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[cedar kiln]

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[cedar kiln]

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[olive heart]

how do u solve 10^3t=5

[glad kestrel]

logs

[olive heart]

obviously

but how

[glad kestrel]

use.. logs

if you don't know how to use logs, look up a video on how to use logs

[ancient valley]

Take the log of both sides

[olive heart]

i did]

[ancient valley]

Show me your work so far

[olive heart]

3tlog10=log5

[ancient valley]

Yep, what is log10?

[olive heart]

1

?

[ancient valley]

I'm showing him how to do so

[olive heart]

i learned it that way

whats next'

[ancient valley]

You can do it either way

What is our new equation?

[olive heart]

3t=log5

[ancient valley]

Yep

If you want to isolate t, what do you do?

[olive heart]

divide by 3

[ancient valley]

Yep

[olive heart]

5/3?

[ancient valley]

Not 5/3

[olive heart]

log5/log3

[ancient valley]

We're dividing the whole equation by 3

Only 3

[olive heart]

thats not on my answer thing

[ancient valley]

What is?

[olive heart]

there are

5/3

log5/log3

3log5

1/3log5

[ancient valley]

With the method we're doing, we get one of those

[olive heart]

which one

[ancient valley]

I ain't gonna tell you the specific one lol

We gotta figure that our ourselves

[olive heart]

oh

[ancient valley]

What will be our new equation of We divide both sides by 3?

[olive heart]

log5/log3

i meant 3

bruh

[ancient valley]

Yep lol

We get t=(log5)/3

[olive heart]

sorryye

[ancient valley]

Now, dividing by a number is the same thing as multiplying by the reciprocal of said number

[olive heart]

yea

[ancient valley]

So, if the number were dividing by is 3, what do we do?

[olive heart]

multiply by 1/3

[ancient valley]

Yep

So, what's the new equation

[olive heart]

wait

what do we multiply 1.3 by

[ancient valley]

No, we Simply change it to 1/3(log5)

[olive heart]

oh

yea

that makes sense

[ancient valley]

Do you see know which answer choice it is?

[olive heart]

yea

got it

[ancient valley]

Ok sweet

[olive heart]

thanks

[ancient valley]

Good job 🙂

[olive heart]

makes more sense

.close

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[kind parcel]

how would you find the center from the asymtopes so I can write the equation in hyperbola standard form?

[kind parcel]

is it even possible?

I've found the a and b

Just can't get center so I can create the equation

a = 9, b = 4

[muted bear]

The center is the intersection of the line

lines

[kind parcel]

oh I see

thank you my friend

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[crisp hill]

for this question im stuck on c and d because im using counting methods to solve it and it makes sense in my head that way

[crisp hill]

but when i look at the answer key

these are the answers

so im just confused on how it becomes 1 million

[drifting fable]

I assume there are 1 million people in nunavut?

or a lot pf ople

people

idk

[crisp hill]

oh

thanks

<@&286206848099549185>

i need help

i have a test tomorrow

please

<@&286206848099549185>

okay

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[mental sorrel]

The hint is confusing me farther? I thought I just have to plug them in?

[royal light]

can anyone help me with this

[slow thicket]

always try 0 first

[royal light]

(0,0) works ig but i dont think thats the point they want me to find

[slow thicket]

are you sure

you can also factor out y^3

[royal light]

this is the full question

[slow thicket]

0, 0 works

simple doesnt mean wrong

[royal light]

ig so

[cedar kiln]

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[eternal crescent]

Can you solve this using substitution

[royal loom]

no

solve using power reduction formula on sin^2

and go from there

[eternal crescent]

I have not learnt that yet

I got the first integral by finding the volume by discs

But I think I have to do it by shells

[cedar kiln]

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[crimson sedge]

$f(x)=\sqrt{1-x}$

[wraith dagger]

okokok

[crimson sedge]

find domain

so 1-x greater than equal to 0

-1

x is greater than equal to -1 ?

[gloomy nacelle]

move the -x over to get both positive terms

[crimson sedge]

how

[gloomy nacelle]

1-x greater than or equal to 0, so 1 greater than or equal x

Therefore x less than or equal to 1

[crimson sedge]

so add x

[gloomy nacelle]

Yeah

[crimson sedge]

https://tenor.com/view/kms-im-gonna-gus-gusjohnson-gif-18194767

[nova snow]

You can also do this if you know what the graph of sqrt(1-x) looks like

[crimson sedge]

so ( infinity, 1]

[gloomy nacelle]

Yeah

[crimson sedge]

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[outer barn]

Need some help on this

[obsidian coral]

Recall that x intercepts also means roots/zeros

[outer barn]

Ok I got it

It’s 2

Part 3

What is the maximum amount of turning points?

It’s 1 right?

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[kindred ivy]

how do i use FCP on this

[wind pewter]

i cant rread that

[vagrant elbow]

,rotate

[wind pewter]

so she has 4 choices times 6 choices times 5 choices ?

[cedar kiln]

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[flat harbor]

not even sure where to start with this problem

[flat harbor]

havent been taught how 2 theta works yet

not even sure where to start

[slow sigil]

It means that whatever the value of theta, 2 theta is that value multiplied by 2

[slow sigil]

See what triangles is theta in

[tropic oxide]

surely the concept of multiplying something by two isn't foreign to you

[cedar kiln]

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[broken knoll]

Yo

[broken knoll]

Have to go with integration

Anyone please

[long arrow]

that's unclear a bit

[wraith dagger]

Modus

[long arrow]

is it x = 8 and x = pi in the last line?

[broken knoll]

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[summer kayak]

I was supposed to give a Taylor polynomial for $f(x) = \sqrt{1+x}$ of degree 1 at around x=0 to approximate the number $\sqrt{\frac32}$

[wraith dagger]

afeAlway

[summer kayak]

I got it to be p(x) = 1 + x/2 which is correct but then the problem comes when trying to approximate that number? How do I do it?

[crimson delta]

well what do you need to plug into x so that 1+x=3/2

[summer kayak]

why? Isn't it supposed it to be $\sqrt{\frac32} = 1 + \frac{x}2$

[wraith dagger]

afeAlway

[crimson delta]

no

[summer kayak]

y

[crimson delta]

$\sqrt{1+x} \approx 1+\frac x 2$

[wraith dagger]

Denascite

[crimson delta]

by taylor

[summer kayak]

so?

[crimson delta]

ok maybe you meant that

so what happens if you plug x=1/2 in

[summer kayak]

where?

[crimson delta]

for x

[summer kayak]

in taylor or f

[crimson delta]

on both sides

[summer kayak]

$\sqrt{\frac32} = \frac54$

[crimson delta]

no

[wraith dagger]

afeAlway

[crimson delta]

yes

but \approx instead of =

so sqrt(3/2) is roughly 1.25

,w sqrt(3/2)

[wraith dagger]

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[crimson delta]

and we see that this is at least somewhat ok

[summer kayak]

but how do you know x=1/2?

[crimson delta]

cause for x=1/2 we get that f(1/2)=sqrt(3/2)

so we are trying to approximate f(1/2)

by g(1/2) where g is the taylor polynomial of f

[summer kayak]

I get it lmao (I have always gotten approximate x=? so I just totally forgot that they might give me y to find x)

Thanks for the help

[cedar kiln]

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[broken knoll]

Yo

[broken knoll]

Can anyone solve this

[tropic oxide]

we don't give out answers here or do your work for you

[broken knoll]

Help me

To get it

Please

[tropic oxide]

are you familiar with the cross product in general?

[broken knoll]

Yes

Even I. The determinant form

[tropic oxide]

don't need that here

do you know that the cross product distributes over addition (and subtraction)

[broken knoll]

No
Not quite

[tropic oxide]

$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

[broken knoll]

Ok
So cross product can do this things

[wraith dagger]

Ann

[tropic oxide]

typo

but yes

you need to also know that the cross product is anti-commutative, and that the cross product of a vector with itself is 0

[broken knoll]

Yes I’m that

Cause sin0=0

Now I can solve this

Thanks @tropic oxide

[cedar kiln]

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[thin matrix]

i’m trying to find x but i got to this point where i need to find 7^3, however this is supposed to be done with no calculator so is there another way to do this?

[slow sigil]

You can do 7^3 without a calculator...

[ebon valley]

just multiply 7•7•7

[thin matrix]

i mean I can do that but we haven’t needed to do any long multiplication all year so it seems wrong

[ebon valley]

it seems wrong, but it isn’t

[thin matrix]

alright thanks

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[pallid abyss]

hello!

[pallid abyss]

i really neeeed help

mean and standard deviation

[cedar kiln]

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[rancid steppe]

Looking for help with question 3

[rancid steppe]

My guess is that it's not linear as it's not got dimension 2**k for any natural number k

And then to make it linear do i just add different codewords to each other mod 2?

Thats the method i used but I'm quite unsure about it I'd really appreciate some help!

[cedar kiln]

@rancid steppe Has your question been resolved?

[rancid steppe]

<@&286206848099549185>

[cedar kiln]

@rancid steppe Has your question been resolved?

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[olive heart]

do i use impulse momentum theorem for this

[latent bloom]

Yeah

[upbeat lotus]

Yoy can use

[olive heart]

alr

[upbeat lotus]

The fundamental formula of force

[olive heart]

which is

[latent bloom]

$\Delta p = F\Delta t$

[wraith dagger]

VulcanOne

[upbeat lotus]

$F=\frac{∆p}{t}$

[wraith dagger]

Offline due to exams

[olive heart]

oh

ill use this

[upbeat lotus]

Where ∆p is change in momentum

[upbeat lotus]

Yeah

[olive heart]

yea

wait

what was delta p again

[upbeat lotus]

.

[olive heart]

oh

[upbeat lotus]

$mv-mv_{0}$

[wraith dagger]

Offline due to exams

[olive heart]

alr

[upbeat lotus]

Yupp

[olive heart]

wait

can u do this problem with me

i wana make sure im using the formula right

so we have mass, initial velocity, and time

[upbeat lotus]

Yes

[olive heart]

i got 722

p=mv

then divide by time

right?

[upbeat lotus]

,w (-9.5×19)/0.25

[wraith dagger]

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[olive heart]

not negative

its positive

[upbeat lotus]

It's negative cause final velocity is 0

[olive heart]

but the answer is positive

on my end

and it says its correct

[upbeat lotus]

Oh

Yeah yeah

[olive heart]

bro

theres no way ur 15

[upbeat lotus]

+722

[olive heart]

learning physics

[upbeat lotus]

Why?

[olive heart]

what grade

are u

[upbeat lotus]

9th

[olive heart]

oh

did u skip a grade

dont u take physics only in 10th grade

nvm

.close

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[upbeat lotus]

I live in india

[olive heart]

oh

that makes sense

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[dire bluff]

How do I find the number of solutions for a math question?

[dire bluff]

Nvm I got@it@

[livid hound]

depends

[vague rapids]

A number of solutions generally means you are talking about diophantine equations

There is no direct way to solve them but what you can do is use therorems, inequalities and mod

You may try to find the general value of all variables in a equation by finding their link of a common variable

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[junior timber]

we tried L N L N L N L

but it’s not right

[slow sigil]

Works?

[vague rapids]

Mayne

Top bottom

And so

Maybe like this

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[jade badger]

Hi so I learn these in french but I used Google translate for yall to understand what I'm asking lol. We just started learning this new chapter and I'm kinda struggling with what I need to do I started it obviously but yeah I don't know what is the next step to take. In the pic I skipped the parts I didn't even know how to begin

[cedar kiln]

@jade badger Has your question been resolved?

[jade badger]

<@&286206848099549185> so sorry to ping but it’s been 20min almost and I haven’t had an answer ^^

[terse dirge]

No worries

[jade badger]

Can you help me ?

[hollow comet]

@jade badger m'envoyez les questions svp

Je peux les lire en français

[jade badger]

Num 11

[hollow comet]

Ok. Permettez H = (x,y)
On connaît que 2HA + 5HB = (0,0)
Ça donne
2(x+7, y-2) + 5(x-3, y+1) = (0,0)
->
(2x + 14 + 5x - 15, 2y - 4 + 5y + 5) = (0,0)

Alors trouvez x et y et ils donnent H

[jade badger]

Comme ça?

[hollow comet]

Je pense. Vraiment je suis logé entre mon travail et plus d'autre choses je n'aide car personne ne vous aide en ce moment mdr

Mais je pense que vous aviez raison, alors désolée mais qqn d'autre doit vérifier

Encore vraiment désolée

[jade badger]

Okay ca marche au pire je redemande un peu plus tard

Merci quand même ^^

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[drifting solar]

I got stuck so I looked up a study page for some help, I started getting on the right track but the study page got this

when I got 5 instead of 1/5

I thought it was just finding the determinant, but is it different for the jacobian??

I honestly missed my lecture on this bc I was sick

nvm I figured it out, I missed that -1 at the top

.close

[cedar kiln]

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[unborn pike]

do uk how to find the no. of ways of choosing 2 men out of 5?

[ionic hornet]

well sir

im ass in pnc

but i have a logic

to propose

they said a commitee of 5 people is to be made

so sample space could be

m: male

f: femal

then

m m m m m

m m m m f

m m m f f

m m f f f

m f f f f

f f f f f

total outcomes = 6

[ionic hornet]

favorable out come = 1

[unborn pike]

well let's say the males r a, b, c, d, e

in that one favourable outcome

u can have a,b

or a,c

or a,d

or a,e

or b,c...

there's way more than one outcome

which is why I would recommend improving ur pnc before attempting questions on it

[ionic hornet]

i get what you are trying to say

like a dude named john

[ionic hornet]

could replace a dude named jhoseph in this list

?? is that what you mean ?

[unborn pike]

yes

[ionic hornet]

but they asked

about men

the person woldnt matter

as long as they are a combination of men and woman = 5

[vague rapids]

It’s (the number of ways of choosing 2 of the 5 men and 3 of the 6 women) divided by (the number of ways of choosing 5 of the 11 people)

[ionic hornet]

ye but no of way of choosing shouldn't matter right cause everyone is a male

in the male list

and not specified by their names

[lyric narwhal]

but they are unique

that's the whole premise of the question

[ionic hornet]

wait how ?

[ionic hornet]

im rtarded

can you pls explain where it is mentioned

[lyric narwhal]

it's understood

[unborn pike]

well u can't have 5 men who r the same

like someone might manufacture similar living objects but

[ionic hornet]

does the phrase "from a group of 5 men:

[unborn pike]

it means 5 unique men

[ionic hornet]

suggest uniquely identifing

[ionic hornet]

ahhy yess

understood

thankj you

how do i close

this

[lyric narwhal]

.close

[ionic hornet]

before ending this

i would like to thank you guys

for your help

really appreciate it

[unborn pike]

no probs, keep in mind u're gonna need combinations for this question

[ionic hornet]

yes sir

thankyou

.close

[cedar kiln]

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[marsh pond]

not quite the full answer for x?

domain restriction

[lusty grotto]

x^2 >= 0

holds for any x

not just for x>=0

[marsh pond]

but how would you show that? i took square root, and got +- 0, which is just 0

[lyric narwhal]

if you simplify the expression you just get root(x^2)

[lusty grotto]

if u take the square root u must introduce an abs value

$\sqrt{a^2}=|a|$

[marsh pond]

oh right

[wraith dagger]

𝕾ilver𝕾oldier

[marsh pond]

i forgot about that

so this would be the domain restriction

[lyric narwhal]

seems correct

[marsh pond]

OK

[lyric narwhal]

if the top part is correct

[marsh pond]

you mean this part?

dunno why it copied full screenshot, glitch with iPad maybe. thought I cropped it

[lyric narwhal]

yes

[lyric narwhal]

if that is correct then the bounds are correct

[marsh pond]

right, now I will check

for all function domains
domain of f(x)
domain of g(x)
and domain of f(g(x))

i think that's what I always have to do for composite functions

[marsh pond]

i keep forgetting...
for f(g(x))
do I need to check for domain of all 3 functions?
or only check for domain of g(x) and f(g(x))?

[royal loom]

for function composition

in order for the entire thing to work

you have to atleast get to the outermost function yes?

so first g(x) has to work

or else you can't input it into f(x)

and then once g(x) works

you have to make sure whatever works for g(x)

fits into f(x)

[marsh pond]

oh, so I'm checking each step of the way

not just beginning and end

[royal loom]

first g(x) possibly limits the domain, and then f(x) possibly limits it again afterwards

imagine the number line

g(x) chops off some interval

and then f(x) chops off another

if there is overlap

that is the domain that is left

[marsh pond]

step 1: check f(x) domain
step 2: check g(x) domain
step 3: check f(g(x)) domain
step 4: combine all 3 domain restrictions from steps 1-3 to get domain of f(g(x))?

[royal loom]

I'd say

Step 1: Check g(x) domain

Step 2: Check f(x) domain

Step 3: Find the overlap (if any)

[marsh pond]

Oh so go from inside first, to outside

Step 1: check g(x) domain
Step 2: check f(x) domain
Step 3: check f(g(x)) domain
Step 4: combine all 3 domain restrictions from steps 1-3 to get domain of f(g(x))

[royal loom]

So I am confused about your steps

step 3 you see

"check f(g(x)) domain"

and then step 4 you have

"get domain f(g(x))"

[marsh pond]

that's the composite function

[royal loom]

those are the same thing though

[marsh pond]

oh

right, so I'm just doing it in 3 steps

Step 1: check g(x) domain
Step 2: check f(x) domain
Step 3: check f(g(x)) domain
Step 4: combine all 3 domain restrictions

[royal loom]

It is always going to be the overlap of the involved functions domains

yes

I mean like think of this

if $$ f(x)=\frac{1}{1-x} $$

[marsh pond]

I guess 4th step is to combine all 3 domain restrictions

[wraith dagger]

AustinU

[royal loom]

then where is the domain restriction

,, g(x)=sin(x)

[marsh pond]

x cannot equal 1

[wraith dagger]

AustinU

[royal loom]

and what happens at x=1?

the result is undefined

[marsh pond]

dne

[royal loom]

okay

so if we had

[marsh pond]

horizontal asymptote

[royal loom]

,, g(f(x)) = \sin{(\frac{1}{1-x})}

[wraith dagger]

AustinU

[royal loom]

then plugging in x = 1

gives sin(undefined)

[marsh pond]

right

[royal loom]

so that can't work

so wherever the domain of the inner function fails

thats out

[marsh pond]

domain restriction is like baggage, it never goes away

[royal loom]

and then you have to check the outer

and in this case sine is good everywhere

[marsh pond]

we can only add to it

alright so it has to work for all 3 functions, whatever we plug in for x

the first two functions on their own

and the composite function combined

or is it just f(x) and f(g(x))?

[royal loom]

if you have 3 functions

or 4

or however many nested functions

I'll make a nasty example like this: p(h(f(g(i(x)))))

lots of functions yea

the domain is

whatever the overlap between the domain of

[marsh pond]

i as in imaginary 😉 but OK

[royal loom]

i(x) , g(x) , f(x), h(x) , and p(x)

wherever all the nested functions domains overlap

that is the domain of the entire thing

because you can take it a step at a time

start at the inside

i(x) has to have a domain

and then that domain has to be in g(x)'s domain

and that has to be in f(x)'s

and so on

and you end up with

[marsh pond]

right but even if the domain works for the huge composite function, i would still need to account for domain restriction of these functions on their own?
i(x) , g(x) , f(x), h(x) , and p(x)

[royal loom]

yes

[marsh pond]

and domain restriction of these as well? or is it redundant, already accounted for with the main composite p(h(f(g(i(x)))))?
h(f(g(i(x))) f(g(i(x)) g(i(x))

[royal loom]

not that you will ever deal with finding the domain of 5 nested functions though...

so you always separate into cases no matter how many times we go over this

the main idea is

x has to be in the domain of the innermost function

or else you would be plugging undefined into the next one

• and then the same thing happens again *

no need to separate into cases and memorize rules or a process

the domain is simply the intersection of the domains of all the individual nested functions.

[marsh pond]

OK

ty

both of these mean the same thing?

[royal loom]

yes

[marsh pond]

when you multiply both sides of an inequality by -1, -2, -3, etc. (any negative number) you flip signs?

[dull oxide]

you could just say R, (-infty, +infty), or all real numbers nvm saw the sign

[royal loom]

no

https://tenor.com/view/wedding-crashers-will-ferrell-what-an-idiot-idiot-stupid-gif-5175269

[marsh pond]

lol

[royal loom]

jk love you SWR

[marsh pond]

no to this?
when you multiply both sides of an inequality by -1, -2, -3, etc. (any negative number) you flip signs?

[royal loom]

no I was saying no to SWR's reply

that is true

[marsh pond]

OK

[dull oxide]

https://tenor.com/view/beating-up-beating-up-lilo-fight-beat-up-beat-down-gif-19482217

[marsh pond]

so the domain is only equal to 0?

[royal loom]

x<= 0

is the domain

https://www.desmos.com/calculator/tmvoalcazi

take a look at this graph avid

listed are all the composite functions

starting from the bottom

x-1 has no restriction

so move on

[marsh pond]

the domain for this

[dull oxide]

domain of what

[marsh pond]

D = [0]

[royal loom]

of whatever this thing was

no

just expand that out avid

[royal loom]

you get sqrt(3x^2)

which is all x

the error is at the step after absolute value

|x| >= 0

when is that true?

for all x

if x is negative absolute value returns positive (greater than 0)

if x is positive absolute value returns positive (greater than 0)

if x is 0 absolute value returns 0 (equal to 0)

[marsh pond]

oh I should have written it -x <= 0?

this part is wrong?

[royal loom]

no you shouldn't have wrote any of the things you circled

[dusk finch]

This is all good

[marsh pond]

😵‍💫

[dusk finch]

just the conclusion D=[0] is wrong

[crystal raptor]

It should be OR rather than AND

[royal loom]

your last proper step says |x| >= 0 Avid

hopefully you should be able to see that the domain of that is not 0

[marsh pond]

but radicand can be 0, and anything larger than 0

i always write domain restriction of radicand as x >= 0

[dusk finch]

He meant it's x>=0 OR x<=0

[dusk finch]

That's correct

[marsh pond]

where did I write "and"?

or how did I write "and"

[royal loom]

the and was implied when you concluded the domain was 0

[dusk finch]

You didnt but you concluded that D=[0]

[marsh pond]

oh

[dusk finch]

This is graph of abs(x). And your domain is all x such that |x|>=0

[royal loom]

if x>=0 and x<=0 the domain is 0

if x>=0 or x<=0 the domain is all x

[dusk finch]

It's easy to see that this property is true for all x

just look at the graph

[marsh pond]

I see where my flaw in thinking is.. I wanna keep x equal to 0

[royal loom]

avid can I say something?

[marsh pond]

yes

[royal loom]

The domain of this should be fairly obvious. Expand the multiplication and you get sqrt(3x^2) knowing sqrt(x) has domain x>0 and x^2 is always > 0 the domain is clearly all x

[dusk finch]

All this was correct

[cerulean sail]

That plus minus on the bottom line

[royal loom]

+- 0 = 0

[dusk finch]

well yes, but it doesnt really influence the result

[halcyon marlin]

gotta follow the rules 💀

[dull oxide]

[marsh pond]

so it's wrong to assume x^2 has to equal 0 here?

this could be an order of operations issue for me
BEDMAS = Brackets First

but really we are distributing -3 into the brackets

[dusk finch]

When did you assume that? What do you mean?

[marsh pond]

for a real solution the radicand must be equal to or greater than 0

so 9 - 9 = 0

-3*3 is -9

therefor we don't want to subtract anything else from 0, otherwise we will not have a real solution

[dusk finch]

What about 9-8

would it be real solution?

[marsh pond]

but it's really 9 - 3(3) ++PLUS++ -3(-x^2)

[dusk finch]

it would be sqrt(1) which is real

[royal loom]

-3(-x^2) = 3x^2

[marsh pond]

yes

[royal loom]

plus 3x^2 (is strictly positive)

[halcyon marlin]

and 3x^2 always positive

[marsh pond]

-a(b)
can be rewritten as + (-1)(a)(b)

[dusk finch]

+(-3(-x^2))=+(3x^2) which is positive

always

[marsh pond]

yeah that's why I look at this part and think to myself "I don't wanna be subtracting a positive number from 0"

but that's not what's going on, is it?

I mean, the subtraction part

I'm actually adding because two negatives make a positive

[halcyon marlin]

its only 0 after you distribute. i feel like you're overcomplicationg things

[marsh pond]

yeah, I just need to do the algebra. follow order to operations

and stop "guessing" what x "feels like" it should be

the algebra tells the truth

when I think of domain I think of this "big picture" to it all, which i guess can be useful for composites. maybe it's even more useful for range, to think rationally about it all.

but really, i'm starting to realize, the domain is where i really need to just follow the algebra

algebra will lead me to the correct answer of what the domain should be 100% of the time

[halcyon marlin]

algebra just aids intuition

you're still at the end of the day just saying what feels right

[marsh pond]

but it always gives the correct answer

my intuition was wrong

I was looking at what the inside of the brackets should be, without realizing the algebra would distribute -3

[halcyon marlin]

because seeing |x| >= 0 for all x is way more intuitive than w.e you were trying to do

and thats all algebra does

[dusk finch]

my advice is start using graphing calculator. I use it everytime I am doing something with algebra. Graph all the functions that you get in contact with, focus on how the domain behaves, what's the end behaviour, what are the asymptopes etc... you will get much more comfy with algebra and you will be able to intuitively guess a lots of limits, range domain etc...

[marsh pond]

I understand what this means now. absolute is piecewise. piecewise for absolute is OR, not AND

if it was AND we would have problems with piecewise definitions overlapping each other

do you bother to include 0 in your piecewise definition of an absolute value?

instructor today said it's not needed, I didn't quite grasp what was meant

[dusk finch]

abs(x)=
x for x>=0
-x for x<0

[marsh pond]

i guess because 0 is gonna be neither positive nor negative, it's just 0

[dusk finch]

It could also be -x for x<=0

[marsh pond]

now when I see an absolute, I look at | abcde |
as solving the same thing twice

solution 1: (abcde)
solution 2: -(abcde)

I just put a negative sign in front of the brackets for the second solution and distribute if needed

that seems to be a good way that works for solving absolutes

[dusk finch]

it is

but it doesnt quite work for inequalities

you have to be more careful there

[dusk finch]

If it was $\left|x\right|\le0$ It would be AND

[wraith dagger]

MathIsAlwaysRight

[marsh pond]

I guess I could flip that -x >= 0 to instead say x <= 0

[marsh pond]

wait thats what I wrote lol

but in my situation the AND is referring to the radicand

greater than or equal to 0

oh I think I see what you're saying

[dusk finch]

I mean to apply this trick in inequalities, you will have to distinguish between x>=0 and x<=0. In the first case it's OR and in the second case it's AND

[marsh pond]

hmmmm...

[marsh pond]

so it's not quite this simple

[dusk finch]

It's perfect for equalities

in equations its simple

[marsh pond]

you will have to distinguish between x>=0 and x<=0. In the first case it's OR and in the second case it's AND
I'm not sure what you mean here, aren't they both AND?

=
<=

[halcyon marlin]

just write the definition of |x| instead of doing w.e that is

[marsh pond]

you mean instead of this?
solution 1: (abcde) solution 2: -(abcde)

[halcyon marlin]

ye write the piecewise function

[marsh pond]

arrows point to the domain of each composite function

would these be correct?

phew! I was worried I needed to point to f(x) domain somehow, for the first composite function, f(g(x))

[dusk finch]

Wait no one is incorrect

[marsh pond]

Oh so I do?

[dusk finch]

You will have to restrict the domain here before modifying

[marsh pond]

so both domains are the same?

(-Inf, 3] for both composite functions?

[dusk finch]

Yep

[marsh pond]

or wait

I'm confused

this is referring to g(f(x))

I already restricted this domain

lol

[dusk finch]

a

[marsh pond]

so it's fine

[dusk finch]

all the red underlined functions have domain (-inf, 3]

[marsh pond]

yes

[dusk finch]

Well I was kinda confused with this

[marsh pond]

ah

yeah I have to combine the domains

[dusk finch]

D=(-inf, inf) and there is also arrow from D=(-inf, 3]

[marsh pond]

the ones I circled in light blue are the final answers

[dusk finch]

all right

[marsh pond]

just to show where the domain composition is coming from

so there is absolutely no need to include f(x) domain restriction (-Inf, 3] into the composite domain restriction for f(g(x))

[dusk finch]

I guess that you made one domain out of original expression and another domain from simplified expression

[marsh pond]

so there is absolutely no need to include f(x) domain restriction (-Inf, 3] into the composite domain restriction for f(g(x))
in fact, this would make the answer incorrect if I did so

I only need to check g(x) domain, and f(g(x)) domain

not f(x) domain

[dusk finch]

yep

[marsh pond]

that's what I needed to know! thanks!

[dusk finch]

What you need to do is to check that range of g(x) is in domain of f(x)

[marsh pond]

now it's a lot more clear to me, as per the original question lol

way up there somewhere

[marsh pond]

hmmmm...

those are connected somehow? like an inverse function?

where domain = range, range = domain

or am I misreading this

[dusk finch]

What do you exactly mean by this?

To find domain of f(g(x)) you need to check domain of f(g(x))?

[dusk finch]

Just remember this. I dont exactly know what you mean by that, but your process of finding the domain is correct

[marsh pond]

there would not need to be a third blue arrow from f(g(x)) domain to point to the f(x) domain, which I highlighted in black

the domain restriction for f(g(x)) only needs to include the domain for g(x)

and the new domain that f(g(x)) creates

but no need to include f(x) domain with this composite function f(g(x))

@dusk finch that makes sense with the way I worded it?

so instead of looking at f(x) domain sqrt(9-3x) we are looking at f(g(x)) domain sqrt(9-3(g(x))?

I'm never sure if I still need to include all original domains, from all original functions, or just the inner function(s) domain(s), and the newly created composite function domain

I will double check with professor today

.close

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[sinful lark]

ok i need a little help

[sinful lark]

on my assignment

This is the first question

that i’m stuck on

[muted bear]

what "test" does a graph need to have to be a function

[sinful lark]

Vertical line test

[muted bear]

exactly

so that one is correct

[sinful lark]

the first one?

[muted bear]

yes

[sinful lark]

Ok

[muted bear]

its looking for the one that is not a function

[sinful lark]

But we need to find one that isn’t a function

[muted bear]

since that one does not pass the vertical line test, it is not a function

[sinful lark]

So the first one doesn’t pass the line test

[unborn pike]

a function is one-one and onto

[muted bear]

thats a bijective functino

[unborn pike]

since they havent mentioned the domain, v can ignore the onto

a function is a bijective relation

[muted bear]

no

[sinful lark]

[muted bear]

sinx is a function

excuse the big words, let me clear this up

[sinful lark]

<@&286206848099549185>

[hidden sky]

The first one is x = y^2 so I guess it doesn’t describe as y a function of x

[sinful lark]

Confused

i just learned functions yesterday i know no equations on them yet

[hidden sky]

Alriggot forget what I said

Basically you know functions give an output for every x you input right?

[sinful lark]

yes

[jaunty mural]

(not true, I'd advise you check up on this)

[sinful lark]

Well anyways

[hidden sky]

These functions give a single output

[sinful lark]

no x can have 2 y basically

WELL

that was worded dumbly

[sinful lark]

X can’t have 2 y’s

[hidden sky]

Yea

[jaunty mural]

think of a function like a factory.

It takes an input and spits out an output

For each input there is only 1 output

Different inputs can have the same output, though

[hidden sky]

Now you can see in the case of option 1 it seems that putting in x gave two different y outputs

So it cannot be a function of x

[sinful lark]

Yes

Okay

But for the second one

How do i know that is a function

it’s irrelevant to the actual question of the assignment but its the only one that is confusing me

[hidden sky]

It gives a single y output for every x no?

[jaunty mural]

https://i.imgur.com/S3CDb4v.png

[hidden sky]

Of course taking into account this

[jaunty mural]

This isn't necessarily the best possible drawn diagram

[jaunty mural]

but it hopefully conveys this point

[sinful lark]

Yes dis one

[unborn pike]

Im sorry abt the mistake, will brush up on my concepts. thanks for pointing it out.

[hidden sky]

You can ignore the fact that all these functions seem to be chopped up, they only care about it locally

[sinful lark]

i confused by this

[jaunty mural]

lets draw a bigger diagram

so an example of an x is 2

[sinful lark]

i see

[kindred hull]

https://ars.els-cdn.com/content/image/3-s2.0-B9780128235508000184-f14-10-9780128235508.jpg

[sinful lark]

i think

[jaunty mural]

we see that x = 2

gives us y = 2

and nothing else

there are no other y's

when x = 2

So there is a single output for x = 2

[hidden sky]

That’s the catenary which isn’t the same as a parabola I think

[kindred hull]

its not the same curve, i just did it for the columns

[sinful lark]

so it’s a function

[hidden sky]

Yes hence its a function of x

[jaunty mural]

Well you need to check all x's

to be sure it is

(by looking at the graph)

[hidden sky]

Note the ‘of x’ , it depends on x

[jaunty mural]

you will see each x has only a single y

[sinful lark]

Yes

[jaunty mural]

that's why it's a function.

[sinful lark]

Okay

[jaunty mural]

So a good way to see some examples of functions

is go to desmos

[jaunty mural]

like i did

[sinful lark]

okay

[jaunty mural]

and go f(x) = something with x's in it

and these will all be functions

or y = something with x's in it

[sinful lark]

Okay

[hidden sky]

A function of x 👍

[jaunty mural]

you only start not getting functions if you type in y's on the right

[sinful lark]

Hmm okay

[jaunty mural]

so even this is still a function. only x's

What I'm saying here is just to get some examples on desmos

[sinful lark]

Ok

i have a few more questions i need help with

if mastery connect will load

[jaunty mural]

open a new channel for each and hopefully ppl can help

[sinful lark]

okay

I will be back later

Thanks for your help

.close

[cedar kiln]

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[sinful lark]

i need help doing 3 problems

[sinful lark]

the first one with the s and the diamond is the one i really am focused on

the other two just need to be double checked.

[gloomy nacelle]

Well a function of x can only have one y-value per x-value

[sinful lark]

yes

Lol the name in one of them oopsi

[gloomy nacelle]

So does that hold for the diamond?

[sinful lark]

i don’t think so

[molten compass]

You can use something like the vertical line test

To make sure

[gloomy nacelle]

^ this

[obsidian coral]

If that says exit ticket, isn't that based on your own knowledge? Like are you suppose to get outside help?

[sinful lark]

i don’t understand how to do that without a graph

[sinful lark]

I don’t really think it matters

[slow thicket]

I mean yeah

but ideally same thing as hw

[gloomy nacelle]

Oh

[sinful lark]

I don’t understand how to do it so i asked for help

[slow thicket]

juse use the veritcal line test as someone said

you can do it for a graph or for just points

[hidden sky]

just put a pencil on your computer screen and see if the functions passes through it twice lol

[sinful lark]

Ok

i will try

okay

it does pass through twice

[gloomy nacelle]

Yeah

Which tells you it’s not a function

[hidden sky]

you could, use your imagination as well to visualise it

[sinful lark]

Yes

[sinful lark]

yeah i know i saw visually as well towards the start

[sinful lark]

so it can’t be that one

Could it possibly be the one at the bottom that looks like a book

[hidden sky]

bingo

[sinful lark]

OK

So we solved that one

for the other two i think i’m correct but not sure

[gloomy nacelle]

Double check third question

[sinful lark]

is the problem with this one

[slow thicket]

use the same test

get a vertical line

[molten compass]

So does it pass or not

[sinful lark]

it does

?

[molten compass]

Nice

What does that mean

[sinful lark]

so it is a function

[molten compass]

Good

Try this with the others

[hidden sky]

🎉

[sinful lark]

for the double one it isn’t a function because it passes through twice

and for the one that is pointing to the right isn’t one because it also passes through twice

[gloomy nacelle]

Yes

[sinful lark]

So the one that’s going up is the function

[gloomy nacelle]

Good

[sinful lark]

Okay

Okay

and for the final one

this one

[obsidian coral]

For it to be a valid function, every y should have one x value

[sinful lark]

Well

that’s true

[obsidian coral]

Meaning in other words, the x values shouldn't repeat

[sinful lark]

And the double one has the same x value

[obsidian coral]

So then the appropriate answer is?

[sinful lark]

C

[obsidian coral]

That's it

[sinful lark]

Okay

let me input my answers really quick

YAYAYAYAYA

i got all of them right

thank you for your help everybody

https://tenor.com/view/thank-you-sticker-thanks-sticker-line-sticker-cat-sticker-orange-cat-gif-26476683

.close

[cedar kiln]

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[main jetty]

whats the difference between what we do in part a) and part b)? the answers look the same

this is my work

a lil messy

but would this match with part a) or part b)?

[crimson sedge]

method (a)

e_1 = [1,0]

e_2 = [0,1]

[cedar kiln]

@main jetty Has your question been resolved?

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