help-10

[hushed moat]

Especially if you've tried it a lot

Try to restart again with a clear mind so you don't go down the same path again

[dull forge]

thanks again! 🙂

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

I don't understand the third line

third fourth and the rest

how do we derive it for theta?

like when we take the derivative of f_x for theta what do we get?

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

What language is this

I don’t understand

it just asks

$\pdv{f}{\theta*r}$

[warm shale]

Cüneyt

[timid silo]

I couldn't type it but

you got it

Ok yeh

relations are given as you see

I see

z=f(x, y) etc

One sec

I don’t understand lol

okay I understand it until the part $\pdv{f}{r}$

[warm shale]

Cüneyt

[timid silo]

but how do we take the derivative of "df/dr" for theta

teacher has got f_xx and stuff like that

where did they come from

I guess I better rephrase the whole question in latex

Is this partial differential

Cause I don’t know that

I cannot type it properly

but I will

just wait

[warm shale]

Cüneyt

[timid silo]

yes

hell yeah

at last

I'll close this and paste this another channel

.close

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[cunning citrus]

i need help with this

[cunning citrus]

idk how to explain this to my friend

[mental cloak]

If the oil price is 100€ and the price goes up by 20% it now cost 120€. But if it goes back down to 100€ you have only a decrease by 17%. since the difference of 20€ is less percent of 120 then of 100

[cunning citrus]

yeah thats what i told him

nvm its fine i thihnk i know how to

/close

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[fervent axle]

How would I solve this?

[fervent axle]

I assume I use 1-cos^2x instead of sin^2x

And then at Cos^3x-Cos^5x I'm not sure what to do

[nocturne cloak]

Nope, use the other one

1-sin^2(x)

That way you have a cos(x) left over that the u substitution will get rid of

[fervent axle]

Ah ok ty. Might be back with more qs on thyis topic cos I haven't fully grasped it.

.close

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[timid silo]

Hmm

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[hot gust]

.close

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[blissful sonnet]

hey can someone help me with this please:

Let p be an integer greater than 3, show that if p and (p + 2) are prime numbers then (p + 1) is divisible by 6

[mental wren]

How can p and p+1 be prime numbers?

The only case of this is 2 and three isnt it

[blissful sonnet]

11 and 13

[mental wren]

Thats p+2 though

And 13 isnt divisible by 6

[blissful sonnet]

yea i made a mistake srry its p and p + 2

[mental wren]

Ahhh

Okay

[silk galleon]

ah the classic 6n-1 6n+1

[blissful sonnet]

if p is greater than 3 so and prime so p must be an odd number so we can write it as p = 2k + 1 and p + 2 will equal to 2k + 3

but what do i do next

.close

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[shadow trellis]

what does it mean for this expression inside the limit to “not be differentiable at 0”

[mental wren]

it's not defined at 0

contrary |x| is defined at 0 but also not differentiable

[rustic harbor]

To not be differentiable at 0 doesn't mean to not be defined at 0

An interval where the function is differentiable and an interval where the function is defined is very distinguishable

In fact, the function f(x)=|x| is differentiable at everywhere, except 0, so its interval of differentiability would be R*=R\{0}

But its definition set is all R

Well here, for any x different than 0, you have x/x = 1

Since the limit when x goes to a is about looking at a function infinitely close to the a value, but not exactly on it

You'll never be on 0

So by simplifying ln(x²)/ln(x²)

You're getting a constant

That doesn't vary with respect to x

So it's the limit you are searching for

[high lily]

that's not what they asked

[balmy mortar]

u are only referring to the fraction, ignoring limit?

[shadow trellis]

yeah

[balmy mortar]

Definition of differentiable?

huh, why isn't it differentiable at 0 🤔
I'm not rlly with it today

I'm guessing there's a discontinuity there

[balmy mortar]

Oh right, the limit doesn't exist because f(a) doesn't even exist

f(0) is undefined

[shadow trellis]

f(0) gives infinity/infinity

[timid silo]

If $800 is deposited in an account that pays 9% annual interest, compounded semiannually, find the balance after 10 years.

Need help with this

I got 1929.37122 this I’m not sure if it’s right

[shadow trellis]

#❓how-to-get-help

this channel’s occupied

[timid silo]

?

[mental wren]

f(0) gives nothing because ln(0) is not defined

[obtuse pebble]

@shadow trellis Has your question been resolved?

[shadow trellis]

so simply if a function’s not defined at a value, it’s derivative doesn’t exist?

because the function doesn’t exist?

[shadow trellis]

why’s this not differentiable?

[mental wren]

it's not differentiable in 0 because there are multiple tangents to that point

[shadow trellis]

oh okay

ty

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[timid silo]

How do i solve this?

I figured out that the center of the circle is (2,-1)

but radius is not known so i tried using the distance formula and got

and then powered it by 2 to get r^2 which is 40

but that looks wrong

i got radius 5

by just using completing like square

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

How do I approach this problem?

[nocturne minnow]

Do you know least squares method?

@timid silo

[timid silo]

No I'm going through last years exam

I start the class in a week

[nocturne minnow]

General form, Ax = b and it's an overdetermined system

You do $$A^{T}Ax = A^{T}b$$

[warm shale]

dldh06

[nocturne minnow]

Then $x = (A^{T}A)^{-1}A^{T}b$

[timid silo]

Is what I did so far

[warm shale]

dldh06

[nocturne minnow]

So do $$A^{T}Ax = A^{T}b$$ first

[warm shale]

dldh06

[timid silo]

Okay this is similar to a regression

TY @nocturne minnow

. Close

[nocturne minnow]

.close

[timid silo]

.close

[obtuse pebble]

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[arctic tusk]

i think we can use an augmented matrix to represent a system of equations and use gaus-jordan elimination to find the solutions

oh sorry it's been closed

[nocturne minnow]

It's overdetermined

[obtuse pebble]

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[stray elk]

how do i draw a graph for f'

[balmy mortar]

describe to me the slope

at points 1, 2, 3

[nocturne minnow]

How did you number so fast?

[balmy mortar]

ShareX?

[stray elk]

its decreasing in 1, at a stop in 2 and rising in 3

[balmy mortar]

ok

So what u just said

will describe how f' acts

have u heard of sign diagrams?

[stray elk]

no

[balmy mortar]

ah well, what u described to me now

is like a sign diagram

[balmy mortar]

Could you try to draw a graph of f' based on what u said? try

f' represents the slope

[stray elk]

alright, but wint it just look like the other graph

[balmy mortar]

Lets start with (1)

You said 'decreasing'

That means f' is negative

So for starters, that doesn't match up

[balmy mortar]

The bit where I've marked 1

Is positive on your graph

[stray elk]

oh

yeah that makes sense

[stray elk]

what?

[balmy mortar]

I'm commenting about this

You drew that bit positive (when it should be negative)

[balmy mortar]

ignore if u get it 😅

[stray elk]

alright im lost

what do you want me to do again

[balmy mortar]

This is your original function

[stray elk]

yes

[balmy mortar]

Think about which part of it has negative slope

[stray elk]

the part under the x-axis

[balmy mortar]

no

the slope describes how steep

the curve is

[stray elk]

so the part that has a neagtive slope is from:

https://i.gyazo.com/107f01eea9b9dc0b1b63919ff931dbe9.png

here to there

[balmy mortar]

yes

Do you also notice how the slope

is getting less and less negative

until its 0

Hope that helps you draw the graph of f' left of that point

[stray elk]

no not really

so let me try again

[balmy mortar]

Ok so if you have a graph of f'

The line you draw

Can only be in the bit I colored in

for the left part

Do you agree?

[stray elk]

yes

[balmy mortar]

And also, it has to hit 0

at the dotted line

[stray elk]

like this

[balmy mortar]

You've draw the tangent

at that point of the curve

That's different from f'

[stray elk]

oh

how about this

[balmy mortar]

no

let me give you hint

https://i.imgur.com/N5vnTcz.png

@stray elkThe first part should look like this

[balmy mortar]

This cannot be it. You have f' positive in bits where it should be negative

[stray elk]

so i need to draw a graph that decreases until it hits the bottom part, and then i will star to increase

[balmy mortar]

Do you see how the line I've draw corresponds to f'

[stray elk]

its negative

[balmy mortar]

Not only that

the slope gets steeper and steeper for the original function

so f' is decreasing as you towards the left

[stray elk]

so like this?

negativ until it hits zero

[balmy mortar]

what you've drawn is correct

If you have a quadratic

ie. parabola

it turns out the line is straight

f = x^2

f' = 2x

[stray elk]

is there a way i can draw it more snake like

or is that the only answer

[balmy mortar]

your answer isn't wrong.

but they are kindof expecting a straight line

[balmy mortar]

the graph doesnt have to be a quadratic

[stray elk]

alright

thank you

[obtuse pebble]

@stray elk Has your question been resolved?

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[timid silo]

how to transform expression in first image to second image for group theory

[spiral maple]

associativity it looks like

[timid silo]

oo

thnx

.close

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[sour lantern]

does this notation mean that I should calculate the derivative while treating y as a constant and instead of x take x^2 and after all that square it?

[nocturne minnow]

It's not squaring it

It's the second partial derivative

[silk galleon]

$$ \frac{d^2f}{dx^2} = \frac{d}{dx} \left( \frac{df}{dx} \right)$$

[warm shale]

xdk1235

[sour lantern]

thank you guys

.close

[obtuse pebble]

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[sour lantern]

how would $\frac{d^2f}{dx}$ look like?

[warm shale]

Erzis エルジス

[sour lantern]

.reopen

[obtuse pebble]

✅

[silk galleon]

that's not a thing

$\frac{d^2f}{dx^2}$ is just a special way of writing a second order derivative

[warm shale]

xdk1235

[silk galleon]

it's not actually "squaring" anything

so there's no $\frac{d^2f}{dx}$

[warm shale]

xdk1235

[balmy mortar]

$$\dv{x}$$

[warm shale]

Shuri2060

[balmy mortar]

This 'thing' is an operator

that acts on functions

Consider it as one object

$$\dv{x}\dv{x} = \left(\dv{x}\right)^2 = \frac{d^2}{(dx)^2} = \frac{d^2}{dx^2}$$

This is shorthand for applying the operator twice

[warm shale]

Shuri2060

[balmy mortar]

a bit like

$$f(f(x)) = f^2(x)$$

[warm shale]

Shuri2060

[sour lantern]

ahhh

amazing notation

really intuitive

thank you all

https://youtu.be/J08-L2buigM?t=181

glad I'm not alone

.close

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[old night]

i dont know why they doing that

i tried to convert 50 sec to h

and then just times the velocity by time

and then convert to meters

but its not working

[spiral maple]

d=vt, however you need to convert km to m and h to s

[old night]

yeah

[spiral maple]

that's all they did

[old night]

but im getting different numbers

am i doing something wrong

[spiral maple]

You didn't convert properly

1h = 3600s

[old night]

i converted to hours all units

ohh

omg

omggggg

sometimes i go blind mode

thanks for the help

.close

[obtuse pebble]

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[halcyon creek]

I can't figure out why this question involves the second equation which includes f(3)

[tardy epoch]

That's one definition of the difference quotient

[halcyon creek]

Isn't the difference quotient a forumla:
(f(x+h) - f(x))/h?

[silk galleon]

Theyre equivalent

Plug in h = 3 - x

[halcyon creek]

Okay give me a second to write it through

I ended up with
= (x^2 -5x+6) / (4x+4)

hang on that can be simplified right?

= ((x-5) (x-1)) / (4(x+1))

NVM not any cleaner 😦

2 things: System says it's not right and I don't understand how h = 3-x

[tardy epoch]

This is the one for h going to zero. The other is x going to 3.

[halcyon creek]

So would it be instead

(-x^2 +6x -9 ) / (x+1)

[storm peak]

or x+3/(x-3)(x+1)

[halcyon creek]

Isn't the Difference Quotient supposed to end with something linear?

[storm peak]

nope

[halcyon creek]

System is saying none of the answers found have been correct

[storm peak]

now plug into difference quotient term

[halcyon creek]

the (x+h) formula?

[storm peak]

yes

[halcyon creek]

Ok computing...

[obtuse pebble]

@halcyon creek Has your question been resolved?

[halcyon creek]

its a long boi bot give me time

[dusk nexus]

theres two different types of difference quotients, one of them is this and the other is simply just

∆f(x)/∆x
or
f(x)-f(a)/x-a

you get the x+h difference quotient formula by substituting h=x-a

both are equivalent

if you take the limit as h->0 and as ∆x->0, you'll see that the derivatives are equivalent too

it's just that the x+h formula is easier and more convenient for calculations at times

since everything usually cancel

s

[halcyon creek]

I think I took a step out of line somewhere
very little cancelled when combining

= ((hx^2+h^2x-2hx-3h^2-3h) / (x+h-3)(x+h+1)(x-3)(x+1) ) / h

I feel like I made some computation error early in combing and have 6 variables too many

but this whole process is kind of breaking my brain as it's been years since I studied this previously

Help! I got stuck again; because inputting h = 3 - x gets me to an incorrect answer

[tardy epoch]

Why are you using h at all?

[halcyon creek]

because I don't know how to use the second equation given in the prompt

inputting f(x) straight gets me to (f(x) -2)/x-3

and I'm lost

[tardy epoch]

Plug in the definition of f in this expression and you'll get your answer

[halcyon creek]

but it didn't 😦

[tardy epoch]

$$[(x+5)/(x+1) - 2] / (x-3)$$

[warm shale]

riemann

[tardy epoch]

Did you simplify?

You should get a cancellation

[halcyon creek]

to simplify you multiply -2 by x+1 /x+1

and end up with -2x -2+x+5

or -x+3

so (-x+3 / x +1) / x-3

but I don't know where from there because multipling by the reciprocal didn't work

[tardy epoch]

a/b/c is the same as a/(bc)

[halcyon creek]

so -x+3 / (x+1)(x-3)?

[tardy epoch]

Do you see the cancellation

[tardy epoch]

I would put extra parentheses to be careful

[halcyon creek]

so -(x-3) cancels with the x-3 on the bottom

leaving -1/(x+1)?

Yes!

Thank you @tardy epoch for your time and paitence

[obtuse pebble]

@halcyon creek Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[stable kernel]

How would I graph f(x) = (x-3)(x^2 +x+1) ? I know that there's an x-intercept at (3,0) but how would I deal with the second part since there are no real solutions?

[autumn adder]

If youre graphing it on a real plane. An XY plane. Then the graph would intersect the X axis at only one point.

,w graph (x-3)(x^2 + x + 1)

[warm shale]

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[stable kernel]

Ohh so there'd be additional curves but they'd be below the x-axis that makes sense

Tyy!

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[old night]

how did they go from 1 to 2

did they just take the common factor t

but its not working with me

[undone geyser]

yes

[brittle swan]

yes

t² (1/t) = t

[old night]

how is that so if i multiply step 2 by t i wouldnt get step 1

[brittle swan]

they skipped trivial steps like this

[old night]

i see

thanks

.close

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[old night]

how do we use product rule for this one

[timid silo]

U use chain rule

Not product

[old night]

nvm

thanks

[timid silo]

Yep that’s ur answer

[old night]

i dont know

when to use the different methods

[timid silo]

When u see something like x.sinx

Use product

When u see something like x/sinx use quotient rule

When sin(x^2)

Use chain

[old night]

so if we have power is chain, if we have AxB thats product and if we have fraction that quotient

[timid silo]

Chain rule is used when ur variable in caught inside another function

For example

Sin(x^2 +1)

This is where u use chain rule

U can say x is chained inside sin

[old night]

ahaha

i see

alright thanks alot for the help

.close

[obtuse pebble]

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[karmic herald]

This is the rule for complex numbers equation, can in some how tow solution for deffir k have the same answer for example
z_0=i
z_1=i

[dusky zinc]

Hello is there anyone here that could help me?

[karmic herald]

<@&286206848099549185>

[obtuse pebble]

@karmic herald Has your question been resolved?

[karmic herald]

.close

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[misty wigeon]

struggling abit on my hw for centroids any help would be good thank you

[autumn adder]

@timid silo #❓how-to-get-help

[obtuse pebble]

@misty wigeon Has your question been resolved?

[misty wigeon]

<@&286206848099549185>

[faint flower]

centroid is like center of mass from physics

$$X_c = \frac{\int dm * x}{\int dm} $$

[warm shale]

Dev Shah

[faint flower]

Now, did you try and graph the curves? @misty wigeon

[misty wigeon]

yeah graphed the curves and added the range

[faint flower]

Cool, so now, did you get what you need to integrate?

for an x between 0 to 1, you'll have to integral for dm first

For any x, $$ \frac{dm}{dx} = \int_{-y}^{y} (x+t)*dt $$

[warm shale]

Dev Shah

[faint flower]

where y = 1-x^2

with me?

[misty wigeon]

kind of, just need to process it

[faint flower]

I basically changed mass/area to mass/length, because we need dm to only be dependant on x for the original integration

A disclaimer though, idk any theory for centroids specifically, so there might be an easier method than this for parabolas, but this works for any function where area is defined

[misty wigeon]

2/3 for top section?

[faint flower]

I don't have a way to try it out rn, sorry

but that's the method

[misty wigeon]

okay let me know when you can if possible but i think i know how to do it now thank you for your help:)

[faint flower]

Nw! It'll take some time to check tbh, hopefully someone else can do it before me!

[obtuse pebble]

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[timid silo]

How do you simplify $-\frac{\ln{\frac{3}{4}}}{2}$ to $\ln{\frac{2}{\sqrt{3}}

[warm shale]

ZED_118
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[timid silo]

@timid silo can u send a picture

Oh ok

[timid silo]

This is it, idk why it’s not compiling but whatever

[dusk nexus]

put the 1/2 back into the ln

[timid silo]

Write(ln3/4) as

Or yeh that

And then reciprocate it

Since it has a -ve sign

[timid silo]

Wait

Do you mean log((3/4)^(1/2))?

I see

Thanks

Yes

Should I close

How do you get rid of the negative sign?

[dusk nexus]

put it back into the ln

[timid silo]

Nvm I get it

Becomes (4/3)^1/2

[timid silo]

Yep

[dusk nexus]

$\ln{3/4^{-1/2}}$

[warm shale]

sean

[timid silo]

.close

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[nocturne shoal]

how do you do this question

[faint flower]

left the original number of boys, girls and adults be 5x, 8x and 2x resp.

so 5x+8x+2x=150

or x=10

so boys:girls:adults = 50:80:20

after k girls leave, the ratio is 10:9:4, so 50:80-k:20 :: 10:9:4

divide the right ratio by 5, 10:16-(0.2k):4 :: 10:9:4

16-0.2k = 9

k = 35

[timid silo]

Uhh

Oh Oki get it

[nocturne shoal]

easy

tysm

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[violet radish]

We hv
Mn = 128
Mk = 64
Nk = 8
Calculate m n k

[timid silo]

Uhh

Send a picture of the question

[violet radish]

I dont have

[warped agate]

hmm alright
divide 1st eqn by 2nd
u get n/k = 2
from this u get n =2k
sub this into 3rd eqn
u get k=2
therefore n=4
and m=32
therefore mnk will be 32x4x2 = 256
@violet radish

[timid silo]

Divide the 1st and second equation

U get n/k

U have nk

[violet radish]

Tysm

[warped agate]

np

[obtuse pebble]

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[sour lantern]

is this correct?

so far

[warm shale]

Erzis エルジス

[high lily]

you messed up a sign near the end

-(-1) is +1

i.e you should have +1 instead of -1 after evaluating the bounds

[sour lantern]

ahh true

should be + 1

what about the limit?

[sour lantern]

is the limit of ln part zero?

[high lily]

the difference of the lns with the ts, yes

it'll be clearer if you consider the difference to quotient law

[sour lantern]

yes lim at inf of $ln|\frac{1+t}{t}|$ is $0$

[warm shale]

Erzis エルジス

[sour lantern]

thank you

.close

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[upbeat bloom]

@flat anvil

[upbeat bloom]

I can solve a summation with i^2 in it like this

[upbeat bloom]

but how do I factor out the other elements in this case?

[obtuse pebble]

@upbeat bloom Has your question been resolved?

[thick fog]

Was this an actual exercise? I don't see a path to computing this sum without going beyond the scope of your course.

[upbeat bloom]

ok then I may have done something wrong

the original exercise is this one

[thick fog]

I think the intention is for you to just do integration using the fundamental theorem of Calculus and not evaluate that limit at all.

[upbeat bloom]

oh ok, thx

.close

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[timid silo]

How do you complete the square here?

[high lily]

the same way you'd complete the square in other questions

[timid silo]

Ok

[high lily]

consider the part you want to complete the square for, in this case the
x^2 - 4x
and add/subtract/manipulate appropriatrly

which they actually showed in the the work

[timid silo]

.close

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[old saddle]

Hello can somebody please help me

[old saddle]

Sorry

[timid silo]

You need to ask first

As in

Put the problem you need help with

So people who do know how to solve it can help

[obtuse pebble]

@old saddle Has your question been resolved?

[old saddle]

It’s being resolved

[silk galleon]

you have three channels open

[high lily]

.close

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[acoustic mountain]

clai

[acoustic mountain]

How do you draw a sketch of a graph from a derivative

[short spruce]

do you have an example? it kinda depends

[acoustic mountain]

Because you can not know the y-coördinate

Does that mean that you do not draw the x-axis?

In my head it should

[sage geode]

If you aren't given at least one value of the function at some point, you can't really graph it

But if you are

[acoustic mountain]

We know that a negative parabola has (a, 0) and (b,0)

a and b are given but I forgot

[sage geode]

Oh, so you know the roots?

[acoustic mountain]

i.e. a parabola with a mininum

[acoustic mountain]

I do

This parabola is the derivative of the graph I should draw

[sage geode]

Oh

[acoustic mountain]

So my question is: how do you draw the x-values without an x-axis

what is the normal way?

[sage geode]

I guess what you can do it just draw the graph go up where the derivative is positive (the bigger the absolute value, the steeper) and go down where the derivative is negative (same thing here)

[acoustic mountain]

True

[sage geode]

That's all I can say

[acoustic mountain]

But do I still draw the axis?

[sage geode]

Yeah I guess

[acoustic mountain]

Because I can not say if the graph is negative or positive

[flat anvil]

what

yeah you can

if f(x) is decreasing, then f'(x) is negative

***most of the time

[acoustic mountain]

But I only know f'(x)

[flat anvil]

ur asking

if you can draw the graph of f(x)

[acoustic mountain]

And need to know f(x)

[flat anvil]

given f'(x)?

[acoustic mountain]

Yes

[flat anvil]

well there are an infinite set of solutoins

[acoustic mountain]

So you know the shape of the graph

[flat anvil]

all you really need to know is the shape

[acoustic mountain]

Yes

[flat anvil]

cuz all of the graphs are just off by a constant

so it doesn't really matter--you don't need to draw specific values of y

[acoustic mountain]

True

But do I still need to include a x-axis?

[flat anvil]

yeah cuz the x's will always be the same

[acoustic mountain]

Because I can not say if it is above or below the formula?

[flat anvil]

the x is not dependent on the y

the other way around

[flat anvil]

wym by this

[acoustic mountain]

If I draw the x-axis so that the graph is negative

That could be incorrect

Because I do not know for certain that is the case

[flat anvil]

how would you draw the x-axis so the graph is negativE?

oh i c what u mean

well for ur purposes it doesn't really matter

[acoustic mountain]

I hope it doesn't

[flat anvil]

you can just assume that the constant would be 0

for the purpose of drawing it

and then say somewhere else that the graph could be shifted up/down any amount

if you wanted to be really specifiuc

[acoustic mountain]

Maybe I should add a disclaimer or paper or something

To be safe

Thank you

[flat anvil]

you're welcome

[acoustic mountain]

.close

[obtuse pebble]

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[gleaming hare]

how to do

[nocturne minnow]

Solve by factoring

[gleaming hare]

3 x -36?

i did that and got 108

[nocturne minnow]

Is that 3 times -36?

[gleaming hare]

then i didnt know what number multiplied to get -108 and adds to get -3

[gleaming hare]

yes

[flat anvil]

you should first (to make it easier) divide everything by 3

to get:

$x^2-x-12 = 0$

[warm shale]

Yottachad

[flat anvil]

and this is much easier to factor than the last one

[gleaming hare]

mhm

and what if the number werent divisible by the leading coeffecient

[nocturne minnow]

Then you can do the method you were doing

[flat anvil]

yep

[gleaming hare]

ok

[nocturne minnow]

Or quadratic formula

[flat anvil]

$(3x+a)(x+b) = 0$

[warm shale]

Yottachad

[flat anvil]

then you would try to find a and b

[gleaming hare]

how

zpp?

[flat anvil]

mayb

[nocturne minnow]

Fyi factors of 108: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54 and 108

[gleaming hare]

wait how did u get this

[nocturne minnow]

(1, 108), (2, 53), (3, 36), so on

That's the general form because you know if there was no leading coefficient, the form is $(x + a)(x + b) = 0$

[warm shale]

dldh06

[nocturne minnow]

But there is a coefficient of 3, meaning that when you FOIL, there has to be a 3x

[gleaming hare]

ok

[nocturne minnow]

To obtain 3x^2

[gleaming hare]

so where are we so far

[nocturne minnow]

You just factor $x^2-x-12 = 0$

[warm shale]

dldh06

[gleaming hare]

$(x + 1) (x - 11)?$

[warm shale]

BlzrdStrike

[nocturne minnow]

No

[gleaming hare]

then what is it

[nocturne minnow]

Does -11 * 1 equal 12? And does it add up to -1?

[gleaming hare]

$(x - 1) (x + 11)??$

[warm shale]

BlzrdStrike

[gleaming hare]

wait

[nocturne minnow]

If you FOIL, does the math check out, to make $x^2-x-12 = 0$?

[warm shale]

dldh06

[gleaming hare]

$(x - 1) (x + 12)$

[warm shale]

BlzrdStrike

[nocturne minnow]

But does that sum to -1?

[gleaming hare]

$(x + 1) (x - 12)$

[warm shale]

BlzrdStrike

[nocturne minnow]

Still doesn't make a difference

Try FOILing

See that equals $x^2-x-12 = 0$

[warm shale]

dldh06

[gleaming hare]

idk

[nocturne minnow]

Did you try to expand $(x + 1) (x - 12)$?

[warm shale]

dldh06

[gleaming hare]

no

[nocturne minnow]

So then expand it

[gleaming hare]

fine

$x^2 - 12x + x -12$

[warm shale]

BlzrdStrike

[nocturne minnow]

Is that the same as the original problem, $x^2-x-12 = 0$

[warm shale]

dldh06

[gleaming hare]

no

[nocturne minnow]

So then what does that mean you should do?

[gleaming hare]

not do my hw and play video games

jkjk

but

like find a match for it

[nocturne minnow]

So what other factors of -12 could you use?

But it should also add up to -1

[gleaming hare]

hm

[nocturne minnow]

What are the factors of 12?

[gleaming hare]

1 (-12)

-1 (12)

[nocturne minnow]

All the factors of -12

[gleaming hare]

1, -12

2, -6

3, -4

4, -3

6, -2

no more

[nocturne minnow]

12, -1

[gleaming hare]

wait

3, -4

IT WORKS!!!

right?

[nocturne minnow]

Yes

[gleaming hare]

wAYYYYYYYAYYYY

[nocturne minnow]

Because that adds up to -1 and multiplies to -12

[gleaming hare]

ok

so

$x - 3x - 4x -12$

[warm shale]

BlzrdStrike

[gleaming hare]

wait

$x - 3x - 4x = 0$

[warm shale]

BlzrdStrike

[gleaming hare]

so now ZPP

[nocturne minnow]

That's when you expand, but don't forget it's in the form (x + a)(x + b) = 0

[gleaming hare]

what about zero product property

[nocturne minnow]

Where a and b where the factors you found

[nocturne minnow]

This isn't correct yet

[nocturne minnow]

Should look like this

[gleaming hare]

$(x - 3) (x - 4)$

[warm shale]

BlzrdStrike

[gleaming hare]

= 0

[nocturne minnow]

Close

Remember, you said it was 3 and -4

So why is it -3 and -4?

[gleaming hare]

$(x + 3) (x - 4)$

[warm shale]

BlzrdStrike

[nocturne minnow]

$(x + 3) (x - 4) = 0$

[warm shale]

dldh06

[nocturne minnow]

To make it more proper

[gleaming hare]

yes

[nocturne minnow]

So zero product property, states that if given AB = 0, then either A = 0 or B = 0

[gleaming hare]

$x + 3 = 0, x = -3/
x - 4 = 0, x = 4$

[warm shale]

BlzrdStrike

[nocturne minnow]

You can just close each one with $

[gleaming hare]

ok

[nocturne minnow]

$$x + 3 = 0, x = -3$$
$$x - 4 = 0, x = 4$$

[warm shale]

dldh06

[nocturne minnow]

But yes

This is it

[gleaming hare]

ok

thx

[obtuse pebble]

@gleaming hare Has your question been resolved?

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[narrow fable]

I have a simple question about this pde. Why is there no resonance in the ansatz?

[narrow fable]

The thing I don't understand is in b, why they don't use the ansatz te^(t)(ACos(t)+BSin(t))

instead of what they do, which is just (ACos(t)+BSin(t))

[spiral maple]

cause the forcing function doesn't have e^t

cos(t) is just trig, so your guess for underdetermined coefficients is a linear combination of cos and sin

[narrow fable]

ahh, I see

Thanks for your help!

.close

[obtuse pebble]

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[karmic vapor]

i just need to know if my answer is correct

[karmic vapor]

i did 14 x 12 x 10 x 8

is that correct^

could i get steps on how to do this question please

[balmy mortar]

So what are the possibilities

for the 2 shapes?

[karmic vapor]

is it 13440?

[balmy mortar]

🟥 ⚫ OR ⬛ 🔴

Do you agree?

[karmic vapor]

yes

[balmy mortar]

So we add these 2 cases.

I don't know why you are multiplying everything together

[karmic vapor]

oh my bad

[karmic vapor]

so what do u add

i mean

from the info

[balmy mortar]

You count how many

are

🟥 ⚫

THEN

You count how many are

⬛ 🔴

Then you add the 2 together. For the total.

[karmic vapor]

so 44?

[balmy mortar]

let me see

I don't agree.

[karmic vapor]

oh

[balmy mortar]

Firstly

Tell me how many are this

🟥 ⚫

Red square, Black sphere

How many ways can she result in this

[karmic vapor]

14 red square and 8 black spheres

[balmy mortar]

Right.

You should imagine these are all different sizes

[karmic vapor]

ok

[balmy mortar]

Or maybe, each of these are labelled

A B C D E ...

So they are different

Now how many choices can she make?

[karmic vapor]

for both?

112?

[balmy mortar]

Yes, she is choosing 1 red square and 1 black sphere

yes, 112.

Now for the other case?

⬛ 🔴

[karmic vapor]

12 and 10

so is it 120

[balmy mortar]

ok.

So the final answer is?

[karmic vapor]

112 + 120

[balmy mortar]

👌

[karmic vapor]

232

[balmy mortar]

Do you understand the process?

[karmic vapor]

thank you so much

yes

[balmy mortar]

👌

[karmic vapor]

.close

[obtuse pebble]

Channel closed

Closed by @karmic vapor

Use .reopen if this was a mistake.

[obtuse pebble]

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[sweet edge]

I was wondering how I'd extend this to R3

#help-13 message

[balmy mortar]

Spherical coords 🤔

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

You describe point with magnitude and 2 angles

[sweet edge]

ah

uhm

Am I expected to use that to solve this question?

(solving for tensions)

[balmy mortar]

Whats the Q

[sweet edge]

this

[balmy mortar]

I have doubts you need it

[sweet edge]

Ah, okay

hm

I can do this when the scenario is flat

but I'm not sure how to translate that into when the situation is 3D

[nocturne minnow]

Find the x, y, z components of each tension, then equilibrium concept

[sweet edge]

what is an equilibrium concept(constant)?

[nocturne minnow]

Sum of forces equals 0

[sweet edge]

yes*

but I'm not sure how to make them so it's like

T1<cos(x), sin(x)> and T2<cos(x), sin(x)> etc.

and this is for x,y only ^

[nocturne minnow]

Trying to recall how to do it

[sweet edge]

okie, n

np*

@ me if u figure it out

I'll just be working on other problems for the moment

[nocturne minnow]

Find the unit vector of each tension

[nocturne minnow]

I figured out what to do

[sweet edge]

Sorry, yeah, i'm trying that rn

I got them, they are just

[nocturne minnow]

How did you get that as the unit vector?

[sweet edge]

well

[nocturne minnow]

Let's take T1

[sweet edge]

okay

<5, -5, 5>

so that's 5 < 1, -1, 1>

and that vector left over is the unit vector?

[nocturne minnow]

Not exactly what I was trying to get you to do

[sweet edge]

ahh

[nocturne minnow]

Express each force in Cartesian vector form

Do you know what that means?

[sweet edge]

using i j and k?

[nocturne minnow]

Yes

[sweet edge]

ah

5i - 5j + 5k

[nocturne minnow]

Not exactly, that's just the distance

[sweet edge]

uhhhh

I'm confused as to what you're asking 😅

[nocturne minnow]

You understand what the unit vector does, right?

[sweet edge]

It describes the direction as a unit length

[nocturne minnow]

It's a mag of 1, meaning if you took some other vector and multiplied it, it doesn't scale it

Right?

[sweet edge]

Yes

[nocturne minnow]

So we what to find the i, j, k form of each T1

I'll walk you through the first one

[sweet edge]

alright, sure

Is this it?

[nocturne minnow]

Not exactly

[sweet edge]

that's a sqrt(75) in the denoms

aw rip

What is that which I did?

[nocturne minnow]

That's the unit vector

[sweet edge]

alright

[nocturne minnow]

If you multiply it by T1

Then that's the overall force vector

[sweet edge]

nice

[nocturne minnow]

So T1 time that unit vector, is what you need

Do the same process for all of the other tensions

[sweet edge]

so that calculation I did, that's correct so far then

[nocturne minnow]

Yes

[sweet edge]

I just repeat it

alright

[nocturne minnow]

Yep

[sweet edge]

I have all the others then

It's just variations of this one

[nocturne minnow]

So for T1, it's $T1\left[\frac{5 \hat{i} - 5 \hat{j} + 5 \hat{k}}{\sqrt{75}}\right]$

[warm shale]

dldh06

[sweet edge]

yes

[nocturne minnow]

So that's T1 in i, j, k form

[sweet edge]

Is this what normal vectors are?

[nocturne minnow]

Yes

I think

[sweet edge]

Okie

anyways, we have our Force for the first tension

and also the other 3

when we sum these, they're equal to the force of gravity (since they need to cancel each other out)?

[nocturne minnow]

So, if you rewrite it, it's 0.577T1 i - 0.577T1 j + 0.577T1 k

Right?

[sweet edge]

that's..

[nocturne minnow]

I just expanded this

[sweet edge]

I'd prefer leaving it as $\frac{5}{\sqrt{75}}\hat{i}...$

[warm shale]

Jukelyn

[nocturne minnow]

That works too

[sweet edge]

yesh

in any case, we have these unit vectors which lets us construct a representation of the forces of tension

[nocturne minnow]

Yes

So, as stated, sum of forces equals 0

[sweet edge]

now we would add all 4 of them and that should be equal to the force of gravity?

or rather, I guess it depends on the POV

[nocturne minnow]

So T1 + T2 + T3 + T4 - 150(9.81) = 0

[sweet edge]

they sum to 0, but grav is neg so it's 0+ 150 (g) = tensions

[sweet edge]

yes

wouldn't this lead to a system of 4 equations though?

[nocturne minnow]

Sorta

[sweet edge]

hmm

[nocturne minnow]

$T1\left[\frac{5}{\sqrt{75}}\hat{i} - \frac{5}{\sqrt{75}}\hat{j} + \frac{5}{\sqrt{75}}\hat{k}\right] + T2 ... - 150(9.81) k = 0$

[warm shale]

dldh06

[sweet edge]

why the k at the end?

[nocturne minnow]

Because that force is in the k direction

[sweet edge]

oh right,

[nocturne minnow]

The weight

[sweet edge]

yes ofc

so, I'm trying to add them but it's getting extremely messy

[nocturne minnow]

Now, system of equations for x, y, z directions