help-10

[crisp marten]

I know that we use it at the end

But comparison hasnt occured to me in this step

Thx

.close

[obtuse pebble]

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[obtuse pebble]

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[bitter echo]

I am enable to find the slope help..........

[high lily]

first identify two points on that line

[craggy tundra]

D

it's going downwards so gradient must be negative

[bitter echo]

(-5,-2) and (-7,1)

[craggy tundra]

and it passes through a point below x=0 so the intercept negative

[bitter echo]

whats gradient slope ?

[craggy tundra]

gradient = slope

[bitter echo]

didnt knew that

damn

ur trick solved it without atually solving

[craggy tundra]

yes

[high lily]

(-5,-2) isn't a point on that lime

[craggy tundra]

here is the gradient formula tho

for future use

[bitter echo]

its clearly there wdym its intersecting isnt it

[bitter echo]

means a lot

ty

[craggy tundra]

(x1, y1) are the coordinates of one point on the line and (x2, y2) are the coordinates of the other line

[high lily]

to me it's clearly not on the line

can you show me where you think (-5,-2) is

[craggy tundra]

note that you cannot have (y1-y2)/(x2-x1)

[bitter echo]

well ik that much

[bitter echo]

i mean the blue line is intersecting those two perfect spots so shouldnt it be ?

[high lily]

no

if you are referring to the intersection with the axes, those are two separate points

[bitter echo]

yeah ik that

so if the intersecting line is above x that means y intercept is + ? @craggy tundra

[high lily]

(-5,-2) is somewhere else

[bitter echo]

oh

[craggy tundra]

yes

[craggy tundra]

your points are (-5,0) and (0,-2)

not (-5, -2)

[high lily]

if you want the coordinates of a certain point, pick one point one the line
and identify it's x and y coordinate

[bitter echo]

but why is that

wait

so (-7,1) is corret ?

[high lily]

wait

it just looks like it is, but isn't

[bitter echo]

i think it is also correct

oh

[high lily]

the graph quality is too poor, but based on the intercepts which they supposedly want you to use, (-7,1) won't be on the line, it'd be very close though

[bitter echo]

alright thanks for the help

.close

[obtuse pebble]

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[thin thistle]

can someone tell me how we got this

[azure wraith]

are u

familiar

with the double angle formula?

[warm shale]

:D

[azure wraith]

u can prove this using the sum of angle for cosine

then u do something with that formula and trig identity to derive that

[thin thistle]

yes i am familiar with this

but is it like a play on formulas

or is it deeper than that

[azure wraith]

its just

play on formulas imo

u use this a lot for trig integrals

[thin thistle]

how did we get it from cos2x = cos^2x - sin^2x and sin^2x + cos^2x = 1

[azure wraith]

u can

equate using cos^2x

[thin thistle]

can u elaborate

oh nvm

[azure wraith]

u can express each equation in the form of cos^2x = ...

then equate

[thin thistle]

its cos2x = 1 - 2sin^2x

divided by 2

😭

[azure wraith]

ye

u can work from that

[thin thistle]

okay thanks

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[tender aurora]

What would be the quickest way to prove this has no solution, I really want to teach my friends how to solve this question as quickly as possible as it is a college entrance test practice question, and part of CETs are time management. Do I just tell them to skip it or what?

[tender aurora]

personally i just tried substitution and found nothing from the choices worked but that kinda takes way too long and how fast you do it mentally or on paper lol

[worthy cargo]

well if u substitute from the choices it’s easy to see A and B don’t work because it makes a denominator undefined and then plugging in 0 is always relatively easy

because if you plug in 0 you
immediately get
-1/2 = 3/2

which is false

so it must be D

[tender aurora]

yeah i found substition to be the fastest, i guess I'll just tell them to substitute

thanks, really needed a second opinion on it lol thanks

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[torn gazelle]

guys i need help

[obtuse pebble]

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

[plucky rivet]

#❓how-to-get-help @tranquil geyser

Oop

[tranquil geyser]

What

[plucky rivet]

You typed too fast

That it lagged my phone out

[tranquil geyser]

[torn gazelle]

How many real roots (counted with their multiplicities) can a monic polynomial P(1)=−3
, P(2)=2, P(3)=−4, P(4)=1 et P(5)=−5 does it have ?

[plucky rivet]

#❓how-to-get-help @torn gazelle

[torn gazelle]

.

[sweet gulch]

You might want to copy and paste that question into an unoccupied help chat

[torn gazelle]

wdym

[plucky rivet]

Are you even reading? Hello

[nocturne bronze]

!da2a

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[nocturne bronze]

!original

[obtuse pebble]

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.

[nocturne bronze]

!nogpt we are here

[torn gazelle]

i already send it

[obtuse pebble]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[plucky rivet]

This isnt your channel

[rocky goblet]

this channel is being used by someone else, you should go to an available channel

[sweet gulch]

Like #help-21｜아리스킨충1 is unoccupied, type your message there and it will reserve that channel for you

[plucky rivet]

If you want help do read first

[tender aurora]

i closed this channel what

.close

[nocturne bronze]

this is already closed

it will take some time to decay away

[sweet gulch]

It's closed, there is a gap time before it goes to open channels

[rocky goblet]

yeah it's closed, it just takes some time after it's closed for it to actually stop you from sending messages

[grizzled portal]

😂😂😂

[rocky goblet]

that way if you like, realise three minutes later that actually you're still confused, you can come back and .reopen it

and also you don't lose any messages you were in the middle of typing when someone else closed it

[nocturne bronze]

your worthiness will be explained when a new help occupied help channel comes

[rocky goblet]

(or if it gets closed by the bot when you weren't done)

[nocturne bronze]

funfact, you can type in a channel without claiming it by using:
.close [message]

[rocky goblet]

more specifically the condition is just putting a . at the start, so . hello works too, it doesn't have to be .close

[nocturne bronze]

english teachers will think otherwise

[obtuse pebble]

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[rare cargo]

can someone check my answer

[latent walrus]

show your working

[restive gorge]

no double working here

[latent walrus]

youre suffering a bit from early rounding

but the process is correct

try keep your values exact in the midst of calculations or the final answer can be slightly off

[rare cargo]

0.9896?

0.9910?

[latent walrus]

i got about 0.9932

was a bit rushed though

[rare cargo]

you are correct

thank you!

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[limber pawn]

I'm stuck on the 'deduce that the three altitudes of triangle ABC are concurrent' part - I've done the first part of i) and can't figure out where this second part of i) comes from.

[obtuse pebble]

@limber pawn Has your question been resolved?

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@limber pawn Has your question been resolved?

[obtuse pebble]

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[timid silo]

how do i go about calculating the second half of this equation? i've gotten the first part by reversing the numbers and doing 350% of 8 or 3.5 x 8 to get $28 of interest but im not sure what to do next

[dire locust]

5 year 4 month

[lone echo]

!noans

[obtuse pebble]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[lone echo]

first off, is it simple or compound interest?

[dire locust]

I = Prt/100
therefore,t=100 i/pr
Where:

P = Principal Amount=350
I = Interest Amount=150

R = Rate of Interest per year as a percent=8%
t = Time Periods involved=?

[obtuse pebble]

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[floral siren]

hi

[floral siren]

what is the difference here? they are the same theorem

what does B_j mean?

whats the difference of B and B_j

[obtuse pebble]

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@fading zodiac Has your question been resolved?

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[zenith mango]

is this done correctly?

[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[cerulean tulip]

let's say you have a plane defined by f(x,y) = z. You know the directional derivative of f(x,y) in 2 arbitrary (different) directions. How do you find the maximum value of the derivative across all directions given that info?

[cerulean tulip]

I am not quite sure where to start

[dark stirrup]

The derivative a specific point, or through the entire domain?

[cerulean tulip]

its a plane so same direction = same derivative no matter where it is right?

[dark stirrup]

Oh so it's of the form z=ax+by+c?

[obtuse pebble]

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[near kraken]

am i being dense

[near kraken]

1000*5^

so

1000*5^1 +2000 = 7000

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[hollow ridge]

how do i do this without cacl

[gloomy vector]

do you know what something raised by a fraction means

[obtuse pebble]

@hollow ridge Has your question been resolved?

[hollow ridge]

it means that like it can be written as a sqrt as well, uhhh

no

[uneven otter]

You could write it as $(16^\frac{1}{4})(16^\frac{1}{4})(16^\frac{1}{4})$

[warm shale]

Calc III Victim

[hollow ridge]

well yes

[uneven otter]

should b easy from there

[limpid eagle]

you could also rewrite 16 as 2^4

[uneven otter]

o ye thats better

[limpid eagle]

and use the rule $(a^x)^y = a^(xy)$

[warm shale]

AwesomeRat

[limpid eagle]

ok

I don't get latex

but

you can get the point

[uneven otter]

a^{xy} should work

[limpid eagle]

ahhh

that would make more sense

[hollow ridge]

latex has too much of a learning curve

[uneven otter]

idts

[hollow ridge]

rip

i could not learn it

[uneven otter]

I learnt it from seein other ppl use it on discord lol

[hollow ridge]

wth

[uneven otter]

I used to think it was complicated but its rlly not

[limpid eagle]

eh

it's not horrible

[hollow ridge]

alr bro

[limpid eagle]

just here and there you gotta look it pu

up

[hollow ridge]

i got 8

2 ^ 3

[limpid eagle]

mhm

[hollow ridge]

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[crisp ember]

hi could i get help with the start, i have come across a few of these questions where the derivative is geiven and a function at o is equal to zero but i always forget hoe to solve them

here is what i have done

[safe ravine]

Ok

[crisp ember]

any ideas @safe ravine

[safe ravine]

Do u know minima maxima

In first part they have asked when potential becomes minimum

[crisp ember]

yeah, do a double derivative

[safe ravine]

I mean they asked to show

[safe ravine]

No

[crisp ember]

oh

[safe ravine]

We begin simply

[crisp ember]

ok

[safe ravine]

A function has a maxima or minima at a point

if

1. f'(x) is zero

[crisp ember]

the derivative is 0

[crisp ember]

yeah ive done that in my working

[safe ravine]

for maxima f''(x) should be -ve and for minima f''(x) should be +ve

[crisp ember]

yeah

[safe ravine]

Now as u have found the points

Where f'(x)=0

To prove maxima minima

Just check the above condition at that points

[crisp ember]

what should i do with the -1

[safe ravine]

This

[crisp ember]

cos it only says 1 in the question

[safe ravine]

See

Do only what they have asked in the question

👍

[crisp ember]

ok

[safe ravine]

Ok do the 1st

If u have any doubt in 2nd let me know

[crisp ember]

im jst trying to do the doubke derivative

ok proved x=0 and x=1 are min and max

[safe ravine]

Okok

@crisp ember u want to find double derivative

Use product rule

[obtuse pebble]

@crisp ember Has your question been resolved?

[short spire]

That only shows local maxima/minima and not global..

[safe ravine]

For global

We consider end points of interval

And even critical points

[short spire]

Yes (if the domain is closed and bounded).

[safe ravine]

But the question asked simple things

So told it simply

It said to prove
(Mainly it wants u take the second derivative)

[short spire]

Well you don’t even have to take second derivative if the point is so to check for global maxima and minima

Just pick the largest and smallest one between the candidates

[safe ravine]

Ik that

Bit

But*

No domain given

[short spire]

I didn’t even read the question and yeah I see your point now, since we’re only given the derivative then that’s okay in this instance I suppose

[short spire]

Wouldn’t matter, you can still show

[safe ravine]

;-;

So what u want

I should directly put it in the f'(x) for it to be zero

We don't have f(x)

[short spire]

Well if you can show that the derivative after the maximum is always negative and likewise positive at the left (up to the minimum) then I would say that’s a better route to really ensure it’s not just a local maximum

And you can argue through the given point aswell

[short spire]

I don’t want anything, I agreed with you in this case that it’s sufficient (if one is lazy)

[safe ravine]

Okok

Oh

Sry beo

Bro* got lazy enough

[obtuse pebble]

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[smoky bear]

Suppose that I have a very small sample (n = 4) and from it, I get a sample proportion p'=1.0 (all successes). What would be an appropriate statistical test for me to do for me to say with statistical significance that it is greater than another proportion (p=0.75)? I've tried deriving the p-value by calculating the exact null distribution and I get a p-value of 0.0. I'm not sure this is correct though--

[smoky bear]

I've constructed the distribution through a discrete binomial probability distribution for numbers 0-4 (possible values given the sample size)

Sorry if my statistics jargon is a bit off;;; It's been a while since I've formally learned stats

[viral blade]

well ig you want the probability of getting a result at least as strong (all successes) under the assumption that p=0.75

so that would be 0.75^4

[smoky bear]

hmm i think i didnt include that in when i was minusing from the 1 since its a greater than

means that p > 0.05 though, a shame

oh well, much thanks+

.close

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[onyx girder]

how did they find the imaginary part

[sudden sage]

the isin(theta) and -isin(theta) in numerator cancel out

[onyx girder]

so should i convert it to cartesian form first

[onyx girder]

^

im not sure where the 0/2+0 is coming from exactly

but i do see that the isin cancel out if i do it in cartesian form

[sudden sage]

yeah

[timid silo]

why did they write 0/(2+0)? Specifically, why 2+0 in the denominator. It doesn't matter though, as the numerator is clearly 0, but I think they send the wrong message.

[onyx girder]

oh ok

i think ill just ignore it then since there is a clearer method

thanks

.close

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[onyx girder]

.reopen

[obtuse pebble]

✅

[onyx girder]

similar question sorry but is this a mistake they made eitheta = costheta it seems

shouldnt the real part be 1/2+2costheta in this case

nvm

.close

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[weak nymph]

hey, can someone explain to me why in step 3 the 3 becomes a -3 in the equations?

[weak nymph]

(im trying to do the elimination method for this)

[ionic oar]

they subtracted 3 from both sides

/moved the 3 over

[weak nymph]

where do they do that

and why

[ionic oar]

final step for step 2

[ionic oar]

to make it solvable for y

for example if i gave you the equation 2x + 3 = 1, how would you solve it?

[tidal isle]

also just saying, ChatGPT is an idiot. It often makes mistakes and presents them like facts. And while it is becoming better with the newer models, i wouldn't trust it with anything, you can't verify to be correct yourself

[tidal isle]

this is still correct tho

[ionic oar]

dang didnt realise it was gpt

[obtuse pebble]

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[glacial ocean]

can someone tell me how i make that like the perfect answer

or like u know

it is supposed to be 2,7... but i can't figure out how

[obtuse pebble]

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[north bolt]

I’m not sure if this is the best place to ask but I need to take a screenshot of a road from google maps and turn it into a set of co-ordinates. I was thinking of using GeoGebra but I’m unsure on how to scale it ?

[ionic oar]

pretty sure the scale is on the bottom right corner or something

[north bolt]

Yeah but how can I scale the screenshot once I insert it into GeoGebra

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[tame vigil]

what are the possible things you can use the expected value formula for a discrete random variable?
as in $\sum_{i=1}^{n} x_i p(x_i)$?

is it sufficient to say "this can be used to compute the average trials per success for anything with a consistent probability"

consistent probability
being the key phrase for "possible things" you can use it on

[warm shale]

AggressiveNooby

[obtuse pebble]

@tame vigil Has your question been resolved?

[obtuse pebble]

@tame vigil Has your question been resolved?

[obtuse pebble]

@tame vigil Has your question been resolved?

[brazen viper]

I think you're thinking a little too narrowly about what a discrete random variable can be, considering you're framing it in terms of successes. You're phrasing it as if you are constructing a geometric distribution (or maybe poisson? it's unclear) out of a repeated trial of bernoulli distributions. But you can also use the expected value formula to find (for instance) the expected roll of a die. Or the expected point value for a game of plinko, and so on.

[tame vigil]

hmm

ah ok

so i have a question

suppose you do $\sum_{i=1}^{inf} x_i p(x_i)$

[warm shale]

AggressiveNooby

[tame vigil]

that would compute like

the average trials per success in general

[brazen viper]

ok, this would be an example of a discrete random distribution, but one that has infinite possible outcomes.

[tame vigil]

right

so basically im tryna ask

suppose you had a trial with a 1% chance of a success

[brazen viper]

We can construct such a number by choosing a uniformly distributed value on (0, 1], then cubing it, inverting it, and then taking the floor.

[tame vigil]

and you could run that trial infinity times

[tame vigil]

this would compute the average trails per success, which in this case, would be 100

right

[brazen viper]

You would agree that this would give you a valid discrete probability distribution on the natural numbers

[tame vigil]

sure

[brazen viper]

and it would be equal to 1/1^2 + 1/2^2 + 1/3^3 and so on?

the expected value, in this case, assuming I constructed it correctly

would be pi^2 / 6 (basel problem)

and would represent the value that you expect to pull, on average

[tame vigil]

ah i see

thanks

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[violet stream]

I figured out everything in order to be able to graph it though I do not know how to use the information I solved for to graph it and I have questions as how do I know if I'm above the x axis or below the x-axis when the concavity is down or up or when it's decreasing and increasing. Thank you for your time and help

[obtuse pebble]

@violet stream Has your question been resolved?

[violet stream]

<@&286206848099549185>

[winter isle]

Hi

To know if the function is above or below the x-axis, you must do a sign table

To know if the function is increasing or decreasing, you must do the sign table of the derivative of the function. If f'(x)>0, the function is increasing and if f'(x)<0, the function is decreasing. If f'(x)=0, it depends of which signs the derivative have. If f'(x)<0 then f'(x)=0 then f'(x)>0, the function has a minimum. If f'(x)>0 then f'(x)=0 then f'(x)<0, the function has a maximum.

[violet stream]

so if f'(x) is increasing it's above the x-axis? if its decreasing f'(x) it's below the x-axis?

[winter isle]

That's it, if f'(x)>0 : f(x) increases and if f'(x)<0 : f(x) decreases

[violet stream]

does the same apply for the concavity?

or how does the convaity work

[winter isle]

Yes, more or less

You need the second derivative

If f''(x) < 0 : f(x) has its concavity down and if f''(x) > 0 : f(x) has its concavity up

[violet stream]

oh ok

[winter isle]

And it is the same thing for inflection points

The second derivative must be equal to zero with a change of sign around

[violet stream]

oh ok I see now

[winter isle]

Good, macte animo !

[violet stream]

I apperciate you taking the time Thank you

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[arctic shard]

<@&286206848099549185>

[timid silo]

Why do you need help?, ur kenny

[arctic shard]

I need help with my geometry

I can send you a screenshot

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[dawn vale]

im stuck

[wary badger]

write sin^2 in terms of cos

[fossil crag]

hint: get rid of sin^2

[wary badger]

oops

sorry

[dawn vale]

i know thats what i need to do but i dont think i know how to

[wary badger]

well if sin^2+cos^2=1 what is sin^2

[fossil crag]

by the looks of it, you clearly know the identity you must be using

[dawn vale]

oh

im dim

1-cos^2

ive seen my self miss easy things like this alot recently

any tips

[obtuse pebble]

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[sleek trail]

OpenOpen!

[sleek trail]

HepHelp

Sorry for the words copying it is a glitchglitch

[timber fox]

No ProblemProblem

What do youyou need help withwith

[obtuse pebble]

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[latent frost]

Solve the equation sin(2x) = -cos(x)

so my idea was to

but umm, seems like my solution aint it

[wary badger]

you may have divided out solutions

by dividing out the cosx

[latent frost]

yeah think so too

so how do I avoid that?

[wary badger]

you should add cosx over

then factor and set each factor to zero

[latent frost]

so 2sinxcosx + cosx = 0

[wary badger]

because the right hand side will be zero

[latent frost]

cosx(2sinx + 1) = 0

[wary badger]

mhm

[latent frost]

and then say sinx = -1/2

[wary badger]

so you have cosx=0 and sinx=-1/2

right

exactly

[latent frost]

aight

got it

tyty

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[solemn isle]

can anyone help me find some videos that explain how to solve these types of questions where there are graphs that relate to other graphs for polar eqautions

[limpid eagle]

https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DaSdaT62ndYE&psig=AOvVaw2S79G7Jh2_KyYgadHk75_b&ust=1723262947830000&source=images&cd=vfe&opi=89978449&ved=0CBQQjhxqFwoTCMjTufCE54cDFQAAAAAdAAAAABAE

it's not definitive of the relation

but it teaches you how to transform polar to cartesian and vice versa (I think)

[solemn isle]

cartesian?

i forgot what that was ):

anyway thank you

you are awesome

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[fallen inlet]

Howw?
I’ve tried x / 2 * 2/9 / x

[worn yoke]

the geometric mean of two numbers is the square root of their product

[fallen inlet]

So

Do I multiply them and then square root

[worn yoke]

yes

[fallen inlet]

Okay Tysm

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[slender dirge]

I can’t seem to figure out what I did wrong.
Pls help

[timid silo]

is the triangle right, first of all?

you are solving for a, correct?

You only use Pythagorean theorem when the triangle has a right angle

If it’s implied then

You foiled (a-5)^2 wrong

and (a+2)^2 wrong

as well as (a+3)^2

I think your foiling was the main error here

[magic geyser]

(a+b)^2=a^2+2ab+b^2

(a-b)^2=a^2-2ab+b^2

[timid silo]

I think from there his steps are correct, his foiling is incorrect that’s all

Like the process

not the numbers or coefficients

[slender dirge]

Ye it’s right angled

[timid silo]

Yea, just fix your foiling, if you need any help you can ping me

The process is right, it’s just your foiling at the start that kinda messed it up

[slender dirge]

oh so the a-5^2 should be a^2 -10a + 25?

@timid silo

[magic geyser]

correct

[timid silo]

Yep

[slender dirge]

Thanks!

[timid silo]

Yw, make sure to fix the other foiling

and then try seeing where that gets you

[obtuse pebble]

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[dense storm]

Hello! I don't really understand strong induction, I was wondering if someone could help me break it down?

[dense storm]

Strong induction from discrete mathematics for clarification

[obtuse pebble]

@dense storm Has your question been resolved?

[dense storm]

hi if we could just like talk about it in general, because i think i udnersatnd the overarching idea but i dont understand e.g how to determine the base cases

[obtuse pebble]

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[obtuse pebble]

@dense storm Has your question been resolved?

[cunning burrow]

You determine base case from the question provided

Sometimes you might need be a bit clever an determine how you going represent the problem but generally speaking it is straightforward

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[stone holly]

2nd ques

[stone holly]

I have find f(x) is inc function now what

[past sand]

Then you already know it doesn't have two roots, right?

[stone holly]

yea

it will only cut once

[royal swallow]

then a and c are down

[past sand]

If a>0, what's f(0)?

[stone holly]

something more than 0

-a

[past sand]

So f(0) < 0

[stone holly]

yes

[past sand]

So if f is increasing, the root is on the right of that

[stone holly]

yup

so only 1

when a less than 0

[past sand]

Yes

No

[stone holly]

what are approach to solve such ques?

[past sand]

When a > 0

[stone holly]

oh yes

I meant that

forgot abt -ve sign

[past sand]

the main thing is to find where the function is increasing or decreasing

[stone holly]

that part is clear

after that is an issue for me

I am facing problems there

[past sand]

Idk, draw the graph to get an idea of what to look for

It really depends on the question

[stone holly]

true

I thought of keeping x= -infinite here

just like in 1

which didn't work

as it wouldn't give us idea of a

[past sand]

In this case they are asking for positive roots, so of course you'll need to do something with 0

[stone holly]

ooo

true

actually this is my weakest part of AOD

[past sand]

Whether you actually compute the root (a function of a) or just check f(0), it's up to you

[stone holly]

does points of inflection has anything to do with such kind of ques?

[past sand]

Not really

Local extrema do

[stone holly]

fair enough

[past sand]

Like x^2 - 1 has a local minimum at x=0 and it is negative, and the limits at both -inf and +inf are positive (+inf), so that function has two roots

Of course you can just solve x^2 - 1 = 0 but for more complex functions the same principle applies, as long as the function is continuous

[stone holly]

uhm okay

got it

thank you

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[wintry cloud]

I need to find a sum, but simplifying it doesn't bring me anywhere

[old lily]

Are the C^k_n binomial coefficients?

[wooden cipher]

based on how ive seen this notation be used, a lot of the terms are 0

[wintry cloud]

yup

[old lily]

i think you can integrate or smth

expanding (1+x)^n

[wintry cloud]

This is a very beginning of a calculus book, I think it must be solved without integrating. And.. I don't know how to integrate yet.

[rose scroll]

how does kCn work when k is smaller than n?

shouldnt k always be larger than n to compute the binomial coefficient?

[wintry cloud]

that's the formula

[rose scroll]

yeah, but arent k and n swapped in the problem?

[wintry cloud]

nope, in this notation n is at the bottom

I guess

I think this can help me solve it

[fossil crag]

use that formula

and see how (k+1) and k! get multiplied together...

[wintry cloud]

that brought me to the sum of n!/(k+1)!(n-k)!

but I don't know where to go next

[wintry cloud]

Now I am trying to apply this formula

[fossil crag]

try to not change the denominator

and change the numerator

to find some ...C...

(hint : what do k+1 and n-k sum to)

[wintry cloud]

using this I got: n(2^n + 2n), but not sure if it's correct

[fossil crag]

I don't think it's correct

[wintry cloud]

hmm, okay, will try again

[fossil crag]

go back to n!/(k+1)!(n-k)!

[wintry cloud]

got you

[fossil crag]

which aCb has the exact same denominator?

[wintry cloud]

(k+1) C (n+1) I guess

[fossil crag]

yes exactly

can you write down (k+1) C (n+1)?

[wintry cloud]

that's

(n+1)!/(k+1)!(n-k)!

[fossil crag]

yep

so maybe you can see the link

between (k+1)C(n+1)

and n!/(k+1)!(n-k)!

[wintry cloud]

((k+1) C (n+1)) / (n+1) ?

[fossil crag]

yesss

[wintry cloud]

now I should probably use this formula and get the denominator out of the sum?

[fossil crag]

yep

[wintry cloud]

so the answer is 2^(n+1) / (n + 1) ?

[fossil crag]

almost

you'll notice that you don't have all the terms in the sum...

[obtuse pebble]

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[slim mortar]

I know that 1/3 of those numbers are divisible by 3 as the some of their digits are divisible by it but the others I don’t know how to prove it. We can express it as a sequence a_n=100a_(n-1)+1, and a_1=101. I don’t know if that helps

[fossil crag]

notice the second number has 101...

if you wanna find a way to show all of those are not prime

notice that a_n = .../99 through geometric sum

[slim mortar]

999….9?

[fossil crag]

a_n = 101...01 with n 1s is probably the best way

so a_1 = 1

a_2 = 101

a_3 = 10101

etc

notice that 101 = 100 + 1

10101 = 10000 + 100 + 1

1010101 = 1000000 + 10000 + 100 + 1

etc...

[fossil crag]

so use geometric sum formula

[slim mortar]

how we do it?

[fossil crag]

1 + r + r^2 +... + r^n

never saw that?

[slim mortar]

Ohh

R=100?

[fossil crag]

yes

[slim mortar]

Ahh but I only know how to do it when r smaller than 1

Maybe it is done the same

[fossil crag]

oh I'm not talking about infinite sum

I'm talking about finite sum

[slim mortar]

Is it with the same equation?

[fossil crag]

well yes

a + ar + ... + ar^(n-1)

[slim mortar]

Ahh

So that’s were we got /99

[fossil crag]

yes

ok so it's $a_n =\frac{100^n-1}{99}$

[warm shale]

rafilou2003

[slim mortar]

So we need to prove that 100^n-1 =0 (mod99)

[fossil crag]

no need

we already know a_n is an integer

it's a given

(also 100 = 1 mod 99 so not much trouble)

[slim mortar]

Oh my bad

[slim mortar]

Yes

[fossil crag]

alright so just to verify

a_1 = (100 - 1)/99 = 1

a_2 = (10000 - 1)/99 = 9999/99 = 101

so seems good

[slim mortar]

Yes

[fossil crag]

we wanna show

that for n > 2

a_n is not prime

any ideas?

(what happens when n is even)

[slim mortar]

It can be factorised into 100^n/2+-1

[fossil crag]

yeah

so (100^p-1)(100^p+1)/99

[slim mortar]

Yes

[fossil crag]

and oh look

that's a_p * something else

(100^p-1)/99 * (100^p +1)

[slim mortar]

Yes so it is composite

[fossil crag]

yes we just have to make sure that both of those factors are > 1

it's the case because p > 1

so 100^p - 1 > 99

otherwise imagine p = 1

the first factor would be = 1

and no composite

[slim mortar]

Oh that’s a good point

[fossil crag]

so

when n is even > 2 we're good

so otherwise n > 2 is odd

hint:

99 = 9 * 11

and 100^n = (10^n)^2

see if you can find the composite

[slim mortar]

We got 10^n+1 that is divisible by 11 so therefore 10^n-1 is divisible by 9 because it is 2 less

And we have to terms multiplying

That are not 1

[slim mortar]

Could we prove with this proof for the even as well?

[fossil crag]

well no

when n is even

10^n+1 is not divisible by 11

10^n + 1 = (-1)^n + 1 mod 11

[slim mortar]

Ahh

[slim mortar]

When n is odd wouldn’t it be the same?

Oh nothing

Didn’t thought of the parity there

[slim mortar]

Thank you for your help!!!!

.close

[obtuse pebble]

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[obtuse pebble]

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[dense schooner]

im not really sure what P(x,y)=0 means? can someone give an example

[slate sparrow]

its a function where the inputs are x and y

this specific one requires the coefficient of y^2 to be 9

so something like

P(x,y) = x^2 + 2y + 9y^2

it can have as many terms of x and y

you know one of the terms is 9y^2

[dense schooner]

how do i get 9y^2 from y=6cos(6t)

[slate sparrow]

well you know what y is

so try find out what 9y^2 is

square it and multiply by 9, see what you get

[dense schooner]

ohh i see

[stoic yacht]

you could also get a polynomial using the pythagorean identity, that's a common technique with these problems

if x = 3sin(6t), then sin(6t) = x/3, and cos(6t) = y/6

and you know sin^2(6t) + cos^2(6t) = 1

[dense schooner]

right

[stoic yacht]

then just multiply whatever you get by whatever constant you need to have the coefficient be 9

[safe ravine]

Jub they simply want u to eliminate t

[dense schooner]

so like this then

[safe ravine]

And get a equation in terms of x and y

Just eliminate t

You can use the relation sin^2t+cos^2t=1 for this

[stoic yacht]

not exactly

[safe ravine]

Hi neil

[stoic yacht]

if sin^2 + cos^2 = 1, then (x/3)^2 + (y/6)^2 = 1

[safe ravine]

Right

Wait she might be trying

Ok nvm

This gives us a crispy vertical ellipse

b>a

,w draw (x/3)^2+(y/6)^2=1

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[safe ravine]

Crispy 😋

[dense schooner]

yes i have (x/3)^2+(y/6)^2=1

how do i make it so its 9y^2

[safe ravine]

9y^2

Ok i see

We have x^2/9+y^2/36=1

[dense schooner]

right

[safe ravine]

Multiply by a suitable no so we get 9y^2

[dense schooner]

324

[safe ravine]

Okok

,w 324/36

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[safe ravine]

Okok

,w 324/9

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[safe ravine]

Ok

So we have

36x^2+9y^2=324 right?

[dense schooner]

right

[safe ravine]

So we get 9y^2+36x^2-324=0

[dense schooner]

yeah

[safe ravine]

Plug it in and chdck

Check*

[dense schooner]

ok thats what i thought

[safe ravine]

Ok

[dense schooner]

just multiply every term by 324 then subtract right side

[safe ravine]

Yes

[dense schooner]

okk

how do i do the second part of the question

[safe ravine]

Okok

I came

[dense schooner]

do i just disrregard the polynomial we just found and just find dy/dx

[safe ravine]

See u found

The equation in terms of x and y

[dense schooner]

yeah

[safe ravine]

Now u need to find the equation of tangent

And substitute

[dense schooner]

ok i will try

[safe ravine]

dy/dx,x and y in terms of t

see we have

y=6sin(6t) and x=3cos(6t) now try

[dense schooner]

should just be -2√3 right

[safe ravine]

What?

[dense schooner]

hmm its not

that was my answer

did i differentiate wrong lol

[safe ravine]

Let's see

,w diff 6sin(6t)

[dense schooner]

wow omg 3/18 is 1/6 not 1/3

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[safe ravine]

Ok i see

[dense schooner]

i just substituted the wrong t

wait not it was 6t

it was right

[safe ravine]

so dy/dt =36cos(6t)

,w diff 3cos(6t)

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[safe ravine]

So dx/dt=-18sin(6t)

Now divide to get dy/dx

And put t=π/18

[dense schooner]

i did that

[safe ravine]

dy/dx=36cos(6t)/-18sin(6t)

That is dy/dx=-2cot(6t)

,w -2cot(6t) at t=π/18

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[safe ravine]

-2/√3

Oh wait

Crap

@dense schooner sry

y=6cos(6t) and x=3sin(6t(

Sry

-36sin(6t)/18cos(6t)

,w -2tan(6t) at t=π/18

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[safe ravine]

@dense schooner it's correct

You are correct

Now write the equation

Of tangent

[dense schooner]

ohhh thats why it was wrong lol

i got the slope but not the eqn

[safe ravine]

Okok

Write the eqn

[dense schooner]

i got it

-2√3x + 12

tysm

[safe ravine]

Welcome

Np

Keep grinding

[dense schooner]

ofc

.close

[obtuse pebble]

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[safe ravine]

np

[obtuse pebble]

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[gleaming berry]

,w system invertible y[n] = x^2[n] - 2

[gleaming berry]

,w y[n] = x^2[n] - 2

,w y[n] = (x^2) [n] - 2

,w y[n] = x^2 [n] - 2

,w y(t) = cos(t)

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[gleaming berry]

,w y[n] = x^2*[n] - 2

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[gleaming berry]

,w y[n] = x^2 * [n] - 2

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[safe ravine]

So?

What is the point?

[obtuse pebble]

@gleaming berry Has your question been resolved?

[gleaming berry]

I wanna know if that invertible system or not

[safe ravine]

Which system

A system of equations is invertible is not invertible

1.No solutions
2.infinite solutions

I spoke it clear

Check if your system has solutions or not?

If it has are they finite or infinite

[gleaming berry]

It has

[safe ravine]

Finite

Then invertible

Enjoy

[obtuse pebble]

@gleaming berry Has your question been resolved?

[obtuse pebble]

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[west delta]

im not sure as to why order matters in the way you transpose the polynomial to have roots 1/a - 2 instead?

[west delta]

the sol says that u first take the f(x+2) transformation first and then take reciprocal

but i did the other way around and got a diff polynomial, does anyone know as to why the two polynomials differ?

[obtuse pebble]

@west delta Has your question been resolved?

[woeful torrent]

Consider
$$ f(x) = \sum_{n=1}^3 f_n x^n = \prod_{n=1}^3 (x - r_n) $$
which has roots $r_n$. Clearly
$$ F(x) = f(x+2) $$
has roots $R_n = r_n - 2$. So, by transforming first, our desired goal becomes
$$ \sum_{n=1}^3 \left(\frac{1}{r_n-2}\right)^2 = \sum_{n=1}^3 \left(\frac{1}{R_n}\right)^2 $$
which is just
$$ \frac{ \sum_{n=1}^3 R_n }{ \prod_{n=1}^3 R_n } = \frac{F_1}{F_3} $$
In other words, transforming first allows the machinery to correctly construct the roots $R_n = r_n - 2$.

[obtuse pebble]

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[warm shale]

Shuba

[obtuse pebble]

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[tribal scaffold]

hi

[tribal scaffold]

There's few parts to the question and I've gotten down to this, lmk if my handwriting is too messy

It's a question on polynomials

Here's the full question, I don't really know how to solve c

O ok

[safe ravine]

What

[tribal scaffold]

Srry if I offended u in anyway for u to act this way

[safe ravine]

No bro

Like huge stuff

!nospam

!spam

Aah we don't have these commands

It's simple but let's see

@tribal scaffold is ur doubt resolved?

[tribal scaffold]

I don't know how to prove part c :[ , I've tried couple of methods but didn't work

I'm in the right channel right??

[safe ravine]

@tribal scaffold

U know x coordinates of M right?

[tribal scaffold]

yes

[safe ravine]

Find y too

Simple

[tribal scaffold]

Yeah that's what I'm trying to

[safe ravine]

As M is mid point of chord AB

[tribal scaffold]

yeah

[safe ravine]

And AB are intersection points of circle and parabola

See u find y coordinates of A and B

Then simply add them and divide by 2

U will get the y coordinate of M

[tribal scaffold]

Uhh wait 1 sec lemme try

[safe ravine]

Ok

[tribal scaffold]

y is prob negative of the x value but I don't really know how to prove it, and the question wants to use something similar to part a and b

[safe ravine]

See

A and b

Parts have done what

They have solved simultaneously

To get x

Then u can put the x in any of the curves to get y

After getting y they simply added both the values and divided by 2

[tribal scaffold]

That's what I did at first but it doesn't equal to negative of x

[safe ravine]

What

[tribal scaffold]

So I thought I'm doing something wrong

[safe ravine]

How do u know it's equal to -x

If u know it is

Then u are doing some calculation error

[tribal scaffold]

Cuz PQ has gradient of -1

Then OM needs a gradient of -1 too since its asking to prove its paralellel

And if I get m and origin, the gradient should be -1 right?

So the y/x of M should be -1

[safe ravine]

Ook

[safe ravine]

Aah

[obtuse pebble]

@tribal scaffold Has your question been resolved?

[stoic yacht]

<@&268886789983436800>

[obtuse pebble]

@tribal scaffold Has your question been resolved?

[tribal scaffold]

I give up

[obtuse pebble]

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[plucky trail]

How to solve this using linear algebra

Status is I don't know where to begin because I have no experience with matrices or linear algebra

And someone said this would be simple with linear algebra

Answer is 25

[static furnace]

first step would be to replace each symbol with a variable

and create a system of equations to solve

[plucky trail]

ok h for heart, c for carrot, w for water

systems of equations