help-38

[knotty locust]

if you let u=tan(x)

from what you have in your paper there

you will get u^8(1+u^2)

so try that again

[thorn shale]

integral of u^(8) + u^(10)

which answer is

u^(9)/9 + u^(11)/11 + c?

plug in tan fr u

this is what i got

apparently it's wrong

[knotty locust]

looks correct to me...

oh

it is because you included the +C

[thorn shale]

hmm

[knotty locust]

see your homework has it for you already

on the outside

[thorn shale]

oh yes

it supplies the c for me

thank u!

[knotty locust]

no problem

[thorn shale]

one quick question..

it there a general rule

when u can realize u messed up?

and subbed for something you shouldn't have?

i feel my integrating skills aren't bad.. but my issue is like

identifying the wrong thing

[knotty locust]

my general rule would be if making a substitution makes it looks more complicated you probably shouldn't be making that substitution

[thorn shale]

and then not knowing if im right or wrong then i get stuck

[thorn shale]

got it

[knotty locust]

every substitution you make should be for a reason, to simplify the integral

[thorn shale]

like in my first case

[knotty locust]

also

[thorn shale]

it was a integral of multiplcation

if it was u^(8) + sec^(2)x maybe it wouldve been viable?

[knotty locust]

doesn't look like it to me

because you have now two variables

so you made it more complicated

instead of more simple

[thorn shale]

ahh

[knotty locust]

There is a general method for integrating tan and sec together

I'll put it below

To integrate

[thorn shale]

ty!

[knotty locust]

,, \int{\tan{x}^{k}*\sec{x}^{j}}

[solid kiln]

AustinU

[knotty locust]

1. If J is even and larger than 2, rewrite sec^j(x) as sec^(j-2)(x) * sec^2(x) and use the identity sec^2(x)=tan^2(x)+1 to rewrite sec^(j-2)(x) in terms of tan(x) and substitute u=tan(x).

^ that is the case for the integral we just did

[thorn shale]

breaking it up

i see

[knotty locust]

2. If k is odd and J >= 1 , rewrite tan^k(x)sec^j(x)=tan^(k-1)sec^(j-1)sec(x)tan(x) and use tan^2(x)=sec^2(x)-1 to rewrite tan^(k-1)(x) in terms of sec(x) then let u=sec(x).

3. If k is odd where k>= 3 and j=0 rewrite tan^k(x)=tan^(k-2)tan^2(x) = tan^(k-2)(sec^2(x)-1)=tan^(k-2)(x)sec^2(x)-tan(k-2)(x). It may be necessary to repeat this process on the last term

4. If k is even and j is odd, use tan^2(x)=sec^2(x)-1 to express tan^k(x) in terms of sec(x) and use integration by parts to integrate odd powers of sec(x)

or use the power reduction formulas

[solid kiln]

AustinU

[knotty locust]

and the similar formula for tan you can find on googl

[thorn shale]

ty for the explanatnion!

appreciate it

[knotty locust]

yeah no problem

feel free to DM me if you need more help in the future

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[visual patrol]

Help to find area for the cone and sphere

[rugged latch]

!status

[trim jolt]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[visual patrol]

I have begun but got stuck

[rugged latch]

Can u show what u have alrdy tried?

[visual patrol]

I’ve tried this

Didn’t get the right answer

[rugged latch]

Alright let’s start from the top

[visual patrol]

Alright

[rugged latch]

So first we wanna break it up into 2 figures

The hemisphere and the cone

A hemisphere is just a half of a sphere

So what’s the formula for a hemispheres volume?

[visual patrol]

A=4x3.14 to the power of 2

oh i forgot to r

[solid kiln]

Stephen

[rugged latch]

So what would half of that be

[visual patrol]

157

[rugged latch]

Half of this

[visual patrol]

261.67?

[rugged latch]

No, Im not asking u to evaluate it yet

Just find the formula for the volume of a hemisphere

[rugged latch]

It would be half of this

Don’t plug in any numbers yet

[visual patrol]

3πr2

is that the formula for the hemisphere?

[rugged latch]

3*pi * r^2 ? No

[visual patrol]

the 2 is power btw

[rugged latch]

I’ll show u

So we have $\frac 43 \pi r^3$ as sphere volume. Half of that would be hemisphere volume which is $\frac 43 \pi r^3 \cdot \frac 12$

[solid kiln]

Stephen

[visual patrol]

Alright

[solid kiln]

Stephen

[rugged latch]

Simplify it

It’s just multiplying the fractions

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[charred jackal]

anyone know this

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[charred jackal]

is this right

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[novel flax]

Prove that $2^{n-1} \le n!$ for all positive integers n

[solid kiln]

chromium

[scarlet escarp]

what have you tried chromium

[novel flax]

bru

instant reaction

you're literally spying on me now

not fair

anyway

[scarlet escarp]

i was waiting for y’all to finish cause i like this question lol

[novel flax]

oh, yea so for this question

no induction is allowed

because it's just a different topic

and i have absolutely no idea how to approach this

maybe

splitting the $2^{n-1}$ into $2^n * \frac{1}{2}$

wait what

i'm on crack

[solid kiln]

chromium

[novel flax]

is that right

smth about that seems off

[scarlet escarp]

so we cant do induction?

[novel flax]

nop

did that last year

none of this topic has contained any induction

[scarlet escarp]

if you’ve done it previously it could be considered common knowledge

[novel flax]

and it will always say "prove by induction"

[high heron]

2^n means you've multiplied 2 n times, and n! means you've multiplied all the naturals n or less together

[novel flax]

yes

now what lol

[high heron]

also the bigger of a (positive) number you multiply by, the bigger the result

[novel flax]

ye but i need some formal proof

some algebraic manipulation or smth

[high heron]

so on the lhs you've multiplied n-1 twos, and on the rhs a bunch of numbers all greater than or equal to 2, and then an extra 1

[novel flax]

.close

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[still comet]

Claim

[still comet]

f(x+ 2) = xf (x)
f (2) = 2
f (8) = ?

[marble wharf]

Can you find f(4) first

[jagged phoenix]

what substitutions have you tried

[still comet]

Okay

Wait

Hang on

[still comet]

f(4) should be 4

[marble wharf]

Yes

[loud kayak]

Now find f(6)

[still comet]

Okay

Does f(8) = 8?

Or am I just being dumb

[hazy light]

What did you get f(6) as?

[still comet]

I haven’t done that but okay

[hazy light]

Well, you should find f(6)

[runic raptor]

is function monotonous?

[marble wharf]

Harder question and why is that relevant

[still comet]

6?

[marble wharf]

No

[still comet]

Oh

[marble wharf]

6=4+2

So f(6)=f(4+2)=?

[still comet]

OH YEA, 24

[marble wharf]

No

[still comet]

16?

[marble wharf]

Are you guessing? Share your thought process

[still comet]

In f(2) can = 1+1 , and f(x) = x f(x) , so f(2) = 1(1+1) = 2
Following this f(6)= 4+2 , thus 4(4+2) = 24,
As of typing this I get why this is wrong

.close

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[foggy swift]

Poor Dolly's T.V. has only 4 channels, all of them quite boring.
Hence it is not surprising that she desires to switch (change) channel after every one minute. Then find the number of ways in which she can change the channels so that she is back to her original channel for the first time after 4 min.

The logic they used I am unable to understand it.

[foggy swift]

Wait

[foggy swift]

After 3rd minute, she has 2 choices to switch the channel

But after 3rd minute it should have only 1 choice right

[frozen plover]

I don't think it's necessary that she switches to unique channels every time

Poor dolly only switches channels like a mindless robot

[foggy swift]

🤡.

[plush skiff]

pearson isee

[foggy swift]

Nah Cengage

[plush skiff]

oh yeah

forgot

will be doing that in like 2 months

[foggy swift]

Oh

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[split chasm]

can you do a rough sketch of what you'll have after the transformation?
label the new turning points

[midnight kite]

From the graph you can decipher the equation of f(x) as (x-3)²x

Now f(x+2) + 4 = (x-1)²(x+2) + 4

Now you just have to find k such that the equation (x-1)²(x+2)= k-4 has 3 solutions

You can do that by plotting L.H.S

I don't think so....

Ah wait f(x) = a(x-3)² x for a ig you can use the point (1,4) which is given

Which gives a as 1

You don't need to know much, a cubic is a polynomial with 3 roots, all 3 of which are shown so you can directly write the equation

I suppose you know curve sketching using derivatives?

[split chasm]

yeh

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[subtle prairie]

@wraith hinge

You can take log to base 1/3 in both sides

Or apply change of bases

[wraith hinge]

Since you asked, suppose we take the log: $\log(4) = y\log\left(\frac{1}{3}\right)$

[solid kiln]

45

[wraith hinge]

then we have $y = \frac{\log(4)}{\log\left(\frac{1}{3}\right)}$

[solid kiln]

45

[wraith hinge]

And using the change of base rule, we can combine this fraction into one logarithm.

That is, $y = \log_{\frac{1}{3}}(4)$

[solid kiln]

45

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[wraith hinge]

i am confused about what the following is asking

[wraith hinge]

specifically the end of it

like what is f[s] whats that notation even

yeah no im sorta struggling to even get what the question is asking to prove

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[cunning cliff]

f[S] probably means the image of S

f[S] := {f(s) : s in S}

Hopefully this notation should have been defined beforehand.

I would consider drawing a diagram to understand the question

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[broken cave]

I m doing any kind of mistakes till here?

[drowsy dirge]

,w rotate

[broken cave]

,rccw

[broken cave]

,w 5481408÷21

[solid kiln]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[drowsy dirge]

Not until now

But wait

Yes.

One place.

[broken cave]

Some kind of mistake is there in my work

But how

I am not able to figure it out

[drowsy dirge]

Only one digit can go down.

Not two.

So you have to add a zero in the quotient to bring down two digits

[broken cave]

Oh that's a rule

[drowsy dirge]

That's how it works.

basically

You could do

[broken cave]

I never tackled that kind of rule I don't know why how

Thanks

[drowsy dirge]

21 * 0 there

And then

4 would remain

[broken cave]

Hmmm

[drowsy dirge]

and the 0 would go down from the dividend

[broken cave]

Got it thanks

[drowsy dirge]

In the next step.

Welcome!

[broken cave]

.close

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[broken cave]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 < 1×1

[solid kiln]

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[earnest breach]

Did I do this correctly so far?

Or do I also need to multiply the n in the denominator by the conjugate?

[charred storm]

idk you split up the fraction in a weird way imo

[subtle prairie]

I feel like the first step is sort of off

[charred storm]

i would just use the conjugate $\frac{1}{\sqrt{n+1}}+1$

[solid kiln]

Køter

[earnest breach]

oh

wait

I see my mistake

[subtle prairie]

Or you can directly use limit rules, ie apply to both numerator and denominator later inside the root

[earnest breach]

why would you do that? The problem right now is the denominator

[subtle prairie]

If you do so, you will se how it get cancels off

[earnest breach]

I'll try it

[earnest breach]

Like this right?

[subtle prairie]

Oh it’s n to zero

My bad, it thought it’s infinite limits

[earnest breach]

ah

[subtle prairie]

Either way it won’t work, only if x^2 was inside root then it will work for infinite too, mb

[earnest breach]

.

Not looking so good!

Should I multiply by the reciprocal ?

Not sure what to do with the n yet

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[earnest breach]

<@&286206848099549185>

.close

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[rapid kiln]

do ye help with physics here?

[rugged latch]

If someone wants to, they can

Post ur question

[rapid kiln]

i seem to be getting this right

idk whats happening

its a very easy question

[rugged latch]

Alright so what did u do first

[rapid kiln]

converted 77F to C

i got 25

then took 3 away since its the difference

then converted 22C to K

by adding 273.15

but it seems to be wrong

[rugged latch]

Hmm yes I did the same thing and got ur answer, but when I did it the other way around, I got a different answer, which is kind of nonsensical

I’m not sure why that would be

[stoic saddle]

if I said that 3 degrees c is 276k does that make it clearer?

a one unit change in celsius or kelvin are identical, the only difference are the zeros

[rapid kiln]

so do i convert everything to Kelvins then do the questoin

[stoic saddle]

not required

[rapid kiln]

what did you mean by this then

[stoic saddle]

just the premise that you are working out the difference between 295 and 276

or 22 and 3.

[rapid kiln]

yea but it still get the same answer

[stoic saddle]

295k?

between 295 and 276?

[rapid kiln]

.close

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[violet nest]

what formulas do i use to get answer? i used online calculator to get this but dont understand how

[long basin]

,tex \linearization

[solid kiln]

Umbraleviathan

[long basin]

But like

You don't even need a formula

Not everything needs a formula; if you understand the relationship between a function and its derivative, as well as properties of a tangent, then a formula is not necessary

[violet nest]

i dont understand relationship between function and derivative

[long basin]

The derivative of a function is the slope of the function

[violet nest]

or props of tangent

[long basin]

And a tangent line intersects curve only once, locally

That's by definition

If you've done tangents with circles

Or even tangent by dictionary definition

Same applies

And then the derivative is the slope of a function. At a location, x_0, f'(x_0) will tell you the instantaneous slope of f(x) at x = x_0

So you know the tangent must only intersect at a singular point

[long basin]

And by extension of the statement I responded to, this is also the slope of the tangent line

Point slope form is gonna carry you

[violet nest]

i do not understand "x_0, f'(x_0) will tell you the instantaneous slope of f(x) at x = x_0"

what is x_0

[long basin]

Some variable

$x_0$

[solid kiln]

Umbraleviathan

[long basin]

Or rather

A random constant

[violet nest]

f'(x_0)

[long basin]

The derivative evaluated at x_0

[violet nest]

is that a place on a graph?

[long basin]

No it's literally an evaluation of the derivative

Like

If f(x) = 2x^2 + 5x

What would f'(2) be

[violet nest]

16 + 10

26

[long basin]

No that's f(2), not f'(2)

So you're not even familiar with the notation...

Hm

$f'(x) = \frac{d}{dx} f(x)$

[solid kiln]

Umbraleviathan

[violet nest]

what is f'(x)

[long basin]

I told you

f'(x) is the first derivative of f(x)

[violet nest]

f'(x) is derivative formula?

[long basin]

No, that's just how we notate it

Or the many ways we notate a derivative

You gotta get familiar with the notations

I'm very concerned that you're doing linearizations, but aren't familiar with notations or fundamentals. You have any notes you took? Maybe what the teacher's been saying has been janky

[violet nest]

i was sick and missed class so i have to try to learn myself

[long basin]

But the underlying aspect of the derivative is that the derivative is the slope of a function. When evaluated at some location, x = x_0, then it tells you the instantaneous rate of change.

I'd go on Khan Academy and view their intro to derivatives

Because it seems you've missed a fuck ton from being sick

Which sucks. Your teacher should've helped you with catching up

[violet nest]

before i go check out khan. can you tell me what numbers i put into derivative formula?

what numbers mean what variables in the formula

[long basin]

x0 is the x-value of the point of tangency

[violet nest]

im confused where i put what numbers

[long basin]

Analyze it a bit

I have to go

[violet nest]

.close

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[strong horizon]

can anyone help me with part 2?

[strong horizon]

Sort of hard to understand what's being asked

Please, anyone clarify. 😦

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[strong horizon]

.close

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[plucky arch]

This is the solution

But i did something which is wrong, and i don't know why.

Number of way of selecting $C^5_{1}*C^n_{1}*C^{4+n-1}_{1}$

So this $C^5_{1}*C^n_{1}*C^{4+n-1}_{1} = 1750$

[solid kiln]

Samarth

Samarth

[plucky arch]

Which gives quadratic $n^2 + n - 350= 0$

[solid kiln]

Samarth

[plucky arch]

I don't know why my solution is wrong

[wraith hinge]

@plucky arch i think its wrong because when u are doing it ur way

You are selecting one boy and then selecting another boy from a reduced group

Not sure lemme think

5C1 x 4C1 is different from 5C2 isn't it

[trim jolt]

@plucky arch Has your question been resolved?

[plucky arch]

I am selecting 1 boy 5C1 , and than 1 girl nC1 and than both boy and girl together 4+n-1C1

[blissful viper]

Yo Samarth

[plucky arch]

@wraith hinge

Yo san

[blissful viper]

R u Nepali

[plucky arch]

Indian

[blissful viper]

Thats my brother's name haha

[plucky arch]

4+n-1 because there are now 4 boys left and n-1 girl left

[wraith hinge]

No thats not the same

[plucky arch]

Cool haha, small world

[wraith hinge]

When u pick one boy and then another boy

U are inherently ordering or arranging them

Or when u pick one girl and then another girl

It's not the same as pickin 2 girls or 2 boys from a group

[plucky arch]

Oh so it's like permutations.makes sense

[wraith hinge]

Yes

Here u are not worried about the order

[plucky arch]

Yeah correct.

Thank you

[wraith hinge]

[plucky arch]

.close

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[wraith hinge]

Aaaaaa

[wraith hinge]

My mind is going brr

I am not very familliar with deravatives

Only know basics of it

But as far as i know

Integration is area under graph

how does area under the graph equals to sum here

[frozen plover]

Don't think too much about it

Also I did warn you a few days ago you're gonna have to integrate

Only definite integration finds area

[wraith hinge]

[frozen plover]

Okay that doesn't help since this is definite integration

[wraith hinge]

What about the limits

[frozen plover]

Pick whatever limits are convenient for you

[wraith hinge]

How do we decide from what to where we gotta integrate

[trim lichen]

oh, generating function fuckery

[frozen plover]

Depends on the question

[trim lichen]

i don't think you should be looking at this through a geometric light @wraith hinge

[wraith hinge]

Oh hmm

[frozen plover]

Since in this case you want the binomial coefficients only, we integrate from 0 to 1

[wraith hinge]

Show me da light

[frozen plover]

Integration is just convenient here since it divides the polynomial by the new power

[wraith hinge]

Bt Y

[frozen plover]

It's the best way to solve the question

[wraith hinge]

Y not 0 to n

[frozen plover]

You can do that

[wraith hinge]

[frozen plover]

But then you're gonna put n = 1 anyways

[wraith hinge]

[trim lichen]

because integrating from 0 to n doesn't get you the expression you want

[wraith hinge]

bt there should be a logic behind it that i am not getting

Like we are just integrating to generate an expression

[trim lichen]

yes

yes we are

[frozen plover]

I think you have it in your mind that integration is solely for calculating area

[wraith hinge]

Bruh

[trim lichen]

we are indeed just integrating to get the expression we want

$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$

[solid kiln]

Ann

[frozen plover]

Like Ann said, don't think of it geometrically

[wraith hinge]

This is again cheating

Like

[trim lichen]

no, this is not cheating.

[wraith hinge]

Aisudhsnamajdhdbs

[trim lichen]

this is a perfectly legitimate method.

[frozen plover]

No it's not, Newton died with no bitches for this

Be grateful

[trim lichen]

generating function fuckery is commonplace in math and esp. combinatorics

[wraith hinge]

yes yes why not

[frozen plover]

I like the ring of that

Generating function fuckery

[trim lichen]

well, generatingfunctionology is a book title

[wraith hinge]

Welp very new for me as i have never seen integration abused like this except in physics but itss okkkk

[trim lichen]

this is not abuse

[frozen plover]

It's ingenious

It says Method I, why don't you look at Method II

[wraith hinge]

Flip

[frozen plover]

Oh wait it is Method II

Yeah

[trim lichen]

i do suggest the very aptly named book Generatingfunctionology by the way

i've heard good things about it.

[wraith hinge]

Hmm leme see method 2

[frozen plover]

Is it a textbook?

[wraith hinge]

Oh hmm how much it costs tho

[frozen plover]

There's an online PDF for free

[trim lichen]

libgen

[wraith hinge]

HOLY MOLY

Lilke

100 bucks

[trim lichen]

libgen that shit

no seriously

[wraith hinge]

Wt is libgen

[frozen plover]

Here you go

[frozen plover]

You seem new to the internet

[wraith hinge]

Well my internet is just limited to youtube, discord and whatsapp

Nothing else

Nothing more

First method is this tho

Using identity

Very simple to understand

[frozen plover]

Go for whatever you think is best for you

I don't know much about the properties of binomial coefficients myself

Calculus is king

[wraith hinge]

Queen of science

[frozen plover]

That's mechanics wdym

[trim lichen]

barred access to everything else?

like you actually physically cannot visit any other websites?

[wraith hinge]

Well i have limited myself to these places only

[trim lichen]

so it's a self-imposed limitation.

and you refuse to lift it from yourself?

[frozen plover]

I'll just hook him up in DMs or whatever :/

[trim lichen]

heres the thing lmao

it's only 2MB

[frozen plover]

Oh well

[wraith hinge]

Yes, acc to me Instagram, facebook, xyz platform and other social media platforms, most of them atleast just wastes your time making reels whole day and what not smh

[frozen plover]

How allowed is this on the server

[wraith hinge]

Downloaded it 👍

You can delete it now if you want as someone might see it

[trim lichen]

i don't care.

[wraith hinge]

And may complain

[trim lichen]

ok, sure. but the internet by itself is not limited to that.

[wraith hinge]

Very well

[trim lichen]

according to your own words, you restrict yourself from wikipedia as well.

why do you restrict yourself from wikipedia?

[wraith hinge]

Well i am not afraid to visit other platforms including wikipedia, but yes i trust textbooks more than wiki as everyone is allowed to manupulate data there. So yeah

[trim lichen]

okay.

so you place some trust in textbooks.

[wraith hinge]

Reputed ones precisely

[trim lichen]

https://en.wikipedia.org/wiki/Library_Genesis

if you know your textbook's title, you can find it on libgen.

and not have to pay a penny for it.

[wraith hinge]

Wait

Wha

[trim lichen]

The site enables free access to content that is otherwise paywalled or not digitized elsewhere.

you know where i got that pdf of generatingfunctionology?

i got it off libgen, just now, shortly before i uploaded it here.

[wraith hinge]

👁️👄👁️ free books

[frozen plover]

"shadow library"

Sounds like something out of Kung Fu Panda

[wraith hinge]

Why did i get so many books then smh

[frozen plover]

I couldn't

[wraith hinge]

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[mortal gale]

Stuck on another Complex problem

[mortal gale]

For Vi ,I tried rearranging to get mod(z-1/z-3)=4 but not sure where to go from there

I understand how |z-1| is the length of line segment between complex number z and 1 on around diagram

Not sure where to go

[kindred pier]

Consider a more brute-force approach solution

Let z=x+iy

Then consider what the problem would look like

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@mortal gale Has your question been resolved?

[mortal gale]

Ah I see

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@mortal gale Has your question been resolved?

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[orchid canyon]

Hello, I am looking for a way to prove that the orthocentre of a triangle is outside of the triangle when the triangle has an angle greater than 90 degrees, specifically using vectors. I have used desmos to see that it happens in all cases, but I do not know how to prove it. For the task I am doing, i need to do it making use of vectors. Thanks

[orchid canyon]

(stolen from quora but shows what i mean)

[inland orbit]

Orthocenter?

In your picture, you've constructed the circumcenter, not the orthocenter

The orthocenter is the intersection of the altitudes

The circumcenter is the intersection of the perpendicular bisectors

[orchid canyon]

my bad, either one works. I need to prove if either the orthocentre or circumcentre are outside, so either is fine

[inland orbit]

Hmm, okay

[trim jolt]

@orchid canyon Has your question been resolved?

[trim jolt]

@orchid canyon Has your question been resolved?

[trim jolt]

@orchid canyon Has your question been resolved?

[orchid canyon]

😦

[trim jolt]

@orchid canyon Has your question been resolved?

[acoustic flare]

Greetings from a random guy~ (who as well have not received help after posting a question lol)
So firstly I am sorry that I do not have relevant knowledge about vectors, but I can try to solve it with the use of coordinate geometry and you can see whether you can gain inspiration~

Letting origin O be one vertex of the triangle, and extending it towards the x+ direction we have the second vertex of (x1, 0). Let us name this point A and the third vertex named B.

Now let <AOB be the obtuse angle. With the use of simple graphical manipulation, we can observe that B is located in the top left quadrant. Hence, the altitude at B does not intersect the triangle at anywhere except B itself!
And let's consider the altitude at A and O. However, as there is no right angle in the triangle, we can deduce that both altitudes do not intersect the triangle at B. Hence the intersection of 3 altitudes, which is the orthocentre, must occur outside of the triangle.

The case will be the same if you let the other 2 angles as the obtuse, because you can simply rotate the triangle to fit the original model

[orchid canyon]

Thanks 🙂 that helps

[orchid canyon]

.close

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[drifting rose]

this is a really easy quesiton

[drifting rose]

2B thats it

I feel like I know what the answer is but the answer page shows such a wacky number I feel like its a mistake

:/

<@&286206848099549185>

[kindred pier]

!show

[trim jolt]

Show your work, and if possible, explain where you are stuck.

[drifting rose]

oh uh

since its reflected across the y axis and horizontally dilated, the scale factor in the equation would be

-1/3x^5

but its not -1/3

its -1/243 for some reason

.close

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[trim lichen]

!onechannel

[trim jolt]

Please stick to your channel.

[trim lichen]

closed the other one

also, don't just post&dash

[frozen plover]

$\Delta = \frac12 ab \sin C$

[solid kiln]

NEONPerseus

[frozen plover]

Are you aware of this formula?

[trim jolt]

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@sonic tartan Has your question been resolved?

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[broken cave]

What is √3/2 in terms of sin

Like $\sin^{-1}(√3/2)$

[solid kiln]

Arnab

[broken cave]

Write it in terms of π btw

So
$\sin(π/12-x/4)=\sin(π/3)$

[solid kiln]

Arnab

[broken cave]

Now just cut the sin's off

π/12+x/4=π/3

,w π/12-x/4=π/3

[solid kiln]

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Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[broken cave]

My fault

It is -π

So x=-π

Oh

Sin is positive in which quadrants?

[broken cave]

So this could also be = 90+30 which is 120

π-30

$\sin(π/12-x/4)=\sin(π-60)$

,w sin(120)

Wtf

[solid kiln]

Arnab

[broken cave]

Now alright

Ok so

We are not writing it as sin(π/3)

Cause it would give a negative value for x

We are just writing it as π-60

Which is ultimately 120° and sin(120°)=√3/2

Which satisfies the conditions

,w π/12-x/4=π-60

[solid kiln]

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Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[broken cave]

Yeah it is a positive value

Also 2π/3 is also possible

$\sin(π/12-x/4)=\sin(2π/3)$

[solid kiln]

Arnab

[broken cave]

You can write this too

Welcome

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[frank silo]

$\frac{dy}{dx} = y^2 -4$
I'm trying to solve this ODE but I don't know how to solve for y

[solid kiln]

Sean C

[frank silo]

its separable so I move the $y^2 - 4$ to the other side and integrate both sides

[solid kiln]

Sean C

[frank silo]

after integrating I get $\frac{y-2}{y+2} = Ce^{4x}$

[solid kiln]

Sean C

[light crystal]

Write y - 2 as (y + 2) - 4.

Then (y - 2)/(y + 2) = 1 - 4/(y + 2). You can then simplify and write the explicit solution.

[frank silo]

ok I will try then report back

[light crystal]

,w y' = y^2 - 4

[solid kiln]

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Requested by stabulo#9301

[trim jolt]

@frank silo Has your question been resolved?

[frank silo]

My textbook has this as the answer

(28)

my initial condition is $y(0) = -6$

[solid kiln]

Sean C

[frank silo]

I think I did something wrong for the first part? ie. $\frac{y-2}{y+2}$ or $Ce^{4x}$ might be wrong

[solid kiln]

Sean C

[frank silo]

I'm gonna work though it from the start then post my work

[frank silo]

This is the work I have, I'm not sure where to go from here

<@&286206848099549185>

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[slim stone]

Hello I’m trying to solve for bc, I know the law of sines and cosines and know it likely used here but I’m stuck

[trim jolt]

@slim stone Has your question been resolved?

[slim stone]

<@&286206848099549185>

[tepid wasp]

easy

so easy dude

use pythagoras theoreme

[slim stone]

it is not 90 degrees

my drawing is pretty bad so it might look like it

[tepid wasp]

it is?

[slim stone]

it is not 90 degress

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@slim stone Has your question been resolved?

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@slim stone Has your question been resolved?

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[full dock]

Can someone walk me through the steps of completing the square?

[full dock]

4x^2 + 9x - 12 = 0

I know you half 9 and square it but I’m not doing the whole thing correctly

[serene latch]

@full dock divide by 4

and do that

divide everything by 4

[full dock]

What

Why

Then we get decimal for 9

[serene latch]

@full dock that's fine

[long basin]

Completing the square works nicely if the leading coefficient is 1. Dividing by 4 enforces that that leading coefficient will be 1

As a result, the x-term will have a coefficient of 9/4, but that shouldn't be an issue nor concerning

[full dock]

How about something like 3x^2 + 7x - 12 = 0 where we can’t divide anything

What should we do

[serene latch]

What do you mean can't divide anything

[long basin]

Divide by 3

3x^2 + 7x - 12 will have the same roots as x^2 + 7/3 x - 4

[full dock]

I was never taught to divide the equation first

[full dock]

Can we do one where you don’t divide the equation

[long basin]

Won't be fun

Because then you'd have to divide 9 by 4 anyways instead of halving it

Might as well divide by 4

[full dock]

I was watching completing the square on organic chemistry tutor

and he never really divided the equation first unless it was something like

3x^2 + 3x - 9 for example

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[stoic sky]

Need some calc help

[stoic sky]

I need help converting the y values to x

since its bounded about the y axis

lmk if anyone couldhelp?

hello?

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[solid kiln]

yomiko

[slim knot]

so for y

[solid kiln]

yomiko

yomiko

yomiko

[slim knot]

but they restrict the graph at -2<y<2

how does that work?

[median radish]

y = 2 is the upper bound

[slim knot]

what about the lower bound?

@median radish

[median radish]

the value of the function reduces between t = 0 and t = pi

[slim knot]

but when its either 0 or pi its 2

[median radish]

so, even if the function is equal at t = 0 and t = pi doesn't mean that it's range is just 2.

[slim knot]

how would i know if its not just 2?

[median radish]

because function has value which are less than 2

for eg, for t = pi/2

y = -2

[slim knot]

oah

[median radish]

range refers to all the values the function can output

so there's no reason that evaluating the function at initial and final bound of the domain would give you the range of the function.

[slim knot]

so how would i find the range of the function if i don't know the shape?

i would have to just draw it out first?

[median radish]

well, for the standard functions like sin, cos and tan, you can have it memorized. For sin and cos, range is [-1, 1]

[slim knot]

the range would look like this?

[median radish]

yip

[slim knot]

yipee

thanks

:)))))))

.close

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[karmic ocean]

Doing this end of topic assessment and theres some notation ive not seen before. It states v={ and then a couple equations and inequalities, whats it saying? Are these all things that apply to v? In which case t has to be something specific rather than a changing variable?

[trim lichen]

look up piecewise functions

[karmic ocean]

Ohhhhh ok

So its equal to the first equation, when t is between 0 and 4, but the second equation when t is grater than 4

[median radish]

yes

[karmic ocean]

Ah ok sweet, cheers guys! ❤️

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[jade swallow]

can someone explain to me how to do the problems on this sheet? I’ve done a little less than half but I don’t know where to go from here

[granite cove]

For which one do you need help?

[latent meteor]

guessing the unanswered ones lol...

[jade swallow]

Well all the blank ones really but primarily I think 7, 10, and 16

[granite cove]

What's ln 1?

[jade swallow]

0?

[latent meteor]

0?

ln 2x+3 = -ln3

im guessing

wait

no?

[granite cove]

Yeah

[latent meteor]

im guessing

[granite cove]

Yes. Well done. But let the one to open the channel solve these

[latent meteor]

my fault

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[wraith hinge]

Let the action in Hamilton's principle is given by
$$
S(\mathbf{q})=\int_{t_0}^{t_1} L(\mathbf{q}(t), \dot{\mathbf{q}}(t), t) \mathrm{d} t
$$
and the trajectory of the system is constrained to lie in a surface ${Q}$. What is the direction of $\delta S / \delta \mathbf{q}(t)$ relative to the surface ${Q}$? I have a hard time trying to understand what the problem is asking for. Isn't it just perpendicular to the normal vector of surface $Q$?

[solid kiln]

Amateur Mathematician

[zinc ginkgo]

take a picture or screenshot of the question

[wraith hinge]

this is the question

[wraith hinge]

except the last two sentences

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@wraith hinge Has your question been resolved?

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[slow sonnet]

How do you do question d??????

[lilac wind]

solve for t

[slow sonnet]

$t= 6cos^-1(d(t)-1/7.5)/pi$

[solid kiln]

Dracu

[slow sonnet]

what does that mean physically?

At a particular time there is is a certain depth of water?

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[wraith hinge]

Relatively simple question that I keep messing up.

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@wraith hinge Has your question been resolved?

[verbal gulch]

okay so what messes you up

what have you done so far

[wraith hinge]

nothing

I keep of cancelling h

[by division]

and getting g-g

Which is 0

Then there is no g value to subject...

<@&286206848099549185>

I need some help

[zinc ginkgo]

subtract gh from both sides first

[wraith hinge]

Ok Thank you

.close

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[twin mortar]

Can anyone tell me why \frac{d}{2}-d is -d/2. I always end up with 1/2

[slender shard]

If you multiply -d by 2/2 to make the denominators equivalent, you should see why.

[keen axle]

if you have half of a thing and take away the thing you get negative half of that thing

[twin mortar]

but how is this just not 1/2

[slender shard]

$\frac{d}{2} - \frac{2d}{2}$

[twin mortar]

or even just 0

d-d is 0

[solid kiln]

Kookiemon

[twin mortar]

so you have to multiply the -d with 2?

[slender shard]

Yes, the denominators need to be equivalent before you can add or subtract fractions.

[twin mortar]

it says here its this tho

how does -d/2 become positive?

[slender shard]

Can you show the whole thing?

[slender shard]

It did not become positive. The negative symbol was moved out front and became the minus sign.

[twin mortar]

ohhhhhhh

thanks for the help

[slender shard]

yw

[twin mortar]

.close

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[wraith hinge]

How did they simplify?

[lament jewel]

um

[wraith hinge]

Like how did they move the y down

[lament jewel]

they multiplied it by 1=y/y

so the numerator and denominator is multiplied by y

u can do this when y is non zero

[wraith hinge]

oh i see. thanks

[lament jewel]

np

💕

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[silent kiln]

guys

I have no idea how to start this

im going to cry

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[silent kiln]

.close

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[unkempt jay]

Hi having a bit of trouble with this question: An apiary of honey bees increases its own numbers by 5%. After the births occur 50 more bees are added to build the population. p_t is used to denote the size of the honey bee's population after t years how I find that value of t when the population is 1500 bees given that p_0 is 500?

[unkempt jay]

Here's the actual question

or more simply how do i solve for t in the equation p_t+1 = ap_t + b I can do the rest

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[supple fiber]

im confused here is there a theorem that could help

[willow vessel]

its a paralleogram

think about this

imagine a paralleogram

[supple fiber]

all parallelograms r equal to 360?

[willow vessel]

what is r?

[supple fiber]

part of 19?

[willow vessel]

no

draw a paralleogram on a page

mention its sides PQRS

and also mention the length of sides given

[willow vessel]

now think what is length of RS now?

[supple fiber]

23?

if its congruent to pq

[willow vessel]

yea

simple

[supple fiber]

OHH

mb

thanks

.close

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[twin mortar]

Can someone tell me what im doing wrong?

[twin mortar]

you have to simplify it

i ended up with 1/2x+2

but its not any of the alternatives

i know the answer is 1/2x+1

but i get 1/2x+2

[midnight kite]

Well you can take 3 common from the numerator and cancel it with the 3 below

You get 2x+1 in the numerator

[midnight kite]

$6x +3 = 3(2x +1)$

[solid kiln]

hanzo

[midnight kite]

Check

You factorised the denominator wrong as well, or even wrote it wrong in the first line

[twin mortar]

how was it wrong

[midnight kite]

You did 4x² + 4x + 1 = 2x(2x+2)

Try multiplying that and see if it matches the L.H.S

[twin mortar]

wdym

oh

i dont understand

[midnight kite]

You even wrote it as 4x² + 4x +x there, it doesn't match with the question

[twin mortar]

i wrote 4x^2+4x+1?

it looks like an x because i do a line to see what i took away

if you see

the 1 above is also crossed with a line

because i took away 1 for 1

what is says on the most above is this

the second one it says this

and the third

@midnight kite

but the answer is

which means i have +1 too much

[twin mortar]

i ended up with this

[midnight kite]

Look you can't just cancel 1 and 1 like that

You can only cancel things from numerator and denominator that are in multiplication with the complete expression

Like you cancelled 3 from numerator and denominator, it was multiplied to the complete numerator and the complete denominator

[twin mortar]

so if its + or - i cant cancel or what

[midnight kite]

You can't, its not addition or subtraction

[twin mortar]

so only if its divided or multiplied with something?

[midnight kite]

$\frac{3+2}{2+4} = \frac{3}{4}$

[solid kiln]

hanzo

[midnight kite]

If we did like your method we will end up like this

As you can see its not correct

[twin mortar]

Yeah

[midnight kite]

Yes

So now lets try to factorise the quadratic expression in the denominator

$4x²+4x +1$

[solid kiln]

hanzo