help-33

[marsh citrus]

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[terse vale]

very lost here

im not sure what to do next

[main idol]

You didn't fix your partial fraction decomposition from last time

[terse vale]

i know it takes this form

but i dont understand how this translates into an inverse laplace

you showed me this earlier

but how do these turn into an actual y(t), especially that first term

[main idol]

Do you not have a table of Laplace transforms

[terse vale]

i do

[main idol]

Do some algebra on this expression until it matches some identities from your table then

[terse vale]

oh i see

i just got it

thanks

.close

[marsh citrus]

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[marsh citrus]

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[idle stratus]

how am i supposed to even approach this? (this is for UIL calc apps)

i'm assuming it might be something similar to heron's formula where it's just some hyperspecific situation where one thing just works

[waxen dust]

What are AB, BC, and BD?

[idle stratus]

like diagonals of the sides

[waxen dust]

Exactly

[idle stratus]

how do i get to the volume from that

is there something obvious i'm missing

[waxen dust]

Diagonals of which shapes

[idle stratus]

rectangles

[waxen dust]

Yes

And how do we find the diagonal of a rectangle

[idle stratus]

hypotenuse thingies

[waxen dust]

Also known as Pythagorean Theorem, I assume?

[idle stratus]

yeah yeah

[waxen dust]

Okay

Let's label the length, width and height as l, w, h

Is it clear which is which

Or do I draw it

[idle stratus]

draw it pls

[waxen dust]

Usually it's like this

It doesn't really matter

[idle stratus]

ok

[waxen dust]

You can use x, y, z as well

You will get the same result

But anyway

How do we find BD, for example

[idle stratus]

BD would be sqrt( l^2 + w^2)

[waxen dust]

Exactly, but let's use squares first and we will see if we even need square roots later

[idle stratus]

alright

[waxen dust]

I will label BD as d_1, i. e. the first diagonal. You can keep it BD, as you wish

[idle stratus]

ok

[waxen dust]

$d_1^2 = l^2 + w^2$

[idle stratus]

k yeah

[waxen dust]

Okay

And for the other two

AB, or d_2, and BC, or d_3?

[idle stratus]

so d2^2 is l^2+h^2

and d3^2 = h^2+w^2

[waxen dust]

Exactly

Also, we know what d_1, d_2, d_3 are right

It's given

[idle stratus]

yeah

[waxen dust]

Okay

So if we write it all out we get

[
\begin{cases}
l^2 + w^2 = 21.5 \
l^2 + h^2 = 20.8 \
w^2 + h^2 = 10.6
\end{cases}
]

[waxen dust]

I just replaced d_1, d_2, d_3 with 21.5, 20.8, 10.6 respectively

Following?

[idle stratus]

yeah

[waxen dust]

Well and this is a system of three equations with three unknowns

And we can attempt to solve it

There's a really neat trick that saves so much time

[idle stratus]

is it the matrix stuff or

i've heard those are fast for these types of questions

especially for calc apps

[waxen dust]

Well, could be, definitely

But an even faster one

Let me ask you this

How many times does each letter appear

[idle stratus]

twice each

[waxen dust]

Okay

So if we added all equations together, we'd get each letter (squared) exactly twice

Right

[idle stratus]

yeah

[waxen dust]

So let's do it

$l^2 + w^2 + l^2 + h^2 + w^2 + h^2 = 21.5 + 20.8 + 10.6$

[waxen dust]

Right?

[idle stratus]

yeah

[waxen dust]

Okay now simplify this

Also sorry, I completely forgot to ask. The reason why we are even doing this. How do we calculate the volume of a cuboid?

[idle stratus]

just multiplying all the dimension lengths right

[waxen dust]

Yes

That is why we are trying to find l, w, h

Because volume is just l * w * h

[idle stratus]

i'm like trying to look at it to find something but i can't lowk

[waxen dust]

Oh I just meant algebraic simplification

$l^2 + w^2 + l^2 + h^2 + w^2 + h^2 = 21.5 + 20.8 + 10.6$

[waxen dust]

for example, l^2 + l^2 = 2l^2

[idle stratus]

oh so just like
$2(l^2+w^2+h^2)=53.9$

[elfin berry]

boneless

[idle stratus]

oh it includes the text

oops

[waxen dust]

Exactly

Though 52.9

[idle stratus]

i haven't used the bot in a while

yeah

[waxen dust]

No problem

We can divide both sides by 2 as well right

[idle stratus]

yeah

[waxen dust]

$l^2 + w^2 + h^2 = 26.45$

[waxen dust]

Anyway

We get each of the dimensions (squared) from this pretty easily

Look at the original system of equations

[idle stratus]

oh so just like subtract any one of those and it'll isolate one of the squares

[waxen dust]

Exactly

For example from the first equation

We have l^2 + w^2 = 21.5

So we get 21.5 + h^2 = 26.45

And you have h

And then just do it twice more for the other two

And you have all 3 dimensions

Then you can compute the volume

[idle stratus]

ok yeah

so it's

add the sides and divide by 2
subtract each of the sides from that and then multiply them all

or the

diagonals i mean

[waxen dust]

Well first

This is just a trick that works for this specifically

As we notice we have these "piecewise sums" (l^2 + h^2, etc.)

[idle stratus]

yeah

[waxen dust]

And each variable appears exactly once in two equations

So adding together all equations makes sense

But otherwise, in other systems, this usually doesn't help

But it's still nice to know

And after adding all equations together

We just do some algebraic manipulation

Like adding like terms, factoring out 2

Pretty much simplifying, like any other equation

And then you notice you can rewrite parts of the equation with the help from the 3 original equations

Which was exactly the point of us adding all equations together

We notice that, for example l^2 + w^2 = 21.5

And since our equation is l^2 + w^2 + h^2 = 26.45

We can replace (l^2 + w^2) with 21.5

To get 21.5 + h^2 = 26.45

And get h that way

And this works out for all 3 of the variables

As we already know the sum of the squares of the other two (from the 3 original equations)

[waxen dust]

But again this

If you have a system like

x + 2y = 5
x - y = 4

Adding both equations, we get
x + 2y + x - y = 9
2x + y = 9.
Which doesn't help much right, just another equation

[idle stratus]

yeah

[waxen dust]

But yeah that's kind of the idea

[idle stratus]

ok

[waxen dust]

We wanted to find the three dimensions of the cuboid

And we had the diagonals of the rectangles which form the cuboid

And we know how to calculate those diagonals

And from that we just construct a system of 3 equations with 3 unknowns

And solve it

🙂

[idle stratus]

alright, that helps a lot

[waxen dust]

Hope I've been of help

Anyway, I wish you all the best in math and life!

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[idle stratus]

ok, thanks

.close

[marsh citrus]

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[waxen dust]

Oh, also to add on. This doesn't mean you have to solve the system this way! It was just a trick I wanted to show. You can of course solve it via any other method you know

Like elimination for example

It just takes much longer

[idle stratus]

yeah

[snow prairie]

Messi 🐐

[marsh citrus]

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[hearty crescent]

hello im really want learn linear algebra but im really bad in math like i dont know math at all and im really asking for help to give me a plan of lessons just name of lessons that i need to learn math first and that math can help me to learn linear algebra and I appreciate your help .

[red nimbus]

I suppose some level of logic and set theory would help

pre algebra as well

[hearty crescent]

first pre algebra then algrebra then linear algebra

alright thank you guys i appreciate it

[cunning fiber]

For future reference, help channels are for specific questions and not general ones. This question would be better suited for the discussion channels (either #prealg-and-algebra or #linear-algebra, depending on if you conflated solving basic linear equations with the field of linear algebra as a whole). You can also dig the message histories of #book-recommendations and the discussion channels for resources that people have mentioned.

[hearty crescent]

alright ty

[marsh citrus]

@hearty crescent Has your question been resolved?

[marsh citrus]

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[remote barn]

You are a colorblind K-9 police dog training to become a bomb sniffer. As a dog you can only see blue and yellow, while red and green both appear grey. In order to hone your abilities, your trainer wants you to test the claim "If a bag is blue, then it has a chicken inside." There are a total of 5 bags. You were only allowed to look at three bags, while the other two bags you were only allowed to sniff.

• The first bag appears blue.
• The second bag smells like chicken.
• The third bag appears grey.
• The fourth bag smells like beef.
• The fifth bag appears grey.
  What is the least amount of bags you can pick for security inspection to verify the truth of the claim, and which bag/s do you choose?

I'm kinda confused but is it 2 because you just check if the blue bag has chicken and check if the chicken bag is blue, the first and 2nd one, although I have a slight suspicion 3 might be right because you need to check the 4th bag too if it's blue

[gritty kelp]

what if the first bag has a chicken inside and the second one is blue ?

[remote barn]

then we don't know about the 4th bag

if the 4th bag was blue, and it had beef then the statement is false

also we don't have to check the 2nd one

I'm imagining it rn any bag can have chicken, if it has chicken inside does not necessarily mean it's blue

oh ok

.close

[marsh citrus]

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[marsh citrus]

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[pallid tapir]

. another example of implication inspection that's helpful: suppose you're ensuring everyone drinking alcohol at a bar is over 21. which of the following do you need to verify:

• someone who is 18
• someone who is 25
• someone who is drinking water
• someone who is drinking alcohol

[remote barn]

the 1st 2nd and 4th one?

[remote barn]

I'll close it later whatevs

[pallid tapir]

haha sorry I couldn't warn you in time to preface with a dot

[remote barn]

nah it's fine I'll just close it

[pallid tapir]

well, suppose you know someone is 25

[remote barn]

I have another question that I don't really wanna picture

[pallid tapir]

if they're drinking water, no biggie right?

[remote barn]

yeah

[pallid tapir]

what if they're drinking alcohol?

[remote barn]

if they're 18 and drinking alcohol, no go, if they're 25 and drinking alcohol, it's fine so don't we have to check the 2nd one or no?

[remote barn]

still fine

[pallid tapir]

it's assumed that the 2nd person had been carded and is known to be 25

only the drink is not verified

yet, in both cases (water v. alcohol), they are fine

[remote barn]

oh so we don't have to check for that

[pallid tapir]

yes

[remote barn]

oh ok

[pallid tapir]

why do we need to check the 18 y.o.'s drink and the age of the alcohol drinker?

[remote barn]

cuz bad

[pallid tapir]

"if you're 18, you can't drink alcohol" is logically equivalent to "if you're drinking alcohol, then you can't be 18"

(ik the prompt involved being over 21)

[remote barn]

so just the 1st and 4th one

[pallid tapir]

so "if the bag is blue, then it contains chicken" is logically equivalent to "if the bag doesn't contain chicken, then the bag is not blue"

an implication "if x, then y" is only false if the premise (x) is satisfied, but the result (y) is not

[remote barn]

Let a∈Z and n∈Z+, the Jacobi symbol is equal to the product of the Legendre symbols (a/n) = (a/p1)^b1 x (a/p2)^b2 x ... (a/pm)^bm, where the prime factorization of n=p1^b1 x p2^b2 x ... x pm^bm
Evaluate the following expression:
(201/52)-2(402/52)+3(603/52)-...+51(10251/52)

Is this 3327

[remote barn]

new problem instead of the old one eee

[remote barn]

ye

I was confused a bit

p->q is only false if q is false and p is true

[pallid tapir]

you should close this channel and open a new one so your new question is pinned at the top

[remote barn]

I don't wanna open a double channel I'd seem like I ask too many questions 🥀

ugh fine

.close

[marsh citrus]

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[remote barn]

. now I gotta wait for this

[bitter cave]

. you don't have to.

[marsh citrus]

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[strong dagger]

im having trouble showing why f is not S measurable

[strong dagger]

We just want to show that f^{-1}(B) for some borel set is not a co-countable set of R

Im assuming this is either gonna be B = {0} or B = {1}

but I cant wrap my head around it

like if B is anything containing 0 and 1 then we get all of R for the pre-image and thats co countable

if B doesn't contain either 0 or 1 then we also get all of R vacuously? Which wouldnt work either

so it has to be B = {0} or B = {1} that gives a counter example I think

Let's see so if B = {0}

then f^{-1}(B) = { x | f(x) = 0}

Is f(x) = 1?

[indigo nest]

On the right track

[strong dagger]

I think im not understanding what this sup is doing here

[indigo nest]

Write out the set explicitly

[indigo nest]

Do you understand what f_t is

[strong dagger]

yeah and if were taking sup over t, then for any f_t(x) there would be some x in R with x = t, t > 0

and so wouldnt we always get 1

when t > 0

but we can ignore t <= 0 since we are looking at the sup

[indigo nest]

Let's think of what we are plugging into the function

We don't care about t, we care about x

[indigo nest]

In this case the function only takes values 0 and 1, so we can see the sup as "is there any t for which f_t(x) is 1? If so, f(x) is 1"

[strong dagger]

and x comes from R, so taking t = 1 we get f_t(1) = 1 and so f(x) = 1?

[indigo nest]

Yup, so 1 is in f^-1({1})

Now look at how f_t is defined and answer this question:

For what values of x does there exist a t such that f_t(x) = 1

[strong dagger]

x > 0

o

[indigo nest]

So f^-1({1}) is

[strong dagger]

the positive numbers?

[indigo nest]

Yes

And is that cocountable

[strong dagger]

no

[indigo nest]

There we go

[strong dagger]

hm I see I hadn't thought about it that way

I didnt think to think of it in terms of existence of t and x

I'll keep that in mind

[indigo nest]

Happens, you get experience by doing more and more problems

[strong dagger]

Thanks for the help Xavier

.solved

[marsh citrus]

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[strong dagger]

im so confused

[strong dagger]

I've never seen Z_+

and I dont understand what 2^Z_+ is

oh

its the power set of Z_+

did they just mean to put Z^+?

[indigo nest]

Z_+ is positive integers

[strong dagger]

oh ok

[indigo nest]

I guess they mentioned it that way cuz N is ambiguous

[strong dagger]

I've never seen it like that I thought Z^+ was universal

[indigo nest]

I've never seen Z^+ cuz the superscript is usually reserved for powers like Z^n

But yea, convention ig

[strong dagger]

I see, thanks ill give it a try now

[crystal lintel]

i have seen Z^+ plenty

quite a high aura symbol if you ask me

[marsh citrus]

@strong dagger Has your question been resolved?

[marsh citrus]

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[worn prairie]

ax+by+cz+d = 0 - (i) \\
ax_{1}+by_{1}+cz_{1}+d=0 - (ii) \\
ax_{2}+by_{2}+cz_{2}+d=0 - (iii) \\
ax_{3}+by_{3}+cz_{3}+d=0 - (iv) \\

\text{ From the above equations after eliminating a,b,c,d we get } \\

\[
\begin{vmatrix}
x & y & z & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{vmatrix} = 0 \]  \\

\text{ Please help me bridge the gap between these two statements.}

[worn prairie]

Can someone help me understand how we got from step 1 to step 2

[elfin berry]

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

[marsh citrus]

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[cyan ledge]

help solve me 3

[red nimbus]

try polynomial division

[gloomy merlin]

In part 2 you established that x = -2 is a factor of P

[cyan ledge]

can you elaborate

[cyan ledge]

Yes

a =2

[gloomy merlin]

So check if x=-2 is also a factor of Q

[indigo nest]

I'm sorry to interject but, (x+2) is the factor. x = -2 is a root/zero

[gloomy merlin]

Yeah my bad

[cyan ledge]

(x+2) is actually the factor of Q(x)

[gloomy merlin]

Just evaluate Q(-2) basically

[gloomy merlin]

Then u r done

[cyan ledge]

but (2x-1) is a factor of P(x)

where is the common factor between the

[gloomy merlin]

It doesnt matter

You just need one factor between them

x+2 is found in both P and Q

[cyan ledge]

after doing the calculations, P(x) = K(x+1) (2x+1) (2x-1) (x+2)

since x+2 is found in both P and Q

the common factor is x+2

right?>

[gloomy merlin]

Yes

[cyan ledge]

oh alright

thanks mate

need help with this one too

the first one

[remote barn]

what can be inside natural log first

[gloomy merlin]

What id the domain and range of y = ln(x)?

[cyan ledge]

0, infinity

domain of the natural log function y = ln(x) is 0, infinity

[remote barn]

what is another way to word that

[gloomy merlin]

What is the domain and range of (2+x)/(2-x)

[cyan ledge]

dont quite know, what is it?

[remote barn]

positive numbers

[cyan ledge]

oh alright

[remote barn]

for what values of x is (2+x)/(2-x) positive

[cyan ledge]

wait let me cfind it

[cyan ledge]

(-2, infinity) intersect (-infinity, 2)
= (-2,2)

[remote barn]

yes so what is now the domain

[cyan ledge]

-2,2

[remote barn]

how do you find the range then

find the range first of (2+x)/(2-x) within the bounds of the domain we set

and then check what happens when we evaluate it with natural log(note that the natural log of infinity is infinity)

[cyan ledge]

y = f(x) = ln 2+x/2-X

or, e^y = 2+x/2-x

[cyan ledge]

okay

so x = 2(e^y -1) / e^y +1

For all real values of y the values of x will be real

Range of the given function is ℝ?

so domain is -2, 2 and range is ℝ

is it correct

[remote barn]

yes

[cyan ledge]

oh alright

thanks mate

[cyan ledge]

can you ellaborate this approach tho

[remote barn]

ln0 is -inf since e^-inf approaches 0
lninf is +inf since e^inf approaches inf

[cyan ledge]

Okay

[remote barn]

we have (2+x)/(2-x) yes? the range is 0 to infinity within the domain we set of (-2, 2)

[cyan ledge]

yeah

[remote barn]

since it's logarithmic, it is increasing, so from we can just find the ln of the range to figure out the range

[cyan ledge]

got it

thank you

[marsh citrus]

@cyan ledge Has your question been resolved?

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[gloomy merlin]

<@&268886789983436800>

[waxen dust]

[marsh citrus]

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[echo scaffold]

I don’t get how those tans equal 2 when I did it I got 1-tan^4x

[obtuse umbra]

look at the numerator

[jade sigil]

-tan^2x+tan^2x=?

[late geode]

how did you get 1 - t^4

[obtuse umbra]

only simplify the numerator

and you'll see

[late geode]

it appears you may be doing something like
(1-t^2) * ( 1 + t^2)
which is very different from what's there

[echo scaffold]

Ohh

Wait

That’s +

Not cancelling

I’m confused

This is the answer key

[late geode]

just basic addition / removal of ()

don't overthink

[echo scaffold]

Yea I totally over thought that one

I feel like I did smth wrong

I’m very confused

[normal fox]

Do u need help

Maybe just review ur answers to check if u did anything wrong @echo scaffold

[stable dune]

nothing wrong

[echo scaffold]

I think I may be good sorry I got like no sleep

[normal fox]

Perhaps

[elfin berry]

Tiresome. 呸

[normal fox]

@echo scaffold ?

Does that help idk

[echo scaffold]

yes

Ty

[normal fox]

If u are helped and done please .close

@echo scaffold ?

@echo scaffold is you are done please type in .close

Thank you

[echo scaffold]

I’m confused on what’s happening in the rec row

[normal fox]

Hm

[jade sigil]

cscx is just 1/sinx

sinx * 1 / sinx = 1

for sinx≠0

[normal fox]

@echo scaffold does that help for the Rec.?

[echo scaffold]

I’m confused where it came from it seems so random

Oh wait maybe not idk I’m failing this test 😭

[normal fox]

OH WAS IT A TEST?

[echo scaffold]

No my test is today

[normal fox]

When is the submition

[echo scaffold]

Why is there an extra csc in the middle in the answer key in the distribute row

[normal fox]

prob in the rec row is in the simplification
$$(1 - \sin x)(1 - \csc(-x)) = (1-\sin x)(1 + \csc x).$$
Distributing correctly gives
$$1 - \sin x + \csc x - \sin x \cdot \csc x.$$
Since $\csc x = \frac{1}{\sin x}$, the term $\sin x \cdot \csc x = 1$. So the expression simplifies to
$$1 - \sin x + \csc x - 1 = \csc x - \sin x.$$
The mistake was writing
$$= 1 - \sin x + \csc x - \sin x \cdot \frac{1}{\sin x},$$
which is redundant; the simplification should directly use $\sin x \csc x = 1$.

prob in the rec row is in the simplification
$$(1 - \sin x)(1 - \csc(-x)) = (1-\sin x)(1 + \csc x).$$
Distributing correctly gives
$$1 - \sin x + \csc x - \sin x \cdot \csc x.$$
Since $\csc x = \frac{1}{\sin x}$, the term $\sin x \cdot \csc x = 1$. So the expression simplifies to
$$1 - \sin x + \csc x - 1 = \csc x - \sin x.$$
The mistake was writing
$$= 1 - \sin x + \csc x - \sin x \cdot \frac{1}{\sin x},$$
which is redundant; the simplification should directly use $\sin x \csc x = 1$.

[elfin berry]

Tiresome. 呸

Tiresome. 呸

[echo scaffold]

I’m so confused

[normal fox]

Genuinely ur test hurts my brain

[jade sigil]

distribute 1*cscx=cscx

[late geode]

where

[normal fox]

@echo scaffold is ur question answered 😭

[echo scaffold]

I know I am distributing wrong but I don’t know how else

[wispy marlin]

lol

[normal fox]

This is starting to hurt my brain

[jade sigil]

do you know the FOIL method?

[echo scaffold]

No

I know the box method

should I use that?

[late geode]

more or less the same thing

[jade sigil]

^

[late geode]

distribute each term in the first set of () to the second set

[echo scaffold]

I need school to be cancelled im not doing this test today 😭

[wispy marlin]

so basically u just want to simplify everything down to sin and cos

[stable dune]

delete school

[echo scaffold]

You know my school was half cancelled for some unexpected phone call a couple months ago I need that to happen again

Or another pipe burst

How would I get rid of the one

[late geode]

what happened to the last term

sin(x) * 1/sin(x) isn't 0

[echo scaffold]

Oh is it 1

Now we have 2 tho

2- blah blah blah

[late geode]

how are you getting 2

[echo scaffold]

Oh wait

[late geode]

note the - sign

[echo scaffold]

It is a -

Honestly I feel like this extra pratice is not helping me and I should instead sleep for like an hour and a half 😭

I’m a senior im accepted into college I’ll get at least a c

[stable dune]

take some rest

[normal fox]

@echo scaffold r the questions done if yes pls type .close

[marsh citrus]

@echo scaffold Has your question been resolved?

[marsh citrus]

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[low scaffold]

how do we solve this?

[jade sigil]

try substitution

[low scaffold]

okay

Im just wonderig why its called a plane

[magic dune]

Any genius

[low scaffold]

for the cartesian equation

[jade sigil]

plane is flat

if you plot all points that fit the equation they form a flat surface extending infinitely

[low scaffold]

!occupied

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[low scaffold]

which direction

[low scaffold]

also sub what in? we dont hae the x,y and z values

[jade sigil]

look at the parametric equation

can you do something with it

[low scaffold]

u can expand it

[marsh citrus]

@low scaffold Has your question been resolved?

[low scaffold]

<@&286206848099549185>

[loud mica]

What's the problem

Reduce the equation to
X=a + (Lambda)b

Then

You can split the equations

And

[low scaffold]

i have no idea how to sovle this

[low scaffold]

Im confused with this

[loud mica]

Wait lemme pull out my textbook

[jade sigil]

try to rewrite these separately

[loud mica]

Yea I was gonna say that

[low scaffold]

and then?

[jade sigil]

.

[low scaffold]

simultaneous equations?

[loud mica]

By splitting of equations you'll get
X=-1+(lambda)1
Y=-4+(lambda)-4
Z=0+(lambda)4

[jade sigil]

yes and because they intersect you want to equate them together

[loud mica]

Compare the equations with r=a+(lambda)b
You'll get
Three a values
Three b values

[low scaffold]

wait so do i write 4 equations seperately

because there is a cartesion equation too

[loud mica]

Then you can put those equations seperately in
R=a+(lambda)b
And format them into a deternimant

Solve the determinant and get three x+y+z equations and then solve for Lambda

@low scaffold

[low scaffold]

👍

While yall wait

Can u check answer on

Apparently it’s wrong

something like this?

[loud mica]

Yea but by this method you should get x y and z values

[jade sigil]

plug your x,y,z into cartesian

[loud mica]

How did you not get those vals?

[low scaffold]

there was no need for me to do this?

[loud mica]

I'm confused mate

[jade sigil]

the line and planes are the intersecting ones, these x,y,z values are from the line

you want to put it into the cartesian equation (plane), which gives the intersection of the line and plane

[low scaffold]

ohhh

[low scaffold]

liken this?

[jade sigil]

Good job

[low scaffold]

but I thnk thats not the final answer

actually nvm

[low scaffold]

what about this question

I dont understand MVT

[reef fable]

Dude so many questions

[unborn condor]

MVT: you have a function, with the graph.
It is continuous, aka, you dont need to lift the pen from the paper to draw it.
It is differentiable, aka, your pen movement is smooth when drawing it.
And you have the value at the endpoints.
Since you have the endpoints, you can draw a line from one endpoint to the other.
MVT states that somehwere in your drawn curve, there's a tangent parallel to the line you have just drawn

[low scaffold]

I seee

what about the formula?

[unborn condor]

the formula just states the slope of said line

since you know the line between A and B, and that all parallel lines to it have the same slope

the MVT just states the existence of said parallel line

to actually compute it, you need to find the points in the curve that have derivative equal to said slope

and then you have the slope, and a point, for each of those lines that exist

[low scaffold]

ah I see

[edgy crescent]

Are you having trouble applying mvt here?

[low scaffold]

yeah A bit

wait

[edgy crescent]

Okay lmk if you need a hint

[low scaffold]

that have derivative equal to said slope
and then you have the slope, and a point, for each of those lines that exist

I dont understand this

[edgy crescent]

Okay let me frame it differently

The Mean Value Theorem (MVT) states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at c equals the average rate of change over the interval [a, b].

[low scaffold]

ah right, yes I understand that

tahts the tangent line

[edgy crescent]

Yeah

What it basically mean that if we apply the formula for tangent between two points, m= y2-y1 / x2-x1
It comes out same as that of a tangent at a point between the two on the curve (when we derivate the function and put the coords of that random point )

[low scaffold]

isnt this formula same as y2-y1/x2-x1

[edgy crescent]

Yeah!

[odd agate]

<@&268886789983436800>

[glossy delta]

https://tenor.com/view/stink-stinky-eww-gross-disgusted-gif-17704588793983568874

bro fell for a scam

[odd agate]

let's not send random GIFs in here please.

[low scaffold]

can I get a hint?

[edgy crescent]

what would you take your function f(x) as?

[low scaffold]

wait wdym by take ur function

[edgy crescent]

see here

he has written that "formula" in terms of f

f is a function isnt it

[low scaffold]

ah right

ok I think I should be able to do it from here

[edgy crescent]

so well here we can assume f as?

[low scaffold]

hm

[edgy crescent]

just tell me what you would assume f as

f(x) =??

[low scaffold]

thats f(x)

[edgy crescent]

Nope

[low scaffold]

or is that the first derivative

[edgy crescent]

no no

we need to assume such a f(x) so that we can define both numetr and denomiantr in terms of it

[low scaffold]

hmm

[edgy crescent]

well looks like you are stuck

we can assume the function to be f(x) = x^2/3

we can put x as 8.019 and 8

[low scaffold]

well

It needs to follow this

[edgy crescent]

Yeah

[edgy crescent]

This is actually defind for all functions that exist

[edgy crescent]

try from here

[low scaffold]

wait im confused why do we do that

[edgy crescent]

Do you see any similarity between 8.019)^2/3 and 8^2/3

[low scaffold]

same 1 s.f and same exponent

[edgy crescent]

right!

Now since they both have same exponent, we can make a funciton where all x have the same exponent

[edgy crescent]

thats what i did here

[low scaffold]

where all x have the same exponent?

[edgy crescent]

yeah we can make a function where we can say that we are raising the number to the powr 2/3

[low scaffold]

ahh yes

what about the same 8

[edgy crescent]

We dont have a method for "8"

and also the 8 part is not a major issue right?

[low scaffold]

why isnt it a major issue?

i mean ur right

because were raising x to 2/3

[edgy crescent]

Yeah

Just adding 8 isnt going to make as much of a mess as power to 2/3

[gray oriole]

is this question pending @low scaffold

[low scaffold]

noo I understand now

alright well thanks alot yajant

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[edgy crescent]

Your welcome
Have a great day!!

[marsh citrus]

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[full fossil]

im a bit confused. so by the four color theorm (if g is a planar graph, the chromatic number chi(G) is at most 4) the chromatic number can't be more than 4 but im getting atleast 5 for the chromatic number.
I also used brook's theorem and a corellary, and got another bound where delta(G) \ge chi(G) \ge clique number, which is 4 \ge chi(G) \ge 6, which is still true but it just doesnt satisfy the four color theorem.

[full fossil]

nvm i colored it completely wrong lol

.close

[marsh citrus]

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[marsh citrus]

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[shy grotto]

Help with 4b

[main idol]

first step is already incorrect

y = f(x) = (the rest remains unchanged)

[turbid mica]

you can't take out the - 1 inside the exponent like that

[covert scarab]

The usage of the implication symbol here is vv unorthodox

[turbid mica]

its ok im sure my professor understands freshman math notation

[covert scarab]

😭

[shy grotto]

Fam

[shy grotto]

What that mean

[turbid mica]

actually nah he said next exam hes taking points off for using implication arrows wrong

[covert scarab]

As one should

[turbid mica]

==> is implies in math

[shy grotto]

o

Yeah

[turbid mica]

if a ==> b that means if a is true then b is true

you wrote y => ... which is odd

[shy grotto]

I thought they all meant the same thing

[turbid mica]

no it's not an equal sign

[covert scarab]

Just forget about implication for this question and only use equality

[shy grotto]

😭

K just tell me what I did wrong

🙏

[covert scarab]

You essentially just want to make x the subject of the equation

[whole sleet]

Have any questions about this?

[covert scarab]

So let f(x)=y

[shy grotto]

Yeah

[covert scarab]

solve for X in terms of y

[shy grotto]

Switch them up right?

[covert scarab]

You as pointed out your first line is wrong

[shy grotto]

X=7-2(y-1)^2

[rough sand]

Yeah but depends where you switch them though

[covert scarab]

It's 7-2(....

Look at the original question

[shy grotto]

I forgot to put mb

[covert scarab]

You forgot that in the first like of your working as well

[shy grotto]

Kk

X=5(y-1)^2

[covert scarab]

Ehhh

[rough sand]

Oof

[shy grotto]

There’s no way we do not add them up

What

[rough sand]

You can’t minus the 2 from the 7

[shy grotto]

Oh wait

Bedmas

[rough sand]

Cause the 2 is multiplying the like brackets

[shy grotto]

Yeah that

[rough sand]

Idk the actual terms for what you call that but yeah

[shy grotto]

This?

[covert scarab]

Don't just swap x and y. Just solve the equation like your solving for x

So you'll get x=...

A function maps X to y so the inverse is mapping y to x

[main idol]

made this mistake again

[shy grotto]

Here

[rough sand]

https://blogs.ams.org/matheducation/2016/11/28/inverse-functions-were-teaching-it-all-wrong/
there is this cool article I found a week or so ago

[main idol]

!redir

[marsh citrus]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[rough sand]

Either way you get the same answer

[shy grotto]

Bro what am I what doing with what I’m listening to

[rough sand]

O sorry mb

[main idol]

you did 7 - 2 * stuff = 7 * 2 * stuff

the subtraction magically turned into multiplication

[shy grotto]

Magically💀

[main idol]

and 2(y-1)^2 is not (2y - 2)^2 because of exponent comes before multiplication in bemdas

[shy grotto]

Do exponent fist right

Makes sense

Should’ve just told me that

Lowk

🙏

[covert scarab]

You know "bedmas" already dough...

[main idol]

you said yourself bemdas. should have listened to yourself

[shy grotto]

Here

I got that

[covert scarab]

your subtraction turned into multilpication again

dont expand the power

makes live vv complicated

[shy grotto]

Brah just give me the answer atp😭😭

[shy grotto]

What am I doing💔💔

[shy grotto]

wtf is 7 supposed to do???????

So 2 cancelled the brackets

Right

But

Now my equation is a mess bruh

[rough sand]

Just move the 7 to the left

To get it out the way

[shy grotto]

Do we just willingly decide because we feel like it, it moves?

Breh I don’t get taht

[rough sand]

No we are minusijg 7 from both sides

[rough sand]

Ok but do it at the start of the question

Cause idk where you went wrong but that doesn’t look right, I don’t have my glasses

😭

[shy grotto]

Fraud

[rough sand]

Trust me bro, it’ll be prettier if you restart

[rough sand]

Ok so restart, subtract 7 from both sides then divide both sides by -2

[stable dune]

here do not vanish the - sign here

[shy grotto]

.close

[marsh citrus]

Channel closed

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[marsh citrus]

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[trim frigate]

So I’m a math tutor (clearly not a very good one) for grade 10 math at my school and it’s been a long time since I’ve done this kind of math so I just need a little help on this question I circled

[trim frigate]

This is my work from when I was in grade 10 but I don’t remember how I got the answer

[brave marsh]

You can go one type at a time.
Start with multiplying the constants together.
Then the s factors.
Then the t factors.

When you get two factors multiplied of the same "type", you add their exponents.

[trim frigate]

Okay

Ohhhh alright I see now

With the same base you add them

[brave marsh]

Yep

[trim frigate]

It’s been so long thank you very much

[brave marsh]

And for divisions you subtract them

[trim frigate]

Yes yes

Thanks!

.close

[marsh citrus]

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[marsh citrus]

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[strong finch]

Find redundant columns of
1 4 7
2 5 8
3 6 9

[strong finch]

I did RREF and all have pivots

[reef fable]

wdym redundant column

[tired ore]

!show

[marsh citrus]

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

[strong finch]

It says it’s a column that can be made from another column in the vector

[tired ore]

<@&268886789983436800>

[torpid anvil]

how'd column 1 become 1 0 0

[strong finch]

Wym

[torpid anvil]

where's the first half of your work?

[tired ore]

ok got it

[strong finch]

This is the whole thing

[tired ore]

the first transition is wrong

in the first matrix in your picture the last row is wrong

9-21 = -12

[torpid anvil]

sum wrong with ur row 3

[tired ore]

also this can be done a little easier by just doing r2 = r2 - r1 and r3 = r3 - r1, but just doing rref is okay too

[strong finch]

What’s that do so r2=1 1 1 and r3=2 2 2

[tired ore]

then the matrix is
1 4 7
1 1 1
2 2 2

[strong finch]

And then I do RREF?

[tired ore]

you can make last row zeros

[strong finch]

How

[tired ore]

subtract second row twice

r3 = r3 - 2r2

[strong finch]

Ok and then RREF?

[tired ore]

yes

i mean

'redundant columns' is sort of unclear I mean if you can make a column out of other two one of them is redundant, but it doesnt matter which one

and there isnt like an exact one

[strong finch]

Wym

It says that if a column doesn’t have a pivot 1 then it’s redundant

[tired ore]

I am saying you should be asking how many redundant columns are there and not find redundant columns

[strong finch]

That was what the question asked

[marsh citrus]

@strong finch Has your question been resolved?

[marsh citrus]

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[carmine hamlet]

<@&268886789983436800> crazy scammers influx

[marsh citrus]

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[carmine hamlet]

<@&268886789983436800> damn

[marsh citrus]

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[neat grove]

How does top turn into bottom?

[fervent rampart]

product rule in reverse

[neat grove]

Huh

This is the whole thing

[fervent rampart]

maybe the simpler question would be to ask how to get from bottom to top

[slim lagoon]

E=mc2

[neat grove]

Mb for the handwriitng ng

[opaque agate]

inappropriate, please don't do this

[fervent rampart]

and then just note that it goes both ways

[neat grove]

What

I dont understand

[fervent rampart]

do you understand how to go from bottom to top?

[neat grove]

Just product rule

But why did he do that

[fervent rampart]

he went from top to bottom because the bottom is easy to integrate via FTC

[neat grove]

Bro but

How am i supposed to spot that

[fervent rampart]

the entire point of the integrating factor is to let you do that

[neat grove]

But my integrating factor is x^4 + 5

[fervent rampart]

it's specifically designed to turn the equation into that form

[neat grove]

Why do we just add the y

OHHH

I GET IT

Ye ty

Js jogged my memory

.,•+{

.close

[marsh citrus]

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[marsh citrus]

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[neat grove]

I got the eigen values but to find matrix P how fo ik which collums to eliminate for the ijk thing

[neat grove]

Or does it not matter

[humble nebula]

what's P?

you don't need to diagonalise it here

no one's asking you to

[jade sigil]

im assuming eigenvector but not sure

[humble nebula]

here it looks like you're trying to take a cross product?

that's not what they're asking you for

[marsh citrus]

@neat grove Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[neat grove]

.reopen

[marsh citrus]

✅ Original question: #help-33 message

[neat grove]

Part b wants me topl

[neat grove]

Part b asking for it

[humble nebula]

oh, have you finished part a already?

[neat grove]

Yeah

[neat grove]

Js proved the equation

And got eigen values 2 -4 6

[humble nebula]

so you need to find the eigenvectors then

[neat grove]

Yes

I did

But then for part b i need to find P

And i needa crossproduct

And u usually need to remove a line for them

But idk how to do that here

Usually its obv but here idk how

[humble nebula]

could you send a picture of the question here? I don't know what it's asking

[neat grove]

K

Mb

[humble nebula]

oh I see

[humble nebula]

why do you think you need a cross product?

you could just find the third eigenvector as normal

[neat grove]

Do find the eigen vectors

U need cross product

At least in my syllabus

Heres the ms

I dont get why they seperated it the way they did

[humble nebula]

Oh oko

I get it now, I think

they subtracted $\lambda I$ from the matrix, removed a row (because it's linearly dependent) and then took the cross product to get a normal vector, which must lie in the kernel...?

[elfin berry]

soup_norm

[neat grove]

Yh but

Why did they remove

The rows

Like

Why did they pick each row

Cuz usually the eigenvalues match and make one of them 0

Since its smth like this

[humble nebula]

in this case, the choice of row doesn't matter

[neat grove]

Oh really

Alr

When does it matter then

[humble nebula]

on second thought, that doesn't seem to be what is happening

[neat grove]

Or does it never

Idk the specifics

All they do is

Get the equation in a

[humble nebula]

sorry, ignore that, I'm not sure what method they're using. Could you please look at the textbook and see what the method is called?

[neat grove]

Find eigen vectors

I dont have the textbook lol

Idk what the method is called tbh i js do it

[humble nebula]

Or your teachers' notes, maybe?

Can you remember when this was taught?

[neat grove]

No

[humble nebula]

oh

[neat grove]

🙈

[humble nebula]

actually it does look like it

[humble nebula]

it shouldn't matter, because all three rows are in the same plane

[jade sigil]

seems like it's cross product of first two rows of A-2I

resulting in vector orthogonal to both rows

ive never done it this way lol

[humble nebula]

yeah, and then the question is why that vector must lie in the kernel

this should be equivalent to A - 2I being an orthogonal projection

[neat grove]

Yh but i needa do it like this to get the makrs

Marks

So the rows dont matter here?

[humble nebula]

which row you pick doesn't matter

[jade sigil]

doesn't matter since they are linearly independent

[humble nebula]

or, if you pick the wrong one, then you just need to change your choice of deleted row

[neat grove]

How do i find D then

[humble nebula]

in the new basis (i.e. after you conjugate by P), the matrix will be a diagonal matrix, and it will have the eigenvalues on its diagonal

so you just need to take the fifth power of each of the eigenvalues

[neat grove]

Ah

And then

Each fifth power

Has to match

Which rows. I used

Or wha

[humble nebula]

could you rephrase that?

[neat grove]

Like

2^5

-4 ^5 and 6^5

How do i know which area to put them in

So it needs to be like

x 0 0
0 y 0
0 0 z

But how do i know which is which

[humble nebula]

oh

it matches which rows/columns you put the eigenvectors in when you make P

[neat grove]

But it doesnt tho

In the ms it goes

2 then -4 then 6

But the ms and i both did for P

For 2 remove seocnd row for -4 remove 1st and for 3rd remove last

[marsh citrus]

@neat grove Has your question been resolved?

[marsh citrus]

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[carmine belfry]

how do I get over my pets death

[gloomy merlin]

Sorry for your loss, but this really isn't the place to ask something like that

.close

[marsh citrus]

Channel closed

Closed by @gloomy merlin

Use .reopen if this was a mistake.

[glass silo]

.(please don't close these troll ones, but ping us and let us deal with them )

[gloomy merlin]

Oh I thought it was genuine

But yeah fair enough

[jade sigil]

sorry for your loss.

[glass silo]

Yea just recently we do want to keep an eye out, and also means we can keep track of these people

[glass silo]

And you know this isn't a math question, you're very lucky I'm not dealing with you

[carmine belfry]

humanise yourself a little bit

[glass silo]

Don't try that with me

[reef fable]

dont think he is trolling but anw

[marsh citrus]

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[glossy delta]

I went wrong somewhere e

[glossy delta]

this is the question

plsssssssss help

did i expand wrong

the mark scheme says this:

oh my farking god

that 1 mistake with the first quadratic ruined it

.close

[marsh citrus]

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[marsh citrus]

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

[obtuse plume]

I'm given these functions, and this graph. How could I link the correct function to the graph? From what I've seen people often use f(x,y) = C.

If we would take the first function we'd get C = x + y

Which implies that y = C - x.

However this still doesn't make me any smarter as I'm a bit unsure what the graph is telling us. What is the bar on the right, what does each curve mean, and what plane are we looking at?

[bleak ledge]

ugh

<@&268886789983436800>

thank you

[bleak ledge]

y=C-x isn't the shape of any of the level curves

so that means taht x+y isn't the answer

you see there, that its a circle

the level curves

and how they're alternating between red and blue?

what does that mean 🧐

[obtuse plume]

That it must be a trig function?

[bleak ledge]

yes, and it should be radially symmetric

since the level curves are circles

which just really means something like x^2+y^2

now, what fuction has that property

curious

[obtuse plume]

function 4, I'd assume

[bleak ledge]

yes, there you go

are you satisfied with the reasoning?

it will only be helpful if you think it works

[obtuse plume]

For the most part, I understand we can kind of see it's shape as you described with it oscillating up and down, but what I'm a bit stuck on still is how we determine whether it will be function 3 or 4

[bleak ledge]

well lets see fuction 3 and assume a random level curve for it, say: sin(x-y)=C

which means arcsin(C) = x-y

now arcsin(C) is just a constant so it may as well be K=x-y

which is a straight line

do you see any straight lines in the level curves?

[obtuse plume]

Ah no, okay, so as I understand the inside of the function kind of tells us how it oscillates

[bleak ledge]

do you understand what i just said to you though

i get that you know how it is a trigonometric function but the level curves are also circles

[obtuse plume]

I do yeah

[bleak ledge]

so its not 3

now how is it 4?

well

can you try this one yourself?

<@&268886789983436800>

jesus christ can these guys fuck off

[burnt abyss]

omg mrbeast

[obtuse plume]

So for that same reason we could say that arcsin(C) = 2x^2+2y^2

Which implies

K = 2x^2+2^y2

and from personal experience I know that x^2+y^2 is shaped as a circle

[bleak ledge]

just factor out the 2

which gives x^2+y^2=K/2=r^2 for some r

so you see how its a circle?

like in the level curves?

[obtuse plume]

I do now yeah

So can we utilize that for each trig function? that the cos^-1 and tan^-1 just becomes a constant and replace it K for example as you did?

[bleak ledge]

for this case yes

because the given range is

-1 to +1

[obtuse plume]

So this would actually be the graph for function 3.

K = arcsin(C)

K = x-y
y = x-K

Or actually just K = x-y

[bleak ledge]

yes