help-42

[tidal grotto]

$\sqrt{4-(x+1)^{2}}$

[potent lotus]

Moosey

[tidal grotto]

u-sub with x+1, then trig sub

[dire lantern]

oh I didn't learn trig sub yet

I am pretty sure

my teacher set this question up

with us not knowing how to do trig sub

Is there another approach?

[tidal grotto]

ok

you can do u-sub (complete substitution)

u=sqrt(1-x)

[dire lantern]

wydm by complete substitution

[tidal grotto]

solve for x

[dire lantern]

ohhh

[tidal grotto]

and substitute it for every instance of 'x'

you way want to find du though first

[dire lantern]

alr

i have

$u\sqrt{x+3}$

[potent lotus]

Fenagon

[dire lantern]

$u\sqrt{-u^{2}+4}$

[potent lotus]

Fenagon

[dire lantern]

and then with du instead of dx

$2u^{2}\sqrt{-u^{2}+4}$

[potent lotus]

Fenagon

[hollow totem]

what happened here

[dire lantern]

well I asked for a different approach to solve the integral

and then I did what he told me to

[hollow totem]

It seems like you did u=sqrt(1-x)?

[dire lantern]

yeah

[hollow totem]

Oh I see

[dire lantern]

From here, what should I do? By Parts?

[hollow totem]

No because then you still have to integrate sqrt(-u^2+4)

I got nothing other than trig subs

If it were a definite integral with very particular bounds there might be another way

[dire lantern]

dang

[hollow totem]

The only other approach I could imagine is to literally sketch out the area determined and then use geometry to determine the area

which would work but it's not a simple area

[dire lantern]

lol its probably not that, he did say this question was a very very special/difficult case

[hollow totem]

Then it's probably a trig sub

[dire lantern]

alr thanks then

[hollow totem]

u = sin(θ) I think if you're curious

cos(θ) would possibly also work

[dire lantern]

wydm

[hollow totem]

I mean it is just normal substitution

It's just a really weird one

[dire lantern]

I didn't learn trig sub

how does it go

[hollow totem]

Well it's just normal substitution

You're substituting sin(θ) for u

it's not "u-substitution" only because the variable name u is already in use

[dire lantern]

oh

then I could technically use sin(v)

or something like that?

its just substitution with trig functions?

[hollow totem]

Yeah

It has its own name because it shows up surprisingly often and it looks very weird at first

[dire lantern]

wait so how would it work out for this question

[hollow totem]

I think we should use u=x+1

[hollow totem]

Then we end up just getting $\int\sqrt{4-u^2}~du$

[potent lotus]

thewizardofOU

[hollow totem]

after that we do u = sin(v)
du = cos(v)dv

oops

my bad

u = 2sin(v)

du = 2cos(v)dv

$\int \sqrt{4-4\sin^2(v)}~\cos(v)~dv$

[potent lotus]

thewizardofOU

[hollow totem]

Note the subtlety of making this far more convenient by writing u in terms of v, not v in terms of u

[dire lantern]

You used this one right

[hollow totem]

Yeah

Oops

$\int \sqrt{4-4\sin^2(v)}~2\cos(v)~dv$

[potent lotus]

thewizardofOU

[dire lantern]

so

would this one come out to be

2cos^2(v) dv

multiplied by 2

[hollow totem]

Yes

[dire lantern]

what happens afterward

[hollow totem]

then you need to use 2cos^2(v)-1 = cos(2v)

[dire lantern]

oh I see

so 2sin(2v) + 2v + C?

[hollow totem]

close

the integral of 2cos(2v) is not 2sin(2v)

[dire lantern]

oh wait I made a mistake earlier

it should be sin(2v) + 2v + C

[hollow totem]

Correct

[dire lantern]

oh I see how the arcsin comes in

thanks a lot

.close

[calm coral]

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[restive dove]

This is on an answer key my pre cal professor posted from an old quiz, why is that false? cause -5 x -5=25?

[velvet osprey]

sqrt(25) means only the positive square root

[restive dove]

when would it be plus or minus?

[velvet osprey]

as in

"when would the radical symbol refer to both the positive and the negative root at once?"

is that what you are asking?

[restive dove]

yes

[deep pike]

Assume x=√25
x-√25=0
Since it's a linear polynomial
It can have utmost 1 root
By convention we take the positive one

Basically √(x²)=|x|

However if the equation is x²=k² where k is a constant
Then x=±k

[restive dove]

ahhh alright

so should I just take the positive one in most cases?

[deep pike]

Square root always gives a positive result
Not most of the cases.. its always

As long as it's defined of

[restive dove]

ah okay thank you

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[supple basin]

how do i do b, c, and d?

[velvet osprey]

differentiate the series that defines f(x) termwise

[calm coral]

@supple basin Has your question been resolved?

[calm coral]

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[remote mural]

hello everyone

I'm having quite a big problem with this question

first thing its centered at x and not c

secondly

its in a weird fraction form los cosx-1/x^2

This my attempted work

[pseudo wedge]

why is the second term squared?

$f(x)=1+\frac{0}{1!}(x-0)-\frac{1}{2!}(x-0)^2+\frac{0}{3!}(x-0)^3...$

[remote mural]

I thought you had to put them over like factorials

[pseudo wedge]

oh yea mb

[potent lotus]

kheerii

[remote mural]

and mb for writng them wrong my brain is fried

anyways

I can get the nth sum for it

but once I find the taylor series for cosx

what do I do fr

this question completely got me lost

[pseudo wedge]

just input it in place of cosx

[remote mural]

yes but will it be in terms of n?

like

here

how do I get rid of the n's

so I can input it and get a real value

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

.close

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[calm coral]

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[normal wasp]

in a quadriteral the three sides are 2cm 3cm 4cm also inscribed circles radius is 1,2cm i need to find the area of the quadriteral

[normal wasp]

.close

[calm coral]

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[remote mural]

is this not root 17

[remote mural]

apparently the answer is 3

nvm

.close

[calm coral]

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[remote mural]

F(x) = sin (x to the power of 2). E to the power of cos (2x)

[remote mural]

Help?

[potent igloo]

what's the question?

[remote mural]

..

[potent igloo]

This is a statement. This is not a question

[remote mural]

Idk my brother needs help

He’s giving me money

[cloud topaz]

[\map Fx = \sin(x^2)e^{\cos(2x)}]

[potent igloo]

Why doesn't he just ask himself on here?

[remote mural]

Find the derivative of the function

[potent lotus]

Pure

[remote mural]

How should I know

Ask him

[potent igloo]

Okay put him on

!status

[calm coral]

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

[remote mural]

His user is jacob09tam

No

It’s Itz2ezybro

[potent igloo]

he should join this server and ask his question

[remote mural]

Ok then bye

[remote mural]

Ikr such a lazy guy

[potent igloo]

sounds like he got banned if he can't join

[remote mural]

That’s mad

.close

[calm coral]

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[calm coral]

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[calm coral]

@simple stump Has your question been resolved?

[calm coral]

@simple stump Has your question been resolved?

[calm coral]

@simple stump Has your question been resolved?

[simple stump]

.close

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[tired comet]

i came up with this problem myself im just curious what a potential solution might look like

[calm coral]

@tired comet Has your question been resolved?

[tired comet]

<@&286206848099549185>

[tired comet]

can someone help

[fathom shuttle]

May i direct you to discord.gg/mods

They are probably better equipped for this

[tired comet]

ok

[calm coral]

@tired comet Has your question been resolved?

[calm coral]

@tired comet Has your question been resolved?

[calm coral]

@tired comet Has your question been resolved?

[calm coral]

@tired comet Has your question been resolved?

[limber nacelle]

lol

[white atlas]

[limber nacelle]

orz ppl in MODS

[white atlas]

i joined so i'm orz ppl now

[limber nacelle]

yes

you will win GOLD in 2024 IMO now

[white atlas]

did you win any considering you're there too?

[limber nacelle]

No ?

lol

im not that gud

[white atlas]

2024 IMO gold incoming then

[limber nacelle]

idk if ill even make it to PRE tst

anyways cya

[calm coral]

@tired comet Has your question been resolved?

[fickle musk]

why isn't my brain functioning

[fervent tendon]

take a nap

maybe u burned out urself

[whole hinge]

Please don't have an offtopic conversation in this person's help channel

[tired comet]

nobody answered the question

[calm coral]

@tired comet Has your question been resolved?

[tired comet]

.close

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[remote mural]

How could i start to solve this last row only using the ctg data?

[dapper sable]

Well, how is tangent related to cotangent?

[remote mural]

uuuhm its reciprocal to it?

[dapper sable]

Right, so take the reciprocal of the cotangent to get the tangent.

After that, you have the fact that taking the tangent of an angle gives you the slope of the angle.

So, like tangent of 45 degrees shows that the slope is 1.

[remote mural]

So should i 1/tan 1.2345 ?

[dapper sable]

1.2345 isn't an angle, so you can't take the tangent of it.

[potent lotus]

Chai T. Rex

[dapper sable]

Fill in cot(alpha).

Then use a calculator.

Or do long division.

Whatever.

[remote mural]

Oh so it was this simple all along, thank you!

.close

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[clever hazel]

Pls help😭

[dapper sable]

OK, so g(x) = -7.

So, under the second line, write -7 = 2/3 x - 5.

Because you can replace g(x) with -7 in g(x) = 2/3 x - 5.

Then, you can isolate x.

Does that make sense?

[clever hazel]

yeah

thanks

[dapper sable]

No problem.

[clever hazel]

.close

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[tropic trail]

I need to find 4th iteration of this function

[tropic trail]

am i doing it right?

[calm coral]

@tropic trail Has your question been resolved?

[dense coral]

i ain wanna be rude but clean up your numbers please

[calm coral]

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[glad berry]

period of

[fading kindle]

Do you know the period of a general function like sin(kx)?

[glad berry]

Yes

It will be 2π×3/5

6π/5 right

[fading kindle]

Yes

[glad berry]

Same for cos

2π×7/3

For tan π×7/2

6/5 14/3 7/2

Now lcm?

[fading kindle]

Know just take lcm

[fading kindle]

Yeah

[glad berry]

,w lcm of 6/5 14/3 7/2

Now lcm?

[potent lotus]

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

Requested by yooda5656

[fading kindle]

Just calculate it

Do you know how to find lcm of fractions?

[glad berry]

36 140 105

2×2×3×3, 2×2×5×7, 3×5×7

It will be 4×9×35

[fading kindle]

Let me check

[glad berry]

1260?

[mortal orbit]

well now you need to scale it back

[glad berry]

Ohh hi rafilou

[mortal orbit]

this is a multiplicative constant * lcm

[fading kindle]

Yeah

[mortal orbit]

by what multiplicative constant did you multiply to get to those numbers?

[glad berry]

I multiplied their denominator

[mortal orbit]

yes

[fading kindle]

For lcm of two fractions just take the lcm of numerator and divide the hcf of denominator.

So for 6/5 and 14/3 it is going to be (14)*(6)/1

Which is 84

And the do the same for 84 and 7/2

[glad berry]

84 and 7/2

84/2 =42

[fading kindle]

Yes

No

It will be 84/1

Hcf of 1 and 2

[glad berry]

It will be 2

So 42

[fading kindle]

Why 2?

[glad berry]

Hcf of 1,2?

[fading kindle]

Highest common factor of any number with 1 is 1

Lcm would be 2

[glad berry]

I see

So it will be 84 π then

[fading kindle]

Yes

[glad berry]

Let's check with wolfram?

π

Right?

[fading kindle]

For that yes

I wanted to ask Modulus of that

[glad berry]

π/2 then?

[fading kindle]

Yes

[glad berry]

So??

[fading kindle]

For what?

[glad berry]

I meant i didn't understand your proposal?

[fading kindle]

I wanted to ask a question in which the lcm of separate periods wouldn’t work.

But I forgot that one.

Just wanted to tell that sometimes it doesn’t work and function could get simplified down

[glad berry]

Cos(|sinx|-|cosx|)

This that one

[fading kindle]

Yes

Pi/2 for it as |cosx| becomes |sinx| after pi/2

Same for | sinx |

|sinx|+ |cosx| was what I was thinking.

[glad berry]

Sin(π/2+x)+ cos (π/2+x)
|Cosx| + |sinx |

.close

[calm coral]

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[glad berry]

.reopen

[calm coral]

✅

[glad berry]

Answer is given 42π

@fading kindle

[fading kindle]

Let me go through it again

[glad berry]

Yes sure

[fading kindle]

I think for 3 number you have to take lcm of all three numerators and hcf of all denominator.

That should give 42.

But let me verify the formula

[glad berry]

Yes please verify

[glad berry]

.close

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[remote mural]

i am done with (a)

[remote mural]

what does it mean with the "identify an appropriate geometric series"

[past zinc]

$\text{Geometric series:}\\\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}\text{ }\text{ }\Leftrightarrow \text{ }\left| x \right|<1$

[potent lotus]

Joanna Angel

[remote mural]

what about maclaurin series then

if im doing geometric series

[past zinc]

geometric series is also Maclaurin series

[remote mural]

can't i just do maclaurin then

[past zinc]

i repeat, it is the same for you here

formally, this is guaranteed by the uniqueness theorem of Power Series Representations

[calm coral]

@remote mural Has your question been resolved?

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[bronze adder]

$x^6(x + 1)^2(x^2 + 1)^2(x^{31}- 1)^2(x - 1)^{-2}$

[bronze adder]

i'm in the business of binomial coefficients

need to extract the coefficient of x^30 from this

(it's a generating polynomial)

[potent lotus]

Norbert Baudin

[bronze adder]

the problem is the the negative index i suppose

[leaden thunder]

[fluid flame]

are you sure your expression is accurate

[leaden thunder]

use uhhh (1+y)^{-1} = 1 -y + y^2 - ...

[bronze adder]

i suppose i could differentiate the geometric series?

[graceful dust]

does this have an x^30 in it

[bronze adder]

yeah i think?

[bronze adder]

considering the negative index there should be right

[leaden thunder]

is |x| < 1

[bronze adder]

the value of x doesn't matter so sure

[graceful dust]

the context is generating functions?

[bronze adder]

i only need the coefficient

yes

[graceful dust]

.

there isn't going to be an x^30

[bronze adder]

if x^6 is multiplied with the constant from every positive index expansion and x^24 from the negative one

[graceful dust]

there is an x^31 term in there

wait

[bronze adder]

why does that pose a problem

[sterile seal]

You could also assume x=/=1 and do (x^31-1)/(x-1)=x^30+x^29+x^28…+x+1

[graceful dust]

maybe i am imagining it wrong

[bronze adder]

you're not gonna believe where i came from

i should just give the original generating function i used i suppose

is it easier to use this?

[graceful dust]

yes

[fluid flame]

that seems a lot easier yes

[bronze adder]

so what would I do

[graceful dust]

it's basically a problem of finding numbers that add to 30 now

with only specific numbers allowed

[bronze adder]

well that kind of defeats the purpose of the function cuz that's where i'm coming from

[graceful dust]

[fluid flame]

no but if all of your coefficients are 1 in all of your factors that's way easier to count

[bronze adder]

uh how so

[potent lotus]

Norbert Baudin

[bronze adder]

squared btw mb

[fluid flame]

ok that's only a tiny bit harder to count

[bronze adder]

we're basically looking at

4 numbers that add to 30

one is between 2 and 6

and the other is between 3 and 7

and the other two between 1 and 30

[fluid flame]

0 to 30^

[bronze adder]

oh right 0 to 30

[graceful dust]

there are identities like $\left(\frac{1}{1-x}\right)^n=\sum_{r=0}^\infty{n-1+r\choose r}x^r$ you can use that give the coefficients in $(1 + \ldots + x^{30})^2$ explicitly

[potent lotus]

chmonkey #1 simp

[bronze adder]

right that would be helpful i suppose

if i just rewrote everything as geometric series

[bronze adder]

yeah i think this is all i needed really thanks

.close

[calm coral]

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[remote mural]

what is uv in an integration by parts formula

[remote mural]

it does not make sense

[past zinc]

$\text{the integration by parts theorem can be written in the popular form:}\\\int_{}^{}g'\left( x \right)f\left( x \right)\text{ }dx=g\left( x \right)f\left( x \right)-\int_{}^{}f'\left( x \right)g\left( x \right)\text{ }dx$

[potent lotus]

Joanna Angel

[past zinc]

hence it is not difficult to guess what the function g(x) is

[calm coral]

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[main musk]

for what values of a does $a^{-a^{a^{-a^{a^{\cdots}}}}}$ coverge?

[potent lotus]

The Great D

[fringe reef]

i mean

for what values does $x^{x^{x^{\cdots}}}$ converge?

[potent lotus]

artemetra

[fringe reef]

set x=a^{-a}

and you'll find it

[calm coral]

@main musk Has your question been resolved?

[main musk]

@fringe reef $x^{x^{x^{\cdots}}}$ converges for $e^{-e} < x < e^{\frac{1}{e}}$

[potent lotus]

The Great D

[main musk]

however careful with $a^{-a^{a^{-a^{a^{\cdots}}}}}$

[potent lotus]

The Great D

[main musk]

its not the same as $a^{(-a)^{a^{(-a)^{a^{\cdots}}}}}$

[potent lotus]

The Great D

[main musk]

do you see the difference?

[fringe reef]

ooh

i see it

still

$a^{-\left(a^{a^{-\left(a^{a^\cdots}\right)}}\right)}$

[potent lotus]

artemetra

$a^{-\left(a^{a^{-\left(a^{a^\cdots}\right)}}\right)}$
```Compilation error:```! Missing { inserted.
<to be read again> 
                   \def 
l.53 $a^{-\left(a^{a^{-\left(a^{a^\cdots
                                        }\right)}}\right)}$
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)```

[main musk]

yes

so is it easy to find the converging interval of a

[main musk]

@fringe reef I have concluded that it converges for any number a

or at least all reals

or at least all postive real numbers

[calm coral]

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[normal wasp]

I got a trapezoid the parallel sides are 16 and 44 the non parallel sides (legs) are 17 and 25 need the area of the trapezoid so i just need to find the hight

[hollow kestrel]

My first instict would be to construct two right triangles and then set up a system of equations. Might get messy though.

[normal wasp]

Look

[hollow kestrel]

You can also use herons formula

[normal wasp]

I have done this its a bad drawing but you get the point right

?

[hollow kestrel]

Yeah you parallell transport the 17-side until it meet the other corner.

[normal wasp]

Ok so the base of the triangel would be 14 right?

A-b/2

[hollow kestrel]

Hmmm

It's not isosceles is it?

[normal wasp]

No

[hollow kestrel]

Then no

[normal wasp]

Mmm

Then I don’t know the base??

[hollow kestrel]

Try letting x and y be the two bases of the right triangles.

Then use everything you know about x and y.

[normal wasp]

25^2-h^2 =x^2 and 17^2-h^2 equals y^2

??

[hollow kestrel]

Yes and one more

[normal wasp]

One more ??

[hollow kestrel]

Yes, x+y=something

[normal wasp]

Hhmm

Idk

What is it?

@hollow kestrel

[hollow kestrel]

It's 28, right?

It's the base of the triangle that wasn't isosceles.

[normal wasp]

Explain pls

[hollow kestrel]

Hmmm, so we have divided the base of the triangle into two parts of length x and y, right.

Considering what you came up with so far, this should not be difficult to see

And you computed the base of the triangle to be 28

So x+y=28

[normal wasp]

Why is it 28 tho?

Ohh 44-16

Right?

[hollow kestrel]

Yeah. Take the bottom side of the parallellopiped(44) and remove the base of the parallelogram (16).

Left is the base of the triangle

[normal wasp]

Ok thanks hight should be 15 right

And the araea is 450

[hollow kestrel]

Sounds believable.

How did you get it?

[normal wasp]

Area?

[hollow kestrel]

No the height

[normal wasp]

Solved the system

[hollow kestrel]

Great, do you want me to check?

[normal wasp]

Mhh it would be good to if you don’t mind I think i did it right but you never know

[hollow kestrel]

Okay, cheers

[normal wasp]

Tag me when you get the answers

@hollow kestrel i got another geometry problem can you also assist me wit it?

[hollow kestrel]

Oh sorry I misread your reply

[normal wasp]

I got a trapezoid the middleline is 5 and the diagnals are perpendicular to each other need the area of the trapezoid

[hollow kestrel]

But you were correct!

[normal wasp]

Its ok

Can you help me with this?

[hollow kestrel]

Okay

Sounds hard

Does the middle line intersect the diagonals?

*i mean di they coincide in a point

[normal wasp]

Isosceles trapezoid forgot that

[hollow kestrel]

Hmmm, one idea is to think of the diagonals as the x and y axis of a coordinate plane.

[normal wasp]

I don’t think its meant to be solved coordinates

Wait if the trap is isosceles legs would also be 5 no?

[quaint sphinx]

do you have a diagram of a trapezoid in front of you?

(for this problem, you do not need a good diagram. but in geometry, having one always helps)

label the two bases of the isosceles trapezoid as x and y

hint: ||you are given the midline is 5 units. what does this tell you about x and y?||

hint: ||try calculating the area of the trapezoid algebraically||

[normal wasp]

X+y/2 equals 5

[quaint sphinx]

yes that information is good to have

[normal wasp]

Is this correct?

[quaint sphinx]

i do not believe so

[hollow kestrel]

How do you prove it

[normal wasp]

I can

Prove it

So middle line is a+b/2

So a+b is 10

Since the trapezoid is isosceles the legs should also be 10 so they are both 5

If im wrong pls correct me

[hollow kestrel]

I'm not convinced.

[quaint sphinx]

that is not how isosceles trapezoids work

isosceles trapezoids mean the 2 slanted legs have the same length

they can be as long or as short as possible

[normal wasp]

Yea you guys are right

Its wrong

Ok so all we have is middle line is a+b/2 equals 5

Whag do i do now?

[quaint sphinx]

did you read the second hint?

[normal wasp]

We need the hight right?

[quaint sphinx]

no

do you have a diagram?

[normal wasp]

A+b/2*h no?

The area

[quaint sphinx]

i must ask you this three times

[normal wasp]

I got a bad one

[quaint sphinx]

alright that is sufficient

so you have the two diagonals drawn

[normal wasp]

Yes

[quaint sphinx]

see if you are able to find a way to express the area of the trapezoid without knowing its height (so only using x and y)

two facts will help you greatly here:

1. the trapezoid is isosceles
2. the two diagonals are perpendicular

[hollow kestrel]

I think Mqnic has a different solution in mind than me. But you can definitely solve this by finding the height. The height is actually part of an iscoleles triangle, which you can use.

[normal wasp]

Do the diagnal cut each other in half?

[quaint sphinx]

no

[normal wasp]

The diagnals make 4 right triangles right

[quaint sphinx]

yes

[normal wasp]

Do i use that?

[quaint sphinx]

yes

that is a very good observation to make here

and in my opinion is part of the easiest/most elegant way to solve this problem (unless i overlooked something important)

[normal wasp]

Idk whag to do now tho im stuck

[quaint sphinx]

are you familiar with the square of a sum?

[normal wasp]

Yes

@quaint sphinx

[quaint sphinx]

ah okay you can rewrite the area in terms of a square of a sum

[normal wasp]

Mhhh

Do i say x is the leg or

I don’t get it

[quaint sphinx]

oh my mistake i thought you already had it written out

[quaint sphinx]

so x is the hypotenuse of a triangle... what's special here?

[hollow kestrel]

Independently of Mqnics solution, I can present a geometric solution:
Drop an altitude from a vertex, then show that it is the cathesus an 45 45 90 - triangle. Lastly, show that the other cathesus in that triangle is equal to the middle line.

Of course this is just a scetch.

[calm coral]

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[calm coral]

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[smoky bobcat]

could someone verify my solution

I wrote that

if B1,B2,B3,4 are the midpoints of A1A2, A2A3, A3A4, A4A1 then by connecting B1B2, B2B3, B3B4, B4B1 we will obtain a paralelogram with B1B2 | B3B4 and B1B4 | B2B3

(i can't use ||)

the parallel sides of a paralelogram are equal therefore B1B2 = B4B3

[calm coral]

@smoky bobcat Has your question been resolved?

[smoky bobcat]

<@&286206848099549185>

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@smoky bobcat Has your question been resolved?

[smoky bobcat]

.close

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[normal wasp]

so so i got a pyramid with a triangle as a base the lateral edges are sqrt70 sqrt99 and sqrt126 the latral edges are also perpendicular to each other i need the vlume

i alr found the sides of the base and the area too i need to find the hight of the pyramid now

[normal wasp]

<@&286206848099549185>

[calm coral]

@normal wasp Has your question been resolved?

[normal wasp]

<@&286206848099549185>

@quaint sphinx

can you help me please

[normal wasp]

.close

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[next wren]

Let $R > 0$. We consider the surface $S$ defined by the intersection of the surfaces
\begin{align*}
    C^+ &= \{(x, y, z) \in \mathbb{R}^3 : z \geq 0, z = \sqrt{R^2 - x^2 - z^2}\}, \\
    T &= \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 = Rx\}.
\end{align*}

Compute the surface area $\sigma(S)$

[potent lotus]

tobi

[next wren]

okay so I have been stuck on this problem for a while

first of all $\sigma(S)$ is a 1-D manifold so the surface is a line Integral?

[potent lotus]

tobi

[next wren]

I want to calculate the length of this intersection

<@&286206848099549185>

[mighty loom]

Hello , Can you help me with the differential Geometry question? My question is how can I find the frenet frames of the Darboux vector?

[calm coral]

@next wren Has your question been resolved?

[calm coral]

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[next wren]

same question as before

[calm coral]

@next wren Has your question been resolved?

[next wren]

<@&286206848099549185>

[mighty loom]

Nope @calm coral

[calm coral]

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[kindred patrol]

This is solving for A and I need help with starting the problem. I've tried substituting with different equations but I can't figure out how to go from there.

[calm coral]

@kindred patrol Has your question been resolved?

[calm coral]

@kindred patrol Has your question been resolved?

[bright lagoon]

I have an idea

[kindred patrol]

oh i forgot to mention

in our class alpha is a variable

so u can substitute A for x if that makes it less confusing

[bright lagoon]

$\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \cos{\left(\frac{A}{2}\right)}\csc{(A)} = 1$

I got that

you want to use

double angle identity for sine

this will get you an expression in tangent and cotangent

then write the cotangent in terms of tangent....

...and multiply through by tangent

you will have a quadratic.

$\sin{(2A)} = 2\sin{(A)}\cos{(A)}$

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

then...

$\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \cos^2{\left(\frac{A}{2}\right)}\csc{(A)} = 1\
\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \frac{1}{2}\frac{\cos^2{\left(\frac{A}{2}\right)}}{\sin{(\frac{A}{2})}\cos{(\frac{A}{2})}} = 1$

[kindred patrol]

wait sorry the cos(A/2) is squared

so cos^2(A/2)

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

sorry if this is confusing, I had a lot of typos

double angle formula using A/2 in the argument means sin(A) = 2sin(A/2)cos(A/2), get it?

$\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \frac{1}{2}\frac{\cos{\left(\frac{A}{2}\right)}}{\sin{(\frac{A}{2})}}} = 1\
\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \frac{1}{2}\cot{\left(\frac{A}{2}\right)} = 1\
\frac{1}{2}\tan{\left(\frac{A}{2}\right)} - \frac{1}{2}\tan^{-1}{\left(\frac{A}{2}\right)} = 1\
\tan{\left(\frac{A}{2}\right)} - \tan^{-1}{\left(\frac{A}{2}\right)} = 2$

[kindred patrol]

OHHH THATS SO SMART

this is going to sound really stupid

but you're able to change cotanget to arctangent right?

so it would work the same with cosine to arcsine?

[potent lotus]

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

[bright lagoon]

$\tan^2{\left(\frac{A}{2}\right)} - 1 = 2\tan{\left(\frac{A}{2}\right)}\
\tan^2{\left(\frac{A}{2}\right)} - 2\tan{\left(\frac{A}{2}\right)} - 1 = 0$

[potent lotus]

Melvin Eugene Punymier

[bright lagoon]

it's not arctangent

I used the same notation

arctangent just happens to have the same notation as tangent-to-the-negative-1-power (you actually should express that quantity as (tan(x))^-1, always)

(I could have put it in parentheses first)

oh well

cot(x) = (tan(x))^-1

that's what I did

anyway you should be able to solve this now

[kindred patrol]

yup thank you

[bright lagoon]

np

[kindred patrol]

.close

[calm coral]

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[neon jetty]

How i got this wrong pls help

[kindred estuary]

Do you know your expoenent rules?

[neon jetty]

yes some of them

[kindred estuary]

How would you handle the numerator?

What exponent rule would you use?

[neon jetty]

multiply?

[kindred estuary]

Those are the rules, which one should you apply first?

[neon jetty]

quotient rule?

[kindred estuary]

Start by simplifying the numerator, like I suggested

What rule would you use first in simplifying the numerator?

[neon jetty]

I guess power of produce rule

[kindred estuary]

Good, if you apply that to the numerator, what would it look like?

[neon jetty]

(16a^4b^2)

[kindred estuary]

Almost

[neon jetty]

what did I do wrong?

[kindred estuary]

You have a^-1

[neon jetty]

oh

a^-4

[kindred estuary]

Good, so you have $\frac{16a^{-4}b^2}{ab^{-3}}$

What next?

[potent lotus]

CaptainNova22

[neon jetty]

yes

quotient rule

[kindred estuary]

And if you apply that, you get what?

[neon jetty]

🗿

I hate negative numbers

16a^-5b^5

[kindred estuary]

Not quite

[neon jetty]

there

[kindred estuary]

Good

[neon jetty]

now?

what do I do

[kindred estuary]

Now you have a^-5, what rule can you apply to that?

[neon jetty]

Negative exponent

so it is 16b^5\a^5

Damn that was easy

🗿

[kindred estuary]

Yes that's it

[neon jetty]

thanks bro

.close

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[urban radish]

Hello, I am trying to solve this, I tried putting mutliple equation... I1, I2, etc, but got the wrong answers every time

the question says "Determine the function equation of the depicted 4th-degree function. Read the necessary points and their properties from the graph"

[urban radish]

I thought that I understood it before, but it seems like I actually didn't 😭

NOW I REALLY GOT IT 🏃‍♂️

.close

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[winged portal]

Is this correct? Also omega is the sample space or the universal set in my course

[marsh valley]

It looks about right.

[calm coral]

@winged portal Has your question been resolved?

[winged portal]

Is about good enough

[marsh valley]

Yes. There's nothing wrong with it.

[winged portal]

Perfect thank you!

.close

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[cedar shell]

ok so

[cedar shell]

i have the info

N0= 100 N(5)= 60

i need to get N(34)

with formula N(t)= NO * a^t

Now what i would usually do is set a wquasion so 60/a^5 = 100/a^2 if it was N(2) and not N0 because if ii do 100/a^0 i get a completely wrong answer

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[narrow rose]

If I linearise at (2,2), does it mean the origin should be at (2,2) instead of (0,0)? For 5c) (2,2) is the equilibrium point

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[lost ice]

how do i prove this?

[lost ice]

apparently the fundamental theorem of algebra says this must be true

Every non-constant polynomial p(z) with real or complex coefficients has a complex root

[civic dirge]

Any polynomial has at least one root in C. The polynomial they gave you doesn't matter

[lost ice]

okay im fucking dumb

of course

thank you

but since we're here here

apparently this is false

i feel like this is super obvious as well

[remote mural]

x+y=1
x+y=2

[civic dirge]

x = 0
x = 1

[lost ice]

hmm this isnt obvious to me. what about the complex coefficients?

[civic dirge]

1 is a complex number

[lost ice]

-i*i ?

[nimble badge]

z= (1+i)x + (1+i)y
z= (1+i)x + (1+i)y + 1

[civic dirge]

1 is the same as 1 + 0i

All real numbers are complex numbers as well

[lost ice]

ouuuuuu

gotcha

[calm coral]

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[neon jetty]

How would I solve it without guessing it?

[neon jetty]

<@&286206848099549185>

.close

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[vocal goblet]

not sure where to start

[upper sparrow]

Might be worth noting the axioms of a vector space that you must satisfy...

[vocal goblet]

would those be like associativity, commutativity, etc.

and then what would i do

[upper sparrow]

Prove that those operations satisfy it (note that, for example, the "zero vector" might be different from what you'd expect)

Most notably, you might want to start off by showing that you have a commutative group under that "addition" operation

[vocal goblet]

ok so lets say their are 8 axioms for the vector spaces

for R^+, x boxplus y, and c boxdot x

do i use all the 8 axioms to prove that they're in an R-vector space

[upper sparrow]

Yep - some of those might be relatively easy to deal with(!)

[vocal goblet]

okok

also this might be a dumb question

what is this:

like what is it saying

and how would i prove commutativity for this?

[upper sparrow]

All strictly positive real numbers - you're considering those as the vector space (over the real numbers, with those new operations)

[upper sparrow]

You want to prove commutativity for the "addition" operation: in which case, show that (or better, explain why) $x \boxplus y = y \boxplus x$

[potent lotus]

@upper sparrow

[vocal goblet]

ohhh

so the R+ is just the restriction for the x,y,c variables?

[upper sparrow]

for x and y - you want those strictly positive, but c is allowed to be zero or negative (as you're considering R to be the scalar field)

[vocal goblet]

right bcz it says c E R

without the +

okay so essentially, i just need to verify that boxplus and boxdot follow the axioms of a vector space

given that x,y are R+ and c is R

alright, thank you so much

[upper sparrow]

lots of those are gonna be just saying "that's true of real number multiplication/exponentiation, innit" tbf

[vocal goblet]

even better

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[graceful pendant]

How to third proportional in in proportions?

[graceful pendant]

If I want to find the third proportional to 2 and 4 is it like this

Our book has error in it and its wrong

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@graceful pendant Has your question been resolved?

[remote birch]

(2/4)=(4/x)

[calm coral]

@graceful pendant Has your question been resolved?

[graceful pendant]

I multiply them already

Sorry my english bad

[calm coral]

@graceful pendant Has your question been resolved?

[upbeat cargo]

2:4 = x:4 means 2/4 = x/4, in this case, you multiply both sides by 4 (since x/4) to get: 4*(2/4) = 4*(x/4). the 4's cancel: 2 = x

For the second: 2:4 = 4:x, you can reverse the order to get: 4:2 = x:4 or 4/2 = x/4, then again, multiply by 4 to get: 4*(4/2) = 4*(x/4). 16/2 = x. x = 8

@graceful pendant

[graceful pendant]

Its solving proportion by 3rd

I arrange them in ad=bc

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[fringe reef]

$\sqrt{ab}=\sqrt{a}\sqrt{b}$

[potent lotus]

artemetra

[fringe reef]

they broke down 416 as 16*26

[fringe reef]

and used this rule

@jade junco does that make sense?

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[fringe reef]

factorizing

16 is a perfect square

[potent lotus]

marulk

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[lavish kernel]

What did i do wrong here?

[inner zealot]

Nothing so far. You just need to evaluate g^-1(sqrt(3)) and then solve f(x).

[lavish kernel]

Wait if i give x sqrt3 then arctan is 60 degrees so g-1(sqrt3) is 75 degrees,

OHHH

ok thank you

[inner zealot]

yw

[lavish kernel]

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[iron oracle]

hi does anyone remember what rational expression gave a graph that looks like this?

[iron oracle]

i forgot how the expression would look like

[quaint sphinx]

1/(1 + x^2) could work

[iron oracle]

yes thank you so much

[quaint sphinx]

welcome

[iron oracle]

can you explain why it dips like that

[quaint sphinx]

as x approaches positive or negative infinity, denominator grows really large which makes function go to 0

and at the center when x = 0, it's at its highest point

[iron oracle]

ok thank youu

.close

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[fringe wadi]

what happened here?

[neat solstice]

Is Z a random variable with normal distribution?

[fringe wadi]

yes

[bright lagoon]

this doesn't look right

[fringe wadi]

good to know

[neat solstice]

I am sure that it's some rules with the cumulative distribution function \Phi of the normal distribution

[bright lagoon]

maybe you should explain what we're looking at in your first post next time

is this somebody's solution that you were told is correct, or are you saying you know something went wrong

[fringe wadi]

random variable in normal distribution

how did he turn to its next form

[bright lagoon]

ok the picture must be cut off

[neat solstice]

Recall that, if $\Phi(z)$ is the CDF of the normal distribution, namely $\Phi(z)=P(Z \leq z)$, then you have $\Phi(z)=1-\Phi(-z)$.
So you can say:
\begin{equation}
\begin{split}
P(-k\leq Z \leq k)&=P(Z \leq k)-P(Z \geq -k)=\
=&P(Z \leq k)-(1-P(Z \leq -k))=2P(Z \leq k)-1
\end{split}
\end{equation}

[fringe wadi]

what are these properties called

yeah finally thank you

[remote mural]

so ur helping him but not me

[potent lotus]

cristorenzo99

[neat solstice]

@remote mural in what help room is your problem?

[remote mural]

27..

[fringe wadi]

you sure?

[neat solstice]

I am sure. Do you know why $\Phi(z)=1-\Phi(-z)$?

[potent lotus]

cristorenzo99

[fringe wadi]

yeah first simplify it and remove that weird symbol so i can understand

you mean p(Z<any) = 1 - p(Z>any)?

[neat solstice]

I mean P(Z<z)=1-P(Z<-z) where Z is a random variable with normal distribution with mean 0 and variance 1

[fringe wadi]

yeah i don't know much but what i know is P(Z<z) = 1 - P(Z<-z)
don't ask me why the signs change when the negative disappear

i know why but i don't know at the same time

I don't know where to find the properties

[fringe wadi]

they're equal

because

they're the same

what are you writting bro

[neat solstice]

I'll explain why, perhaps a graph of a gaussian curve could help:
\begin{equation}
P(Z<z)=1-P(Z\geq z)=1-P(-Z \geq z)=1-P(Z<-z)
\end{equation}

because the normal distribution is symmetric, Z and -Z have the same distribution, namely $P(Z \geq z)=P(-Z \geq z)$ and you already know that $P(Z<z)=1-P(Z\geq z)$

[potent lotus]

cristorenzo99

[fringe wadi]

why is Z and -Z equal

[neat solstice]

Because the graph of the gaussian is symmetric with respect to the y axes

[fringe wadi]

oh

ok

to y

[neat solstice]

Yes

[fringe wadi]

thanks

[neat solstice]

You are welcome!

[fringe wadi]

.close

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[remote mural]

what exactly is the importance of making the codomain of a function not equal to its range at all times?

[remote mural]

you can say all real valued functions are mapped to C if you wanted to, but we obviously dont

but we still say the codomain is R, even when the range is not necessarily that

what are the implications of defining the codomain differently?

[pallid halo]

well practically speaking, if you have some complicated function it may be difficult or at least annoying to work out the image (range), but it may be trivial to at least say that it only produces real values

so its codomain is R, and its range is some subset of R that you can't be bothered to write down explicitly

[remote mural]

I understand that, but is there anything else to it then that

Because, in my mind, the only thing that could be affected is the surjectivity of the function

which I will be honest, with how arbitrary the codomain of a function can be, whether a function is surjective or not seems totally dependent on whatever you choose your codomain to be

[pallid halo]

for function composition, you care more about the codomain than the range:
for example, if f and g both have domain R and codomain R, i know i can compose them either way (f.g or g.f)

even if i don't know the ranges

[remote mural]

this question is independent of my last's

[pallid halo]

this is true, but sometimes you want to specify the codomain ahead of time (say for an entire class of functions) and then ask for a particular one whether it's surjective onto that codomain
think for example of linear maps from R^n to R^m

they're all surjective onto their range, but what we really care about is, which ones are surjective onto R^m

[glass heart]

it would be pretty awful to not have function spaces like {f:R->R continuous} because you want the image of each f

[pallid halo]

or putting it another way, how awkward would it be if you couldn't have theorem hypotheses like "let f : R -> R"

[glass heart]

you have to know when you need to be specific and when not. and specifying codomain precisely is nearly never an issue

[calm coral]

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[remote mural]

how do i find all the roots for this?

[remote mural]

i set it up as i = (a + bi)^6

but like i see no efficient and intiuitive way to go from here

[left umbra]

do you know euler's formula?

[leaden thunder]

i = sqrt(-1)

not i = -1

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[remote crown]

$$\begin{align*}
& y \geq 0 \
& x^2 + y^2 \leq 2
\end{align*}
$$

[potent lotus]

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

[remote crown]

how do i make a plot in python that satisfies these two inequalities

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[fossil quest]

Need help for trig

[fossil quest]

1 sec

how come he didnt rationalize like this

[drifting seal]

this is ugly

they first got a common denominator

[fossil quest]

ik

ohhhhhh

lemme try it now

ty

[drifting seal]

lmk if u have any other questions

[fossil quest]

@drifting seal did I do it correct ?

[drifting seal]

hmm

how’d u get 1+sqrt3+sqrt3+3

oh i see

u squared it

ok

bottom should be 1-3

not 3-1

[fossil quest]

how come over here its 3-1

[drifting seal]

(a-b)(a+b)=a^2-b^2

because they have sqrt3 -1

u messed that up

it should be sqrt3 -1

not 1-sqrt3

[fossil quest]

ok i got it

but the finial is correct right

[drifting seal]

no

(4+2sqrt3)/2≠2+2sqrt3

(a+b)/(c)≠a/c +b

it’s a/c + b/c

u have to factor the 2 out of the numerator and get

2(2+sqrt3)/2

then the 2 cancels

and u get 2+sqrt3