help-39

[weak surge]

professor says: "find the limit as x approaches the right side of -1"

do you go with +(absolute) or -(absolute) version of the limit?

the answers will be different

[ebon mesa]

in this case its not specified what side of -1, so both work ig

[weak surge]

no the limit on both sides dne

one side is +10, the other is -10

but I'm not sure which is which

professor says: "find the limit as x approaches the right side of -1"

[ebon mesa]

x+1 will be negative when x<-1 and positive when x>-1
so left is -10
right is +10

[wind wigeon]

-1- means left limit

[weak surge]

oh right

[wind wigeon]

So in this case when |x+1| = -x-1

[weak surge]

i'm looking for -1+

[wind wigeon]

No you're looking for both

[weak surge]

which means x>-1

[wind wigeon]

If they don't match

Your limit doesn't exist

[weak surge]

no for this

professor says: "find the limit as x approaches the right side of -1"

so if x>-1

that means +10 for the limit

if x<-1
that means -10 for the limit

and the limit does not exist, as it is different on both sides

[ebon mesa]

but the limit doesnt exist as said

yeah

[weak surge]

but it does still hold a limit from both sides?

what do we call that again? "full limit" if both sides match?

or i guess, "the limit exists"

limit from right exists, limit from left exists. but the limit does not exist

lol probably wording this wrong

[last summit]

we can say that

the right side limit exists, or the left side limits exists, or the limit at point a exists. If the right side limit doesn't equal the left side limit (but both individually exist) then we say that the limit at point a does not exist (although the other limits do) this would be a jump discontinuity

[pearl pond]

@weak surge Has your question been resolved?

[pearl pond]

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[stark abyss]

I'm stuck on this question rn, any help would be appreciated!

[stark abyss]

for part b i think that they would make a basis since a3 + a4 is in the span of the other 3 vecotrs

and a3+a4 is also LI to a1+a2

not sure if i'm on the right track though

.close

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[bronze linden]

I need help with some clarification

https://media.discordapp.net/attachments/885604469671354401/1080611104377536615/image.png

[bronze linden]

https://cdn.discordapp.com/attachments/885604469671354401/1080611323026604134/image.png

[limber oasis]

2(a+b) = 2a + 2b
(a+b)^2 != a^2 + b^2
Same thing

Just because a property holds for one operator doesn't mean it holds for another, very different operator

[limber oasis]

Also that doesn't always hold. Notably if y = 0

[bronze linden]

So is the first one right?

except the y = 0 and others

The limit will apply to thhe exponent?

Because I'm getting these problems

but the hting is I only know what the limit of g(x) is

not g(x) itself

so I'm assuming the limit will apply to the exponent?

resulting in for a 0^0 right

[limber oasis]

Sorry currently busy in another channel. Didn't forget though

[bronze linden]

it's alright i thikn i got it clarified however why is infinity^infinity not indeterminante?

and 0^infinity

[limber oasis]

inf ^ inf = inf

[bronze linden]

0^infinity would be more of does not exist?

[limber oasis]

Should just be 0

By epsilon delta on whatever example you'd be working with

[bronze linden]

alright thank you

while you're here

could i ask

https://cdn.discordapp.com/attachments/885604469671354401/1080616157633052774/image.png

if this is wrong?

becaause I'm trying to find the derivative of

https://cdn.discordapp.com/attachments/885604469671354401/1080614912260321420/image.png

e^x^2

and I'm getting 2x*e^(2x)

but symbolab is saying it's e^x^2 * 2x

[limber oasis]

e^x^2 = g(f(x)) not f(g)

Order matters

Symbolab is right

[bronze linden]

wait i don't understand why f(g(x)) doesn't work

the x in f(x) is simply replaced by e^x

and then it's raised to exponend 2

is it not the same

[limber oasis]

f(g(x)) = (e^x)^2 = e^2x != e^x^2

Remember how to read power towers

[bronze linden]

i actually don't know how to read power towers what's the rule?

[limber oasis]

Top right to bottom left

[bronze linden]

my ideaa was bring the most outer exponent in

[limber oasis]

Precisely because the other way round is useless

As then you just untower it

[bronze linden]

i was so confused because in e^x^2 i tried to bring 2 down like for example (6^2)^2 = 6^4

but i guess it's not on the same level

thanks

[limber oasis]

Never take 2 and 2 for things like that. It makes it all the same because 2+2=2*2=2^2

[bronze linden]

alr

[limber oasis]

e^x^2 = exp(x^2) = e^(x^2), not the other way round

[bronze linden]

aok

thank you

.close

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[olive orbit]

How do I get 1/2pifC = 2.850 to look like C = 1/2pif2.85?

[old geyser]

Divide by C's coeffient

[pearl pond]

@olive orbit Has your question been resolved?

[olive orbit]

I'm trying to work that now

1/2pifC / 2pif = 2.850 / 2pif is this going in the right direction?

I keep ending up with C = 2.850 * 2pif/1 instead of C = 1/2.850 * 2pif

[pearl pond]

@olive orbit Has your question been resolved?

[olive orbit]

how do I make it so that its C = 1/2.850 * 2pif and not C = 2.850 * 2pif/1?

[olive orbit]

hi

[olive orbit]

<@&286206848099549185>

[karmic fern]

what do you even have

$\frac{1}{2} \pi f C = 2.85$?

[jolly parrot]

Saccharine

[karmic fern]

I think you probably have something like $\frac{1}{2\pi f C} = 2.85$ because it looks a lot like the reactance of a capacitor lol

[jolly parrot]

Saccharine

[olive orbit]

yes, that is what it is from 🙂

[karmic fern]

okay so do you understand that writing something like this is impossible for anyone to decipher without knowing the context?

[olive orbit]

I understand now.

[karmic fern]

as for rearranging, you can multiply both sides by C/2.85 to get the desired equality

but I wonder how you did it

[olive orbit]

I divided LHS by 1/2pif to isolate C but RHS got messed up

[karmic fern]

again

you need to write more clearly

use parentheses where appropriate

did you divide by 1/(2pi f)

or 1/2 pi f

[olive orbit]

one second, I need to redo it

yes, I divided by 1/(2p f)

[olive orbit]

when I multiply both sides by C/2.85 I get : C = ${2 \pi f 2.85}$ and not C = $\frac{1}{2\pi f 2.85}$

[jolly parrot]

PepsiLeper

[karmic fern]

don't see how that's possible

what do you get when you multiply the left side by C/2.85

[olive orbit]

I get $\frac{C}{2\pi f C 2.85}$

[jolly parrot]

PepsiLeper

[olive orbit]

and then I can cancel out the C's

[pearl pond]

@olive orbit Has your question been resolved?

[olive orbit]

haha, after taking a break I just realized that is pretty much already the desired answer

Thanks for your help 🙂

[pearl pond]

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[iron dragon]

Hello, what is the multivariable calculus behind converting point clouds into solids/ meshes?

[iron dragon]

I am interested in the multivaraible calculus behind laser triangulation 3d scanning, and this is part of it

[lilac jackal]

.close

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[midnight haven]

I need help on this

<@&286206848099549185>

??

<@&286206848099549185>

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[midnight haven]

.close

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[worn parrot]

what does the subscript 0 mean in R0^+?

[abstract locust]

positive reals including 0

/ non-negative reals

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[warm pivot]

I need help

[pearl pond]

Please don't occupy multiple help channels.

[toxic lichen]

.close

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[warm pivot]

.reopen

[pearl pond]

✅

[toxic lichen]

!onechannel

[pearl pond]

Please stick to your channel.

[toxic lichen]

.close

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[warm pivot]

oh ok

[pearl pond]

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@midnight haven Has your question been resolved?

[neon glacier]

how do you get the midpoint of 2 points

and whats the slope between points J and L

yeah]

whats the midpoint

[ashen hill]

the slope is not 0

oh wait

ur on the 1st question

nvm

ok

[neon glacier]

you get the midpoint of both points

ill give you a hint its the average between the 2 points

[ashen hill]

@midnight havenshould I help you on question 4?

or should i wait till u get your answer for #1

ok

ok

so

alright

1. Plot the point (5, -6)

and draw lines that pass through those points

so y=6

and x=-5

now the question is focusing on the line that is perpendicular to the y axis

so we should focus on y=6

now

keep in mind

when you see a line that is perpendicular to the x axis

or just x=something

the slope is always undefined

but

the slope of y=something is always 0

so the answer would be C

do u understand?

ok

good

i think u should in case if you need that as a reference for future exams

i plotted the point (5, -6)

thats it

i only graphed the lines x=5 and y=-6

[ashen hill]

because

if you look at the graph above

those lines pass through the point (5, -6)

m doesnt represent x

it represents the slope

if a constant is equal to x the slope is undefined (Example: x=6, x=7, etc.)
if a constant is equal to y the slope is 0 (Example, y=5, y=8, y=9, etc.)

yes it does

ok

well i approached the problem differently

but just look at the graph

which line is perpendicular to the y axis

is it y=6 or x=-5

its when 2 lines intersect to form a 90 degree angle

let me find a pic of it

thats more tied to geometry tho

people use this definition

and are used to it

ok

let me just draw it

[ashen hill]

since y=6 is perpendicular to the y axis (see graph above), you would focus on the graph of y=6 instead of x=-5

yes

you are welcome

Glad that helped! 👍

[pearl pond]

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[median cove]

is there anything wrong with this proof, is there any mistake

[median cove]

its the answer to part (ii)

[timid spindle]

I think your proof is overcomplicated, you might instead just want to actually construct a bijection between the two sets

I.e. send an equivalence class to the image of one of the guys inside it (this is well defined as by definition they all have the same image)

[median cove]

ahh i see

but what ive done

is it correct

[pearl pond]

@median cove Has your question been resolved?

[median cove]

well

[pearl pond]

@median cove Has your question been resolved?

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[odd saffron]

How do I solve this?

[shell crest]

plug -2 in for x

[odd saffron]

Ohhhh

Thx

Im dumb

[shell crest]

ur not alone buddy

[odd saffron]

Lmao

Can I @ you

I might need help later

I have a big test tmmr

[shell crest]

i DM u

[odd saffron]

Ok

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[midnight haven]

I need help with a math question

[digital dove]

post it

[midnight haven]

Is this clear enough for you?

[tight storm]

What are your thoughts?

[plush bramble]

How old are you?

[tight storm]

That's a slightly mean question

[plush bramble]

It isn't

[tight storm]

This question looks like it could appear on a GCSE Foundation Maths Paper here in the UK. I don't think asking questions like this is especially kind toward people who are clearly doing this in their own time and trying to get better

[plush bramble]

OP has left

.close

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[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[plush bramble]

most people don't know geology here

this is not taught in most math classes
https://en.wikipedia.org/wiki/Mohorovičić_discontinuity

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[plush bramble]

you know people here don't know and that's why you ask? makes sense

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[round mountain]

how can i simplify this further

[vital estuary]

(a-b)^2=(b-a)^2

You can factor the second factor in the numerator

[round mountain]

ohhh

got it thanks!

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[primal spindle]

In simple terms, can someone explain to me why 3 and -3 are excluded

Is it because they are present in the domain?

[buoyant panther]

in the original problem, is x^3 + 6x^2 - 9x - 54 the denominator?

[primal spindle]

Yes

[buoyant panther]

then, try to substitute x = -3 or x = 3

into the rational function given

you'll get division by zero

[primal spindle]

That part I get

It says it’s excluded from the zeros

[coarse dawn]

My assumption from that screenshot is that 3 and -3 are holes

[coarse dawn]

The zeros are the values after you simplified the rational function

[primal spindle]

For the numerator

And I got 6, -3, and 3

But it says that -3 and 3 are excluded, and the only reason I can think of is because they’re already present in the domain part setting it to zero

[coarse dawn]

Yes and because (x + 3) and (x - 3) are both in the numerator and denominator, they get canceled out

[primal spindle]

But that isn’t consistent with -3 and 3 being answers in the denominator

[coarse dawn]

Like x/x, sure one of the zeros is x = 0 but that's with the unsimplified expression. Simplified, you get 1, so x = 0 isn't a zero

[primal spindle]

That’s not consistent

-3 and 3 are answers for the denominator

But they are “excluded” from being in the numerator answer

[coarse dawn]

It's both in the numerator and denominator

[primal spindle]

Yes

And they’re only excluded from being in the numerator

[coarse dawn]

It gets excluded because it gets canceled out

[primal spindle]

But it isn’t

[coarse dawn]

Yes it does, with the terms in the denominator

[primal spindle]

Because by that logic they would be canceled in the denominator

You’re wrong

They are answers in the accepted denominator

It says it right there

[coarse dawn]

They're accepted because that's what the values would be to that make the denominator 0, meaning the denominator can't be those values

[primal spindle]

So this is reinforcing what I proposed here:

[coarse dawn]

Then the zeros of the numerator is after you simplified the rational function

[primal spindle]

^^

[coarse dawn]

More because of this

[primal spindle]

But it isn’t simplifying because then the denominator would be simplified in the same

They both contain -3 and 3

So it can’t be simplifying

[coarse dawn]

To find what the denominator can't be, you use the not simplified denominator

To find the roots, you simplify

And then set the numerator equal to 0

It's because during that simplification, those values will cancel out and they are holes

[primal spindle]

The denominator answer is -6, -3, and 3

The numerator answer is 6, -3, and 3

-3 and 3 are now excluded from the numerator answer only

Does this reinforce my original thought?

[primal spindle]

Which is this

[coarse dawn]

Because they would get canceled out in both the numerator and denominator creating holes

[primal spindle]

But they aren’t canceling because they are still existent in the proper denominator answer

I asked yes or no

Does that reinforce my original thought

[coarse dawn]

No

Because that's not the reason it gets excluded

[coarse dawn]

.

[primal spindle]

The reason I think it gets excluded is because those set the denominator to undefined

[coarse dawn]

It gets excluded because they will get canceled with the terms in the denominator

[primal spindle]

But that makes zero sense

You realize that

Because then they wouldn’t be in the denominator

Canceling and simplifying sends them both to 1

If that’s the case the denominator doesn’t have -3 and 3

[coarse dawn]

Notice here, how you wrote 3 and -3 in both the numerator and the denominator, that means it exists in both, and when simplified, it equals 1

[primal spindle]

ITS STILL IN THE DENOMINATOR ANSWER

[coarse dawn]

Because those are the values that would make the denominator equal to 0

[primal spindle]

so it isn’t “simplifying”

Because those values are still -3 and 3

[coarse dawn]

.

[coarse dawn]

.

[coarse dawn]

.

[primal spindle]

You realize how radically semantical that is when teaching someone right

And also not consistent with the concept of simplifying

Good day

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[quartz crag]

I’m a bit confused on how phase portraits work

I understand that it would blow up in the vector (2,1)^T since te^t dominates

But how do you know the direction the curve will take?

[edgy scaffold]

Please help, can someone please explain where the h went in the bottom left

[regal herald]

#❓how-to-get-help but it went to 0, thats what limits do

open your own channel

[pearl pond]

@quartz crag Has your question been resolved?

[pearl pond]

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[fluid arch]

Need help finding a critical number

[fluid arch]

Work:

(Only found 1 critical number, not sure how it's even possible to find the next)

[feral sedge]

when you go from cos(6x) = -1 to 6x = arccos(-1) that's a mistake

there's more than one solution to cos(6x) = -1

[fluid arch]

How

[feral sedge]

here's the graph of y = cos(x)

notice there's lots of points where y=0?

[fluid arch]

yeah

[feral sedge]

lots of points where y=-1 too

[fluid arch]

What would I have to do with the equation to find that

[feral sedge]

here's the graph of arccos(x)

[fluid arch]

Because I thought you had to get the derivative, set = 0, and isolate x to find critical numbers

[feral sedge]

right

but here you can't really isolate x

[fluid arch]

cos(6x) = -1

Do I just plug 0 and 2Pi

[feral sedge]

what do you mean plug in?

[fluid arch]

for x

[feral sedge]

you can try

you'll see neither of them satisfy the equation

would you know how to find all the solutions to cos(u)=-1

[fluid arch]

No

[feral sedge]

could you find all the solutions between 0 and 2pi?

[fluid arch]

pi/3, pi/4, pi/6

[feral sedge]

how are those solutions?

[fluid arch]

no idea, just got the radians in between 0 and 2pi

and assumed they were

[feral sedge]

,w cos(pi/3), cos(pi/4), cos(pi/6)

[jolly parrot]

Results provided by WolframAlpha

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Requested by monikanicity#0205

[feral sedge]

try sketching the graph, or using the unit circle

[fluid arch]

oh i didnt know you wanted the points

[feral sedge]

just the x values

[feral sedge]

this isn't meant to be a point, i'm plugging in the 3 values you mentioned and showing none of them are solutions

right so

[feral sedge]

you see if the angles are u

then the arrows point to cos(u)

[fluid arch]

what is u

6x?

[feral sedge]

sure, yeah

i'm just asking about the question cos(u) = -1 for now

[fluid arch]

What is even the solution

= -1?

[feral sedge]

?

[fluid arch]

what am i even looking for

[feral sedge]

you're trying to find the values of u that make cos(u) be -1

[fluid arch]

ok

cos(pi)?

[feral sedge]

right

that's the only value between 0 and 2*pi, right

[fluid arch]

Weird because it doesn't strike me that it's in in the first quadrant

[feral sedge]

pi isn't in the first quadrant, if that's what you mean

[fluid arch]

yeah exactly, so it's not in between 0 and 2pi

right

[feral sedge]

?

first quadrant goes from 0 to pi/2

[fluid arch]

oh ok

so 0, 2pi covers the whole unit circle

[feral sedge]

yeah

[fluid arch]

so cos(pi) and cos(-pi) = -1

[feral sedge]

both true, yes

there are other values outside of [0,2pi) as well that will work

[fluid arch]

uhh what are they?

[feral sedge]

use the fact that the cos(x) function repeats itself to find them

[fluid arch]

do you mean degrees

well

nvm

3Pi

5Pi .. 7pi .. so on

all coterminals of Pi which can get -1

though not sure what these values have to do with anything

[feral sedge]

they're all solutions to cos(u) = -1

so

$\cos(\pi+2\pi n)=-1$

where n is any integer

[jolly parrot]

monikanicity

[feral sedge]

now we want solutions to cos(6x)=-1

so what are the possible values for 6x

and then from that, what are the possible values for x, and which ones are between 0 and 2pi

[fluid arch]

i have no clue, none of the values i put into x will get -1

6(1) = cos(6) ~= -1

3pi, 5pi, 7pi ... = -1

[feral sedge]

do you have possible values for 6x?

[fluid arch]

yeah 1/2

[feral sedge]

?

you solved cos(u)=-1 right

so if cos(6x) = -1

what are the possible values for 6x

[fluid arch]

6x
6(1/2*pi) = 3pi
cos(3pi) = -1

[feral sedge]

ah

so x = pi/2, right

that's one possible value for x

[fluid arch]

5pi/2

though 5pi/2 is just the same as pi/2

[feral sedge]

5pi/2 isn't the same as pi/2

well

they're coterminal angles

[fluid arch]

yeah

[feral sedge]

it is true that x=5pi/2 is a solution

but the point is

the possible values of 6x are pi + n*2pi, right

so -3pi, -pi, pi, 3pi, 5pi, etc

[fluid arch]

yeah

[feral sedge]

so for example

if 6x = pi

then what's x

[fluid arch]

pi/6

[feral sedge]

right

since cos(pi) = -1

cos(6x) = -1 when x = pi/6

so x = pi/6 is a solution

oh cos

not sin

[fluid arch]

Well

I got that at the beginning

[feral sedge]

right

the values of 6x that work are -3pi, -pi, pi, 3pi, 5pi, etc

so what are the values of x that work

[fluid arch]

are you sure 3pi work for 6x?

[feral sedge]

yeah the problem is there were other solutions that got erased by taking arctan

[feral sedge]

cos(3pi) = -1

so yes

[fluid arch]

oh i thought it was 6(3pi)

[feral sedge]

?

3pi works for 6x, but not for x

if that's what you mean

[fluid arch]

no i mean like

cos(6*3pi) = cos(18pi) which ~= -1

[feral sedge]

yea so 3pi doesn't work for x

[fluid arch]

yeah doesn't work for x when multiplied by 6

So the solutions are:

Pi/6
Pi/2

[feral sedge]

so far

those are the solutions that correspond to 6x = pi and 6x = 3pi

[fluid arch]

right

[feral sedge]

ye

there's more values for 6x that work though

6x = 5pi

6x = 7pi

6x = 9pi

6x = 11pi

[fluid arch]

i get it

Thanks

[feral sedge]

you're welcome

[pearl pond]

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[barren remnant]

can somewone tell me how this is inccorrect assume all values are positive btw

[near echo]

they probably wanted the x and y under the same root

[undone olive]

^

[barren remnant]

thanks

]

[pearl pond]

@barren remnant Has your question been resolved?

[tough token]

@barren remnant close your last channel before opening a new one

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[pearl pond]

@midnight haven Has your question been resolved?

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[copper kestrel]

"Using the known MacLaurin series, construct the first 3 nonzero terms"

am I doing this right?

I feel like I can only get 2 nonzero terms or 4

should I be subtracting 1 inside the parantheses like that?

it's what makes sense to me, cuz each term is like its own thing

[haughty drift]

where is the -1 coming from?

[copper kestrel]

the function?

idk im very new to this

[haughty drift]

for the first term, sure, but you have to differentiate f(x)

[copper kestrel]

oh

so there isn't a -1 for the rest of the terms

[haughty drift]

No for f'(x) you get -2x*sin(x^2)

[haughty drift]

You calculate the first few derivatives until you get 3 non-zero terms

[copper kestrel]

well i already know that's how you would do it for maclaurin

but this is power series

it wants me to use the known series' terms

[haughty drift]

You wrote maclaurin

[copper kestrel]

like $cosx = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$

[jolly parrot]

Kihei

[copper kestrel]

yea

i mean the directions was to use a known maclaurin series

to create the 3 terms

[haughty drift]

Ok

[copper kestrel]

so like my thought process was

[copper kestrel]

to replace each of the x in these, with x^2 and then subtract 1

[haughty drift]

yea subtract 1 from the whole thing

[copper kestrel]

ohh

[jolly parrot]

b3s4d

[copper kestrel]

so is it

$-\frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^12}{6!}$

[jolly parrot]

Kihei

[copper kestrel]

goofy ah 12

[haughty drift]

that's correct yea

[copper kestrel]

ty

now for the nth term uh

is it

[haughty drift]

You have to write it as a summation

[copper kestrel]

$(\sum _{n=0}^{\infty } \frac{(-1)^n x^{4n}}{(2n)!}) - 1$

[jolly parrot]

Kihei

[copper kestrel]

is that how ud write it?

[haughty drift]

looks correct

maybe it's a bit cleaner if you start at n=1 and delete the -1 at the end since the first and last cancel each other out

[copper kestrel]

hm okay

I see

ty for the help

.close

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[pearl pond]

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[teal siren]

How do I rewrite this as the quotient of a polynomial?

[teal siren]

And I can’t figure out this one either

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[teal siren]

.reopen

[pearl pond]

✅

[midnight haven]

a^2-b^2 = (a+b)(a-b)

Apply this

To the first part

[pearl pond]

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[tame coral]

A six sided die has sides of either 1, 2, 3. The sum of their sides is 13. When rolled twice, the probability of rolling a sum of 5 between them is 1/3. How many sides of 1, 2, and 3 does this die have.

I've got it down to two cases that could be:

3 3 2 2 2 1
3 3 3 2 1 1

I'm pretty sure it's the first, but not sure how to show it w probabilities.

[pearl pond]

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[tame coral]

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[safe dawn]

Hello how do I evaluate this using theorems

[manic sparrow]

huh

[safe dawn]

The plus sign in the constant is bothering me

Do I just ignore it

Or do what you normally do on those thats without those signs

[undone olive]

i think its just cause we cannot take the square root of negative numbers

[safe dawn]

Ohhhhhhhhh

[manic sparrow]

well square root of a fraction is equal to square root of the numerator and denominator so you have √((x-3)/[(x-3)(x+3)]) and then you cancel so you get √(1/(x+3)) and now you can plug in 3 so you get √(1/6)

[safe dawn]

Ohhhh

Okayy thanksss

A lot

.close

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[open mist]

don't do people's work for them... you should know this after being in this server for this long

[undone olive]

please do not give out solutions

[safe dawn]

.reopen

[pearl pond]

✅

[safe dawn]

Sorry I didn't notice that

.close

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[safe dawn]

Hello there

[safe dawn]

Is this correct

This is the question

[light helm]

final result is fine, but bad notation

[safe dawn]

How do I fix the bad notation?

[light helm]

$$=\lim_{x\to -3} \text{(stuff)}$$
instead of
$$\lim_{x\to -3} = \text{(stuff)}$$

[jolly parrot]

ℝamonov

[safe dawn]

Ohhh

Alrrr thankss

[light helm]

and write the limit every time until you plug in the indicated value

[safe dawn]

Okay got it

[light helm]

you've also written
2(-3) -3
twice for some reason

[safe dawn]

Yeah idk why I did that

Anww thankss

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[vapid harness]

yo

[vapid harness]

how to solve this diffrentiation eqn

ik first i have to calculate the integrating fraction

but i cant integrate the integrating fraction

anyone here

?

[pearl pond]

@vapid harness Has your question been resolved?

[pearl pond]

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[pearl pond]

@fierce solar Has your question been resolved?

[pearl pond]

@fierce solar Has your question been resolved?

[pearl pond]

@fierce solar Has your question been resolved?

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[next blade]

it says that this span is a subspace of R^3

is that true

[snow skiff]

check it

[next blade]

how

[snow skiff]

is it closed under + and scalar multiplication

recall definition of span

[next blade]

span is like c1 times first variable

• c2 times second variable

right?

[snow skiff]

yeah its every possible linear combination

[next blade]

and defintion of subspace does it mean it can span up the whole R3 space

or just a part of R3

[midnight haven]

the entire three-space

the linear combination of vectors can span anywhere in the three space

[snow skiff]

[next blade]

so only 2 vectors is enough to span up whole R^3?

[snow skiff]

could it span 3 space for this list? hint: ||can a list of length n span a space of dim m where n<m?||

[next blade]

well it has 3 variables too

it only has 2 vector

so I think it can

but basically if u can write it as a span

it is a subspace right?

[snow skiff]

the definition of a subspace is a subset with closure under vector addition and scalar multiplication

[next blade]

what u mean by closure

well u can add them and scalar multiply

[snow skiff]

has this not been covered in your textbooks

or lectures

or notes

[next blade]

it has

[snow skiff]

then you should go and revise it

[next blade]

but it was confusing

[next blade]

so this can't span up whole R3?

[snow skiff]

think about dimension

[next blade]

ye amount of variables

is the dimension right?

so it has 3 variables

so I think it can span R^3 actually

[next blade]

correct?

[snow skiff]

no

[next blade]

can't you give me a hint

[snow skiff]

I did

see above

[next blade]

so is it in 2 dimensions?

I think it is in 3

[snow skiff]

If you think it is in 3

prove that the span of that list is R^3

[next blade]

I gotta prove linear independent?

[snow skiff]

no I mean

if you think that your list has dimension 3

show that R^3 = span of your list

[next blade]

I have a quesiton about this so the basis for R^n can only have up to n vectors right?

[next blade]

or only 2 vectors?

so to span up R^2 you can only have 2 vectors as basis

[snow skiff]

if you remove the linearly dependent thing it could be a basis

[next blade]

well it is required to be linear independent

to become the basis

but my question was more in general cuz I see like to span up whole R^2 u need 2 vectors and max 2 vectors(cuz of linear independece)

so for R^3 is that the same case?

[pearl pond]

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[tight flint]

what went wrong

[coarse dawn]

Something weird happened with your subtraction

A trick is (a - b) = -(b - a)

[coarse dawn]

So do the bigger minus smaller, then make it negative

[tight flint]

.close

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[hollow marten]

do i add 5200 as total cost after finding the quadratic function?

[hollow marten]

i found r(x)= x(250-2(x-1))

[pearl pond]

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[hollow marten]

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[proper shadow]

can i have help on this question pls

[midnight haven]

Let the center be O

[sturdy imp]

search up alternate segment theorem

[midnight haven]

What is angle DOA?

[sharp agate]

do you know alternate segment theorem?

[midnight haven]

Ignore this. Alt segment theorem makes it much easier.

[proper shadow]

Whats the answer

[sharp agate]

via alternate segment theorem,

bad= 31

now just use properties of isosceles triangle and angle sum

[proper shadow]

i still dont get it

[sharp agate]

do you know about base angles in an isosceles triangle?

[pulsar lark]

Isn't this exterior angle theorem also?

Sum of opposite interior angles = exterior angle?

[sharp agate]

this is not exterior angle this is the angle made by the tangent line

[proper shadow]

can i pls have the answer ill try to see where im wrong

[sharp agate]

ok

[pulsar lark]

No

Against rules

[sharp agate]

youre right

so basically base angles of an isoscles triangle are equal

[pulsar lark]

Grimm pretty much said bad = 31

[proper shadow]

ik

[sharp agate]

so, bda=?

[proper shadow]

31

[sharp agate]

yeah

so now use angle sum

[proper shadow]

whats the angle sum again?

[sharp agate]

180

all angles in a triangle =180 degrees

[proper shadow]

ik

so like what do u do

[sharp agate]

so we know that abd is a triangle and bad=31 and bda=31

[proper shadow]

yh

[sharp agate]

we also know that all angles add up to 180

[proper shadow]

yh

[sharp agate]

there are three angles in a triangle

[proper shadow]

yh

[sharp agate]

31+31+x=180

what is x?

[proper shadow]

118

[sharp agate]

thats it

[proper shadow]

i put that answer

[sharp agate]

did you give reason

[proper shadow]

doesnt ask for one

it just say to give the size of that angle

[sharp agate]

it said it was wrong?

[proper shadow]

yes

[sharp agate]

in the question it asks to give reason

[proper shadow]

k i done it thx

[pearl pond]

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[midnight haven]

im back

[midnight haven]

@coarse dawn

i forgot how to divide bro

equations

[coarse dawn]

Please don't ping me

Also google

You can easily look up videos

[midnight haven]

sorry man

forgive me

😭

.close

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[karmic fern]

ok ive confused everything in my head now so I just want to confirm:
suppose f: X -> Y is surjective. then it has a right inverse g: Y -> X. g is always injective right?
suppose f: X -> Y is injective. then it has a left inverse g: Y -> X. g is surjective right?

seems like injectivity and surjectivity have some kind of duality

[last summit]

so basically this is left as an exercise for the reader

[karmic fern]

the first appears to follow from the fact that f cannot have more than one output per x

and the second appears to follow from the fact that f cannot have less than one output per x

[last summit]

I'm just going to blab at you, so I might be wrong but

it looks to me that the second isn't true

if X -> Y is injective

there can be elements in Y that aren't in X

[karmic fern]

okay

suppose X and Y are disjoint

[last summit]

and for Y -> X to be surjective

[karmic fern]

you have $(g \circ f)(x) = x$, correct?

[jolly parrot]

Saccharine

[last summit]

yeah I take that back already

[karmic fern]

g circ f is X->X, and g circ f is surjective, so g must be

so now we have identified a bijection between the sets of injective and surjective functions right

followup question

the left inverse of the right inverse of f is just f itself right

[last summit]

if X-> Y is injective this means that all elements in X map to an element in Y. For Y -> X to be surjective all elements in X must have an element in Y that maps to them. These don't seem to confirm eachother

[misty patrol]

what is a left or right inverse

[last summit]

my thoughts about statement 2

[karmic fern]

oh yeah this is just the defn of right inverse and left inverse

[karmic fern]

no that's f is defined on all of X

[last summit]

?

[karmic fern]

f is defined on all of X means that all elements in X have an output in Y

[last summit]

and those outputs are distinct

for it to be injective

[karmic fern]

well yeah distinct is the injectivity part

so do u think statement 2 is false or true

[last summit]

from my brief thougths about it, it doesn't seem to be necessarily true

[karmic fern]

ok but what about my argument here

g left inverse of f means that

$(g \circ f)(x) = x$

[jolly parrot]

Saccharine

[karmic fern]

and since g circ f is from X to X

g circ f is surjective

and therefore g must be surjective

[misty patrol]

what preamble/texit config is that

doesn't look normal

[karmic fern]

honestly idk I forgot

idk how to find it

I set it a long time ago

[last summit]

f: X -> Y being injective means that all X map to Y. So X and Y are 1-to-1 g: Y -> X being surjective means that all X have an input Y that produces them. Although the sets have a one-to-one correspondence, that doesn't garuntee g's surjectivity. Because there can be an element in X that isn't produced from an element from Y being inputted to g, in the case where say two elements of Y map to the same element of X. This leaves an outlier element of X with nothing to map onto it. and, two elements in Y can map to the same element in X because g isn't necessarily injective

I don't understand your argument or notation the best, but that's my best articulated opinion on this

[near echo]

f: X -> Y surjective doesn't guarantee a right inverse

X = any set with cardinality >= 2, Y a set with 1 element

[karmic fern]

right inverse only means that f(g(y)) = y

oh wait

uh

[near echo]

did you mean left inverse by chance?

[karmic fern]

so the right inverse is not necessarily unique

no I mean right inverse

[near echo]

oh

[karmic fern]

where g is a right inverse of f iff f(g(y)) = y for all y in Y

[near echo]

i am flipping them

that's my b

wait no

you're saying f o g, g being the right inverse

[karmic fern]

yes

g is the right inverse of f

[near echo]

then yeah i was flipping them

let me read the whole convo

if f is injective, then it does have a left inverse (which is surjective)

that's easy to see if you restrict the codomain of f to the range of f

[near echo]

the thing is that g is not arbitrary

it is specifically the left inverse of f

the question is more about existence

actually

you are correct

but what you are talking about sounds like a right inverse

a left inverse of f can be defined as g : Y -> X s.t. g(y) = x, where f(x) = y, and some arbitrary element otherwise

this is well defined by the one-to-one-ness of f

this is necessarily surjective since f has an input for every element in X

as for a right inverse of f : X -> Y surjective, then we can just take g : Y -> X, g(y) = any (single) x such that f(x) = y

this has an input for every element since f is surj., and is one-to-one

you basically partition the set X into equvalence classes [x]_y = { x' in X | f(x') = y }

[pearl pond]

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[shut linden]

For the first claim, where did they use minimalist if d to deduce r=0

[lavish totem]

we have $g^r = h(g^{d})^{ - q}$ where $h = g^m$

[jolly parrot]

wiitabix

[lavish totem]

and d = gcd(m,n)

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[opal pecan]

Im trying to find the phase shift, would it just be -5pi/3? If not how do i find it

[plush bramble]

factor out a 4 from the argument

[opal pecan]

So i factor it out and rewrite it should look like this?

[plush bramble]

yes

[opal pecan]

So phase shift would be -5pi/12 instead?

[plush bramble]

depends on your class/teacher/book

https://mathbitsnotebook.com/Algebra2/TrigGraphs/TGShift.html

[opal pecan]

Alright, ty

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[opal pecan]

If my graph is (-5pi/12, pi/12) whats the mid point or how do i find it

[real scarab]

the midpoint between two values a and b

is always (a+b)/2

[opal pecan]

so this should be my mid point?

[real scarab]

you can simplify more

but yes

[opal pecan]

thanks

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[peak dove]

im not geting the correct answer

[tall flint]

show your work

[peak dove]

theres a -1

i did that but forgot to put down on paper

[tall flint]

the 56,788 isn't the amount of the payment, it's the accumulated value at the end of the period

you're solving for the monthly payment

[peak dove]

so how would i go about that

[tall flint]

by solving the equation 56788 = P(1+...) for P

1+... being the formula that I don't feel like typing in plaintext

[peak dove]

ok

so thats the formula

[tall flint]

that's the equation you're supposed to solve yes

[peak dove]

could i also use this formula

i got 916.12 btw

[tall flint]

it's asking for the accumulated value, not the present value, so not directly

[peak dove]

ok. so when ever it asks for accumulated put that value for S as in s = R ....

[tall flint]

i don't know specifically what variable names your class uses

it seems here they use R for the recurring payment instead of P, so fine

[tall flint]

future value and accumulated value mean the same thing

[peak dove]

how about for withdraws, where would the value go

like for wich variable would it go in to

[tall flint]

would that not be R?

[peak dove]

ok i see, i was thinking that R always had to be putting in money like into an account or payment

thank youy

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[trim barn]

unsure on how to prove this with the subset

[pearl pond]

@trim barn Has your question been resolved?

[tropic saddle]

take an element of the left side and show it is in the right side

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[silk owl]

n, m = ?

[buoyant panther]

use

[jolly parrot]

Modus

[plush bramble]

maybe simplify the fraction first

[pearl pond]

@silk owl Has your question been resolved?