help-23

[lean otter]

good

=0 *

that is the characteristic polynomial

[dire escarp]

yes

[lean otter]

now you know what to do right

[dire escarp]

get the 3 λ values

[lean otter]

yes get the eigenvalues by simplifying that

lmk if u get stuck

[lean otter]

and what do we do to these eigenvalues when we get them

how do we get the eigenvectors

[dire escarp]

find the plane each eigenvector is on?

like xz plane would be (1, 0, 1)

[lean otter]

you substitute them into

the matrix (A- Iλ)

or in your case the B - Iλ matrix

[dire escarp]

oh yea

i dont have much of a problem getting the eigenvector i think

[lean otter]

[dire escarp]

wait what about

[lean otter]

what about what

[dire escarp]

algebraic multiplicity and geometric multiplicity

am is how many lambda values right

[lean otter]

ah good

[lean otter]

yes

[dire escarp]

then the other is....?

[lean otter]

and geometric multipicity is the no of free variables in your B-λI matrix

[dire escarp]

um

[lean otter]

do u understand

[dire escarp]

no 💀

[lean otter]

actually

the number of times an eigenvalue repeats

for example

[dire escarp]

like square would be 2 but in this case it'll be 1 for all 3?

[lean otter]

if you got λ=5 and λ=5 and λ=1

algebraic multiplicity of 2 for the eigenvalue 5

[dire escarp]

but λ=1 would be am 1?

[lean otter]

yas

[lean otter]

no worries

[dire escarp]

from what i think rn i think gm is how many dimensions its on

xy would be 2 and xyz would be 3

[lean otter]

it will when you're solving each (A-Iλ) matrix

[dire escarp]

or is that wrong

[lean otter]

you'll see this in the number of paremeters in your eigenspace

[dire escarp]

like for each λ?

[lean otter]

yes

[dire escarp]

does that mean gm can only be 2 or 3

[lean otter]

no you'll see gm in your solution space

would you like us to see with your question?

[dire escarp]

like my question?

[lean otter]

yea

[dire escarp]

yea ofc

[lean otter]

okay

what eigenvalues did you get when you solved the characteristic polynomial

[dire escarp]

1 -1 and 2

[lean otter]

good

now we're going to find the eigenvectors for each of these

lets start with λ=1

write down the matrix for me and row reduce it

[dire escarp]

this right

[lean otter]

[lean otter]

you generally want your matrix in this form before solving

This helps us identify the free variables.
$$\begin{pmatrix}
1 & 0 & *\
0 & 1 & *\
0 & 0 & *
\end{pmatrix} $$

btw tip that i hope you wont have to use, if you ever get a full rank matrix when you reduce your A-Iλ you did something wrong

[flat frigate]

sen

[dire escarp]

full rank is all li right

and no zeros

[lean otter]

you'll see what i mean when you do more exercises

[lean otter]

a full rank is like the identity

1 0 0
0 1 0
0 0 1

pivots in every row

anyways

[dire escarp]

yes

[lean otter]

pls row reduce your matrix

[flat frigate]

sen

$$\begin{pmatrix}
0 & 3 &  0\\
0 & 1 & 0 \\
-2 & 1 & -2
\end{pmatrix} $$

[dire escarp]

so i make that into identity matrix?

[lean otter]

have you have done row reduction?

[dire escarp]

like 1st row - 2nd row?

[lean otter]

like have you done row reduction in your class

and why we do it

[dire escarp]

row echelon right

gaussian

[lean otter]

and the types of row operations

yes

$$\begin{pmatrix}
-2 & 1 & -2 \
0 & 3 & 0\
0 & 1 & 0
\end{pmatrix} $$

[flat frigate]

sen

[lean otter]

what row operation did i do

[dire escarp]

swap?

[lean otter]

yes

now reduce further

row 2 and 3 are dependent

[dire escarp]

yea

[lean otter]

so one of them will be a zero row

lmk if u get lost

[dire escarp]

1 0 1
0 1 0
0 0 0

[lean otter]

okay perf

one sec lemme confirm

good

$$\begin{pmatrix}
1 & 0 & 1\
0 & 1 & 0\
0 & 0 & 0
\end{pmatrix} $$

[flat frigate]

sen

$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 &  0\\
0 & 0 & 0 
\end{pmatrix} $$

[lean otter]

so we have

[dire escarp]

yea

[lean otter]

what are the free variables

[dire escarp]

um

[lean otter]

assuming x1,x2,x3

[dire escarp]

like non zeros?

[lean otter]

or x,y,z

the ones with no leading pivot column

[dire escarp]

x+z=0
y=0

[lean otter]

x1 is a pivot correct?

[dire escarp]

yes

[lean otter]

x2?

[dire escarp]

no

[lean otter]

why not

[dire escarp]

oh wait yes cause theres 1

[lean otter]

look at the second row

what about x3?

[dire escarp]

no

[lean otter]

right

so

x1 + 0 + x3= 0
0+ x2 + 0 = 0

right

[dire escarp]

yea

[lean otter]

since x3 is a free variable

lets let x3 be some variable s

now give me x1,x2,x3 in terms of this s

[lean otter]

do you understand this method

[dire escarp]

er

is x1 also a free variable

[lean otter]

or we can use x,y,z if thats easier for you

[lean otter]

is it from your matrix?

[lean otter]

ok lets make them x,y,z then

is x a free variable?

[dire escarp]

yes

[lean otter]

why

$$\begin{pmatrix}
1 & 0 & 1\
0 & 1 & 0\
0 & 0 & 0
\end{pmatrix} $$

[flat frigate]

sen

[lean otter]

its a pivot so its cant be

you get?

[dire escarp]

oh

so only z

[lean otter]

yes

[dire escarp]

ok yea ik now

[lean otter]

ok now back to this do this for me

but with z

[dire escarp]

like x=z??

[lean otter]

no

since z is a free variable lets let it be some variable s, s in R

what is x,y,z

in terms of this s

[lean otter]

this reduced matrix tells us
x + z = 0
y = 0

[dire escarp]

x+s=0?

[lean otter]

so what is x

[dire escarp]

-s

[lean otter]

and y?

[dire escarp]

0

[lean otter]

and z?

[dire escarp]

s

[lean otter]

ok now write them properly

x= blah etc

[dire escarp]

x=s like that?

x=-s y=0 z=s

[lean otter]

yes

$$\begin{bmatrix}
x \
y \
z \
\end{bmatrix} = s\begin{bmatrix}
-1 \
0 \
1 \
\end{bmatrix}$$

[flat frigate]

sen

[lean otter]

right?

[dire escarp]

yea

[lean otter]

now recall what i told you when trying to figure out geometric multiplicity

[lean otter]

.

how many parameters did we use

[dire escarp]

so theres only 1 s here

is it 1

[lean otter]

we only used s

if we had more free variables

we could have used more letters

s, t etc

[lean otter]

so a geometric multiplicity of 1

[dire escarp]

i see

[lean otter]

now this is our first eigenvector

now you can find second and third on your own

good job

[dire escarp]

hehe ok

[lean otter]

anymore questions?

[dire escarp]

nooo thats all

very clear now

[lean otter]

[dire escarp]

<3

tysm :)

[lean otter]

np np

[dire escarp]

should i close this then?

[lean otter]

ye you can

[dire escarp]

ty for your help once again

been a good hour

.close

[safe radish]

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[safe radish]

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[tardy copper]

Hi, idk what I am doing wrong. I followed the formula exactly, any idea where I am messing up?

[split fulcrum]

is this goodnotes 6

[spice flower]

mm reduction of order 🙂

[tardy copper]

@split fulcrum yup

hated it at first but got used to it super fast

@spice flower yessirrrr

I did the propblem withou the formula

and got it right

[spice flower]

im working through it now gimme sec

[tardy copper]

kk

[split fulcrum]

i don like how you gotta pay for it now

[tardy copper]

you always did no?

[split fulcrum]

good notes 5 you didnt

[tardy copper]

@split fulcrum 5 was paid too

[split fulcrum]

naw cuz it let me use it for free

[tardy copper]

School ipad or acc gives u both for free

:D

but it was always paid

[spice flower]

where did the - go on the top of the fraction exponent?

[tardy copper]

@spice flower u found my mistake lmao

[spice flower]

I was like I can't seeee

[tardy copper]

lmaoo u good

this math is kinda making me make super tiny errors

[spice flower]

yes it is fiddly

Always trust the way without the formula though.

formulas can easily be used incorrectly

[tardy copper]

@spice flower Yeah i prefer that way better but

my prof uses it so

just trying it out

[spice flower]

no problems then 🙂 gl

[tardy copper]

while i have u here

[spice flower]

ok

[tardy copper]

are you good with finding geniral solutions for undefined coefficients

[spice flower]

it's been a while since my diff eq but I can try go for it

[tardy copper]

like these types of problems:

[spice flower]

plz zoom 🙂

[tardy copper]

I understand all of it accept for the formulas used for geniral equations

[spice flower]

your approach looks fine

[tardy copper]

Its taken from this video:
https://www.youtube.com/watch?v=P3fc6v191mA&t=253s

I understand everything accept for how to actualyl get the formula

like the ax^2+bx+c

im really lost on that

[spice flower]

ok so

are you clear on why we have the Ax^2 term?

[tardy copper]

nope

I understand it has something to do with x^2 being g(x)

[spice flower]

It is because the G(x) in this equation is x^2. So we are saying that we have some multiple of that function term in our particular integral.

[tardy copper]

Hmm ok

but how does he tack on the rest of the linear equation

Is there any way you can hop in a vc rewally quick lol

Ive been reading about this for an hour and just getting more lost

[spice flower]

So when we differentiate 2 times as you can see the Bx term and the C term disappear

So we need to account for these in our particular integral also

give me second whilst I try and find a clear explanation for you. As I said it has been a while

[tardy copper]

no worries honestly its a hard concept to explain

[spice flower]

Well here

try it without and tell me what you get?

so just use the Ax^2 term

[tardy copper]

ok one sec

[spice flower]

you will see by doing

[tardy copper]

ok so let me restart the problem:

[spice flower]

I will do it with you

[tardy copper]

ok bet thank you

[spice flower]

first find the complementary function.

[tardy copper]

bam

thats y_c

ok so for y_p

this is where I kinda get lost

[spice flower]

so now you need your particular integral

[tardy copper]

is it ok if we go backwords a little, im still a little fuzzy on the difference between the particular solution and the complimentary one

I understand the superposition rule allows them to just be added together for the geniral solution

but im lost on the rules of what it actually is

if that makes sense

[spice flower]

sure

so the complementary function is the solution to the homogeneous case. In this example that is $$y''+5y'+6=0$$

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

ok, I understand that

[spice flower]

Now we have the nonhomegeneous case with $$y''+5y'+6= x^2$$

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

Ok

[spice flower]

and so we need to account for the x^2

[tardy copper]

wait just so I have all my bases covered

[spice flower]

or any function on the other side of the =

[tardy copper]

y=0 is always a solution

but we just dont consider it as one we are looking for

[spice flower]

to the homogeneous case yes

[tardy copper]

and for the homogenious equation, we find the geniral solution that matches that equation which has 3 cases

[tardy copper]

hmm ok, do we have to explicitly state this every time or is it just taken as the trivial solution

[spice flower]

trvial solution to the homogeneous case

now lets work on are particular integral

[tardy copper]

ok

[spice flower]

So you write down what you have figured out as the $$y_p(x)$$

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

hmm?

The general formula?

[spice flower]

the function we are choosing as a particular integral

[tardy copper]

ok, so the ax^2.... one

[spice flower]

so write it out in full

[tardy copper]

ok

[spice flower]

now find $$y_p^{''}$$ and $$y_p^{'}$$

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

Ok:

[spice flower]

now what do you think we do now?

[tardy copper]

I understand the part from here, we plug it into the equation and solve for abc

[spice flower]

go for it then

[tardy copper]

but im still lost on how to find the original equation in the first palce

ok one sec

[spice flower]

I have looked at your original solution. As you can see, if we did not have the Bx and C, we would have missed part of the solution

[tardy copper]

so its just intuition?

[spice flower]

let me ask you this then as a final point. If the G(x) was $$x^2e^x$$ what would the particular integral be?

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

Im not sure

[spice flower]

have a think and use what you have just done

its similar with a minor adjustment

[tardy copper]

ax^2e^c+bxe^x+ce^x?

[spice flower]

correct. Easier to write as $$(Ax^2 + Bx + C)e^x$$

[flat frigate]

𝓜𝓐𝒯𝒯𝓗𝓔𝓦ツ

[tardy copper]

hmm ok, but why is that? what are the rules for picking these equations?

[spice flower]

if you practise questions and look at examples, you will get used to it.

[tardy copper]

Like for example:

https://www.youtube.com/watch?v=AgyeJEO2a-k&t=467s

In this example, he tries ae^(-t)

[spice flower]

Because it all comes down to the fact that if they are not included, you end up missing parts of the solution.

[tardy copper]

which doesnt work

[spice flower]

he should use Ae^2t

than less

[tardy copper]

so he just simpally adds in ate^(-t)

[spice flower]

If e^2t is part of the complementary function, then yes, an adjusment needs to be made. This is a special case...

[tardy copper]

Ok wait I think i might understand it better if I understand exactly what is going on with the solution and how y_c and y_p are solutions

Like how does the complimentary function work on both? And also why does the superposition rule work?

Its ok if you do not have time for this btw I understand it may be a long explanation

[spice flower]

no I am kind off rusty, so my clarity is off, I apologise.

Do you have any ODE books?

[tardy copper]

no worries

I just have my teachers notes, we were not given books

[spice flower]

ok close this and dm me

[tardy copper]

.close

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[fleet pine]

in the 3rd step, where does the second 1/x+y come from? (implicit differentation)

[lean otter]

[
\bmap [\bigg] {\dv x}{\map \ln{\map f x}} = \f{\map {f'}x}{\map f x}
]

[fleet pine]

this one i mean

[cedar lily]

They distributed the 1/(x+y)

[lean otter]

oh then ^

[fleet pine]

but how and why?

[cedar lily]

Distributive rule iirc says a(b+c) = ab + ac

So a = 1/(x+y), b=1, c = dy/dx

[fleet pine]

ah i see

but is it necessary in this case?

[cedar lily]

Maybe there's another way to do it but it seems to have helped them in isolating the dy/dx term

[fleet pine]

ah okay thank you

also, why does the second dy/dx dissapear?

in this step

[safe radish]

@fleet pine Has your question been resolved?

[fleet tendon]

disappear how?

[safe radish]

@fleet pine Has your question been resolved?

[fleet pine]

well, why do you have only one dy/dx in the step after, instead of two

[fleet tendon]

it does, they just factored it out in front

if you distribute dy/dx in the parenthesis, then you should get the same thing as the left-most expression

[fleet pine]

ah thanks

.close

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[whole acorn]

i think so.

[cold aurora]

Yes, you are.

[safe radish]

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[jovial scroll]

can i perform the last step

[jovial scroll]

i.e a(c) + b(d) = (a+b)(c+d) ?

[cold aurora]

No.

[jovial scroll]

.close

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[astral lodge]

Hello, can someone help me with this problem?

[astral lodge]

"The triangle ABC has AB = 24 cm, BC = 36 cm, and AC = 32 cm. Point M belongs to AB in such a way that AM/MB = 3/5, and point N belongs to AC in such a way that NC = 20 cm.""

a) Determine the position of the line MN relative to side BC of triangle ABC.

b) Calculate the length of segment MN.

[fleet condor]

did you draw it

if so can you send the drawing

[astral lodge]

no, unfortunately

I will and I'll send in 1 min

I hope it s readable

[cold aurora]

!status

[safe radish]

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

[astral lodge]

2

[fervent shoal]

using the properties of similar triangles, we know that AM/AB = AN/AC = MN/BC

maybe that helps

[cold aurora]

!show

[safe radish]

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

[astral lodge]

AC = 32cm|
| = > AN = 12 cm
NC = 20cm|

[astral lodge]

don t I have to prove in someway?

[cold aurora]

Ah. Well, that's something.

[astral lodge]

I'm not sure

but

AN/NC = 12/20

simplified by 4 = 3/5

[cold aurora]

Yes

[astral lodge]

as in

[cold aurora]

Yes

[astral lodge]

AM/MB = 3/5

[cold aurora]

Yes. AM/MB = AN/NC

Any clue as to where you have seen that?

Or seen something like that?

[astral lodge]

thales?

[cold aurora]

YES

[fervent shoal]

we never had to

[astral lodge]

resulting MN || BC

[cold aurora]

Exactly

[astral lodge]

I was confused cause I thought I have to write in a way, or to prove that AN/NC

So

[cold aurora]

So, now, you can use similarity of triangles.

[astral lodge]

oh

I don't really know to use it, I mean, to write the ratios

[cold aurora]

Are you familiar with the criteria of similarity?

[astral lodge]

um, AAA being 3 angles, SAS being 2 sides and 1 angle and SSS being side side side?

is that what u mean?

[cold aurora]

Yes

[fervent shoal]

SSS is a congruency, right?

[cold aurora]

works here as well.
Of course, here, they mean ratios are equal.

[astral lodge]

hmm

so

I have triangle ABC ~ triangle AMN

[cold aurora]

Not ABM

[astral lodge]

sorry, ABC

is AB/AM = BC/MN = AC/AN?

[cold aurora]

Yes

[astral lodge]

24/x = 36/x = 32/12

BC/X ( MN ) = AC/AN

36/X = 32 / 12

x = 36*12
------
32

?

I think I'm wrong somewhere

it gives me a fraction

or wait

nvm

@cold aurora, sorry for tag, do you have any idea?

[cold aurora]

Why 24/x ?

Do you not know AM?

[astral lodge]

AM/MB = 3/5

[cold aurora]

Use that.

[astral lodge]

not really

how?

[cold aurora]

You know AB.

So MB = ?

[astral lodge]

5 am = 3 mb

[cold aurora]

From figure, what do you know about AM, MB, and AB?

[astral lodge]

AB = 24, 5 AM = 3 MB

[cold aurora]

You know more.

Like you know AN + NC = AC

[astral lodge]

yea, but I knew NC was 20

[cold aurora]

Similarly.

[astral lodge]

So I just subtracted

[cold aurora]

You know AM + MB = AB

AB = 24

You can write MB in terms of AM

[cold aurora]

Using this

[astral lodge]

umm, I am a bit confused, having AB/AM = BC/MN = AC/AN can't I do derived proportions ?

and have BC/MN=/AC/AN

cause I already know 3 of them

.close

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[safe radish]

@pulsar vine Has your question been resolved?

[safe radish]

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[full agate]

show that log3, log6 and log12 are consecutive terms in an arithmetic sequence

[full agate]

all of them are base 2

this is the answer

why does doing this work?

[earnest mica]

in an arithmetic sequence a,b,c a+c=2b

[peak estuary]

if a1+a3=2a2, that means a3-a2=a2-a1

[earnest mica]

bcuz c-b=b-a

[peak estuary]

aka common difference. aka arithmetic sequence

[full agate]

thats interesting

havent seen that before

but makes sense

thanks

.close

[safe radish]

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[cold aurora]

Hello.

Prove or disprove the following statement.

"Translation of a cartesian point without going into the homogeneous coordinate system is a linear transformation."

[cold aurora]

Now, i have a really shaky understanding of this topic. I think that by linear transformation, they mean - there exists a matrix with which you can multiply the coordinate matrix to get the translated coordinates.

Is that right?

I can do this very easily by going into the homogeneous coordinate system. Without using homogeneous coordinate system, i'm not so sure.

Any useful input whatsoever is appreciated. Thank you.

<@&286206848099549185>

[fleet tendon]

what does "without going into the homogeneous coordinate system" mean?

[cold aurora]

Well, homogeneous coordinate system is a technique of representing a n-dimensional vector into n+1 dimensional vector by addition of dimension w.
w is usually taken as 1.
Basically, point (1,2) becomes (1,2,1). It's as if we put the point in z=1 plane.

Then i can easily multiply with this matrix:
[1 0 Tx]
[0 1 Ty]
[0 0 1]

This gives me translated coordinates. Tx is the amount by which x coordinates have to be translated. Similarly Ty

[fleet tendon]

https://tenor.com/view/why-huh-but-why-gif-13199396

[peak estuary]

because it allows translation to be a linear transformation

among other nice things

[cold aurora]

So without this, it can't be a linear transformation?

[peak estuary]

the issue with normal n dim space is that the zero vector has a bit of a special role

which you don't want in certain contexts

[fleet tendon]

that just sounds like a bs excuse lol

[peak estuary]

no, because it doesnt send zero to zero

which a linear transformation always does

well projective space in general is quite nice

for all kinds of geometry stuff

which builds on these coordinates

[fleet tendon]

why not just deal with affine transformations then?

don't they allow translation

[cold aurora]

Oh. I think i get it.

It's because no matter what 2-D matrix we choose to multiply with coordinate matrix, (0,0) will stay (0,0) and not be translated. Right?

[cold aurora]

That is my next question after this. Lol

[drifting mountain]

translation is not a linear transformation

[peak estuary]

sending zero to zero is one of the properties of a linear transformation

[cold aurora]

Thank goodness. So i did it correct.

[peak estuary]

if L is linear, that mean L(u+v)=L(u)+L(v). pluggin in v=0 gives you L(0)=0

[cold aurora]

Probably using wrong reasoning though.

[fleet tendon]

You just have to give one counter example

[cold aurora]

Ah.

[fleet tendon]

and that's the zero argument

[cold aurora]

Next question was -
Translation of a cartesian point without going into the homogeneous coordinate system is an affine transformation.

Is that true?

[peak estuary]

well its a trade off. you use one dimension more but get quite a bit of other transformation which suddenly are linear. which previously weren't even close to linear

[peak estuary]

what do you think

[cold aurora]

Also - please correct my understanding of affine transformation.
I understand it as "one linear transform followed by a translation". How wrong is that?

Based on my this understanding, i concluded that it would be false as well.

[peak estuary]

that's correct

then why is it false

[cold aurora]

Because translation part isn't linear transform so we can't write that as TX so overall, we don't get (T+T')X ?

[peak estuary]

?

Ax+b

[cold aurora]

Sorry. Let me write again.

[peak estuary]

A matrix

b vector

[cold aurora]

Okay. I'll follow that one.

So, if x is my vector -
Ax is a linear transform. b is the translation vector.
So i thought that it's an affine transform if
we can club that Ax + b as Tx where T is single matrix which gives the same effect as Ax+b.

[drifting mountain]

that's linear

[cold aurora]

Maybe that's not what affine transform means. Does it mean that as long as there is an A such Ax + b does the translation, we'll call it affine?

[peak estuary]

affine means first do linear, then translate

just like you said

and if we look at Ax+b that's exactly what happens to the x

first linear part, we get Ax

[cold aurora]

Oh. So it is an affine transformation with A as identity matrix ?

[peak estuary]

and then we translate by b and get Ax+b

yes

[cold aurora]

Damn it. I thought that and then i messed it up.
Anyway, thank you all so much.

[peak estuary]

you're welcome

[cold aurora]

I understand that my wordings were poor. Thank you for bearing.

I should probably study linear algebra for better understanding of this?

[peak estuary]

yes

[cold aurora]

Okay. Thanks.

.close

[safe radish]

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[smoky fulcrum]

how do i calculate the first number

[mystic flint]

find the roots

add and multiply

separately

then

[spice flower]

why did you square them?

[mystic flint]

yes

as i was saying

it should be

x^2

-(-b/a)x+(c/a)

where -b/a is the sum of roots

and c/a is the product

waitttttttttttt

wait

no

[safe radish]

@smoky fulcrum Has your question been resolved?

[smoky fulcrum]

?

[mystic flint]

i am seriously

lost

[safe radish]

@smoky fulcrum Has your question been resolved?

[safe radish]

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[slate thorn]

how do i evaluate this?

[light shoal]

hint: what impact does the value of f(0) have when evaluating this limit?

[slate thorn]

makes it either 1 or 3?

well it would approach 1

oh

nvm

[gentle latch]

how can this

simplify

to this

[slate thorn]

this would be the graph right?

[slate thorn]

$\sqrt[3]{z^{5}} = z^{5/3}, z*\sqrt[3]{z^2}=z^{2/3} \implies \frac{\sqrt[3]{z^{5}}{z*\sqrt[3]{z^2}} = \frac{ z^{5/3}}{z* z^{2/3}}$

[flat frigate]

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

[gentle latch]

wha

[slate thorn]

one sec

@gentle latch

$\sqrt[3]{z^{5}} = z^{\frac{5}{3}}, z*\sqrt[3]{z^2}=z^{\frac{2}{3}} \implies \frac{\sqrt[3]{z^5}}{z*\sqrt[3]{z^2}} = \frac{z^{\frac{5}{3}}}{z*z^{\frac{2}{3}}}$

[flat frigate]

Joshii

[slate thorn]

$\implies \frac{z^{\frac{5}{3}}}{z*z^{\frac{2}{3}}} = \frac{z^{\frac{5}{3}}}{z^{\frac{2}{3}+1}}$

[flat frigate]

Joshii

[slate thorn]

can you do the rest from there?

@gentle latch

[gentle latch]

ye

[slate thorn]

do you understand it?

[gentle latch]

it was a part of a bigger question

hold up

[slate thorn]

also please read #❓how-to-get-help next time

[gentle latch]

do u want me to send the bigger question

ok

[slate thorn]

@light shoal

[light shoal]

correct

[safe radish]

@slate thorn Has your question been resolved?

[safe radish]

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[onyx pike]

i thought my algebra was good but it isn't

[onyx pike]

so i have u=((c/n)+v)/(1+(cv/nc^2)) and i need it in the form of

u = (c/n) + v - (v/n^2)

starting with this

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[safe radish]

@onyx pike Has your question been resolved?

[safe radish]

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[ancient nimbus]

hi can someone help me with this? its quite confusing

[ancient nimbus]

do I expand?

[ember sphinx]

what are you trying to do with it?

simplify?

[ancient nimbus]

yes

[ember sphinx]

ok

see what the top and bottom have in common

so instantly i see the top has an x^4 and the bottom has an x^2

since x^4 contains an x^2

x^2's

so we're left with 9x^2(x-y)^3 on the top and 27y^3(x-y) on the bottom

yay or nay?

[ancient nimbus]

yay

[ember sphinx]

ok

now lets see if we can find naything else

9 and 27

you think they have anything in common?

[ancient nimbus]

9

[ember sphinx]

yep

[lean otter]

whats up

[ember sphinx]

so cancel out the 9's (27 = 9 x 3)

[lean otter]

wait

nvm

yall got it

or do you guys need help?

[ancient nimbus]

if i simplify 9x^2 and 27y^3 itll become 9x^2 the common factor

[ember sphinx]

no

one contains a y^3

the other is an x^2

x is not y

[ancient nimbus]

oh sorry my bad

so therefore if simplified its 9x^2(x^2)/9x^2(3y^3)

[ember sphinx]

no

we cancled out the 9's

[ancient nimbus]

this then?

[ember sphinx]

so we left with x^2(x-y)^3 / 3y^3(x-y)

you shouldnt have a 9 at the top

9/27 = 1 /3

[ancient nimbus]

oh okok

[ember sphinx]

now

[ancient nimbus]

im confused about

(x-y)^3

[ember sphinx]

what you confused about

Yea

Ok now on that

we know the top has an (x-y)^3

and the bottom has an (x-y)

since everything in the top and bottom of the fraction are times by eachother (apart from within these brackets)

we can cancel out ONE (x-y) to get rid of the bottom one

and left with (x-y)^2 on the top

[ancient nimbus]

ok

the equation will be x^2(x-y)^3/3y^3

?

[ember sphinx]

(x-y)^3 or (x-y)^2?

[ancient nimbus]

(x-y)^2

[ember sphinx]

yea

so final would be

x^2(x-y)^2/3y^3

[ancient nimbus]

hm let me try to solve it myself if i got it step by step

brb

[ember sphinx]

ok

[ancient nimbus]

how did the (x-y)^3 turn to (x-y)^2?

ohh nvm got it

thanks

.close

[safe radish]

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[fierce gust]

topic: trig integrals

[fierce gust]

this is what I have so far

what is the next step?

[crystal palm]

so far so good

i would expect that we could somehow do this with partial integration, but maybe that would take some effort

[fierce gust]

photomath says to use sub

but im confused on where they are getting cos(t)

[crystal palm]

oh yeah that is a good idea

think about it this waay

u=f(t)

then du/dt=u'(t)

so dt=du/u'(t)

meaning with a substitution, we have to divide by it's derivative

so in this case, we want to reduce the number of sines

and if we use cos as a substitution, we will have to divide by its derivative

which happens to include a sin

[safe radish]

@fierce gust Has your question been resolved?

[safe radish]

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[soft marsh]

how does taking the deravitive of a function determine if it has an inverse or not

[plucky elk]

Are you working on an actual problem

[soft marsh]

i understand that the derivative is 2x + cos(x)

but isnt this always increasing? so should it not have an inverse, but the answer says it shouldn't

[plucky elk]

No it's not always positive

,w plot 2x + cos(x)

[flat frigate]

Results provided by WolframAlpha

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

Requested by rie.mann

[plucky elk]

It's negative for x < -1

[soft marsh]

I don't understand how this proves that an inverse does not exist if anything should it support it? since each x has its own y value

[plucky elk]

each x has its own y value
You haven't proven that yet

Use the definition of one-to-one if you think it is

[soft marsh]

do i try this on the original function or the one i derived?

[plucky elk]

Whatever the question asks for

[soft marsh]

so i tried on the original and solved for 0 and was proven wrong

I see now how this does not have an inverse

so then when would we have to use the derivative test? to determine where it does have one to one in a given interval?

[soft marsh]

like with number 8 where it is an inverse?

because then we get 1+cos(x) = 0 and then I would determine where it is only increasing and only decreasing?

[plucky elk]

Inverse functions exist when the function is monotonic

So you have to restrict your domain to those intervals

https://proofwiki.org/wiki/Inverse_of_Strictly_Monotone_Function

[soft marsh]

ohhhhhhhhhhhhhhhhhhhhhh

i got it

i think

i will practice more thanksss

.close

[safe radish]

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[obsidian oxide]

im quite confused on how to proceed with this

[obsidian oxide]

im not sure how to even start it

i mean, i know i must do the base case and then proceed with the inductive step but im not sure how to do it without being given a statement i guess

[stoic dune]

If you've payed n dollars, is there a way to swap coins and pay n+1 dollars instead?

[obsidian oxide]

what do you mean swap coins?

[quasi bison]

the statement is "a bill of n dollars is payable using 3s and 7s only"

[stoic dune]

Actually, I think my way of thinking has problems. Hmm.

[obsidian oxide]

like do you mean that if we're able to pay 12$ then we should be able to pay n+1$ by changing the amount of 3$ and 7$ bills?

[stoic dune]

But yes that's what I meant

[obsidian oxide]

i know but i meant a statement similar to the one given in part a

[quasi bison]

what like symbol soup????

i mean idk like

$(\exists a \in \bN)(\exists b \in \bN)(n = 3a + 7b)$ or something? good luck symbolpushing your way through that one tho.

[flat frigate]

Ann

[obsidian oxide]

i've only ever seen my teacher do it with the summation at the front there

[quasi bison]

this is not a summation at all

this isnt about any kind of summation

not all induction problems are about summations

[obsidian oxide]

i know

i just mean that i have only ever seen my professor do like 2 problems and they both had the summation at the front

thats why im wondering how to do it if its a word problem

i actually thought it would be 3a+7b=n but i realized thats probably not the best way to do it lol

[stoic dune]

It's the same idea. "Assuming the proposition is true for n, prove it's true for n+1"

[obsidian oxide]

i see

but then what do i do the inductive step on

i think i understand the base case part, i just have to do n=12 and show that 3(4)+7(0)=12 , right?

[quasi bison]

yes

or in words, that's "$12 can be paid with 4 threes"

[obsidian oxide]

ok thanks

does that mean i have to use the "3a+7b=n" for the inductive step if i say "n=12 -> 3(4)+7(0)=12"

[quasi bison]

"have to" ...

[obsidian oxide]

as in, am i restricted to using that expression?

[quasi bison]

think you might be imagining rules or requirements that just aren't there.

so im gonna say no you're not

[obsidian oxide]

i see

though i'm not sure i understand how to proceed with the inductive step, would i just need to show that the proposition is true for 13, 14, 15, 16 etc.

also, the hint in the question states that i would need to prove more than one base case once i do the inductive step

so im assuming that i would show that the proposition is true for 12, 13, 14, 15, 16 etc. in the base case

[safe radish]

@obsidian oxide Has your question been resolved?

[stoic dune]

Inductive step always looks like this:
"Assuming the proposition is true for n, prove it's true for n+1"

You want to induct over the cash amount.

So, assuming you can pay n dollars, prove you can always pay n+1.

.
Let's do a numerical example. Let's say I want to pay $20 with a 7, 7, 3, 3.

Then I'm told I actually need to pay $21.

How might I swap out my coins to get the amount I want?

@obsidian oxide

[obsidian oxide]

use 7, 7, 7

[stoic dune]

In general, what is one way we can pay an extra dollar?

[obsidian oxide]

switch between $3 and $7 bills

[stoic dune]

Can you be more specific? Swap how?

What did you do above, to get from $20 to $21?

[obsidian oxide]

i took two $3 bills and replaced them with one $7 bill to get the extra $1 needed

[stoic dune]

Right. That's a way we can generate n+1, given we have n.

Does it always work?

[obsidian oxide]

what do you mean thats how we can generate n+1 given n?

& yes it should always work

[stoic dune]

Can I get from $21 to $22 this way?

$21 is a 7,7,7

[obsidian oxide]

yes you can

one 7 and five 3's

[stoic dune]

When I say "this way" I mean with the "swap two 3s for a 7 strategy"

We're looking for a strategy that takes us from $n to $n+1

[obsidian oxide]

oh ok

so in order to get the +1 to bill "n", we take two 3s and put in a 7

[stoic dune]

That only works if we have two 3s, though

You did something different to get from $21 to $22

[obsidian oxide]

one 7 and five 3's

so it will be a different combo every time

[safe radish]

@obsidian oxide Has your question been resolved?

[obsidian oxide]

@stoic dune

[safe radish]

@obsidian oxide Has your question been resolved?

[safe radish]

@obsidian oxide Has your question been resolved?

[safe radish]

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[safe radish]

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[lean otter]

What's wrong with my working

[lean otter]

$2^{(4 \cdot 4)^{x}} = 16^{2^{x}}$

[flat frigate]

Mushaar

[lean otter]

Using exponent properties

put both in base 16

[violet viper]

You don't need to

[lean otter]

yeah i know you can solve it with log much easier

[violet viper]

But do it for the other 16

[lean otter]

i'm trying to see what's wrong with this process

[thin bridge]

can you type out your work in its entirety

or take a pic

[quasi bison]

least answered question in the universe

[lean otter]

so continuing from here $16^{4^{x}} = 16^{2^{x}}$

[flat frigate]

Mushaar

[thin bridge]

that's where you screwed up

[lean otter]

oh wait i see

nvm

thanks

.close

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[lean otter]

So over here, after log base change formula I get

$\frac{1}{log_B A} = log_B A$

[flat frigate]

Mushaar

[lean otter]

and then multiplying out, I get log_B (A) = 1,-1

oh wait I see

it's the second case

so them multiplied is 1

.close

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[livid scaffold]

hello is this correct or do I need to cancel the /cm^3 in final answer

[livid scaffold]

do I need to put this?

to cancel cm^3 cuz I still don't get it

do I even have to cancel it?

[safe radish]

@livid scaffold Has your question been resolved?

[safe radish]

@livid scaffold Has your question been resolved?

[safe radish]

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[serene summit]

.reopen

[daring cairn]

!help

[daring cairn]

Problem Solving. Solve the following problems. Use significant figures when showing your final answer.

is it correct

[lean otter]

how do i call for help?

[fleet condor]

I forgot how sig figs work but that seems like the wrong amount

[daring cairn]

just type "!help"

[daring cairn]

ohhh

[lean otter]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[solid shell]

Go to a channel that is vacant (category above) and type something in an empty one

[daring cairn]

here

[fleet condor]

what are the multiplication and addition rules for sig figs

[daring cairn]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[daring cairn]

hehe is the ans

.close

[safe radish]

Channel closed

Closed by @daring cairn

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[devout raven]

Sorry, @plucky elk u said this was wrong bc wrong formula but channel closed so I couldn’t ask, but how is it the wrong formula, isn’t aT= r’ dot r”/r’mag

[devout raven]

And aN is r’ x r”/ r’mag

Top is mag as well

[plucky elk]

yea you didn't show enough of your work to arrive at your answers and i can barely read your handwriting

[devout raven]

Does this help for the top one

[plucky elk]

no

[devout raven]

Damn

[safe radish]

@devout raven Has your question been resolved?

Channel closed

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Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[foggy edge]

''Determine the area of the regular poligon, knowing L=2cm''

just need to confirm if the answer is 9.659078631!

[obtuse plover]

I put in the side length into googles septagon area calculator and I got something else

[foggy edge]

oh right

I see what I did wrong

14.54 isn't it?

[obtuse plover]

Ye

[foggy edge]

cheeers thank you

.close

[safe radish]

Channel closed

Closed by @foggy edge

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jade lotus]

1/2+2/5t-1=1/5t+t

[drifting mountain]

!show

[safe radish]

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

[bold aurora]

What are we doing with it? What did you try?

[jade lotus]

I need help solving it

[heady osprey]

is that 2/(5t) or (2/5) t

[jade lotus]

2/(5t)

[gentle oak]

Is it $\frac 12 +\frac{2}{5t} -1=\frac{1}{5t}+t$ or $\frac 12 +\frac{2}{5t-1}=\frac{1}{5t}-t$ ?

[flat frigate]

Joseph.P

[gentle oak]

Ok so get the t’s to one side and the constant on the others side

[jade lotus]

How do I do that

[gentle oak]

Add/ subtract t

[jade lotus]

So I move t to the left

[gentle oak]

Ye or to the right if you want

[jade lotus]

Ok what next

[gentle oak]

1 second

[jade lotus]

Oo

[gentle oak]

Ok so you got $\frac 15\cdot t-t=\frac 12$ correct ?

[flat frigate]

Joseph.P

[jade lotus]

Where did the 2/5t go?

[gentle oak]

2/5t-1/5t=1/5t

[normal moss]

2/5t - 1/5t

[jade lotus]

Ok

What next

[gentle oak]

Reduce it to the same denominator

[jade lotus]

Using 10?

[gentle oak]

10 ?

[jade lotus]

Ye

[gentle oak]

Show me I don’t understand what you did

[jade lotus]

By * 10 by everything

Hello?

[gentle oak]

Sorry I don’t see what you’re doing but if you think you can do it like this do it just that I didn’t do the same as you

[jade lotus]

Do I reduce it by itself?

Let’s go back to what you were saying

[gentle oak]

Ok so you have 1/5t -t=1/2 right ? If so multiply -t by 5

And also to not make errors write 1/5t as t/5

[jade lotus]

Ok

[gentle oak]

Now you can multiply both sides by what ?

[jade lotus]

by 1/2?

[gentle oak]

Actually you can but there’s a simple way

[jade lotus]

Ok what is that way

[gentle oak]

By 5

[jade lotus]

5/1

[gentle oak]

Can you write what did you get exactly

[jade lotus]

1/2 *5/1

[gentle oak]

What is it equal to ? In term of t ?

[jade lotus]

2.5

[gentle oak]

This is 5/2

[jade lotus]

Ok

[gentle oak]

What I want is $k\cdot t=\frac 52$ with $k$ is a real number

[flat frigate]

Joseph.P

[jade lotus]

That’s what I had before but I thought it was wrong

What else do we do next?

[gentle oak]

So for you what’s the value of k

[jade lotus]

I don’t know

-1?

[gentle oak]

You said you had it before

No

Don’t worry

[jade lotus]

Ok, apparently what I thought I had was right was wrong

[gentle oak]

I’m going to explain it in a better way

[jade lotus]

Ok

[gentle oak]

So $\frac t5 - t=\frac 12$

[jade lotus]

Ok

[flat frigate]

Joseph.P

[gentle oak]

That’s what we got by putting all the t’s on one side and the constant on the other

[jade lotus]

Ok

[gentle oak]

And you have $\frac t5 -\frac{5t}{5}=\frac 12$ right ?

[flat frigate]

Joseph.P

[jade lotus]

Yes

[gentle oak]

So it’s $\frac{-4t}{5}=\frac 12$

[flat frigate]

Joseph.P

[gentle oak]

What can you do from now ?

[jade lotus]

You take the 5 and times it by 1/2

[gentle oak]

Yes

What else

[jade lotus]

And times it by the T as well

[gentle oak]

No when you multiply by $5$ you get $-4t=\frac 52$

[flat frigate]

Joseph.P

[jade lotus]

Oh yes

5*-4 =-20

[gentle oak]

Where did the t go ?

[jade lotus]

-4+ t =5/2

[gentle oak]

It’s not -4+t but -4 time’s t

[jade lotus]

Ok

[gentle oak]

Did you solve it