help-17

[lucid palm]

need help with boolean expressions
Using the Identities and Laws of Boolean Algebra, simplify the following expressions. List
the specific law used for every simplification.
A∗B∗~C+A∗B+A∗C+A

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[hexed kestrel]

(5x-50)/8 = 20
5x-50=8*20=160
5x=210
x=210/5=42

[blazing wigeon]

Don’t do the work for them

[hexed kestrel]

Y

[blazing wigeon]

Against rules

[silk osprey]

@ancient glacier pick a channel and close the other

[hexed kestrel]

Oki

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[crude shell]

Hi can someone help me with a practice problem, I keep getting two answers

[crude shell]

“Find the area bounded by y=x^3 and y=x^2-2x over (-1,1)

[lavish river]

ur working?

[crude shell]

Working?

[lavish river]

could u show the steps u took to get the two different answers

[crude shell]

Ya give me a sec I keep getting 3/2 and 5/2

I basically used the area between curves formula

When I check with my calculator it says 3/2 in decimal form

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[crude shell]

No

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[analog narwhal]

who can help me step by step and help me see if. didn't anything wrong

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[analog narwhal]

no

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[royal cairn]

i need help with this proof

[royal cairn]

so far i made the ratios x/y, y/z, and z/x and used componendo dividendo to make them in the correct format

For Example:

but when i tried with the other ratios and then plugged it all back into the question it didn't cancel out for some reason

<@&286206848099549185> ^

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[royal cairn]

.reopen

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✅

[bitter pilot]

Because I proved it above as well

[dense moon]

.

that said, open a new channel if you want to continue this discussion

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@untold arrow Has your question been resolved?

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[crimson heath]

I'm confused on number 2 on why a result this extreme would be unlikely. Wouldn't it be likely since rejecting the null hypothesis takes up 5% in the critical region?

[crimson heath]

I am also confused on part c of number one because I'm not exactly sure what a comparison distribution actually does

[sharp lynx]

The cutoff is the boundary at which you go from not rejecting to rejecting. It would have been more precise for the statement to say "a result at least this extreme is..."

[crimson heath]

Why is it unlikely tho

[sharp lynx]

Because the probability of obtaining a sample at least as extreme is (usually) at most 5%

Something happening 5% of the time is pretty unlikely

[crimson heath]

Ohh

Ok I get it

Makes sense

[crimson heath]

Thank you so much for the help

I guess I got confused by the wording

[sharp lynx]

Fair enough

I would also recommend you read the first question again, because both of your hypotheses are wrong

[crimson heath]

Ok

Oh I see, no equipment

Should just be new exercise

[sharp lynx]

Indeed

[crimson heath]

Thank you

[crimson heath]

<@&286206848099549185>

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[inner ether]

Where is the ques?

[crimson heath]

1 c

Question 1 part c

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[pale rune]

@bitter pilot @gentle sleet Not even sure if that answer is correct but uh yeah

[pale rune]

[gentle sleet]

you just have the x= there

shouldn't be there

[past anchor]

,w x^2+14x-11=0

[twin meteor]

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[past anchor]

yeah i think i was already correct

i think the mistake was on the input idk?

[pale rune]

it is- im asking how to input. Sorry if I didn't specify- I've been coming back to this for the same question of "how 2 input" like 3 or more times now lol

[past anchor]

i dont know

[woeful igloo]

you need a comma in answer so you should write x1, x2 instead of x = x1 x = x2 as you wrote there

[past anchor]

it was depending on the question app

does it use latex?

[pale rune]

I have no idea what that means

[woeful igloo]

-7+2sqrt(15), -7-2sqrt(15)

[pale rune]

Just saw this- mb missed it

but you were correct

[past anchor]

yeay

luckyly i dont need to use some website or digital stuff to input my answer

lol

[pale rune]

I hope I understand how to input sometime soon cus this is annoying

[worthy trench]

It tells you. You just have to read carefully

[pale rune]

Idk is it some kind of rule that I'm meant to assume x= is not meant to be inputted? Some of these previous ones also just aren't clear and don't tell you.

[gentle sleet]

it says

enter your answers as a comma separated list
and also the x= on the left there

[pale rune]

RIIGHT

[gentle sleet]

ig the first one you did do

[pale rune]

oops caps

I'm getting mixed up yeah that makes sense

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[spice perch]

would you say the function y*tan(y)=sqrt(lambda+y^2), where 0<y<sqrt(lamda), has a solution for every lamda>0? what about at natural number multiples of pi/2? Isn't tan undefined in these points, making the statement false ?

[uneven flicker]

can you pls write the eqn in a better way

like ss from wolfram alpha

[hard atlas]

well who says that the solution should be y=n*pi/2. only then it would be undefined

[woeful stream]

given lambda, you want to solve for y, right?

if pi/2<sqrt(lambda) then you probably can intermediate value theorem out a y

[spice perch]

yeah I just realzied that mb

i completely misunderstood what it said

[spice perch]

no its 2 functions of y

but I understand where I went wrong

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[ionic marlin]

how can i do this without l hopitals?

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Please don't occupy multiple help channels.

[keen umbra]

@ionic marlin don't occupy multiple help channels

alr I'll start my explanation anyways

we'll deal with the denominator first

recall the formula for a^3 - 1 here

we'll let a=cbrt(x)

then we get x-1=(cbrt(x) - 1)(cbrt(x^2) + cbrt(x) +1)

now try to express cbrt(x) - 1

[ionic marlin]

aaaah

got it

thankyouuu

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[shy ermine]

How do i even find cos(415⁰)

[strong grove]

cos (360+55) :d

[shy ermine]

I know

But 55⁰ isnt compiled of any of the basic radiants i know

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[sullen pasture]

Can someone please explain how to do task C? I have 0 idea

[sullen pasture]

I'm struggling to understand laplace transforms

[kindred iris]

The request is to prove the equation, and գ) and դ), at least any of them

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[mighty sundial]

define touch

[livid prairie]

make a new channel lol

lemme

[kindred iris]

<@&286206848099549185>

[cosmic cloud]

make a new help channel for your question

[kindred iris]

What do you mean

Are you blind

Look higher what i wrote near my nick

[vast shale]

ya

[cosmic cloud]

this channel is occupied

refer #❓how-to-get-help

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@sullen pasture Has your question been resolved?

[flat whale]

Use Laplace transform identities. One that has a derivative and another that multiplies exponential function

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[warm sage]

How do I do this with like sum and product

[lucid meteor]

You know about splitting the middle term ?

@warm sage

[warm sage]

I have no idea what to do

[lucid meteor]

Ok so here you have to split the 3x

Such that adding it you get 3x and multiplying it you get 2x²

[warm sage]

Yeah

[lucid meteor]

for an equation ax² + bx + c
Split bx . Let we split it into kx+ jx. It should be in way that kx+ jx = bx and (kx) (jx) = cax²

[warm sage]

😿

[lucid meteor]

Because complex ig

x² + 3x + 2 = 0
x² + 2x + x + 2 = 0

x(x + 2) + 1(x+2) = 0

[warm sage]

Ooh

[lucid meteor]

(x + 1) (x+2) = 0

You get x = -1 and x = -2

[warm sage]

Yeahhh

[lucid meteor]

Ask me about any step you are confused about

[warm sage]

Well i guess i dont really understand how too start with it

Like why x^2

Turns into 1

[lucid meteor]

I see

Well I would suggest you to watch a video on yt on middle term splitting and some examples with it

You will get it

What happened

[warm sage]

Ohh alr

[lucid meteor]

It cannot be explained verbally well

[pine hull]

Well x² didn't turn into 1 technically

[lucid meteor]

If you still have any doubts after that feel free to ask

[pine hull]

Do you know how to take factors @warm sage ?

[lucid meteor]

Let her understand the whole method first thunder

Most of her doubts will be solved

Maybe all

[pine hull]

No, if she doesn't know how to take factors or what factor is

She won't understand the method

[lucid meteor]

True

[warm sage]

Ill have a look at both of em

[pine hull]

@warm sage I would suggest read about quadratic equations first before solving the above problem

[warm sage]

Alr thank you

[lucid meteor]

And look into some examples as well

[warm sage]

I have a bunch of tasks too solve🙏

[pine hull]

Then as me and FlexO suggested read some theories and examples, then come back and we will be here

[lucid meteor]

You cannot solve them until you understand what you are given in question

[warm sage]

Alright ty, ill watch some youtube

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[vast shale]

this is a question on my study guide for calc, and i know the answer is B but i don't know why 😭

[spiral turtle]

Well, you don't actually have any data, so you kinda have to eyeball it. And you seem to have successfully eyeballed it.

[vast shale]

i got the answer out of luck though, what would i be eyeballing for like if i get a different problem siimilar to it?

[spiral turtle]

you can mentally draw a line between (a, f(a)) and (b, f(b))

and see which point's tangent looks closest to the correct slope

[fading plume]

a is 0, b is plausible, c is too high, d is negative

[vast shale]

where is f(a) and f(b)?

[spiral turtle]

The endpoints

(a, f(a)) is the point on the graph at x=a

[jovial basin]

hey everyone i just joined this server

[spiral turtle]

Hi, if you need help please read #❓how-to-get-help

[vast shale]

like this?

[spiral turtle]

no

[vast shale]

how is it

[jovial basin]

k sure

[fading plume]

lowercase a, b not A, B

[vast shale]

ohh okay, i get it now thank you

[spiral turtle]

yw

[vast shale]

.close

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[dim lodge]

I did 20/e-1 * integral lnx/x
u = lnx
du=1/x
xdu=dx
so u/x * xdu
x's cancel
left with u du
so its just 20/e-1 * integral lnx

[vast shale]

$\int u du = \frac{u^2}{2} + C = \frac{\ln^2(x)}{2} + C$

[dim lodge]

ohh

[twin meteor]

icannotdoanymorecauchy

[dim lodge]

yea i think i need to work on understanding usub cause i think the problem is done when its not

you get u by itself so you can integrate it

[vast shale]

y4s

thats the point of u-sub

[dim lodge]

so usub isnt the actual integration. its just manipulating the equation inside

[vast shale]

yes

[dim lodge]

got it

[vast shale]

but u're using u sub to integrate in terms of u

and then when u're done

u change back to x

[vast shale]

in fact what u're doing aint quite right

[twin meteor]

icannotdoanymorecauchy

[vast shale]

u're integrating f(x) with respect to du

[dim lodge]

well i did du = 1/xdx
i wanted dx by itself
so x(du)=dx

[vast shale]

the u-sub part went correct

but when u substituted u as ln(x) back during integration isnt correct

because the respect u were integrating to was du not dx

[dim lodge]

i forget, "with respect" means im trying to isolate dx

[dim lodge]

i started learning usub 2 days ago 💀

[vast shale]

cant really blame u mate

its integral calculus

[dim lodge]

its not so bad because if i had time i would spend it on watching videos and reading material that explains it intuitively

but i just gotta figure it out as i go along

thanks for the help

.close

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[feral drift]

can someone explain this to me please why there are zeros of two rows up and down are added to the matrix X

[feral drift]

<@&286206848099549185>

[inner ether]

It doesn't effect I think

Cause it will still give the same matrix

That will come after multiply

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@feral drift Has your question been resolved?

[feral drift]

yes but I mean how i come to the last matrix at the end of H, the answer is 4*2 matrix

[inner ether]

U did X matrix * W matrix=Y matrix * H matrix?

H matrix is 4 * 2 matrix

Basic rule of out put of multiplication of two matrix

Did u solve this ques right?

Like if a matrix is given

2*3 why u making it

6*3

It can be solved by just letting H matrix as

[feral drift]

this is convolution * this bold star means convolution not matrix multiply

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[feral drift]

@open sundial

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[tame river]

Im trying to figure out the general rule to follow when solving this problem here,

[tame river]

and so far,

a(-1) , find it in the

1st row of the table.

for m(4) , use the graph.

[loud walrus]

If you found the whole thing is 4

Why u typed -1?

[tame river]

oh mb lol

[tame river]

old, i changed to 4,

well actually,

when i did that i must have forgot to get rid of the -

ok then :) thanks.

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[pale python]

Can anyone explain where the 5.0026 came from or help me figure out how to do this definite integral

I feel like I'm losing my mind because I swear no matter what I do I can't get 5.0026

also 20pi( is outside the integral but chatgpt doesn't show it)

[regal bane]

"is evaluated numerically" means they used some approximation method. They did not solve the integral, and it may be impossible to solve

,w integral of x^4 sqrt(1600x^6 + 1) dx between 0 and 1

[twin meteor]

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[pale python]

what does that mean and how do I solve it because my professor didn't teach this 💀

[regal bane]

"is evaluated numerically" means they used some approximation method. They did not solve the integral, and it may be impossible to solve

(It very likely is impossible to solve)

[gaunt sparrow]

If that is a ChatGPT response, what was to original question exactly? Just in case something went wrong along the way that made this harder to compute than it should be.

[pale python]

i'm confused on how they 20pi * 5.0026

[gaunt sparrow]

,w 20pi * 5.0026

[twin meteor]

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[regal bane]

As always, ignore ChatGPT

[gaunt sparrow]

I think that might just be a ChatGPT quirk unfortunately. Kaynex's wolfram output says it's 5.0059

[regal bane]

You want some kind of software to numerically compute integrals. Wolfram Alpha can do it

[pale python]

so I substitute 1 from the upper bound into the equation but I still cannot get 5.0059

[regal bane]

https://www.wolframalpha.com/

[pale python]

can i do it on a graphing calulator or no?

[regal bane]

Probably some, yeah

[sonic escarp]

Yes, I do it all the time.

[pale python]

ok I'm just going to assume these problems won't be on an exam because how are we supposed to solve if we can only use wolfram alpha 💀

Anyways, thanks for the help @regal bane @gaunt sparrow 👍

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[sterile wyvern]

I need help with this strange problem

[sterile wyvern]

Number 23

This is what I’ve been able to come up with

<@&286206848099549185>

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@sterile wyvern Has your question been resolved?

[sterile wyvern]

So helpful

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[fossil crown]

Hey guys, I need some help on my calc one homework

[fossil crown]

I've been stuck on this problem for the past hour, and would like some pointing in the right direction. Thanks!

[formal yew]

what is this

like how can u simply

definiton of what...

[fossil crown]

that's the definition of the derivative right?

[formal yew]

ye so

whats the deriative

on the right side

[snow trout]

huh

[fossil crown]

would it be 1/4

[formal yew]

ye

[fossil crown]

how would you evaluate it for the left side?

[formal yew]

what is the deriative at a hole

is a a hole continus

continous

[fossil crown]

no

it doesn't exist

[formal yew]

is a function diffierntable when it isnt continious?

ye so

its DNE

[fossil crown]

no so the two sided limit cannot exist either.

thanks!

[snow trout]

@fossil crown keep in mind a limit can still exist with a hole

the limit doesnt exist for this q cuz the lims from two sides differ

so just saying f(x) discontinuous at x_0 is not good enough to say the lim doesnt exist there

[fossil crown]

could i say that the derivative doesn't exist at that point because of the discontinuity?

[snow trout]

ya thatd be correct

discontinuity means derivative dne there
but discontinuity doesn't mean limit dne there

[fossil crown]

yeah alr

thanks!

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[cinder oyster]

n

[cinder oyster]

i feel like sig figs change rules every day

Like i am told to not round until final answer

and not round to sf early

but why are they doing it

like

if i waited till the end

i get a differenta nswer

[drifting jackal]

How different?

[cinder oyster]

uh

1 sec

so

for 0.9893(12.0000)

i get 118716

im keeping in mind the sfs till the end

then when i do

0.0107 times 13.0034

i got 0.13913638

[drifting jackal]

Where is the decimal?

[cinder oyster]

addd them both u dont get anywhere near 12

wot

OH

Omg

[drifting jackal]

0.9893(12.0000) does not equal 118716

[cinder oyster]

11.8716

oops

and then i got

0.13913638

oh

ok

that makes the difference

i need some sleep

thnx

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[cinder oyster]

wioahh

wait

round to nearest 3 sf

how would it be 12.01

oh

wait

ok i have scenario question but i closed this channel i ask it later

.close

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[inner osprey]

consider the possibilities if:
3 dimes are counted, or 2 dimes are counted, or 1 dime is counted, or 0 dimes are counted

[dusky willow]

yea i thought of the possibilities to be

3 dimes
2 dimes 2 nickels
2 dimes 1 nickel 5 pennies
1 dime 4 nickels
1 dime 3 nickels 5 pennies

i think thats all the ways they add up to 30

but im not sure how to put it all together

as a probability

[torn oracle]

what you could do is calculate the total number of possibilities

and then the total number of possibilities that add up to 30

and then divide the latter by the former

[dusky willow]

im kinda stuck here

are the total possibilities just 2^12

4096

and then theres 5 distinct ways of getting exactly 30 cents facing up

[torn oracle]

the total possibilities I would think would be 12!

given that you have 12x11x10x...x1 different ways to select a coin

then you could calculate the number of ways to make each of the possibilities

oh nvm the number that lands heads up

you're right

it would be 2^12

my bad

as for the number of distinct ways you could calculate the probability of each

for example 3 dimes you know would be 3x2x1 total ways to be made

[dusky willow]

wait what

[torn oracle]

think about it

if we are discussing the probability of 2^12

you will be discussing the total number of permutations

so we need to consider the total number of permutations for each case that will make 30 cents

to pick up 3 dimes

you have 3 * 2 * 1 arrangements

[dusky willow]

but wouldnt it be like

for the case that its 3 dimes landing on heads

[torn oracle]

oh wait

you're right

sorry

[dusky willow]

those 3 have to be facing heads up

[torn oracle]

I keep forgetting that it's heads or tails

[dusky willow]

everything else has to be tails

[torn oracle]

💀

yeah

[dusky willow]

and the same for the rest

only the specific ones have to be heads

the rest have to be tails

[torn oracle]

yeah in that case you have to get all 3 heads and the rest tails

[dusky willow]

Oh wait

[torn oracle]

that's only one possibility

[dusky willow]

for the nickels

its not just one way

[torn oracle]

yeah for 2 dimes 2 nickels you would have 3C2 * 4C2 possibilities

because you have 3 dimes and 2 that need to be heads

and 4 nickels and 2 that need to be heads

[dusky willow]

yeah

[torn oracle]

then you would do that for each case

that adds up to 30

[dusky willow]

and then 2 dimes 1 nickel 5 pennies is

3c1 * 4c1 * 1

and so on

[torn oracle]

ye

[dusky willow]

and i add those probabilities up

and divide by 4096

thats my answer

[torn oracle]

I believe so

[dusky willow]

yea that makes sense

its been a while since ive done probability

but i appreciat the help

have a good one

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[torn oracle]

sorry for being a lil confusing

you too

[dusky willow]

nah its fine

im the type of person to make those type of mistakes too

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[daring forum]

How to solve lim->0 sinx/x?

[daring forum]

as a reminder since its been a month or so

[calm pecan]

lhopital's? :D

[daring forum]

Without lohoptial

[calm pecan]

use the taylor series expansion.

[vast shale]

squeeze

theorem

[calm pecan]

,wolf taylor sin(x)

[twin meteor]

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[calm pecan]

or squeeze yes.

[daring forum]

well i know how to do that but my professor hasnt taught that yet so it might seem dishonest

[vast shale]

there are two classic ways

[calm pecan]

i prefer taylor when possible

[daring forum]

oh yeah i forget about squeeze

[vast shale]

1. perimeter of n-polygon
2. similar triangles inside and outside unit circle

[calm pecan]

here it's plain convenient

[daring forum]

-1<sinx<1 but where do you go from there after multiplying by 1/x

[daring forum]

i havent learnt that yet altogether

i think it comes soon after exponentials in my textbook

[vast shale]

i dont think that's appropriate here

[calm pecan]

<= btw

[daring forum]

right

[vast shale]

squeeze theorem doesnt hold here

[daring forum]

how do you mean?

[daring forum]

interesting

[vast shale]

the limit of both functions arent the same

go from here instead ^^

[calm pecan]

Well that's not at all scary lol

[daring forum]

i didnt know that was something to look out for now im curious about that lol

what are the general exceptions for squeeze theorem?

[calm pecan]

well they have to converge no

that's more or less it

[calm pecan]

also...

...you sure you did right

it should be minus x in the denom no?

[calm pecan]

am i super confused lol

[calm pecan]

hm but

presumably i just forgot how to squeeze lol it's fine

[vast shale]

i hope this helps

[calm pecan]

fucking pngs

[calm pecan]

hmm

righttt

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[noble mortar]

Find the curves that satisfy each of the following geometric conditions:
(g) The polar angle θ equals the angle ψ from the polar radius to the
tangent.
(h) The angle ψ from the polar radius to the tangent is constant.

got 2 questions here
in part (g) θ = 0 is also a solution, i dont get why
im not that familiar with polar coordinates so it would be helpful if someone helped explain that as well, like i get it but as you see, dont understand why θ = 0 is a solution

and in part (h)
i saw a solution online
they said this essentially means that
either θ = k
or tan θ = k
therefore
(1/r)(dr/dθ) = k
dont get this also

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[noble mortar]

<@&286206848099549185>

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[green gulch]

how would I do part a lol

[flat whale]

,calc (-5^2+5)/(5+1)

[twin meteor]

Result:

-3.3333333333333

[flat whale]

Wot

[flat whale]

Show the entire question

And entire page

[silk osprey]

maybe they meant +n^2?

[pallid zenith]

or -(n^2+n)

[silk osprey]

yea

[pallid zenith]

,calc -(5^2+5)/(5+1)

[twin meteor]

Result:

-5

[pallid zenith]

,w Table[ -(k^2+k)/(k+1), {k,0,6}]

[twin meteor]

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Requested by janniku

[pallid zenith]

[silk osprey]

k^2 + k = k(k+1)

[green gulch]

thats the entire q

[silk osprey]

yea it’s a typo then

[green gulch]

oh what

OMG mb I didnt copy the quesiton properly

mb

[silk osprey]

they either meant -(n^2 + n) or just n^2 + n

[green gulch]

my bad

😭

also another q how would you latex the sequence part?

[pallid zenith]

if you include physics, you can use mqty

its a little more straightforward

[twin meteor]

jan Niku

$\mqty( -n^2+n \\ 3 )$ or whatever

[green gulch]

ty

alr so now how would I do part a

also what exactly does n represent here?

naturals?

or real numbers

[oblique kettle]

is this like a vectorial sequence

like a sequence of vectors instead of scalars

never seen this, interesting xD

[oblique kettle]

yeah, natural numbers ig

like with usual sequences

[green gulch]

ohh lightwork then

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[oblique kettle]

for b, since this is a vectorial sequence
the sequence is convergent if the modulus of the sequence is finite
like if lim n -> oo | Vn | != oo

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[vast shale]

so...

i think that i should find these two lines right?

[outer warren]

yes

[vast shale]

and is the radius of the circle one?

[inner crater]

hi

[vast shale]

if so, then i can just use Y-Y1 = m (X-X1) from
(0,1) and (-1,1)
(0,1) and (1,-1)
is it correct?

[outer warren]

no

where are (-1,1) and (1,-1) coming from

[vast shale]

oh

sorry i just thought it's gonna be like this

xD

then where is the intersection point on the circle?

[outer warren]

the tangent lines that'd pass through the indicated points would be vertical lines

(which wouldn't pass through (0,1)

start with getting a general equation of a line that passes through
(0,1)

[vast shale]

how to find a line that passes through 0,1 without knowing the intersection point or slope (gradient)?

or should we find the slope first?

[outer warren]

leave the slope unknown

that's ultimately what you want to determine

[vast shale]

(0,1)

y-1 = m (x)
y = mx+1

just for illustration, I will ignore the gradient values

after i subtitute the y=mx+1 to x^2+(y+1)^2=1

= x^2 (1+m^2)+2mx=0

[oblique kettle]

youre in the right direction to solving this

but no, you would find another quadratic

[outer warren]

didn't expand correctly

[vast shale]

i just take the discriminant = 0
and i find m = +- 2

[oblique kettle]

yee

[vast shale]

ouh

wait

[oblique kettle]

yeah, thats what you would do exactly

force it to be 0

[vast shale]

(1+m^2)x^2 + 4mx + 3 = 0

and

[oblique kettle]

yes, thats the quadratic i found, noice

[vast shale]

m = + - 3

oh

[oblique kettle]

m² = 3

[vast shale]

so it should be

y1 = 3x+1
y2 = -3x+1

[oblique kettle]

m² = 3

!!!

[vast shale]

root of 3 ?

[oblique kettle]

yep, +- sqrt(3)

and not +- 3

[vast shale]

ah

y1 = sqrt(3) x +1
y2 = -sqrt(3) x +1

[vast shale]

alright tysm ❤️

[oblique kettle]

you figured it out by yourself though 😛

subbing the line into the circle, then putting d=0

[vast shale]

idk I suddenly remembered my 2nd grade high school lessons

about circle

[oblique kettle]

mhm yeah

its true like in general

not only circles

if you want 2 things to intersect

sub one into another

the solutions to the resultant equation will automatically be the potential intersection points

[vast shale]

at first, i was so confuse of this because i don't have one of the two variables haha

[oblique kettle]

yeah xD

[vast shale]

alright tysm have a good day bye 😄

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[vocal sleet]

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[honest orbit]

When I try to split tan^4 into two tan^2, I still have an extra tan after applying my du

[honest orbit]

my u is sec(6x) and my du is 6sectan(6x)

[brisk moss]

thephalluspulverizer

[honest orbit]

wassup

what is this 😭

[brisk moss]

exactly what it says

[honest orbit]

anyways

[honest orbit]

perhaps u may assist me?

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[brisk moss]

sorry i was never any good at math

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[safe jewel]

They want it written in this format this is another answer

But how did I get the one I did wrong?

@lucid meteor

@flat whale

[flat whale]

!noping

[vocal sleet]

Please do not ping individual helpers unprompted.

[safe jewel]

😂😂

[gentle sleet]

that one you solved it wrong

[safe jewel]

How’s it wrong)

[gentle sleet]

you forgot to sqrt the 25

[safe jewel]

So 14 and 5

Ohhhh

3/4-5sqrt14

[gentle sleet]

also from using the quadratic formula it should be -3/5
I'm not sure where you did the mistake though

[safe jewel]

@gentle sleet this?

[safe jewel]

Feel free to look at the steps

[gentle sleet]

sqrt(14)/5

sorry, connection issues

[safe jewel]

So this isn’t right?

[pine hull]

w+3/5 = ±√14/5
w = -3/5±√14/5

[safe jewel]

So this is what ur saying

[gentle sleet]

ah, ic

[safe jewel]

So is it this

Or -3/5

[pine hull]

-3/5 bro

[gentle sleet]

the negative one

[safe jewel]

Where did that come from

I sent my steps

[pine hull]

(w+⅗) =±√14/5

[safe jewel]

The 3/5 is positive

[pine hull]

Subtract both side ⅗

[safe jewel]

How’s it negative

[safe jewel]

Is that squared tho

[pine hull]

a+⅗=b
So
a = b-⅗

You did a = b+⅗

[safe jewel]

So how come

The 3/5

Doesn’t affect the 14/25

[pine hull]

[safe jewel]

Oh shit

I see

It’s going to the right side

Of the =

[safe jewel]

I see

Do I use same format as I did for 1

Here

[pine hull]

You don't know how to find the roots of quadratic equation?

Using formula?

Of discriminant

[pine hull]

Complete the square is one method

[safe jewel]

I use this method

[pine hull]

There is another with formula

[pine hull]

Okk

[safe jewel]

Would I use this method for the next 2

Or is it not the same

[pine hull]

Yes

Every quadratic equation can be solved

With that

[safe jewel]

Which do you reccomend that one or the -b+- sqrt 4(a)(c) over 2a

[pine hull]

This is one liner , you just put values of a,b,c and find x

Complete square is a lengthy process

[safe jewel]

So which one is easier and better

[pine hull]

But whichever feels good for you

[pine hull]

Easier and faster

[safe jewel]

They both do but what’s easier and more simple

[safe jewel]

Which one

[pine hull]

Bro see which message I have replied to

[safe jewel]

So the -b+- sqrt of 4 one

[pine hull]

-b± one is faster @safe jewel

[safe jewel]

I should go with

[safe jewel]

Dumb professor doesn’t use that one

[pine hull]

Well completing square helps your brain to get sharp

[brisk moss]

quadratic formula makes your brain even sharper

[pine hull]

[pine hull]

Its easy way, just put a ,b,c no extra brain needed

Completing square you need to think what terms to add to make it complete square

Sometimes fractions too

[brisk moss]

think about how much you could learn in the time saved not doing that

[pine hull]

[blazing wigeon]

Completing the square is a necessary skill for your integration paraphernalia

[safe jewel]

So. I shouldn’t use the easier one

[pine hull]

Once you feel confident about using completing square method, then use the easier one to save time

[safe jewel]

Bet that up little homie

[vocal sleet]

@safe jewel Has your question been resolved?

[vocal sleet]

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[onyx mirage]

Is it possible to determine the Ratio R/Ρ?

[onyx sparrow]

Is R the radius of black circle?

[onyx mirage]

Yes

[vocal sleet]

@onyx mirage Has your question been resolved?

[onyx sparrow]

Ok I give up for the night ping me if anyone gets it though it's interesting 😔

[vocal sleet]

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[lavish slate]

i dont understand at all

[bitter pilot]

commplete the square in the denominator

[heavy yoke]

partial fraction decomposition seems promising

[onyx sparrow]

Aren't y'all overcomplicating =_=

Can't you just factor out x, separate into two fracs, then they're just ln

[lavish slate]

i ddid that but the answer pulls out a 1/4x and a 1/4(x+4) and idk where the extra 4s came from

[onyx sparrow]

What the

[cyan shadow]

how would you separate into two fracs

[onyx sparrow]

Holy shit

I'm high

Sorry

[heavy yoke]

that's what you would get from perfoming the partial fraction decomp

[onyx sparrow]

Ignore me lmao

[lavish slate]

i got it

thx

idk why uit didnt occur to me to do that

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[tawny marten]

Hi. I'm stuck on the following problem: "$x(t)=(t^2 - t) * sin(2 * t)$, $y(t)=t$, $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Find the distance between x and y in $C[a, b]$ and $C^{(1)}[a, b]$ spaces". \
For $C[a, b]$ the distance is $\rho(x, y) = max_{a \le t \le b}|x(t) - y(t)|$ \
For $C^{(k)}[a, b]$ the distance is $\rho(x, y) = \sum_{i=0}^k (max_{a \le t \le b}|x^{(i)}(t) - y^{(i)}(t)|)$ \
For the first problem, I declared a function $f(t) = x(t) - y(t)$ and tried to find its local extremums by finding the roots of its derivative. \
$f^{(1)}(t) = 2 * t^2 * cos(2 * t) + 2 * t * (sin(2 * t) - cos(2 * t)) - sin(2 * t) - 1$ \
I tried solving it as a quadratic equation, but $t = \frac{2 * cos(2 * t) - 2 * sin(2 * t) \pm \sqrt{4 + 8 * cos(2 * t)}}{4 * cos(2 * t)}$ didn't help much. \
Wolfram Alpha returns some weird approximations as the answers and not fractions with pi or something similar: $x \approx -1.006$ and $x \approx -0.2023$. \
I've also tried asking latest GPTs, but they just calculate the distance in 0 and $\pm \frac{\pi}{{2, 3, 4, 6}}$ and do some strange assumptions that those are extremums or something like this. \
It is worth noting that I've plotted these functions on Desmos: https://www.desmos.com/calculator/lphyvkxufs and from the plots it can be seen that the maximums of $|f(t)|$ are achieved in $t = -\frac{\pi}{2}$ and $t = \frac{\pi}{2}$ and that the maximum of $|f(t)| + |\frac{d}{dt}(f(t))|$ is achieved in $-\frac{\pi}{2}$, so it seems like I don't need the exact roots of derivatives, but I don't understand, how should I prove that the solutions are achieved in the bounds of t and not somewhere in between?

[twin meteor]

Gragonite

[vocal sleet]

@tawny marten Has your question been resolved?

[tawny marten]

<@&286206848099549185> ?

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@tawny marten Has your question been resolved?

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@tawny marten Has your question been resolved?

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@tawny marten Has your question been resolved?

[flat whale]

Why are you adding the two here
|f(t)| + |f'(t)|
to infer the max occurs at t=-pi/2?

[tawny marten]

It's the distance in $C^{(1)}[a, b]$

[twin meteor]

Gragonite

[flat whale]

it's not according to your definition

[tawny marten]

Ah, you are right, I've made a mistake there. However, max(|f'(t)|) is also achieved in -pi/2: https://www.desmos.com/calculator/kbk2lr2ydh, - so this mistake doesn't change anything

[vocal sleet]

@tawny marten Has your question been resolved?

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@tawny marten Has your question been resolved?

[tawny marten]

Okay, I gave up and used approximations from wolfram alpha

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[obsidian ore]

trying to look for a pattern here

[obsidian ore]

quadratic ax^2+bx+c: Σα = -b/a, Σαβ = c/a
cubic ax^3+bx^2+cx+d: Σα = -b/a Σαβ = c/a Σαβγ = -d/a

can i extend this and alternate between positive and negative

-b/a c/a -d/a e/a -f/a g/a and s on

sorry if im not making myself clear

[split wind]

yes of course!

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

[obsidian ore]

oh my god thank you

@split wind this applies even if im not just looking for the products of roots?

if im looking for something like this

i can still apply my logic above?

[split wind]

yep

[obsidian ore]

omg thank you this has saved me so much time

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[pliant oyster]

I have that $A_4 = {\frac{n-1}{n+1} : n \in \mathbb{Z}_{\ge 0} }$

[twin meteor]

Curry

[pliant oyster]

I want to prove that the least upper bound of this set is 1 in the rationals

What I've done is

Suppose there exists alpha in Q such that alpha is an upper bound and that alpha is less than 1

I want to find some n such that (n-1)/(n+1) > alpha

so I did some rearranging and got that (1+alpha)/(1 - alpha) < n

I know n needs to be a part of Z greater than or equal to 0

But I'm not sure how to find an n here

Is there a better bound I can use to find n?

[cedar burrow]

The greatest upper bound would be larger than 1, since 2, 3 etc are also upper bounds. What you could prove is that the least upper bound is 1

[pliant oyster]

oh sorry that's my typo

yeah that's what I'm trying to prove

Apologies I have edited the message now

But this method is for attempting to prove that 1 is the least upper bound in A_4

[cedar burrow]

Were you given this task from a teacher or are you trying to solve it on your own?

[pliant oyster]

It is a part of a problem sheet

but I am trying to solve it on my own

[cedar burrow]

Right

[pliant oyster]

I wanted to get some guidance on where I could go from here

As I'm trying to find some n to disprove my negation

If I can find an $n \in \mathbb{Z}_{\ge 0}$ such that \ $n > \frac{\alpha + 1}{\alpha - 1}$

[twin meteor]

Curry

[dense eagle]

ur basically done here

alpha+1/alpha-1 is just some number

and so since the naturals are unbounded

you can find some n large enough (where it's larger than that fraction)

[pliant oyster]

ohh

so the archimedian property

just calling this value k I can say k is apart of Q which is a subset of R

So by the archimedian property there exists some n in Z_+ such that It is greater than k

That makes sense thank you

[dense eagle]

nw!

[pliant oyster]

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[soft lodge]

"Use the error formula to give an estimation of the error of your approximation"

I have computed the 4rth degree of the taylor polynomial, but what am i supposed to do next?

[frigid sorrel]

Clearly u have to apply an error formula u were given, as said in the question

[soft lodge]

Ye i understand that, but that is like computing the fourth degree polynomial and then put in 0,1?

as our x?

[tough hamlet]

x would be 0.1 your a would be e, so you could solve for e then

that would give you a formula for your error

[soft lodge]

Well

Thats not really it i think

[tough hamlet]

Actually you are right, because your a should be 0.1

[soft lodge]

Are you sure?...

how do i rotate it

[tough hamlet]

I dont know how to rotate, but that is super annoying to read

[soft lodge]

We need to have our a as (e) so that the ln(a) becomes ln(e) which is 0

[tough hamlet]

Can you solve it for e, when you put x=0.1?

[soft lodge]

1**

[tough hamlet]

I cant really read what you have.

[soft lodge]

I will get. ua better picture

The formula (7.6.2)

And my approximation of the third degree with x=0.1 and a=e is 1.1253333…

[tough hamlet]

Sorry its not clear to my from the question why you used 7.6.2

It asks for taylor series, then the error

[soft lodge]

Error formula

It ask about the Taylor polynomial of the third degree and then an approximation at x=0.1 and a=e

Then “ Use the error formula to give an estimation of the error of your approximation”

I have my approximation

Which is already calculated from the “first question”

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.close

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[viscid bridge]

hi guys, i’d like to confirm something.
when we’re given the equation of a curve that intersects a line, we can find the point of intersection by substitution right?

so i’m going to make x as the subject for the line, and substitute it to the x of the curve to find y. then substitute the answer to the y of the line.

it’s correct right?

[vast shale]

!original

[vocal sleet]

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.

[arctic mantle]

yes

[viscid bridge]

okay i found this hard question and i suddenly forgot everything about intersection of graphs

1.) A curve has equation y = 2xy + 5 and a line has equation 2x + 5y = 1.
The curve and the line intersect at the points A and B. Find the coordinates of the midpoint
of the line AB.

pls help guys, why can't i do substitution here

[dull bear]

You can, your way is fine

[viscid bridge]

okay i’m going to do it again

[dull bear]

(remember to make sure that the coordinates of the points you find are in the right order, and that you want the midpoint of the line segment between A and B in the end )

[viscid bridge]

okaayy

i was stuck in the substitution thingy before

[dull bear]

Awww let us know if you get stuck again, I'm sure you'll be fine with it though

[viscid bridge]

okkk thankyouu

i found ittt

[viscid bridge]

thank you so much!!

[viscid bridge]

thank youuu!!!

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[carmine ether]

how do i sketch z^2=(z*)^2

[carmine ether]

i let z=x+iy

and z*=x-iy

subbed it in

squared both sides

simplifed and got to y=2ix

ive still got an i in my cartesian equation so yh idk

[lavish river]

just look at their arguments

[carmine ether]

idk to simplify it ig

i said y=0 was a solution

so y^2=2ixy?

idk how to visualise that tho

y^2-2ixy=0

y^2 is real

-2ixy is imaginary

0

ohh alright

so to draw that on an argand diagram😭

would it just be

the re(z) axis

[lavish river]

theres 2 lines though

the imaginary line too

just look at the arguments

magnitude is the same trivially

for z=z* its the real line

but when its squared we have more solutions, being the imaginary line

[lavish river]

really you would probably use quadratic equation and idk

[gentle sleet]

this simplification isn't right

[lavish river]

or there is a mistake i didnt break it down

[gentle sleet]

I assume that you've done (-iy)^2 = +y^2

which isn't right

yea, I've triple checked myself because of that lol

[lavish river]

did no one use arg :(

[gentle sleet]

I thought of that at first

but they went the other way so

[lavish river]

i was looking the algebraic way but it felt overly complicated