help-0

[obsidian cave]

so what are the labels exactly

[fiery spire]

labels are the class, for example cat and dogs

the shifting is not related to label

I am only testing the effect of shifting concerning adversarial suppression

[obsidian cave]

seems like a unique task

[fiery spire]

somehow

It was a very interesting discussion

Thank you very much @alpine sable and @obsidian cave

Very grateful for the help

Have a good night or day !

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[azure bison]

How can I prove a limit is equal to L using the comprehension/sandwich theorem?

[alpine sable]

by proving the sequences in both sides also tend to L

[azure bison]

If I have lim x - 0 of (x * sen( 1 / x^2)) is equal to 0

what can I do? Ive only seen how to use this when they give me the criteria to both sides

not the middle one only

[alpine sable]

use the fact that $-1 \leq sin(x) \leq 1$ for all x

[ocean seal]

AimaneSN

[azure bison]

Hmm Ill try it out and see if I can do it with this. Thanks!

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[azure bison]

does this mean that -1 and 1 are the sandwich functions?

if what I have is the limit of sin(x)

[tacit arch]

as x approaches what?

[azure bison]

0

[azure bison]

this

[bitter vault]

Use the fact that $|x \sin(1/x^2)| \leq |x|$

[ocean seal]

1345631

[azure bison]

[noble sinew]

If -1<=sin(y)<=1 for all y

[noble sinew]

It holds for y=(1/x^2) also don’t you think?

[azure bison]

yeah exactly

[noble sinew]

So what is the problem?

[azure bison]

its that if f(x) was only sen(x) then then the limit of the sequences of both sides would be -1 and 1 right?

but then what happens if its x * sen(x)?

[noble sinew]

Try replacing sin(1/x^2) with -1 and 1

And compute the limit

As multiple people have said

[azure bison]

what do you mean by this?

[noble sinew]

If 1>=sin(1/x^2)

Then x>=x * sin(1/x^2) for x>0

[azure bison]

Ah I see, so x * -1 <= x * sin(1/x^2) <= x * 1

[noble sinew]

Well something with signs flipping for x>0 and x<0

But yes that is the idea

[azure bison]

So the signs would flip if it was -x right?

-x * 1 > -x * sin(1/x^2) > -x * 1

its the same thing though I would presume

as long as the other two functions have the same limit

[noble sinew]

Nothing with -x?

Its just if x<0 then when multiplying both sides of an inequality by x since x is negative

Inequality flips

[azure bison]

What does that mean then?

if in this case x was -1 instead of 0?

[noble sinew]

What?

x is x???

[azure bison]

I dont know I dont understand what you mean by inequality flips

[noble sinew]

1<=2 agree?

If we multiply both sides by -1 we get -1<=-2 which is false

The mistake is we need to flip inequality when multiplying by a negative number

So its -1>=-2

[azure bison]

So thats if the limit was approaching a negative number?

I would have to flip the functions on the sides

[noble sinew]

It approaches 0

Can approach 0 from negative or from positive can’t it?

[azure bison]

yeah

Do I ever have to worry about that when doing this though? How does that change what im doing?

[azure bison]

this

[noble sinew]

If x<0 your inequalities are false

[noble sinew]

.

[azure bison]

So its wrong?

[noble sinew]

.

[azure bison]

I feel really stupid for not understanding what you are trying to say it's probably so easy

[noble sinew]

If x<0 x is negative, agree?

[azure bison]

yes

[noble sinew]

So what happens when we multiply an inequality on both sides by a negative number?

[azure bison]

this

[noble sinew]

So can you write the correct inequality for x<0 now?

Using that?

[azure bison]

x * 1 < x * sin(1/x^2) < x * -1

?

[noble sinew]

Yes

[azure bison]

So since x is approached by both sides and one side is negative I have to take that into account for this, as if x was negative?

[noble sinew]

lim exists if lim x approaches from left is equal to lim approaches from the right

So its just using definition and splitting up in the two cases in the definition

[azure bison]

I think I get it but I will have to study this further because it's kinda confusing still. But thanks a lot for the help and patience sir

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[spark pebble]

can somone help me with this

[alpine sable]

$Arclength=\theta \cdot \frac{\pi}{180} \cdot r$

[ocean seal]

AimaneSN

[alpine sable]

where r is the radius of the circle

and $\theta$ is expressed in degrees

[ocean seal]

AimaneSN

[spark pebble]

so it would be 160 x pi/180 x 9 ?

[alpine sable]

Correct, they asked you to express the answer as some fraction x pi

[spark pebble]

do I solve the equation now?

or just leave it as is

[alpine sable]

wdym

[spark pebble]

nvm

that was a dumb question

so this is what I got

so now I round to the nearest 10th?

[alpine sable]

you just need to simplify the fraction that is multiplied by pi, I mean $\frac{160 \cdot 9}{180}$

[ocean seal]

AimaneSN

[alpine sable]

no you need to leave it as a fraction

[spark pebble]

oh

[alpine sable]

but a simplified one

a fraction multiplied by pi

[spark pebble]

160 times 9/180 = 1440/180

and now I simplify that fraction?

[alpine sable]

Yes, it can be simplified

[spark pebble]

oh alr

[alpine sable]

it's equal to 8

[spark pebble]

8/1

right

8

so that's the answer?

[alpine sable]

it's $8 \cdot \pi$ I think

[ocean seal]

AimaneSN

[spark pebble]

yep

thanks man

appreciate it

[alpine sable]

Np

[spark pebble]

w

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[elder umbra]

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[floral jolt]

This isnt really a specific math problem, but if I had a convex polygon and I wanted to find the minimum area bounding square, would it work to find the centroid and extend it out to the farthest point to make half of the diagonal of the square?

[floral jolt]

If that doesnt work are there any other solutions?

ik for the minimum bounding rectangle it has to share a side with the polygon but i dont’t know how i would apply that to a square

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@floral jolt Has your question been resolved?

[tight locust]

The general case of a convex polygon is tough to deal with

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@floral jolt Has your question been resolved?

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@floral jolt Has your question been resolved?

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[dim bane]

Need help converting this series to a summation

This is what I think it is but not sure how to verify

[sudden hinge]

what does the sum give for n = 0?

something divided by 4 right? so that cant be right

[dim bane]

-x^2/4

[sudden hinge]

So we need to somehow get rid off the /4 and make the exponent 3 for n = 0

How could you do that?

[dim bane]

The exponent in the denominator?

[sudden hinge]

nah we want the first term to just be x^3

x^(n+2) doesnt work if you start from n=0

[dim bane]

So if I change it to x^(n+3) I get -x^3/4

Closer

[sudden hinge]

so notice the denominators are
1^2, 2^2, 3^2, 4^2, ...

so you just need to tweak your n+2 a tiny bit in the denominator

[dim bane]

n^2

and start the index at 1, not 0

[sudden hinge]

or just write (n+1)^2

[sudden hinge]

that way you exactly get that sequence

[dim bane]

💪

[sudden hinge]

now there are 2 problems left

• the signs of the actual sum is "+ - + - ..." but you have "- + - + ..."
• you're increasing the exponent of x only by one after each step of the sum

[dim bane]

right

So take away the negative sign on the (-1)

[sudden hinge]

no

(-1)^(n+1), you start at n=0 so your exponent starts being odd

remember (-1)^n = 1 if n is even and (-1)^n = -1 if n is odd

[dim bane]

Trying to see it

@sudden hinge Believe I have the sign issue fixed. Not sure about the exponent problem

[vale wigeon]

your exponents form an arithmetic sequence with initial (zeroth) term 3 and common difference 2

[dim bane]

Thanks @sudden hinge @vale wigeon Believe I figured it out.

[vale wigeon]

(-1)^(n+1)

also clear out some of those redundant parentheses

[dim bane]

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[loud oriole]

what does my teacher mean by direction angle in each case

[silver adder]

uhh is that the answer on the right

[loud oriole]

yes this is the correction sheet

[silver adder]

ive never had to use these before

i think theyre just the angles between the vector and each axis

[loud oriole]

so how exactly would i get one of the answers

which is what i cant figure out

[silver adder]

well first you want to normalize the vector

[loud oriole]

ok

so sqrt(1+4+4)

[silver adder]

that is the length

[loud oriole]

ok wait lol

we never did normalisations

oh u mean 1/|u| * u

[silver adder]

yup

[loud oriole]

so (1/3 ; 2/3 ; -2/3)

[silver adder]

by dividing each component by length it normalizes the vector, aka makes it length 1

you have to do this here so the coordinates are in the correct range to use arccos

the unit vector represents direction

[loud oriole]

ok i see

so basically i normalize then i add arcos to each value

nice, thx my g

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[silver adder]

heres a visualization of the angles

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[barren elbow]

How do you do this

[silver adder]

triangular shelf systems are where its at

[clear stump]

is it a perfect triangle

I mean like all degrees the same

[barren elbow]

i would assume so yeah

since each of the shelves are parallel

[clear stump]

they could be parallel without that but assuming that it is then find the height of the triangle

[barren elbow]

They are parallel it says in the question

[clear stump]

I mean could be different angles

[barren elbow]

oh I see

use pythagoras?

[clear stump]

You would need more info

[barren elbow]

so how do I find the height then

[clear stump]

sin

You know the height

the length of the side

[barren elbow]

whats the angle tho?

ohhh i see what you were saying about not assuming its equilateral

but what angle do I use then

[clear stump]

60

lets assume all are 60

[barren elbow]

how can we do that though

[clear stump]

Just say that all are 60

No angles were given

you are allowed to use sin cos and tan?

[uncut plank]

Use similar triangles

[barren elbow]

It's non calculator

[barren elbow]

How would you do that

[clear stump]

oh yeah

[barren elbow]

omg

is it just 9?

[uncut plank]

There are 3 similar (right) triangles in this picture. What did you do?

[barren elbow]

a : b : c is 2:3:1?

[uncut plank]

Yes

[barren elbow]

18 / 6 = 3

3 * 3 = 9

[uncut plank]

Exactly

[barren elbow]

thank you so much

[uncut plank]

Np

And also type .close when you are done

[barren elbow]

.close

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[severe jungle]

How do i do this

[atomic acorn]

u know newton leibnitz rule?

or called leibnitz rule

[severe jungle]

no

I don't

what is that?

[atomic acorn]

oh

you need that

[severe jungle]

oh how do you use it

[atomic acorn]

i doubt you can do it without it

[severe jungle]

how do you use it

[severe jungle]

oh..

we never learned that

[atomic acorn]

if you dont know bout this i suggest you read about it from somewhere

[severe jungle]

is there no alternative method

[atomic acorn]

there could be

but i cant think of one rn

coz you clearly cant integrate the function

[severe jungle]

yeah

wow ok that actually works, thanks!

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[alpine sable]

can u help me define the maximun and minimum number y and x will take?

[alpine sable]

i found that x<y<root of 1+y^2 but i cant find the numbers for x

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

[opaque hinge]

Holy cow

That’s commitment lol

[alpine sable]

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[daring prism]

i have 2 questions about this

[wary hull]

Mannn fuck you you stole my channel

[daring prism]

sorry dude...take it i didnt mind

[wary hull]

Nah you're good bro

Cya

[abstract fractal]

No, this is your channel

[daring prism]

understod, then i keep, with the channel, my question is about the first exercise, cant say if the red ? are true or false

and the second one, i think there is no diagram there that really match what the excersice says

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[shell vault]

• The point P (-15, -8) is on the terminal side of the angle ϴ in normal position. Determine the
  value of trigonometric ratios for ϴ

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[foggy acorn]

y'''+2y''+y'+2y = 0

while i already know the general solution, i do not know how f_1(x) = e^-2x, f(x) = cos(bx) and g(x) =sin(bx) are supposed to be used as a different way to get the general solution
the following info are given:
f_1(x) = e^-2x satisfies the equation
f(x) is a solution for b ∈ R
g(x) is a solution for all positive b that are values for f(x) to be a solution
f_1(x), f(x) and g(x) are linearly independent

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[foggy acorn]

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[tacit arch]

3 is enough to form a fundamental set.

https://tutorial.math.lamar.edu/classes/de/HOBasicConcepts.aspx

You just need to show the Wronskian isn't identically zero

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[blissful plover]

what is this question asking for ?

a)

[worn fox]

what happens to g(x) as x gets really really big

[blissful plover]

yea i understand that

but is it asking like 0+ or -0

something like that

[hollow shale]

x → infinity

[worn fox]

0 or 0+, doesnt really matter

we have no more information than you about what the question wants

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[empty kindle]

how to show that converges or otherwise 😄

[empty kindle]

think i should divide with another function

but i dont know with what

[marsh rapids]

I'd say it converges

Because you can compare to the series

,w integral from 1 to inf of (x 2^x +7)/(3^x + ln x + 1) dx

[ocean seal]

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[marsh rapids]

Yep

See how bringing it down to a series can be useful ?

[empty kindle]

hmm

[placid zinc]

It really looks like it should converge, as it acts "like" x(2/3)^x

[marsh rapids]

You can make that rigorous actually

But you probably know that

[placid zinc]

But I'm at a bit of a loss for how to actually obtain that

[marsh rapids]

Equivalents all the way

Theory of series of positive terms

[marsh rapids]

Want details ?

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@empty kindle Has your question been resolved?

[empty kindle]

well

since

3^x grows faster than x^3

(2/3)^x will decrease faster than 1/x^3

and after some time

this will be true maybe

but integral from 1 to infinity of 1/x^2 converges

therefore

then the first func also converges 😄

idk

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[marsh rapids]

Lacks rigor but that's an idea

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[junior atlas]

Prove or disprove the following conjecture: There are three distinct interiors a, b and c greater than 1 such that a^b = b^c
Anyone can help me with the question?

[marsh rapids]

You just need to find an example to show it's true. Try to find one

[junior atlas]

oh yeah, i find now. Thank you

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[elder hamlet]

how many isoceles triangles can be made from the verticies of an octagon

[open folio]

56

[elder hamlet]

isoceles?

[open folio]

Oh...

I thought you meant like this

Hmm thats going to be trickier

[elder hamlet]

total number of triangles is just 8C3 so

yea idk how to calculate isoceles

[open folio]

Me neither lol

[elder hamlet]

<@&286206848099549185>

[vestal wedge]

Do you want every points of the triangle is the point of an octagon?

[elder hamlet]

yea the vertices

[vestal wedge]

Then I believe there are total 24 of them

[elder hamlet]

hm so 24 isoceletes triangles?

[vestal wedge]

by choosing one of the red lines, and connecting their end points to one of the blue dots

[elder hamlet]

You can do 4 rotations of that right?

[vestal wedge]

yeah

[elder hamlet]

so 24

ok ty

.close

[lone heart]

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[green hazel]

can someone explain this to me

why is only 3 true

I though 1 is true

infinity to 22 power has to be greater than 2 ^infinity

[worn fox]

What makes you think that?

[green hazel]

Is it wrong

[dawn birch]

yes

[velvet pelican]

try considering their logs instead

[green hazel]

so 2 ^ infinity is greater than x^22

[dawn birch]

a polynomial will never grow faster than an exponential in the long term (as long as base > 1)

[green hazel]

thx

I will remember that

What about ln function

[worn fox]

Look at their graphs

Logs grow very slowly

[green hazel]

v

why a aned not d

[worn fox]

Polynomials are always continuous

[green hazel]

also does f(3) have to be defined

I thought if the limit for left equals right than the function is continuos

[worn fox]

f(3) has to be equal to those limits for it to be continuous there

[green hazel]

ok

[alpine sable]

Are you done with the question?

[green hazel]

I want to know

Is A wrong because f(3) is not defined

Explicitly

[alpine sable]

Just my comment in the earlier u can also solve it without knowing that fact u get a infinity / infinity form using lhoptal rule u will get 1/3x 3x becomes very big so 1 over that is just 0

[alpine sable]

@worn fox already answered that ig

[alpine sable]

^

[lone heart]

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[spark pebble]

I dont understand why 5 is Subtraction Property of Equality

<@&286206848099549185>

[green hazel]

90 degrees plus x = 180 degrees

Subtract 90 degrees

subtraction property

It's just a stupid definition

I don't like how the proof is like that lmao

[spark pebble]

Loll

alr I think I get it

so the subtraction property

isn't it like

you have to take away the same amount from both sides to keep the equation equal?

[green hazel]

yes in general

but this is isolation of 1 variable

You're solving for x

how do you isolate x

subtract 90 degrees from the x side, and subtract 90 degrees from right side

So the equation is equal

[spark pebble]

ohhh

right right

[green hazel]

when u solve for something

[spark pebble]

so x=90

[green hazel]

Do the opposite operation

If it's adding

Subtract to isolate

[spark pebble]

okk

[green hazel]

if it's negative

add to isolate

Same with multiplication and division

[spark pebble]

alrr

[green hazel]

9x=18

solve for x

divide by 9 on both sides

x=2

[spark pebble]

ohh

ook

[green hazel]

you have to isolate

variables

That is algebra

very important

So practice it a lot

[spark pebble]

alrr loll

thanks man

[green hazel]

np

[spark pebble]

so I wouldn't subtract since its multiplication

right?

[green hazel]

no

there is nothing to subtract

opposite operation

addition is opposite of subtraction

and vice versa

[spark pebble]

yep

[green hazel]

multiplication is opposite of division

and vice versa

[spark pebble]

okk

thanks man

appreciate ti

[lone heart]

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[plush moon]

how would i do this question

[plush moon]

^

[dawn birch]

what progress have u made so far @plush moon

[plush moon]

i've finished the other questions

this is the only one i dont know how to do

ik that i would have to organise the x and y terms

to be on opposite sides

then differentiate

but i don't even know how to get to that stage from here

[upbeat peak]

$\frac{ln(y)dy}{xe^{2x}dx}=3 \implies ln(y)dy = 3xe^{2x}dx$

[ocean seal]

Elyazid

[upbeat peak]

now you need to integrate both sides

the left side with respect to y and the right side with respect to x

[plush moon]

thank you so much

ill do that now

i didn't even realise it was that easy lmao

ig im just tired

thx again

[upbeat peak]

no problem 👍

[lone heart]

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[spare fern]

why isn't x greater than or equal to 2?

please dm or something cause I'm very confused with Inequalities in general, I would appreciate if someone can take some time off and help me out, I've been trying to understand this for over a week now.

[tight locust]

Shit i misread

Nvm

[uncut plank]

First solve 2x + 3 ≥ 7 and then solve 2x + 9 > 11

[lone heart]

@spare fern Has your question been resolved?

[gray isle]

show all work

[tacit arch]

Probably the OR that's tripping you up

[lone heart]

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[dawn quail]

What is the rule for flipping signs? Negative to positive and positive to negative, I’m a bit confused on when it’s legal to do so

[dawn quail]

Straight addition subtraction only gives very different answers, but addition subtraction with multiplication or division gives exact same answer

[gray ingot]

when it's multiplication or division you only flip signs once,
eg
ab = c

(-a)b = -c
or
b(-a) = -c

(-a)(-b) != -c

[gray isle]

um both those equations at the bottom are weong

[wary stream]

[dawn quail]

Oh right sorry need brackets before division

[gray ingot]

still it would be wrong if you changed signs of both the numerator and denominator

flipping the sign is basically multiplying by -1 on both sides

[gray isle]

if there are sufficient/appropriate parentheses it's fine

[dawn quail]

Yeah, that part makes sense to me..

I guess it’s why it only works with mult or div to flip all signs

Without mult or div you can’t flip signs like that

To get the same result

In other words..

[gray isle]

$\frac{-10-2}{6} = \frac{-10-2}{6} \cdot \frac{-1}{-1} = \frac{10+2}{-6}= -2$

[ocean seal]

ℝamonov

[dawn quail]

It’s how he was able to flip the final answer, it threw me off when I saw that but I understand now it’s only possible by multiplying by -1/-1

[gray ingot]

yes

[dawn quail]

Ty!

How do I close help without waiting for bot?

.close

[lone heart]

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[dawn quail]

Nvm found it

[lone heart]

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[lunar girder]

Hey guys

[lunar girder]

.close

[lone heart]

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[blissful plover]

how do i make find the top numerator of the function while making sure local max occurs at x = 0.5 ?

[half stream]

How would you find a vertical asymptote?

This will lead you on the path to getting the solution

I think

[blissful plover]

what ?

[half stream]

Sorry I haven't done this in a while

[blissful plover]

i already found the vertical asymptote wdym

lol its ok

[half stream]

Well if you reverse engineer it

You should also consider the y-int and the hole at (1,UND)

If I remember correctly, a hole occurs when a factor of the numerator cancels out with a factor of the denominator

So that means a value of the numerator would be (x-1), as well as the denominator

[real solar]

A horizontal asymptote occurs when the numerator is 0

Maybe that helps

[blissful plover]

yea im aware but that doesn't really help with the other ones i have to consider

[real solar]

But its only 0 at one point

[blissful plover]

idk how to make the numerator so that i can set the local max at 0.5

[lone heart]

@blissful plover Has your question been resolved?

[lone heart]

@blissful plover Has your question been resolved?

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[shell widget]

@blissful plover Were u able to solve the question?

[shell widget]

.close

[lone heart]

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[blissful plover]

Nope but no one was responding lol

[shell widget]

Your function is not defined at x=1

so it's something of the form y= (x-1)/(x-1)(x+5)(x-6)

But if you put x=0

Do you get y=-0.1?

@blissful plover

[blissful plover]

huh wdym y=-0.1

y intercept ?

[shell widget]

Look at your graph

There's a point given (0,-0.1)

[blissful plover]

yea i see

[shell widget]

That means at x=0, y=-0.1

If you put x=0 in y= (x-1)/(x-1)(x+5)(x-6), do you get y=-0.1?

[blissful plover]

how do i put that on the equation ?

[shell widget]

Just replace x by 0

[blissful plover]

wdym lol

so how would it look

[shell widget]

y=(0-1)/(0-1)(0+5)(0-6)

[blissful plover]

oh u mean sub in 0 for x

[shell widget]

yes

[blissful plover]

no i dont get y = -0.1

something a lot less

[shell widget]

Yes

So we need to scale our function

We will have y = a(x-1)/(x-1)(x+5)(x-6)

And we need to choose a now

So again, put x=0, y=-0.1 in the equation and now find "a"

This will allow us to scale our function in such a way that it meets the given requirements

@blissful plover Did you find "a"?

[blissful plover]

nop

[shell widget]

You just need to solve the equation -0.1 = a(-1)/(-1)(5)(-6)

[blissful plover]

a is 3

ohh

omg i got the y intercept

y = 3(x-1)^2/(x-1)(x+5)(x-6)

[shell widget]

Don't really need the y-intercept here

[blissful plover]

i had to square tho

[blissful plover]

ok

[shell widget]

Oh wait

Nvm lol

[blissful plover]

so now what do we do abt the 0.5 lol

[shell widget]

wdym?

[blissful plover]

i think thats all wee need now

other than the 1, undefined

[shell widget]

Why did you square x-1 in the numerator?

[blissful plover]

because

the degree needs to be the same

it has HA at y=0

no ?

[shell widget]

y=3(x-1)/(x-1)(x+5)(x-6) meets our requirements

[blissful plover]

wait wtf ur right

why did you put the (x-1) at the bottom tho ?

i guess that made us not rquire squaring the top

[shell widget]

Because we have 0/0 when you plug in x=1

Hence undefined at x=1

[blissful plover]

wait i dont get it lol

if we make all the x's 1, the numberator and denominator both become 0 ?

[shell widget]

Yeah

You just need to look at the x-1 in the numerator and denominator

[blissful plover]

ohh

[shell widget]

But yeah, you get 3(0)/(0)(5)(-6) = 0/0

[blissful plover]

wow ur not even a helper and u helped me 😂

thanks so much

[shell widget]

now you have your function, f(x) = 3(x-1)/(x-1)(x+5)(x-6)

it should be easy to show that there's a local max at x=0.5

yeah np

Good luck

[blissful plover]

wait but last question

if we put (x-1) on the denominator

does that mean there's also a vertical asymptote at 1 ?

[shell widget]

No

[blissful plover]

how come

[shell widget]

If it was 1/(x-1), then yes

But since it's (x-1)/(x-1)

[blissful plover]

oh

cancels out ?

[shell widget]

As you approach 1 from either side

They cancel out

But at 1, it's undefined

[blissful plover]

ohhh ok i see

ok im gonna ask more questions but if ur busy u dont have to help me

im already thankful

[shell widget]

Np

If I can help, I will. otherwise someone else will, hopefully.

[blissful plover]

https://cdn.discordapp.com/attachments/490557019623915520/972397418467450900/IMG_3710.jpg

i have to fill in the chart with the previous graph

using this

its not even telling me like

to solve for increasing or decreasing ?

[shell widget]

Sign of slope of tangents is the same thing

If the sign is negative, function is decreasing, if sign is positive, function is increasing

you can find the sign of function easily from the graph

for the second part, find y'

[harsh swallow]

sign of the function is when is the function positive and when is it negative

[blissful plover]

ohh

like the intervals ?

[harsh swallow]

the second one is when is the function going down and when is the function going up

the slope

[blissful plover]

wait so

[harsh swallow]

if going from left to right

when do you have to go uphill when downhill

[blissful plover]

basically the entire

part is positive

and this part is negative

[harsh swallow]

yes

but you have to correctly put that into intervals

[blissful plover]

how would i express those as an interval though ?

[harsh swallow]

well you know that -5 is one of the asymptotes

and that's where it switches from positive to negative

[blissful plover]

oh and the other part is negative infinity

[harsh swallow]

yes

[blissful plover]

so something like uh

-∞ > -5

or is it

-∞ < -5

[harsh swallow]

no

[blissful plover]

neither ?

[harsh swallow]

neither

you can write intervals in multiple ways but the most common is

[blissful plover]

oh

[harsh swallow]

-infinity < x < -5

or

(-infinity, -5)

[blissful plover]

oh yea i forgot the x

lmao

wait how do i show that

[harsh swallow]

there's subtle differences between the two but you don't need to worry about that 😄

[blissful plover]

-infinity < x < -5 is positive

well we know -5 < x < 6 is negative

https://cdn.discordapp.com/attachments/490557019623915520/972398605656809542/unknown.png

this right

[harsh swallow]

not -6

[blissful plover]

yea just 6

mb

so would i put in the table

sign of slope of tangents | positive: -infinity < x < -5 , infinity > x > 6 negative: -5 < x < 6

like this ?

[harsh swallow]

that's the sign not sign of slope

[blissful plover]

what

im confused now

[harsh swallow]

you wrote the sign of the function

not the sign of the slope

[blissful plover]

ah

wait

so

sign of function | positive: -infinity < x < -5 , infinity > x > 6 negative: -5 < x < 6

like this yes ?

is there anything wrong there

[harsh swallow]

that's good

[blissful plover]

how do we do this

[harsh swallow]

is it on paper?

[blissful plover]

yuh

[harsh swallow]

is the graph on paper?

do you have a ruler?

[blissful plover]

wait

ye sbut

i put it on desmos

why

and yes i have a ruler

[harsh swallow]

because the easiest way i can think of to know the sign of the slope

is to hold the ruler along the curve

and see if the right side is above the left

[blissful plover]

oh or else i havet o do the chart right

[harsh swallow]

if so

[blissful plover]

with 0.001

or smth

[harsh swallow]

then it's positive

if not then it's negative

if one is above the other then there's no slope

if the ruler is horizontal then the slope is 0

[blissful plover]

wait which curve

theres 3

[harsh swallow]

it's all the same function

[blissful plover]

i mean yea but

like where along the curve

[harsh swallow]

starting from the left

everywhere

[blissful plover]

oh shoot off topic but

is that 0+

[harsh swallow]

yes

[blissful plover]

oh okay

[harsh swallow]

as x goes to infinity

g(x) goes to 0

from the positive side

[blissful plover]

yuhh tyty

ok anyways back to the thing

but

the graph is so smal

i cant use the ruler 😭

[harsh swallow]

you could also do it in your head

imagining a ruler along the function from left to right

lets start with the left most bit

what's the sign of this bit?

the sign of the slope

i mean

[blissful plover]

obviously

[harsh swallow]

yes it goes up from -infinity to -5

then from -5

what does it do

[blissful plover]

just keeps going up

wdym

[harsh swallow]

until?

[blissful plover]

-5 ?

or u mean infinity

[harsh swallow]

no i mean from -5

[blissful plover]

oh

[harsh swallow]

on the right side of -5

[blissful plover]

-5 to 6

[harsh swallow]

what does the slope do

[blissful plover]

it goes positive til x=0.5

then drops back down negative

[harsh swallow]

exactly

[blissful plover]

to 6

[harsh swallow]

so we know that -infinity < x < -5 is positive slope

and

-5 < x < 0.5 is positive slope

0.5 < x < 6 negative

and the last one?

[blissful plover]

6 < x < infinity

[harsh swallow]

the arrows are the wrong way around

[blissful plover]

yes

[harsh swallow]

but what's the sign there?

[blissful plover]

positive

[harsh swallow]

no

[blissful plover]

wait

no i

its going negative

right

[harsh swallow]

yes

[blissful plover]

cuz its left to right

[harsh swallow]

yes

so we have

positive: -infinity < x < -5, -5 < x < 0.5 and negative: 0.5 < x < 6, 6 < x < infinity

[blissful plover]

sign of slope of tangents | positive: -5 < x < 0.5 , -infinity < x < -5 negative: 0.5 < x < 6 , 6 < x < infinity

[harsh swallow]

for the slope

[blissful plover]

oh yeah

[harsh swallow]

there we go

you didn't need me

[blissful plover]

no i def did

[harsh swallow]

you got it all covered

[blissful plover]

ur helping me so much

😭

im almost done my work

but rest is like solving

which im completely new to

2 solve questions and 2 word problems

i hate word problems

wait nvm solving is super easy

am i supposed to solve for x ?

it doesn't mention

[harsh swallow]

yes

[blissful plover]

x +3 = ( x+1)(x-1) + 2x(x+1)

x =1 , x = -4/3

since the equation is undefined for 1, the answer is just

x = -4/3

yes ?

[harsh swallow]

ye

s

[blissful plover]

ok yea i finished all solving

these 2 and im done

i need help

<@&286206848099549185>

nvm

.close

[lone heart]

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[serene thistle]

Abstract algebra map of representation question. How do I find rho(1) and phi(1)?

[serene thistle]

Question

Answer

[vale wigeon]

Here are some linear maps between the two representations in the last question.

[lone heart]

@serene thistle Has your question been resolved?

[serene thistle]

@vale wigeon could you further clarify.

[vale wigeon]

what's there to clarify? this question refers to the previous question.

so presumably rho and phi come from there too.

since you didn't show the previous question, i cannot clarify anything about the previous question. i would need to see the previous question to clarify something about it, but i can't see it because you haven't sent it.

[serene thistle]

@vale wigeon I didn’t even read the “last question” part. So I have spent ages trying to figure out where they got them from, I thought they appeared from nowhere. Anyway, thanks.

[lone heart]

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[dense dock]

One of the vertices of a regular hexagon is located at the point (0: 2). Its two sides belong to the second quarter and the rest to the first. Find the coordinates of the point of intersection of the large diagonals of a hexagon if its side length is 8 cm.
The official answer is 4(2-sqrt(3); 6) and 4(2-sqrt(3); -2) But I got (4; 2+4sqrt(3)). Can U help me? Also, I don't get why we have two solutions.

[lone heart]

@dense dock Has your question been resolved?

[lone heart]

@dense dock Has your question been resolved?

[lone heart]

@dense dock Has your question been resolved?

[alpine sable]

no idea

your solution seems correct

[livid jolt]

4(2-√3 -2) is (+,-) so 4th quadrant
but the hexagon is only in 1st and 2nd
that answer is not possible

[dense dock]

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[sinful umbra]

i need some help

[sinful umbra]

how do i graph this

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[quaint orbit]

Hi just wanted to check whether my understanding and approach to this question is correct - (a) is asking for the average rate, so does that mean after differentiating, I have to add up all the rates at each min from 0min - 5min, then divide by 5? and for (b), do i just get the rate at the 5th minute?

[quaint orbit]

This is my working for (a) and (b)

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@quaint orbit Has your question been resolved?

[rigid smelt]

for a you kinda got the idea right, but not really a good execution

technically what you are doing is finding the average value of a function

but that's not really how you execute it

and also, for a, it doesnt have anything to do with the instantaneous rate at which the water is flowing out

you can just take the volume of water leaked out after 5 minutes and divide it by the time it took to leak out, that's your average rate

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[rugged dust]

Hi. I've been trying to find a solution for literal hours, and I'm gonna go insane over here.
I'm looking for a bijection g: {1,2,3} -> {1,2,3} such that

f o g does not equal g o f

where (in my example, could be any other) f: {1,2,3} -> {1,2,3} | (x mod 3) + 1

[rugged dust]

I'm trying to prove that the Group (S, o) where S is the set of all Bijections possible on {1,2,3} is not commutative

[prime badge]

there are 6 bijections, how can you spend hours trying every one

[ivory pivot]

try to compute (1,2,3)->(2,3,1) and (1,2,3)->(3,2,1)

[ivory pivot]

true

[rugged dust]

I've been trying to find a function that does (1,2,3)->(3,2,1)

[ivory pivot]

find ?

[rugged dust]

Ehhh communicating about math in English is hard, sorry

[prime badge]

you don't need to find it

[ivory pivot]

this is the definition (1,2,3)->(3,2,1)

[prime badge]

it exists

you're allowed to pretend you found it

[rugged dust]

oh

[ivory pivot]

you have to find a good book

[rugged dust]

you're absolutely right

[ivory pivot]

hahah joke

[rugged dust]

welp that's the biggest brain lag i've had all week, thanks

.close

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[rugged dust]

.reopen

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✅

[rugged dust]

Uhhh in fact I thought I needed to "find" it in order to be able to really show that f(g(x)) != g(f(x)) so I don't know how I would go about that

I can say I have bijection f {1,2,3} -> {3,2,1} and g {1,2,3} -> {2,3,1}, but then I have no way to construct a concatenation, or do I?

[ivory pivot]

you may as well not do all the construction of fg and gf. you could verify that 1 goes to different output so the function are different (e.g. f(g(1))=f(2)=2, g(f(2))=g(2)=3)

[rugged dust]

Ahh, I see

that helps alot

okay, thank you!

.close

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[ivory pivot]

you deserve it for your work hours

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[atomic acorn]

Translate?

[ivory pivot]

we don't know the angel language

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[alpine sable]

sorry, wrong one.

[ivory pivot]

what was your attempt

?

[alpine sable]

i'll tell you in a second.

well the problem is i was only introduced to the formulas s=v*t and v=v-vand v=v+v and i dont understand how to really use them here.

i havent done equations where a specific time isnt given that is necessary to be found.

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[dawn quail]

This last part is bugging me a bit

Plus or minus

The square root of 13 is 3.6

Give or take

[hard patio]

there are 2 solutions

one involving the + and the other involving -

[dawn quail]

But shouldn’t there be parenthesis?

I feel like it’s wording it add 3.6 or subtract 3.6 to get the same answer

[hard patio]

there are two values of y

one that adds, the other that subtracts

[dawn quail]

Oh

So it’s supposed to be like that

[hard patio]

yes

[dawn quail]

I thought the plus and minus was referring to how to get the square root itself

You can use positive or negative numbers but both will become positive

[hard patio]

??

[dawn quail]

(-2)(-2) = 4

(2)(2) = 4

[hard patio]

yes

exactly

[dawn quail]

But for this it is actually saying there are two answers

[hard patio]

there are

[hard patio]

i already said that

[dawn quail]

6 - 3.6

6 + 3.6

I’m assuming these land on the y coordinate on the Cartesian plane?

To give the curve

X or Y axis?

Because this variable is using Y for some reason.

Oh Ok

So it would be sideways

Horizontal infinity

Quadrant I and IV infinity

.close

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