help-13

[cedar kiln]

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[sick fossil]

is there a way to prove that a function is one to one without the horizontal line test?

[crimson sedge]

whats g(x) when x is infinity

[sick fossil]

positive?

and vice versa

[crimson sedge]

you gotta be more specific

[sick fossil]

wdym

[crimson sedge]

it is +infinity

[sick fossil]

oh

[crimson sedge]

now use the fact that g(x) is an odd function

and also the fact that x+sqrt(x^2+1) is an increasing function

[sick fossil]

what about part d?

swap x and y?

how do I get rid of the natural log?

[crimson sedge]

e^y = x + sqrt(x²+1)

probably

[sick fossil]

isn't first step replace x with y?

[crimson sedge]

yeah but i changed the form

[sick fossil]

to find inverse

[crimson sedge]

before i do that

[sick fossil]

oh

is it easier this way?

[valid locust]

Hi

[crimson sedge]

i hope so

try it

[sick fossil]

move over the square root first ig?

so it becomes e^x -sqrt(y^2 +1) = y ?

or keep both y on the same side?

hello?

<@&286206848099549185>

trying to do 2d

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[cinder saddle]

I have 3 problems like this (this is the first one) I am confused on how to solve it

[slate lintel]

usually we want to assign variables to things, so here you could say that you purchase C pounds of the cheaper chocolate and E pounds of the more expensive chocolate

and then translate the sentences into equations

[cinder saddle]

4.1c+8.4e=5.4(25.8) ?

[slate lintel]

ok good that's one equation

[cinder saddle]

is it the correct equation?

[slate lintel]

yep it looks good'

but you need another one

since we have two variables

we need two equations

[cinder saddle]

do i need to have just one variable in the second equation

[slate lintel]

nope, you can have both of them

it just has to be, like, different than the first one

[cinder saddle]

how much different can i just switch the order?

[slate lintel]

essentially different haha

so no, switching the order isn't good enough and multiplying both sides by something won't work either

need to go back to the problem and look for another constraint

you've dealt with the cost

what about the like

actual amount of chocolate

[cinder saddle]

do i use the same variables

[slate lintel]

yep, you'll want to use C and E here

[cinder saddle]

and would it still be 4.1c and 8.4e?

[slate lintel]

not necessarily

check this out -- you want to buy 25.8 pounds of chocolate right?

and you know that you're buying C pounds of cheap chocolate and E pounds of expensive chocolate

can you make that into an equation?

[cinder saddle]

25.8/c+e???

[slate lintel]

what about just C + E = 25.8

[cinder saddle]

ok

[slate lintel]

does that make sense how i got that?

[cinder saddle]

since the pounds of c and the pounds of e add up to 25.8

[slate lintel]

yeah

so now we've got two equations

4.1c+8.4e=5.4(25.8)
c + e = 25.8

there's a bunch of ways to solve these but the simplest is substitution

take the second equation (it's way simpler) and solve for C

and then plug that into the first equation and solve for E

[cinder saddle]

e=25.8-c

[slate lintel]

ok that's fine, you solved for E first but that's alright

now plug that into the first equation and solve for C

[cinder saddle]

4.1c+8.4(25.8-c)=5.4(25.8)

[slate lintel]

yep

[cinder saddle]

4.1c+216.72-8.4c=139.32

-4.3c+216.72=139.32

-4.3c=-77.4

c=18

and then 25.8-18=7.8

so e=7.8

is my math correct

I also have this problem

I think I can solve it the same way im gonna try to do it

[austere hull]

can you change your pfp

because it's nsfw

fuck my brain

yeah i got it

[cinder saddle]

for my math i have one more problem i got to see if it works with the same kind of formulas

i solved that one

now i got this one

[austere hull]

can you apply the same formula if you doing same topic?

[cinder saddle]

i did that for the second one but now its mph

[austere hull]

whats mph

[cinder saddle]

miles per hour

i dont like miles i wish it were kilometers

[slate lintel]

you can pretend it's meters per hour and they're just going reallllllly slow

[cinder saddle]

how do i do it with miles and time and stuff

[slate lintel]

anyway this problem works fairly similar to the last one, have one variable for the amount of time he went Fast, and one variable for the amount of time he went Slow

[cinder saddle]

would it be 14f and 7s?

[slate lintel]

14f and 7s are the distances yeah

[cinder saddle]

ok this one was similar to one i did earlier

there were other mph ones but for some reason i was confusing myself

thank you for helping me

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[lost osprey]

Whas the domain for that one? i'm puttin it in desmos and thats where I got my answer

[gilded mesa]

Ok sorry I can see

So you can't put 0 only because there function is undefined

1/0 form

Your answer is right

[lost osprey]

I think I know what I missed

[livid hound]

you need to consider the intersection of the domain of f(x)

[lost osprey]

where it hits x right

the x axis*

[gilded mesa]

Why intersection

[lost osprey]

(-inf, -2) U (-2, 0) U (0, inf)

right?

[gilded mesa]

No

[lost osprey]

aight now im lost

[livid hound]

f(x) needs to be defined for g(f(x)) to be defined

[gilded mesa]

Ya right

I forget that

Sorry

[lost osprey]

so this is right?

[livid hound]

yeh

[lost osprey]

aight

thanks

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[twilit oxide]

Am I missing a formula that I can use to find X from P?

What’s this?

Yea I only have angle

I don’t have either sides

Except for that 10 m between the poles

So I’m kinda confused

oh yeah i can actually get all the angles

cause its right angled anyways

alright thanks man

wait hold on

i dont have the hypotenuse or the adjacent though?

@crimson sedge

do I??

its in the form of x

mm hmm

aight man

whats thetha tho

interesting

well problem is i cant find any one of the sides

are you solving for them rn? @crimson sedge

but you see there are still 2 unknowns eh

idek what SR is

Oh

aight that makes sense now

ima just make em both x

aight thanks g

[cedar kiln]

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[candid mason]

So the riemann integral of 0 is 0. The question as result has a contradiction i.e. false statement.

Also integral is indefinite??

[lucid stump]

Yea

[slate lintel]

this is an initial value problem

what they meant was I ' (x) = e^sin^2 etc

[dusty hazel]

Yes.

And I(0) is given to find the constant.

Because indefinite

[cedar kiln]

@lucid stump Has your question been resolved?

[lucid stump]

<@&286206848099549185>

[vagrant elbow]

have you thought of anything?

[wraith dagger]

NEON

[vagrant elbow]

ignore the second part for now

and for the first part

integrate by parts

[wraith dagger]

NEON

[vagrant elbow]

you'll see something cool happen (hopefully)

[lucid stump]

Um
I haven't done by parts for such complex eqns yet

[vagrant elbow]

hmm

give it a shot

try computing the anti derivative of that

(it might need a substitution)

[cedar kiln]

@lucid stump Has your question been resolved?

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[cerulean fiber]

Trying to find area using integrals and I got a negative result. Can't tell what the mistake is

[cedar kiln]

@cerulean fiber Has your question been resolved?

[cerulean fiber]

The exercise says to find the area between the:
y = cos2x
y = 0
x = π/4
x = π/2

<@&286206848099549185>

[undone cliff]

The answer is supposed to be negative. It looks like your math is right, you just drew the graph wrong

You drew the graph taller but the 2 is on the inside of cos not outside so it actually causes a horizontal shrink

[cedar kiln]

@cerulean fiber Has your question been resolved?

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[crimson sedge]

Quick question

Is 1000 m2

1 km2

[runic vector]

yes

[crimson sedge]

Okay thanks so the volume doesnt

Matter

No thats area

[runic vector]

kilo = 1000
centi = 100

[runic vector]

volume is cubed

[crimson sedge]

nooo

What

you're asking if 1000m^2=1km^2?

Ye

S

it's not

Oh

What is it

1km=1000m
1km^2=1000m*1000m

[runic vector]

oh mb, ray is right

[crimson sedge]

Oh so when its area or volume

I have to times it

yes

So 1km2 is

What

1000000m^2

Wow okay ty

Thats a lot of zeros

yeah it's easier to calculate if you just write
1km=10^3m
(1km)^2=(10^3m)^2
=10^6 m^2

Okay thank you!!

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[ripe badge]

The answer to this question is Solve for X and X is 45 degrees, but I don’t know how my teacher got that answer.

[mental salmon]

You need to find x or x is already given?

[ripe badge]

I need to find X

My teacher said X was 45 but I don’t know how she got that

Can you help explain it to me pls

[cedar kiln]

@ripe badge Has your question been resolved?

[cedar kiln]

@ripe badge Has your question been resolved?

[ripe badge]

.close

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[maiden venture]

A sailor is with his boat at the point A=(1,3,5) and sails in the direction v=(3,1,1)^T. He intends to make a right angled course change to be able to pass through the harbor entrance at the point P=(2,4,7). In what plane pi is the water surface located?

[maiden venture]

I don’t really know how to solve this

[lethal jackal]

it's basically asking you what plane the boat is sailing in

[maiden venture]

Yes okay but how do I solve it

[lethal jackal]

you can find three points in the plane and then use that to find the plane

that's probably the most straightforward way of doing it

[maiden venture]

Okay!

How do I find the three points?

[lethal jackal]

well you already have two...

[maiden venture]

Oh okay so can I do (3,1,1) * (x,y,z) - (2,4,7) = 0?

[lethal jackal]

I don't think that works

(3, 1, 1) is not the normal vector to the plane

you have two points in the plane that are given by the problem

how might you find a third, given the remaining piece of information?

[maiden venture]

Hmm I don’t really know

[lethal jackal]

well, first what's the remaining piece of information you're given?

[maiden venture]

That it sails in the direction of v = (3,1,1)^T?

[lethal jackal]

yes

so would you agree that if I sail from point A (1,3,5) in that direction for a while, I'll end up at A + (3, 1, 1)?

[maiden venture]

Yes!

So I can combine them and get the third point?

[lethal jackal]

yes that gives you a third point on the plane

[maiden venture]

Do I just ignore the T or do I do something with it?

[lethal jackal]

T usually means transpose, so it's probably just some shorthand of writing vectors

[maiden venture]

Oh okay

If I combine them I get the point (4,4,6)

[lethal jackal]

can you find the equation of a plane with three points?

[maiden venture]

Do I just combine the three?

[lethal jackal]

what do you mean by combine?

[maiden venture]

Take A + P + (4,4,6)?

[lethal jackal]

that wouldn't make a whole lot of sense

that wouldn't even give you a point on the plane

[maiden venture]

Oh okay

[lethal jackal]

do you know what the general form of the equation of a plane is?

like what do all plane equations look like?

[maiden venture]

Ax + by + cz = d

[lethal jackal]

how were you taught how to find the equation of a plane? there are a few different approaches here

[maiden venture]

I don’t really know how to find the equation

[lethal jackal]

hmm I guess I'll go over the linear algebra solution

well, let's first start with lines which might be easier to understand

Do you know what it means when someone says "the equation of the line is y = 2x + 4"?

[maiden venture]

Yes!

[lethal jackal]

do you agree that it means that if you take any point (x1, y1) on the line and plug it into the equation, you'll see that it checks out?

For example, (1, 6)

[maiden venture]

Yes I do

[lethal jackal]

so suppose I tell you that the general form of a line's equation is Ax + By = C and that the points (1, 6) and (10, 24) lie on the line.

[maiden venture]

Oh wait I think I have learned about this is it not that 1 in A is x1 and 2 in P is x2 and so on

[lethal jackal]

Would you agree that
A * 1 + B * 6 = C
A * 10 + B * 24 = C
?

[maiden venture]

Yes I would

[lethal jackal]

do you know how to solve systems of linear equations?

[maiden venture]

Yes I think so

[lethal jackal]

okay, suppose I subtract these two equations (the first from the second)

now I have 9A + 18B = 0, right?

[maiden venture]

Yes!

[lethal jackal]

doing a little more solving, I get A = -2B, right?

[maiden venture]

Yes

[lethal jackal]

plugging this back into the first of the original equations (I could also choose the second, but whatever), I get -2B + 6B = C, and therefore C = 4B, right?

[maiden venture]

Yes right

[lethal jackal]

in every case, you will be left over with one "free variable", which you can basically pick freely. In this case, it's B

so if I just pick a value for B, like 1

Then I get A = -2 and C = 4, right?

[maiden venture]

Yes okay

[lethal jackal]

So plugging this back into the general form of the line's equation (Ax + By = C), you get that the line has an equation of -2x + y = 4

and indeed you can check that (1, 6) and (10, 24) lie on this line

[maiden venture]

Oh yes okay that makes sense

[lethal jackal]

now suppose I picked a value of B = -3 instead

I would get the equation: 6x - 3y = -12

you can check that this is just a multiple of the previous equation

so basically the "pick freely" part comes from the fact that you can take multiples of an equation for a line and you still have the same line

does that make sense?

[maiden venture]

Yes it does!

[lethal jackal]

now lines and planes are actually very similar in the mathematics, so much the same thing applies

suppose we have Ax + By + Cz = D

and we know that the points (1, 3, 5), (2, 4, 7), and (4, 4, 6) lie on the plane

[maiden venture]

Yes exactly!

[lethal jackal]

Can you write out the three equations that must be satisfied?

[maiden venture]

Yes I can

A* 1 + B *2 + C *4 = d

A * 3 + B * 4 + C * 4 = d

A * 5 + B * 7 + C * 6 =d

[lethal jackal]

you do not get that

you know that the point (1, 3, 5) has to satisfy the equation Ax+By+Cz=D

[maiden venture]

Oh I’m sorry

I did it the other way around

[lethal jackal]

so what equation in A, B, C, D do you get?

[maiden venture]

A * 1 + B * 3 + C * 5 = d

[lethal jackal]

So you get the equations
A + 3B + 5C = D
2A + 4B + 7C = D
4A + 4B + 6C = D
correct?

[maiden venture]

Yes correct

[lethal jackal]

have you learnt about reduced row echelon form or matrices at all?

[maiden venture]

Not really yet no

[lethal jackal]

have you learnt about solving systems of equations by elimination?

[maiden venture]

Yes we have

[lethal jackal]

okay

so I'm going to eliminate D from all of the equations and get two equations without D:

Suppose we number the equations for ease of reference:
A + 3B + 5C = D (Equation 1)
2A + 4B + 7C = D (Equation 2)
4A + 4B + 6C = D (Equation 3)

Then eliminating D, we get:
A + B + 2C = 0 [Equation 2 - Equation 1] -- Equation 4
2A - C = 0 [Equation 3 - Equation 2] -- Equation 5

Do you follow so far?

[maiden venture]

Yes

[lethal jackal]

I can use these two equations and eliminate C now:
5A + B = 0 (add 2 times Equation 5 to Equation 4)

do you agree so far?

[maiden venture]

Yes I do

[lethal jackal]

So suppose I pick the free variable to be A this time (picking B is more convenient, but I picked B last time, and I want to show that it works with any of them)

we rearrange that last equation to B = -5A

[maiden venture]

Yes okay!

[lethal jackal]

now I pick a value of 100 for A, again not picking the most convenient value, but it'll suffice

Therefore, what is B?

[maiden venture]

-500?

[lethal jackal]

yes

now, knowing that A = 100, B = -500, can you find C?

[maiden venture]

Can I put in those values in A + B + 2C = 0 or does it have to be one of the original equations?

[lethal jackal]

you can put it into any of the equations listed, but only equations 4 and 5 will let you solve for C

[maiden venture]

Okay then I get c = 200

[lethal jackal]

and can you find D?

[maiden venture]

-400

[lethal jackal]

okay, so you get the equation 100x - 500y + 200z = -400 after plugging these variable values back into Ax + By + Cz = D, right?

[maiden venture]

Yes right

[lethal jackal]

you can check that all of the mentioned points are indeed in the plane

[maiden venture]

Yes!

[lethal jackal]

do you know how to find a normal vector of a plane from its equation?

[maiden venture]

Yes I do

[lethal jackal]

what do you get from that?

[maiden venture]

N = (100, -500, 200)

[lethal jackal]

do you agree that, no matter the captain's choice, a ship must sail on the surface of the water?

[maiden venture]

Yes

[lethal jackal]

so in other words, the ship must sail perpendicular to the normal vector, right?

[maiden venture]

Yes right!

[lethal jackal]

you can check this too

[maiden venture]

Yes I can!

[lethal jackal]

you can check that (100, -500, 200) is perpendicular to (3, 1, 1)

so the answer is correct

[maiden venture]

Yes it equals 0!

[lethal jackal]

a worthwhile exercise is seeing if you can solve the problem again but making different choices, like maybe eliminating C first and then A and picking a different free variable

it will all work out the same

[maiden venture]

Yes okay I will do that!

[lethal jackal]

there is another method where you find the normal vector with the cross product, but it only works with planes

[maiden venture]

But that is a valid answer?

Oh okay

[lethal jackal]

yeah 100x - 500y + 200z = -400 is a valid answer to the question "what is an equation of the plane that the ship sails on?"

[maiden venture]

Okay thank you so much!

[lethal jackal]

if you make more natural choices, you might get something like x - 5y + 2z = -4

[maiden venture]

That was very well explained!

Yes I will try to solve it again now

But thank you

[lethal jackal]

np

[maiden venture]

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[dire geode]

Not enough context to tell

[glacial moth]

is there no standard order of precedence?

the text also mentioned linear congruence so idk if that narrows it down

[dire geode]

Not with modulo, no

Depends on the text/author unfortunately

[upper abyss]

I would personally take it as (...) mod 2^n

Well, is there any other context?

[glacial moth]

its about cryptography

So I think it is (...) mod 2^n

[daring kelp]

if they are less than 2^n then it's definitely (...) mod 2^n

[cedar kiln]

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[gloomy crater]

how do u find the taylor polynomial if its not centered at 0

[cedar kiln]

@gloomy crater Has your question been resolved?

[upper laurel]

taylor polynomial about $x=\color{yellow}2$ is\
$\sum_{n=0}^\infty\frac{f^{(n)}(\color{yellow}2\color{black})}{n!}(x-\color{yellow}2\color{black})^n$

[wraith dagger]

matt07734

[upper laurel]

you can think of this as the maclaurin polynomial but translated 2 to the right

[cedar kiln]

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[upper laurel]

from that, the coefficient of $(x-\color{yellow}2\color{black})^{\color{cyan}4\color{black}}=\frac{f^{(\color{cyan}4\color{black})}(\color{yellow}2\color{black})}{\color{cyan}4\color{black}!}$

[wraith dagger]

matt07734

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[burnt seal]

how 2 solve this ?

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[hollow crown]

Can someone tell me how to get the answer for this

[gloomy crater]

dont u just find the equation of the normal line then add 1

for the first problem

[hollow crown]

Only the problems I circled

[gloomy crater]

ye

for the first one circled

[hollow crown]

I got the answers but I dont know how to get them

[gloomy crater]

is the answer y=1/3 x+1 for the first one in red

[hollow crown]

yup

How did u get it though

[gloomy crater]

do u know how to get the slope of the line perpendicular to y=-3x

[hollow crown]

no but im assuming its the opposite so y=3x?

[gloomy crater]

opposite and reciprocal

[hollow crown]

How can I turn 3 into reciprocal?

[gloomy crater]

If the slope of the orignal line is $m$, then the slope of the perpendicular line is $-\frac{1}{m}$

[wraith dagger]

mXd

[hollow crown]

oh so 3=3/1

so reciprocal is 1/3

[gloomy crater]

ye

so the equation of the perpendicular line is y=x/3

and then u have to shift it up 1

so u get y=x/3+1

[hollow crown]

Okay thanks

Now how could I get the answer for the second one?

[hollow crown]

How could I get the second one since its not at origin anymore

[cedar kiln]

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[cedar kiln]

@hollow crown Has your question been resolved?

[distant kindle]

One of your options would be to take two points on your original line, apply the transformation to those points, then get the equation of the line that passes through your two new points

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[pine sun]

Could I get help on this? There's a video assigned to it but it isn't really clear..

[crimson sedge]

well like

a composition of functions is given by [
(f\circ g)(x) = \m f{\m gx}
]

[pine sun]

right

but u and s aren't defined.. usually they are so I'm lowkey lost

[crimson sedge]

So to make things easier on you

lets make u = g(x) alright?

so we have f(u)

[pine sun]

jep its easier 4 me too lmao

[crimson sedge]

now we want a function in terms of u

this is like the usual function

ya know

,align
\m fu &= \s u \
\m fu &= \f1u \
\m fu &= 4u^4 + 3u

[crimson sedge]

choose your poison basically

is this clear?

[pine sun]

how did u get those values though 😭

[crimson sedge]

they are just examples

im like

[pine sun]

aa okeei lmao

[crimson sedge]

pulling them out of my ass

[pine sun]

fair enough

[crimson sedge]

so like

[pine sun]

they're diff rules or

[crimson sedge]

we said u = g(x)

[crimson sedge]

no

they are just examples

[pine sun]

aa okei

aight

[crimson sedge]

so like

we said u = g(x)

so imma plug in g(x) in all of them

[pine sun]

okeei

[crimson sedge]

,align
\m f{\m gx} &= \s{\m gx} \
\m f{\m gx} &= \f1{\m gx} \
\m f{\m gx} &= 4{\m gx}^4 + 3\m gx

[crimson sedge]

i did not do anything crazy here

im just subbing in g(x) for every u

[pine sun]

yeah

[crimson sedge]

so thats your composed function

SO

the idea is

we want to retrieve our f(u) first

you have several choices

for your first question [
\m hx = \f1{(x^2+3)^8}
]

[crimson sedge]

we want to first try to find a 'g(x)' in there

or like, the inside function

[pine sun]

x^2

?

[crimson sedge]

i mean, could be

you are not wrong!

[pine sun]

not the easiest?

😭

[crimson sedge]

yeah but like

go further

a tiny bit

include more to the function

[pine sun]

x^2+3?

[crimson sedge]

yeah!

that works!

[pine sun]

and then f is ()^8?

[crimson sedge]

not quite!

but we will get to that

[pine sun]

LMAOO

[crimson sedge]

so like

lets sub in u = x^2 + 3

what do u get?

[pine sun]

in what

wait

no idea

[crimson sedge]

for $\ds \f1{(x^2+3)^8}$

[crimson sedge]

just like, hide the x^2 + 3 behind u

what do u get

[pine sun]

(u)^8??

im lost

[crimson sedge]

almost, but what happened to the fraction?

[pine sun]

imean

1/(u)^8

[crimson sedge]

yeah

exactly

[pine sun]

so thats the value of u?

(x^2+3)

[crimson sedge]

so your outside function is 1/u^8

[pine sun]

aa okeei

[crimson sedge]

and your inside function is x^2 + 3

[crimson sedge]

yes

[crimson sedge]

so like i used u just so i dont confuse you

but like you can make it x

so lets verify this works

[pine sun]

right

[crimson sedge]

for $\ds \m fx = \f1{x^8}$ and $\m gx = x^2 +3$ we have: [
(f\circ g)(x) = \m f{\m gx} = \f1{{\m gx}^8} = \f1{(x^2+3)^8}
]

[crimson sedge]

so yes, turns out we are right

also, alternatively, you could have said

for $\ds \m fx = \f1x$ and $\m gx = \p{x^2 +3}^8$ we have: [
(f\circ g)(x) = \m f{\m gx} = \f1{\m gx} = \f1{(x^2+3)^8}
]

[crimson sedge]

both are correct

this goes to show that the functions you choose are not unique

you could choose multiple functions as composition

Anyways, I am done with what i am saying, are you lost? @pine sun

[pine sun]

aaa okeei I get it

[crimson sedge]

noice

[pine sun]

I'm not lost but just to confirm

[crimson sedge]

try doing b on your own

alright go ahead

[pine sun]

LMAO

[crimson sedge]

whats so funny?

[pine sun]

so B would be

u= 2 and s= sqrt(x-1)?

[crimson sedge]

you are very close, but not quite no

the s has to go somewhere in u

[pine sun]

aa okeei

[crimson sedge]

but like u = 2 is just a constant

its always 2

it never changes

[pine sun]

I thought so

[crimson sedge]

does this make sense to you

you need to add an argument for u

[pine sun]

yeah it does

so u is the sqrt😭

[crimson sedge]

somewhere s can go into

[pine sun]

orr

[crimson sedge]

like think about it

[pine sun]

i have no idea on what it could be

[crimson sedge]

you have 2 raised to something

how do you denote "raised to something"

[pine sun]

idk

[crimson sedge]

can you tell me what an exponential function is

[pine sun]

so u is 2(??)

[crimson sedge]

no

2 raised to something

[pine sun]

isnt it like e something

ln something

I don't remember all that well

[crimson sedge]

what do you write when i tell you "2 raised to 3"

[pine sun]

3^2

[crimson sedge]

no

the opposite, 2^3

[pine sun]

2^3

[crimson sedge]

yes

[pine sun]

mb

u would just say 2 cubed

[crimson sedge]

so now if i told you "2 raised to x"

[pine sun]

2^x?

[crimson sedge]

yes

so your u is?

[pine sun]

2^sqrt?

[crimson sedge]

no

just 2^x

[pine sun]

just x?

[crimson sedge]

yes

[pine sun]

so u always has to have an x so s can go into the x?

[crimson sedge]

yes

exactly

[pine sun]

aaa okeei lmao i get it now

[crimson sedge]

[pine sun]

thanks sorry 4 being a lil difficult

you're good at explaining 🔥

[crimson sedge]

its alright

thank you

thanks for trying

[pine sun]

thank you😭😭

have a good rest of ur day!

[crimson sedge]

you too

cya

you can close with .close

[pine sun]

jep

.close

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[south tulip]

I've got no problems in this
But please clarify to me what does the highlighted part mean

[south tulip]

Uh
The part was supposed to be highlighted

There

[mental trail]

it means the explanation on the right

"the vector space of continuous functions on the interval (-inf, inf)"

[south tulip]

So you denote continuous functions like this

[mental trail]

yes

[south tulip]

Alrighty

[mental trail]

C for continuous

sometimes we add a little 0 on top

C^0

to say that the 0th derivative is continuous

(0th derivative being the original function)

[south tulip]

Thanks for the quick reply, wasn't expecting it so fast

[south tulip]

Oh

[south tulip]

Wait but the 0th derivative is the function itself right?

Right, it is
Thanks

.close

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[candid lotus]

What formulas do you know ?

[timber plume]

all of them

all of the suvat formulas

[quick hollow]

How to do

[timber plume]

bruh what

find a help channel not in use

[candid lotus]

I believe you can use KE = 1/2mv²

And PE = mgz

You find the energy with mgz

Then you use it to find V in the other equation

@timber plume

[timber plume]

i dont think we're supposed to do that

we havent used energy or ke at all

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@timber plume Has your question been resolved?

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@timber plume Has your question been resolved?

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@timber plume Has your question been resolved?

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[fleet lake]

Ik it’s graded already but I need to study that question for an exam, anyone know what I did wrong?
WITH QUESTION C

[vapid dove]

you simplified incorrectly

it should be 8

not 27

if i were doing it i would solve the fraction part first, by the rules of exponents, it becomes 2w^3

[fleet lake]

ooh

yea i did the opposite i think

[vapid dove]

yep, then 2^3 becomes 8 and w^3 becomes w^9

you can do it your way too, you just have to watch out for simplification stuff

your way would become 6^3/3^3

[fleet lake]

where do u get the w^3 from?

[vapid dove]

the quotient rule

[fleet lake]

oh yea

mb didnt see it

alr i see

so if another question is similar to this, i should do ur method as its way easier right?

[vapid dove]

depends on the question

but 9 times out of 10 probably

2^3 is easier to calculate than 6^3 right

[fleet lake]

100%

[vapid dove]

good luck!

[fleet lake]

ok appreciate it man

🙏 \

@vapid dove 1 more question if u distribute an exponent like the 3 there to an exponent its just 3x2 but if its to a base it would be 4^3 for example?

[vapid dove]

uh

yes correct

[fleet lake]

oh okay

thank you\

[vapid dove]

yup

[fleet lake]

yea i think thats where i got confused

[vapid dove]

expoent rasied to expoent just becomes product of two exponents

but coefficients/bases are going to be raised by that power

[fleet lake]

oohh ok

[vapid dove]

the two you are talking about are power of a power and the regular exponent

(regular like 2^2 or 4^9)

[fleet lake]

oh ok

ok yea I understand now

I appreciate it man

[cedar kiln]

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[autumn lagoon]

i tried powering both sides in the first equation

and changing coses and sec by 1+tan and 1+ cot

but im stuck

<@&286206848099549185>

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[autumn lagoon]

<@&286206848099549185>

plsss

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[crimson sedge]

hello

[crimson sedge]

got 3 hours, needa cram all of those xD

is that doable?

I have some knowledge on them

[void sand]

depends on how much you already know

[crimson sedge]

Thats true

Im just gonna focus on the lhopital rule

is that a difficult concept?

honestly

I just cant get that much done in this much little time

Ill find a way to get a mark somehow

thanks

.close

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[void sand]

not especially

hardest thing is making sure the conditions are met before applying it

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[gloomy crater]

Find the point on $y=e^x$ closest to $(4,-3).$

[wraith dagger]

mXd

[mighty shuttle]

!status

[cedar kiln]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[raven shard]

are you allowed to use calculus on this problem?

(honestly i don't even think you can reliably do this without calculus but i thought i'd ask anyway)

anyway, the easiest way to do this problem is probably by using the distance formula; see if you can figure out how to apply that and ask if you need further assistance

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[brazen vortex]

The first term on a geometric series is 20 and a common ratio of 0.2.

Calculate the sum to infinity of the series.
i have an MCQ with this question, but my answer isn't any of the options

[wispy bronze]

!status

[cedar kiln]

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1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wispy bronze]

1. I don't know where to begin.

[raven shard]

what answer do you receive?

[brazen vortex]

S(inf) = a/(1-r)
20/(1-0.2) = 25

[raven shard]

that... seems right?

it seems either the question was flawed or you might have misunderstood it, but the problem and solution you've provided are definitely aligned

[brazen vortex]

those aren't any of the options, though, maybe my teacher is wrong?
there's 110, 100, 80, and 120

[brazen vortex]

the question i posted was just a copy and paste of what was sent, so i think there might be a problem with the question

[raven shard]

ah i see

🤔 the answer choices use a common ratio of 9/11, 0.8, 0.75, and 5/6

[brazen vortex]

yup, worked it backwards to determine that too, well, thanks anyway, sorry for wasting your time, i'll let my teacher know c:

[raven shard]

no worries misunderstandings are what they are 😅

and welcome 🙂

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[wet gorge]

How can I compute nCr(x,y) % x == 0 for any x,y efficiently? x does not have to be prime and x and y do not have to be coprime.

[cedar kiln]

Please don't occupy multiple help channels.

[wet gorge]

!status 2

[cedar kiln]

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 was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[slate lintel]

didn't someone link you a stackexchange post about this earlier?

[wet gorge]

It wasn't relevant

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[royal loom]

I was wondering if anyone could check my proof, I'm concerned that maybe I don't include enough explanation about which properties of a ring/whatever preserve each of my steps, but I also don't want to make it excessively wordy, if that makes sense

[slate lintel]

it's a good balance imo

i like that you don't consider 0 a zero divisor

makes things a lot simpler

[royal loom]

It wasn't my decision but thanks

XD

We consider 0 a subring though

that truffles some feathers

Okay thanks for the feedback

.close

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[cedar kiln]

Please don't occupy multiple help channels.

[slate lintel]

yeah AUSTIN

[royal loom]

I FORGOT

[potent wave]

Is it about abstract algebra?

Or more advance?

[royal loom]

Yes, I was wondering if the reasoning in my proof was 👌

[royal loom]

just abstract algebra, rings

[sturdy rose]

when will i do all this math 😋

[slate lintel]

this is usually mid college

if you're a math major

[sturdy rose]

highly doubt i will become a math major, the unis suck, the teachers suck and ofc the pay is like peanuts

[royal loom]

but the math doesn't suck! That's the appeal after all of a math major

[slate lintel]

say a+b is nilpotent here, not just "this"

[royal loom]

I think I meant thus

but yeah

[slate lintel]

yeah but what you mean is like (a+b)^{m+n} is 0, and therefore (a+b) is nilpotent

[solid juniper]

you say “so 0 is in N” twice

[royal loom]

0 is in N

but I have to prove that that 0

is the additive identity

in N

[slate lintel]

austin...

[potent wave]

I already taken class about class algebra and this is my first time i hear about nilpotent element

[royal loom]

yeah

okay hayley

whatever

[slate lintel]

the worst part is i believed it too until i looked at it the second time

[solid juniper]

lol ok

[royal loom]

a^(mn)b^(mn)

my bad

😭

since m,n are naturals

nm geq than m (or n)

and so we can say it's

a^(m+j) b^(n+i)

for j, i in 0, ....

so it's 0 * a^j

[sturdy rose]

my man being put down for not adding exponents, gotta love this server

[royal loom]

0* b^i

to fix it

[slate lintel]

i wouldn't do this to anyone else

[solid juniper]

i would

[slate lintel]

ok yeah true so would i

[sturdy rose]

okay i take this back

[slate lintel]

but they mostly have green or purple names

[royal loom]

Does my fix work :o

[slate lintel]

[slate lintel]

yeah i mean this is the key

[slate lintel]

we are friends :c

[sturdy rose]

[royal loom]

Okay so ignoring my horrendous exponentiation

[solid juniper]

it looks fine after the comments in this channel

[royal loom]

how's the rest XD

[slate lintel]

i don't like this sentence idk

just like

you say the same thing five times

just say it once?

[royal loom]

I prefer to make my TA suffer

I think

[slate lintel]

i actually just don't understand the last sentence at all

[royal loom]

if n^k=0

-n^k=-0=0

so -n^k in N

[solid juniper]

it’s a bit weirdly written but it’s ok i think

[slate lintel]

right....

[royal loom]

so n^k+(-n^k)=0

[slate lintel]

what makes you think n^k = (-n)^k

[royal loom]

and we have additive inverses

[slate lintel]

or like

[royal loom]

n^k=0 => -n^k=0

[slate lintel]

what you really need to show is that (-n)^k = 0

which i guess you do

sort of

[royal loom]

what why

[slate lintel]

because you need to show that the additive inverse of n is in N?

[royal loom]

I have to show there is m

such that

m^j=0

and

n^k-m^j=0

so use m=n and j=k

[slate lintel]

i mean no you just have to show that (-n) is in N

like

since n is in N

and you need it to be closed under inverse

[royal loom]

since n is in N we know n^k=0

consider (-n)^k

this is either n^k or -n^k

which is either 0 or -0

[solid juniper]

if you’re worried about that, n^k = 0 implies (-1)^k*n^k = 0

[royal loom]

and thus (-n) in N

[slate lintel]

ok yeah i wasn't sure if that was allowed in an arbitrary ring but it makes sense bc this is a commutative ring

[royal loom]

the fix is in

okay cool

ty @slate lintel @cursive gorge

[slate lintel]

yeah thanks chmonkey

[solid juniper]

the even odd case thing is unnecessary

[royal loom]

k is a natural number and thus is even or odd

https://tenor.com/view/funny-emo-wolf-werewolf-transform-gif-27196401

[solid juniper]

this is a simpler way to put it

[royal loom]

https://tenor.com/view/alpha-wolf-sigma-male-gif-26562640

.close

[cedar kiln]

Channel closed

Closed by @royal loom

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.

[odd sierra]

plz i just need to get started with it thats really what i am lost with

Like do i multiply 5x^2 and 5y^2 on the numerator and denominator

for the x/y-y/x

and the same with the bottom

Ok i try and show

Uh

DOes anything cancel out

Like when im multiply

Ill just send the pci its prob wrong

[slate lintel]

you should be able to cancel quite a few things

[odd sierra]

the set up shoud like this?

look*

idk what would cancel here assuming this is right

[slate lintel]

what i would probably do here

is kind of treat the numerator and denominator independently

and for each of them, come up with a common denominator (so like for the top that would be xy)

and combine them into one fraction

[odd sierra]

Ohhh

thats sounds much better

ok i try thrn show

[slate lintel]

and use this $\frac{\sfrac{\rsq}{\osq}}{\sfrac{\bsq}{\gsq}} = \frac{\rsq}{\osq} \times \frac{\gsq}{\bsq}$

[wraith dagger]

hayley!

[odd sierra]

yes

Ok perfect thnx

ok i feel like i keep making mistake sry

5 more mins

i will be done

i keep messing up

I show

[cursive gorge]

Lmfao, I got pinged instead of @solid juniper

[solid juniper]

[odd sierra]

wot

o

.close

[cedar kiln]

Channel closed

Closed by @odd sierra

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.

[dense dawn]

Is this correct?

[cedar kiln]

@dense dawn Has your question been resolved?

[hearty arch]

looks good to me

[cedar kiln]

@dense dawn Has your question been resolved?

Channel closed

Closed by @dense dawn

Use .reopen if this was a mistake.

[cedar kiln]

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.

[maiden venture]

Given are that plane pi: x-y+z =-4 and the line L: (x,y,z) = (2,1,3) + t(5,-2,1) a) determine the point of intersection P between L and pi. b) state the orthogonal projection L1 of L onto the plane pi.

[maiden venture]

I have done a and I got (-3,3,2)

How do I solve b?

[burnt vapor]

You could find two points in L1 and then find the equation of the line

[maiden venture]

Oh okay great, how do I find the two points?

[burnt vapor]

Choose a point Q in L, and then find the orthogonal projection of Q onto π. That will be in L1

[maiden venture]

Okay so I can choose (2,1,3) from L?

[burnt vapor]

Yes

Any point

[maiden venture]

And do I find the projection of q by using a formula?

[burnt vapor]

Yes, you can use a formula if you know it

[sharp stratus]

hello

oh its taken my bad

[maiden venture]

We have a formula that says u’ = u•v/ |u|^2 * v

[burnt vapor]

That works when u and v are vectors

[maiden venture]

Oh okay maybe it’s easier not to use a formula then

[burnt vapor]

You could find a line l such that π and l are perpendicular and Q is in l

[maiden venture]

Okay I am not entirely sure how to do that

[burnt vapor]

Do you know how to find a perpendicular line to a plane?

[maiden venture]

I have done it before but I am not sure if I remember

I think I am supposed to use n = (1,-1,1)

[burnt vapor]

Why 2?

[maiden venture]

Oh sorry I meant 1 my z looked like a 2

[burnt vapor]

Yes, you have to use that

That is a normal vector to the plane

[burnt vapor]

So this line l will have as a direction n, because it is perpendicular

[maiden venture]

Yes okay

So do I do (2,1,3) * (1,-1,1)?

[burnt vapor]

Why?

[maiden venture]

I don’t really know