help-17

[torn oracle]

My fault

[vast shale]

yeah

starting to make sense, ty

.close

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[royal kestrel]

is this correct

[royal kestrel]

and if it is how can i simplify that further

the f'(x) in red says 2e^(2x) js might have erased it

[elder scaffold]

pro what kinda derivative is this shit?

[royal kestrel]

💀

idk man

[elder scaffold]

u could not pay me to solve this shit

[hoary blaze]

aw hell nah

[royal kestrel]

LMAO

[fading plume]

my prof gave us one of these on the final

usually you don't need to simplify

[royal kestrel]

ah ok

is it right though?

[fading plume]

i ain't reading allat

check photomath

[elder scaffold]

yeah i'd use some form of online calculator

[royal kestrel]

okay ill do that now

[stone gazelle]

This may have been made easier by simplifying with logarithms first

Let me see

[royal kestrel]

my denominators were different to wolfram

if its (g(x))^2 its just ((x^2 + 1)^3 * (1 + sin(x)^5)^2

so why is it 4 and 6

[elder scaffold]

holy fuck i have brain damage

i thought u were integrating the entire time

[royal kestrel]

noooo integrating it wouldve been so much worse omd

[elder scaffold]

yeah was about to say, no fucking way am I integrating that

also i think u typed it in wrong?

u typed in e^2 * x rather than e^2x

[royal kestrel]

i wrote that but is still coming up with e^2 * x

ill js write (e^2)^x

[elder scaffold]

brackets in the exponent?

e^(2x) should work fine

[royal kestrel]

ah okay yes

denominator is still 4 and 6

[lapis river]

Is that calculus?

[royal kestrel]

yes

[elder scaffold]

i mean i cant be asked to do this myself but i assume there must be some cancellation of sort somewhere

[royal kestrel]

oh you mean theyd cancel out like some (x^2 + 1) and (1 + sin(x)) term in the numerator and denominator

[elder scaffold]

yeah something like that maybe

cant say for sure tho

[royal kestrel]

is there a website where i can check if my answers correct just in a different form

[elder scaffold]

i think wolfram alpha can simplify expressions?

[stone gazelle]

,w differentiate (e^(2x))/( ((x^2)+1)^3 (1+sinx)^5)

[royal kestrel]

(x^2 + 1)^3

in the denominator

[twin meteor]

Results provided by WolframAlpha

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

Requested by shsgd

[elder scaffold]

i have a headache cuz im sick af so i cbf, but gl

[royal kestrel]

,w simplify ((2e^(2x))(x^2 + 1)^3 * (1 + sin(x))^5 - (e^(2x))(6x(x^2 + 1)(1 + sin(x))^5 + (x^2 + 1)^3 * 5cos(x)(1 + sin(x)))/((x^2+1)^6 * (1 + sin(x))^10)

[twin meteor]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by yummy_taro

[stone gazelle]

@taro it can be simplified a lot pre differentiation

Let me write up a solution

[royal kestrel]

okay ping when

[stone gazelle]

Will probably take me bout 10 mins so bear with me

[vocal sleet]

@royal kestrel Has your question been resolved?

[stone gazelle]

Begin with the equation we want to differentiate
[ y=\frac{e^{2x}}{(x^2 + 1)^3 (1+\sin x)^5} ]
Take the natural log of both sides and simplify using logarithm rules
[ \ln y = \ln (\frac{e^{2x}}{(x^2 + 1)^3 (1+\sin x)^5}) ]
[ = \ln e^{2x} - \ln ((x^2+1)^3(1+\sin x)^5) ]
[ = 2x - 3\ln (x^2 + 1) - 5\ln (1+\sin x) ]
Differentiate both sides with respect to $x$, note that $y$ is a function of $x$ and must therefore be differentiated implicitly (chain rule)
[ \frac d{dx} \ln y = \frac d{dx}( 2x - 3\ln (x^2 + 1) - 5\ln (1+\sin x) ) ]
[ \frac 1y \frac{dy}{dx} = 2 - \frac{6x}{x^2+1} - \frac{5\cos x}{1+\sin x} ]
So we have
[ \frac{dy}{dx} = y\cdot \frac{2(x^2+1)(1+\sin x) - 6x(1+\sin x) - 5\cos x(x^2+1)}{(x^2+1)(1+\sin x)} ]
Now simplify the numerator
[ \frac{dy}{dx} = y\cdot \frac { (1+\sin x)(2(x^2+1)-6x)- 5\cos x(x^2+1)}{(x^2+1)(1+\sin x)} ]
[ = y\cdot \frac { 2(1+\sin x)(x^2-3x+1)- 5\cos x(x^2+1)}{(x^2+1)(1+\sin x)} ]
Sub in the original $y$
[ \frac{dy}{dx} = \frac{e^{2x}}{(x^2 + 1)^3 (1+\sin x)^5}\cdot \frac { 2(1+\sin x)(x^2-3x+1)- 5\cos x(x^2+1)}{(x^2+1)(1+\sin x)} ]
[ = \frac{ e^{2x}(2(1+\sin x)(x^2-3x+1)- 5\cos x(x^2+1)) }{ (x^2+1)^3(1+\sin x)^5 \cdot (x^2+1)(1+\sin x) } ]
Simplify as needed
[ \therefore \frac{dy}{dx} = \frac{ 2e^{2x}(1+\sin x)(x^2-3x+1) - 5e^{2x}\cos x(x^2+1) }{ (x^2+1)^4(1+\sin x)^6 } ]

@royal kestrel

hope this helps

[royal kestrel]

this line here

is it supposed to be ln((x^2 + 1)^3

[stone gazelle]

yeah mb, also missed some other 2s, let me fix it all

[twin meteor]

shsgd

[stone gazelle]

I think that's everything

[royal kestrel]

howd you go from this to this

oh wait by factoring out 1 + sinx

okay

[stone gazelle]

yeah

[royal kestrel]

wow okay much easier way to solve it than what i did

thank you sm

[stone gazelle]

logs are very powerful

maybe look up some videos on logarithmic differentiation if you want a more in depth explanation

[royal kestrel]

okay ill do that

thank you again for this

u have a good day

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[daring marsh]

would the answer be x or

x^2+1 / x-1

[plucky trout]

For which one

[daring marsh]

for (f/g)(x)

[plucky trout]

Yes

[daring marsh]

the x^2 over x-1

[plucky trout]

Isn't $x^2+1=(x+1)(x-1)$

[twin meteor]

w

[plucky trout]

Idk I can't remember difference of square formula

Wait I'm dumb nvm

[daring marsh]

ohhh I see

[plucky trout]

That's wrong

[daring marsh]

ohhh

[plucky trout]

It's x^2 -1

Not plus

[daring marsh]

oh how

[plucky trout]

$x^2-1=(x-1)(x+1)$

[twin meteor]

w

[plucky trout]

In ur equation

It's plus one

[daring marsh]

yes it should be

[steep finch]

yooo this made me giggle

[daring marsh]

x^2+1/x-1

then I am not sure if that's the answer or if I cross out the x and one

[steep finch]

anyone else doesnt like this notation

[daring marsh]

I am not sure what you mean

[stone gazelle]

divide them directly using long division or synthetic division

[grim lotus]

? how else would u write it

[steep finch]

$\frac{f(x)}{g(x)}$

[twin meteor]

Obotron

[steep finch]

which is how you will see it literally everywhere else

especially in calculus

[grim lotus]

(f/g) is it's own function

[grim lotus]

u define it like this

[steep finch]

i understand that: in that case you would denote: $h(x) = \frac{f(x)}{g(x)}$

[twin meteor]

Obotron

[grim lotus]

nah more letters is unnecessary

[daring marsh]

okay

[steep finch]

yeh its fine. just lazy notation

[grim lotus]

like say u want (f/g)'

[steep finch]

h'(x)

[grim lotus]

derivative of f/g

💀

[steep finch]

it doesnt matter but its not popular notation for a reason

[west nebula]

this is the answer

it cannot be simplified further

[daring marsh]

okay thank god I would hate to try to synthetically divide

[west nebula]

nevermind i trolled

[twin meteor]

misha

[fading plume]

that's arguably less simple

[west nebula]

right?

i am now just confusing myself

[daring marsh]

this is what happened when I entered into mathaway

I think you can technically do synthetic division but as problem it want x^2 over x-1

[stone gazelle]

I have a feeling f is supposed to be x^2 - 1

[west nebula]

the answer in the first picture should be it

[west nebula]

what are you saying?

[stone gazelle]

although I guess since you are looking at composition of functions you can probably just leave it as x^2 + 1 / x-1

[stone gazelle]

ignore that, I meant it was possible f was a typo

[daring marsh]

yeah I think so

[west nebula]

ohh alright

[stone gazelle]

but it doesn't ask to simplify so probably not

[vocal sleet]

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[charred violet]

hello

[charred violet]

i need help understanding trigon

anyone

[finite mist]

Send the question

[finite mist]

understand?

[charred violet]

yes

[finite mist]

Which part?

[charred violet]

im still confused

[finite mist]

trigon in geometry or trigon in calc?

[charred violet]

how to input the formulas

both?

[finite mist]

It would be nice if there's a specific problem that u stuck on

Cause "input the formula" is a bit vague

[charred violet]

im finding one

i cant find one in my paper

can u find one instead

[finite mist]

Dont u have homework or assignment?

[charred violet]

yea but its schoolwork so teacher took it back

[finite mist]

Search the internet or ur book

[vocal sleet]

@charred violet Has your question been resolved?

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[charred violet]

my bad i need to go

[charred violet]

.close

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[grand frigate]

okay

[vocal sleet]

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[tight sedge]

help pls

[tight sonnet]

use diffrence of squares formula

[tight sedge]

so 6?

[tight sonnet]

yeah

[tight sedge]

oh

that was simple

[tight sonnet]

lol

that was fast

[tight sedge]

thanks anyway

.close

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[tight sonnet]

np my guy

[tight sedge]

ill be a regular here soon

.reopen

[vocal sleet]

✅

[tight sedge]

i think im bugging today

[hollow lily]

?

[finite mist]

use rise/run

[tight sedge]

wdym?

[strong grove]

id just check the ratio of each triangle side to the corresponding side of T

[hollow lily]

T has a height of 2 and a length of 3

[strong grove]

if they are equal then its similar

[tight sedge]

so c would be similar

[hollow lily]

How do you conclude that?

[tight sedge]

it has a scale factor of 2

[hollow lily]

Look at the T

It has a Rise of 2 and a run of 3

[tight sedge]

yes

[hollow lily]

To see what shapes are similar

we compare the ratios

of their rise/run

[tight sedge]

D

[hollow lily]

Again, how do you conclude that?

[tight sedge]

wait can something be similar if its rotated?

[hollow lily]

Yes!

It's still the same.

well not the same

but it has the same ratio.

[tight sedge]

so it cant be d or a or c

[hollow lily]

Why not A?

[tight sedge]

its a and f

[hollow lily]

Alright let me make some gounds for you.

The ratio of triangle T

[tight sedge]

ok

[hollow lily]

is 2/3

[tight sedge]

ye

[hollow lily]

2/3 is the same as 20/30

[tight sedge]

yes

[hollow lily]

and 60/90

just different numbers of the same ratio

[tight sedge]

ye

[hollow lily]

now these other triangles might have the same ratio.

You check that by looking at how much they rise, and how much they run.

Try seeing the rise and run on triangle A

[tight sedge]

4:6

[hollow lily]

Yes!

2:3 has the same ratio 4:6

as*

[tight sedge]

e has 6:9

so that is one aswell

[hollow lily]

Nice!

[tight sedge]

i think taht is it

that

and it was

[hollow lily]

Woohoo.

[tight sedge]

woop woop

thanks once again

[hollow lily]

no probs!

[tight sedge]

be back soon

.close

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Channel closed

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[tight sedge]

.reopen

[vocal sleet]

✅

[hollow lily]

?

Oh.

[tight sedge]

ye

i thought one could be A but im 100%

[hollow lily]

I mean, you can factorise the equation or simply multiply into the parantheses.

[tight sedge]

i tried but it doesnt seem to work

[hollow lily]

Or something smarter.

[hollow lily]

Show me what you tried.

[vast shale]

Ok

[fading plume]

keep in mind the question wants two answers

[vast shale]

Just factor

[fading plume]

A is one of them

[hollow lily]

Chill man, I know the guy.

[vast shale]

?

[hollow lily]

Micos...

[amber veldt]

My bad

[vast shale]

I mean

U used a spoiler

I don’t see anything wrong

[fading plume]

if you give it away they can't find the solution organically

which arguably defeats the point of learning

[vast shale]

True

BUT

[hollow lily]

Doorbell.

[tight sedge]

ill be back anyway

with more

[vast shale]

Nobody told op to click on the spoiler

[hollow lily]

We're not here to discuss amongst how to help.

[tight sedge]

fr

[hollow lily]

Just let it go.

[vast shale]

What did I even do

[hollow lily]

Nothing man.

[vast shale]

Ok

Good

[hollow lily]

Alright bibalek.

[tight sedge]

yes

[hollow lily]

Can you show me what you intuitively think is the way to get the answer?

To help me see your thought process.

[tight sedge]

well factorisation

[hollow lily]

Oh, so you know that!

[tight sedge]

ye

[hollow lily]

Great.

You did it?

[tight sedge]

now i have

[hollow lily]

And...

what did you get?

[tight sedge]

it logged me out so i got new questions

[hollow lily]

oh.

[hollow lily]

Ah ok.

[tight sedge]

so we can still work it out

[hollow lily]

Sure.

[tight sedge]

would one be D

and B

no

ok

[hollow lily]

Do you want a trick?

[tight sedge]

go for it

[hollow lily]

Alright.

In factorization.

you can obviously find the roots.

[tight sedge]

ye

[hollow lily]

and there are some special qualities of these roots.

for example...

$a^2+b+c$

[twin meteor]

the catmullah

[tight sedge]

ye

what about it

[hollow lily]

my bad, my discord glitched

ok.

[tight sedge]

u good

[hollow lily]

The c-value in your original equation

[hollow lily]

that's exactly it.

[vast shale]

The other guy is wasting your time @tight sedge

[hollow lily]

Mean.

[tight sedge]

nah

[vast shale]

But true

[tight sedge]

he helped with other questions

[vast shale]

It took me 10 seconds to find that image

[tight sedge]

lots of them

[vast shale]

lol

[hollow lily]

No matter what, you don't disrespect other people like that.

[vast shale]

What was disrespectful

[tight sedge]

"hes wasting ur time"

that

[vast shale]

What I stated was a factual description of events

[vast shale]

That’s not what he quoted lmao

[tight sedge]

nearly the same thing

[vast shale]

No

[hollow lily]

Alright.

Thank you doorbell.

[vast shale]

Reposting for reference

[hollow lily]

Now please leave.

[vast shale]

Yes

Sorry

[tight sedge]

u good

[hollow lily]

Basically, this is telling you, that the two values shown here:

Will, multiplied together, give the c-value of a 2nd degree polynomial.

So, from all the answers there.

From A to H

The possible ones are more easy to find with this rule.

And to narrow it down even more.

[tight sedge]

ohhhh

[hollow lily]

The product of those values put together, also give the b-value

[vast shale]

*sum

[tight sedge]

man

[hollow lily]

My english isn't great doorbell.

[vast shale]

Ok

Mb

[hollow lily]

Please respect my attempts.

And leave.

[vast shale]

Just wanted to clarify since product is used for multiplying in English

And we are adding

[hollow lily]

Thank.

You.

Now leave.

[vast shale]

I helped you and you still mad

Lmao

[hollow lily]

You keep popping up like an annoying ad.

Please just leave.

[tight sedge]

ur getting invloved to much

[hollow lily]

it's really simple.

Forget about all of this, and we will too.

Great.

[hollow lily]

You understand the system now?

From the picture that doorbell sent, you can try and find the answers now.

[tight sedge]

ye i do

all of them equal - 32

so that doesnt work here

wait

[hollow lily]

And now, find the sum of them to find the b-value.

Let me help with the first answer.

-8 + 4 = -4

The b-value of the original function is not -4

so this must not be the answer.

[tight sedge]

its not b

[hollow lily]

Yep!

[tight sedge]

is one of them c?

[hollow lily]

Yep!

The second one is identical to c

[tight sedge]

its c and h!!

[hollow lily]

ehh

c and...

[tight sedge]

damn

[hollow lily]

h is:

-8 + 4

[tight sedge]

f

[hollow lily]

or 4 + (-8)

which is 4 - 8

[hollow lily]

Oh, thought it was a reaction

Yes!

[tight sedge]

yesss

[hollow lily]

C and F

[tight sedge]

thats wouldve been so much quicker without doorbell

[hollow lily]

Lmao.

[tight sedge]

well

ill be back soon with more bye

thanks

[hollow lily]

Bye! Np!

[tight sedge]

.close

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[scarlet sphinx]

help

[scarlet sphinx]

Sea f(x) = x 2L(x) + 1. Entonces f 0 (x) = (a) 2 (b) x(2L(x) + 1) (c) 2xL(x) + Divide[1,x] (d) 2xL(x) + 3x

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@scarlet sphinx Has your question been resolved?

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[daring marsh]

How do I solve this problem? Do I plug in -1 for x?

[finite mist]

No?

Since g(x) = -f(6x) -6

And we want the point (-1, 4)

So we could think

f(-1) = 4

So

g(x) = f(6x)
So x= -1 becomes x = -1/6

And the g(x) = -f(6x) -6 = -4 -6 = -10

So the point is (-1/6, -10)

[daring marsh]

how doess it become -1/6

then how does this become the equatiuon>>> And the g(x) = -f(6x) -6 = -4 -6 = -10

[vocal sleet]

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[dull sphinx]

basically its just saying

make a function for s(a)

and linearize it (with taylor polynomials) at a=0

I have s(a) = h - sqrt(h²-a²)

If I linearize it with an linearized function s1(a) = s(0) + s'(0)*a

I just get s1(a) = 0

but the solution book says s1(a) = a²/(2h)

which approaches it really good

Im surprised

but how did they even get that

they even have the same s(a) function like me

how did they approach that to a²/(2h)

man this really gets me

s(0) is 0, and s'(a) is a/[sqrt(h²-a²)] so s'(0) is also 0

Im so confused

function s (in blue), my approach (in black) at a=0, schoolbook solution (in red)

my approach fails, their approach is very close

what am I doing wrong

ok nvm, just turns out my question book is bad

if I approach it with s(a) = s(0) + s'(0)*a + s''(0) x a²/2 (which is NOT linearized)

I get that solution book answer

bad book

thanks btw

.close

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[silver dagger]

Hello, I have a little problem, I have to calculate the domain and range of a quadratic function but I don't know how to use parentheses or brackets, could someone help me?

[vast shale]

Yes

Do you need help with notation?

Or actually finding the domain and range

[silver dagger]

Just the notation.

[vast shale]

Ok

Are you able to get the end points of the domain and range

[silver dagger]

Yes I can, but do you mean do it right now?

[vast shale]

No that’s not necessary

So basically

Parentheses means exclusive

Brackets means inclusive

Sorry

Lemme edit

[silver dagger]

Ok.

[vast shale]

Ok

U get what I means

[silver dagger]

But what means exclusive and inclusive?

[vast shale]

Ok

Say we have a function 1/(x-5)

What is the domain

Of that function

[silver dagger]

Any value that x can take.

[vast shale]

And what are those values

[silver dagger]

Real numbers?

[vast shale]

Yes but what real numbers

[silver dagger]

For example 10.

[vast shale]

There is one specific real number that x can’t take

[silver dagger]

I don't know, sorry.

[vast shale]

Oh

Are you in algebra one or two

[silver dagger]

I could say that one.

[vast shale]

Ok

So you don’t know what an asymptote is yet

that’s ok

We can try to work around

[silver dagger]

That's right.

[vast shale]

In that function

f(x) = 1/(x-5)

What happens when you try to compute f(5)

[silver dagger]

∞

[vast shale]

No

Who told you 1/0 = infinity

That’s wrong

[silver dagger]

so it's mistake?

[vast shale]

Yeah

Diving by 0 is undefined

[silver dagger]

Ohh ok.

[vast shale]

So

You see now that x can’t = 5

[silver dagger]

I understand.

[vast shale]

Ok

[silver dagger]

So x can be any value except 5?

[vast shale]

Yes

Perfect

You got it

So

Can x be 4.9?

[silver dagger]

Yes.

[vast shale]

Can x be 5.1?

[silver dagger]

Yes.

[vast shale]

Cool

So you see how x can get as close to 5 as possible without being = to 5

[silver dagger]

But how can I express the domain without including the 5?

[vast shale]

Parentheses

Hahahahaha

We’ve come full circle

This is good

[silver dagger]

Ok.

[vast shale]

So

Let’s start from the left side of 5

What is the left end point of the domain

[silver dagger]

-∞

[vast shale]

Yes

Ok

Does f(x) ever actually = -infinity

[silver dagger]

No.

[vast shale]

It gets as close to -infinity as possible

But never actually = it

[silver dagger]

Interesting.

[vast shale]

So when that happens

We use parentheses

So for the left endpoint of the domain

We can write (-∞

[silver dagger]

It should be (-∞

[silver dagger]

Yeah.

[vast shale]

Yes

Ok

Now to the right side

What is the right end point

[silver dagger]

∞

[vast shale]

Yes

Can f(x) = infinity

[silver dagger]

No.

[vast shale]

Good

So how do we type it

[silver dagger]

(-∞, ∞)

That's right?

[vast shale]

ur close

Because there is one number between -∞ and ∞ that f(x) does not =

[silver dagger]

Ohhh.

So we write (-∞, ∞] because of the 5?

[vast shale]

No

Adding brackets didn’t do anything to the 5

[silver dagger]

Ok.

[vast shale]

So what do we do when we get really close to a number but never reach it

[silver dagger]

We use parenthesis.

[vast shale]

Yes

So let’s read it left to right

we start at -∞

And go right

Until we hit where x can’t =

[silver dagger]

Ohhhh.

I get it.

[vast shale]

Good

So in that case we need to use parenthesis

[silver dagger]

(-∞, 5)?

[vast shale]

YES

but

That’s only half the domain

Can you write the other half

[silver dagger]

(5, ∞) I guess.

[vast shale]

Yes

So

We need to join those

Whenever we have to split the domain because x can’t = something

We use U to join those fragments

U for Union

So can you write the full domain

[silver dagger]

Sorry, I have to go, maybe I'll ask that again later.

But thank you for the explanation.

[vast shale]

Nooooo

U were one message away from the answer

lol

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[mystic valley]

how do i set up the formula to solve this

[mystic valley]

for EB

[vocal sleet]

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[mystic valley]

.close

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[bleak berry]

do you still need help?

[mystic valley]

yes

[carmine patio]

😭😭😭 I was so confused I saw my name

[mystic valley]

lol

[bleak berry]

(AE)x(EB) = (DE)x(EC)

since you only know AB, give AE an arbitrary value of

[carmine patio]

This the only Tobi thing I got rn

[bleak berry]

'x'

[carmine patio]

Anyway have fun with ur math bye

[mystic valley]

i know

but

i only know AB rn

[bleak berry]

^ read this

[mystic valley]

mb

[bleak berry]

just assign either AE or EB to 'x' or any variable

[mystic valley]

ohhhh

[bleak berry]

and you know that AB + EB = 17

[mystic valley]

didn't think of that

[bleak berry]

yeah i think u got it now

gl

[mystic valley]

i should got it

lemme do this rq and then i might need help with another

[bleak berry]

minor typo do not set AB to x

[mystic valley]

wait

so i have two x?

[bleak berry]

you should

[mystic valley]

cause idk ae or eb

so i put x for both

[bleak berry]

well not quite

lets just say AE is x

then x + EB = 17

so EB = 17-x

now you only have one variable

and so x(17-x) = 6(10)

[mystic valley]

ohhhh

yea

[bleak berry]

and you can solve from there

[mystic valley]

bro i have like

no

smartness

when it comes to making formula

[bleak berry]

dw u gain intuition from doing alot of problems

[mystic valley]

wait brb

wait what tf do i do if i get a quadratic

ohh nvm

mb im slightly dumb

[bleak berry]

you should get a quadratic

[mystic valley]

yes

[bleak berry]

its just that your two roots will be the measure of the segments

[mystic valley]

i realized that

ok lemme factor

[bleak berry]

ok swag

[mystic valley]

wait what adds up -17 and multiplies to 60

ohhh 20

[bleak berry]

-12 and -5

[mystic valley]

wait what

oh nvm

20 and 3 wouldn't get you negative

[bleak berry]

yeah -20 and 3 would multiply to -60

[mystic valley]

yea

ok 12 and 5

ok that question is done

i need help on this one

bottom one

i tried it but

can't find a way to get ED

and PB

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[mystic valley]

bru

[mystic valley]

i forgot i closed this

[mystic valley]

^

[bleak berry]

find PD first

EP is the same as PD as they are both the radius of the circle

from there just multiply PD by 2 to get ED or just add them

[mystic valley]

it's safe to assume P is the centroid?

does the symbol of the circle and the dot in the middle mean that

[bleak berry]

i believe so

but your PD is incorrect

[mystic valley]

oh

so what did i do wrong

i don't think it's wrong

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[waxen hawk]

How to convert y=5 into a function of y

[scenic ravine]

it's already a function of y

[waxen hawk]

It is y=5 * x^0

It is a function of x

[fervent wasp]

No, it’s y

[waxen hawk]

You have to isolate the x variable on one side, so it will be a function of y

[waxen hawk]

But x isn’t isolated

[fervent wasp]

It is isolated

Anyways

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[quick wind]

Can someone answer all of this for me?

[uneven chasm]

This isn't a place to get answers

If you need help, what are you stuck on?

[lavish wing]

Why do so many people here think we just calculators

😭😭

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[warm vine]

since the initial statement isnt true

i guess its the second one

the other one

idk

i just concluded that using the initial statement

nice

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[pulsar sluice]

How do you do this?

[steep fjord]

do you know the curve length formula

[pulsar sluice]

sqrt (1 + y') right?

integral from a to b

[steep fjord]

y' is squared

[pulsar sluice]

oh right

[steep fjord]

yeah

so

you y is that

so what's y'

[pulsar sluice]

wait y is the whole integral?

[steep fjord]

yes

[pulsar sluice]

o

would it be just sqrt(t^3 - 1)

[steep fjord]

right

cs ur taking the derivative of the antiderivative

[pulsar sluice]

oohh okay

[steep fjord]

then js plug that into the curve length formula

[pulsar sluice]

ahh okay got it ty

[steep fjord]

np

[pulsar sluice]

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[unreal pagoda]

what did i do wrong

[vocal sleet]

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[left olive]

For two bases A={a1, a2, a3} and B={b1, b2, b3} in a vector space the following applies:
a1 = 2b1+b2+2b3
a2=b1+b3
a3=b1+2b2

a) What are the coordinates in base A for the vector [2; 3; 4] which is in base B?
b) What are the coordinates in base B for the vector [2; 3; 4] which is in base A?

[left olive]

for clarification this is how the vector [2; 3; 4] in base B is given in the actual assignment

So basically i wanna know how to convert coordinates from one base to another with the convertion equations given

i did something like this but i dont really know if this is correct at all

[vocal sleet]

@left olive Has your question been resolved?

[left olive]

no help 🗿

.close

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[young cosmos]

How do i notate that a vector is a unit vector?

[scenic ravine]

$v \cap$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

wait

that's wrong

[young cosmos]

Thats what i thought

Is it to the 0 power?

[scenic ravine]

no

just a minute

$\hat{v}$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[young cosmos]

Ok

Thx

.close

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[tough elm]

I need some guide/direction with question 4a how do i show that that there is no finite cyclic group with 5 generators.

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@tough elm Has your question been resolved?

[tough elm]

<@&286206848099549185>

[shell bolt]

Sup

[tough elm]

Hello, kinda need help with the highlighted question here. I really have no idea where to begin with this one.

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[silk wind]

"Determine the area A and the coordinates (µx, µy) of the center of gravity of
sector D of the ellipse if"

[silk wind]

am i thinking right that you start with integrating x from 0 to 1 and integrate y from 0 to x?

so that A = 1/2?

[worn bobcat]

Note that D is a general ellipse

So its area for example will depend on a and b

[silk wind]

Hmm okay so I am thinking wrong?

[worn bobcat]

probably

[silk wind]

Oh okay

[worn bobcat]

Why did you want to integrate x from 0 to 1?

[silk wind]

Well I am not really sure I was looking at the conditions of D but I was thinking wrong

[worn bobcat]

Do you know how to do a multivariable substitution or have you integrated in polar coordinates before for example?

[silk wind]

We have at some point gone through multi variable substitution so I think I could do that

[worn bobcat]

yeah, it's possible to do integration in elliptical coordinates and I think that would be really helpful here

So you'd have like
x = ar cos(theta)
y = br sin(theta)
etc.

[silk wind]

Ohh okay

[worn bobcat]

so like (a cos(theta), b sin(theta)) is some point on the boundary of the ellipse

and if you scale that by some let's say r between 0 and 1 you get all the points of D in a very natural way

or umm you could get all the points inside the ellipse and then by doing some additional thinking you'll also get D

one thing I'll warn you about in advance because I've done this almost exact same problem before is that you need to be very careful about the angle

because it's temping to say that the angle would be pi/4

but it's not

[silk wind]

ohh okay thats good to know

[worn bobcat]

because the theta parameter doesn't really represent an angle, it's just like a scaling parameter pretty much

[silk wind]

Ohh I see

[worn bobcat]

so really what you need is some theta such that

a cos(theta) = b sin(theta)

and that's pretty easy to solve for theta

[silk wind]

So that we get theta = arctan(a/b)?

[worn bobcat]

yeahh exactly

[silk wind]

Ohh okay i see

So does that represent like the upper bound of D?

[worn bobcat]

well if you have these (r, theta) coordinates then D in r theta land will be the rectangle [0, 1] x [0, arctan(a/b)]

[silk wind]

Hmm okay

[worn bobcat]

idk if you're familiar with that notation

[silk wind]

Hmm not very but I think I understand

[worn bobcat]

but it's like
0<=r<=1
0<=theta<=arctan(a/b)

[silk wind]

Yeah okay

[worn bobcat]

because the cross denotes a cartesian product

and a cartesian product of two intervals in R forms a rectangle in R^2

set theory shenanegans

[silk wind]

Ahh yes that I have heard of

Is the area for an ellipse not A = pi * a * b?

[worn bobcat]

yup that is the area of an ellipse

so you can plug that in for A

but it's not really gonna help you when you need to integrate x over the region D

but it saves you from having to integrate 1 over the region D

although it would probably be a good exercise to try to integrate 1 over D and see if you get pi * a * b

and if you do then you're probably doing something right

[silk wind]

Yes okay

But when integrating x over D do we use acos theta as x?

[worn bobcat]

r*(a cos(theta))

remember that you need to be able to scale the point by r

[silk wind]

Oh right r as well

[worn bobcat]

yeah. r and theta might be slightly poor choice of symbols because r reminds us of a radius and theta reminds us of an angle but in an elliptical setting they're slightly different

[silk wind]

Yeah that is true

[worn bobcat]

I think my lecturer had like x=as cos(t) and y=bs sin(t) or something

but either way it doesn't really matter as long as you don't get confused yourself

[silk wind]

Yeah I think r will confuse me less haha

[worn bobcat]

alright then you should stick with that

[silk wind]

Can we substitute ar cos(theta) with something?

[worn bobcat]

I'm not sure if I understand your question

but I want to remind you that when you do a change of coordinates you need to remember to multiply by the absolute value of the determinant of the jacobian matrix associated with the coordinate transform

[silk wind]

Ohh yeah right thanks

[worn bobcat]

and that's how dxdy becomes drdtheta

[silk wind]

I will continue tomorrow but thank you for the help!

[silk wind]

Ohh okay I see

[worn bobcat]

inga problem ^^ ha en bra kväll

[silk wind]

Detsamma haha

[worn bobcat]

tack

[silk wind]

.close

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[harsh heart]

Hey, I would like to show that a function is surjective and therefore I would like to find the codomain. Is calculating the image of the bounds enough?

[harsh heart]

I mean if the images of the bounds of the domain fits with the bounds of the codomain so is the function injective?

[eternal quarry]

You have to show that f^-1(y) exists in the domain for each y in Im(f)

Or that codomain equals image

[eternal quarry]

You mean f: [a,b] → Y and you calculated f(a), f(b)?

[harsh heart]

Okay so for example to show that cosh(x) is bijective de R+ on [1 ; infinity[, I can just find cosh(0) and lim x->infinity cosh(x)?

[harsh heart]

Yes

[eternal quarry]

This implies injectivity only if f is continuous and strictly monotonic

[harsh heart]

Ok so if the function is continous and stricly monotonic, I can just calc f(a) and f(b)?

Why it has to be strictly monotonic?

[eternal quarry]

If it isn't strictly monotonic then f(a) = f(x) for x > a is possible

And so it isn't injective

[harsh heart]

Ah yes of course

But if I want to show that f is surjective I can just calc f(a) and f(b) and see if it fits with the codomain right?

Or it has to be continous

and monotonic

[eternal quarry]

I am not sure. Showing that f^-1(y) always exists is 100% enough

[harsh heart]

Okay don't worry, thank you and have a great evening 🙂

.close

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[faint forum]

Hello

[faint forum]

Can someone help me in factorising please.

[tawdry canopy]

post your question

[faint forum]

As Im doing Vector Geometry and I just dont know how to factorise.

Ill send now,

=6/4b+3/4a

I learnt this but I completely forgot

I know it equals 3/4(2b+a)

but is there like a method I can use to get there

and not my common sense

because there are harder ones

or is it just how many eg, on the question I sent how many a's go into b

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[twin meteor]

faiyrose

faiyrose

faiyrose

faiyrose

faiyrose

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[vocal sleet]

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[stiff fjord]

I don't understand letter A and B for the graph.

[stiff fjord]

I get that it is an infinite limit where in part A, x is approaching -infinity and you have to see what f(x) equals but from what I got as an answer versus what I was told the answer was are two totally different things. To me it looks like where x is approaching -infinity, f(x) = 1, but my professor said the answer to that one was -5. For part B, to me it looks like x is approaching infinity at f(x) = 0 but she said it was 2. That is all that I know so far and why it is not making sense.

[outer warren]

can you circle / indicate which part of the graph you're looking at for each part

your prof is also wrong for part a)

[stiff fjord]

yea so for part A for example, I am looking at the middle line where it seems to be approaching -infinity, when looking at which x coordinate that is, it seems to be 1. for part B, i am looking at the left-most line where it is approaching infinity, by looking at the corresponding x coordinate for that, it is 0.

[outer warren]

please circle/mark the location on the graph

[outer warren]

you're looking at the wrong places

[stiff fjord]

where should i be looking?

[outer warren]

where x→-inf, x→inf
you're looking around where x is 0 and 1

you should be considering what is happening as the graph:
extends to the left
extends externds to the right

[stiff fjord]

okay so i should be looking here instead for the first one because its like a book (reading it left to right)...

[outer warren]

yes

and what's the y value there?

[stiff fjord]

-5

[outer warren]

no

[stiff fjord]

ohhhh

its 2

[outer warren]

no

[stiff fjord]

0

[outer warren]

yes

[stiff fjord]

since its on the axis

so the limit x approaches -infinity at 0? and for the second one, the limit x approaches infinity at 5?

[outer warren]

no

and wrong wording

as x approaches -inf, f(x) approaches 0

$\lim_{x\to -\infty}f(x) = 0$

[twin meteor]

ℝαμΩℕωⅤ

[outer warren]

how are you getting 5 for the other limit?

[stiff fjord]

that makes sense, yes.

[outer warren]

again it seems like you're looking at the wrong value

you're after the y value the curve approaches

[stiff fjord]

i was getting 5 for the other limit because i was looking at the first line where it was curving up and going to infinity. the closest value to that on the y axis was 5. but maybe I should have been looking at the third line where x approaches infinity it is at the HA of 2.

the difference being y approaches infinity on the first line when it curves up, not x

does that sound correct?

[outer warren]

wdym first line

oh,

yeh

for x→inf, you're interested on what's happening way over on the right side

[stiff fjord]

yea, so the green line is the first line i was talking about, the orange is the third line i was talking about

so it should be:
x->-inf f(x)= 0
x->inf f(x)= 2

[outer warren]

yes

[stiff fjord]

i think thats where i was getting confused was with where to look and me looking at where y was going infinity or -infinity rather than x

thanks so much!

.close

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[rain hollow]

The distance from points A and B to l line is 20cm and 16cm. Need to find the distance from AB lines middle point to the l line

[vocal sleet]

@rain hollow Has your question been resolved?

[rain hollow]

<@&286206848099549185>

.close

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[raw crag]

Use L'Hospital

[spiral inlet]

I'd try a double angle identity for cos(2x)

Can you show what you did?

[heavy flare]

Maybe devide up and down with 2x

[spiral inlet]

It would be nice to have everything in terms of sinx

[heavy flare]

You get more manageable limits

[spiral inlet]

or this works, but put everything in terms of cosx then

If you put everything in terms of sinx, I'm pretty sure you'll see the way forward

go from here

cos^2(x) = 1-sin^2(x)

Yep

🎉

Not just sin^2, 1-sin^2

keep factoring