help-13

[swift bridge]

okay two things one

can you help me find a derivative mine just doesn't seem right haha

and two

did you play league customs

in the boss status discord server

heres the answer i got

-2cos(x) * -sqrt2x

[scenic vale]

I mightve I dont remember but the name sounds familiar

and then your derivative is wrong

it should be 2cosx - sqrt(2)x

you have an extra negative sign in your first term and the terms should be added not multiplied

remember that dy/dx (a + b) is equal to the dy/dx (a) + dy/dx (b)

[swift bridge]

i dont understand that

[scenic vale]

if you have two terms you can take the derivatives of them seperately

like if you have d/dx(x^2 + cosx) or something, it is equal to d/dx(x^2) + d/dx(cosx)

so your equation is equal to d/dx(2sinx) + d/dx(-sqrt(2)/2 * x^2)

the derivative of 2sinx is 2cosx

the derivative of -sqrt(2)/2 * x^2 is equal to -sqrt(2)x

so you add the two to get 2cosx - sqrt(2)x

is the first derivative

then you take the derivative of that to get the second derivative

[cedar kiln]

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[obtuse coral]

What abt #3 c and d? How do I do those?

@vestal fern

Could you plz also help me w/ these?

[cedar kiln]

@obtuse coral Has your question been resolved?

[dire geode]

Try plugging in x values and find the corresponding y value

f(0)=1

So 3 * f(0) = ?

[obtuse coral]

What abt for d?

It’s a reflection across the y-axis, right?

Does this look right ?

[cedar kiln]

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[devout turtle]

idk what 2 do can anyone just assist me with just 1 of these so i get an idea

[livid hound]

apply sohcahtoa

[devout turtle]

to what

to the triangle?

[livid hound]

yeh

[devout turtle]

i think i got it now

i jus related the answer to the answer from the txtbook

thanks

.close

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[formal pasture]

how do

i start this

i think it needs the sum or diff of cubes

[humble karma]

64 = 2^6, 512 = 2^9

[crimson sedge]

you gave the answer 😭

[humble karma]

How is that the answer lol

[crimson sedge]

lol

[humble karma]

This is merely a hint. They still have to apply their formula correctly.

[formal pasture]

correct

thing is

im confused

how to apply

n stiff

😭

[humble karma]

Plus they already STATED that they think it should be a sum/diff of cubes.

[formal pasture]

ik the formula

[humble karma]

Write both terms as something cubed

It looks to me like the powers I gave you are divisible by 3 ...

[formal pasture]

i dont get this fraction stuff

[humble karma]

If 64 = 2^6, can you not write 1/64?

[formal pasture]

how’s that’

the same thinf

[humble karma]

1/x = x^{-1}

[formal pasture]

😭

[humble karma]

ok.

1/x = x^{-1}

So 1/64 = ?

[formal pasture]

yes

would be

64^-1?

help

[humble karma]

Yes.

Now I told you that 64 = 2^6. Hence 64^{-1} = (2^6)^{-1} = ?

[formal pasture]

64

[humble karma]

use exponent rules

[formal pasture]

1/64

[humble karma]

You can multiply exponent of exponents.

[formal pasture]

so ur saying

two to the power of six to the powrr of -1

[humble karma]

yes.

[formal pasture]

2^1/6?

[humble karma]

No you multiply the exponents

[formal pasture]

2^-6

[humble karma]

What is 6 * (-1)

Yes

Ok

So you have 2^{-6}*x^12 + 2^9.

Can you write the left term as a cube (divide exponents by 3) ?

[formal pasture]

yes

dont u divide

by two

to get cubic

[humble karma]

Last time I checked something cubed is (something)^3

[formal pasture]

-6/2=-3

yes

and its to power of -6 rn

[humble karma]

You don't want to have a cube at the end, you want to express it as something^3

For instance

[formal pasture]

didnt u say the 2^-6

[humble karma]

2^15 = (2^5)^3

[formal pasture]

needs to be expressed at cubic

[humble karma]

You want to express the left term as something cubed

You're looking to the something

[formal pasture]

yes

[humble karma]

If you'd rather, just take the cube root of both terms

[formal pasture]

yea

cant we take common

[humble karma]

But that is the same as dividing the exponents by 3

[formal pasture]

root

out

theyre all divisble

cant i take x^3 out

oh wait

nvm

[formal pasture]

yea

[cedar kiln]

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[gentle orchid]

Why is generalised modus ponens also called lifted modus ponens—what exactly is being lifted here?

[cedar kiln]

@gentle orchid Has your question been resolved?

[north wyvern]

doesnt it refer to the order of logic

i.e: modus ponens is in first-order logic
general modus ponens is in second-order logic
so its "lifting" an order of logic

assuming i know what you are talking about

[gentle orchid]

that sounds reasonable. the context in which generalised modus ponens appeared for me is in first-order logic compared with propositional logic

I'm assuming--not entirely sure--that first-order logic is higher order compared to propositional logic. Hence modus ponens that is used in propositional logic, can be generalised to apply to higher-order logic

[north wyvern]

isnt prop first-order?

oh

rookie mistake!

my bad

[gentle orchid]

no prob! i don't know either that's why i'm asking haha

[gentle orchid]

i think this works, propositional logic is zeroth-order logic as it has no free variables. thanks!

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[livid dust]

I left this question to last because it seemed intimidating

Now I'm here, and it still looks intimidating

I'm guessing it involves something about gaussian integers, but I don't know exactly where to start

[cedar kiln]

@livid dust Has your question been resolved?

[livid dust]

Think I found a way, looking at it as a pythagorean triple with y=2^k

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[livid dust]

solved it myself

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[chrome turtle]

pls help i going insa e on this why the hell when i resolve the above component only i get a different result than on resolving the mg one

[cedar kiln]

@chrome turtle Has your question been resolved?

[neat dune]

Are you on a ramp

[chrome turtle]

yea w wedge

a

with theta inclination

[neat dune]

Well then it's cus the net force on the vertical axis isnt actually equal to 0

it's 0 along the axis perpendicular to the ramp

[chrome turtle]

ohhh yea

damn im dumb

thanks

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[desert blade]

What does the mod mean in this problem, and how do I go about interpreting it?

[neat dune]

Modular arithmetic, 2^k divides i - t

So like, sum of nCi, for i between 0 and n such that 2^k divides i-t

Hmm unless it's saying 2^k is t?

[desert blade]

Where are you getting i-t from? It looks like it's the sum of t * nCi

[neat dune]

Cus tbh congruences are usually written as $i \equiv t \mod 2^k$

[wraith dagger]

992qqoloy

[neat dune]

The t thing is part of the subscript but nCi isn't

What is $k$ referring to exactly? I think that would help answer the question

[wraith dagger]

992qqoloy

[desert blade]

k is just an integer value

[neat dune]

And t isn't 2^k? It's its own thing?

[desert blade]

This is the full problem. t is 0, 1, 2, .... etc.

[neat dune]

OK so $i \mod 2^k = t$ is most probably just a different way of writing a congruence relation

[wraith dagger]

992qqoloy

[desert blade]

Can you explain that a bit further?

What does a congruence relation entail?

[neat dune]

If you have $a \equiv b \mod c$

[wraith dagger]

992qqoloy

[neat dune]

That means the same as c divides $a-b$

[wraith dagger]

992qqoloy

[neat dune]

so for instance $7 \equiv 1 \mod 3$

[wraith dagger]

992qqoloy

[neat dune]

So the only way I can see to interpret this is for instance, if $n =16$ and $k =2$, then $a_3$ would be $\binom{16}{3} + \binom{16}{7} + \binom{16}{11} + \binom{16}{15}$

[wraith dagger]

992qqoloy

[neat dune]

cus 3,7,11,15 are the numbers between 0 and 16 that are equivalent to $3\mod 2^2$

[wraith dagger]

992qqoloy

[desert blade]

Ohhh okay, so it's like a summation, but only choosing the values of i that fit with the mod thing?

[neat dune]

yeah

[cedar kiln]

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[forest violet]

Hi, I've attempted this question

I don't have the answers and I'm not even sure if it is possible with the given information

[manic bronze]

what type of lvl math is this?

im curious

[forest violet]

yr10

im in australia

[manic bronze]

oh 10th grade???

[forest violet]

ye

[manic bronze]

thats insane

i dont even know what the hell that is

[clear umbra]

calculate angle BDA, then you have CDA, angle z will be equal to CDA

[forest violet]

isn't CDA a line

180 degrees

[clear umbra]

CDB i mean

sorry

[forest violet]

tyty

i got it

.close

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[hot dune]

2cos²(3x-pi/4)=1 0<=x<=pi/2

[hot dune]

i don't know where i did wrong bc answer key says a different answer

and I'm confused cause the answr is pi/3 in fraction form but I get decimals so is there a way to convert or something

[umbral dew]

okay but what we have to do here?. like to find the general solution of this?

[hot dune]

yes all angles between 0<=x<=pi/3

i already found one, pi/6

solve for x

[umbral dew]

there is a formula If$\cos^2(\theta)=\cos^(\alpha)$ then $$\theta = n\pi+ \aplha$$

[wraith dagger]

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

[umbral dew]

okay im not doing it now

i always typed it on my pc. in phone it sucks

okay so we have a formula

[hot dune]

but there's no cos (a) in the question srry i don't understand

and what is n

[umbral dew]

that is cos^2 x = cos^2 y, then x= n pi + y

n here is integers

[hot dune]

What is cos²y

• alpha?

i haven't been taught this

[umbral dew]

okay

you missed the postive part

you solved for the negative part, now make another case and take root 1/2 positive

[hot dune]

i solved the positive part tho

The pic i sent is for the negative part

[umbral dew]

oh I see

wait im brushing my teeths

right back in 2 min

[hot dune]

nvm i solved it

There's still another angke to solve

Answer key says its 0, how

pi/3, pi/6 and 0 how did they get 0

[umbral dew]

so when you're taking postive part, cos inverse 1/root2 should give you pi/4, so pi/4 from each side gets cancelled out and then 3x=0 therefore x=0

,w \arccos(\frac{1}{\sqrt{2}})

[wraith dagger]

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[hot dune]

i got 3x-pi/4=pi/4

[umbral dew]

yes

good

wait

no

[hot dune]

But that doesn't give u 0

[umbral dew]

,w \arccos(-\frac{1}{\sqrt{2}})

[wraith dagger]

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[umbral dew]

okay

yea

so in the negative part only

cos inverse - 1/root 2

gives you 3pi/4

[hot dune]

That's a wrong answer according to the answer key

[hot dune]

u can't use -1/root 2 tho cause the negative sign just tells u which quadrant it's in so it should still be positive 1/root 2

they got 0 by -pi/4 but where did the negative sign cone from

[umbral dew]

,w cos(3pi/4)

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[umbral dew]

,w cos(-pi/4)

[wraith dagger]

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[umbral dew]

,w cos(pi/4)

[wraith dagger]

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[umbral dew]

cos(-pi/4) is the same as cos(pi/4)

and that is why we're getting 1/root 2 from both

[hot dune]

aight i think i solved it thanks

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[hollow cloak]

Have you done squeeze theorem?

[scarlet heron]

yes

i know that sin(x)/x as x aproachhes infinity is 0

but for this im not too sure

[hollow cloak]

Probably sin of that

I mean if you want to check, you can plot it out

[lyric narwhal]

direct substitution gives u finite/infinity, which is always equal to 0

[scarlet heron]

what is sin(inf) tho

oh its between one and negative one

so its still finite

[lyric narwhal]

indeed

[scarlet heron]

and sin of that is still finite

and finite over inf is zero

[lyric narwhal]

[scarlet heron]

oh ok thanks so much haha

have a nice day

:)

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[crimson sedge]

wouldnt the distance from the intercept to the vertex be 1.25

[crimson sedge]

the person doing the sum seems to be getting 7.25

but -17/4 is -4.25

and -3 is the vertex

so 1.25 should be the distance between right

leading the other interceptto be at (-7/4,0)?

[lyric narwhal]

yes

you're right

[crimson sedge]

ah so the answer i got is correct right

[lyric narwhal]

yeah, should be -7/4

unless i'm missing something

[crimson sedge]

thought so too

pretty sure he missed the -3 here

so it shouldve been 17/4-3

[lyric narwhal]

should be $-3-(\frac{17}{4}$

[wraith dagger]

kheerii

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[indigo shard]

Given an angle, how could I find a positive integer such that the cosine of the product n times the angle is minimal?

[tropic oxide]

given a fixed angle θ, minimize cos(nθ) for n in N?

[indigo shard]

yes

the angle given is between the value -180 and 180

I understand that from the cosine table I could try to convert the angle to one such that cos(nθ) becomes 0, but what if that's not possible ;-;

[lyric narwhal]

that doesn't seem very trivial

depends on the angle

[indigo shard]

in the problem, If I am given 16, the solution is n=17 and cos(272)

so I guess I just have to compute the answer with every possible number until I find the minimum value?

💀

[lyric narwhal]

well

you want an angle that is as close to an odd multiple of 90°

as possible

if the given angle is an integer, you could probably do smth with modular arithmetic

[indigo shard]

what could I do with modular arithmetic? :0

[lyric narwhal]

use Bezout's Identity

[cedar kiln]

@indigo shard Has your question been resolved?

[indigo shard]

thanks, I;m looking that up

is there something I could do if the number is not an integer though?

If I am given 88.817841970012523233890533447265625, I should get 154618822656 as the answer since the cos of the product is close to 0

[cedar kiln]

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[rocky dragon]

What do i do on g

[rocky dragon]

Or wait maybe ill get it

Its just first the things inside

And then the others

[hollow pilot]

Are you trying to solve for all the variables?

[rocky dragon]

The thing i should do is simplifying

[midnight hatch]

There’s no equality how would you solve nothing

[rocky dragon]

I think i get g h gonna be a bigger problem

[hollow pilot]

Yea thats why im confused

[midnight hatch]

Distribute the -1 and reduce

[rocky dragon]

Oh right there is no equality

Thats weird

[midnight hatch]

They ask you to simplify so not particularly

[rocky dragon]

What does the last word mean im not really good at understanding english words for math im german

[midnight hatch]

Just saying it’s not weird

-(a + b) = -a-b

And -(-c) = c

Use this to simplify

[rocky dragon]

Huh

What

I dont get it i cant do anything on g

[midnight hatch]

What do you want to do ?

[rocky dragon]

Simplify it but idk how

[midnight hatch]

Expand and reduce ?

[rocky dragon]

What

@clear umbra do you know what to do on g

Its simplifying

But idk what to do

Jts basically nothing i can do

<@&286206848099549185>

[dreamy sleet]

you can distribute the - over the parens

[rocky dragon]

What does that mean

[dreamy sleet]

for example, $$ -(3b - 4c) = -3b + 4c $$

[wraith dagger]

rysrobrgldvoelr👻ep>vneae=u

[dreamy sleet]

do you know how to open up the brackets?

[rocky dragon]

Nope

Thats why im struggling

A

Wait

But how do i open the first one with none

@dreamy sleet

a+2b-3b+4c+5c+6a?

Now to make it even easier

[dreamy sleet]

Where did this come from?

[rocky dragon]

O opened

I

All

[lusty ginkgo]

hm, check againcarefully

with opening the brackets

[rocky dragon]

what

Whats wrong

What did i do wrong

@lusty ginkgo

[lusty ginkgo]

second bracket

[rocky dragon]

Oh

-3b+4c

[lusty ginkgo]

yeaaa and then just simplify everything

[rocky dragon]

One sec

a+2b-3b+4c+5c+6a?

So

[rocky dragon]

-1b+9c+7a

[lusty ginkgo]

well it's nicer to write it in alphabetical order but thats just me being picky

[rocky dragon]

My main problem is just that we had a new theme and i forgot about everything

[rocky dragon]

[lusty ginkgo]

same w -ve signs

[rocky dragon]

Thank you

[lusty ginkgo]

ah.. practice lots ig

ask ur teacher for help too 👍

[rocky dragon]

Got this day and tmr and then im writing my exam

[lusty ginkgo]

good luck 🙂

[rocky dragon]

Always do that

[rocky dragon]

Thank you!

Close.

close.

.close

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[median pumice]

x+y= 5
x^3 + y^3= 35
x and y?

[median pumice]

<@&286206848099549185>

[primal socket]

!15mins

[cedar kiln]

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

[primal socket]

Do you know how to solve a system of two equations with two unknowns?

[median pumice]

yes but not with the powers

[primal socket]

Why should it be different?

[median pumice]

idk

i only know how to solve basic simutanious equations

[primal socket]

Well

How would you go about solving this?

Let me rephrase

[median pumice]

i factorise it

[primal socket]

If we didn't have the powers there

How would you solve it?

[median pumice]

oh wait

hmm

i would move

y to one side in the first equation

[primal socket]

Okay

Then?

[median pumice]

then put - on the second equation

like multiply the second equation with -

[primal socket]

You mean (-1)

[median pumice]

yeah

[primal socket]

Okay

[median pumice]

so then i would get the answer

by like

changing the subject

i think im wrong

[primal socket]

Hm

Do you know how to solve systems in general?

[median pumice]

wdym by that

[primal socket]

You've probably learned 2 different methods

[median pumice]

yeah

uh subsitutionb

[primal socket]

Yes

[median pumice]

and ex smth

[primal socket]

Elimination

[median pumice]

yeah

[primal socket]

Solving by substitution and by elimination

[median pumice]

ik subsitiution

i forgot elmination

[primal socket]

Solving by substitution means isolating a variable in one equation and plugging it in the other

[median pumice]

yes

[primal socket]

Solving by elimination means getting rid of one variable and getting the other that way

In our example

[crimson sedge]

a^3+y^3=[a+b][a^2-ab+b^2]

[primal socket]

We can easily solve this by substitution

[median pumice]

what

[primal socket]

You see that we can isolate either x or y easily in the first equation right?

[crimson sedge]

here is a theorem

[median pumice]

yes

[primal socket]

So why don't we do that and plug it into the second equation?

[crimson sedge]

u can have a try

[median pumice]

u mean without the power right?

[primal socket]

Nevermind this

[primal socket]

No

[median pumice]

oh

[primal socket]

The example you were given

[median pumice]

yeah?

[primal socket]

$x + y = 5 \ x^3 + y^3 = 35$

[wraith dagger]

USS-Enterprise

[median pumice]

what does the dollar sign mean

oh its a command

[primal socket]

Math mode for latex

[median pumice]

kk

[primal socket]

Anyway

You can solve this easily by substitution

[median pumice]

yeah but the thing is im not that good at it

[primal socket]

Well now's as good as time as any to learn it

So what did we say earlier

Isolate a variable in one of the equations and plug it in the other

Would we isolate a variable in the second or first equation

[median pumice]

like add them?

[primal socket]

Which looks easier?

[primal socket]

No

[median pumice]

first one

[primal socket]

Okay

And which variable would you like to isolate

It doesn't matter

[median pumice]

x

[primal socket]

Okay, what do we get?

[median pumice]

x=5-y

[primal socket]

$x = 5 - y \ x^3 + y ^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

So

If x equals 5 - y

[median pumice]

yeah

[primal socket]

Can we not plug it in the second equation

And solve for y?

[median pumice]

right

i forgot what plugging is

[primal socket]

Substituting

Writing (5-y) instead of x

[median pumice]

i dont get it

[primal socket]

Okay

The second equation

We've got $x^3 + y^3 = 35$

[wraith dagger]

USS-Enterprise

[median pumice]

plugginf = subssituting?

yes

[primal socket]

This is the same as $x \cdot x \cdot x + y^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

And we know x = (5-y)

So we can we not write (5-y) instead of x and solve for y?

[median pumice]

OH

u factorised it?

[primal socket]

No

$x^3 = x \cdot x \cdot x$

[wraith dagger]

USS-Enterprise

[primal socket]

That's exponentiation

[median pumice]

no i mean like u factorised x

yeah

[primal socket]

Means repeated multiplication

[median pumice]

that thing

ohhh

kk

so we got that then?

[primal socket]

We've got $x^3 + y^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

And x = (5-y)

[median pumice]

yes

[primal socket]

So just write (5-y) instead of x

And solve for y

But note it's getting cubed

So we get $(5-y)^3 + y^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

Or $(5-y) \cdot (5-y) \cdot (5-y) + y^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

See how I just replaced x for (5-y)?

[median pumice]

yes

[primal socket]

We had x * x * x, now we have (5-y) * (5-y) * (5-y)

$(5-y)^3 + y^3 = 35$

[wraith dagger]

USS-Enterprise

[primal socket]

So just solve this for y

[median pumice]

i was doing it in note pad and i got this x = 5 - y
if x = 5 - y then
(5 - y )^3 + y^3 = 35
(5-y)(5-y)(5-y) + y^3 = 35
125 - 75y + 15y^2 - y^3 + y^3 = 35
5(25 - 15y + 3y^2) = 5(7)
25 - 15y + 3y^2 = 7
15y + 3y^2 = 7 - 25
y(15+3y) = -18
15 + 3y = -18/y
3y = -18/y - 15

im pretty sure its wrong

[median pumice]

wait

woudlnt y get canceled

nvm

[primal socket]

15y + 3y^2 = 7 - 25

Is your first mistake

You've got -15y not 15y

So we get $3y^2 - 15y = -18$

[wraith dagger]

USS-Enterprise

[primal socket]

Not sure what you went to do then though

This is a quadratic equation

[median pumice]

ohhhhhhhhhhhhhhhhhhhhhhh

[primal socket]

First we've got coefficients 3, -15, -18 so we divide by 3

$y^2 - 5y = -6$

[median pumice]

yeah

[wraith dagger]

USS-Enterprise

[primal socket]

And we move everything to one side

$y^2 - 5y + 6 = 0$

[wraith dagger]

USS-Enterprise

[primal socket]

And solve this

$(y - 3) \cdot (y - 2) = 0$

[wraith dagger]

USS-Enterprise

[primal socket]

$y_1 = 3 \ y_2 = 2$

[wraith dagger]

USS-Enterprise

[primal socket]

And then you plug y1 and y2 into the original equation and get two x's

So this system has 2 pairs of solutions

[median pumice]

oh aight so i just have to substitute it

after this process

[primal socket]

Yes

First pair is (y1, x1), second is (y2, x2)

So (3, x1) and (2, x2)

x1 is what you get after plugging in y1 and x2 is what you get after plugging in y2

[median pumice]

X=2

And x=1

TYSM I GET IT NOW

[primal socket]

How'd you get 1

x = 5 - y, so x1 = 5 - 3, x1 = 2, x2 = 5 - 2, x2 = 3

[median pumice]

oh righttt

sry

[primal socket]

$(x_1, y_1) = (3, 2) \ (x_2, y_2) = (2, 3)$

[wraith dagger]

USS-Enterprise

[primal socket]

That's your solution

[median pumice]

i thought that the other way

yeah

[primal socket]

So yeah

You solved this by substitution

[median pumice]

OH i havent learnt simutanious equation for like this

[primal socket]

ah

Should refresh your knowledge

On both substitution and elimination

They are quite useful

[median pumice]

yeah

[primal socket]

Anyway

[median pumice]

ima go learn elimination

[primal socket]

Hope I've been of help 🙃

Good luck!

[median pumice]

YOU HAVE BEEN OF HELP

tysm bro

[primal socket]

No problem

Have a good day 😄

Just please .close if that's all

[median pumice]

aight byeee

oh kk

.close

[cedar kiln]

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[wide nest]

hello can some one help me about this solution

[cedar kiln]

@wide nest Has your question been resolved?

[cedar kiln]

@wide nest Has your question been resolved?

[cedar kiln]

@wide nest Has your question been resolved?

[limpid plume]

!status

[cedar kiln]

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

[limpid plume]

@wide nest

[wide nest]

2

[wide nest]

there are 2 deltas and i dont know how to pick the correct one

[limpid plume]

Just asking, are you allowed to use continuity?

[wide nest]

this part no

[limpid plume]

Ok got it, lemme think

[wide nest]

maybe yes

i dont know exactly

this is the first part of limit

[limpid plume]

Okay so. Start with a delta, and use | x - 9| < delta

Expand that to get

9 - delta < x < 9 + delta

From this I think you can get somewhere

4 - delta < x - 5 < 4 + delta

Don't take this literally, I'm just suggesting

[wide nest]

but i couldnt undertsand the fingding of the correct delta

should i choose the small one

[limpid plume]

Problem is it might not work...

I could find epsilon in terms of delta but idk if it has a delta for every such epsilon

Nvm, it certainly doesn't

Wait

I think I got it

[heavy compass]

what class is this?

[wide nest]

thomas calculus math101

[limpid plume]

I don't see any other way (read: simple) way to proceed except proving sqrt(x - 5) is continuous in [5, infty)

Then the limit of the function at that point is the value of the function at that point

Continuity is easy to prove, in fact it can be shown it's even uniformly continuous

This is how you show its continuous

[cedar kiln]

@wide nest Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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

the last one is wrong

I gotta put it in algebraic notation

[cedar kiln]

@shadow prism Has your question been resolved?

[shadow prism]

<@&286206848099549185>

[cedar kiln]

@shadow prism Has your question been resolved?

[shadow prism]

I'm finna throw hands at yall

[heavy compass]

hi

try what are the options of the last one?

[cedar kiln]

@shadow prism Has your question been resolved?

[cedar kiln]

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

,tex lim$_{x \to 0} \frac {1 - e^{x}}{ln(2 - e^{x})}$

[wraith dagger]

@brazen wigeon

[brazen wigeon]

so um

i literally have no clue what to do

i think i have to get rid of ln, but idk how

[crystal raptor]

lhoptial allowed?

[mental trail]

do you know $\lim_{y \to 0} \frac {\ln(1+y)}{y}$?

[wraith dagger]

rafilou2003

[brazen wigeon]

i dont

[brazen wigeon]

all i know is that it is likely a hole

[mental trail]

use hôpital then

[brazen wigeon]

i have no clue what that is 😭

[mental trail]

if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ and $\lim_{x\to a}\frac{f'(x)}{g'(x)} = l$, then $\lim_{x\to a}\frac{f(x)}{g(x)} = l$

[wraith dagger]

rafilou2003

[silver fable]

Just divide by x top and bottomm

[brazen wigeon]

what is L at the end?

[silver fable]

You'll have derivative of a function

[mental trail]

just the finite limit of f'(x)/g'(x)

[brazen wigeon]

so if there is likely a hole at a specific X value, then the derivative of top/bottom functions represent the x/y variables of the coordinate of the limit?

(tried my best to rewrite your rule from math to english)

[silver fable]

$\lim_{x\to 0} \frac{ \frac{1-e^x}{x} }{ \frac{ln(2-e^x)}{x} }$

[brazen wigeon]

both of those are equal?

[mental trail]

farc

[wraith dagger]

Adam Chebil

[silver fable]

ty

[brazen wigeon]

what is this equal to?

[silver fable]

It's just the derivative at x=0

[brazen wigeon]

and does it work (althoguh likely unnecessary) with almost any X value?

[silver fable]

I think it does work

Try it

[brazen wigeon]

sadly a certain stat of my body has shot up a lot in the past year
stat name: ||laziness||

just needed help with this quesiton

ty for you help

[silver fable]

Derivative of e^x is e^x

-e^0

[brazen wigeon]

oh

ya i see now

[silver fable]

And here's the derivative of ln(2-e^x)

Just evaluate that at 0

And divide

[brazen wigeon]

its just 1/(2 - 1) = 1

[silver fable]

Yeah you just did lhopital rule there but not directly
It's helpful when u're not allowed to use lhopital rule in exams like me

[brazen wigeon]

do you often find yourself memorizing rules just so you can derive them during exams?

[silver fable]

U mean derivative rules?

[brazen wigeon]

no i mean literally any kind of rule

smth like sin rule from geometry

idk

but do you do this often/ at all?

[silver fable]

I rarely memorise i just understand the intuition behind it (not the exact proof)

[brazen wigeon]

so just the english logic?

[silver fable]

Mathematical logic xd

And L'Hôpital's rule is really simple u just derive top and bottom and evaluate
But since I'm not allowed to use it in exams i just divide by x like i did there (if limit as x approaches 0 of course) divide by x-a in general

So doesn't need memorising

[brazen wigeon]

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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

Hi i have problem set up as a system of equations

x+y+z=200
10x+53y+84z=7850
.04x+.04y+.03z=272

yet i put it into a calculator and get answers in the thousands

[dreamy sleet]

What did you get

[brazen vector]

the assignment is for us to use matrices for the equations

[dreamy sleet]

Is there any difference between a share of a stock and a stock

[brazen vector]

umm not sureee

[dreamy sleet]

maybe you have the formula for dividend yield wrong?

[brazen vector]

perchance

maybe because it's supposed to equal a price?

[dreamy sleet]

You solved it right

[brazen vector]

if its only suppose to equal 200 shares total why am i getting numbers in the thousands

[dreamy sleet]

they add to 200, it’s just that some of them are negative

[dreamy sleet]

there is no problem with the solving of the equations, so the equations are probably wrong

[dreamy sleet]

or maybe the dividend yields given are for monthly dividends and you actually need the annual dividends

[brazen vector]

i thought i had it when i multiplied my price by dividend yield and submitted it but it was wrong because they only equaled to be 199.9998 shares

so whatever

.close

[cedar kiln]

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[dreamy sleet]

What equations did you use

Maybe you rounded some intermediates

[brazen vector]

.reopen

[cedar kiln]

✅

[brazen vector]

Ohmygod

i forgot to take the extra zero out of .4 instead i put .04

but i understand what to do now

.close

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[dreamy sleet]

Where

[dreamy sleet]

I’m curious, where did you go wrong

[brazen vector]

i had to set the dividend yield equation to price

so i mutipled 10 x 4%

53 x 4%

84x 3%

and put those answers back in the equation

[cedar kiln]

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[ebon drift]

how would you simplify this?

[flint plinth]

do you see any relation between the numerator and the denominator?

[ebon drift]

well yeah

if the signs were flipped for one of them then it'd be 1

[flint plinth]

ok

so if the signs are not flipped, what should it be?

[ebon drift]

would it be -1

[flint plinth]

yes

[ebon drift]

oh you can multiply the numerator or denominator individually by a negative?

[flint plinth]

one easy way to show that would be to factor out a -1 from the numerator

[ebon drift]

oh ok

[ebon drift]

so you're not multiplying a numerator or denominator by negative you're factoring

[flint plinth]

yes correct

[ebon drift]

?

Ok alright

that makes sense

Thank you

.close

[cedar kiln]

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[cedar kiln]

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[wanton summit]

hello

[crimson sedge]

yes?

[wanton summit]

it will be helpful if someone explain this slide to me i am confusing betwwen equation

of utility function and expected value

[cedar kiln]

@wanton summit Has your question been resolved?

[neat dune]

What's confusing u about it

[wanton summit]

risk premium

[cedar kiln]

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[fleet granite]

.close

bruh

[fleet granite]

.close

[cedar kiln]

Channel closed

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[fleet granite]

here

[tropic oxide]

the channel was already closed, what were you doing?

[fleet granite]

Oh i thinking

[old laurel]

Help with the derivative of trignometry

[fleet granite]

what?

[tropic oxide]

!occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[cedar kiln]

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[sterile gust]

I dont even know where to begin

[spice kraken]

notice that f is decreasing

[sterile gust]

yes

[spice kraken]

that means if f(x)>f(y), x<y

[sterile gust]

yes

[spice kraken]

cancel the f using that property

f's

[sterile gust]

yeah so the inequality keeps getting opposite?

[spice kraken]

yes opposite each time you cancel

[sterile gust]

Okay

we get f(x) > -x

finally

But is there a way to solve inequalities for a cubic?

[spice kraken]

just plug in positive x

[sterile gust]

positive?

where

[spice kraken]

no need to solve it as we just check positive integer

[sterile gust]

yeah

so trial and error?

[spice kraken]

yep

[sterile gust]

Okay Ill do it

Thanks

[spice kraken]

no problem

[sterile gust]

.close

[cedar kiln]

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[cedar kiln]

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[lapis laurel]

at what point on parabola y=9-x^2 does the tangent line draw the smallest area triangle with the coordinate axes?

[exotic olive]

I would recommend starting with a diagram

[lapis laurel]

like drawing?

[exotic olive]

Yep!

just to get your head around the situation

[lapis laurel]

well its a downward parabola

[exotic olive]

Yep!

Now I would define a variable point on the graph

[lapis laurel]

like a and f(a)

[exotic olive]

Ye

just a question is there a domain restriction on the graph?

[lapis laurel]

no restriction are mentioned

[exotic olive]

Hmm ok

[lapis laurel]

now i would crete the tangent line at the variable poiknt?

[exotic olive]

yes

[lapis laurel]

so it is y-f(a)=f'(a)(x-a)

[exotic olive]

Yes

(Note: The parabola is symmetric so we can disregard the negative values of a

[lapis laurel]

yes that is through

[exotic olive]

ok so now what do we need to find the area of the triangle?

[lapis laurel]

well it is going to be a right triangle since one angle is 90 because of the axes. So the area is going to be width X height X 1/2

the width can be solved by getting the x-intercept?

[exotic olive]

yep so find the x-intercept in terms of the variable you used for the variable point

[lapis laurel]

so y=0 and i solve what x is then

[exotic olive]

exactly!

[lapis laurel]

how do i solve x?

is this possible? i didnt substitute f(a) and f'(a) this time

actually i did a mistake ill try again

[exotic olive]

Yep

sorry I am also practicing piano so my reply is a lil delayed

[lapis laurel]

no hurry

[exotic olive]

small mistake, gradient is actually -2a

[lapis laurel]

ah yes

here are both axis intercepts and they should be correct

and heres the function for the area

and now to find the lowest value i have to derivate it

[exotic olive]

Exactly!

I would then also test that a is actually producing a minimum using second derivative test

[lapis laurel]

how wouldnt it be quite hard to derivate that a second time?

[exotic olive]

True...

you could use a gradient table instead

[lapis laurel]

well here I simpligied the denominator

of the derivative

[exotic olive]

Instead of using the expanded form it would have been easier to use quotient rule

I'll send a picture of my working

[lapis laurel]

the derivative is thiså

[exotic olive]

,rotate

[lapis laurel]

so you differentiated it f'g-fg'

/g^2

[exotic olive]

yup

and then I factored out (a^2 + 9)

[lapis laurel]

ok so you started the differentiating with gf'-g'f

but i dont really grasp what happens after that

[exotic olive]

$4a^2(a^2+9) - (a^2+9)^2$ \let $y=a^2+9$ which gives $4a^2y-y^2$ \then factor out y $y(4a^2-y)$ \then sub back $a^2 + 9$ to get $(a^2+9)(4a^2-a^2-9)$ \then simplify $(a^2+9)(3a^2-9) = 3(a^2+9)(a^2-3)$ let $\frac{d}{da} = 0$\ thus you get $3(a^2+9)(a^2-3)=0$ \which then gives $3(a^2+9)(a-\sqrt{3})(a+\sqrt{3}) = 0$ \which means that $a=\pm\sqrt{3}$

[wraith dagger]

SKELEROY

[exotic olive]

then the last step of the gradient table

I subbed in values of a that are around sqrt(3) into the derivative to figure out that the curve is concave up when a=sqrt(3)

[lapis laurel]

ok i think i finally understood what you did there, did you substitute x^2+9 with y for simplicity so that the equation is essier to factor

[exotic olive]

Exactly!!!

that way I don't have to deal with massive numbers

[lapis laurel]

practical, so to get the answer for the question i just chuck the value cbrt 3 in the area function

[exotic olive]

yes, additional question, is a=-sqrt(3) a valid solution?

[lapis laurel]

well thats also another point on curve so seems logical because even though its on the negative side of the axis it still creates a triangle with the axis. So to answer the original question "at which point on the parabola 9-x^2 does the tangent draw the smallest triangle" there are two points

cbrt 3 and 6 and -cbrt 3 and 6

[exotic olive]

wait a sec why are we doing cuberoot?

[lapis laurel]

oops meant sqrt

[exotic olive]

haha all g

[lapis laurel]

well gotta go so thank you very much for helping

[exotic olive]

No worries!

[lapis laurel]

.close

[cedar kiln]

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[ebon oxide]

How to solve 1<=(2x^3/(3x^2-1))<1.1

[tropic oxide]

that looks like it'll be tricky

but also sus!

!xy

[cedar kiln]

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.

[ebon oxide]

My calculator doesn’t work so I need alternative ways 😭

[tropic oxide]

no but where did this inequality actually come from

that's what i'm asking

[ebon oxide]

I’m doing the last part on the second page

Also, I am allowed to use a calculator, it’s just that I can’t directly solve the inequality

[cedar kiln]

@ebon oxide Has your question been resolved?

Channel closed

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[crimson sedge]

why cant i simplify root(4x^2 + 9) to 2x+3

[buoyant latch]

The root function isn’t linear

ie. it doesn’t have the f(a+b) = f(a) + f(b) property

[crimson sedge]

what does that mean? is it because the 9 would have to be 9x for this to work?

[buoyant latch]

I said what it means

It doesn’t have that property

[cerulean harness]

U can only breke down to 2x+3 if (4x^2 + 9) has sqaure

[calm sierra]

root doesnt split over sums

[buoyant latch]

It’s like asking why isn’t 5-3 the same as 3-5

It just doesn’t work that way

[crimson sedge]

I see okay

[cerulean harness]

√(a+b) not equal to √a + √b

[crimson sedge]

okay thanks

.close

[cedar kiln]

Channel closed

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[faint vine]

I have an exercice where: Given a serie Un like this: {Un+2 =(2/5)Un+1 - (1/25)Un
{ U0 = 0 and U1=1

We have a serie Wn = 5^n * Un
Prove that Wn is a arithmétique serie with reason r=5

I tried Wn+2 - Wn+1 but i cant get a constant r value its weird and tried Wn+1- Wn and also couldn't get it, any help please?

[wintry pier]

thiz channel will soon close, take another one

[faint vine]

Okay

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cyan swallow]

Its a 9x^2 btw

[cyan swallow]

I do not understand how to factorise this at all or how to expand the fully factorised version

Like there is parts i understand but i cannot connect them. I guess i need someone to walk me through it all with words, no metaphors though.

Ping me if you respond. Thank you.

[clear umbra]

,rccw

[clear umbra]

@cyan swallow well do you see a connection between 3x and 9x^2?

[cedar kiln]

@cyan swallow Has your question been resolved?

[cyan swallow]

Lcf

3 and x

Also, 3 x -3 = -9

Can you walk be through with words on how it was factorised

And how it could be expanded from that further factorised point

[clear umbra]

so you know the common factor for 3x and 9x^2

then just factor it out

so you have 3x((a+b)^4 - 3x(a+b)^2)

[cyan swallow]

Yea

[clear umbra]

then (a+b)^4 and (a+b)^2 also has a lcf of (a+b)^2

so you factor that out too

[cyan swallow]

Yea

[clear umbra]

so whats left is that second line there