help-10

[white flame]

<@&286206848099549185>

[restive ridge]

!15m

[nocturne minnow]

You have to wait at least 15 minutes before pinging helpers

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[white flame]

yeah mb

i forgot

i already have a answer if u dont mind sharing urs and if i got it wrong explain it to me pls

[high lily]

there is no question. just a pic of a solid

[reef grotto]

wait you're right lol i just realised

[white flame]

the question is find the lateral area and surface area

mb guys

[high lily]

show your work

[reef grotto]

well just add up the triangles and the base

[nocturne minnow]

Post your work, if you have an answer

People here can check over it

[white flame]

alright

[obtuse pebble]

@white flame Has your question been resolved?

[white flame]

IS IT RIGHT <@&286206848099549185>

[obtuse pebble]

@white flame Has your question been resolved?

[white flame]

No one is helping me what to xo

<@&286206848099549185> pls guys

[obtuse pebble]

@white flame Has your question been resolved?

[obtuse pebble]

@white flame Has your question been resolved?

[daring rock]

@white flame Looks right to me

[hexed agate]

Just don't forget the units

[supple basin]

Yes

[obtuse pebble]

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[supple basin]

.reopen

[timid silo]

can someone explain what it is asking for lol

[harsh latch]

I gottchu

Is this a multi-part question

And what class is this for? There's some ambiguity here

Either, they want you to show a breakdown of a simple interest model and account for compounding, or you have to view this as a decision based model and account for the opportunity cost inherent to the situation

[timid silo]

im confused on how to apply IRR (internal return rate)

[harsh latch]

Do you know what irr is?

Both mathematically and spiritually

[timid silo]

its how much percent you increase your money by?

[harsh latch]

Not exactly, irr is a hard idea to wrap your head around.
You're solving for a break-even rate of your cash flows.
So if I reinvested that money I should be able to earn my irr (that's the underlying assumption, there's also an adjusted irr that sets the reinvestment rate at a more realistic rate and is more common practice in industry)

Now let's put this into the scope of your problem

You're going to buy, so its 95K today. That's one cash flow option.

Your other option is 25K now and the rest over 7 years. Seems like you know how to solve for this value with accumulating interest.

Now, what is the break even rate you would have to earn on that money for financing to make sense? That's irr

https://tenor.com/view/boom-bomb-explosion-iron-man-tony-stark-gif-15429820

[timid silo]

so i do net present benefit = net present cost?

[harsh latch]

Yea, but they're both outflows so I'd call them costs on the test

[timid silo]

ok but for option 2 the net present benefit is 100k and the net present cost is 95k

like i dont understand how u apply an annual interest rate to that

[harsh latch]

for option 1 the NPCost = ?
for option 2 the NPCost = 95K

[timid silo]

option 1: NPCOST = 25k + A*(P/A, 5.5%, 4) = 100k

[harsh latch]

yeah, solve

[timid silo]

wait sry

shouldnt NPcost1 = 100k

so confused lol

[harsh latch]

yeah that's right

But what's A

[timid silo]

the annual cost

[harsh latch]

value

[timid silo]

its unknown

[harsh latch]

You must make it known

lol you're right tho. Just keep going

[timid silo]

im still confused, like once i get A then what

[harsh latch]

Let's assume you solved for A.
Now you have to solve for your irr.
Let's call i=irr and v=1/(1+i)

Then 95v^0 = 25v^0 + Av^1 + Av^2 + ... + Av^n, here n = 4. Solve for i

Solving for i isn't easy. Most systems use some variation of Euler's method to find it

Guess and check, excel, or a financial calculator are also commonly recommended options

[timid silo]

hmm ok ill try that thx!

and what about the 2nd option?

is it just 95v^0 = 100v^0

that makes no sense

[harsh latch]

bruh, I already included the 2nd option

look at the values above and connect them to the problem

[timid silo]

oh fuck

[harsh latch]

lol

[timid silo]

ok this is making some more sense now tysm

[harsh latch]

yeah np my guy

[timid silo]

so basically we should end up with one rate of return for the entire problem?

i thought there would be a rate of return for each option

[obtuse pebble]

@timid silo Has your question been resolved?

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[wanton dagger]

So this is what I did

Ans is 0.57, so I’m not sure where I went wrong

[scarlet locust]

is this correct?

[wanton dagger]

Hmm.. let me recheck

[scarlet locust]

yes it is correct nvm

[wanton dagger]

I think it’s fine

Yh

[shadow badge]

$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

[warm shale]

Era238TM

[shadow badge]

you forgot r

[wanton dagger]

Oops!

Yeah I got it, thanks

.close

[obtuse pebble]

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[torn slate]

Help

<@&286206848099549185>

[kindred spoke]

This is a mess, please write your solution and question in another paper

[torn slate]

Ah nvm

.close

[obtuse pebble]

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[sly vine]

how was this simplified

[sly vine]

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

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[next birch]

https://i.imgur.com/6dfS9Jw.png

[next birch]

B,C

Unable to figure out how to Simplify it so that I can un square root it

[obtuse pebble]

@next birch Has your question been resolved?

[next birch]

<@&286206848099549185>

I know I need to multiplay √2+√3

But not getting what

[remote dragon]

For B multiply and divide by √2

[next birch]

So uh

it will be root(2√2+√6)/√2?

[remote dragon]

Uh it will be √(4+2√3) /√2

4+2√3 is just (1+√3)²

[next birch]

Got it

thanks bro

So multiply by 2 in root

and after removing root divide by √2

.close

[obtuse pebble]

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[timid silo]

What is a spring constant for k?

[nocturne minnow]

Two things, one that's a physics question which you should ask in the physics server linked in #old-network
Two, you need to elaborate, because that's not enough info to help you out

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[grizzled finch]

Im solving a differential equation using the power series method

The solution should look like this

But I don’t know how to get to there
P.S. I used s instead of y as the independent variable

When i tried to solve it, i got stuck because i couldnt find a way to equalize the exponentials and the index simultaneously

[obtuse pebble]

@grizzled finch Has your question been resolved?

[next birch]

Damn

what university are u in man

[obtuse pebble]

@grizzled finch Has your question been resolved?

[mint bronze]

Hi again. Your technique is right, but there is a small sign mistake on the sum containing K

It should be negative all the way through

With regards to getting their result, we need to equate every coefficient of s^w to 0. Right now the exponents on the sums do not match up

To fix this, separate all the sums in s^w as one sum, and separate all the sums in s^(w+2) as another sum. We then need to shift the dummy variable w in the w+2 sum backwards by 2 so that the y exponents match up

We can do this by starting the sum at w=2 instead of w=0, and subtracting 2 from all instances of w in this sum (there is no need to do anything to the first sum).

Notice that by putting w=2 as the first term, all the -2s bring everything back to 0 that needs to be.

You can then get the stuff multiplying a_(w-2) which simplifies to their answer

Hope that was useful

[grizzled finch]

omg, hello 😂

thanks for your reply

[mint bronze]

No problem

[grizzled finch]

i feel a bit guilty, youre always helping me out lmao

lemme try again

[mint bronze]

It's no problem, it refreshes my memory on these things

[obtuse pebble]

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[mint bronze]

You might want to reopen maybe, or not up to you :p

[grizzled finch]

.reopen

[obtuse pebble]

✅

[grizzled finch]

just to make sure i understand correctly

"We can do this by starting the sum at w=2 instead of w=0, and subtracting 2 from all instances of w in this sum (there is no need to do anything to the first sum)"
is that what you meant

[mint bronze]

Yeah, your working is right, what I did was just combine them into two sums rather than having 6 sigma signs. One for w=0 and another for w=2

It is just to make it a little neater

[grizzled finch]

but how can i combine them when their initial index still hasnt match?

[mint bronze]

You combine the first three and the second three because they do match

[grizzled finch]

oh right

[mint bronze]

Also, the last line seems to have dropped the s^w+2

[grizzled finch]

because w = 2 and w =3 for the first in the bracket is zerp

[mint bronze]

One moment I just noticed the w=1 sum starts

So we only need w=0 to begin with

For all sums. We will then shift only the s^w+2 sum starts forward by 2

The rest should still be w=0

I know that w=0 makes those terms 0

But will be a little neater to keep those there for the moment so we don't confuse things when we start shifting the variables forward

[grizzled finch]

okok

can we start over

[mint bronze]

Ye

[grizzled finch]

what should i do right now

[mint bronze]

,,\text{Put in }Y=\sum_{w=0}^{\infty} a_w s^w \text{ in the equation, but keep all sums starting from w=0}

[warm shale]

Social Capital Gainer

[grizzled finch]

but differentiating the series will automatically shift the index, so how do i keep all of them starting from w = 0

oh wait

[mint bronze]

Yes, even though that will make some terms 0, keep that in

[grizzled finch]

i see it

like that

[mint bronze]

Yes, thats what I have

Now, for neatness, we will combine all the sums in s^w as one sum starting from w=0 and all the sums in s^w+2 into another sum also starting from 0

[grizzled finch]

like that?

[mint bronze]

Yes that's right

[grizzled finch]

great

[mint bronze]

Now for the second sum, we will start it from w=2 and shift all the instances of w back by 2 (in other words, subtract 2 from all places where w is) so that the start from w=2 does not change the sum

[grizzled finch]

like that?

[mint bronze]

Almost, the w on thr middle term is still w and not w-2

[grizzled finch]

oh right

my bad

[mint bronze]

To add an extra step to make the step after clearer, we can split the first sum into two sums, one going from w=0 to w=1 and the second starting from w=2 and going to infinity. You can take out all the 0 terms from this

[grizzled finch]

ok

[mint bronze]

Yea, when w=0 the first 2 terms are 0 when w=1, only the first term is 0

[mint bronze]

Ye basically, that piece doesn't matter just now anyway

[grizzled finch]

right

[mint bronze]

We can combine the sums from 2 to infinity and those have all consistent s^w so we can equate the coefficients to 0 and so on to finish

[grizzled finch]

holy shit

we got it

thanks a lottttt bro

[mint bronze]

Nice well done!

[grizzled finch]

I owe you once again

btw

just curious

have you graduated

[mint bronze]

No problem at all. By the way, small note for speed, you can factorise out w from the first bracket and (w-2) from the second to get the result without having to expand a quadratic. It is good to look for repreated things like that which you can use

[grizzled finch]

i see i see, that's good thinking

[mint bronze]

Yeah, I have graduated

[grizzled finch]

so are u an acadameia in physics?

[mint bronze]

I'm a physical chemist actually

[grizzled finch]

ahhhh

nice

i aspire to be one as well

[mint bronze]

Nice, it is a very interesting subject, good choice

You can kind of float between physics and chemistry research which is nice flexibility, but that depends on other factors too

[grizzled finch]

i want to specify in quantum chemistry and theoratical chemistry

but im not sure how deep in physics i need to go into

[mint bronze]

Oooh, this is stuff I've looked at

[grizzled finch]

do i have to learn quantum field theory as well lol

[mint bronze]

The more physics you know from qm and some electromagnetism will help a lot as you are probably finding, as my experience has been that it is taught in more mathematical detail from the physics side as if you are learning it in a chemsitry department, you know, there is chemistry to learn too, it can't just be physics. Quantum field theory is not something I know, and tbh probably comes in useful in some niche aspects, but much more useful is qm and em from a physics and maths perspective. You might use stuff like feynmans quantum electrodynamics, but this is fairly approachable do look at as much books as you can to get a feel for things

[grizzled finch]

i see i see, that does not sound as scary as i thought then

im trying to read as much qm books as i can, but the one in the chemistry department usually dont go into detail

but the one in the physics department often require a deep understanding in vector spaces and matrices

[mint bronze]

It is not scary, all of it is manageable with enough practice and perseverance. I can guarantee that. As long as you spend as much time as you need, and for me that was sometimes a lot of time, you will become confident in what you are encountering

[grizzled finch]

i will make sure i do that

thanks a lot for your advice

[mint bronze]

No problem at all. You know, physics books can be pretty maths heavy and sometimes you will need to read a lot of extra material to understand it fully and this is not often time efficient. That is one of the biggest things I learned, if you have the time, by all means read the maths and the physics, but if you have to produce something quickly, the more stripped back approach may give more clarity in the here and now, which might be what you need in that moment. The deeper you go into rabbit holes, the more there is to explore and you might encounter other tricky stuff that slows you down potentially. Just something to bear in mind that you perhaps already know

[grizzled finch]

that is indeed a problem i face quite often, it took me a while to accept that the legendre polynomials is the solution without any good explanation. The reason why i want to do phys chem instead of other branch of chem is becuz i want to accept the least amount of premise possible, but i guess i need to reinforce my maths and physics background first to do so

[mint bronze]

Ye, honestly, that was very much my motivation too. But like you say, I have to make compromises (which I sometimes didn't want to make) and accept something, or not understand something fully. Like I shouldn't have to research a bunch of stuff because I don't understand one tiny thing that is bugging me. It uses time I didn't have. So it really just depends on the circumstances.

[grizzled finch]

Exactly, i guess that's the curse of curiosity

[mint bronze]

You could put it that way hehe

[grizzled finch]

btw, you dont have to if you dont want to, do you mind if i add you on discord so that i can ask you questions about phys chem directly in the future

[mint bronze]

Yeah that's no problem at all, but just to say, I am quite rusty, so I will have to be honest if there is something I am not familiar with as what you know will likely go beyond what I know soon. But I am more than happy to take a look, by all means

[grizzled finch]

That's great, I hope i wont bother you too often lol

[mint bronze]

No please don't worry about that, I'll let you know if I cant look at something at that moment

[grizzled finch]

can you add me, cuz your profile said you dont accept frd request at the moment

[mint bronze]

Oh ye sure

[obtuse pebble]

@grizzled finch Has your question been resolved?

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[wanton dagger]

For part e)

[wanton dagger]

Do I differentiate and then plug in 3?

[drowsy girder]

Yea

$\frac{dx}{dt}$ , t = 3s

[warm shale]

Pluton

[wanton dagger]

Right, thanks

.close

[obtuse pebble]

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[wraith wren]

hey

[wraith wren]

i need help

with pÿthon

[devout solar]

Why did you opem 2 help channels @Minatozaki#8791

.close

smh

seems that he left

[timid silo]

It's under maths help category

Not programming

[obtuse pebble]

@wraith wren Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@fickle hill Has your question been resolved?

[obtuse pebble]

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[timid silo]

How many factors greater than 1 do 120, 210 and 279 have in common?

Is there any fast way to solve this question?
At first I thought we had to use GCF but that's not right

The correct method is to write all factors and do crossing and stuff but that's time consuming

[timid silo]

Is there an efficient way to go about this question. Can someone pls help?

[high lily]

prime factorisation

[timid silo]

Thanks

[timid silo]

.close

[obtuse pebble]

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[agile plinth]

I have a question i have started the answer but i don't know what to do from now on so if anyone could give me a tip or help me out i'd appreciate it here is what i wrote.

[agile plinth]

$f(x) = x^3 + 3x^2 - 45x$ Find the extreme points, draw with the help of these point the functinos graph.

[warm shale]

killua

[agile plinth]

so this is what i did:

so i have x1 and x2

and i don't know what to do with them

[obtuse pebble]

@agile plinth Has your question been resolved?

[agile plinth]

<@&286206848099549185>

[obtuse pebble]

@agile plinth Has your question been resolved?

[agile plinth]

.close

[obtuse pebble]

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[full nymph]

forgot algebra lol. can someone correct me. we aren't supposed to be touching cubic equations so let me know if it involves cubics

[full nymph]

sending solved

[slim leaf]

@full nymph would help if you just factor out (x+3)(x+2) from the first line

you get (x+3)(x+2)(4 - 3(x-1))

on the RHS

[full nymph]

im sorry where?

[slim leaf]

first line of your working

[obtuse pebble]

@full nymph Has your question been resolved?

[valid crown]

if you found the roots of polynomial in denominator

and changed it's form into (x+3)(x-1)

then why didn't you shorted it to 1/(x-1)

[obtuse pebble]

@full nymph Has your question been resolved?

[full nymph]

.cloaw

.close

[obtuse pebble]

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[agile ravine]

how would I check if the three poles are enclosed by gamma?

.close

[obtuse pebble]

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[static patio]

In how many ways can 8 people form couples of two?

[drowsy girder]

$${8 \choose 2}$$

[warm shale]

Pluton

[drowsy girder]

Only 1 couple?

[static patio]

i thought exactly this

but answer is

idk why

[drowsy girder]

Is it maybe
$$\frac{{8 \choose 2} * {6 \choose 2} * {4 \choose 2} * 1}{4!}$$

[static patio]

yea, it is

[static patio]

how did you come up with 4! ?

[warm shale]

Pluton

[drowsy girder]

Well you see you got 4 terms up

And basically it means there will be some repeated combinations

So i just divided by 4!

[static patio]

uhm

it applies only when i choose groups of same size ?

[drowsy girder]

I mean its just logical lets say we have people

They are
AB CD EF GH

We can get
that in 4! Ways

Using the thing in numerator

[static patio]

uhm

got it

thanks

.close

[obtuse pebble]

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[high lily]

what have you tried

[analog heath]

$20(1)^3+10(1)^2-3a(1)+a^2$

[warm shale]

Rapiid

[obtuse pebble]

@analog heath Has your question been resolved?

[lament geode]

x+1 is a factor, so you'll put in the value (-1) not 1 cuz x+1=0 and then for an eq in a to solve

[obtuse pebble]

@analog heath Has your question been resolved?

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[vocal escarp]

Does Inverse Sin still follow the Unit Circle?

[vocal escarp]

.close

[obtuse pebble]

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[wanton kraken]

quick question

what's the 2 on top of x, which is diagonal to the other 2

and wtf is the 2 above the 2 at the last x

this is in my Viet formula lesson

oh wait

is it squared?

[ocean raft]

yes

[wanton kraken]

it's squared ain't it

oh ok ty

.close

[obtuse pebble]

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[umbral river]

Can someone help me with theres questions and explan it?

[wanton kraken]

take a free channel this is about to get closed

[umbral river]

Okay, thank you for saying that

[obtuse pebble]

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[frosty river]

Remember the formula of the area of the triangle: half the base times the height.

How can you calculate a height using angle and side?

Using a trigonometric ratio since the height is perpendicular to the base and it defines two right subtriangles.

[lilac grove]

Ye ye ik I can use A=(1/2)ab×sin(C)

But how

[high lily]

have you drawn a diagram

[lilac grove]

Yes and Two sides aren't on an included angle

[high lily]

can you show a pic of your diagram

[lilac grove]

Ok

[high lily]

what type of triangle do you have

[frosty river]

The triangle is isosceles

Isosceles triangles have a well-known property

[lilac grove]

c=a

?

[frosty river]

No

It is isosceles because 10=10.

[lilac grove]

Nvm

Angles add up to 180

[frosty river]

Isosceles triangles have a property on their angles

[lilac grove]

Wait so

32+x+y=180

[high lily]

forget about setting up an angle sum equation

for now focus on properties of isosceles triangles

and you can use the vertex labels for the respective angles

instead of introducing even more variables

you are told that two of the sides are congruent
does that tell you anything about some of the angles in your triangle

[lilac grove]

Same size?

[high lily]

some angles will have the same size

which angles will have the same size

[frosty river]

Two equal sides imply two equal angles

[lilac grove]

Down angles are equal?

[high lily]

all angles in the triangle are interior angles

some angles will have the same size
which angles will have the same size
do you just pick two at random and hope for the best?

[frosty river]

That would make all isosceles equilateral

[high lily]

or are there theorems / rules that tell you exactly which ones are congruent to each other

[lilac grove]

Angle A and C?

[high lily]

no

are you implying that if i were to rotate the triangle, that would affect which angles are congruent?

and that if i were to rotate it in such a way that no lines were considered horizontal then no angles would be congruent because where the fk is the base?

[lilac grove]

💀

[high lily]

perhaps it would be best to first read up a bit on isosceles triangles

understanding the properties will tell you which angles will be congruent, (given congruent sides)

[frosty river]

You might visualize the angles if you imagine an isosceles triangle.

Imagine a base and the opossite vertex equal distance two the oder two vertices. What happens with the angles when you move the vertex through the mediatrix of the base as the triangle changes?

You can use GeoGebra or simply look for the properties online

[lilac grove]

c is the same as b

[high lily]

do you mean lowercase c,b or capital C,B

[lilac grove]

lowercase

[high lily]

well that's given

the sides b=c =10cm is given

[lilac grove]

In an isoceles triangle, a=c, and A°=C°, and in this case c=b, as 10=10, so a should be 10 aswell?

[high lily]

the question is what does that tell you about the angles

[lilac grove]

Is all I can find after reading

[lilac grove]

Both sides have same size

So congruent

Angle shoukd be same?

[high lily]

which angles

which specific angles will be congruent

[lilac grove]

A°=B°

[high lily]

no

try not to overthink this

and it's probably better to use the word description of the property

as the use of variables in a=c, and A°=C°
conflicts with what's in your question

[lilac grove]

I think angle is A equal to C

I'm confused

[high lily]

(if you understand it then it shouldn't have been an issue)

well supposedly in the documentation you're looking at, (not your question)

you are presented with a triangle where a=c, and in that case A° = C°

[lilac grove]

if i rotate this

Sides can kinda change

I kinda get this

[high lily]

note that is NOT the same as the triangle you have

[lilac grove]

c=b here is a=b in the documentation

[high lily]

rotation doesn't affect angles or length of sides in your triangle

is it a=c or a=b in the documentation

anyway this is getting nowhere

in short

[lilac grove]

So here

C°=B°

[high lily]

in an isosceles triangle,
angles opposite congruent sides are congruent and vice-versa

yes, in your triangle, since
b = c, then <B = <C

[lilac grove]

If I rotate my device

I can see how C=B

Clearly

[high lily]

the point is that you shouldn't be relying on rotation

as i implied earlier

[lilac grove]

Yes that was for conforming

[high lily]

no

it doesn't confirm anything

are you implying that if i were to rotate the triangle, that would affect which angles are congruent?
and that if i were to rotate it in such a way that no lines were considered horizontal then no angles would be congruent because where the fk is the base?

[lilac grove]

So if C=32

(32×2)+y=180

y=116°

[high lily]

don't use x or stuff that looks like x for multiplication in the presence of x

[lilac grove]

Oh

Alr

So A=116°

[high lily]

yes

[lilac grove]

A=(1/2)×10×10×sin(116)

A=50×sin(116)

[high lily]

don't use x or stuff that looks like x for multiplication in the presence of x
try not to use x for multiplication in text in general

[lilac grove]

A=45cm²

[high lily]

use \* instead

[lilac grove]

Oh ok

[high lily]

A=45cm²
how were you asked to round

[lilac grove]

Damn that answer is correct

I wasn't asked to round

The book has 44.9 cm²

As answer

[obtuse pebble]

@lilac grove Has your question been resolved?

[obtuse pebble]

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[atomic bronze]

hello, to prove this by induction, what is our n=0?

[atomic bronze]

where (M, d) is a metric space

[alpine raven]

nothing lol

better to start at n=2

[atomic bronze]

So d(a_1, a_2) >= d(a1, a_2)?

[alpine raven]

yea

[atomic bronze]

Ok because online Ive read that the base case is d(x_1, x_1) <= d(x_1, x_1), which would also make sense

[obtuse pebble]

@atomic bronze Has your question been resolved?

[atomic bronze]

Can i write d(a_1, a_n) <= d(a_1, a_2) + d(a_2, a_3) + ... + d(a_n-1, a_n)

d(a_1, a_n+1) <= d(a_1, a_2) + d(a_2, a_3) + ... + d(a_n, a_n+1)

is this a proof?

[obtuse pebble]

@atomic bronze Has your question been resolved?

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[hollow vigil]

hello do u guys accept physics problems?

[cedar lichen]

There's a channel specifically for physics in #old-network . You can ask it here, but don't hold your breath on getting an answer

[obtuse pebble]

@hollow vigil Has your question been resolved?

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[unkempt orbit]

Can I have help with these. I marked all I know

[nocturne minnow]

,rotate

[nocturne minnow]

That's a test?

[unkempt orbit]

Oh don’t worry. I have my final this week. These are all old tests

[spice rune]

for the 1st one just prove that ΔAXM = ΔCYM (there's no C in the picture but you should get what I mean)

[unkempt orbit]

Oh okay

[spice rune]

is there anything below 2? like that ∠L = ∠P

[unkempt orbit]

It must of not printed or scanned. I think mn is congruent to no is given

Same with what you said

[obtuse pebble]

@unkempt orbit Has your question been resolved?

[obtuse pebble]

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[smoky crypt]

im reviewing old stuff from finals and im totally blanking on how to solve this, ik that if vectors are perpendicular their dot product is 0 but idk what to do with the points

[obtuse pebble]

@smoky crypt Has your question been resolved?

[smoky crypt]

okay i figured out that the plane through those points is <4, -3, 5> but idk how to find the vector perp to it

[halcyon tulip]

I think you can join two points and form AB and AC, then do their cross product.

These points lie in plane I am assuming

[smoky crypt]

i just realized i did ab x bc instead hol up

okay i got the right answer ty

[halcyon tulip]

Yw

[smoky crypt]

.close

[obtuse pebble]

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[thick ibex]

forgot how to do this so need a brief explanation

[haughty echo]

Plot graph of sin theta

Then move it down by 1

[thick ibex]

@haughty echo could you give an example of an answer? im not trying to mess this up

[haughty echo]

Umm

[tardy epoch]

Is it because it's an exam?

[thick ibex]

no

PBL

[tardy epoch]

Pbl?

[thick ibex]

Project Based Learning

so basically a project

im pretty sure i remember now but i need to be sure

im not trynna present and somehow i messed it up

[haughty echo]

Lets take an example, plot a graph of y=x

And a graph of y=x-1

You see any similarity in it?

[tardy epoch]

Try it and just show your work here

[thick ibex]

ye

[haughty echo]

@tardy epoch can you help me in a question?

[haughty echo]

What's that?

[thick ibex]

means yes

[haughty echo]

What do you observed?

They have a constant gap in both lines everywhere

Yea?

[thick ibex]

ye

[haughty echo]

Hm that is same principle which applies here

If you subtract a constant from a function

Then graph moves down by a gap of constant

[thick ibex]

like this?

[summer spire]

not very good sketches

try and plot the points that they would reach

[thick ibex]

lol but general concept is right?

[summer spire]

no

your first graph

what is the period of your graph

[earnest elk]

Hey Denton

Hows it going?

[summer spire]

hey

exams are over so more loose

how are you doing

[earnest elk]

nice

[thick ibex]

2/3 pie

[earnest elk]

good

[summer spire]

no

[summer spire]

math is going well?

[thick ibex]

did i mess up the period lmao?

[summer spire]

yea

[thick ibex]

im gonna kms

[summer spire]

both your fundamental sine and cosine graphs should be with a period of 2pi

[thick ibex]

dont you get period like this?

[summer spire]

what is A

[thick ibex]

amplitude

[summer spire]

no

also. you dont hae any amplitude

for your first thing

so idk how you came to the 3 at the denom

[thick ibex]

you get amplitude my doing this

[summer spire]

yes

[thick ibex]

this is how i got mine

[summer spire]

we are discussing the period

[thick ibex]

and you get period by getting amplitude

no?

[summer spire]

okay what is your max and min

in this graph then

[thick ibex]

first?

[summer spire]

refer to the image I sent you

[summer spire]

yes

[thick ibex]

1 and -2

[summer spire]

how

so youre saying the principle axis is -0.5?

[thick ibex]

what

bruh i just learned this yesterday

i have no idea what you talking about

[thick ibex]

im lost

[summer spire]

what about this doesnt make sense

lets start with that

then we'll go into any questions you have

[thick ibex]

it makes sense

maybe im confused about the laws of getting amplitude

[summer spire]

okay cool, previously you stated that your period is dependent on your amplitude

which is not the case

[thick ibex]

oh

so either me or my teacher is wack?

[summer spire]

you are correct about this

like the mx and min thing

however, you didnt find the correct max and min

you somehow got 1 and -2

[summer spire]

in this graph its clearly 0 and -2

[thick ibex]

yes but im confused on when to turn it into negative or positive

[summer spire]

it wouldnt really matter

[thick ibex]

LMAO

im slow

[summer spire]

since your amplitude is relative to the principle axis

like it goes both ways

[thick ibex]

oh ok

[summer spire]

at least at this level

[thick ibex]

can you explain what you mean by principle axis?

[summer spire]

principle axis is the line that is at the middle of the max and mins

either ways

please sketch the correct graph and send it again

if youd like to at least

I like your enthusiasm to learn

[thick ibex]

oh ok

alright here ill try again

???

this fine?

or am i still lost?

[summer spire]

okok

lets start with the basics

cos theta

[summer spire]

okay just note

that at 0 its correct yes

but at 90 degrees

you should be at your principle axis

so make sure it goes through there at least

same with 450 degrees

also makei t more curvy if you can

so it looks more legit

[thick ibex]

alright give me a bit

something came up

[summer spire]

no worries

[obtuse pebble]

@thick ibex Has your question been resolved?

[obtuse pebble]

@thick ibex Has your question been resolved?

[summer spire]

dude ask if you dont get something

dont hog the channel for yourself

[thick ibex]

oh alright ill ask later

how do i close this?

-close

!close

.close

[obtuse pebble]

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[lusty cedar]

really?

that is not i its iota

[nocturne minnow]

i^2 = -1

Simple

[lusty cedar]

well

before that you must question how on earth do we have a line with imaginary length

we can't

see we literally can't draw a line with imaginary length

[nocturne minnow]

You can't, gotta take everything in consideration

[lusty cedar]

well i'd say misleading

[obtuse pebble]

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[nocturne minnow]

Also it is labeled as "mathematical joke"

[obtuse pebble]

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[chrome stratus]

I am having a bit of a brain fart. How does r(0) become r(2)?

[chrome stratus]

Alright nevermind, I'll ask a different question.
I've already found the equation, but how do I calculate the points to be used?

[obtuse pebble]

@chrome stratus Has your question been resolved?

[chrome stratus]

<@&286206848099549185>

.close

[obtuse pebble]

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[violet bolt]

how do i do this question

[vernal wren]

Do you know that Sin(180-x) = sin(x)?

[violet bolt]

uhh i think yeah

i think thats what i tried to do on the second line of working on the first question

i wasnt sure if it was right

[vernal wren]

It's right, so your answer should he 'q' and not sin180-q

[violet bolt]

ohh okay, and same for the second question?

[vernal wren]

Cos(180-x) = -Cos(x)

So cos(50) = -p

For the other parts, use the formula
Sin²x + cos²x = 1

[violet bolt]

this part confuses me a bit, wouldnt cos50 be in the first quadrant of the unit circle so its positive ?

[vernal wren]

Yes, but cos(130) is negative

So -cos(130) is a positive value

[violet bolt]

ohh i think i get it

[obtuse pebble]

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[karmic bluff]

How do I solve this, step by step? I don’t know where to start.

[frosty river]

Simply use the first condition for f(x)

[karmic bluff]

How?

[haughty abyss]

substitute x for x-2 for f(x-2) and x+1 for f(x+1)

[frosty river]

$f(x-2)-f(x+1)=k(x-2)+m-k(x+1)-m=k(x-2)-k(x+1)=k(x-x-2-1)=-3k$

[warm shale]

jnkena

[frosty river]

Do you understand this?, does that help?

[karmic bluff]

I don’t no.

[haughty abyss]

so you have f(x)=kx+m

what would f(2) be?

[karmic bluff]

f(2)=k•2+m?

[haughty abyss]

yes exactly

[karmic bluff]

Aha?

[haughty abyss]

what about f(x-2)

[karmic bluff]

f(2-2) then? nah?

[haughty abyss]

what no

f(x-2)=k(x-2)+m

you just have to change x with x-2

im bad at explaining stuff lol

[karmic bluff]

Nah you’re all fine, i’m just so terrible at it

Instead of them x’s i put x-2?

[haughty abyss]

yes

same thing for f(x+1)

$f(x-2)-f(x+1)=k(x-2)+m-(k(x+1)+m)$

[warm shale]

Revreal

[karmic bluff]

Okay that’s confusing

[haughty abyss]

what's confusing about it

and f(m)=km+m

I don't really know how to explain it to you

@karmic bluff https://www.mathsisfun.com/sets/function.html
maybe reading this will help you understand functions more

[obtuse pebble]

@karmic bluff Has your question been resolved?

[timid silo]

Hi

i am little confuse of the follow explication of a solving problem

so it's using the r*u^(r-u)*u'(x) but i don't get it cuz it was suppose to get like 3(x-3)^2 right?

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

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[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[slim leaf]

@timid silo Are you confused about how they found the derivative of f(x)?

[timid silo]

yeah

[slim leaf]

You can write 1/(x+2) as (x+2)^(-1) and 1/(x-3)^(3) as (x-3)^(-3)

Then apply power rule.

[timid silo]

the power rule is the one with x^n=n(x)*n-1 right? then don't i get -3(x-3)^(-4)?

[slim leaf]

d/dx (x^n) = nx^(n-1), yes

Yes, you get -1(x+2)^(-2) and -3(x-3)^(-4)

which you can rewrite as -1/(x+2)^2 and -3/(x-3)^4

[timid silo]

thx, no i get it

so i get the fraction and i transform it into a function with negative power cuz i take it as inverse fraction and then i take the faction as i get it and turn it back in fraction after i derivate it

[slim leaf]

In other words, you shift the variable with the power to the numerator so you can apply power rule

and then after applying the power rule, its your choice to keep the variable with the power in the numerator or denominator

-1/(x+2)^2 and -1(x+2)^(-2) are both equal and correct

[timid silo]

oh so if i want i can let it as -(x+2)^(-2)

thx

[slim leaf]

Yes

[timid silo]

do i close it?

.close

[obtuse pebble]

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[timid silo]

Which formula will use here??

what exactly is this?

[timid silo]

It's a practice for sharpen your brain
These type of q's would be asked in some our national exams

yeah, but what's the question?

[warped agate]

india ?

[timid silo]

what are we supposed to solve here?

[warped agate]

bro find the ?

[timid silo]

No

pattern recognition?

Yes

Find out the missing word

are there more than 1 answers?

[warped agate]

it depends on the way of thinking

[timid silo]

i found a pattern but Im not sure if its the right one

[warped agate]

r u getting 16 ?

[timid silo]

yep

[warped agate]

coz I am

[timid silo]

i am too

[warped agate]

lol

[timid silo]

How you find it

It is correct one

[warped agate]

one sec bro i type

[timid silo]

Okay

subtract the third number of a column by the second of a column in the same row and then multiply it by the first number of the column in that row

[warped agate]

[ 3rd no - 2nd no(column) ]*1st no(row)

loll

[timid silo]

yep

rdt where you from?

[warped agate]

india wbu

[timid silo]

macedonia average iq here is about 60

[warped agate]

i see

[timid silo]

nvm you have the disadvantage

[timid silo]

Hmm thanks

[warped agate]

LMAO
why did u search that up xDD

[warped agate]

kool vro np

[timid silo]

cause I thought macedonia would be top1 in worst iq

[warped agate]

welp ur wrong

[timid silo]

Ariamu thanks btw for the "brain sharpener" question was fun

[timid silo]

😇 I will ask again if I stuck in any other question....

im not even a helper I just joined here an hour ago because I had a problem too then i randomly clicked on this channel and saw the picture xd

Hmmm

.close

[obtuse pebble]

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[unkempt heart]

How come y = (x^2 - 3)/(2x - 4) = (x/2) + (1) + (1)/(2x - 4)

It's for an oblique asymptote

Sticker was an accident btw

<@&286206848099549185>

[brave bramble]

Long division, basically

[unkempt heart]

Would I split up x^2 -3 into (x-3)(x+3)?

I'm not sure how to start solving it

Would I have to simplify the denominator first @brave bramble to x-2?

Then divide the 2 after?

[brave bramble]

https://www.mathsisfun.com/algebra/polynomials-division-long.html

[unkempt heart]

It's the coefficient that I'm confused about @brave bramble

Is it okay if I get someone else to explain it to me with the @ Help after more minutes? Or should I wait longer since I've already used it @brave bramble ?

.close

[obtuse pebble]

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[obtuse pebble]

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[glass parrot]

In the equation y=4x+2 where y is meant to represent the distance an ant is from its home in feet, and x is the time in minutes that has passed. Would the 4 represent how many feet the ant travels per minute?

[blissful pollen]

Pretty much

[timid silo]

yeah and 2 would represent the distance it has already traveled beforehand

[glass parrot]

ye

ok

ty

[hexed agate]

type .close @glass parrot

[glass parrot]

.close

[obtuse pebble]

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[cold abyss]

so for this problem, wouldn't we just need to get the degree of the denominator higher when you add g(x)? I tried (-3x^4-3x^3-3x^2)/x^2+x-2 in that form but the answer was incorrect...

[hexed agate]

what is the answr

[cold abyss]

oh nvm, got the answer

.close

[obtuse pebble]

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[small kestrel]

Heya! I'm preparing for my exam and there is this exercise where I have to construct all circles that touch the given circles a, b, and the line c. I am also supposed to determine the number of such circles (that touch the cirsles a, b, and the line c), but I reckon that is 2. However, I know the first step is to draw a line through the intersections of the circles and therefore create a line p perpendicular to the line c - on the line p, there will be both centers of the two circles I am supposed to construct. Would anyone be able to help me on how to solve it?

[small kestrel]

Just a few more notes: O_a is the center of the circle a and lies on the circle b, O_b is the center of the circle b and lies on the circle a, and the line O_aO_b is parallel to the line c.

[obtuse pebble]

@small kestrel Has your question been resolved?

[small kestrel]

<@&286206848099549185>?

[glass parrot]

Could you elaborate a bit more as to what your problem is

[small kestrel]

Okay, so as you can see in the picture, there are two circles: a and b, and one line c. The centers $O_a$ of the circle a, and $O_b$ of the circle b lie on a line that's parallel to the given line c. My job is to find all the circles that touch the circles a and b, and the line c at exactly one point each. I know there will be two but I have no idea how to find them.

[warm shale]

Smethisko

[glass parrot]

is this what you are describing?

[small kestrel]

uhm.. basically? i mean i sent the picture

long story short: "Draw all the circles that touch the circles a and b, and the line c."

[glass parrot]

well you said a line parallel

[small kestrel]

yeah, if you draw a line through the centers of the given circles, you will get a parallel line

[glass parrot]

right

[small kestrel]

so i know the circles will look something like this

but i need to get the exact points, of course

[misty terrace]

construct?

so by compass and straightedge?

[small kestrel]

yup

euclidean constructions

[glass parrot]

oh ok well here

[misty terrace]

do circles a and b have the same radius?

[small kestrel]

yeah

[misty terrace]

and do they have to intersect?

[small kestrel]

yes, the input image is exactly the one I sent

[misty terrace]

i see

sounds like a fun problem, lemme see

[small kestrel]

as I said, I know the first step will be to draw a line that connects the intersections of the circles

[misty terrace]

yeah

[glass parrot]

I'm assume you just want to know how to legally draw these circles

So from the given circles drawn

You can make a 3rd circle with the same radius as the others on either of the intersect points

And from there, use the circle's diameter you just drew and use that as a radius to double the circle which will encompass the entire figure

[small kestrel]

no, no, i have to draw these ones

[glass parrot]

wdym

[misty terrace]

how do you make it tangent to the line c though?

that's the heart of the problem

[glass parrot]

ohh tangent

[small kestrel]

the circles i'm supposed to draw have to touch circle a, circle b, line c

[glass parrot]

mb i thought he just wanted to connect

[small kestrel]

TANGENT! THAT'S THE WORD I WAS TRYING TO FIND

[misty terrace]

https://cdn.discordapp.com/emojis/853555811842261063.webp?size=48&quality=lossless

[glass parrot]

right so you want your largest circle to be tangent to c?

honestly i dont know how to do they unless I know the distance from O_a to the line C

[small kestrel]

yeah, if you take a look at the picture i send (the one im replying to), then both red and green circles have to be tangent to c

[misty terrace]

rip the reply

[glass parrot]

i need a distance tho

[small kestrel]

SHOOT

[misty terrace]

i'm working on it though :p

[glass parrot]

did it give one

[small kestrel]

this one

[misty terrace]

no

[glass parrot]

cuz idk where the placement is

[misty terrace]

it's a construction

[glass parrot]

mmm

[small kestrel]

well you can measure the distance by dropping a perpendicular line to c from the point O_a, but as it's euclidean construction, you don't have any exact number

you will just have the distance in your compass

do you think it would be possible to construct it by dilatation?

[glass parrot]

u mean like this?

[small kestrel]

i mean that's the first step to the solution, i guess

[glass parrot]

do you get a ruler/straightedge

[small kestrel]

you do

[misty terrace]

straightedge only

[small kestrel]

you have a ruler and a compass

[misty terrace]

(note that the ruler is without markings)

[small kestrel]

exactly

[misty terrace]

hence why i like to call the tool a straightedge :p

[small kestrel]

oh so the proper term for the infinitely long ruler without marks is straightedge?

[misty terrace]

i think so

[small kestrel]

kay, great to know, thanks!

so yeah, you have a straightedge and a compass

[misty terrace]

but 'ruler' will get you by fine as well

although i thought there is a specific type of 'ruler' construction?

[small kestrel]

.

[misty terrace]

dilation in what point?

in the tangent point with the line isn't going to help because it's the exact same figure then :(

[glass parrot]

there is

i forget the name

[small kestrel]

neusis constructions?

[misty terrace]

oh yeah exactly

[small kestrel]

if we dilate/dilute (idk the proper term ) both of the circles so much that they will only become two points...?

[misty terrace]

that won't reveal anything about the target circle then though :(

[small kestrel]

true ://

WELL! it will, as it will just become a thing of drawing a circle through three given points, won't it?

[misty terrace]

what dilation are you thinking of then

https://cdn.discordapp.com/emojis/944581311354511390.webp?size=48&quality=lossless

[small kestrel]

i hope it's the same thing in english and in czech does "dilatation" in english mean reducing or increasing the radius of a circle?

while keeping its centerpoint in the exact same place

[misty terrace]

yeah

alternative name is homothety if you want to be cool

oh yeah small note, it's "dilation"

[small kestrel]

well here is my thought process:

1. we draw this green line through the intersections of the circles

2. we dilate the circles until they reach the point where they are equal to their centers (they become a point)

OH NO

[misty terrace]

rip

[small kestrel]

it would move the center of our newly drawn circle

[misty terrace]

yeah

exactly

[small kestrel]

...

[misty terrace]

https://cdn.discordapp.com/emojis/778966798653587467.webp?size=48&quality=lossless

[small kestrel]

okay i feel dumb

[small kestrel]

thanks ❤️

[misty terrace]

https://cdn.discordapp.com/emojis/842306526776524800.webp?size=48&quality=lossless

hmm this problem is kinda tough :(

[small kestrel]

it is and I have no idea how to approach it 😭

[misty terrace]

lemme think about this more :p

[small kestrel]

sure! i'm not in a hurry

[misty terrace]

@small kestrel i've actually seen this problem before it turns out, but that problem asked for a way to 'construct' the circle using the overpowered tools of geometry software

here are my notes back then on it if you want to read it