help-23

[barren vale]

we need to use thesep

but like if you use chain rule you have to integrate tan

[thin bridge]

you have to integrate tan regardless

[barren vale]

if i u sub after splitting tan i dont need to

but i cant do that with inverse chain rule

[thin bridge]

wdym splitting tan

[barren vale]

splitting it into cos and sin

[thin bridge]

that's literally the technique for integrating tan

[barren vale]

oh

[thin bridge]

(without applying integration identities)

and that integral IS an application of chain rule

[barren vale]

what im trying to say is

i can sub in 1/x

to cancel out x^2

then solve for tan u

but i cant do that using inverse chain

[thin bridge]

not sure what you mean

wdym by inverse chain here

what specifically can't you di

[barren vale]

i cant solve it using inverse chain rule

[barren vale]

aka using these

shouldnt it be f’(x)

with u sub i can just sub in 1/x

but idk how to do it without u sub

[thin bridge]

recognise that -1/x^2 is derivative of 1/x

which would've come up when you did it using sub

like all the work is pretty much the same

understand what's happening when you do the u-sub, what gets cancelled, the result etc...

[barren vale]

yeah ik that

but the chain rule doesnt work

because the cos is on the bottom

[thin bridge]

wdym doesn't work

[barren vale]

wait like 10 mins

cant use my computer rn

[thin bridge]

when you come back, can you show the full work/how you'd do it with sub

[hard crest]

i don't know why your teacher doesn't want you to use u-substitution it's like the core of integration

[barren vale]

he wants us to use it only when necessary

idk why

but he has the highest bc average in the country

also, i think i fugred it out

you dont split tangent when skipping u sub

[thin bridge]

my recommendation would be to do a crap ton of problems involving u-sub

until you get to a point where you are comfortable withing doing chain rule tipe questions without explicitly subbing

[barren vale]

yeah

.close

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[final halo]

"Thanks Ramonov!"

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[tawdry salmon]

I'm confused

[tawdry salmon]

I have $$\frac{4}{x+h - 2} - \frac{4}{x-2}$$

[flat frigate]

geoxcaliber

[tawdry salmon]

I get a common denominator right

How do I think about getting the common denominator

[plucky elk]

depends what the question is

[tawdry salmon]

I'm just simplifying

[tawdry salmon]

LOL

[plucky elk]

yes you can find a common denominator

[tawdry salmon]

its x + h - 2?

[glass portal]

multiply 4 by (x-2) and multiply 4 by (x+h-2) 😉

[plucky elk]

there's nothing common to each denominator

[tawdry salmon]

What about the x

[plucky elk]

what about it

[tawdry salmon]

its common

[plucky elk]

a common factor is something you can factor out

not something that just appears in both

[glass portal]

the x and -2 are two terms that comprise ONE factor

ditto: x, h, and -2

Do this: 4/A-4/B

common D?

.... A*B

[tawdry salmon]

the first four gets the x -2

[glass portal]

yup

[tawdry salmon]

isn't that 4x -8 / x+h - 2

how does that help

[glass portal]

well, "what you do to the top, you do also to the bottom" 🙂

[tawdry salmon]

4x -8 / (x+h - 2)(x-2)

this too much work aha

[glass portal]

that's it

now do the same to the second fraction

[tawdry salmon]

But its not gonna cancel the denominator

[glass portal]

I'll bet some reduction occurs in the numerator

well, do the subtraction first

$$\frac{4*(x-2)}{(x+h - 2)(x-2)} - \frac{4*(x+h - 2)}{(x+h - 2)(x-2)}$$

[flat frigate]

AndJ

[glass portal]

$$\frac{4x-8}{(x+h-2)(x-2)} - \frac{4x+4h-8}{(x+h - 2)(x-2)}$$

[flat frigate]

AndJ

[tawdry salmon]

ok now you can subtract

without caring for the denominator

4h + 16 / that

[glass portal]

4h

(the 8s cancel)

[tawdry salmon]

-8 - -8?

[glass portal]

right

actually -4h

[tawdry salmon]

oh so -8 + 8 rofl

[tawdry salmon]

how so

[glass portal]

$$\frac{(4x-8)-(4x+4h-8)}{(x+h-2)(x-2)}$$

[tawdry salmon]

iswtg this shit making me hate math

[flat frigate]

AndJ

[tawdry salmon]

ok

Now I want to remove the denominator

I will just multiply it on top

[untold forge]

I don t understand , where did 1/n come from and where did n disappear plus 1/u = t can I always use it?

[tawdry salmon]

Something didn't go right

[safe radish]

Please read #❓how-to-get-help

[tawdry salmon]

Is this right?

Definitely not cancelling

-4h * everything in the denominator

Oh I was right

I plugged it into mathway

its just a -4 on the top @glass portal

Do you know where we went wrong

[plucky elk]

they didn't expand the denominator

[tawdry salmon]

what do you mean?

[tawdry salmon]

its same as hre

[plucky elk]

you expanded it here

[plucky elk]

yes. do you know where this came from?

[tawdry salmon]

im not talking about expanding

[tawdry salmon]

I mean this

here we get end result -4h

but they get -4

[plucky elk]

show the entire mathway input

[plucky elk]

that's not what you told us

[tawdry salmon]

but

[plucky elk]

.

[tawdry salmon]

what difference does it make??

your still reserving the h for the end

you subtract

and then you divide by h no?

[plucky elk]

they didn't

[tawdry salmon]

it shouldn't be diferent

[plucky elk]

yes it should be

[plucky elk]

this expression has an h in the denominator

[plucky elk]

cancels the h here

[plucky elk]

to get this

[tawdry salmon]

is this answer the same

[plucky elk]

it's important not to mislead people helping you

[tawdry salmon]

as this

[plucky elk]

i answered that already here

[tawdry salmon]

I wonder if this is the same as

$$\frac{1-x}{h}$$

[flat frigate]

geoxcaliber

[plucky elk]

Why do you think it would be

Plug in any two random numbers h and x and see if they give the same value

[safe radish]

@tawdry salmon Has your question been resolved?

[feral relic]

uhhh maybe try turning the denominators into perfect squares and tumble around?

nope forget

either you'd have to burn to simplify or just leave it, nobody bothers what it is

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[devout raven]

Why is this wrong

[devout raven]

Isn’t the vector direction correct just the scalar is off

How would we get the ortho vector, I know how to get the scalar of it

Do I j multiply it by a unit vector

[dapper venture]

I think proj_a(b)+orth_a(b)=b

[devout raven]

Ye that’s right

I c

Would my way work

Of doing the ortho scalar * it’s unit vector

Something like 3.638*<4/sqrt17,1/sqrt17>

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@devout raven Has your question been resolved?

[devout raven]

<@&286206848099549185>

[dapper venture]

how did you get 1,-1/4> anyways

[devout raven]

Ortho slope of a

Inverse recip

So I got the unit vector of that

And multiplied it by the scalar I got for the ortho

[dapper venture]

how did you find the scalar though

[devout raven]

By finding the scalar projection of b onto a

Then I have the magnitude of b

Pythag

[dapper venture]

umm

it should work

why the effort though

[dapper venture]

it's easier this way

[devout raven]

Yeah fair

Idk I guess I j didn’t think of it like that idk that makes sense tho

It would also work to j do bmagnitude sin theta and get theta from adotb=amag*bmagcostheta and solve for theta right

Idk for me it’s j hard to think about why there’s a 90 degree angle made between that component of the triangle

[dapper venture]

?

it's orthogonal

[devout raven]

But how come like

B- bproj =ortho

[dapper venture]

orthogonal is by definition 90 degrees

[devout raven]

Right right I agree

[dapper venture]

orth is the vector you get if you go from a to b orthogonally

[devout raven]

Nvm I’m being dumb

When we go the max distance or find the horizontal vector aka projection

Any connection that makes a triangle has to be right triangle

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[devout shale]

I was hoping someone could check my work for this question

I am willing to clarify anything if needed

I suppose the relevant definition here if needed is that a set is disconnected if it is the union of two nonempty sets neither of which intersect the closure of eachother, otherwise the set is connected.

[feral linden]

Correct

[devout shale]

like all well and good?

[feral linden]

Yeah

[devout shale]

@hardy lion do you concur

[hardy lion]

Uhh

Lemme see

Hold on i got lost

I can confirm the disproof for statement 2 is correct tho

[feral linden]

Or simply this: connected <-> not disjoint union of two non-empty open subsets
S_1 union S_2 disconnect-> S_1 union S_2=C disjoint union D for non-empty open C, D.
Then S_1 = S_1 intersection C or S_1 intersection D. Meaning S_1 is contained in either C or D, same for S_2.
Both contained in C->D is empty, contradiction
S_1 in C, S_2 in D->S_1 intersection S_2 is empty, contradiction

[hardy lion]

I think the proof works

[devout shale]

very reassuring garlic ty

well ty both

.close

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[dull sequoia]

.roepen

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[proud zinc]

I’m just lost on what to put, do I put 5 units down 2 units left, or am I missing something

[thin bridge]

do you know your transformation rules?

it's actually a bit ambiguous as what they want

whether they want you to just do a word description or simplify the result

for the former, what f(x) is equal to isn't that relevant

and you'd focus on identifying whats being done to f(x) in g(x) = 3f(x) -4

in the latter, you'd sub in f(x) and simplify

[safe radish]

@proud zinc Has your question been resolved?

[proud zinc]

ohh so it’s 3 times fx?

[thin bridge]

3f(x) is 3 times f(x), but they'd want you to describe what transformation that's do to f(x)
stretch? dilation? shift?

[proud zinc]

stretch vertical right?

[thin bridge]

yes

[proud zinc]

and it goes down 4 units

so scale factor of 3 vertically and goes 4 units down?

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[scarlet wagon]

How would I determine every x and y that satisfies this equation

[feral linden]

There is no question in the picture

Only a definition

[scarlet wagon]

oh

[feral linden]

Send a full picture

Or you want to prove this unique equals R^2?

Which one you want

[scarlet wagon]

first

[feral linden]

Prove union equals R^2, intersection equals empty by definition

You know the definition of two sets being equal?

[scarlet wagon]

not sure

[feral linden]

Google

[scarlet wagon]

so if they have the same elements

is there more to it than that

[feral linden]

No

So prove

Union of those =R^2

[scarlet wagon]

idk how

x^2 + y^2 = 0^2 if x and y are 0

but how would i get the x and y values for every other r

[feral linden]

prove
Any element of the LHS is an element of the RHS
And
Any element of the RHS is an element of the LHS

[feral linden]

In this

[scarlet wagon]

we haven’t gotten into proofs yet so idk how to do that

[feral linden]

Got to go

[scarlet wagon]

ok

x and y can just be any real number

[safe radish]

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[oblique abyss]

Hey guys, I'm struggling to understand the implications of this proof/maths objects terminology.

I feel like I understand the concept of invariance, but the wiki description states

A wallpaper is invariant under an infinite number of translations, members of a group, of which the operation denoted by
∘\circ is the function composition.

I feel like under an infinite number of translations could mean a lot of very similar but precise things

[oblique abyss]

What, precisely, does "Under an" imply?

[quasi bison]

it means that those actions (here translations) keep it the same

"action X keeps property Y the same" = "Y is invariant under X"

[oblique abyss]

It's really strange wording. I think I'm getting caught up in the grammar, thinking about not just invariance but all the other possible mathematical properties I know are out there

Okay, I think I get it. I feel I may have been getting my statements mixed up, assuming that the above statement was what defined wallpapers, instead of it being a sentence describing a single fact/property of wallpapers

You can do an infinite number of translations to a wallpaper without making it into a different type of wallpaper

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[drowsy venture]

How do I solve part C

part a and b were obvious

[safe radish]

@drowsy venture Has your question been resolved?

[fast jungle]

oh

[drowsy venture]

I think I understand how to do it now

[fast jungle]

What can I say

[drowsy venture]

since its a unit vector I can just multiply the compnents by 1/2?

[fast jungle]

My oh is very magical

[drowsy venture]

what?

[fast jungle]

this was the reason u got the answer

[drowsy venture]

lmao no

[fast jungle]

Oh dang

[plucky python]

what unit vector are you talking about?

[fast jungle]

I thought for a sec I had genius energy

[drowsy venture]

4/5j + 3/5k

thats a unit vector no?

it has a length of 1

[plucky python]

yea

[plucky python]

yes

[drowsy venture]

so I can just multiply it by 1/2

alright thx

[fast jungle]

dont forget to close the channel @drowsy venture

[drowsy venture]

right

.close

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[tawdry salmon]

You can only cancel terms in the numerator & denominator if its present in every term?

[tawdry salmon]

-4x -2h -h

All divided by h

Why can’t I cancel the h’s

[fast jungle]

because its subtraction

[final halo]

yes because for example $\frac{a+c}{c} \neq a + 1$

[flat frigate]

ΣΑCu

[tawdry salmon]

So if it was multiplication

I would be able to divide the h

[tawdry salmon]

So clearly ur doing something to both sides, so that all terms must have c in it to cancel

What r u doing? Dividing each term by c

[final halo]

more so you are doing the same thing to both the top and bottom of the fraction

i.e. dividing the top and bottom by the thing you want to cancel

[tawdry salmon]

Thats what I mean by sides ye

Hm ok

[tawdry salmon]

So why does it matter that its substraction

[flat frigate]

ΣΑCu

ΣΑCu

[ashen ice]

https://www.youtube.com/watch?v=9YwPjwdLrbw
You should watch this i think

This is an example of when you cant "cancel terms" i dont know if that answers your question

[tawdry salmon]

Can you explain where both 1/c are from

[tawdry salmon]

Thanks I’ll check it out

[final halo]

im just rewriting 1

anything/anything = 1

[vital dirge]

the cos(x+y) = cosx + cosy physically hurts me

[final halo]

but letting "anything" = 1/c as thats the thing we want to cancel in the top and bottom of the original fraction

[tawdry salmon]

Your dividing by c

How does that relate to 1/c

[ashen ice]

For "anything" not equal to 0 of course

[final halo]

okay alternative: you're happy that c/c = 1?

so we can write $ac = ac \cdot \frac{c}{c}$

[flat frigate]

ΣΑCu

[ashen ice]

Yes, especially after you learned the correct formula

[final halo]

actually forget that lol

[tawdry salmon]

Is this true $$\frac{1}{x} + x / x = \frac{1}{x}/x = \frac{1}{x} \cdot x/1$$

[final halo]

i dont know how to motivate it other than this is exactly whats happening when you want to cancel things like this

[flat frigate]

geoxcaliber

[devout shale]

no

[tawdry salmon]

Theres two x’s in the numerator

I’m just focusing on the first term 1/x and how jt cancels out

[tawdry salmon]

What

[devout shale]

you just wrote 1/x+1 = 1

and asked if it was true

no

[final halo]

okay alternative: $\frac{ac + bc}{c} = \frac{c(a+b)}{c} = \frac{c}{c}\cdot (a + b) = 1 \cdot (a + b) = a + b$

[tawdry salmon]

How did u make it c(a + b)

When theres no factor of c in ab

[final halo]

typo good spot

[tawdry salmon]

This true @final halo ?

[flat frigate]

ΣΑCu

[final halo]

i have no idea what you've written because your fractions arent formatted properly

[devout shale]

no it isn't true

[final halo]

you can, and should, nest \frac if you want to display nested fractions

just remember all the {}

[tawdry salmon]

So I am familiar with $$\frac{c(a+b)}{x} = c/1 \cdot \frac{a+b}{x}$$

[flat frigate]

geoxcaliber

[final halo]

sure

[tawdry salmon]

But I don’t get how you can take out c/c

Whats the idea

[lean otter]

c/c = 1

[tawdry salmon]

like u can take out anything in numerator no problem

[lean otter]

c/1 = c

[tawdry salmon]

Thats a fraction

[final halo]

its how fractions work: $\frac{xy}{z} = x \cdot \frac{y}{z} = \frac{x}{z} \cdot y$

[tawdry salmon]

Top divided by x

[flat frigate]

ΣΑCu

[lean otter]

i mean what always helps is trying with with plugging in numbers

you get a better feel for it

[tawdry salmon]

Yes u can take out multiplied terms from numerator but how can u take out c/c

[lean otter]

you can always do that

or wdym

[final halo]

here, x = c and z = c and y = a + b, and ive used the last form

[tawdry salmon]

Is this true $$\frac{\frac{1}{x} + y }{x} = \frac{1}{x}/x = \frac{1}{x} \cdot x/1$$

[flat frigate]

geoxcaliber

[tawdry salmon]

Focusing on just dividing the first term by x

[final halo]

most certainly not as your y disappeared

[tawdry salmon]

?

[final halo]

well then you should just have removed the y entirely or not used an equals sign

[tawdry salmon]

Equals sign

I mean just ->

[final halo]

but $\frac{\frac{1}{x}}{x} = \frac{1}{x^2}$ not 1

[flat frigate]

ΣΑCu

[tawdry salmon]

What in the

Isnt it x^1 - x^1

[devout shale]

because the bottom "fraction" is x/1

[tawdry salmon]

x^0 = 1

[devout shale]

keep change flip

becomes 1/x * 1/x

[final halo]

how are you getting subtraction?

[tawdry salmon]

I was just thinking of dividing common bases

[final halo]

if you're talking about subtracting exponents, then it would be -1 - 1 = -2

[tawdry salmon]

I dont see how this is wrong

What math went wrong

[final halo]

what math did you do?

[tawdry salmon]

1/x divided by x

x is x / 1

U flip to reciprocal and make it multiplication

You get 1

[final halo]

you flip x / 1 you get 1 / x

so you're doing 1/x times 1/x

[tawdry salmon]

Shit lol ur right

So you can’t cancel the x there then

Its just 1/x^2

[final halo]

because there is no x on top of the main fraction, there is a 1/x

[tawdry salmon]

I have a question on canceling fractions

So let’s say you have

1/x-5 - 1/x-9 = 0

Ok maybe not

Can’t conceptualize it

Will come back later

[safe radish]

@tawdry salmon Has your question been resolved?

[safe radish]

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[fast jungle]

Bruh

[bold ferry]

.close

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[lean otter]

$x * (x-4) * 3\
x^{2} - 4x *3\
x^{2} - 12x$

[flat frigate]

Chocolate

[lean otter]

bracket in multiplication, why the bracket did not go, after we multiply in the first line

[flat frigate]

Chocolate

[lean otter]

what I expect is, after x * x and x * -4, the bracket will be gone

the remaining expression which is x^2 -4x would be multiplied with 3

[final halo]

the whole of (x-4) needs to be multiplied by 3 at some point, if you first decide to multiply it by x to get (x^2-4x), that whole thing still needs to be multiplied by 3

[buoyant shadow]

there's no way to make it simpler

like you're expecting some rule like, the bracket only goes when this

but it's not the case here

i don;t think there's anything like that

[lean otter]

a * b * c should be equal to b * c* a right ? than why in this case it's different

[final halo]

that is true but i dont see how thats different here

[worthy hemlock]

It's because (x - 4) that -4 is "attached" with the x still

[final halo]

if you want to move things around then it might be easier for you to rearrange it to $3 \cdot x \cdot (x-4) = 3x \cdot (x-4)$

[flat frigate]

ΣΑCu

[lean otter]

left to right => x^{2} -4x * 3
right to left => 3x-12*3x

[final halo]

but this

you are dropping the brackets around the quantity that still needs to be completely multiplied by something

[lean otter]

because I already multiplied the bracket in the previous step

does bracket never dropped?

[final halo]

multiplying a thing in brackets by something does not mean you should drop the brackets

if there is nothing more for it to be multiplied by then you can drop them

[lean otter]

what if I never dtop it, would it be a mistake?

[final halo]

not a mistake, just a pain to keep writing over and over and might make things look more "untidy"

[lean otter]

thank you guys

bee[it/it's] is typing ...

[fast gazelle]

there's never really a situation where you "drop brackets"
it's just that sometimes you don't need brackets to disambiguate, because the interpretation without brackets is the same as what you want

[lean otter]

I always thoughts we drop the brackets after the first time we multiply it, however I just learned that is wrong

thank you Bee

the name Bee is unique, I can remember it from here or somewhere else, I guess you helped me in the past many times

.close

[safe radish]

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[tawdry salmon]

Gadamn it

[tawdry salmon]

What did I do wrong here

Have mercy on my handwriting

[worthy hemlock]

What's the original question?

[tawdry salmon]

The top

1/x+h-9 - 1/x-9

All divided by h

[worthy hemlock]

And I'm asking for the original question

What was the question that started it all?

[tawdry salmon]

Whats that gonna do

Ok lemme get

[worthy hemlock]

Maybe you made an error writing the original down?

[worthy hemlock]

Not understanding what you were trying to do with the first step

[tawdry salmon]

Remove the denominators?

Multiplied by both denominators

What am I supposed to do

[worthy hemlock]

So you multiplied the first fraction by (x - 9)/(x - 9)? And the second by (x + h - 9)/ (x + h - 9)? Basically getting a common denominator, correct?

Because don't forget that it's - (x + h - 9) which is -x - h + 9

Not sure where -x + h - 9 is from

[tawdry salmon]

Huh

Let me check

Ur right

Top just becomes 1

[worthy hemlock]

Does that resolve everything now?

[tawdry salmon]

But still wrong hmmm

Wait

[worthy hemlock]

So you have $\frac{\frac{1}{x + h - 9} + \frac{1}{x - 9}}{h}$, correct?

[flat frigate]

dldh06

[tawdry salmon]

Ok now its right it was a -1 at the top and not 1

I have a question about subtracting fractions

[worthy hemlock]

Sure

[tawdry salmon]

So in this scenario there are no common factors?

So to get a common denominator

I have to multiply both

What does a common factor between denominator look like

[tawdry salmon]

Eg I would think x would be a factor

[worthy hemlock]

For a common denominator, you normally want the LCD, lowest common denominator. But one for sure way to get a common denominator is just multiply the denominators together and that's a common denominator

For example, $\frac{1}{2} + \frac{1}{4}$, the lcd is 4 but a common denominator is just 2 * 4 = 8

[flat frigate]

dldh06

[worthy hemlock]

So a valid common denominator for said example is 8 but it's not the lowest

[tawdry salmon]

Gotcha

When dealing with variables and all that

How do I know when its a factor

I know theres some denominators

Where u dont multiply the whole thing

Only by the piece thats not common

[worthy hemlock]

You mean like $\frac{1}{x(x+2)} + \frac{1}{x + 2}$?

[flat frigate]

dldh06

[tawdry salmon]

I think

What would you multiply the right with?

[worthy hemlock]

x/x

[tawdry salmon]

$$\frac{x}{x} \cdot (x+2)$$?

[flat frigate]

geoxcaliber

[worthy hemlock]

That (x + 2) is still the in denominator

[tawdry salmon]

Ah

Wouldnt that be x / x^2 + 2

[worthy hemlock]

No

It's $\frac{x}{x} \cdot \frac{1}{x + 2} = \frac{x}{x(x+2)} = \frac{x}{x^2 + 2x}$

[flat frigate]

dldh06

[tawdry salmon]

Why do you think of multiply x and x + 2 as x(x+2)

Is that when there is more than two terms always

And the left you also have to multiply by x/1?

How do you figure out what to multiply brute force thinking or there a trick

[worthy hemlock]

Because (x + 2) has a plus sign

x * x + 2 is not the same as x * (x + 2)

[tawdry salmon]

But it doesn’t appear like it has parenthesis

I thought u just combine denominators. So naturally it becomes x * x + 2

[worthy hemlock]

This is the example

That $\frac{1}{x + 2}$ has a hidden part of parentheses

[flat frigate]

dldh06

[worthy hemlock]

So it's basically $\frac{1}{(x + 2)}$

[flat frigate]

dldh06

[tawdry salmon]

Gotcha

[tawdry salmon]

What about these

[worthy hemlock]

No because x/1 is not multiplying by 1

The logic for a common denominator is that you are ideally multiplying by 1

(x)/(x) = 1

[tawdry salmon]

Right right we already have our common denominator anyways

How did you figure out whats a common factor and what to multiply

[worthy hemlock]

.

[tawdry salmon]

Whats the problem with multiplying with 1/x and not x/x?

Bc its not 1 so you have to do it on both sides?

[worthy hemlock]

Because x/x is the same logic as multiplying by 1

You want to multiply by 1 to leave it "unchanged"

[tawdry salmon]

Got it thank you for helping me @worthy hemlock

.close

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[safe radish]

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[haughty nymph]

as you can see my attempt im just confused on what i did wrong exactly

[tidal imp]

Ok

[tidal imp]

Show your work first

[haughty nymph]

I didnt write it down but it was mainly following this

[tidal imp]

Solve it on paper

If you don’t memorize stuff fully, don’t do it mentally

[haughty nymph]

i see what i did wrong

instead of multiplying 3 and -4 i added them

but the correct equation is y=3x+16

.close

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[safe radish]

@quiet dew Has your question been resolved?

[safe radish]

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[mint veldt]

I don't understand how to approach this problem. What do they mean smaller and larger x value here?

[topaz oar]

They are asking at what certain interval is the point 110

[mint veldt]

ohhhh

so like

the whole curve would get smaller if the height was 110m right?

so would the interval where the ends of the curve are right?

huh

If so, I'm not sure how to approach that

Is this a question about derivatives or ?

[mint veldt]

^

[topaz oar]

one sec

[mint veldt]

okie

[topaz oar]

this will be your graph

[mint veldt]

uh hu

following

[topaz oar]

hol up even my brain stopped responding

this is the graph when y=110

so the smaller value will be -41.439

of x

and the larger value 41.439

Im not entirely sure about this but yeah

[mint veldt]

I was more looking to see how to do this without desmos

[topaz oar]

i dont think theres a way to do without desmos

[mint veldt]

really

[topaz oar]

what do you suggest

?

[mint veldt]

Xd

bro

that's why I'm here

I have no clue

My only thought here is using integreals or derivatives

that's this whole section

derivatives, computing the area between two curves, and trig hyperbolic functions

[topaz oar]

this is the graph when you integrate it

this one when u diffrentiate it

[mint veldt]

hmmm

that can't be it then right

sigh

[topaz oar]

well i still havent got to areas b/w graphs

sorry

[mint veldt]

np

I think I'll just have to skip this.

My professor didn't go over it

.close

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[lean otter]

The problem asks to show how this equation is similar to y=log(10) 40x. I know the Log Rules, but don't know how the apply them here. The handwriting was for another problem.

[twilit spindle]

$\log_{b}(a)-\log_{b}(c)=\log_{b}(\tfrac{a}{c})$ :)

[flat frigate]

MrFancy

[lean otter]

Thats the Quotient Rule, can you point a hint on where to start

[karmic hedge]

Use them

[lean otter]

I don't know how

Can you guys give an example problem of some sort?

[karmic hedge]

log(2)-log(3)=log(2/3)

[lean otter]

So would log102+log108 be log10(16)?

I'm just confused. Sorry if it seems aggressive.

[split fulcrum]

Bad notation

[lean otter]

Sorry for it.

I'm guessing I have to () my arguments more regularly

Will do.

[split fulcrum]

You are going to make people think 102 is the term inside the log

when you should be doing log_10(2) + log_10(8) to indicate base

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Yes

[safe radish]

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[wide hawk]

I don’t get it

[quasi bison]

,rccw

[quasi bison]

@wide hawk think you may have cropped a bit too much there

there isn't a question here

[wide hawk]

Oh my bad

Here

@quasi bison mb there

[quasi bison]

,rccw

[quasi bison]

that's blurry as shit

also this looks like it's on a tablet...

why not just take a screeenshot lmao

[frigid locust]

the question is
f : x -> 2x
g : x -> 4x - 2
find (f o g)(x)

[wide hawk]

No

[frigid locust]

whats the question then?

[frigid locust]

this isnt a question tbh

[tidal imp]

4x-3 ig

[wide hawk]

B

Not a

[tidal imp]

You know what the -1 means?

[wide hawk]

Yes

Inverse

But it’s not inside the brackets

It’s outside

[tidal imp]

Ok, we do it step-by-step

[wide hawk]

It’s a different answer

[frigid locust]

just do part a and take the inverse and u get ur answer

[tidal imp]

You have f o g already, right?

What did you get

[wide hawk]

Inverse of a

Is

x+6/8

Now I need to

[tidal imp]

Take note we’re given the output of the inverse and we’re asked to find the input

[wide hawk]

Yes ik

Nah

Didn’t get it

[tidal imp]

If I said f(x) = x-3 and I said f(k)=5, how would you solve it

[wide hawk]

What do I find

[tidal imp]

K

[wide hawk]

K would be 8

[tidal imp]

Yeah, you just add 3, right?

But what’s the inverse of f in this case?

[wide hawk]

Yea

Which equation

[tidal imp]

The one I gave

I promise it’ll be related

[wide hawk]

Oh bruh

I think I see something

[tidal imp]

what did you see

[wide hawk]

Bruh

Got it

It was that simple

It’s just

[tidal imp]

you might have observed if f(k)=5, f^(-1) (5) = k

[wide hawk]

2=x+6/8

Which is 10

Ain’t it

I ignored the fact that it was =

[tidal imp]

Yes, although there’s a much simpler solution

[tidal imp]

From here, if (f o g)^-1 (k)=2, then (f o g)(2)=k

You didn’t even need to find the inverse

[wide hawk]

But what does it mean

When the -1 is outside the box

@tidal imp

[tidal imp]

It means you take the inverse of whatever composite function you got

[wide hawk]

So it’s usually in the second part of questions correct?

[tidal imp]

I mean idk, I’m not your teacher so I can’t promise that

Not every quiz will have progressive questions

There’s a chance I’ll just give the functions and then ask you to find the inverse of the composite function straight away

But you know what to do

[wide hawk]

Okay thanks

[tidal imp]

No problem

[tidal imp]

You have any other questions? If not, you can close the channel

[safe radish]

@wide hawk Has your question been resolved?

[safe radish]

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[quasi bison]

,rccw

[quasi bison]

let me reiterate what i said earlier

substituting R = -6 and S = -2 into your expression,

you would have had $\frac{(-6)^3}{-2}$

[flat frigate]

Ann

[tight marsh]

alright

[quasi bison]

but by coincidence, the exponent 3 is an odd number, therefore (-6)^3 = -(6^3) and your mistake was not fatal.

but then you blabbered on about the odd-numbered exercises.

and you called out for me, but didn't ask a question that i could answer.

[tight marsh]

ye sorry

[quasi bison]

do you have a question to ask

[tight marsh]

so -6 * -6 * -6

correct?

[quasi bison]

so the numerator is (-6) * (-6) * (-6), yes.

[tight marsh]

-216

so thats now -216/-2?

[quasi bison]

yes

as it always has been

[tight marsh]

oh

wait so thats the final answer

[quasi bison]

no

there is still work to do

[tight marsh]

wait but we could still keep it as a fraction

or divide it

[quasi bison]

this fraction can be simplified

[tight marsh]

alright then

108

final answer?

[quasi bison]

yes

[tight marsh]

ty

.close

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[safe radish]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[terse jackal]

1 a , b, c

<@&286206848099549185>

[dapper venture]

!show

[safe radish]

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

[dapper venture]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[terse jackal]

?

[lean otter]

First of all, if you've attempted these questions but have failed, please show us your work so that we can see where you are stuck
Second, please don't ping helpers until 15 minutes have passed without help

[terse jackal]

Okay

Anyone

?

[final halo]

.

[safe radish]

@terse jackal Has your question been resolved?

[terse jackal]

No

[terse jackal]

Okay

Im not sure

<@&286206848099549185>

[mellow carbon]

yes hi hello

[terse jackal]

Hello

[mellow carbon]

what help do you want

[terse jackal]

.

Idk if i did it correct

And if so can you explain it to me in depth

[mellow carbon]

sure

one sec

number 3, right?

can you post the question again so that we don't have to scroll up?

[safe radish]

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[mellow carbon]

are you still here?

[terse jackal]

Yep

[mellow carbon]

bot shouldn't have closed it lmao

anywho

which question number are we looking at specifically?

[terse jackal]

1

[mellow carbon]

so

x^2+4x=x^2+4x+4-4=(x+2)^2-4

so that's correct

you need help with b and c also?

[terse jackal]

Yep

[mellow carbon]

for b you have 2x^2-8x

take a 2 common, we get 2(x^2-4x)

can you break x^2-4x similarly as in (a)?

salmoon?

[terse jackal]

Yes

Im trying to figure it out but i cant

[mellow carbon]

okay np

x^2-4x=x^2-4x+4-4

can you try now?

[terse jackal]

Ok

Give me some time please

[mellow carbon]

sure, tyt

it's very similar to (a) if you look closely

[terse jackal]

Is this correct

[mellow carbon]

2(x^2-4x)

you missed a x with the 4

[terse jackal]

O

[mellow carbon]

try it again

[mellow carbon]

how did you do this?

do you see an error?

[terse jackal]

Idk

Im just stressed

[mellow carbon]

okay let's do this

[terse jackal]

Ok

[mellow carbon]

you have x^2-4x=(x^2-4x+4)-4

do you agree?

[terse jackal]

Ye

[mellow carbon]

then that is just (x-2)^2-4

thus our equation is now 2[(x-2)^2-4]=2(x-2)^2-8

that's your required answer

[terse jackal]

Wait

B is (x-2)^2

?

[mellow carbon]

wdym?

[terse jackal]

.

What is B

?

[mellow carbon]

the question you mean?

or small b in the equation?

[terse jackal]

Yes

[mellow carbon]

so you need to write it as a(x+b)^2+c, right?

[terse jackal]

Yes

[mellow carbon]

we got 2(x-2)^2-8

thus b=-2

a=2

and c=-8

[terse jackal]

Oh okay

[mellow carbon]

do you understand?

how we did it?

[terse jackal]

Yes

I did

[mellow carbon]

now for (c)

[terse jackal]

Ill be afk 10 minutes

[mellow carbon]

the channel will timeout lmao

if this channel dies, then open a channel and ping me

[terse jackal]

Back

[mellow carbon]

hi

[terse jackal]

Hello

[mellow carbon]

where were we

[terse jackal]

At c

[mellow carbon]

oh look 3 ws in a row

W

[mellow carbon]

right right

(c) is x^2+8x+7

right?

[terse jackal]

Yes

(X+4)^2

[mellow carbon]

we first see that x^2+8x+7=x^2+8x+16+7-16

[mellow carbon]

yes almos

[terse jackal]

I wanted to continue but its alright

😂

[mellow carbon]

oh aaaaa

sorry

;-;

[terse jackal]

Its okay 😊

[mellow carbon]

you can finish up if you want

[terse jackal]

Ill try

[mellow carbon]

go on

i am looking

[terse jackal]

(X+4)^2-(4)^2+7

[mellow carbon]

+7

[terse jackal]

(X+4)^2-16+7

[mellow carbon]

yeah

[terse jackal]

(x+4)^2-9

[mellow carbon]

yeah

there you go

[terse jackal]

Wooooo....

Thank you so much

My charge is low now

I need to go

How do i close the channel

?

[mellow carbon]

.close

you do

[terse jackal]

.close

[safe radish]

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[feral linden]

You are saying you want a rigorous proof

Instead of an explanation by graph

?

Sure give me few mins

Exactly what I am going to do in next few mins

[quasi bison]

In proof there is said since g'(x)!=0 for all x€(a,b), from Rolle theorem we can conclude g(a)!=g(b)
it's just the contrapositive of rolle's

if g(a) = g(b) then g' vanishes somewhere

if g' doesn't vanish anywhere then g(a) ≠ g(b)

"if P then Q" is equivalent to "if not Q then not P"

[feral linden]

You let

h(x)=f(x)-(f(b)-f(a))/(g(b)-g(a))g(x)

h(a)=h(b) you can use Rolle now

h’(c)=0 for some c in the middle and you can obtain the result

[safe radish]

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[terse crow]

Factorization

Ughhhh

[timid pasture]

Bring everything to one side

[terse crow]

Okk

Then

@quasi bison

@timid pasture

[timid pasture]

What?

[white umbra]

stop pinging random people

[terse crow]

Next step

[timid pasture]

What did you get?

[terse crow]

2y² - 5y

= zero

[timid pasture]

Alright, then just take y common out

[terse crow]

Mhmm

Y (2y -five )= 0

[frigid locust]

Just cancel y on both sides and it becomes a simple equation

[timid pasture]

[terse crow]

😮‍💨😮‍💨

[timid pasture]

No, don't

[terse crow]

Ok

[frigid locust]

Why not?

[terse crow]

Oh well ok

[timid pasture]

You're basically done after taking common

Since you only had to factorise

[white umbra]

y could be zero, and you can't divide by zero

[terse crow]

Oh point

[timid pasture]

Or are you supposed to solve?

[frigid locust]

Chances are low but good point

[terse crow]

answer is 0,5/2

[timid pasture]

Chances???

[white umbra]

yeah this isn't a matter of chance lol, one of the answers is 0

[timid pasture]

Oh cool so since product is zero one of them have to be 0

[calm bridge]

you are just removing a solution by cancelling y

[terse crow]

Thank you guys

[frigid locust]

Oh i didnt solve it my bad

[terse crow]

I am closing channel

By2

[timid pasture]

[white umbra]

.close

[safe radish]

Channel closed

Closed by @white umbra

Use .reopen if this was a mistake.

[terse crow]

.close

[safe radish]

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[quasi bison]

what does the B stand for again

can you write out the rule in its entirety

well you can define g(a) = 0 can't you

that'll make it continuous on (a-ε, a+ε), an unpunctured nbhd

and then if g were to vanish anywhere else rolle would kick in aand you would get a zero for g'

be equal to zero

[safe radish]

Channel closed

Closed by @lilac totem

Use .reopen if this was a mistake.

[safe radish]

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.

[round willow]

can anyone teach me how to solve these?

[wary dune]

what is $x^{\frac{1}{3}} * x^{\frac{1}{3}}$

[flat frigate]

dabbingpotato

[lament delta]

idk if it's clear to you

you have to reach the same power of the square root so that you can put the x and the y under the same square root

[lament delta]

Btw, this is the solution of the first picture that you send

[safe radish]

@round willow Has your question been resolved?

[lament delta]

This is the solution for the second question

you don't have to write power 2 of the square root because it raises the square root, which also makes it simplify

[safe radish]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

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

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

[somber widget]

hello, i need help!

[hasty falcon]

do you know trigonometry?

[somber widget]

not confident

[frigid spruce]

lol, what does that mean

[somber widget]

not confident in trigonometry?

[hasty falcon]

okay you have to just know the basic definition to solve this

for approach try to find what angle DAB is and then somehow find way to use definitions of either sin, cos or tan to solve this

[somber widget]

is angle DAB also 14 degrees?

[hasty falcon]

it is

[somber widget]

oh that sums up a lot

sorry i didnt know if DAB was 14 degrees

i got confused

[hasty falcon]

why is it 14?

I don't want you to have any confusion in this regards..try to convince yourself