help-10

[vestal steppe]

What do you think $\frac{5^{-1}}{5^{-1}+4}$ is equal to, using your logic?

[warm shale]

Average Maths Student

[austere gazelle]

5/5+4

Because the negative exponents have the switch to the opposite side of the fraction

So like the botton five moves to the top

And the top one moves to the bottom

[vestal steppe]

So you are saying $\frac{5^{-1}}{5^{-1}+4}=\frac{5}{5+4}$ is true?

[warm shale]

Average Maths Student

[austere gazelle]

Yeah even though ik it's not

But that's what I get

[vestal steppe]

Well yeah that's what I'm saying - it's false

It's false because you can't swap it like that

These are the index laws

You can see on the top right rule, you have the so called 'swapping rule' for negative indices

[austere gazelle]

Yeah Ik all of those

[vestal steppe]

Notice how the swapping rule only applies when a^-m is by itself.

[austere gazelle]

Yeah but like my teacher taught it to me so that when say 7^-1 is over 4^-1 it becomes 4/7

[vestal steppe]

That's fine

[austere gazelle]

Si what's the difference

[vestal steppe]

Because there's no addition sign anywhere in your sentence

[vestal steppe]

There's no addition sign - it's just division

[vestal steppe]

But if you look here

There is an addition sign

[austere gazelle]

What does the addition sign do then

[vestal steppe]

It's 5^-1 + 4

[vestal steppe]

It changes the denominator completely, it means you can't apply that rule

In general, once you see an addition sign in a fraction, there are so many things you can no longer do

anyways

To circumvent this issue

[austere gazelle]

Yeah as I was saying earlier it becomes like 1/5+4=4/1

Which isn't the same as 5/1

[vestal steppe]

nooo this is not true 😭

[austere gazelle]

Why?

[vestal steppe]

1/5 + 4 is not 4/1

,calc 1/5 + 4

[warm shale]

Result:

4.2

[vestal steppe]

what okay

there

,calc 4/1

[warm shale]

Result:

4

[austere gazelle]

Ik 4/1 =4

[vestal steppe]

Yeah

So you should know that 1/5 + 4 is not 4

What did you do to get one into the other

[austere gazelle]

Oh wait i realized how I messed up

It's 21/5

Not 20

[vestal steppe]

there you go

That's right 👍

[austere gazelle]

Which as I was saying earlier is non compatible with 1/5

[vestal steppe]

non.. compatible?

I don't think it matters that its non compatible

[austere gazelle]

I just realized that now

Lmfao

[vestal steppe]

oof lol

[austere gazelle]

It's 4am

[vestal steppe]

Get some sleep

[austere gazelle]

After this question

So like, I still dunno what to do

[vestal steppe]

tell me what you have right now

As in

[ \frac{5^{-1}}{5^{-1}+4}= ] What changes did you make to this fraction?

[warm shale]

Average Maths Student

[austere gazelle]

(5/5+4)-(5^-1/21/5)

[vestal steppe]

okay yep

Let's keep working on the second fraction

How would you simplify it even further?

[austere gazelle]

1/5 times 21/5?

[vestal steppe]

close; no

You're forgetting something here

[austere gazelle]

Imeant 5

I meant like 1 over all that

Not 1/5

Mb

I meant 1 over 5 times 21/5

[vestal steppe]

oh

Yeah

[austere gazelle]

Do the 5s cancel?

Cause like 5×21/5

I swear they cancel out

[vestal steppe]

yep

No no they do

5 times 21 and then you divide by 5 again, so they cancel out

[austere gazelle]

So now we have (5/9) - (1/21/1)

Right?

[vestal steppe]

Yep

[austere gazelle]

Do I really have to multiple 9 by 21 amd 21 by 9?

[vestal steppe]

Well

You could

You could also recognise that 21 and 9 are a very interesting set of numbers

[austere gazelle]

Divisibke by 3?

[vestal steppe]

Yep!

[austere gazelle]

How does that help?

[vestal steppe]

What's 21 divided by 3

[austere gazelle]

Do you mean multiple them both by 3?

[vestal steppe]

Nope

I do mean division

[austere gazelle]

But 3 and 7 can't work

[vestal steppe]

Why not

What's 9 times 7

[austere gazelle]

Because they're different numbers

9×7=63

[vestal steppe]

What's 21 times 3?

[austere gazelle]

63

Bro that's what I said

[vestal steppe]

WHAT

[austere gazelle]

I said multiply by 3

[vestal steppe]

They're the same number

no

You don't multiply by 3

Because 9 times 3 is 27

And 21 times 3 is 63

Here they actually are different numbers

[austere gazelle]

I'm so confused

How does diciding by 3 help?

Cause like then I have to divide 5 by 3

[vestal steppe]

wait wait wait

Let's backtrack

We have

5/9 - 1/21

We want to just subtract the two fractions

[austere gazelle]

Yeah

[vestal steppe]

How do you subtract two fractions together?

What do you need to do?

[austere gazelle]

The denominators have to be the same

[vestal steppe]

Okay good

How do we make the denominators the same?

[austere gazelle]

Multiply each side by the others denominator

[vestal steppe]

Yes you could but that could be very annoying

So instead

Can you think of a smaller number to multiply each fraction by?

[austere gazelle]

No

[vestal steppe]

really?

[austere gazelle]

Yeah

[vestal steppe]

Not after seeing this

[vestal steppe]

Or this?

[austere gazelle]

Yep

[vestal steppe]

okay fine 😭

[austere gazelle]

That made no sense to me sorry man

[vestal steppe]

I want you to do something then for me

Nah you're good

5/9 - 1/21

What happens if you multiply the first fraction by 7/7?

What do you get?

[austere gazelle]

63

35/63

[vestal steppe]

Perfect

Can you tell me what happens when you multiply the second fraction by 3/3?

[austere gazelle]

3/63

Ohhhhhhhhh

[vestal steppe]

Yess so the idea is

We actually found a pair of numbers

That multiply to a certain product

That is, 9 times 7 is 63

and 21 times 3 is 63 too

So all we need to do is multiply the corresponding fractions together

[austere gazelle]

I understand

[vestal steppe]

This works because you already showed that 9 and 21 have a common factor of 3

So you just need to multiply that other smaller number

[austere gazelle]

If they're both divisble by a number then you do funny thing my sleepless brain can't explain at the current moment and then boom

[vestal steppe]

I mean nothing wrong with doing 21 * 9

[vestal steppe]

Yeah good enough for me

[austere gazelle]

Like those 2 numbers that you divide by are the 2 numbers you multiply by

Or technically 1 number you divide by

So now then we have 32/63

But the answer is 32

So like

Huh?

@vestal steppe

[vestal steppe]

https://media.discordapp.net/attachments/903430332324405288/1505167176229851217/20260516_034951.jpg?ex=6a09a3ab&is=6a08522b&hm=b9b2e27d384d1d771da8852430ca2e5ad0a30a6dbb6b0754e69e5dcf97ae87a3&=&format=webp&width=1576&height=1182

What does the queston say?

[austere gazelle]

Ohhhhhhh

Thank you

Thanks for putting up with my tired ah brain

See you around

.close

[obtuse pebble]

Channel closed

Closed by @austere gazelle

Use .reopen if this was a mistake.

[vestal steppe]

Nah you're chill

Good work!

Take some rest

[austere gazelle]

You too

Good work on your end

W teacher

[obtuse pebble]

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[half forge]

helllppp;

[versed stratus]

with what

the proof?

okay

[frail vigil]

I haven't worked with inverse trig much, but would reccomend taking tan() of both sides

[versed stratus]

that's it basically

than and iff at eveyr step and mention you're working on the princiapl branch

[frail vigil]

Also you can set x and y appropriately as given above, and then find values of A and B in terms of x and y

[half forge]

ahh

thanks

[frail vigil]

Np

!done

[obtuse pebble]

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@half forge Has your question been resolved?

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[untold apex]

can someone do this for me

[untold apex]

guys?

[versed stratus]

!15m

[obtuse pebble]

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.

[untold apex]

yeah actually

I tried factor the denominator

[versed stratus]

lemme think

[untold apex]

I also did till the partial fractions part

[versed stratus]

lemme see

no won't work here I think

[untold apex]

<@&286206848099549185>

[karmic bronze]

!15m

[obtuse pebble]

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.

[frail vigil]

!15m

[obtuse pebble]

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.

[frail vigil]

@versed stratus already said

[untold apex]

but my question was not resolved

[frail vigil]

Have you passed 1st grade?

[versed stratus]

thanks for the ping :)

/j

[untold apex]

bruh what?

[versed stratus]

anyways , lemme think

[untold apex]

Im new to the server...

[frail vigil]

The message given by the bot says ping helpers if you question has not been answered for 15 minutes

[versed stratus]

honestly this is probably calling for a. trig sub or something

[untold apex]

I did till the partial fraction part

but dont know how to proceed after that

[versed stratus]

I just realised

[karmic bronze]

try and find a pattern

[versed stratus]

this is just a riemann integral

😭

[karmic bronze]

let n = 1, n = 2, n=3… etc

[karmic bronze]

wait what

[frail vigil]

Oh wait it is

[versed stratus]

had to be, looks too complicated to be solved otherwise

[untold apex]

do u have any idea?fter this

[karmic bronze]

wait what 😭

is it not a telescoping series

[versed stratus]

oh taht works too

yea

[untold apex]

It is a telescoping series

Wait i just got it

U can use the function f(n)=1/nsquaredplus 1

ok guys ty for da help i got it

soo how do i close this?

[versed stratus]

.close

[untold apex]

.close

[obtuse pebble]

Channel closed

Closed by @untold apex

Use .reopen if this was a mistake.

[obtuse pebble]

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[twilit yacht]

help

[twilit yacht]

so look

help.

please comefast

[mystic folio]

What are we finding?

Values of x?

[lone echo]

!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.

[twilit yacht]

it says "solve"

[short crescent]

looks like "Los op" translates to "solve for"

[twilit yacht]

yeah

[frail vigil]

What are you stuck on

[twilit yacht]

i cant know where to start

[frail vigil]

Try find the equations for f, g and h

[twilit yacht]

yea ok, ik, but how

[flint abyss]

is this for an exam

[twilit yacht]

no

homework

[frail vigil]

U can take points on the graph and form an equation

[lone echo]

you're given the graphs

[twilit yacht]

im late

[short crescent]

find the slope and y-int and put it in form y=mx+b, or find slope and pick a point to put it in point-slope form

[lone echo]

if it says both things are equal, it means that you're in both graphs at the same time

[frail vigil]

Think that's against the rules...

[mystic folio]

I think we don't really need to in this case

[twilit yacht]

no

basically i need help

[lone echo]

if it says a>b, it means a is above b

[short crescent]

no, helping with homework is completely fine

[twilit yacht]

yeah ok, what should i write

[flint abyss]

idth homework is

[frail vigil]

Good point

[lone echo]

!noans

[obtuse pebble]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[frail vigil]

Mb

[short crescent]

nw, actually that is mainly what the channels are for

[lone echo]

you should not write anything until you understand what you're doing, for a start

[twilit yacht]

ok

[lone echo]

start with (a). Do they ask for an equality, or an inequality?

[twilit yacht]

wait lemme use translator

[hardy widget]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[twilit yacht]

inequality

cuz its =

wait

no equality

mb

[lone echo]

.

so look at the graphs of f and h. Is there any point where you can be on both at the same time?

[twilit yacht]

yeah

[lone echo]

where

[twilit yacht]

(4.7)

[lone echo]

so that's the point where both are equal

because they are both in the same place

[twilit yacht]

oh i should write (4.7)

[lone echo]

dont you guys use a comma?

[twilit yacht]

no

[lone echo]

should be (4, 7) or (4; 7) usually

[twilit yacht]

ok

[lone echo]

are you sure it's not a comma?

[twilit yacht]

yea it is

its with comma

mb

and what about b, it says h > g

it can be infinity

uh

[frail vigil]

Yo wsg

What are you stuck in?

[obtuse pebble]

@twilit yacht Has your question been resolved?

Channel closed

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[frail vigil]

Hello. Whats the integral of ${(x^4+1)}^\frac{1}{4}$

[warm shale]

-TimeLord-

[hardy widget]

$$\int \frac{1}{\left(x^4+1\right)^{1/4}} \dd{x}=\int \frac{1}{x \left(1+\frac{1}{x^4} \right)^{1/4}} \dd{x}$$
should do something.

[warm shale]

Civil Service Pigeon

[hardy widget]

Further hint: ||1/x = x^4/x^5||

[frail vigil]

Yo wtf

How do you come up with solutions like this?

[hardy widget]

dividing by the highest power is a common trick

[frail vigil]

Really? I had no idea uptil now

[frail vigil]

(This isn't sarcastic)

[covert fjord]

is it u sub after or what?

[frail vigil]

I took u as 1/x^4

[hardy widget]

yes it's just a substitution

[covert fjord]

ah yeah makes sense

[hardy widget]

yeah, or 1 + 1/x^4

[frail vigil]

Yes that works too

1 min ill solve the integral

[hardy widget]

[frail vigil]

Wait a minute... if I didnt do anything wrong, im getting -w^2/(w^4-1). I did this before but I completely forgot how

Wait 1 min

Im dumb

[hardy widget]

mhm that's good so far

[covert fjord]

is it ||trig sub? w^2 = sec(theta)???||

[hardy widget]

Try it.

[covert fjord]

ukw sure

[frail vigil]

It is i tried

[covert fjord]

oh alr

[hardy widget]

oh I was gonna say just pfd

,w partial fractions x/(1-x^2)

[warm shale]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[covert fjord]

ohhh yeah

[hardy widget]

$$-\frac{1}{2} \int \frac{1}{x^2+1} \dd{x}-\frac{1}{2} \int \frac{1}{x^2-1} \dd{x}$$

[warm shale]

Civil Service Pigeon

[hardy widget]

and atp it's trivial

[frail vigil]

Oh no no I mean trig sub after partial fractions

[hardy widget]

oh

[covert fjord]

yeah i think u could still do it before idk fs tho

[frail vigil]

Wait a minute

[hardy widget]

these are integrals that are not unreasonable to have learnt off tbf

[frail vigil]

How did you get this?

[frail vigil]

Liek x/(x^2-1)

[hardy widget]

I replaced x^2 with x for convenience

[frail vigil]

Fym?

[hardy widget]

just in case wolfram would give me some crazy stuff with the x^4-1 factorisation

though you can do it in your head in 1-2 lines tbf

I'm just lazy

[frail vigil]

Ye prob

I did partial fractions and then did trig sub

And I got the answer

[covert fjord]

nice nice

[green sun]

I meant I got some thing but it waybis to a huge answer in terms of ln

[hardy widget]

$$\frac{x^2}{(1-x^2)(1+x^2)}=\frac{1}{2} \cdot \frac{(1+x^2)-(1-x^2)}{(1+x^2)(1-x^2)}$$

[warm shale]

Civil Service Pigeon

[frail vigil]

Thanks @hardy widget and @covert fjord

.close

[hardy widget]

[obtuse pebble]

Channel closed

Closed by @frail vigil

Use .reopen if this was a mistake.

[frail vigil]

Where u getting ln from?

[green sun]

I substitute u=1+x^(-4)

[covert fjord]

u can pfd 1/(x^2-1) and that will yield ln

[frail vigil]

Technically you get x^2/(x^2-1)

[covert fjord]

u could split into 1 + 1/(x^2 - 1)

[frail vigil]

Kk

[obtuse pebble]

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[winged prairie]

Im currently stuck on A(II). I understand the steps toward u2/u1 giving me 16/20 and whatnot leading to 4/5sin^2pheta, but now i get confused when finding the possible ranges.
Am i supposed to learn that pheta not equaling pi automates it to zero? Why am i supposed to multiply it by 4/5

[haughty stump]

you need to find the possible values of $r$, so you just need to look at the boundaries of the function $\sin^2 \theta$ within your given domain

[warm shale]

thecrumbeler2

[haughty stump]

basically in the range $0 < \theta < 2\pi$, the value of $\sin \theta$ is usually $0$ at $\theta = \pi$

[warm shale]

thecrumbeler2

[haughty stump]

so cause the problem says $\theta$ cannot be $\pi$ (and it’s also restricted from being $0$ or $2\pi$ by the inequalities), $\sin \theta$ can never actually reach $0$

[warm shale]

thecrumbeler2

[haughty stump]

meaning that $\sin^2 \theta$ is always greater than $0$

[warm shale]

thecrumbeler2

[haughty stump]

the highest value $\sin \theta$ can reach is $1$ (at $\frac{\pi}{2}$) or $-1$ (at $\frac{3\pi}{2}$)

[warm shale]

thecrumbeler2

[haughty stump]

when you square these, you get 1

meaning $\sin^2 \theta$ is less than or equal to $1$

[warm shale]

thecrumbeler2

[haughty stump]

u already found that $r = \frac{4}{5} \sin^2 \theta$

[warm shale]

thecrumbeler2

[haughty stump]

so to find the range for $r$, you gotta apply that same $\frac{4}{5}$ multiplier to the boundaries of $\sin^2 \theta$

[warm shale]

thecrumbeler2

[haughty stump]

basically:

Lowest value: $\frac{4}{5} \times (\text{slightly more than } 0) = \text{slightly more than } 0$
\Highest value: $\frac{4}{5} \times (1) = \frac{4}{5}$

[warm shale]

thecrumbeler2

[haughty stump]

and you just end up with $0 < r \leq \frac{4}{5}$

[warm shale]

thecrumbeler2

[obtuse pebble]

@winged prairie Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[wispy wadi]

assume your polynomial is indeed larger than 0

and evaluate it at x = 1 and x = -1

what conditions do you get on a?

[sly elm]

oh ty

[wispy wadi]

my hint is not enough to conclude you'll still have to think a bit, but you're welcome

[sly elm]

a>= -14 and a<=38

right?

[wispy wadi]

yes, so a is in [-14, 38]

can the answer be (C) or (D) ?

[sly elm]

no

[wispy wadi]

exactly, so it's (A) or (B)

[sly elm]

hmm

[wispy wadi]

the last thing you need to find is if a = 38 works or not

if it works it's B, otherwise it's A

[sly elm]

i see

ty

is there not other way ? if they asked numerical

[lucid glacier]

Well if you group some of the terms, you can write the polynomial as:
$x^3\left(x^3+\frac{1}{x^3}\right)-6x^3\left(x^2+\frac{1}{x^2}\right)+12x^3\left(x+\frac{1}{x}\right)+ax^3 \geq 0$

[warm shale]

Sarin

[sly elm]

oh

[lucid glacier]

If you let $t = x+1/x$, $x^2+1/x^2 = t^2-2$, $x^3+1/x^3=t^3-3t$. \\
$\implies x^3(t^3-3t-6t^2+12t+a) \geq 0$

[warm shale]

Sarin

[lucid glacier]

Not sure about what you can do after this though without using calculus

[sly elm]

i see ty

lets differentiate

[wispy wadi]

I think you can factor the polynomial in t

[lucid glacier]

Oh yeah

[sly elm]

ohh

[wispy wadi]

like it's t^3-6t²+9t+a+12

normally

[lucid glacier]

t^3-6t^2+9t = t(t-3)^2

[wide zinc]

Very nice problem ngl

[wispy wadi]

and what you would want to prove is that a+12 is between 0 and 50

[wide zinc]

Btw wsg

[wispy wadi]

I think you can split the cases x > 0 and x < 0 to get conditions on a+12

[sly elm]

nothing xd hru

[wide zinc]

I got cooked on the exam last week , now I'm waiting for the score to be released so I can know how cooked I am

[obtuse pebble]

@sly elm Has your question been resolved?

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[uncut thunder]

yo

[uncut thunder]

how do i do part c

it kinda looks like
R2 = area under graph from (4 to 5.442)

but this is wrong

[brave bramble]

R2 starts at x = 0

Part (b) talks about what the value b is. Part c basically just wants you to make a diagram for it

I don't know what I would put other than what figure 2 has, lol

[obtuse pebble]

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[willow robin]

Need help with 1.b) Log(z1), struggling with calculating the angle

[willow robin]

Ive got r already which is just r = sqrt(a^2+b^2) which is 2

but idk how to continue from then on , since i dont rlly understand the whole angle thing

[restive gorge]

You can draw -1+i geometrically to get a better understanding of the angle

[willow robin]

how do i do that?

ah wait

[restive gorge]

Draw a coordinate system

[wild glade]

argand diagram

[willow robin]

Like this?

[restive gorge]

yes

Now getting the angle is really high school trigonometry

[willow robin]

135 degree right?

[restive gorge]

yes

[willow robin]

thats 3 pi / 4 right?

is that the angle then

[restive gorge]

yes

[willow robin]

thats it?

:p

[restive gorge]

ja

[willow robin]

oh okay well that wasnt that hard whoops

tysm

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

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[cyan pagoda]

Help

[surreal plover]

!da2a

[obtuse pebble]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[safe mauve]

With?

[cyan pagoda]

Oh

[cyan pagoda]

My Internet is bad

Sorry

It won't load

[safe mauve]

:(

[cyan pagoda]

Well ama type it out,

I cant find this answer,
2lgx²y = 3+ lg x - lg y

[safe mauve]

lg means log I assume?

[cyan pagoda]

Find y in terms of x

[cyan pagoda]

Yes

Base 10

[safe mauve]

okay

what have you tried

[cyan pagoda]

Power rule

[safe mauve]

also use parenthesis

2lg(x^2y) or 2lg(x^2) * y

[cyan pagoda]

2 lg(x²y)

= 3 + lg(x) - lg(y)

[safe mauve]

alright

what have you tried

[cyan pagoda]

lg(x⁴) + lg(y²)

[safe mauve]

Yes, that's correct, but we could also leave the 2 outside of the log

[cyan pagoda]

Huh

Why

[safe mauve]

well we want to find y in terms of x right

[cyan pagoda]

lg(y²) + lg(y) = 3 + lg(x)- lg(x⁴)

[cyan pagoda]

Yes

[safe mauve]

So we have 2 (log(y * x^2)) = 3+ log(x) - log(y)

right

[cyan pagoda]

Finally

[safe mauve]

Wow that internet is crazy

[cyan pagoda]

3

[safe mauve]

Yes sorry

[cyan pagoda]

Okie

[safe mauve]

we use the property of the log

[cyan pagoda]

Full bars with 5G+😭✌️

[cyan pagoda]

Yes

[safe mauve]

log(a * b) = log(a) + log(b)

so we have 2 ( log(y) + log(x^2)) = 3 + log(x) - log(y)

[cyan pagoda]

lg(y³) = lg(x^-3) + 3

[safe mauve]

true?

[cyan pagoda]

Yes

[safe mauve]

Next we distribute the 2's

2(a+b) = 2a + 2b

so 2log(y) + 2log(x^2) = 3 + log(x) - log(y)

[cyan pagoda]

Yes

I agree

[safe mauve]

now we can just take all the log(y)'s to one side and the log(whatever x) to the other

do you see how to proceed?

[cyan pagoda]

yes

2log(y) + log(y) = 3 - 2log(x²) + log(x)

What now

[safe mauve]

well 2a + a = 3a true

[cyan pagoda]

2 log y = log y² ?

Oh

[safe mauve]

so just put a = log(y), we get 2log(y) + log(y) = 3log(y)

[cyan pagoda]

I see

[cyan pagoda]

Yes

[safe mauve]

and now?

[cyan pagoda]

3-3log(x)

For the other side

3lg(y)= 3-3lg(x)
lg(y) = 1 - lg(x) ?

then 10^ both side

y = 10/x ?

[digital isle]

Hey

[cyan pagoda]

Hi

[safe mauve]

Well log(y) = 1- log(x)

[digital isle]

Log?

[safe mauve]

indeed take 10^ both sides

y = 10^(1 - log(x))

[cyan pagoda]

Okie

[digital isle]

Cat cat ok

[safe mauve]

which is 10 * 1/x

which is 10/x great job

[cyan pagoda]

Yes

Ohh

[safe mauve]

wow that was quicker than I could do it

[cyan pagoda]

So like

I had to js simply it

Okay

Omfg

Thank yoy

Ypy

[safe mauve]

No problem!

[cyan pagoda]

I basically wasted my time by removing the logs

Okay ty!!

.stop

.close

[obtuse pebble]

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[obtuse pebble]

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[tawdry stag]

hey! i have a math final coming up for algebra two and i could really use some help with the study guide, please feel free to dm me <@&286206848099549185>

[frail vigil]

!15m

[obtuse pebble]

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.

[karmic bronze]

!da2a

[obtuse pebble]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[frail vigil]

A study guide will just be ur syllabus

[gilded sail]

I believe your textbook should have a lot of exercise

[obtuse pebble]

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[tawdry stag]

i dont have a textbook..

[obtuse pebble]

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[frail vigil]

What di you study from?

[tawdry stag]

uhhh my notes

[frail vigil]

Oh

Well you can use practice problems your teacher gave you

Also how is this my help channel???

.close

[obtuse pebble]

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[tawdry stag]

😭😭, dont know

[frail vigil]

<@&268886789983436800>

[calm rose]

beat me to it

[obtuse pebble]

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[safe mauve]

what have you tried

[grand relic]

I wish i knew where to start

[inland sapphire]

Inside the integral term is it cosx?

[grand relic]

cos t

[inland sapphire]

And also can you show a screenshot of the question?

[grand relic]

Huh

[frosty vault]

Does that infinite series look familiar?

[grand relic]

It was given in class

[safe mauve]

First find a

then use the derivative of an inverse function trick

[grand relic]

How to go about finding a

f(0) is plain zero

Cant find a

[safe mauve]

what about f''(0)

[grand relic]

how do i differentiate that 2 times

Im gonna dedge

That series gonna fry me

[safe mauve]

Yeah it's a bit of work

(it's 24 / pi^2)

But I don't think it's needed now that I look at it

You just need the inverse definition and chain rule

[inland sapphire]

Here's a full development process

A little mistake I took the T(x) as the initial e^x taylor series sum expansion but later changed it to s(x).

[obtuse pebble]

@grand relic Has your question been resolved?

[obtuse pebble]

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[tropic terrace]

if you look at an open ball in C, its just a circle without the boundary. can we describe open ball something similarly on the set of single variable real functions defined on [a, b], notation probably C[a,b]? is it just "all functions bounded by the x function in some way (like between x-r and x+r)"?

[tropic terrace]

or is there a better description like the circle?

[obtuse pebble]

@tropic terrace Has your question been resolved?

[gilded needle]

depends on the norm you're using, but assuming the usual supremum norm, it's easy to visualize: draw the graph of the function, then the ball of radius epsilon around that function is the set of all continuous functions whose graphs lie within +/- epsilon of that graph, so anything that lies between the two red curves in this picture

(ignore the annotations in the picture, i just copy/pasted it from some random MSE post)

[tropic terrace]

yeah i see

saying bounded by x +- r works too?

[tropic terrace]

anyway thanks

.close

[obtuse pebble]

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[forest sinew]

simplification?

[obtuse monolith]

yes I'm too confused to understand this

[forest sinew]

how about combining all of those constants first?

looks like you have a $12$, a $- \frac{2}{5}$ and a $3$

[warm shale]

jan Niku

[forest sinew]

what do you get when you multiply all those together

[obtuse monolith]

36?

[forest sinew]

that doesnt sound right

i think 12 * 3 is gonna be 36

but then you also need to multiply in the -2/5

[obtuse monolith]

🤔

[forest sinew]

I'm suggesting to move the constants just because theyre one kind of thing is all 😄

theres no addition or anything in this problem so those parentheses are more suggestions than anything

its just multiplication, which is commutative, so you can move any factor around wherever you want

it may make the problem a bit more approachable

you could start with the x's or y's instead, if you don't like the constants

[obtuse monolith]

ok

.close

[obtuse pebble]

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[forest sinew]

[obtuse monolith]

.reopen

[obtuse pebble]

✅

[obtuse monolith]

sorry

now i understood something i just needed to think a little

[forest sinew]

oh

whatd you understand

[obtuse pebble]

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[kindred silo]

find 3 functions where f(f(x)) = x
ive already got f(x) = x, now i need 2 more

[candid halo]

IIRC, there's only two such functions if you only consider continuous ones

[kindred silo]

theres another trivial solution?

[clear inlet]

any function symmetric about y=x would work

[candid halo]

so you'll have to relax your expectations on what the function looks like

[kindred silo]

hmm nyann

so for example

f(x) = x^2

ohhh i see

[candid halo]

wait nevermind, I might be thinking of something else

[kindred silo]

wait

no i dont see

so i have f(x) = x^2
f(f(x)) = f(x^2) = x^4

[candid halo]

f(x)=x^2 is not symmetric about y=x

[calm charm]

x^2 is not symmetric about y=x

[kindred silo]

ohhh

[calm charm]

damn

[kindred silo]

y=x

not

y

axis

kek

hmm

[calm charm]

uh

are there any functions symmetric about the y axis

[candid halo]

yes that's called being even

[calm charm]

I mean y=x

[candid halo]

yes

[calm charm]

sorry I'm tired lol

[kindred silo]

so y = -x

would work i think?

[candid halo]

that's one

[kindred silo]

woohoo

and then i can do y = -x + 1

yay, thanks

.close

[obtuse pebble]

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[weak furnace]

A house worth 250 million, every year the price increases constantly for 25 million (only valid for 5 years). Determine the equation of the house depreciation price line and the gradient of the line!

[weak furnace]

can someone help me im so confused

help

pls

pls help

<@&286206848099549185>

[whole scroll]

@weak furnace Try going through this tutorial--hopefully it can help answer your question: https://www.youtube.com/watch?v=NtTOWodMINU

[weak furnace]

ok lemme watch

im still confused

[whole scroll]

So, what is our y-intercept (the value at which the house starts)?

I have to go, but hopefully another individual will be able to help build your intuition.

[obtuse pebble]

@weak furnace Has your question been resolved?

[weak furnace]

idk

its not in the question

[obtuse pebble]

@weak furnace Has your question been resolved?

[obtuse pebble]

@weak furnace Has your question been resolved?

[weak furnace]

please help

[visual kelp]

It is in the question - or rather, you can intepret one way that has it

[weak furnace]

im still so confused

[weak furnace]

where?

is it the 250

[visual kelp]

Yes

Consider year 0

[weak furnace]

so

its

250y = 125x + c

?

[warm sleet]

its 25x-y+250=0

and gradient is m = 25

lemme send you pic of solution

[obtuse pebble]

@weak furnace Has your question been resolved?

[obtuse pebble]

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[plain topaz]

how to do without expanding it

[royal basin]

do you know the binomial theorem

[plain topaz]

uh

expanding that like a binomial?

wouldnt that take ages to do

[royal basin]

no you don't need to expand

you are interested in only one term from the expansion

[plain topaz]

yes

[royal basin]

$(2a-1)^5 = \sum_{k=0}^5 \binom{5}{k} (2a)^k (-1)^{5-k}$

[warm shale]

Ann

[plain topaz]

oh wth

i have never

seen that sign before

[royal basin]

bruh

[plain topaz]

hahah

i just got this ques form somewhere harder

so i prob might not have learnt how to do

[royal basin]

$\binom{5}{k} (2a)^k (-1)^{5-k}$

[warm shale]

Ann

[royal basin]

these are the terms in your expansion

[plain topaz]

yea

[royal basin]

adding this up for all values of k from 0 to 5 inclusive will get you the full expansion of (2a-1)^5

but all you want is the a^2 term

[plain topaz]

mhm

[royal basin]

which is the term you get when you take k = 2

[plain topaz]

righh

and so..

(2a)^2(-1)^3 ?

[royal basin]

you forgot the 5C2

[plain topaz]

huh

what is c

[royal basin]

are you telling me you have never heard of n choose k before?

[plain topaz]

nup never

i got this ques from somewhere harder

and it may turn out to be much harder

[royal basin]

huh

...

how did you use to do binomial expansions if you don't know this

[plain topaz]

yea idk

its ok

i ll even know how to when i get older

lol

thanks

i assume i close this

.close

[obtuse pebble]

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[toxic elm]

is my calculation for this question correct?

[obtuse pebble]

@toxic elm Has your question been resolved?

[toxic elm]

.close

[obtuse pebble]

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[languid otter]

Hello, can someone help me this question?

[languid otter]

i needa help for question a

the correct answer should be -8

[royal basin]

am i understanding correctly that the 4 (x-int) and 2 (gradient) were given in the problem while the 2 in blue is what you filled in yourself?

[languid otter]

yes and thats the wrong ans

[royal basin]

why don't you tell me how you found it?

maybe i can pinpoint your error

[languid otter]

nah i just misunderstood it

the lastest equation is on the right

i got positive 8 and idk how to make it back to -

[royal basin]

i see you started with $\frac{0-y}{0-4} = 2$

[warm shale]

Ann

[languid otter]

yep

[royal basin]

this sounds like you wrote down the gradient of the line connecting (0,0) to (4,y)

not (4,0) to (0,y) as you were supposed to

[languid otter]

hmm ye

oh i guess i know what happen

lemme try again and i will come back right quick

wait

its still the same

@royal basin how?

[royal basin]

how what

maybe you made another mistake

you aren't showing your new work after all

[languid otter]

i tried again and its the same mistake again

e

[royal basin]

show your work

otherwise nobody can help you

if you don't show your work then you cannot get help here

[languid otter]

its like 0,0 and 4,y

[royal basin]

but no

it isn't

[languid otter]

then 0-y/0-4

[royal basin]

this sounds like you wrote down the gradient of the line connecting (0,0) to (4,y)
not (4,0) to (0,y) as you were supposed to

no!!!!!!

[languid otter]

why

[royal basin]

your points aren't (0,0) and (4,y)!!!!

[languid otter]

y-y/x,x isnt

[royal basin]

you have the wrong points, yan!!

you have the wrong points!!!

[languid otter]

uh shoot

[royal basin]

how can you plug in the wrong points and expect to get anything right from the result!!!!

[languid otter]

hmm idk

thats a big problem

so its

0,4 and uh 0,y ?

[royal basin]

no

[languid otter]

whhhaaaa?

[royal basin]

the x intercept is a point on the x axis

in your case it's (4,0)

(4,0) is on the x axis and its x coordinate is 4 and it is the x intercept of your line

do you understand this, yes or no

[languid otter]

not really

[royal basin]

do you know what an x-intercept is?

[languid otter]

ye

[royal basin]

what is an x-intercept

[languid otter]

the dot on the x line

eh is it

[royal basin]

the point where your line crosses the x axis

[languid otter]

This is x intercept right????

The up one

[royal basin]

what

[languid otter]

ok nevermind imma figure this out later

back to the point

so its 4,0 and y,0

is it?

[royal basin]

no!!!!

no!!!!!

[languid otter]

what

[royal basin]

it's (4,0) and (0,y)!!!!

(0,y) is the y intercept

it's a point on the y axis

[languid otter]

arent they a straight line?

[royal basin]

who are "they"

[languid otter]

wait ntg

i mess this up from the begining

oki done

thankyou

.close

[obtuse pebble]

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[vast plaza]

.close

[obtuse pebble]

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[karmic plover]

How'd i find the stationary point for this funciton?

i'll take the derivative and set it to zero

but i don't remember how to get the roots of a cubic function

[plain owl]

That cubic would not be that hard to factor.

[karmic plover]

my first idea was to factor x^2 out, i.e. x^2(4x + 12) = 0

but it feels like a useless maneuver

[plain owl]

That -9 will disappear

[high lily]

you didn't differentiate properly

[plain owl]

Now the roots are clear

[karmic plover]

yeah

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[obtuse pebble]

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[forest meadow]

integral of 1/x from -1 to 1

[forest meadow]

I tried to do it using the presented limits, and ended up with the result -iπ. Is there a way to get a real finite value?

Note that you can split the integral into two similar integral -1 to 0 and 0 to 1

[lethal sand]

ehhh how did you get to that?

the integral doesnt converge

[hushed moat]

~w integrate 1/x dx from -1 to 1

[lethal sand]

did you forgot the absolute value sign for the natural log?

[forest meadow]

Oof

I just put ln(-1) =iπ 😅

Forgot about the absolute value

[lethal sand]

anyway, it doesnt converge

[hushed moat]

right, but you have to account for the singularity at x=0

[lethal sand]

as lim_{x -> 0} ln(x) goes to -infty

[forest meadow]

Umm, yeah ln(0) approaches -infinity

[lethal sand]

yes, so where was your -infty?

dont forget about it

[forest meadow]

So it's basically infinity - infinity

[lethal sand]

which is definitely not 0

be careful

[forest meadow]

Unfortunately 😅

[hushed moat]

principal value is 0 right

[lethal sand]

since you are dealing with an indeterminate form

[forest meadow]

Wish I could just cancel them out

[lethal sand]

you have to work it around a bit

[forest meadow]

I am listening

[lethal sand]

basically you are having $\ln|-1| - \lim_{x\to 0}(\ln|x|) + \lim_{x\to 0}(\ln|x|) - \ln|1|$

[warm shale]

waler

[lethal sand]

now you have to put those two limit together

the ln|-1| and ln|1| is just 0, so dont need to worry about it

and then just evaluate

another way around this is that you can notice that the integral of 1/x from 0 to 1 doesnt converge

[forest meadow]

And then cancel ln|x| - ln|x|?

[lethal sand]

so it will be the same for the other integral

[lethal sand]

oh wait, actually, excuse me, a bit of wrong working out there

its actually limit as x tends to 0- and the other is tending to 0+

so you cant add them up like that

sorry, bit of miscalculation there

[forest meadow]

So

[lethal sand]

so basically, since one of the integral does not converge already

the answer will not converge

no matter what the other integral is

[forest meadow]

It's still infinity - infinity

[lethal sand]

yes, but one of the integral diverges

hence the original integral diverges

basically, you have to understand the two integral as a whole

if one of them diverges, the answer is divergence

[forest meadow]

Yeah ok 😊

.close

[obtuse pebble]

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[karmic plover]

how'd i know what type of function f(x) will ~approach as x->+-infinity?

[karmic plover]

not the value of f(x), but its tendency to follow a certain linear eq.

[fervent cradle]

so for things like this you can just divide

(x^2 - 9)/(x - 5) = x + 5 + 16/(x-5)

as x goes to +-inf, obviously that last bit will go to 0

so it'll basically be x+5

[karmic plover]

actually i heard there's a shortcut to do the division, other than the standard polynomial division

[fervent cradle]

ok cool but it should give you the same answer as i got

[karmic plover]

yeah ofc

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[obtuse pebble]

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[karmic plover]

wait is this function odd?

[karmic plover]

i think it is, just want to make sure

[weak furnace]

thank you so much😆

[obtuse pebble]

@karmic plover Has your question been resolved?

[tacit root]

Try substituting (-x) into the equation and see if you get (-f(x))

[warm shale]

ricey

[tacit root]

if you do, then the function is odd