help-19

[mystic saffron]

The 1200 pages 4 book series of a complete lore

Best

BTW this is wrong too

Lmao

[mystic saffron]

<@&286206848099549185> feel free to dm if you have the answer

Goodnight

.close

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[steep mantle]

how do i do number 13 14 15

[fluid tundra]

#13 do you know how binomial expansion works?
#14 some combination of intuition and similar triangles (draw the vertical line that splits the square in two. what do you see?)
#15 count the cases this is possible. do you know how?

[desert tree]

^

[odd edge]

@steep mantle Has your question been resolved?

[steep mantle]

13# not really but is ir the pascals triangle thing
14# i dont see anyrhing special
15# im pretty sure its 20, if the first one rolls 6 it has 10 possible continuation, if it rolls 5 then 6 continuation, 4 then 3, 3 then 1

[desert tree]

for 14 look how some lines cut the shape in half multiple times

[steep mantle]

huh

[desert tree]

line segment DB cuts the square in half (approx*)

and so does line segment CA*

[steep mantle]

wdym segment DA

[desert tree]

DB mb

and CA

my apologies

does that help?

[steep mantle]

um

i still dont know how you get 1/12

[desert tree]

ok so those lines (DB and CA) cut the cube into fourths correct?

[steep mantle]

yea

[desert tree]

the shaded area is only in one of those fourths

[steep mantle]

yup

[desert tree]

now we look at what DM and CM do to that last fourth

those 2 lines cut it into thirds right? approx*

[steep mantle]

idk

i guess

[desert tree]

so whats 1/3 of 1/4

[steep mantle]

1/12

but you cant say "that looks like 1/3"

[desert tree]

i mean its just a quick way to do it

[steep mantle]

what if it asks me to prove it

[desert tree]

itll be a lot of angles and law of sines and pythag thm

pythag thrm gets you DM and CM

isosceles and law of sines gets you the 2 equal angles of triangle DMC

find all the angles in tri DAM with trig

[steep mantle]

hows this?

[desert tree]

idk y youre only given 2

it doesnt feel like enough info

i can list the steps for 14 i drew it out tho

only took 20 mins

[steep mantle]

huh

oh, sure

[desert tree]

pythag thrm for tri DAM gives u DM length

[steep mantle]

whats tri dam

[desert tree]

trig on tri DAM gives u angles ADM, AMD, MCB, and CMB

[desert tree]

triangle formed by the points D, A and M

[steep mantle]

oh

im not supposed to know trig at my level so i dont think the intended solution is to use trig

[desert tree]

oh

[steep mantle]

but i know trig so go on

[desert tree]

angle BDA minus angle ADM gives you angle MDB and its equiv imo angle MCA

im like 90% sure we can just call tri DMC isosceles but it can be proven by the pythag thrm on tri DAM and tri CBM

use tan on angle MDB = x/.5 which gives you the line segment

subtract the portion from the whole for the other side

[steep mantle]

what line segment

[desert tree]

the little line with a single mark on it

split line AC in half and the shorter section of it after its bisected by DM

[fluid tundra]

formula exists (this formula is actually not that well-known)

A = 2/3 * median * median * sin(angle between medians)

and the answer directly follows

[steep mantle]

so 2/3 × 12 × 15 × 1 (max of sin(x) is 1 right)

this?

[fluid tundra]

write on the diagram if you can't then make one of your own

use what you know

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[dapper fable]

What will be the compound interest on a shm of 51200$ for 7/3 years and 15/4 % per annual

[plain badge]

Dude

use the formula

[dapper fable]

A= 51200 (1+ 0.0375)^(7/3)

@plain badge

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[fiery tendon]

is this right?

[zinc glacier]

it told you its wrong

what is your rise and run?

[cursive field]

no actually

he's right

the avg rate of change should b the slope b/w two points

[fiery tendon]

previous answer

[cursive field]

BUT he got the slope wrong

[pulsar hedge]

-3/4

the x

change is 4

not 5

[cursive field]

what?

[zinc glacier]

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[cursive field]

oh yea

correct

[pulsar hedge]

I think this is an error of the oppsie doopsies

[cursive field]

is fine

@fiery tendon you messed up the x and y

the slope is change in y/change in x

not change in x/change in y

check again

[fiery tendon]

Im supposed to (1, 4) and (5, 1) and ten do 5-1 = 4 and 1-4 = -3 which is - 3/4?

[cursive field]

yes

[fiery tendon]

yeah realized that I put the wrong one as the denominator and me putting the 5 is jus math issue

[pulsar hedge]

yes

[fiery tendon]

Also got this one

(-1, 1) and (5, 5) = 5 + 1 = 6 and 5 - 1 = 4 | 6/4 = 1.5? So would the rate of change be 1.5?

ah no same mistake

its 4/6 right

[cursive field]

wait one sec

... yes

correct

[odd edge]

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[steady tide]

hm, pretty sure that is not integrable

or at least, cannot be expressed in elementary functions

[odd edge]

@dapper fable Has your question been resolved?

[dapper fable]

ahh

can we not use leibeinz rule

[summer cradle]

you are probably not meant to find any nice expression for it because it is not representable by elementary functions

perhaps a taylor expansion and termwise integration will provide some intuition or a solution

[odd edge]

@dapper fable Has your question been resolved?

[toxic monolith]

notice that:

[dapper fable]

thanks

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[tiny lagoon]

(a+b)/3 = (ab)^1/2 I'm tryna prove this using a^2 + b^2 = 7ab

I've tried smth like this

a^2 + b^2 = 7ab

a^2 - 2ab + b^2 = 5ab

(a-b)^2 = 5ab

log_a-b(5ab) = 2

I don't know what to do after or is that even helps

[zinc glacier]

you were close

try adding 2ab instead

since you want a+b, it would make more sense to do (a+b)^2

[tiny lagoon]

oh then I can sub it in

I'll try that

hm

so I get (a+b)^2 = 9ab

(a+b)/3 = (ab)^1/2

(a+b)^2/9 = ab

[zinc glacier]

wait

that step is wrong

[tiny lagoon]

oh, why

[zinc glacier]

how did the ab become a+b

ok yeah you fixed it

[tiny lagoon]

mb, I just fixed that

we then sub (a+b)^2 = 9an

9ab/9 = ab

ab = ab??

what does this accomplish though

[zinc glacier]

you are trying to show a^2 + b^2 = 7ab implies log((a+b)/3) = (1/2)(log(ab))

get from the first equation to the second

[zinc glacier]

you did get to this point, which is like one step away

[tiny lagoon]

I think I know what I'm not getting

[tiny lagoon]

since I turned the log statement into a normal exponent statement they are still equivalent stateements

and when we're trying to prove something, we don't normally do anything to the equation we're trying to prove to prove it

so like (a+b)/3 = (ab)^1/2

I would of left (ab)^1/2 unchanged

and tried to make (a+b)/3 into (ab)^1/2 using the other equation

[tidal laurel]

does anyone has the z transform example problems ???

[tiny lagoon]

nvm

I just understood what u meant

thanks @zinc glacier

[zinc glacier]

youre welcome

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[desert marlin]

🐺

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[gloomy mauve]

Could someone tell me my mistake?

[gloomy mauve]

Well never mind

I found it

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[proven cape]

prove that the sequence (1+1/n)^n is bounded.

[proven cape]

I verified that the sequence is non-decreasing but I don't know how to prove this.

pls help

[mystic saffron]

do you know like

the binomial theorem

[proven cape]

i have a sort of dimostration without binomial theorem

[proven cape]

yes

but I would prefer to do it without

[ancient idol]

you can also use Bernoulli's inequality

[proven cape]

i have the proof with Bernoulli's inequality but I don't understand how he got to this point

[ancient idol]

what exactly you don't understand?

[proven cape]

why do we have to consider a sequence b_n=(1+1/n)^n and show that it is monotone non-increasing?

<@&286206848099549185>

[odd edge]

@proven cape Has your question been resolved?

[gray bronze]

i think that proves convergence, not boundedness

https://en.wikipedia.org/wiki/Monotone_convergence_theorem

convergence is a stronger form of boundedness so i guess that works too

[odd edge]

@proven cape Has your question been resolved?

[odd edge]

@proven cape Has your question been resolved?

[odd edge]

@proven cape Has your question been resolved?

[azure kernel]

Yes, because a convergent sequence is also bounded.

[odd edge]

@proven cape Has your question been resolved?

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[mystic saffron]

Weird question, but it's been nagging at me, its not related to any studywork I'm doing

But like, we can express "2X+Y=15" or whatever as a line on a graph, because theres an infinite amount of solutions

But what about multiplication? (Something like "2X*Y=15")

Cause there's infinite answers but its not an unbroken line right? Cause neither X or Y can equal 0, wouldn't that mean there's no intercepts?

Idk, the answer is probably that theres no way to express this on a graph, but i was curious.

[south lichen]

,w plot 2xy = 15

[clever fjord]

Results provided by WolframAlpha

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

Requested by abstia

[south lichen]

Answer is often simple >.>

[mystic saffron]

Is there a word for this type of equation? Since its not linear

And thanks yeah, idk why i didnt check it.. i own a graphing calculator

Actually 2

[green elm]

it's a hyperbola

[mystic saffron]

I see

Thank you

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[viscid rain]

anyone know how to do this properly algebraically?

[green elm]

can you write 100m^4 as a product involving 5, 4 and m^2?

[viscid rain]

oh wow thanks

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[twilit flax]

Hey peeps, I have a question

[twilit flax]

How do you multiply (3 + 2i√5) (3 - 2i√5)

you're supposed to get 17 here, but i forgot how to multiply it

[quasi sparrow]

$(a+b)(c+d) = ac + bc + ad + bd$

[clever fjord]

riemann

[twilit flax]

foil method works on that?

[narrow blaze]

Yes.

[twilit flax]

i try but i keep making the "2i√5" a problem

cause how do i foil that

[nimble blaze]

wdym

[twilit flax]

ere

you're supposed to get 17

but my answer's different

[nimble blaze]

show your work

[twilit flax]

wait

yea i'm confused now cause of "2i√5"

[nimble blaze]

show your work

[twilit flax]

alright

ok i dont get it anymore

[quasi sparrow]

why is your inner negative

oh that's a hyphen

your outer should be negative

[twilit flax]

then

isnt the inner supposed to be negative or not

[quasi sparrow]

why do you think inner should be negative

[twilit flax]

ok nvm

how do you multiply the imaginary number and radical sign

?

that's only my problem, i forgot how to do it

[quasi sparrow]

i is its own number

[nimble blaze]

the same way you multiply other numbers

[twilit flax]

if i were to multiply (2i√5) (2i√5)

would i get 9

[quasi sparrow]

no

[twilit flax]

7 perhaps?

[quasi sparrow]

no

guessing is pointless

learn the rules of multiplication

$i^2 = -1$

[clever fjord]

riemann

[quasi sparrow]

$\sqrt{5}^2 = ?$

[clever fjord]

riemann

[twilit flax]

ah

i forgot about this

[twilit flax]

this would just be 5

i was thinking 2(-1)5

or 6

[quasi sparrow]

you had 4 above

how did 4 turn into 2

[twilit flax]

oh

i multiplied 2i and - 2i

a

yea like i did that in order to get 4

i forgot

or it would become 4(-1)5

[quasi sparrow]

you had 5 here

yes that's right

,calc (2 * i * sqrt(5)) * (-2 * i * sqrt(5))

[clever fjord]

Result:

20

[quasi sparrow]

lost a minus sign in front

[nimble blaze]

off by a - sign

[twilit flax]

thats a 20?

or an example

[quasi sparrow]

this is how you get L in your FOIL

[twilit flax]

but back to this

how is the answer 17 like on my book

[quasi sparrow]

[twilit flax]

i forgot

wait

ok i get it

nvm

[nimble blaze]

(correct answer isn't 17 btw)

[twilit flax]

alright

i get it now

oh

i wasn't supposed to use foil but multiplication

[nimble blaze]

wdym

[twilit flax]

basically

the topic is "Determining the quadratic equation given the roots"

[nimble blaze]

"foil" or otherwise indicates how to initially expand
then multiply normally

[twilit flax]

i try foil and multiply, both are different answers

[nimble blaze]

how so

[twilit flax]

in foil i get 29

multiply i get 18?

[nimble blaze]

how are you getting 18

[twilit flax]

what am i supposed to get

[quasi sparrow]

!original

[odd edge]

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.

[nimble blaze]

how are you getting 18 from "multiplication"

[twilit flax]

wait

[nimble blaze]

valid methods willlead to the same result

getting different results means you messed up somewhere

[twilit flax]

wait

oh yea

i do get -29

29 or - 29

nvm 29

[odd edge]

@twilit flax Has your question been resolved?

[odd edge]

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[cedar geyser]

How would I sinplfy this?

Would it just be 1/(4x^7/2y/2) / (x^6y^-3/2)^2?

[leaden widget]

first, spread the -2 exponent to all the factors of the fraction

[cedar geyser]

How do I spread them?

[leaden widget]

just basic exponent properties, you would multiply by exponent of each factor

[cedar geyser]

So I would get

[leaden widget]

$\frac{4^{-2}x^7y^{-10}}{x^{-12}{y^3}}$

[cedar geyser]

(4x^-7 * y^-10)/(x^-12 * y^3

[clever fjord]

Soosh

[cedar geyser]

Oh right forgot to square the 4

[leaden widget]

now combine like factors and move around as needed to have positive exponents

[cedar geyser]

Should I turn them into lik

1/16?

[leaden widget]

$\frac{x^{19}}{16y^{13}}$

or sorry x^19

[clever fjord]

Soosh

[cedar geyser]

Cause 7+12 right

[leaden widget]

yep

[cedar geyser]

Alr makes sense

Thanks

I factored this as x(x^3 + 8)

Is there a thing for sums of cubes?

.close

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[cobalt pike]

How do I go about this when it's those characters?

[quasi sparrow]

treat a like a number

a=2 if you like

[cobalt pike]

I'm not quite understanding what you're saying, so just plug in 2 for a?

[quasi sparrow]

yea do that

if you can find the expression for x=2, then you can do it for x=a

just don't simplify the algebra when you do any arithmetic

[cobalt pike]

Would this be the answer?

[quasi sparrow]

yea but simplify probably

[odd edge]

@cobalt pike Has your question been resolved?

[cobalt pike]

simplified is 8/(a+7)(a+h+7)?

.close

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[grave yew]

.status

[split ice]

Where you stuck

A) false

B)true

[grave yew]

yeah i got False, True, False

just wanted clarification

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[grave yew]

thanks

[split ice]

C) is true

Wait

[grave yew]

.reopen

[odd edge]

✅

[grave yew]

you said true?

if so why

[split ice]

Like we have 6 columns,4 povit colum9

Columns

[grave yew]

yes

wait no

6 rows

4 pivot columbs

sorry no columns*

😭

6 columns

[split ice]

Equation are rows right

Variables are column

[grave yew]

yes

[split ice]

6 columns,4 povit se got null space

Infinity solutions

[grave yew]

okay

[split ice]

👍

[grave yew]

.close

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[astral rock]

I did log differentiation but it didn't work

[astral rock]

can you help please

[knotty oasis]

do you know the limits of this type?

[leaden widget]

rewrite as: $e^{\ln(1+\frac{r}{x})^x = e^{x\ln(1+\frac{r}{x})}$

[clever fjord]

Soosh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[astral rock]

I did that

[leaden widget]

$=e^x(1+\frac{r}{x})$ and now use product rule i suppose

[clever fjord]

Soosh

[leaden widget]

er sorry its not differentiation nm product rule

[knotty oasis]

You can use the fact that the expression $\left(1+\frac{1}{x}\right)^x$ goes to $e$ as $x$ goes to infinity

[clever fjord]

Gapi

[knotty oasis]

This is also true for any f(x) : $\left(1+ \frac{1}{f\left(x\right)}\right)^{f\left(x\right)}$

[leaden widget]

@astral rock hm not sure if this is easiest way to do it, maybe im overcomplicating it but you can rewrite $e^{x(1+\frac{r}{x})}$ as $e^{r\frac{\ln(1+\frac{r}{x})}{\frac{r}{x}}$

[clever fjord]

Soosh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Gapi

[trail prism]

ohh yeah isnt (ln(1+x))/x = 1 for limit x -> 0

[leaden widget]

and you should be able to evaluate the limit if that fraction

as 1

[trail prism]

thats just e^r then

[knotty oasis]

it f(x) foes to infinity, then this expression goes to e

[trail prism]

this becomes e^r right? and then to again get limit we take logarithm as initially we exponentiated it. so limit just becomes r

[leaden widget]

so do that then you can rewrite some variable u = r/x and evaluate as u -> 0,

you can probably use l'hopitals for that or just think of it as lim u -> 0 ( ln(1+u) - ln(1) ) / u which is definition of derivative

(sliding the ln 1 = 0 in there)

so its derivative of ln at 1 which is 1/1 = 1 : ) yeah kinda roundabout way to think of it but whatever works, there is probably better way

[leaden widget]

yes becomes e^r as required, i'm a bit unclear on what you mean but yeah i'm sure there are multiple ways to think about the limit of that fraction part

[trail prism]

no sorry i thought you obtained e by exponentiating both sides but you just used z=e^ln(z) ok so no need to take log, its evidently e^r by using limit y-> 0 ln(1+y)/y = 1

[leaden widget]

@astral rock you still here? seems original asker left anyway

[leaden widget]

yes

actually yeah thats easily to l'hopitals, better than thinkinf of definition of derivative 😄

[odd edge]

@astral rock Has your question been resolved?

[odd edge]

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[cedar laurel]

Not part of my syllabus but would like to learn to solve for a project I have to do in my courses later on. My teacher taught me anyway and I think I remember the answer at 20%, but not sure how to get there, the answer scheme is written in error so I can't check.

[wooden python]

well let's see

so the respondents draw one of three cards: "say yes", "say no" and "answer honestly"

and for the responses we have 60 people of which 24 said yes and thus 36 said no

we can assume the cards were well-shuffled so 20 people got the Yes card, 20 got the No card and 20 got the Honest card

so subtract 20 from each of the yes and no totals

and you get 4 yes vs. 16 no

[cedar laurel]

Okay thanks.

.close

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[silver pecan]

how do i find the inverse of this function?
usually I'd try to make x the subject and then swap the variables at the end, but this isn't really possible in this case to make just x the subject, or is there?

[marble laurel]

Hmm

[signal oar]

Quadratic formula

[silver pecan]

what

really?

quadratic formula here?

[marble laurel]

Square both sides and you'll see why

[signal oar]

Well that's one possibility..

[silver pecan]

damn

letmme try taht

[signal oar]

Another would be completing the square

(which is basically the same, internally)

[marble laurel]

Ye

Make sure when you're multiplying y^2 and (5n+1) to distribute and then take n as a common factor

And then b = whatever is multiplied by n

[silver pecan]

na ur right quadratic function is the way to go here

lmao never in a million years i expected to use the quad eqn for these types of questions

wait

do i take the positive or the minimum

for the square root

[signal oar]

You get n =

In other words, the expression on the right describes your domain

[silver pecan]

wot

[signal oar]

Now think about what the domain of your original function is

[silver pecan]

the domain is 0 to 2.25 for the inverse

[silver pecan]

0 to 3

[signal oar]

[f(n) = \frac{3n}{\sqrt{5n + 1}}]

[silver pecan]

yes

[signal oar]

The domain of that is 0 to 3?

[silver pecan]

yes

wait no

[signal oar]

Is n a real number or a natural

[silver pecan]

its not

its just the question said f(n) is defined from 0 to 3

the domain is NOT 0 to 3

[signal oar]

Ah

[silver pecan]

(i have to sketch it from 0 to 3 later on)

[silver pecan]

i dont get ur question

[signal oar]

You have n = something in terms of y

If we let y run freely, what values does this expression take on

Like for y = 0 we get n = 5/18

[silver pecan]

ykw, i got this from here, thanks for the help

[signal oar]

For y = 1 we get something more than 5/18 etc.

For y = -1 the same

[silver pecan]

close

close

.close

[odd edge]

Channel closed

Closed by @silver pecan

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[signal oar]

So we get $n \in \Big[\frac{5}{18}, \infty)$

That doesn't include 0 to 3

So we should take the negative root..

[signal oar]

np

[odd edge]

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[marble laurel]

Heyo did you try anything yet?

[odd edge]

@minor rivet Has your question been resolved?

[mystic saffron]

consider the behaviour of the integral as x approaches 0 and as x approaches infinity on i) @minor rivet

[odd edge]

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[spare escarp]

someone help me with this pls?

[mystic saffron]

rationalize

[wooden python]

!status

[odd edge]

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

[wooden python]

also what's the goal here? to find the 9th decimal place of this?

[solemn leaf]

Gasppp

I know this one

[wooden python]

gonna fire a warning shot here

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[solemn leaf]

You just get the bottom thin-

DID I DO SOMETHING WRONG

[wooden python]

even if you do know this problem, let op tell us what he's struggling with first

[spare escarp]

no forget it-

[wooden python]

you didn't YET do anything wrong, im just reminding you of the rules around here.

[solemn leaf]

Oh okay okay, Yeah that's what I was about to do dw

[spare escarp]

you don't give accurate answers someone told me that

[lavish kite]

well you hadn't yet, but yeah it looked like you were about to just give an answer

[solemn leaf]

Got it, thanks

SORRY 😭

[wooden python]

1. "you" as in "y'all", or "you" as in me specifically?
2. who said that?

3. do you still need help with this, yes or no?

[spare escarp]

bruh idk lmao but someone told me do not trust Ann cuz she doesn't give specific explanations and stuff

[spare escarp]

idk if you want to help

[wooden python]

and who told you not to trust me?

[wooden python]

i do.

[solemn leaf]

I have bothered everyone in this server so much and I assure you, they KNOW what they are doing

But

[fluid tundra]

[solemn leaf]

OHHH

💀

[wooden python]

i am actually curious

who is smearing me?

[solemn leaf]

Anyways, I think you can get help by everyone here so-

[spare escarp]

idk it was like month ago or weeks ago

some guy named jason

[wooden python]

anyway

[spare escarp]

I think month ago

[wooden python]

do we maybe wanna get back to the problem

and maybe you can tell me your status

[spare escarp]

are u sure u wanna help?

[wooden python]

yes

i can tell you as much 19 more times

[spare escarp]

thanks 🌟

[wooden python]

if you don't believe me the first two

do we maybe wanna get back to the problem
and maybe you can tell me your status

[spare escarp]

so first square root of 5 - 5
?

as in double right or?

[wooden python]

wording...?

[spare escarp]

√5-5

[wooden python]

i am not sure what you're going for here.

[spare escarp]

√5-5
√5-5

[wooden python]

ok alright. i think i might understand

[spare escarp]

to multiply these or no?

[spare escarp]

idk-

I never used this method before

[wooden python]

you're saying "I want to multiply the top and bottom of this fraction by sqrt(5) - 5. Is this a good idea or not?"

[spare escarp]

idk I've had some but not same as this one

so idk if it needs to be the same

[wooden python]

fuck

look i am trying to make sure we communicate clearly and i understand what you are saying

is that so criminal

on my part

[spare escarp]

bruh idk either lmao the teacher told us to work in groups the end

[wooden python]

fuck

this is kind of frustrating

i was about to say that yes, your idea (as i understood it) is good

but like

when people try to confirm what you're saying

you're supposed to either CLEARLY confirm ("Yes, that's what I am saying.") or CLEARLY deny ("No, this is not what I am saying.")

[spare escarp]

Uh that's what I am saying

[wooden python]

communication, yeah?

[spare escarp]

I guess

ok

[wooden python]

yes, your idea of going from $$\frac{3 - \sqrt{5}}{\sqrt{5} + 5} \quad \mbox{to} \quad \frac{(3 - \sqrt{5})(\sqrt{5} - 5)}{(\sqrt{5} + 5)(\sqrt{5} - 5)}$$ is good.

i have in mind a slight improvement on your idea. but i won't say what it is unless you ask me for it clearly. you can ask me for it clearly by saying ``can you share what you had in mind?'' or something similarly clear.

[clever fjord]

Ann

[wooden python]

@spare escarp

[spare escarp]

yeah

[spare escarp]

we dont know what u talking about

[wooden python]

what part of "yes, this idea is good!" do you not understand?

[spare escarp]

thats what we were talking about

we don't know if they need to be all multiplications etc

[wooden python]

i... really do not understand what you're trying to say here.

you expressed the following idea: multiply the fraction on the top and bottom by sqrt(5) - 5

i said that this idea is good.

and now i've said it again for the second time.

(A) We still don't understand what you wrote up there.
(B) We understand what you wrote, but how do we progress from here?
(C) What was that improvement you mentioned?
(D) Something else.

[spare escarp]

B

[wooden python]

ok right

you'd do the multiplications in num and denom

separately from each other, but with the same principle

i.e. the distributive law

for the bottom, you can also recognize that the identity (a+b)(a-b) = a^2 - b^2 works.

which can save you some time/ink/space/etc.

but the top you have to do the honest way.

[spare escarp]

uh nvm forget it

[wooden python]

??

what's wrong now?

[spare escarp]

one of my group said we don't know if they need to be like this

3sqrt 5 - 15 - sqrt 5 sqrt 5

cuz of 3 times sqrt of 5 and so on

[wooden python]

talking about the top here, yes?

you're missing the last term: (-sqrt(5)) * (-5).

the full expansion will be 3 sqrt(5) - 15 - sqrt(5) * sqrt(5) + sqrt(5) * 5.

i have to say by the way
"uh nvm forget it" sounds to me as if you mean the following: "I've read your explanation, and I think it is useless and I don't understand a bit of it, and honestly I don't even want to hear you explaining yourself."

it sounds very rude, is my point.

i don't know if you're trying to be rude. and i hope you're not.

but that's how it reads on my end.

[spare escarp]

I asked someone in random servers if thats correct and he said Not simplified enough

nvm thanks

.close

[odd edge]

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[wooden python]

fuck

ajtbhglasgmgfh;

if you're going into "random servers" then why here??

[odd edge]

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[spare escarp]

bc I tried to find someone else not you bc I've heard that you trolling and I doubted them at first but they were right

[spare escarp]

.close

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[wooden python]

and you could've been explicit about it?

you could have said "sorry Ann I don't want help from you specifically"

i would be annoyed but i would understand

[ember oak]

What a strange journey

[odd edge]

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[peak lake]

$\lim\limits_{x \to \infty}\frac{e^{\sqrt{\ln x}}}{x}$

[clever fjord]

normalAtmosphericPa=101,325

[peak lake]

Can the numerator be simiplifed?

If not, how do you compute this limit?

can you even compute this limit?

my overall goal is to show that $f(x)=e^{\sqrt{\ln x}}}$ is $O(x)$

[clever fjord]

normalAtmosphericPa=101,325
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[peak lake]

For intergers $x$

[clever fjord]

normalAtmosphericPa=101,325

[plain ruin]

Well

I have an idea

Let's think

√(lnx) means this number is the root of ∞

And it's e^of this root of ∞

So it is indeed less than infinity

Then

Denominator is x

Which is infinity

So when we divide ∞ but less than it by the true ∞

We get 0

I guess

Idk actually

[plain ruin]

No idea

[odd edge]

@peak lake Has your question been resolved?

[odd edge]

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[fallen prism]

for 9b

i thought its 1.25 x 65?

my answers wrong

[fast whale]

what's the answer?

[fallen prism]

162.5m?

its weird

[fallen prism]

isnt 1.25 radians

or no

why is it being converted

[fast whale]

in q7 where the options were in degrees ur book put the ° sign with them

so I suppose 1.25 is not degree but radians

[fallen prism]

yea same

so y did they convert

[fast whale]

is the answer from the official book answer scheme?

[fallen prism]

yea

[fast whale]

pretty sure u can't even use degrees in s=rtheta

[fallen prism]

they have wrong answers

[fast whale]

theta is in rad

[fallen prism]

sometimes

[fast whale]

then the answer is wrong ig

[fallen prism]

is mine

correct

1.25 x 65

[fast whale]

yes

[fallen prism]

yayokay

ty

.close

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[fast whale]

@fallen prism
if u wanna solve in degrees then the formula is
(theta/360)x2πr
for that u convert 1.25 to degrees which u get as 71.62°

(71.62/360)x2π(65) = 81.25
which is the same as 1.25x65

[odd edge]

Please don't occupy multiple help channels.

[fast whale]

ur book did the conversion wrong from rad to degree

[fallen prism]

okk ill try it

tysm

[fast whale]

.close

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[gleaming hedge]

I have to find all complex solutions of this system depending on the values of the complex parameter lambda

[gleaming hedge]

I'm guessing I have to RREF it first?

I get this, am I doing good so far?

.close

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[spare escarp]

Hi someone help pls?

[spare escarp]

Thanks!

[nimble blaze]

an expression by itself is not a question

what are you being asked to do with this?

[spare escarp]

(2√2-3)
(2√2-3)

?

[spare escarp]

it only says

Rationalise the denominator and simplify:

were on 8th page

we got stuck on this

[sullen ferry]

Well if the denominator were not squared would you know how to do it?

[spare escarp]

Yeah we think so

[sullen ferry]

And we have this exponent property that a^n * b^n=(ab)^n

[spare escarp]

👍

yeah

[sullen ferry]

So all you would need to do is multiply by conjugate squared in numerator and denominator. Because a rational number squared is still rational

[spare escarp]

3 times (2 sqrt 2 + 3)^2

?

[sullen ferry]

(2 sqrt 2 -3)^2

[spare escarp]

sorry I mean -

lol

[sullen ferry]

Multiplied in numerator and denominator

[spare escarp]

(2sqrt2)^2 - 3^2 = 8 - 9 = -1

?

[spare escarp]

were not sure if its correct

[sullen ferry]

That is (2 sqrt 2 + 3)(2 sqrt 2 - 3)

[spare escarp]

(2 sqrt 2 + 3)(2 sqrt 2 - 3)

(2sqrt2)^2 - 3^2 = 8 - 9 = -1

?

or maybe (3)^2

?

[spare escarp]

this is our first experience so were not quite sure

[sullen ferry]

So what should we expect (2 sqrt 2 + 3)^2 * (2 sqrt 2 - 3)^2 to be?

[spare escarp]

wait

[spare escarp]

(2sqrt 2 times 2sqrt 2) + (2 sqrt 2 times 3) + (3 times 2 sqrt 2) + (3 times 3)

?

[sullen ferry]

????

[spare escarp]

oh wait it was correct?

[spare escarp]

?

[sullen ferry]

You mean * instead of +

[spare escarp]

times *

[spare escarp]

they have + though

my group said sorry if this got chaos

they said they don't know how to continue

@sullen ferry sorry do you think this is a lot to work on or no?

some people use shorter methods

so idk

[sullen ferry]

So if this is -1

What would (2√2+3)(2√2-3) (2√2+3)(2√2-3) be

-1 *-1 right?

Which is

[spare escarp]

1

[spare escarp]

right?

[sullen ferry]

Yes.

So the denominator becomes 1. It goes away.

Now all thats left is 3(2√2-3)(2√2-3)

[spare escarp]

👍

so we need to multiply all of them by 1 or?

[spare escarp]

👍

[sullen ferry]

Multiplying by 1 keeps them the same...

[spare escarp]

Oh yup lol

[sullen ferry]

So just keep it as 3(2√2-3)^2 or expand it out

[spare escarp]

uh the answer is that wth?

[sullen ferry]

Ye

[spare escarp]

one of our group sended a msg to the teacher and the teacher said its wrong

[sullen ferry]

...

,calc 3/((sqrt(8)+3)^2)

[clever fjord]

Result:

0.088311754568578

[sullen ferry]

,calc 3 ((sqrt(8)-3)^2)

[clever fjord]

Result:

0.088311754568578

[spare escarp]

yup its wrong oof

[sullen ferry]

What

[spare escarp]

bruh nvm 🤦‍♀️ my group said forget it

[sullen ferry]

I think you just need to expand 3 ((sqrt(8)-3)^2)

[odd edge]

@spare escarp Has your question been resolved?

[odd edge]

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[buoyant matrix]

solution check (not a Yossi question)

[odd edge]

@buoyant matrix Has your question been resolved?

[odd edge]

@buoyant matrix Has your question been resolved?

[odd edge]

@buoyant matrix Has your question been resolved?

[odd edge]

@buoyant matrix Has your question been resolved?

[buoyant matrix]

[haughty hornet]

☠️

[odd edge]

@buoyant matrix Has your question been resolved?

[odd edge]

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[river dirge]

If we have n iid geometrically distributed variables, what distribution does the n-th order statistic have? What expectation does it have?

[river dirge]

Sorry, I can't find this online

[tardy lagoon]

The nth order statistics is the maximum of the variables

[river dirge]

Yeah

[tardy lagoon]

Let Y=the nth order statistics, then the CDF of Y is P(Y<=y)

The probability shouldn't be hard to calculate

[river dirge]

$$(1-p)^{n * (y-1)} * p^{n} - (1-p)^{(n-1) * (y-1)} * p^{n-1}$$

[clever fjord]

yoohoo

[river dirge]

does this seem like a reasonable pdf?

how would you calculate the expectation?

hmm

maybe its not so bad actually

[tardy lagoon]

Can you show your steps?

[river dirge]

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[midnight goblet]

can someone explain how to solve this without knowing the original function?

[quasi sparrow]

find where A, B, and C get mapped to

Write down the coordinates of them

[midnight goblet]

oh I didn't think it was that simple, I thought we had to look for the original function.

thanks 👍

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[tribal narwhal]

might need a lil' help, i actually do not know how to write this down

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[hexed edge]

Hi everyone!

[hexed edge]

I am working on this affine equation problem

but when i sub in the values back, they dont really match

Did I do it correctly?

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[dusty halo]

hello

[dusty halo]

I don't understand this

hello?

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[sinful stone]

Hello. I asked a question earlier than immedielty went afk, so I apologize for that.
"Solve the system of equations
and state the solution in parametric vector form"
I'm quite stuck on how I should approach this.

[sinful stone]

I turned this intro a matrice

[split ice]

You know how to solve system of equations by matrix ??

[sinful stone]

And this is what I got simplified,

I'm recognizing the order of the columns goes from x1 to x4, with the last column being the constants

I'm recognizing that x4 = 1, but I don't know where to go from here

[split ice]

X4 is not 1

[sinful stone]

woop

[split ice]

There are 4 pivot column

So there are infinity many solutions

[sinful stone]

Wait? Isn't the third column a free column?

Free variables + pivot/leading variable = number of variables

[split ice]

Question matrix and your matrix different

It's 3 x 4

Yours 3 x 5

[sinful stone]

the 5th row is combined with the = [matrice]

[split ice]

Oh sorry

Ya your x3 column is free

There is no pivot

Ya you are right then x4=1

Apologies

[sinful stone]

this is what the answer states and I am confused

could u help me? I'm forced to go afk in like 3 minutes

[split ice]

So x4=1

By 2nd row we get

X2 + x3 =0

So x2 =-x3

And by 1st row we get x1-x3=0

So X1=x3

Because x3 is a free variable

We need all variables in form of x3

Are you following

So according to you row reduction matrix answer should be

X3 [1 -1 0 1]

[sinful stone]

I’ll ask for more help later today, thanks so far ❤️

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[somber turret]

the n and m are switched places but i wanna know the name of this formula

[spring knot]

permutation

[brisk flicker]

yup

[somber turret]

ty pookies

😘

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[scenic lodge]

i dont know the things i didnt fill in

how do you do that part?

[wild rune]

the thing to know is that

if a root has an odd multiplicity, the graph crosses the xaxis (from positive to negative or viceversa)

if a root has an even multiplicity, the graph only touches the x-axis and comes back where it came from (it doesnt cross it)

[scenic lodge]

so all of them are odd except 5

[wild rune]

yes exactly

[scenic lodge]

ok

how would you find the degree, at first i thought it was 6 then 5 but its both wrong

btw

[scenic lodge]

when you say root you mean?

[wild rune]

root refers to the zero

[scenic lodge]

okok

wait i think i get it now

the multiplicity is given in the equation

like y = (x-1)(x+1)^2

[wild rune]

ahh perfect

[scenic lodge]

so the zeros would be 1 and -1

and the multiplicity of 1 = 1 and -1 = 2

so if you had to graph it

it touches at -1

and goes through 1

and degree is 3

[wild rune]

exactly

[scenic lodge]

so its odd and positive

[wild rune]

yup

[scenic lodge]

so end behavior will be diagonal left up to right

so how would i know the leading coefficient if its in form of y = (x-1)(x+1)^2

[wild rune]

the leading coeficient is equal to the coefficient on the product of the highest degree term of each factor

[scenic lodge]

so it would be (x+1)

which is 1?

[wild rune]

not quite

the equation is the same as $(x-1)(x+1)(x+1)$

[clever fjord]

The Great D

[scenic lodge]

true

[wild rune]

take the x from each factor and multiply them together

you get $x^3$

[scenic lodge]

x^3

[clever fjord]

The Great D

[wild rune]

this has a coeffcicient of 1

thats the leading coeffecicient

[scenic lodge]

humm

im just wondering why it would ever be negative then or how would it ever be anything other than 1

ok wait

[wild rune]

if it was $(2x-1)(-3x+1)(x+1)$

[clever fjord]

The Great D

[scenic lodge]

thats dubm

lol

yea

hahaa

my teacher only gave us 1 example

[wild rune]

no its a very good question

[scenic lodge]

thank you

[wild rune]

your welcome

[median sleet]

Can anyone help me with direct and indirect proportion

[scenic lodge]

other question is , after i et the leading coefficient, the degree, 0's, multiplicities and figure out the end behavior how do i know where to graph in between the zeros

like for example y = (x-1)(x+1)^2... degree = 3, leading = 1, multi = 2 and 1

so like i said before the end behavior is an > 0

so left to right diagonal, positive and it toucjes the -1

but how do you know how far down it goes

after it touches

[scenic lodge]

sorry you have to go to a help channel that isnt being used

[median sleet]

Yh thank u I figured it out I’m new

[scenic lodge]

ok np

not sure what this is asking?

brb

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[weary creek]

Can someone please help me with this one?

[uneven sable]

What did you try?

[weary creek]

I have no idea of what to even do tbh

I know that it should be 1/3*pi*30²*120h

But that isn't right