help-42

[abstract wasp]

how would I turn this from degrees to radians

[potent smelt]

To convert degrees to radians multiply by pi/180

[abstract wasp]

huh?

what are radians anyway?

[potent smelt]

The length of the arc of the angle on a unit circle

[abstract wasp]

mmm

[potent smelt]

So in other words, if you have an angle of 3π/5, and you consider a unit circle centered at the angle in question, the angle would cut an arc in the circle and the length of that arc will be exactly 3π/5

Hence 2π radians is 360 degrees because the circumference of a unit circle is 2π

[abstract wasp]

so this is what I need to do>?

[potent smelt]

So to convert from degrees to radians you have to convert using:

360 degrees = 2π radians
360 degrees / 2π radians = 1
(180 / π) (deg/rad) = 1

And (π/180) (rad/deg) = 1

[potent smelt]

Not quite

[abstract wasp]

wait

[potent smelt]

You took the 180 you had before

Rather than canceling it

[abstract wasp]

yeah so this

mb i didnt realise i did that

[potent smelt]

It's all good

[abstract wasp]

alr thankyou

🙏

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[jaunty grail]

What about this is wrong

[jaunty grail]

I’ve been stuck on this for a while

Nvm

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[jaunty grail]

.reopen

[calm coral]

✅

[jaunty grail]

This one

[normal anchor]

whats your first step? @jaunty grail

[jaunty grail]

Implicitly differentiate both sides

.rotate

,rotate

[jaunty grail]

That’s my first step

[normal anchor]

remember to multiply by dx or dy when you differentiate

i.e. the left hand side should be 2xdx

[jaunty grail]

I thought since we are differentiating only x^2 it would only be 2x

I thought when u r differentiating in terms of y than u do what u said

?

[normal anchor]

oh, i learned it using a different technique. yours is probably correct let me see

[jaunty grail]

Oh ok

Did u get answer

[normal anchor]

hmm

sorry i was busy

[jaunty grail]

This question due for me in 58 mins 😭

[normal anchor]

your first step looks right to me

what did you do after?

[jaunty grail]

I multiply out the terms

The isolate y’

[normal anchor]

yep

[jaunty grail]

But I keep getting it wrong

,rotate

[jaunty grail]

Ignore very bottom

I might’ve messed up my algebra idk

<@&286206848099549185>

[normal anchor]

it looks completely fine to me so i dont know why its wrong, maybe divide the numerator and denominator by 2?

[calm coral]

@jaunty grail Has your question been resolved?

[jaunty grail]

K

It didn’t work

Well it’s all good thanks for you help

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[placid anchor]

hi

[placid anchor]

can you help me with this

[potent smelt]

,rccw

[potent smelt]

For clarification: this deals with 6 digit numbers consisting only of the digits "5", "7", and "9", such as 555555, or 795597. And it's asking for the probability that such a number chosen uniformly at random contains 2 "5"s, 1 "7", and 3 "9"s, such as 557999

Is the above a correct understanding of the question?

[placid anchor]

yes

[potent smelt]

If so, consider the permutations of 557999 vs the total number of ways to make a number consisting only of 5, 7, and 9s

[placid anchor]

so this would be total ways of number right?

[lost thistle]

is the answer 0.0823?

like i solved it as 3^6 for total

and 6!/2!3!

for required case

then divide it?

[potent smelt]

Yup

[lost thistle]

im not sure

[potent smelt]

That's what I was thinking of

[lost thistle]

mhm

[potent smelt]

,w 6!/(2!3!) / 3^6

[potent lotus]

Results provided by WolframAlpha

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

Requested by omnipotententity

[lost thistle]

yup

[potent smelt]

And I get the same decimal approximation as you

[lost thistle]

then ig thats the answer??

[placid anchor]

ok i got it now

[potent smelt]

It should be!

[placid anchor]

thanks

[lost thistle]

wanna solve this question i made?

[placid anchor]

yes

[lost thistle]

here

[restive mortar]

bro format that better please

[potent smelt]

Fr

[lost thistle]

haha

[restive mortar]

also if anyone has time pls help in #help-23

[placid anchor]

i think answer is 2

[restive mortar]

no its 3

[lost thistle]

its 0

or 1

[restive mortar]

obviously it is

[lost thistle]

i made it way before

yep its 0

[placid anchor]

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[ornate star]

J ai fais O1I mais impossible de faure le O2I

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@ornate star Has your question been resolved?

[ornate star]

.help

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Commands:

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[thorn bear]

thats what ive done so far

[calm coral]

@thorn bear Has your question been resolved?

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@thorn bear Has your question been resolved?

[idle pendant]

i think u have to expand (z-1/z)^2 (z+1/z)^4 ? @thorn bear

then use what youve already written

[thorn bear]

right I see

I did something similar to that but I expanded the last line by putting it in the identity form for de moivre's thing and I think its right

thank you

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[calm coral]

@remote mural Has your question been resolved?

[remote mural]

<@&286206848099549185>

[remote mural]

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[kind kraken]

This... doesn't seem right..

[kind kraken]

Is my answer really going to be 0? I feel like I probably have the wrong upper and lower bounds for theta

[crisp aspen]

well -7x is odd, so it cancels out since your domain of integration is symmetric over the y axis

[kind kraken]

Odd.. peculiar? Or odd.. instead of even?

[crisp aspen]

an odd function satisfies f(-x)=-f(x)

so you can think of summing ..., -2, -1, 0, 1, 2, ...

it cancels out around 0

[kind kraken]

So.. it's just me not recognizing something about the function before I get started?

0 is a legit answer?

[crisp aspen]

yeah

[crisp aspen]

something to look out for in the future

[kind kraken]

Interesting... thanks!

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[slim fox]

Jack does 3/5th of the jobs around the house and Jill does the rest. Jack finishes 35% of his jobs properly and Jill finishes 55% of her jobs properly. Find the probability that a job done around the house will be done.

[slim fox]

I'm clueless.

Solved.

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[untold fern]

i need help with geometry proof

[untold fern]

i confused on how to prove a line is an altitude of a triangle

this is the problem

meb is congruent to dec becaude they are vertical angles

im trying to prove angle erb is congruent to drb because congruent supplementary angles are right angles, which proves ar to be an altitude

[spring lagoon]

you mean angle erb is congruent to erc?

[untold fern]

yes mb

i proved triangle mbc is congruent to dcb just now

i realize i often overlook overlapping triangles

still dunno how to prove erb and erc are right angles tho

[spring lagoon]

Ok I found a way to do it but it's a bit long-winded and Im not sure if it's the most efficient way

If you're interested though you can start by showing that triangles ADE and AME are congruent

[untold fern]

its fine, most of the proofs we do are like 12-13 steps

[untold fern]

i was thinking the same, from sss = sss right

[spring lagoon]

yeah nevemrind that should be fine

[spring lagoon]

Oh yes SSS should work

I was actually thinking SAS but that's slightly more work

[untold fern]

but what to do with that information

[spring lagoon]

ok so notice that if we could show triangle ECR congruent to triangle EBR we would be done right?

Is there a way to apply the recent development to trying to prove this

[untold fern]

yes

i see uh

angle aem is congruent to aed

because cpctc

and i think that means angle reb and rec are congruent

because vertical angles

then er = er because reflexive

last things that can be used to prove it is sas i think

because asa would require knowing the right angles already

[spring lagoon]

Yup

I think that's pretty much it

[untold fern]

idk from there, maybe trying to prove mbe is congruent to dce

[spring lagoon]

Ph have we not proved that yet

Shoot that's on me

Hmmm this is actually a little tricker than anticipated without using some outside knowledge which may be circular

[untold fern]

yoo wait what if we look at the bigger picture like proving aec is congruent to aeb

[spring lagoon]

yeah that's fine

you don't even need the entire triangle syou can just look at the angles as well

to rpvoe that triangle mbe is congruent to dce

[untold fern]

easy to prove aec to aeb

i think

o shit ssa doesnt work, how to prove they are congruent

is there a way to prove rab is congruent to rac?

ohh shit we already know that cuz of previous triangle being congruent

triangle arb is congruent to arc because angle side angle,

rb is congruent to cr because cpctc

and im stuck again 😭

uh am i allowed to ping helpers its been like 20 minutes

<@&286206848099549185>

every set of triangles comes out to be ssa which doesnt work and idk how to prove triangle erb is congruent to erc

rec is congruent to reb, cr is congruent to rb

are medians also angle bisectors? that would change everything

OH MY GOD

they are
thanks for your help @spring lagoon sorry for pinging helpers

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[spring lagoon]

Ah, sorry for not being there just now, but glad to see you solved it!

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[blazing coyote]

trying to prove this

using the defn of a definite integral

$\lim_{n \to \infty}\sum_{i=1}^{n} \frac{1}{n+i}$

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

would be the left hand limit

how do. find this given that the haromonic sum diverges

hmm

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@blazing coyote Has your question been resolved?

[blazing coyote]

<@&286206848099549185>

I mean I could convert it into an integral, but that feels like cheating

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[olive matrix]

can someone help me with this question?

[olive matrix]

I found the factors after making it into division

but I'm stuck afterwards

[leaden thunder]

Show what you mean

[olive matrix]

log{(2x-3)(x-1)/(x-3)(x-4)]

oh shoot

I factored wrong

but still

shouldn't it be x<4 and x=-1/2

as the restriction?

@leaden thunder

[leaden thunder]

No

[olive matrix]

or do I have the definition wrong

[leaden thunder]

You combined the logs too early

The arguments for each log term must be positive

log(a) - log(b) means both a> 0 and b > 0

But by combining them into log(a/b), you could have a=b=-1

[olive matrix]

oh okay. thanks

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[native raptor]

I'm not sure how to do 13

[brazen elbow]

can you tell me the linearization of y at some point a?

[native raptor]

Erm not sure I understand?

[potent smelt]

If you have a point on the graph, can you find a tangent line to that point?

@native raptor

If so, how?

[native raptor]

Hmm

[native raptor]

By them having the same graident

[potent smelt]

Yup, do you recall the point slope formula?

[native raptor]

Y2-y1/x2-x1

[potent smelt]

Sorry I meant the point slope form of the line

Not the point point formula for the slope

[native raptor]

Y=mx+c?

[potent smelt]

That's the slope intercept form

[native raptor]

Or y-y1=m(x-x1)

[potent smelt]

There it is!

Ok

[native raptor]

Oh

I did get it to some part

[potent smelt]

So let's say I'm examining the point x = a on the curve f(x)

And I want to find the formula for the tangent line

Can you use the point slope form to give me that tangent line in terms of a, f(a), and f'(a)?

[native raptor]

Y-y1=dy/dx(x-x1)

[potent smelt]

y - f(a) = f'(a) (x-a)

[native raptor]

Oh

[potent smelt]

Ok, now, can you rewrite this into the slope intercept form of the line?

y = mx + b

[native raptor]

Uh huh

So er

Y=y'(x)x+b?

[potent smelt]

y = f'(a) x + (f(a) - a f'(a))

[native raptor]

But aren't the points just (0,0) so they cancel oit

Out

[potent smelt]

Yes. b = 0 for the points we are interested in

This means f(a) - a f'(a) = 0

If you solve this equation for a, you will find the points we want.

[native raptor]

Wait I'm not sure I understand why is b=0 after dy/dx I got like a quadratic

[potent smelt]

Yes, the slope changes as a changes

[native raptor]

Err I'm not sure I understand

[native raptor]

Well I did get smth from what ik y-y1=m(x-x1) gives u the eqn of a line that's tangent to that point? but when I sub m=dy/dx
In the end I got a quadratic which I'm not rlly sure what it represent

[potent smelt]

@native raptor https://www.desmos.com/calculator/pornayvvtk

This is an animation of the tangent line right?

[native raptor]

Hmm yes

Wait the cubic is the expanded form of the eqn?

But I don't rlly understand the f'(a)(x-a) thing

Like I don't understand what it means

[potent smelt]

Here I have included the f'(a) (x-a) thing and a helpful line to give you an idea of the meaning.

Maybe you can adjust f(x) to be a little bit better?

https://www.desmos.com/calculator/pc4afnbjtl

[native raptor]

Hmm the black line?

[potent smelt]

Yes the black line

[native raptor]

Hmm what does that represent tho the distance from the y axis to the curve?

[potent smelt]

It is drawn from the y-intercept of a particular tangent line, to the function that describes all of the y-intercepts of all of the tangent lines.

[native raptor]

Hmm okay I get it now thanks 🙏

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[potent smelt]

https://www.desmos.com/calculator/ojsywb4fez

Better helper lines

@native raptor better late than never

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[ivory crag]

My guess is BCDA but for some reason that's incorrect

[brazen elbow]

show your work?

[ivory crag]

There's no actual function so I'm basing it off the interval and how much area there would be based on the interval

[brazen elbow]

yeah i would say BC is correct since they are negative

DA im not so sure

you can kinda eyeball it and see that with D, it covers the area without the tiny area in [4, 5]

but for A, it covers the area with the tiny area, but its pointless since the negative area from [5, 8] destroys that tiny area

so its bigger loss for A

[ivory crag]

That's right I forgot the negative area would basically cancel out the positive

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[prime topaz]

Hi I dont get how to turn the equations in part b to the form in part a

[pseudo wedge]

try comparing the coefficients of z and zbar to the coefficients in the equation in part (a)

you can deduce what z0 should be, from which you can find r^2

[prime topaz]

I tried and the only thing I could deduce was that the beginning z * zbar is the same

[pseudo wedge]

you mean z = zbar?

[prime topaz]

nah like this

[pseudo wedge]

oh well yeah

now compare the coefficients of z

in the top equation, the coefficient is -z0bar

what is it in the bottom equation?

[prime topaz]

2z?

[pseudo wedge]

no

what's the coefficient

[prime topaz]

um is that meant to be 2

-2?

[pseudo wedge]

well yes

but you sound unsure

[prime topaz]

I think im just confused between the z, z bar and z0 bar

Oh wait I see the coefficient mbmb

Ok hold up so that does that mean the answer to bi) is just |z+2|=0

[pseudo wedge]

not quite

z_0 = -2 is right as you identified

but r is not 0

[prime topaz]

There's no corresponding value for r though?

[pseudo wedge]

there is

what's the constant term in the top equation?

[prime topaz]

Wait is it because z0=-2 so z0*z0bar is 4
But then r^2 has to be -4 huh

[pseudo wedge]

yes, z0*z0bar = 4

[prime topaz]

Oh wait no r^2 is 4 asw isnt it

Ohhhhhhh thank you

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[blazing coyote]

Trying to evaluvate this

I suspect it's $\int_{0}^{1} \sqrt{x}dx$

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

right?

[bronze adder]

yes

[blazing coyote]

cool

thanks

one more problem

I feel like I'm tripping here

[bronze adder]

you should avoid using the same variable in the first line

[blazing coyote]

Let $g(x) = \int_{2}^{x} \sqrt{1+t^3}dt$
\
$g(x+h) - g(x) = \int_{x}^{x+h} \sqrt{1+t^3}dt$
\
$g'(x) = \lim_{h \to 0} \frac{1}{h} \int_{x}^{x+h} \sqrt{1+t^3} dt$

[potent lotus]

A dense set(Ping when reply)

[bronze adder]

yeah looks good then

I mean you did take the long way round instead of just LH

but I guess it was necessary

[blazing coyote]

We haven't proven LH yet

we have proven FTC in a handwavey way

which is why I used it

[bronze adder]

alright

[blazing coyote]

the 2 in the question is confusing me

is this basically saying

$g'(2) = \lim_{ h \to 0} \frac{1}{h} \int_{2}^{2+h} \sqrt{1+t^3}dt$

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

and we know that $g'(2) = 3$

[potent lotus]

A dense set(Ping when reply)

[pseudo wedge]

Yes

[blazing coyote]

cool

Thanks!

[pseudo wedge]

LH is just first principle when the denominator is linear btw

[blazing coyote]

Yeah, I see that now

Thanks!

This stuff is kind of fun

wish I could do LA instead though

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[calm coral]

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[shy nova]

I didn’t wanna write it all again

[dull wagon]

don't forget about the 1/2

[shy nova]

what about it

[dull wagon]

don't forget about it

[shy nova]

I didn’t

I used half angle formula for ctg

and the square root is +-

im having trouble choosing the sign

[dull wagon]

you seemed to only considered the location of the arccos in your reasoning

arccos(4/7) gives you an angle in Q2,
90° < arccos(-4/7) < 180°
what interval will 1/2 * arccos(4/7)
lie in

[shy nova]

45 to 90

?

[dull wagon]

yes

and which quadrant will that be

[shy nova]

first

ohh

makes sense

thanks

.solved

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[calm coral]

Please don't occupy multiple help channels.

[remote mural]

huh

I get x = 1

but the answer's 2

[radiant sluice]

that is a common mistake

that you made

you have to use Work Energy Theorem

[marble pendant]

when a=0, v is non zero

[marble pendant]

Yeah

[remote mural]

when is a 0

how is my anser not correct

😡

..

[radiant sluice]

maximum elongation of spring is when

the velocity of block reaches zero again

after being released initially

[remote mural]

oh

[marble pendant]

!show

[calm coral]

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

[remote mural]

kx = 100

x = 1

I think my answer is wrong

because

[marble pendant]

Use WPE

[marble pendant]

When a=0 v is non zero

[remote mural]

Mhm

[marble pendant]

100 is mg

[remote mural]

Doesn't the block come to rest after some time if there is friction?

and the elongation being 1m? herein?

i get that 1/2 kx^2 = mgx

<@&286206848099549185>

[marble pendant]

Friction?

[remote mural]

yes

if there's friction

[marble pendant]

Was having dinner

[remote mural]

Ok

won't the block come to rest

[marble pendant]

For 9th you can neglect friction

Even in 11th majority of questions will be without friction

[remote mural]

yes

[radiant sluice]

question says there is no friction btw

[remote mural]

YES

[marble pendant]

If there was friction then you are right, x<2m

[remote mural]

But if the block comes to rest due to air friction, what will be the elongation at last

oh

1m precisely?

[marble pendant]

To know the exact value I would need the function for air resistance

Which is usually kv

So the answer will now be exponential

e^ something

[remote mural]

what if there was no friction, and we lower the block slowly until when it is not lowered by the gravitational force due to the force of the spring, what will be the elongation then?

[marble pendant]

You mean neglect gravity?

[remote mural]

yeah

no

friction

I mean

Till the block is not allowed to go down due to the force of the spring

[marble pendant]

I don't understand what you are doing with gravitational forces

[remote mural]

ok

[marble pendant]

Oh I get it

Try drawing the FBD

You will see that the block won't move

[remote mural]

so 1m?

[marble pendant]

Because no force will pull it down

Still 2m

Because the normal force of our hand will have 0 work done

(when normal acts, the block isn't moving)

[remote mural]

thanks, I gotta go cycling

.close

[calm coral]

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[marble pendant]

Kk gn

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[glad sinew]

for b I disagree

because you have 4P4 for 24 ways to order the triplets (egAAA is a triplet)

but 3P3 for the ordering of the triplets

so there can be multiple AAABBBCCCDDD if that makes sense

so wouldnt it be 24(6)/369600?

[radiant sluice]

the triplets are umm indistinguishable for the second question

they are only distinguishable for first question

[glad sinew]

oh what

thats so trippy

the wording literally makes all the difference

[radiant sluice]

yeah it is

sometimes happens

[glad sinew]

thats so annoying

no but when you divide by 369600, thats the num of outcomes where each type is distinguishable

[radiant sluice]

uhh

[glad sinew]

oh I guess not, you're just putting them in different places

[radiant sluice]

yeah i calculated they are not

[calm coral]

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[atomic kernel]

can someone explain this to me?

[atomic kernel]

i cant really visualize this

[rustic geyser]

The essence of calculus by 3blue1brown is excellent at visualising this and more

[atomic kernel]

alright thank you

[calm coral]

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[spare meadow]

Is this solvable easily by hand?

[stiff notch]

<@&286206848099549185>

[cobalt bone]

unless you can do 1000 factorial then no

[cobalt bone]

ok

[spare meadow]

Ok thx

.close

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[cobalt bone]

(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3 = 42

[calm coral]

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[floral stag]

I can't find how to do this question

[floral stag]

i know torque = |r*F|

and F=1i+2j+3k

do you first find cross product? or dot product?

[pallid halo]

well which product is used in the definition of torque?

[floral stag]

i dont know

[pallid halo]

i would suggest looking that up first

[floral stag]

i cant find it and i dont know what you mean

its a vector?

[kindred estuary]

You're asking this, to find torque. Perhaps you should formulate your question something like "is torque calculated by cross or dot product?"

Then you'll see your answer on what to do

[floral stag]

i need help on the entire process, i don't know what to do at all

[kindred estuary]

You asked if you needed cross product or dot product, someone suggested to look it up, did you?

[pallid halo]

^ your first step is to work out what this * stands for, is it a dot product or a cross product?

[calm coral]

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[tired locust]

equilateral triangle ABC has sides of 14cm. A circle of a radius of 2 cm it is tangent to sides AB and AC. Find the distance from the circle's centre to side BC in cm

[tired locust]

i know if you make the centre of the circle o you have to find the distacne of 0 to line bc but I am not usre where to start

[unkempt drift]

try drawing it out

then you should make use of the side length ratios in the two small 30-60-90 degree triangles

[tired locust]

ok

[unkempt drift]

so that you can find the distance from the circle centre to point A

[tired locust]

but how do you aount of the lenght on the angle where the circle is

[unkempt drift]

wdym?

[tired locust]

yeak i dont even know

but i will try to draw it out

see if that help

[unkempt drift]

yeah

I'm using the fact that if a point is the same (perpendicular) distance to 2 lines

then the point lies on the angle bisector btw

that's how you know it's 30-60-90

[tired locust]

o that nice

I understand how get it to a 30-60-90 but i am notsure how to firgure out the distance form the center to the line

[unkempt drift]

which side is the radius 2 in this triangle?

[tired locust]

the hypothunuse

[unkempt drift]

nope

[tired locust]

also can elxplain a little bit more what is means byt he circle is tanget to sides ab and ac

[unkempt drift]

that's actually what you want to find

[unkempt drift]

oh

the circle touches the sides at exactly one point for each side

[tired locust]

yeah

i get that

but then what

you make it into two 30-60-90

[unkempt drift]

yes

[tired locust]

ok

but now what

[unkempt drift]

then this

which side corresponds to the radius = 2?

[tired locust]

the base

[unkempt drift]

you mean a right?

yes that's correct, the side opposite 30 degrees

[tired locust]

ok

[unkempt drift]

great and it would be easier to find the distance to point A first

so that's 2a

[tired locust]

ok

[unkempt drift]

great then after that you just need to draw the line from point A to the midpoint of side BC

use the 30-60-90 triangle again to find that distance

then you're pretty much done

[tired locust]

oooooooooh

[unkempt drift]

that's it

[tired locust]

thanks

[unkempt drift]

subtracting gives you the answer

[tired locust]

yes

[unkempt drift]

no worries!

[tired locust]

.close

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[tall moon]

let there be a set that contains n+1 positive integers that are not greater than 2n, prove that there is atleast one number that divides another number

[tall moon]

i need to use PHP for this, mot sure how tho

oop gtg

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[serene urchin]

I have tried to multiply the brackets to form a quadratic and divide the cubic by the quadratic but im still unsure on how to find the a and b terms

[solid crag]

the cubic has 3 roots

so we have (kx+l) for some k and l, we know that (x)(2x)(kx)=2kx^2=2x^3 so k=1 and (-2)(-1)(l)=2 so l=1

right?

does this make sense

[serene urchin]

yes

[solid crag]

than (x-2)(2x-1)(x+1) write it out and set a=x^2 term and b= x term

since $a_1x^3+b_1x^2+c_1x+d_1=a_2x^3+b_2x^2+c_2x+d_2$ when $a_1=a_2$, $b_1=b_2$, $c_1=c_2$ and $d_1=d_2$

[potent lotus]

SayMyName

[serene urchin]

oh thats pretty neat

thank you

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[old epoch]

for this problem, the solution first says lets x1 ^2 = y and x2 ^2 = y. Then x1 ^2 = x2 ^ 2...

[old epoch]

but why can we assume that x1 ^2 = x2 ^ 2?

becaus then arent we already assuming two solutions

and how do we know it is exactly two?

and not more?

[torn gorge]

I think it is related to the fact that a polynomial (with coefficients in a field) of degree 2 has at most 2 roots

but you know that they are at least 2 as you have two solution for x^2=1, that are +-1

remark: if p=2 then +1=-1, in fact x^2 is 1-1 here

[calm coral]

@old epoch Has your question been resolved?

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[calm chasm]

So I did DSP

Got me inf/inf

So I did L’H

[vernal portal]

what did you get

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

[potent lotus]

w

[calm chasm]

Now I’m not sure what to do

[vernal portal]

direct sub gives you 0 / 0+6

so

[calm chasm]

Yes

[vernal portal]

0/6 = 0

[calm chasm]

But doesn’t it have to be 0/0 for me to do L’H again

[vernal portal]

huh wait

[vernal portal]

yes thats true but if direct sub works it right

[calm chasm]

But the answer isn’t 0

I tried that

[vernal portal]

ye

let me try it myself rq

[calm chasm]

So what do I do from here to find my answer

[vernal portal]

using L'hopital i got $\lim_{x \to \infty} \frac{\frac{10}{10x+4}}{\frac{8}{8x+2}}$

[potent lotus]

w

[vernal portal]

if we simplify the frraction

we get

$\lim_{x \to \infty} \frac{80x+20}{80x+32}$

[calm chasm]

Ohh when I take the der of a natural log is it still like chain rule where I take the der out and multiply it?

[potent lotus]

w

[vernal portal]

look here

[vernal portal]

clearly, this evalutes to 1

[calm chasm]

Oh bruh I’m so dumb i left the 6 when i took the derivative

Smfh

[calm chasm]

@vernal portal ?

[vernal portal]

huh

i dont think of it like that

i suppose you could but

[calm chasm]

Then how did u get 10/10x+4

[vernal portal]

it easier for me to just memorize this

its easier for me to just find the antiderivative of f'(x)/f(x)

[calm chasm]

Cus isn’t the derivative of lnx 1/x so I thought the der of ln(10x+4) would just be 1/10x+4

[pseudo moss]

chain rule babyyyyy

[calm chasm]

Ok yea

[vernal portal]

again... this is a general form

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

[potent lotus]

w

[calm chasm]

.close

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[winged pollen]

Given 3 vectors, A,B,C (x,y,z), and a circle Radius.

I'm trying to find tangent points of the circles.
Circle 1, is place either "above" or "below" B, Circle B will have a tangent that is B.
Circle 2, will have a tangent along the BC line (BC2) and on Circle 1 (BC1). Circle 2 will be perpendicular to the BC line.

So far, I've been able to figure out the location of the Circle 1, by finding a line that is perpendicular AB line and that intersects B. From this, I've taken vector B and added the perpendicular AB * radius. and taken vector B and MINUS the perpendicular AB * radius. Then I've gotten the distance from these two points and choose the smaller one as the location for Circle 1.

What I'm trying to achieve with this is a "smooth" path when traveling from A to B, and then turning towards C after passing the point B.

[calm coral]

@winged pollen Has your question been resolved?

[calm coral]

@winged pollen Has your question been resolved?

[winged pollen]

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[nocturne zephyr]

Im a bit confused on the overall process here

[nocturne zephyr]

I know I have to use the form (lambda * v - A)(v) = 0, and I can get my matrix from there, then I know I use the rows of said matrix to create a system of equations, but where do i go from there?

[calm coral]

@nocturne zephyr Has your question been resolved?

[manic hawk]

you mean once you have v2=0, -v2=0, and 7v1 + 7v2 + 7v3 = 0?

[nocturne zephyr]

yes

I know v2 is fixed so that stays at zero

is the whole idea here just to set v1 or v3 equal to some variable like aplha or beta and to rewrite it into whatever form?

[manic hawk]

yes

remember than scalar multiples of eigenvectors are also eigenvectors

so when you end up with v1 + v3 = 0, you can just choose a non-zero number for one of them

[nocturne zephyr]

Ok so there’s technically more than 1 answer here?

[manic hawk]

,rotate

[nocturne zephyr]

was this what you were meationing, the scalar being -1?

[manic hawk]

technically no, but there are many ways to write the one correct answer

because alpha is any nonzero number

[nocturne zephyr]

gotchaa

[manic hawk]

these two boxed answers are actually the same answer

[nocturne zephyr]

there was one more thing, a lot of online resources Ive seen talk a lot about when solving these kinds of problems you assign a variable to a free variable, what is that exactly? is that just some variable that does not have a set value?

[nocturne zephyr]

gotcha

[manic hawk]

when you get to v1+v3=0, you might say v1=x, then deduce that v3=-x

so your answer will look like (x, 0, -x)

and then you factor out x to get Vectors = x(1, 0, -1)

[nocturne zephyr]

i see, that makes a lot more sense

would anything change if v2 was not zero

its not in this example here, but I ran into a problem where I had a matrix like this

that was the matrix after doing all the (A - I*lambda)

where I ended up with two equations -v1 + v2 - v3 and essentially the same thing for the third row, I'd assume here I would only have to fix one of my V's?

[manic hawk]

you're going to get two parameters

your answer will look like V = a (-,-,-) + b(-,-,-)

[calm coral]

@nocturne zephyr Has your question been resolved?

[nocturne zephyr]

cool thanks!

[calm coral]

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[blazing coyote]

what does it mean to index sets using say $\R$

[potent lotus]

A dense set(Ping when reply)

[calm coral]

Please don't occupy multiple help channels.

[blazing coyote]

or any uncountable set

closed the other help channel

[manic hawk]

It's basically the same as a function with domian R

[blazing coyote]

ah

[swift laurel]

each element of the set is associated with one real number

in general the indexing is a function from the index set to the set of interest

[blazing coyote]

I see

thanks

.close

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[blazing coyote]

How do I prove if $S$ is injective, it is an inner product space

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

like can I assume $

$

$\langle Su, Sv \rangle$ is an inner product space

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

can I assume this

[swift laurel]

all you are given is that S is injective and that <,*,*> is an inner product

[blazing coyote]

okay

Thanks

As S is injective, it follows that $\rangle S(v),S(v)\langle =0 $ iff v=0$. It thus follows $\langle v,v \rangle v=0 iff v=0.$
\
$ \rangle S(v),S(v)\langle \geq 0 \froall v \in V. $ being an inner product space, as $\langle v, v \rangle v_1 = \langle Sv,S v \rangle$ it follows that the ouput of $\langle \cdot, \cdot \rangle \geq 0$

logic like this works, right?

[potent lotus]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

As $S$ is injective, it follows that $\langle S(v), S(v) \rangle = 0$ iff $v = 0$. It thus follows that $\langle v, v \rangle = 0$ iff $v = 0$. \
$\langle S(v), S(v) \rangle \geq 0$ \text{ for all } $v \in V$, being an inner product space. As $\langle v, v \rangle = \langle S(v), S(v) \rangle$, it follows that the output of $\langle \cdot, \cdot \rangle \geq 0$. \
We now have $\langle S(u + v), Sw \rangle = \langle u + v, w \rangle_1 = \langle Su, Sw \rangle + \langle Sv, Sw \rangle = \langle u, w \rangle + \langle v, w \rangle$.
We now have $\langle \lambda Su, Sv \rangle = \langle S(\lambda u) ,Sv) = \langle \lambda , Sv \rangle = \lambda \langle u , v \rangle $

is this fine

[potent lotus]

A dense set(Ping when reply)

[calm coral]

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[celest adder]

how do i even solve this

[gray smelt]

first divide

I don't know if there is any shortcut to it

but you can expect the coefficient to be either 1 or -1

or zero

[celest adder]

how to divide

[gray smelt]

https://www.youtube.com/watch?v=3eowXt_LNbg

[celest adder]

never mind i know how to divide

ok thanks

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[celest adder]

.reopen

[calm coral]

✅

[celest adder]

@gray smelt @gray smelt

need help

how to divide such large degree though

many coefficients are missing too

nvm got it

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[blazing coyote]

I know this is $\N + {0}$

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

but for that I'd have to prove that each side is a subset of the other

and advice

[mortal orbit]

easy to see that each A_i is a subset of N_0?

(N_0 = N U {0})

@blazing coyote then you only have to show that 0 and every positive integer is in some A_i...

[blazing coyote]

ah

okay

We first show that $A_i$ is a subset of $\N_0$. Consider $A_n$ for some $n \in \N$. ${0,1,2,\dots, n} \subseteq \N_0$ as $0,1,2 \dots, n \in \N_0$
\
We now prove that $\N_0 \subseteq \bigcup_{i \in \N} A_i$
\
let $n \in \N$. Then by construction , $n \in A_n$, thus every natural number $n$ is in some $A_n$. Thus $N \subseteq \bigcup_{i \in \N} A_i$
\
BY construction $0$ is also in every set, thus $\N + {0} \subseteq \bigcup_{i\in \N} A_i$

[potent lotus]

A dense set(Ping when reply)

[blazing coyote]

This is just ${0}$ as it's the only element common to all $A_i$

[potent lotus]

A dense set(Ping when reply)

[pseudo wedge]

well it depends if your N starts from 0 or 1

if it's 1, then {0, 1} will be the intersection

[blazing coyote]

oh yeah

true

I forgot that

oops

thanks

[pseudo wedge]

what book is this btw?

[blazing coyote]

Book of proof

by hammack

just revising for my endsems

[pseudo wedge]

nice, all the best

[blazing coyote]

thanls

thanks

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[sharp flare]

What's delta y in this context?

[calm coral]

@potent sorrel Has your question been resolved?

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[ocean cypress]

$$\frac{\dd y}{\dd x} + \frac{y^2+y+1}{x^2+x+1} = 0$$

[potent lotus]

Mr. Gamer 🇵🇸

[ocean cypress]

solve this as an exact equation

[hard valley]

Oki

I did this much

I thought of doing with completing the square method

[ocean cypress]

https://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

read this

then return

(y^2+y+1) dx + (x^2+x+1) dy = 0

that is your exact equation

[hard valley]

Gotcha

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[fluid crag]

@remote mural bro

what you need

[hollow totem]

[copper hemlock]

[dreamy lance]

Do you have a question?

[calm coral]

@vale bobcat Has your question been resolved?

[glad parrot]

Meh

[fluid crag]

wtf

[calm coral]

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[compact drum]

heya

[compact drum]

consider these two functions

now the problem:

For each value of gold, find the tuple (hp,armor) that maximizes ehp

I did solved it by rearranging gold to armor. This way I can substitute this armor into the equation for ehp giving us a function ehp(hp, gold) where gold is given.
taking the derivative of ehp wrt hp and setting that to 0, we can find hp and then armor each depending on gold

as shown here

the issue is that here, armor and gold can be negative

i want to avoid that

what approach would be best here? Could we pose armor,hp>=0 as a constraint and use lagrange multipliers?

[remote mural]

im not rly good at this type of math but why do u make a link with gold if u just want to maximize ehp ?

i mean that u could probably find a tuple (hp,armor) just by studying the ehp function

[compact drum]

lets say you have 100 gold. the question then becomes, how do you best invest your gold?

[remote mural]

ok i get it

[compact drum]

armor and hp will cost you a different amount of gold per stat point

[tall moon]

does hp and armor have to be integers?

[compact drum]

practically yes, but we can ignore that

[remote mural]

gold is actually just a constant right ?

[compact drum]

gold is like a parameter

we can treat it as a cosntant

[remote mural]

then u can express let's say armor in function of gold and hp

and u put this expression in the ehp function

and now it depends only on 1 variable

[compact drum]

yes

i did that

but then after solving we get negative armor at gold=0

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@compact drum Has your question been resolved?

[compact drum]

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[mossy wigeon]

why isnt w=lwl

[mossy wigeon]

ah nvm

.close

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[unique jackal]

[calm coral]

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[digital inlet]

$a_0=-(a_1w+a_2w^2+\cdots+a_nw^n)$
$a1=-(2a_2w+3a_3w^2+\cdots+(n-1)\cdot a_n w^{n-1})$

[potent lotus]

dThirteen

[digital inlet]

a0, a1 being dependent, dimension of the subspace reduces by two? am i correct?

[calm coral]

@digital inlet Has your question been resolved?

[ocean cypress]

sure, but first you have to make sure it's actually a subspace

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[lusty sand]

hi guys so, if you look at this question it seems like it has one obvious way through, but for some reason there is also a different answer which i don't understand where is came from.. can someone please help me out there and explain it to me?

[fluid oxide]

what is the "different" answer?

[lusty sand]

oh sorry let me send you a picture of my answer give me about a minute

@fluid oxide there, this is my answer

[fluid oxide]

"sometimes this answer will not be available in the mcq" crazy

[lusty sand]

Our teacher said that sometimes this answer won't be available in the mcq so he suggested a second way which I don't understand, do you know it's name?

[fluid oxide]

im not sure of any second method

[lusty sand]

Oh no I sent it's photo

[fluid oxide]

you correctly simplified cot(pi - 1/x) into -cot1/x

[fluid oxide]

oh lol

[lusty sand]

That was his work

[lusty sand]

There

Second photo

[fluid oxide]

Hm yeah

do u know trigonometry quadrants?
like

what he did with the ASTC

[lusty sand]

No..

[fluid oxide]

lol

so

you should look up unit circle stuff on youtube

[lusty sand]

whats its name? and whats the methods name so i can look it up and practise

[dull wagon]

their work is incomplete,
their first step was to simplify the initial expression applying the periodic and odd property of cot
they have yet to do any differentiation

[fluid oxide]

its not really a method, its just how periodic functions or trig functions in general works

i can give you a summary but its better to learn it off youtube

look up "unit circle" or "ASTC rule"

[lusty sand]

okay, thank you! can i come back afterwards?

[fluid oxide]

yeah bet

[lusty sand]

okay i'm back

[fluid oxide]

fire

soo did you understand the simplification?

[lusty sand]

yep, just- not his method

i mean the teachers method for that question

[fluid oxide]

ah

well

so

first thing

ASTC rule, are you fine with it?

[lusty sand]

yes

[fluid oxide]

now

$\pi\text{ rad}=180^\circ$

[potent lotus]

@fluid oxide

[fluid oxide]

$\pi-\theta=180^\circ-\theta$, which comes in the second quadrant

[potent lotus]

@fluid oxide

[fluid oxide]

by the ASTC rule, only sin is positive in that quadrant, rest all, including cot, are negative

in this case, we have theta = 1/x

so cot(pi - 1/x) = cot (1/x)

[lusty sand]

hmm makes sense.. so you looked at the angle inside cot

[fluid oxide]

yep

[lusty sand]

okok

[fluid oxide]

i treated 1/x as theta

and yeah thats how the teacher did it

[lusty sand]

how do i know whether sec, cosec and cot are positive or negative

[fluid oxide]

whatever letter / function is positive in the quadrant, the rest are negative

its like

[lusty sand]

i know astc

[fluid oxide]

I. All functions are positive.
II. Sine and therefore its reciprocal, cosec, is positive.
III. Tangent, and therefore its reciprocal, cot, is positive.
IV. Cosine and therefore its reciprocal, sec, is positive.

[lusty sand]

but what about cot sec and cosec??

[lusty sand]

okay let me read this

[lusty sand]

got it

thank you so much i'll write everything you said then close the thread

again, youre really generous for giving me some of your time, really appreciate it <3

[lusty sand]

sorry dont understand this part

where you removed pi

[fluid oxide]

ah i forgot the -

Its $\cot\left(\pi-\frac{1}{x}\right)=-\cot\left(\frac{1}{x}\right)$

[potent lotus]

@fluid oxide

[lusty sand]

nono dw i realised that tiny error, but i'm talking about removing pi

[fluid oxide]

pi is 180deg right?

[lusty sand]

yeah

[fluid oxide]

and 180 - theta is second quadrant?

[lusty sand]

yep

[fluid oxide]

in second quadrant, is tan or cot positive?

[lusty sand]

nope