

[stone sequoia]

So that would be 24^4y^5

24x^4y^5

[high lily]

yes

[stone sequoia]

Okay thank you so much

[obtuse pebble]

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[timid silo]

Why???

[dark mango]

#❓how-to-get-help

[timid silo]

Why power of 2

[dark mango]

i cant see the question

[royal basin]

lots of typos here

the ^2 you circled was actually supposed to be ^(1/2)

and the cos(x) is raised to the power of 1/2 and not x/2 as written

[timid silo]

Oh yea professor

Thanks

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[timid silo]

For this question I got x=3.10 + (pi)n, neZ and x=5.05 + (pi)n, neZ but the answer key says that its x=3.10 + (pi)n, neZ and x=1.91+ (pi)n, neZ

[timid silo]

What did I do wrong?

Thatvwhat i did

[dark mango]

,w 5.05 - pi

[warm shale]

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[timid silo]

but if I wrote 5.05 in my general solution instead of 1.91 would that be fine

[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[eager arch]

needs help to prove that this function is surjective or not?

Ig no but I'm not pretty sure,

anyone please?

[raven spire]

there exists no integer x such that f(x) = 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, ...

hence not surjective

but it's injective

@eager arch

[eager arch]

yes I have proved it's injective but I was confused in second step

[raven spire]

You got it now right? To prove generally you have to put dozens efforts but to disprove you only need to present a counter example

it's fine even if you write something like: f(n) = 7n has range 7Z which are all multiples of 7 but that implies all Z not multiplies of 7 are not in range.

[eager arch]

Ahhh okay!!!
Thnx bro

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[frosty locust]

Could someone check if I have done this right? My answer is x = 3.4

[timid silo]

No

Parentheses

You have to expand the negative for the 1 too

[frosty locust]

How do i do that?

[timid silo]

$-(a-c) \to -a+c$

[warm shale]

♡LexQa♡

[potent compass]

i got x = 3, gimme a sec im taking a picture of my work

[timid silo]

Let them figure it out themselves

[frosty locust]

I’m redoing just now

[potent compass]

sorry its sideways

stupid phone

anyway so you you forgot about the negative

so that 1 would be a positive 1, not a negative 1

[frosty locust]

Ohh I see

I forgot to multiply 2x by 4

[potent compass]

do you understand though?

[frosty locust]

Yeah I understand now

Thank you!

[potent compass]

awesome :]

have a good day

[frosty locust]

You too :))

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[timid silo]

i have a question

[timid silo]

how i simplify sin3a by using the duplication formula

[knotty crow]

wdym by the duplication formula

use sin(A+B), cos(2A) and sin(2A)

[timid silo]

how i do with sin2a

for sin3a

[knotty crow]

sin(3A) = sin(2A + A) = ...

[timid silo]

ah

and if i have sin5a its like sin(sin(2a+a)+2a) ?

[knotty crow]

no

[timid silo]

so its what

sin((2a+a)+2a) ?

[knotty crow]

might be

but for higher multiples of 'a', it's pretty painful to derive formulas, it's better to use complex stuff then

[timid silo]

ok

@knotty crow can you help me for this pls

[obtuse pebble]

@timid silo Has your question been resolved?

[pine sail]

Simplify first 2 terms.

Then factor sin3t

[obtuse pebble]

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[pine sail]

.reopen

[obtuse pebble]

✅

[pine sail]

,tex \trigsumprod

[warm shale]

What the hell am I doing here?

[pine sail]

This is what you use to simplify the first two terms.

.close

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[summer snow]

Ima just pretend I didn’t see this-

[teal turret]

<@&268886789983436800>

[summer snow]

<@&268886789983436800> this is concerning, can I get help over here lol-

[timid silo]

Just don't interact and wait @summer snow

Move on

[distant moth]

gus

[hot hazel]

yeah just ping mods and dont react

.close

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[cobalt valve]

How can i use sin2x and sin3x like identities to derive sin(pi/2)

[devout sable]

wdym derive sin(pi/2)

[timid silo]

I have an inkling they mean deriving the half angle formulas

[devout sable]

uh sin(x/2)?

[timid silo]

Maybe

[distant moth]

sin(3x)-sin(2x)=2cos(5x/2)sin(x/2) 🥸

[cobalt valve]

I meant to find the value of sin(pi/2)

[cobalt valve]

Like to find sin (pi/6) you can use sin(pi/2-pi/3)

And use the sin(x-y) formula

How to do the same with sin2x

[timid silo]

2sinpi/4cospi/4 ?

[obtuse pebble]

@cobalt valve Has your question been resolved?

[weak cradle]

I believe you meant something like this(this could be done using sin(3x) formula too) ?

[obtuse pebble]

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[silver plover]

is this correct?

[final thunder]

,w integral of (x^3+1)/(x^2 +1)

[warm shale]

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[final thunder]

Yes

Different value of C

1/2 is “absorbed” into C

Because 1/2 is just a constant

[silver plover]

i see ok om

cheers pal

[final thunder]

[silver plover]

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[merry geode]

Expectation of that is 300 too.

[zenith stone]

100% chance of getting a 3 and another chance of getting a number between 1-5?

so you will always get a 3?

1+2+3+4+5 devided by 5 =3

3x100 is 300 so on avarage you get 300

what was the question?

[high lily]

it's still not clear where this 100% fits into this

what's the original wording of the question

[zenith stone]

i think he means that there is a even chance to get a number between 1 and 5, once u get that number you will get it 100 times in a row

[obtuse pebble]

@white cave Has your question been resolved?

[obtuse pebble]

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[deep steeple]

If a line intersects a plane at a point P, am I correct in thinking that there can be infinitive many lines that fulfills the same condition?

[versed cave]

Yup you need at least two points to define a single line
Or you need some other kind of condition(s)

[deep steeple]

Ah okay

Does the same thinking apply if the point P is not on the plane, but that the line and the plane are orthogonal?

I think that there can be only two distinct lines then. Drew the normal vector and a point p. Unsure how to analyze these statements algebraically

[versed cave]

I'm not 100% sure, I can't help any further

[deep steeple]

I tried googling this and coulnd't find anything relevant either. I'm thankful for your time, Nonna! I'll ask my professor later

[obtuse pebble]

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[obtuse pebble]

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[proper ridge]

hi. how would this set look like on complex numbers graph and how does -z affects the graph?

[proper ridge]

earlier i have been doing only graph where z was positive so for example i changed |z-2-i| to z-(2+i) and draw a point on 2, 1i

[thick fog]

You can just pull the negative sign out of the modulus to see the effect.

[obtuse pebble]

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[supple topaz]

help

[supple topaz]

prove QO = ab / a+b

AB = a

DC = b

AB |l DC || PQ

plz <@&286206848099549185>

[proper coyote]

Only ping after 15 m

[supple topaz]

oh sorry

but can u help? XD

[unique pollen]

if i understand right i can help you

[supple topaz]

aight thanks

[proper coyote]

No i don’t know how to solve that sorry

[unique pollen]

i can write wait a munite pls

you can easily prove that draw some parallelogram here it is

[supple topaz]

wha-

[unique pollen]

wait

umm

[supple topaz]

im getting a seizure from looking at that

[unique pollen]

sorry for that i am trying

[supple topaz]

its ok

but i dont understand

<@&286206848099549185>

[unique pollen]

i think there is similar triangles

but i couldnt find ah

i am trying

[obtuse pebble]

@supple topaz Has your question been resolved?

[supple topaz]

heeeeeeeeeelp

<@&286206848099549185>

[unique pollen]

i did it

there are so many similar triangles i use first aob cdo and after i use the doa and dpa

[supple topaz]

dpa is not a triangle

[unique pollen]

sorry dba

[unique pollen]

it is not clear i can try again clearly

[supple topaz]

ye i dont understand

[unique pollen]

ok i will come back

[supple topaz]

<@&286206848099549185>

[unique pollen]

first use that

these are same angles cause they are parallel

use similarity here

[supple topaz]

Corner CDO = Corner OBA

same to the other side

corner BOA = corner COD

now what

[unique pollen]

do you now this similarity

there is a ratio

if ab is" a "bo is a ratio of"a"

same thing in dc and do

k is a ratio constant

[supple topaz]

؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟

[unique pollen]

i am sorry i am trying to tell but i dont know how can i explaine this

you know this similar triangles

this is same

[unique pollen]

yes

after that name ao or bo and do and co

[supple topaz]

name them what

[unique pollen]

ratio of a

you can say for that a.k

k is proportionality constant

it is not important

[supple topaz]

where is the nearest bridge

[unique pollen]

i think the problem is mine i couldn't understand

but i didn't understand still you didn't understan where

where is the problem ask that

[supple topaz]

vc?

and screen share?

ill screen share

[unique pollen]

vc allright

where

[obtuse pebble]

@supple topaz Has your question been resolved?

[obtuse pebble]

@supple topaz Has your question been resolved?

[obtuse pebble]

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[clever jacinth]

All i need is the answer to this. I cannot seem to figure it out. Please help me out here.

I don't need to learn it or anything I just need the answer to it, thanks!

[dark mango]

You can just use the finance feature of ur calculator if u dont want to know how to solve it

[clever jacinth]

What is the finance feature and how do i use that?

[dark mango]

the PV PMT thing if u have a TI84 just google it

if u wanna solve it normally just plug in the numbers

and solve for t

using the formula it gives

[clever jacinth]

im not gonna lie I have tried to solve this so many times and just plug in the numbers to the formula but it has not been working.

[dark mango]

So what is A

[clever jacinth]

18000

[dark mango]

what is P

[clever jacinth]

12500

[dark mango]

what is n

[clever jacinth]

2

[dark mango]

and what is r

[clever jacinth]

6.25

[dark mango]

no

its as a decimal

[clever jacinth]

is it 0.0625

[dark mango]

yes

plug in and solve for t

[clever jacinth]

how would I plug it in just start at P/

?*

[dark mango]

replace each variable with the value you just told me

so you told me A = 18000 so replace A in the formula with 18000

[clever jacinth]

I have plugged it all in before and whenever I do it gives me the wrong answer and I get a different question.

[dark mango]

Well show me what you put

like whats ur equation after substituting in the values

[clever jacinth]

18000=12,500(1+0.0625/2)^2t

[dark mango]

ok simplify the brackets

what do you get now

[clever jacinth]

1.03125

[dark mango]

yeah write the whole equation though

[clever jacinth]

18000=12500(1.03125)^2t

[dark mango]

ok now whats next step

[clever jacinth]

should i divide both sides by 12500?

[dark mango]

yes

[clever jacinth]

okay so now i have: 1.44=(1.03125)^2t

right

[dark mango]

ok now what do we do to both sides to remove the ^2

[clever jacinth]

uhhh i actually dont know. do we divide both sides by 2 again

[dark mango]

sorry give me a min i have to do something

[clever jacinth]

all good

[narrow fiber]

To get rid of a square you use the square root

[dark mango]

ok

so how do u bring down exponents in math normally

[clever jacinth]

would I just do like do 1.03125 x 1.03125

or go the square root option

[dark mango]

urm id just use logarithms here

to bring down the exponent

so like

log(1.44)=log((1.03125)^2t)

and now what can u do

[clever jacinth]

im not sure

[dark mango]

u can bring down the 2t

so

like

log(1.44) = 2t * log(1.03125)

and this should be easy to solve

[clever jacinth]

would the answer be 5.9? (since it wants me to round to one decimal place)

[dark mango]

,w 1.44=(1.03125)^2t

[warm shale]

Results provided by WolframAlpha

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

Requested by cantprogram#3779

[dark mango]

oops

1.44=(1.03125)^(2t)

,w 1.44=(1.03125)^(2t)

[warm shale]

Results provided by WolframAlpha

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

Requested by cantprogram#3779

[dark mango]

yeah

[clever jacinth]

thank you so much for the help. Iappreciate it

[dark mango]

🙂

[clever jacinth]

.close

[obtuse pebble]

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[clever jacinth]

did i do this right

lol

[obtuse pebble]

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[pearl oasis]

I'm trying to calculate the following probabilities from the Bayesian network. I've been looking at resources online but I haven't found anything where I understand the formulas/explanations. Thanks.

[obtuse pebble]

@pearl oasis Has your question been resolved?

[pearl oasis]

<@&286206848099549185>

[tardy epoch]

Use Bayes theorem

[pearl oasis]

I know the formula for Bayes theorem but I don't know how to apply it to this problem

I don't know how to use it with these problems

[pearl oasis]

nvm I figured it out

.close

[obtuse pebble]

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[timid silo]

I've got a problem I don't even know how to phrase. To sum it up, say I have a plane $\pi$ and a point $A$ such that the point is not in that plane and the plane does not contain the $z$ axis. Let's say I divide the entire space in two sides, one that is "below" and one that is "above", and that's exactly what I'm struggling with. How could I define it?

[warm shale]

3317

[wild swallow]

plane normal + dot product

[timid silo]

Thanks a lot, I'll look into it.

.close

[obtuse pebble]

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[mental sphinx]

z is complex , the weird word in is And and I need to find z the locus of z on the complex plane.

[mental sphinx]

I tried to say z = a + bi and somewhat in the middle i got stuck

[obtuse pebble]

@mental sphinx Has your question been resolved?

[obtuse pebble]

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[twin sapphire]

what did you try?

[molten cosmos]

I'm not sure how prove this

regular induction starting from e^x

and then for k+1 but

but I was thinking the chapter this question pertains to deals with eigenvalues, eigenvectors and determinants

so I thought if I could show the elements of this set were all distinct eigenvectors with different eigenvalues to some linear transformation T: C(R) -> C(R)

then I can claim the set in linearly independent

[twin sapphire]

why would you need eigen values or eigenvectors?

[molten cosmos]

is that overcomplicating it

[twin sapphire]

just assume {e^x , e^2x .....e^kx} is linearly independant

and then prove

e^(k+1)x

is linearly independant from those

[molten cosmos]

oh ok, I guess I was just making it unnecessarily complex

@twin sapphire I cant just say since {e^x , e^2x .....e^kx} is lin independant
so coefficients a_1 to a_k = 0
therefore 0 + a^(k+1) (e^(k+1)x) = 0
So a^(k+1) = 0 right

does that make sense cause it feels off to me

[twin sapphire]

set things up properly first

i'm not in your mind

what are a_i ?

[twin sapphire]

first off its not really equivalent

to the family being lineraly independant

and second

you should startt with what you want to prove

and use at some point your assumption

[molten cosmos]

Ok I'll rewrite it properly

[obtuse pebble]

@molten cosmos Has your question been resolved?

[molten cosmos]

@twin sapphire

[twin sapphire]

this is not right

[molten cosmos]

ok thats what I was suspicious about

cause then this would work for linear combinations of elements of the induction hypothesis set

[twin sapphire]

if it was right

you could prove anything

is linearly independant

by induction

you didnt use the property of the actual family you have here

at some point you gotta use it

talk about exponential

because if not

your proof stands for any family of functions

even linearly dependant ones

[molten cosmos]

hmm

ok is the property that

[twin sapphire]

theres is no one property

like start with the linear combination

forget about this

just prove

that ak+1 =0

and then yopu will be able to use the induction property

because you will land on the sum up to ak

to do so maybe you can study the limit

of

the linear combination

[molten cosmos]

since n > 0, e^(k+1)x can never be creates as a sum of e^x +.. +e^kx?

[twin sapphire]

divided by e^xk

[molten cosmos]

or is that also wrong

[twin sapphire]

n?

yeah thats kind of the spirit

but you have to prove that

[molten cosmos]

oh sorry n = k

[twin sapphire]

like

$lim( e^{-xk} \sum_{i=1}^{k+1} a_i e^{xi})$

[warm shale]

Benjamin

[twin sapphire]

what is that equal to?

lim when x->inf

i misswrote

its supposed to be

e^-x(k+1)

[molten cosmos]

oh

[twin sapphire]

on the left

so if you distribute the e^xk+1

you get

$a_{k+1} +\sum_{i=1}^{k} a_i e^{(i-k-1)x}$

[warm shale]

Benjamin

[molten cosmos]

oh the right is 0 because its a linearly independent set right?

since i-k-1 < k

wait

yea

wait nevermind

the limit is still there right

[twin sapphire]

yeah

whats the limit of this

given i<k+1

[molten cosmos]

so as x approaches infinity the right side approches 0

so we are left with a_k+1?

[twin sapphire]

yeah

but remember

the whole sum

was assumed to be equal to 0

so ak+1 =0

we started from

assume there exists a_i such that

$\sum_{i=1}^{k+1} a_i e^{xi} =0$

[warm shale]

Benjamin

[twin sapphire]

then we want to prove

that the a_i are all equal to 0

[twin sapphire]

so ffrom this

you multiply both sides by e^x(k+1)

and you evaluate the lim

which is equal to ak+1

but its also equal to 0

so ak+1 must be 0

and then by induction

all the others are also 0

[molten cosmos]

I think I understand

[molten cosmos]

but im not sure why do we do this?

and where did this come from

[twin sapphire]

the idea

comes from the fact

that e^xk+1 is bigger

when x goes to infinity

that all the other combined

so if you divide by e^xk+1

or multiply be e^-xk+1

thats the same

you 'll only get the ak+1

that doesnt go to 0

[molten cosmos]

divide?

[molten cosmos]

I thought u said multiply

[twin sapphire]

i forgot the - here

[molten cosmos]

ah ok

ohhhhh

I get it now

thank you it makes a lot of sense, for some reason I thought I could always assume e^k summed with a power less than k would not add up to k+1

[twin sapphire]

well its true

since they ask you to propve it

[molten cosmos]

does this mean I cant assume some thing like given {x^1,x^2,} is lin independent then I cant just assume {x^1,x^2,x^3} is linearly independent

[twin sapphire]

this is not the same though

because

the x are not in the exponent here

but yeah

if you have a family of continuous functions

such that

each element is assymptotically bigger

that all the previous ones

yeah the proof stands

[twin sapphire]

and since its true for this family then yeah its true

like for example

ln(x), sqrt(x), x, xln(x), x^3 ,e^x x^x

are linearly independant

[molten cosmos]

ok but your point is if I wanted to use this fact in another proof

I would also have to prove it first?

[twin sapphire]

well you dont have to

but you have to have the right kind of functions

or you will be blocked when calculating hte limits

but yeah

[molten cosmos]

ok thank you for the help

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

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[regal shuttle]

how is this not infinity

[dark mango]

wats the exponent?

[civic zealot]

why do you think it's infinity?

[regal shuttle]

the exponent is inf over 12

so wouldnt whatever is inside to the power infinity is infinity

[civic zealot]

but the inside goes to 1

[regal shuttle]

yeah

and 1 to the power infinity is infinity no

[civic zealot]

no

[timber fox]

whats 1*1?

[regal shuttle]

what why not

[solar hornet]

isn't it asking to use the (1+x)^x = e^lnx thing?

[regal shuttle]

ohhhh lol

oops

so yeah it undefined form yeah?

[timber fox]

its one of those indetermite thingies endless shenanigans

[green epoch]

1^inf form you need to assume limit = L and take natural log on both sides

[regal shuttle]

yeah I made the whole function equal to y

then took the ln of both sides

[green epoch]

Rearrange it a bit and you'll see a 0/0 form

[regal shuttle]

gotcha im on it

thanks

got the right answer gracias ppl

.close

[obtuse pebble]

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[obtuse pebble]

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[trail musk]

What does the discriminant tell you about the nature of the roots of a quadratic polynomial with complex coefficients?

[timid silo]

[civic zealot]

if you have a polynomial with complex coefficients then it will always factor into linear terms. So.... no idea

[timid silo]

I don't think it says anything in that case

lol

[civic zealot]

I guess whether the roots are complex or not.

[timid silo]

max to max

u can determine if any of ur roots will be real or not

I think

[civic zealot]

yeah, if the discriminant is a positive real number then the roots are real

[timid silo]

no

[solar hornet]

isn't the answer somewhere here? https://en.wikipedia.org/wiki/Complex_quadratic_polynomial

[timid silo]

this is complex no

the thing won't work anymore

in that case it would be

well

I was wrong

u can't determine in it without computing thr roots in that case

I think

[civic zealot]

oh yeah, because b might be complex, so even if the discriminant is real, the root is complex.
Yeah. it doesn't tell you anything.

[timid silo]

exactly

also a in the denominator

[civic zealot]

true

[obtuse pebble]

@trail musk Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[velvet peak]

Can someone tell me what I’m missing?

[civic zealot]

,rotate

what are you trying to do?

[merry geode]

,rotate

Is no bot working

[velvet peak]

How do I rotate

[civic zealot]

apparently the rotate bot is down anyway.

we're saved from the AI apocalypses for one more day.

[merry geode]

Wolfram is too

[velvet peak]

The apocalypse is not my problem my math teacher is

[timid silo]

Here

[civic zealot]

right, so what are you trying to do?

[merry geode]

You can factorise the numerator in ur final answer, some factors might cancel out.
Or you can take the LCM from the start instead of multiplying the denominators

[velvet peak]

I’m trying to factor it all the way down and simplify

I know I could of used the rule of two squares but this way wasn’t working for me so I’m trying to work it out

[merry geode]

Multiply the second term num and denom both with x-1 in the original expression

What would you get?

[timid silo]

Whats the equation?

Or we just need to simplify

[merry geode]

Simplify

[velvet peak]

Umm

[timid silo]

Will this work?

[drowsy nest]

ya

[velvet peak]

Wow

[timid silo]

Cool then

[velvet peak]

Okay thanks ,but also where did I go wrong on mine?

[timid silo]

Nowhere

[velvet peak]

How come I got stuck simplying?

[timid silo]

You can factor the x+1 term out from the numerator

[velvet peak]

Like cross cancel? You mean in the top right hand corner

[timid silo]

Yes

[merry geode]

$5x^2-4+x = (5x-4)(x+1)$

[timid silo]

Exactly

[merry geode]

Even texit is down

Is no bot working

[velvet peak]

Ahhh

Yes

[timid silo]

Lol

[velvet peak]

Thanks a lot guys

Sorry

[timid silo]

No problemo

[velvet peak]

Much appreciated

.close

[obtuse pebble]

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[dense osprey]

help with number 10

what i did first was get the vertex of f which is (-1, 2)
so i used that to get the vertex of m
which is (0, -5)
so i graphed it and it looks like this

so the increasing interval is supposed to be the left side right since its increasing

so my answer is x<0

but on the answer key, it says the answer is x>0

what did i do wrong

bro go to a different channel

[deft nest]

Differnt channel.

[oblique wolf]

Oh my bad

[dense osprey]

alg

[deft nest]

..

[oblique wolf]

Sorry

[deft nest]

No, it’s fine.

[obtuse pebble]

@dense osprey Has your question been resolved?

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[ruby path]

Can I somehow set up an equation and use summation and product whatever

A little confused as to what C means

[tardy epoch]

yea it's really confusing

$\sum_{i\ge 1}$ means sum all $i=1,2,3,4,5$

$\sum_{j\le 5}$ means sum all $j=1,2,3,4,5$

[ruby path]

Texit died?

[tardy epoch]

dammit bot

[ruby path]

$\ell$

Yup it's dead

[tardy epoch]

but you can factor out Q(alpha_i) from the inner sum

sum_i [Q(alpha_i) (sum_j Q(alpha_j)) ]

then you can factor once more

[ruby path]

I'm confused

Like what I was thinking was setting up a new quintic with roots alpha_i^2 - 2

And doing something with that

:/

.close

[obtuse pebble]

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[tardy epoch]

.reopen

[obtuse pebble]

✅

[tardy epoch]

do you know vieta's formula?

https://en.wikipedia.org/wiki/Vieta's_formulas

[ruby path]

Francois Viet

I know the guy

I know he's a big deal

[ruby path]

idk his formula

I've heard of it

[tardy epoch]

if you square the sum of roots and subtract the cross terms, you could get something like sum_i alpha_i^2

[ruby path]

hmm

I'll look into it

.close

[obtuse pebble]

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[obtuse pebble]

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[icy hare]

a question on oblique asymptotes. If I have for example y=|x|+(1/x) the asymptote is y=|x| but why is this true? I know as x--> ∞, 1/x --> 0, so doesn't |x| --> ∞? Doesn't that mean y--> ∞?

[zenith raft]

when x is really large, |x| + (1/x) is really close to |x|

cus 1/x is small

|x| goes to infinity as x goes to infinity, but that’s saying something else

[obtuse pebble]

@icy hare Has your question been resolved?

[obtuse pebble]

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[hollow fern]

hi, is this correct, and if so how do i progress from here?

[timid silo]

Oops or did you

Did you do product rule? @hollow fern

[dark mango]

they didnt

[woeful folio]

oooooo

[timid silo]

Yeah okay just wanted a double check to keep my intense self doubt in check

[hollow fern]

wait i thought i did do it

oh i only did it for 4xy nevermind

[timid silo]

Yeah

[hollow fern]

so would this be correct ?

[timid silo]

Uhh

U differentiated

Both terms in ur product rule

Like u took the derivative of both x and y in ur first term

Oh wait

Is it the stuff at the bottom? Thought it was a redo

Okay

But yeah seems fine

[hollow fern]

yeah its the bottom one

[hollow fern]

im not sure how to progress from here though

[timid silo]

do you want to solve for y'?

Basically put all the terms that don't have a y' on the right and factor out the two terms that have y'

And the divide by the remaining expression where you factored the y' from

[hollow fern]

yeah y'

[timid silo]

Yeah okay just do what I said

[hollow fern]

oki

and then i sub in 7 for the x and 1 for the y?

[hollow fern]

aidez moi s'il vous plait 😄

[timid silo]

Une seconde 😄

je viens de finir de me doucher lol

[hollow fern]

i think that means im finishing my shower? hahaha

[timid silo]

Yeah lmaoo

[hollow fern]

je ne parle pas beaucoup de francais

seulement un peu

elementary french lol

[timid silo]

Haha

I am a semi native speaker

[hollow fern]

ahh

yeah im just korean lol

english native tho

[timid silo]

Oo I know a bit of Korean

[hollow fern]

tres bien

are you fully french or half?

[timid silo]

Ehh half ig

My family's lineage is weird

Somehow managed to be fluent in Arabic, Turkish, Arabic, French, and Russian because of it hahaha

[hollow fern]

wtf hahahaha

i only know a bunch of random words in russian and a bit of french

well by random words i mean the lyrics to a basta song

[timid silo]

Anyways let's see what u have done

[timid silo]

LMAO

Wait can u rotate ur pics?

[hollow fern]

😄

yeah one sec

[hollow fern]

nani

[timid silo]

Keep the y @hollow fern

I mean wait

You solved for it entirely

Okay I guess u took it the shorter wY

[hollow fern]

yeah i didnt really know what to do lol

[timid silo]

Basically uhh

Okay let me write it all out eq

$7xy^6y' + y^7 + 4xy' + 4y = 0$ is what u have

[warm shale]

♡LexQa♡

[hollow fern]

yessir

[timid silo]

Make it so that it is $7xy^6y' + 4xy' = -y^7 -4y$

[warm shale]

♡LexQa♡

[timid silo]

Factor y'

$y'(7xy^6+4x) = -y^7-4y$

[warm shale]

♡LexQa♡

[hollow fern]

ohhhhhh

[timid silo]

Divide both sides by the brackets expression

$y' = \frac{-y^7-4y}{7xy^6+4x}$

[warm shale]

♡LexQa♡

[timid silo]

U can do the rest I think

[hollow fern]

that makes so much more sense

[timid silo]

Yepp

[hollow fern]

meric beaucoup

merci*

[timid silo]

bien sûr :)

[hollow fern]

ill close it now 😄

[timid silo]

Also really random @hollow fern

[hollow fern]

if that sok

[timid silo]

But I just realised

[hollow fern]

hi

[timid silo]

You joined discord on my birthday hahaha

[hollow fern]

feb 5th? 😮

[timid silo]

Yep

[hollow fern]

ahahaha

the discord gods are calling

[timid silo]

Indeeed

I was the grand age of

13 back then ? 💀

[hollow fern]

o lord

im older than u

🥲

me stupid lol

just saw youre 19 now as well

[timid silo]

Haha

Yes. It is sad that I am becoming an oldie and hitting 20 in like 4 months 😔

Anyways, was nice talking to ya

Have a nice day! :>

[obtuse pebble]

@hollow fern Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[vast wren]

What is the angle DKB in a regular square pyramid with a square base and equilateral triangles on the sides?

[vast wren]

Since all sides of the pyramid are equal I tried drawing DB and forming a triangle with sides x-x-x*sqrt(2)

thus being able to use the cosine theorem

but that just yields the result 90° (the answer is 45°)

[merry geode]

90 is correct

[vast wren]

Hmm, then the answer key is probably false

Thanks for the confirmation

[merry geode]

True. They most likely drew a perpendicular from K to the plane ABCD. Then found half angle K.

And left it at that

[vast wren]

Yeah, seems like that is the case

Anywho, thanks again

.close

[obtuse pebble]

Channel closed

Closed by @vast wren

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