help-13

[crimson sedge]

Comparing both sides?

[near wing]

oh am i comparing?

[crimson sedge]

Like

Both sides are in sin

So we can just remove it roght

Roght

Roght

Right

[near wing]

oh yes lol

[crimson sedge]

Yeah

x= 30? (This one is to get in radian)

[near wing]

2(30)=60+180

bruh im sorryy

[crimson sedge]

Nvm I went off a bit

Here

Again

x = 30??

We got this right?

[near wing]

yes

[crimson sedge]

Substitute in original eqn

2x = 180 + 60

Damn bro I'm tripping wait

[near wing]

its cool i dont blame you

[crimson sedge]

I mean ik the answer

Like 2x = 240

X = 120

That's the first one

Then 270+30 = 2x

x = 150

360 - 60 = 300 = x

[crimson sedge]

I did this directly like

540 + 60 / 2

And then

630 + 30 / 2 = 330

Man ik the answers it's just idk how to explain 😭

Only if we could VC yk 😭👎

I mean I cnat right now even if I want to, only after 12 hours

Till then you'll be asleep

[crimson sedge]

We went to this cuz , if x = [0,360]
2x = [0,720]

Lmao didn't understand a thing right

Sorry man

[near wing]

no your doing good

[crimson sedge]

Stop lying

Okay restart

[near wing]

lemme reread it

[crimson sedge]

180+x is what we needed

[crimson sedge]

Won't understand, it's just vent and direct calc

Here in india trigo is way harder so I had did it within 10 seconds of seeing the question

Okay Listen

@near wing

We want minus sign

[near wing]

mhmm

[crimson sedge]

So it should be of the form sin (180+x)

Or cos (270+x) as

Cuz 270 is odd multiple of 90, it will change to give sin

Correct?

So cos (270+x) will give sinx

With minus sign

Okay?

[near wing]

tes

yes

[crimson sedge]

And do practice quadrants and their signs

Do that forst

First

Need that you do

[near wing]

bruh

[crimson sedge]

Wtf

<@&268886789983436800> what is this botwog doing ??

[near wing]

i aint think i was gonna see this today but aye lol

[crimson sedge]

Nvm KJ

[crimson sedge]

Okay you get that

??

<@&268886789983436800>

Just ignore him

[near wing]

aight i blocked him bruh distracted tf outta me

[crimson sedge]

Good

Okay

So either cos (270+x) or sin (180+x)

Understood this much?

[near wing]

but yeah im getting it a bit more everytime

[crimson sedge]

Good

When you do the questions of quadrnt with trigo signs and all

You'll get what I'm speaking

You need practice , a lot

Then

[near wing]

shit i know i do

lol

[crimson sedge]

In reality in question instead of x it's 2x

Now they told us x was between 0 and 360

Right ?

[near wing]

yup

[crimson sedge]

0 ≤ x ≤360

Multiply all by 2

0 ≤ 2x ≤ 720

Okay?

[near wing]

mhmm

[crimson sedge]

Cuz that's what we are given , a sin of 2x

Now

You know from quadrant

0 = 360

Like same position

Not literally

The axis where 0 comes, 360 will too

Check it

90 = 450

180 = 540

270 = 630

Okay?

You just add 360 to all

To get same place

😭😭

Understood?

[near wing]

yeah i got you

[crimson sedge]

Now

Just as sin(180 + 2x) was out req we also get sin(540 + 2x)

Okay??

[near wing]

yuo

yup

[crimson sedge]

Yeah

Now x = 30

Right

[near wing]

yes

[crimson sedge]

Put in all

You get

Sin (180+60) = sin 240

Sin (540 + 60) = sin 600

Now as what we needed was really x

This is 2x

240 = 2x

X = 120

600 = 2x

X = 300

For the cos ones you just add 30

So that you get sin 60

And shit

You didn't understand much of it man

And I don't have much time left have to sleep yk that's whyyyy

Will you be up till 2 in the night?

Like do you?

[near wing]

bro go to sleep i dont wanna keep you up but i get you mostly its just that it doesnt click to my brain fast enough

[crimson sedge]

Uhmm

[near wing]

i gotta be to work in less than an hr amyways

[crimson sedge]

Well ykw

[near wing]

anyways

[crimson sedge]

Wait a sec

Just skipp the odd multiples part fr

Just use 180 and 360 lines

Cuz you don't have to change shit in them

Just easy

So now we want

Sin (540+60) to get a uh - sin 60

Understood?

And sin (720 - 60) to get - sin 60

[near wing]

yes

[crimson sedge]

That gave us 2x = 300

2x = 480

X = 150,240

And we already got x = 120

No wait shot

Wrong thing

[near wing]

lol damn

[crimson sedge]

This would be uhhhh

Sin (660)

2x = 660

X = 330

So 120 150 330

[near wing]

and 300 between 150 and 330

[crimson sedge]

Yeah wait

[crimson sedge]

I wrote this wrong too

Didn't go in the second lap

Yeah

This would be

Sin (600) to get - sin 60

2x = 600

X = 300

So atlast correct eqn are

[crimson sedge]

This

[crimson sedge]

This

[crimson sedge]

This

And yeha

Sin 360-60

Sin 300

2x = 300

X = 150

Phew

That's all values of x

[near wing]

mm okay i got that part

[crimson sedge]

I used 180° axis and 360° axis

To get negative values of sin

I added 60 to 180° axis / (180+360)= 540° axis

And subtracted 60 from 360° axis / (360+360)° = 720° axis

To stay in the region which was negative for sin

You see?

[near wing]

yessir i got you

[crimson sedge]

Aye aye

😭

I'm bad at explaining but yeah

Just read these last things and you'll get it

[near wing]

i gotchu bro i took a pic

[crimson sedge]

Do DM me of you have any other questions

[near wing]

alright i got you

[crimson sedge]

Like regarding this

[near wing]

thank you very much

[crimson sedge]

I would answer tomorrow with a much more active brain

Rn it don't function properly 👎

[near wing]

i appreciate the time you took out your sleep to help my dumbass once again

[crimson sedge]

Welcome man

[crimson sedge]

Aye aye no worries

[near wing]

goodnight bruh

[crimson sedge]

I'll sleep now peacefully ig atleats I completed the question 😭😭

[crimson sedge]

Goodnight man

[near wing]

yessirr

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[royal nexus]

I would like insight on my derivative so far if its all ok i will continue on my own thank you!

[dire geode]

unclear what your variables are. looks like a d

or is that supposed to be an alpha?

[royal nexus]

no its d

[deep oriole]

if the top is what you are supposed to differentiate then it looks fine

[royal nexus]

this is the problem

[deep oriole]

however you can simplify the last line

[royal nexus]

thats what im trying to do here but im not sure how to go about in the denominators

[deep oriole]

expand out the square on the bottom

and realise what you can multiply top and bottom by

[royal nexus]

oh ok got it

ill work on it

$\frac{-100d^{2}}{d^{2}\left(d^{2}+10000\right)}+\frac{50d^{2}}{d^{2}\left(d^{2}+2500\right)}$

[wraith dagger]

Eggy 🍳

[royal nexus]

this is what i got

[deep oriole]

and you see where you can simplify?

[royal nexus]

yea

$\frac{-100}{\left(d^{2}+10000\right)}+\frac{50}{\left(d^{2}+2500\right)}$

[wraith dagger]

Eggy 🍳

[deep oriole]

yep

[royal nexus]

aa i see ok

thank you!

[deep oriole]

np

[royal nexus]

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[late heath]

interval [0,2pi), solve following equation 3tan^2 x + 11tanx = 4

[late heath]

baslically i factored it as (3tan x - 1)(tan x +4) = 0

and found two possible answers, but im not sure what should i do next 😦

[smoky idol]

well, first off, did you mean tan^2(x)?

[late heath]

oh, yes. sorry.

[smoky idol]

no worries. I was just wondering cause tan^x is pretty wild xD

So, your factoring is the right thing to do

[late heath]

i do know i have to convert it something like value +- pi, but not sure how to that...

[smoky idol]

What did you do after this?

[late heath]

well found two answers, tan x = 1/3, tan x = -4

and im stuck and out of ideas, what should i do next 😦

[smoky idol]

Do you know what inverse tan is?

[late heath]

yes, you mean i should convert it to inverse tan?

arctan 1/3 + pi?

[smoky idol]

nono, I mean that you should use inverse tan

oh wait maybe that's what you meant by converting

then yes, you want to use arctan to get x, and then add and subtract pi to fit in your interval

[late heath]

honestly, im so lost at here. im not even sure what interval stands (like what does it actaully affects for) :(.

so, for first answer does it leaves two answers like
arctan 1/3 and arctan 1/3 + pi?

[smoky idol]

okay so, you seem to be aware that tan is a periodic function, am I right?

[late heath]

Q1/Q3

yes

[smoky idol]

Alright, and the specificity with periodic functions is that they repeat after a certain time. That time is called the period. In the case of tan(x), period = pi. So when you find an answer to your equation, an x value, you can technically add or subtract an infinite amount of pi, and still get valid answers. However, they gave you an interval, so you don't want infinite answers, you just want x values that fit in that interval

[late heath]

oh

oh

omg

tysm, you are the legend. very straightforward and gave me lightening.

[smoky idol]

Glad I could help you :D

[late heath]

.close

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[dapper crane]

Guys May I know why this formula used for?

[dapper crane]

Pls explain

[teal rune]

looks like the quadratic formula.
Not sure about the delta there though... (triangle)
there should be b^2 - 4ac

[smoky idol]

people refer to b^2-4ac as delta sometimes

It's often taught to check delta first, as if it's less than 0 you know that there are no real roots

[teal rune]

ah, the discriminant then.. never seen it that way

(deleted)

(I'm not sure how to do a +- symbol in TeXit?)

[lunar lynx]

\pm

[teal rune]

Alright thank you

[teal rune]

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[wraith dagger]

lanz 🍇

[teal rune]

This is your formula in it's 'original form'.
The triangle (delta) signifies b^2 -4ac in this case.
It is used to find out x in quadratic equations.

[dapper crane]

Got it thank you friend

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[distant adder]

I know nobody is going to want to help with this but I was trying to work through some of the details in a paper by Knuth and couldn’t work out the step notated by the blue arrow https://media.discordapp.net/attachments/886483914410049556/1135765501797224671/IMG_2040.png

[distant adder]

@distant adder

for refernce, this is from Ch 22 of Knuth's "Selected Papers on the Analysis of Algorithms" - I have managed to get the second line to the following:

$$ \frac{\sqrt{2 \pi m}}{e^m} \sum_{1 \le k \le m-1,;; j \ge 0} \frac{\sigma_j}{m^j} \frac{k^{k-1} e^{-k}}{k!} \left( 1 + k/m\right)^{\frac{1}{2} - j} \frac{1}{k+m-1}$$

[wraith dagger]

rubixcyouber

[distant adder]

although i am not sure how to collect the $\sigma_j$ into the $\tau_j$ and eliminate the factor of $\frac{1}{k+m-1}$

[wraith dagger]

rubixcyouber

[distant adder]

if anyone else wants to uh

pls help

[dire geode]

Where's your actual question

[distant adder]

how do i simplify from the 2nd line to the 3rd line

ive only gotten from the second line to the above

which is not the third line

[cedar kiln]

@distant adder Has your question been resolved?

[dire geode]

Looks like they just switched order of summation and factored out m^j from the denominator.

The 1/2 in the power of (1+k/m) comes from the square root in the 2nd line

[distant adder]

How do they get rid of the 1/(k+m-1) factor

[dire geode]

k+m-1 = m when k=1

[distant adder]

I got up to here, I’ll think about this in the morning

[distant adder]

What order did they switch

[dire geode]

There's only two summations

[distant adder]

Oh I thought you meant like the upper and lower bound

Still don’t really get the tau terms

Thanks for the help

[cedar kiln]

@distant adder Has your question been resolved?

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[obtuse coral]

How do I do this?

[vestal bear]

those functions are the same

[south tundra]

Not really

[deft gull]

no

[south tundra]

The domain of the first is {x: x >= 0 and x+1 > 0} and the other one's is {x: x/(x + 1) >= 0}

[deft gull]

ya

[south tundra]

Basically, solve each system of inequalities and see if the solutions match up

Although it should be obvious that [ \frac{x}{x+1} \ge 0 \cancel\Leftrightarrow \begin{cases} x \ge 0 \ x + 1 > 0 \end{cases} ]

[wraith dagger]

A Lonely Bean

[slate lintel]

the important part here is that you can't take the square root of a negative number

[cedar kiln]

@obtuse coral Has your question been resolved?

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[subtle horizon]

note: $$\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2} =\frac{1}{x^2+1}$$

[wraith dagger]

Brian

[subtle horizon]

How it this possible

thought this was only equals to the integral of inverse tan

[vestal bear]

magic

im confused on your question

arctan is inverse tan

[subtle horizon]

Yes but in this case they are differentating

not integrating

[vestal bear]

,w integrate arctanx

[wraith dagger]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Aug 1, 2023, 6:07 AM

[vestal bear]

,w differentiate arctanx

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Arctic#1647

[vestal bear]

i dont think integral of arctanx is that

i believe there is a ln component to it and you need to do integration by parts

[subtle horizon]

ok

.close

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[fast relic]

Uhh not sure where to start

[cedar kiln]

@fast relic Has your question been resolved?

[fast relic]

<@&286206848099549185>

[tropic oxide]

p sure it's just angle DAC you're asked for

[fast relic]

Yes

[fast relic]

But how do you find it?

[tropic oxide]

take one side length as 1 and use trig to express everything else in terms of it

[fast relic]

This is the best I can reach up to

Basically nothing

[fast relic]

What do you mean by taking one side as one?

[fast relic]

<@&286206848099549185>

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@fast relic Has your question been resolved?

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[subtle horizon]

Can someone please confirm ofc this is correct

This is the question

$$\int\frac{x^2 -x +6}{x^3+3x} dx $$

[wraith dagger]

Brian

[subtle horizon]

But my question is

why cant you do

this

[tulip stream]

,rotate

umm nice copy image discord

,rotate

[subtle horizon]

$$-1/2\int\frac{x+1}{ux} du$$

[wraith dagger]

Brian

[tulip stream]

the integral on bottom left is lacking a du btw

the one that is multipled by -1/2

[subtle horizon]

True

but my point is what do i do now

[tulip stream]

split like shown

or turn every x into u

[subtle horizon]

How and why

[subtle horizon]

How and why

[tulip stream]

you did the substution $x^2 + 3 = u$ correct

[wraith dagger]

Cyrenux

[tulip stream]

and you are taking integral with respect to u

if you have done substituion transforming a variable to u , then there shouldnt be any remaining of the variable that was being transformed

[subtle horizon]

so no x

[tulip stream]

also i didnt notice but 2nd integral needs to be multiplied by -1/2 too i guess

2nd integral from bottom

but you will see that now

$$\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$$

[wraith dagger]

Cyrenux

[tulip stream]

use this fact to split

[subtle horizon]

wait i come soon

[tulip stream]

tag me when back then

[subtle horizon]

explain

[subtle horizon]

so me wrong

[tulip stream]

just make sure you transform every x variable to u BEFORE taking the integral

you havent took the integral yet

[subtle horizon]

but i have

i have to use tan

[tulip stream]

i still see integral and du

[subtle horizon]

that what integral means

wait me come

[tulip stream]

also i dont have any detailed explanation for that

you can ping helpers 15 mins later for a more detailed explanation

[cedar kiln]

@subtle horizon Has your question been resolved?

[spice kraken]

yo

sup

[subtle horizon]

yes

look

$$\int \frac{x+1}{x^2+3} dx = \frac{1}{2}\int \frac{1}{u} du + \int \frac{1}{x^2+3} dx$$

[wraith dagger]

Brian

[spice kraken]

what do you need

hmm

[subtle horizon]

How is 1/2 positive

why

shouldn't be negative

b and c are negative ones

what i mean is

when i was doing partial fraction

[spice kraken]

ok

what is u though

[subtle horizon]

i used $$\frac{A}{X}+\frac{Bx-c}{X^2+3} $$

[spice kraken]

$\frac{A}{X}+\frac{Bx-c}{X^2+3}$

[wraith dagger]

Brian

WhereWolf

[spice kraken]

ok

[subtle horizon]

Yes

and when i multiplied that by the quadratic $$x^2 -x +6 $$

[wraith dagger]

Brian

[subtle horizon]

and did the comparing method

by adding all the x^2 x together and so on togehter

[spice kraken]

yee i saw that

[subtle horizon]

i got b to be -1 and c to be -1

is that true

[spice kraken]

yes

a=2,b=-1,c=-1

what's wrong

[subtle horizon]

Then how is this right

[spice kraken]

$\int \frac{x+1}{x^2+3}=\int \frac{x}{x^2+3}+\int \frac{1}{x^2+3}$

[wraith dagger]

WhereWolf

[subtle horizon]

subsittution

[subtle horizon]

I did not get that i got $-x-1 $

[spice kraken]

$\int \frac{x}{x^2+3} = \frac{1}{2}\int\frac{2x}{x^2+3}=\frac{1}{2}\int\frac{1}{u}du$

[subtle horizon]

as in $$ Bx+C$$

[wraith dagger]

WhereWolf

Brian

[subtle horizon]

Not ansewering the negative question

[spice kraken]

hmm

yeah there should be a -1

[subtle horizon]

Can you demonstrate

[spice kraken]

ok

$\int\frac{x^2 -x +6}{x^3+3x} dx=\int\frac{2}{x}dx+\int\frac{-x-1}{x^2+3}dx=\int\frac{2}{x}dx-\int\frac{x+1}{x^2+3}dx$

[wraith dagger]

WhereWolf

[subtle horizon]

tyyyyyyyyyyyyyyyyyy

[cedar kiln]

@subtle horizon Has your question been resolved?

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[subtle horizon]

$$\int \frac{1}{x^2+3} dx = \frac{\sqrt{3}}{3} \int \frac{1}{1+u^2} du $$

[wraith dagger]

Brian

[subtle horizon]

Can someone please explain how root 3 was gottne

[buoyant latch]

What’s relationship between x and u

[subtle horizon]

u = x^2 + 3

[buoyant latch]

Ok what’s the next step

[subtle horizon]

???

[buoyant latch]

The root 3 is much further down the line

There’s a bunch of working before it comes out

[subtle horizon]

Any hints

[buoyant latch]

Well use the substitution

[subtle horizon]

i did that

[buoyant latch]

Show your work

[subtle horizon]

ignore -1/2

[buoyant latch]

Where did x+1 come from

Why does it say du but the integrand is all x’s

[subtle horizon]

to simplify

[subtle horizon]

the is reason

[buoyant latch]

Can you write it properly

I absolutely cannot follow what you are doing

[subtle horizon]

what should i write

[buoyant latch]

First write the question

Then the substitution

Then do all the relevant substitutions

[cedar kiln]

@subtle horizon Has your question been resolved?

[subtle horizon]

I have sent

[buoyant latch]

Ok we’re at $\int \frac{x+1}{x^2+3}, dx$

[wraith dagger]

Frosst

[buoyant latch]

So what was your first step

[subtle horizon]

use tan inverse integral

[buoyant latch]

On what

[subtle horizon]

that is $$ 1/x^2 +1$$

[wraith dagger]

Brian

[buoyant latch]

What about the x+1 on the top

[subtle horizon]

for the bottom part

no

we do not do taht one

[spice kraken]

bro what are you solving

I'm confused

[buoyant latch]

Actually

[subtle horizon]

this

[buoyant latch]

There’s a faster way

[deep oriole]

why dont you just split it up

[exotic furnace]

You can’t just multiply integrals btw, it’s an involved process

[spice kraken]

seperate the fraction

I thought you solved this already?

[subtle horizon]

go stuck again

[spice kraken]

bruh

[buoyant latch]

Have you seen $\ln|f(x)|+C=\int\frac{f’(x)}{f(x)}, dx$

[wraith dagger]

Frosst

[spice kraken]

nah don't use formula for these stuff

[buoyant latch]

This will get you the answer the fastest

[spice kraken]

it's just a u substitution

[buoyant latch]

Yeah sub x = denominator

Lol

[subtle horizon]

what

[buoyant latch]

(It literally is)

[subtle horizon]

why not u

[buoyant latch]

I mean u = denominator

Same thing

[subtle horizon]

but why

[upper ruin]

You can use whatever letter/name you want

[buoyant latch]

Because of this

[exotic furnace]

Yeah this, just don’t say x=x^2+3 lol

[subtle horizon]

How does this affect anything

[upper ruin]

Of course not the same variable you are substituting

[buoyant latch]

Look at the bottom

Look at the top

Both polynomials

[spice kraken]

$\int\frac{x+1}{x^2+3}dx=\int\frac{x}{x^2+3}dx+\int\frac{1}{x^2+3}dx$

[buoyant latch]

Look at the degree

[wraith dagger]

WhereWolf

[buoyant latch]

Top is 1 less than the bottom

[subtle horizon]

ok

[spice kraken]

brian we've been through this

[buoyant latch]

If we differentiate the bottom we can probably get a bunch of the stuff on top

[spice kraken]

where are you stuck again

[subtle horizon]

the root 3 part

[spice kraken]

ok

[subtle horizon]

this

[spice kraken]

so focus on $\int\frac{1}{x^2+3}dx$

[wraith dagger]

WhereWolf

[spice kraken]

right?

[subtle horizon]

so what now

[spice kraken]

we want to perform a u substitutioon such that it becomes $\frac{1}{u^2+1}$

[wraith dagger]

WhereWolf

[subtle horizon]

why use u

[spice kraken]

it's just a dummy variable

[subtle horizon]

what is diffrence

[upper ruin]

@subtle horizon !

[subtle horizon]

we shall use

x

It is ismpler

[deep oriole]

bruh

[spice kraken]

because $\int \frac{1}{u^2+1}dx = tan^-1(u)+C$

[upper ruin]

Noo, since the integral is in x

[wraith dagger]

WhereWolf

[deep oriole]

x = x^2 + 1

easy

ofc

[spice kraken]

bro don't use the same alphabet

[upper ruin]

Did you understand the method of substitution for integrals

[subtle horizon]

yes

[upper ruin]

It doesn't look like actually

[subtle horizon]

why we subsiturte in the first place

[spice kraken]

because it's easier to calculate

[subtle horizon]

x=u

is not doing anythign

[spice kraken]

no

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[exotic furnace]

lmao

[subtle horizon]

.repopen

.reopen

[upper ruin]

Indeed, we didn't tell you to substitute u = x

[cedar kiln]

✅

[spice kraken]

from $\frac{1}{x^2+3}$ to $\frac{1}{u^2+1}$

[wraith dagger]

WhereWolf

[spice kraken]

what should we substitude

[subtle horizon]

me thingk

x^2 + 3

[spice kraken]

no

we don't have x on numerator

[deep oriole]

isnt it easier to explain that first you take out a fraction and then substitute

[spice kraken]

true

[subtle horizon]

What fraction

[spice kraken]

$\frac{1}{x^2+3}=\frac{1}{3(\frac{1}{3}x^2+1)}$

[wraith dagger]

WhereWolf

[spice kraken]

ignore the constant 3

what to substitute? $\frac{1}{\frac{1}{3}x^2+1}}$

[wraith dagger]

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

[spice kraken]

we want the denominator to be u^2+1

[subtle horizon]

wait

(x^2)/3 is equals to u

[spice kraken]

?

[subtle horizon]

so we can use tan

[deep oriole]

u^2

[spice kraken]

yes

u^2=x^2/3

[upper ruin]

Hence, u = ?

[subtle horizon]

(x^2)/3

[upper ruin]

That's u²

We want u

[subtle horizon]

so it is tan inverse ((x^2)/3)

[upper ruin]

So that we can calculate the differential

[deep oriole]

no

[subtle horizon]

x/root 3

[deep oriole]

oh wait i thought you were saying u was that

[subtle horizon]

u squared

[upper ruin]

Nice, now you need to compute the du (as a function of dx), do you know how to do it?

[subtle horizon]

then use tan inverse u

[upper ruin]

Nope, don'overthink ahead

[spice kraken]

we set u =x/root3 now what about du and dx

[upper ruin]

How can u itself be something else as a function of u, it doesn't make any sense

[subtle horizon]

we diffrentiate u^2 +1

[spice kraken]

nooo

?????

[subtle horizon]

,w differentiate u^2 +1

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Brian#8701

[deep oriole]

wtf

[spice kraken]

?????

[upper ruin]

You said you know how substitution works, that doesn't seem you know it actually! @subtle horizon

That's why I asked you

If you have t = x² how do you calculate dt? @subtle horizon

[subtle horizon]

dt/dx

then cross multiply

to get dt

[deep oriole]

"multiply"

[upper ruin]

Ok, so in the end what do you get ?

[subtle horizon]

2xdx

[spice kraken]

nicee

[upper ruin]

Oh nice

Therefore you need to apply the same here

To get du = ...dx

[spice kraken]

we know u =x/root3

[subtle horizon]

why this

[spice kraken]

that's how you do u substitution

[subtle horizon]

so we represent dx as du

by diffrentiating the u^2 +1

[upper ruin]

The opposite, but yes

[upper ruin]

Forget that for now

You did it correctly when I asked you of t = x², do the same here, but you have u = x/√3

[subtle horizon]

du /(-x3^-1.5)/2

[spice kraken]

?

[upper ruin]

You have $u = \frac{x}{\sqrt{3}}$

[wraith dagger]

Alberto Z.

[upper ruin]

@subtle horizon , how do you find du/dx ?

[subtle horizon]

-0.5x3^-1.5

[spice kraken]

what are you doing?

[spice kraken]

???

[subtle horizon]

normal diffferentiation

[spice kraken]

please explain your thought process

[subtle horizon]

Bring root 3 up

the power becomes negative

[spice kraken]

you got it backwards

[subtle horizon]

then use that power to times variable

then minus one to power

[spice kraken]

You have $u = \frac{x}{\sqrt{3}}$

[wraith dagger]

WhereWolf

[upper ruin]

The denominator is just a number, think of it as 2 or 1 if you find it easier

[subtle horizon]

1/root 3

[spice kraken]

$\frac{1}{\sqrt{3}}$ is just a constant

[subtle horizon]

is ans

[wraith dagger]

WhereWolf

[spice kraken]

finally

now do the substitution

[subtle horizon]

then dx becomes

1/root 3

[upper ruin]

du = 1/√3 dx

$du = \frac{1}{\sqrt{3}}dx$

[wraith dagger]

Alberto Z.

[upper ruin]

This, right? @subtle horizon

[subtle horizon]

yes

[upper ruin]

Alright

[upper ruin]

Now, recall that you were solving this integral

So what's the integral like now?

[subtle horizon]

why we look for du

[spice kraken]

we're doing u substitution

bro

you need to relearn that

[subtle horizon]

but there is dx

[upper ruin]

Yes definitely

It seems you think we are doing things randomly

[subtle horizon]

no

first

we neeed to use tan

so we need a x^2 -1

right

!!

[spice kraken]

yes

that's why we substitude

but you don't know how to substitude

for practice try to solve $\int\frac{2x}{x^2+1}dx$

[wraith dagger]

WhereWolf

[subtle horizon]

so we represent the 1 over 3 x^2

as U

for the tan

[spice kraken]

no

[subtle horizon]

then we need to represent the dx

[spice kraken]

if $u=\frac{1}{3}x^2$ then $\frac{1}{x^2+3}=\frac{1}{u+1}$

[wraith dagger]

WhereWolf

[spice kraken]

it doesn't work

du would be 2x/3dx

we don't have that x on the top

[upper ruin]

@subtle horizon this the whole process (with complete steps) for that integral

[subtle horizon]

sorr u^2

[spice kraken]

try to understand the steps

you can do this

[upper ruin]

I hope my handwriting can be read 🙈

[cedar kiln]

@subtle horizon Has your question been resolved?

[cedar kiln]

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[white rain]

For

(1+tan(q))(1+tan(X))=-√2

Prove q+X =-67.5 degrees

Help idk how to start

[upper ruin]

I already saw this question

Isn't it the same you posted some days ago?

[lilac totem]

Basically I can give you a hint

$67.5 = 45 + 22.5$

[wraith dagger]

therealtdp

[lilac totem]

We know tan 45

And tan 22.5 is also a known angle

[cedar kiln]

@white rain Has your question been resolved?

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[bright sorrel]

Need help

[bright sorrel]

X is a random variable such so that DX=4, then find D((X-6)/2)

where D is variance

I tried this:

D(X-6)/D2=D(X+(-6))/2=4/2

=2

but the answer is 1

[spice kraken]

it should be 2 though I think

[upper ruin]

It's 1 yes

[bright sorrel]

me too, but I think it has something to do with diving with 2 a property which I haven't learned at least yet

[upper ruin]

Remember var(aX) = a²var(X)

[bright sorrel]

yes

then what

[upper ruin]

2 in the denominator becomes 4

[bright sorrel]

wait how

2 is a constant

there's no variable to it

no?

@upper ruin

[spice kraken]

oh yeah

variance is standard variation squared

[bright sorrel]

wdym?

[spice kraken]

$\sigma_{cX}=c\sigma_X$

[wraith dagger]

WhereWolf

[bright sorrel]

D(aX)=a^2DX

yea?

[spice kraken]

so D(X/2) = 1/4D(x)

[bright sorrel]

ah yes

So this can become

D(X/2-3)=1/4D(X-3)

[spice kraken]

yes

and the constant don't affect variance

[cedar kiln]

@bright sorrel Has your question been resolved?

[bright sorrel]

.close

[cedar kiln]

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[cedar kiln]

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[glossy jewel]

my teacher said that we can't direct say that lim x tends to infinity f(infinty)=infinty(which means we can't use l'hopital here) and after that he did something to prove that we can for this ques can say that and after that he use l'holp how'z that?

[glossy jewel]

that's what he have done

hello

?

hello anyone here?

fuck it i will figure out my self

.close

[cedar kiln]

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[pseudo nest]

In a race, the lead runner is 60m ahead of the chaser with 200m to go and is running at 4ms-1. The chaser is running at 5ms-1.
a) Find the minimum constant acceleration required by the chaser to catch the lead runner.
b) If the lead runner is actually accelerating at a constant rate of 0.05 ms-2, find the minimum constant acceleration required by the chaser to catch the lead runner.

[pseudo nest]

idek where to start

[placid flower]

so you will need to create the movement equations

and fill in the appropriate parameters and solve for acceleration

the difference in b) is that Runner 2 also has a acceleration

[cedar kiln]

@pseudo nest Has your question been resolved?

[pseudo nest]

what about the time?

[placid flower]

in both cases by solving the Runner2 equations for time

we know everything

[cedar kiln]

@pseudo nest Has your question been resolved?

[cedar kiln]

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[odd seal]

This looks weird. Think I did it wrong.

[sinful ocean]

Can you take a better quality picture cause i cant really read that

[cedar kiln]

@odd seal Has your question been resolved?

[odd seal]

could just be my bad handwriting

[sinful ocean]

I dont really get what are you doing from k^3=3m+4k-6

Also -(4k+4) not equals to -4k+4

[odd seal]

I'm not supposed to do that?

[sinful ocean]

Maybe its right what youre doing but i dont understand where did you get the last some lines

[odd seal]

ok so fixing it to -4k-4 I simplified it down to 3m+3k^2+3k-3 -> 3(m+k^2+k-1)

[sinful ocean]

Why 3k?

[odd seal]

because 4k-4k+3k=3k

[sinful ocean]

Oh i didnt see the 4k on the beginning the line sry

[sinful ocean]

Ok and after that?

[odd seal]

well after that then it's finished right? I proved that it's a multiple of 3

[sinful ocean]

Oh wait yeah than you did it right

[odd seal]

ok, just thought it looked a bit messy. All the previous induction ones only had m as a variable

[sinful ocean]

Well ok

[odd seal]

alright thanks

t!rep @sinful ocean

[dense orbit]

No reputation points!

You've already given a reputation point today!
You can send another rep after <t:1690848000:t>.

.liluzibird

[odd seal]

hmm

no rep :(

.close

[cedar kiln]

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[crimson sedge]

Hi

[crimson sedge]

$ \lim

$\lim$

[wraith dagger]

Kangaroo

[crimson sedge]

Sorry

[south tundra]

Try latexing the question in #latex-testing first and then send it here

[crimson sedge]

$\lim_{x \to -\infty } \sqrt{4x^2-5x-2} +2x$

[wraith dagger]

Kangaroo

[crimson sedge]

that is my problem

[south tundra]

What have you tried so far?

[crimson sedge]

conjugate

and apply properties

[south tundra]

What did you get?

[crimson sedge]

$lim_{x \to -\infty } \frac { 4x^2-5x-2 +2x} {\sqrt{4x^2-5x-2} +\sqrt{2x}$

[wraith dagger]

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

[crimson sedge]

$lim_{x → -∞} [ (4x^2 - 5x - 2 + 2x) / (√(4x^2 - 5x - 2) + √(2x)) ]$

Oh Chat gpt change it all

[south tundra]

Did you get something like this?

[crimson sedge]

No

haaha

[south tundra]

[south tundra]

Should be -4x^2 in the nominator instead of +2x

[crimson sedge]

oh why?

[south tundra]

and -2x in the denominator instead of +sqrt(2x)

[crimson sedge]

it is not a-b/ (root a + root b)?

[south tundra]

\begin{align*}\frac{(\sqrt{4x^2-5x-2}+2x)(\sqrt{4x^2-5x-2}-2x)}{\sqrt{4x^2-5x-2}-2x} = \frac{\sqrt{4x^2-5x-2}^2-(2x)^2}{\sqrt{4x^2-5x-2}-2x} = \frac{4x^2 - 5x - 2 - 4x^2}{\sqrt{4x^2-5x-2}-2x} = \frac{-5x - 2}{\sqrt{4x^2-5x-2}-2x}\end{align*}

[wraith dagger]

A Lonely Bean

[south tundra]

Hence $-\frac{5x + 2}{\sqrt{4x^2-5x-2}-2x}$

[wraith dagger]

A Lonely Bean

[crimson sedge]

So the limit is -5/4 no?

[south tundra]

Nope

[crimson sedge]

Now I understand why a lot of probelms I can´t did them

But that is the answer on my book

[south tundra]

[crimson sedge]

XD

[south tundra]

I am looking at the graph and it is definitely not -5/4

[south tundra]

Anyway, now you can divide top and bottom by $x$ and you get $-\frac{5 + \frac2x}{\frac1x\sqrt{4x^2-5x-2}-2}$

[wraith dagger]

A Lonely Bean

[crimson sedge]

but here you didn´t multiply the divisor?

[south tundra]

Hm? I just multiplied top and bottom by the conjugate

[crimson sedge]

Yeah, I understand it right now

Sorry

[south tundra]

We can bring the 1/x into the square root, but we have to be careful, because x is negative here, so 1/x = -1/sqrt(x^2)

[south tundra]

Meaning that this is equal to [ -\frac{5 + \frac2x}{-\sqrt{4 - \frac5x - \frac2{x^2}}-2} ]

[wraith dagger]

A Lonely Bean

[south tundra]

Or just $\frac{5 + \frac2x}{\sqrt{4 - \frac5x - \frac2{x^2}} + 2}$

[wraith dagger]

A Lonely Bean

[crimson sedge]

But that is not -5/4 weird

Anyways thank you I want to understand It by myself, thank yoy

.close

[cedar kiln]

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[cedar kiln]

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[simple bane]

how do i solve this

[void sand]

my first thought was to try sine law

[simple bane]

$\frac{sinA}{a}=\frac{sinB}{b}$etc?

[wraith dagger]

紅卫兵

[simple bane]

but what would opposite of angle A be? like a?

[void sand]

actually

there might be a better way to do this

[simple bane]

please elaborate 🙂

[void sand]

A + B + C = 180 degrees

so A + 2A + 3A = 6A = 180 degrees

so A = 30 degrees

[simple bane]

oh sorry

[void sand]

then we can use sine law

[simple bane]

i get angles and lenghh messed up

but i dont know the length opposite of A

[void sand]

what do you mean

"The side opposite of angle A measures 3 units"

[simple bane]

what would the side name be?

[void sand]

i'd call it 'a'

[simple bane]

is this a right triangle?

[void sand]

this one happens to be yes

is a 30-60-90 triangle i believe

[simple bane]

so

so c would actually be hypotenouse

[void sand]

yes

[simple bane]

c would be = 5, $3^2+4^2=25$

[wraith dagger]

紅卫兵

[simple bane]

.close

[cedar kiln]

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[cedar kiln]

Please don't occupy multiple help channels.

[obsidian umbra]

did i make a mistake?

[slate lintel]

I mean clearly yes

since you don't have a theta on the left

the last step doesn't make any sense, I don't see what you did

[obsidian umbra]

cancelled out the costheta/costheta and sintheta/sintheta

but i think that was wrong

[slate lintel]

you don't have a costheta/costheta

nor a sintheta/sintheta

[obsidian umbra]

ok

i kinda see it now

but i don't know what the next step is

[slate lintel]

if you want to divide the top and the bottom by something you can but it needs to be the whole thing at once

[slate lintel]

wait what are you trying to do

[crimson sedge]

Ever learnt sin(A+B) formula ?

And cos(A+B)?

Apply it

I'll go to sleep

Bye

[obsidian umbra]

verify the trig function

trying to get the left side to equal the right side

[crimson sedge]

And uhh also

What was the question cuz

[devout hamlet]

$sin(3 \pi)cos(\theta ) + cos(3 \pi )sin(\theta ) \neq (cos(3\pi ) + sin(3\pi ))(cos(\theta ) + sin(\theta ))$

[crimson sedge]

Tan(3π + x) = tan x

[wraith dagger]

Yojda

[crimson sedge]

Like based on qudrant

[slate lintel]

your third line was right
what's sin(3π)?

I thought you were going to bed Bromsson

[obsidian umbra]

sin(3pi) = 1?

[crimson sedge]

No

[slate lintel]

,tex .unit circle

[wraith dagger]

Hayley

[obsidian umbra]

im tryna memorize the unit circle 😭

i suck

[crimson sedge]

Just read NCERT and remember the values @obsidian umbra

There's a useful table given

[obsidian umbra]

oh its 0

sin(3pi) = 0

[crimson sedge]

OH DAMNNN YOU READ THE SERIES TOO??

[crimson sedge]

Yeah put it in your eqn you'll get the answer

Like I went to bed and was like how did I miss it

[obsidian umbra]

ok

[crimson sedge]

Yeah

Correct

Now all that remains is -sinx / -cosx

Which is tanx

Goodnight

[obsidian umbra]

thanks

.close

[cedar kiln]

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[crimson sedge]

64 ÷ 8 ÷ 4 = 2 or 32 ?

[crimson sedge]

which is the answer here?

i ended up in this kinda situation when i was simplyifying

[slate lintel]

slightly ambiguous notation but i'd say that's (64 ÷ 8) ÷ 4

but you should probably know how you meant it since you were the one that wrote it i'd think

[crimson sedge]

yeah

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[whole lance]

Just need answers checked

<@&286206848099549185> just need work checked (:

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[gritty viper]

,w integral of ye^(-xy-y) dx from 0 to inf

,w integral of ye^(-xy-y) dy from 0 to inf

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Requested by a_mog

[gritty viper]

Idk why the squared in the denominator went away for f_x

at the end

https://integral-calculator.com might be able to do this

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[nova pendant]

Hi

[nova pendant]

What is the easiest way to do this

All my ways gave crazy numbers

[dry heart]

uh just try to write some composition of functions

like f(g(h(x)))

!nosols

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