precalculus

[willow bear]

@languid dust how did you arrive at your answer for (b)? show your work.

[languid dust]

at first i put 0

bc 2(0)

then i was like maybe it's 2

so i just yoloed it

xd

[willow bear]

okay let's put it this way

do you know what $g \circ f$ means?

[obsidian monolith]

Ann:

[willow bear]

@languid dust

...

[shrewd urchin]

Welp me

Leave i figured that out

[willow bear]

nope
[in reference to a deleted message]

[mint widget]

Ah okay

Man you’re always here helping out - I appreciate it 👌

[willow bear]

welp, it looks like i got ghosted by @languid dust !

[languid dust]

oh sorry i got it lmao

[willow bear]

you shouldn't have left me hanging like you did.

[shrewd urchin]

Lol why u people delete messagw

[wary plover]

find x intercepts for y = -5 sin(4x + π/3) on interval [-π/6,π/2]

assuming you get to 0 = sin(4x + π/3), what do you do next?

[willow bear]

well

do you know when sin(θ) = 0?

[shrewd urchin]

AT =0

[primal karma]

and....?

[shrewd urchin]

And

0+npi

n intger

$\sin(n\pi)$ =0 where n is an integer

[primal karma]

well now we just need to know what happens if you change the period

[obsidian monolith]

Krishna:

[valid gazelle]

180

@shrewd urchin

[willow bear]

it can't be 180° @valid gazelle that'd make the sum parallel to them both

[shrewd urchin]

Yes ann is coorect

Draw a diagram

and see yourself

Ok lets say that i have a system of linear equatio in 2 varaiable and they dont have any solution

so can i writee that

the solution to the system is not defined $\mathbb{R \times R}$

[pale kettle]

\times

[obsidian monolith]

Krishna:

[shrewd urchin]

is it ok

????

[pale kettle]

Not really

Well first of all, the grammar isn't great

But in general, the solution to the system not being defined over R x R isn't really what's happening

You saying that implies that there's some other set S so that the solutions are defined over that set

But that's not true

[shrewd urchin]

ok

but R times R describes the set of all points on cartesian plane

yes you are Right

i found my mistakee

[willow bear]

you will often see these as $\bR^2$ and $\bR^3$

[obsidian monolith]

Ann:

[shrewd urchin]

Oh thank you

[shrewd urchin]

$A \times B={(1,3),(2,3),(1,4),(2,4)}$

[willow bear]

\{ ... \}

[obsidian monolith]

Krishna:

[willow bear]

yes good

[shrewd urchin]

Ok so n(A \times B)=4 \implies P(A \times B)=2^4$ but i can think of only 11 subset

[obsidian monolith]

Krishna:
Compile Error! Click the reaction for details. (You may edit your message)

[willow bear]

$n(P(A \times B)) = 2^4$ yes

[obsidian monolith]

Ann:

[shrewd urchin]

yes sorry

[proud sparrow]

Can you list them out?

[shrewd urchin]

yes

[proud sparrow]

Systematically?

[shrewd urchin]

${\phi,}$

[proud sparrow]

I have a feeling that you are missing subsets with 3 or 4 elements

[obsidian monolith]

Krishna:

[shrewd urchin]

No

There are total two set having 3 elemts and are subset of A X B

woops i got what i forgot

[willow bear]

\phi used for empty set

please no

it has its own symbol in latex

[stuck lark]

err. brackets pls

[willow bear]

\{ ... \}

[stuck lark]

ty

[obsidian monolith]

RokettoJanpu:

[shrewd urchin]

what do they mean by distinct elements

[proud sparrow]

Well, it means x is not equal to y, y is not equal to z, z is not equal to x.

$x\neq y, x\neq z, y\neq z$

[obsidian monolith]

Element118:

[shrewd urchin]

A={x,y,z}

B={1,2}

Right ?

[willow bear]

yes

[willow bear]

blurry

[shrewd urchin]

Ok

So

In iv)

I first found a unit vector along the direction $\vec{v}$

[obsidian monolith]

Krishna:

[shrewd urchin]

Ok i am done with this

v)

How do i show that they are parallel

?

[proud sparrow]

Well, there are many ways

[shrewd urchin]

$\vec{a}=1/2\vec{b}$ is this enough

[obsidian monolith]

Krishna:

[proud sparrow]

$\vec{a}=\frac{1}{2}\vec{b}$ would be enough.

[obsidian monolith]

Element118:

[shrewd urchin]

Yes

I mean this is enough

Are there any other way to show the same

[proud sparrow]

Well, cross products

[shrewd urchin]

Ye

[proud sparrow]

finding the unit vector of each?

Dot product and showing that the magnitude is the product of the 2 magnitudes?

Showing that projections of any other vector on these 2 vectors are the same length?

(dot product again)

[shrewd urchin]

Ye ok

[proud sparrow]

Showing linear dependence?

[shrewd urchin]

This os for phusics ok

[proud sparrow]

In short: many ways

[shrewd urchin]

Will learn vector next year

Ok lol

[willow bear]

idk why they wrote $\frac{10}{\sqrt{5}}$ instead of $2\sqrt{5}$

[shrewd urchin]

Unit vectpr method was on my mind top

[obsidian monolith]

Ann:

[shrewd urchin]

Ye

They did not rationalize the denominatpr

[proud sparrow]

Both are the same.

[shrewd urchin]

I know

[proud sparrow]

ah context here was previous question

[shrewd urchin]

Ye ok

[proud sparrow]

probably to save space?

or maybe the textbook writer was bored

[shrewd urchin]

The book is written by some non passionate proffesors in a month

1 chapter is writtten by 1 professor

Only chemistru textbook is goood

And bio alsp

Also*

I have to show three vevtprs to be mutually perpendicular

Do they mean i have to show that the thtee vectprs are orthoganally

So a | b and b__|__ c implies a__|b|__c?

Thst is prependicylat rign

Perpendicular

[proud sparrow]

No it doesn't

a perpendicular to b, b perpendicular to c, but it can be a not perpendicular to c.

Example would be a=c=i, b=j.

[proud sparrow]

You would need to do it for every pair of vectors

[shrewd urchin]

a p c ,a p b , b p c

p =perpendicular

[proud sparrow]

shouldn't be a problem to show all 3, right?

[shrewd urchin]

So i can say that a p b p c

[willow bear]

perpendicularity isn't transitive

[shrewd urchin]

Transitive?

[proud sparrow]

Like equality is transitive

a=b, b=c means a=c

[shrewd urchin]

a=b b=c then a=c this is transitove kaw of addition

Yes ok

[proud sparrow]

Parallelness is transitive. a//b, b//c means a//c

[shrewd urchin]

ok

[proud sparrow]

$a \perp b$ and $b\perp c \n \nRightarrow a\perp c$

[obsidian monolith]

Element118:
Compile Error! Click the reaction for details. (You may edit your message)

[shrewd urchin]

Mutually perpemdicular means ?

[icy current]

means its 90 degrees to a line

[willow bear]

this is transitove kaw of addition
no

it's the transitive property of EQUALITY

[shrewd urchin]

Lol

Yes sorry

[uncut mulch]

what have you tried?

[shrewd urchin]

Done

[fringe stream]

why post it here then?

[willow bear]

sounds like he got done with it in the 6 minutes between the two messages there

[short sorrel]

but if you can get the answer in <6 minutes, why would you post here instead of just... doing it yourself

¯_(ツ)_/¯

[fringe stream]

[shrewd urchin]

I try it for 10 mins then post it here

[fathom field]

@shrewd urchin

[stuck lark]

@fathom field 👍🏽

[shrewd urchin]

Lol done with it

Thanks though!

[wary plover]

this one is on the tip of my tongue but I cant quite recall

[proud sparrow]

split it into things you can find the value of

[wary plover]

I know that

im probably forgetting my exponent rules

[wary plover]

I have to isolate a 2^x I think? to make it a 3

[limber bone]

solve for x

[wary plover]

but I dont remember how to do that

[limber bone]

using logarithms

[wary plover]

fairly certain this problem does not involve logs

[limber bone]

ok then

you have to

from knowing that 2^x = 3

get to 2^(-3x+3)

we can do this problem by reverse

[wary plover]

good idea

[limber bone]

see whats 2^(-3x+3) like in terms of 2^x

or like how are they smiliar

so by using exponent laws

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

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

[wary plover]

I always forget that exponent rule

[limber bone]

ye you will get used to it

ok so now

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

[wary plover]

considering I havent taken precal in years I doubt it

[limber bone]

dont worry

so anyways

if 2^x = 3

whats (2^x)^-3 ?

[wary plover]

im still trying to read the formatting

[limber bone]

what do u not understand

in the formatting

2^x is 2 to the power x

[wary plover]

I cant tell whether you're stacking exponents or using multiple instances of 2

too many ^

[limber bone]

i am stacking

this is an exponent property or law

a^(bc) = (a^b)^c

so i did that

with the 2^(-3x)

[wary plover]

can you use a rich text and post screenshots

or the bot

[limber bone]

idk how to do latex sorry

and too lazy to write tbh xd]

[wary plover]

?

i am so lost

[stuck lark]

You’re on the right track

You can also write that as

$(2^{x})^{-3} \cdot 2^3$

[obsidian monolith]

RokettoJanpu:

[valid flint]

is the ^(-3) suppose to be on x or (2^x)

[wary plover]

I have no clue

[valid flint]

it looks like its on x, but anyways lets get some context

what does the problem want you to do

[wary plover]

solve for x

[valid flint]

ok wait its on (2^x)

just scrolled up

so rules of exponent, you familiar?

[wary plover]

fairly

[valid flint]

$2^{-3x+3}$, can you break it up to isolate a $2^x$?

[obsidian monolith]

John Doe Smith:

[wary plover]

oh so we're just using product and power rules but in reverse to split things up

I never thought of it that way

but I still dont understand how to get a fraction answer from there

[valid flint]

well just write interms of 2^x first

[wary plover]

in the same way mo did?

[grizzled orchid]

$2^x = 3 \implies 2^{-3x} \cdot 8 = 3^{-3} \cdot 8$

you already got this right?

[wary plover]

what

[grizzled orchid]

oh

uhh

okay can you type out what exactly you have done right now?

reading through chat i'm a bit confused as to how far you've gotten in this problem

[wary plover]

I havent gotten anywhere, ive just been reading like 3 different suggestions for how to proceed now

[grizzled orchid]

oh

okay

we have $2^x = 3$ right?

[obsidian monolith]

hegel:

[grizzled orchid]

and we want to find out what $2^{-3x + 3}$ would equal

[obsidian monolith]

hegel:

[grizzled orchid]

i think this was discussed above, but the first step here is to isolate your 2^x

let's first try breaking apart the $2^{-3x + 3}$ into two different terms

[obsidian monolith]

hegel:

[grizzled orchid]

you know the exponent rules, right?

[wary plover]

yes

[grizzled orchid]

so we know the rule for addition in the exponent, right?

$a^{b + c} = a^b \cdot a^c$

[obsidian monolith]

hegel:

[grizzled orchid]

you get this right?

[wary plover]

yes

[grizzled orchid]

so how first could we split up $2^{-3x + 3}$?

[obsidian monolith]

hegel:

[grizzled orchid]

try thinking about what our "a," "b," and "c" would be analogous to in this expression

[wary plover]

was the way mo did it correct?

if so can we fast forward to that?

I could understand that

same exponent rules

[grizzled orchid]

Mo ended up at $(2^x)^{-3} \cdot 2^3$ right?

[obsidian monolith]

hegel:

[wary plover]

yes

[grizzled orchid]

so we have 2^x = 3, right?

like, that's our given

[wary plover]

uh huh

[grizzled orchid]

so we can substitute 3 in to our equation

right?

$(2^x)^{-3} \cdot 2^3 = 3^{-3} \cdot 2^3$

[obsidian monolith]

hegel:

[wary plover]

yeah

[grizzled orchid]

so now we just evaluate, right?

[wary plover]

oh fuck

I messed up with the exponent

dont even ask

did something really dumb

ok thakns

should have been an easy problem

[grizzled orchid]

👍

[wary plover]

I cant figure out why im wrong

[stuck lark]

Domain = set of x values you’re allowed to plug into the function

[wary plover]

affirmative

[stuck lark]

We should find out what x values we CAN’T plug into f(x)

We see square roots in f(x)

Remember what kind of numbers we can’t take the square root of?

[wary plover]

...........

never mind

now im making up imaginary rules

[stuck lark]

Just ask yourself, what numbers cannot we take the square root of?

[wary plover]

i know

its obvious now

[stuck lark]

So you got the interval right, but you’ve got to determine whether the endpoints are in the interval too or not

[shrewd urchin]

If f is relation from A to B and $(a,b)\in f$ then f(a)=b

[obsidian monolith]

Krishna:

[proud sparrow]

Well, that's the definition

[shrewd urchin]

What do they mean by a,b in f

[proud sparrow]

A function is a special type of relation

[shrewd urchin]

I knew the machine model of function

that was a simple

why is this book overcomplicating the defination of function

[proud sparrow]

It's rigourising it.

Down to set theory

[shrewd urchin]

Ok

Yes in preface it was written that set theortic language will be usded

[proud sparrow]

^

[patent beacon]

@shrewd urchin
(a, b) can be thought of the arrow that takes a to b.

f is then the set of all arrows from the set A to the set B.

(a, b) ∈ f means that (a, b) is an arrow in f.

The reason you'd want to think of functions in terms of set theory, is now your functions can apply on any set. Numbers no longer have to be included

[shrewd urchin]

ye i know the arrow diagram

i cant understand the graph

[willow bear]

uh

yeah there's a bit missing from it

[shrewd urchin]

??

[willow bear]

there

10-second paint edit

this is what the graph of the floor function (or the greatest integer function, as your book calls it) looks like

[patent beacon]

For example, f(2.5) = 2, since going to x = 2.5 on the x-axis, then going up, lands me on a blue line at y = 2

[willow bear]

is it the little circles that are throwing you off @shrewd urchin

[shrewd urchin]

yes

kaynex i understand the funcrion but it graph is complicated

[willow bear]

yeah ok so

[patent beacon]

Blue circles just mean "function is not defined here"

[willow bear]

these circles aren't meant to be taken as literal circles

[proud sparrow]

the circles are just "not this point"

the function is "not at this point"

f(1) is 1

not 0

[willow bear]

a hollow circle like this means that whatever curve ends in one contains every point except the one marked by the circle

[shrewd urchin]

I better watch a video on u tube

[willow bear]

k fine

[proud sparrow]

a well

[shrewd urchin]

finally got it

[heady jewel]

isnt this also named step function

[willow bear]

no

step's different

[hardy abyss]

@heady jewel step function tends to mean finitely many steps

[heady jewel]

k

[shrewd urchin]

In videos the speaker also told it called step function

[patent beacon]

Normally "the" step function is
0, x < 0
1, x > 0

[wary plover]

I don't remember how to find the crossing

[limber bone]

substituite

[wary plover]

yes I know

I don't know how to complete the substitution

[stuck lark]

did you find the equation of the horizontal asymptote?

[wary plover]

3x^2+2x+8=3 I think

[stuck lark]

that's not the right substitution

i think you already found the asymptote, y = 3, now you have to find x when f(x) = 3

[wary plover]

yes

so why is it not right

[stuck lark]

what's f(x)?

[wary plover]

the big thing im not going to type out

I thought if the denominator has no factors you dont consider it when substituting

the Y I mean

[obsidian monolith]

RokettoJanpu:

$f(x) = \frac{3x^2+2x+8}{x^2+10-9}$

[stuck lark]

plug in f(x) = 3

[wary plover]

that's what I did

[stuck lark]

then why do you have 3x^2 + 2x + 8 = 3?

[wary plover]

because I just asked you about the denominator

[stuck lark]

you don't just suddenly toss out the denominator when substituting

[wary plover]

I swear putting blank equal to blank you toss the denominator in some scenario

I cant remember what though

[stuck lark]

so look at the equation above

the substitution you're supposed to do is f(x) = 3

so change the left side of the equation to 3

then you solve for x

[wary plover]

yes?

that's where I started

[stuck lark]

i have no idea where you got 3x^2 + 2x + 8 = 3

[wary plover]

we literally just talked about it...

[stuck lark]

you should get this after the substitution

$3 = \frac{3x^2+2x+8}{x^2+10-9}$

[obsidian monolith]

RokettoJanpu:

[wary plover]

hm

trig now

im pretty good with trig usually but im not sure how to solve this one quickly

without guessing and checking

[uncut mulch]

expand and simplify

[wary plover]

I always know what to do

just not how to do it

is there an identity here?

factoring?

[uncut mulch]

you'll see which ones you'll need

[wary plover]

?

where should I begin

can I apply the exponent to both values

[uncut mulch]

expand and simplify

[wary plover]

that doesnt help

[uncut mulch]

(a - b)^2 =

[wary plover]

I ended up with cos^2 - 4sin^2 = 45/6 earlier but im not sure if that was legal

it doesnt seem right

[uncut mulch]

tell me what you get from squaring that first term

[wary plover]

a^2 - 2ab + b^2?

[next willow]

Your spider sense should be on

[wary plover]

this problem is melting my head

[uncut mulch]

what do you get when you square the first term in the question?

[next willow]

Isn't it tingling?

[wary plover]

no

[next willow]

Mm well that's equal to
a^2-ba-ba+b^2

[wary plover]

I just said that

[next willow]

Ah this is an older question

I'm sorry

[wary plover]

im so confused

[next willow]

I was hoping you would recognize (a-b)^2 lol

[uncut mulch]

(-6cos(x)-3sin(x))^2=?

[wary plover]

I just said that I did

are my messages not getting through?

I dont get it

[next willow]

Is that the original question so?

[wary plover]

????????????????

[uncut mulch]

its part of it but this term seems to be causing issues

why aren't you able to square that? it has a similar structure to (a -b)^2

if you want, it can be rewritten as
( (6cos(x)+3sin(x))(-1) )^2=?

[wary plover]

sinx cosx = 1/4

which still leaves me befuddled as to what x is

even if I missed something I dont see how to get x to equal that

[uncut mulch]

what trig that identies involve sin(X)cos(X) do you know

[wary plover]

why do we keep going backwards

I already did all this

I just need to know where I made a mistake

which I clearly did since x has to be one of 3 choices

none of which work out there

[uncut mulch]

are you familiar with the dbl angles?

[wary plover]

unfortunately

oh god not the cos ones

[uncut mulch]

and can that be applied here

[wary plover]

I dont have cos^2-sin^2 here though

[uncut mulch]

what is sin(2x)?

[wary plover]

god dammit

π/12

I hate review material

especially when it's clear I need it

almost done

I cant figure out why this one is incorrect

[uncut mulch]

what is the domain of log(x)?

[wary plover]

ummm

0< ?

I think

oh wait domain

infinite?

[uncut mulch]

can X be 0?

[wary plover]

good point

that makes sense

[uncut mulch]

can X be negative?

[wary plover]

so it's x = 2

that's the last thing I got wrong

im gonna take my second attempt now

think Ive got a handle on it

[viscid thistle]

I am deeply struggling with this

I have to find the domain of this problem in interval notation

Number 7

[blazing parrot]

okay so when is the function not defined

[odd yarrow]

@viscid thistle so f(x) is undefined when the denominator is 0

the denominator is x^2+5x+4

factor the quadratic equation and solve for the roots by setting each factor = to 0

and the x-values you get, -4 and -1 are used for this

Domain: x == -4, x == -1

[viscid thistle]

Thank you so much

Worked it out I understand it now

[upper flint]

Im working on 4 right now

Could anyone assist me?

[stuck lark]

remember how to foil?

[upper flint]

Waitttt i forgot. sqrt 3 × sqrt 3 is 3 right

[stuck lark]

ya

[upper flint]

but then when i multiply sqrt 3 and sqrt 7 is it sqrt 21 or?

[obsidian monolith]

RokettoJanpu:

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$

[upper flint]

so the F part of foil in this would be 45

because 5×3 is 15 and sqrt3 × sqrt 3 is 3. so 15×3 is 45

[stuck lark]

ya

[upper flint]

I got 10sqrt21 - 39

im kind of lost here

[grizzled orchid]

what is the problem asking?

to simplify?

[upper flint]

yes

[grizzled orchid]

do you know what the conjugate is?

and how it works?

[upper flint]

not at all

[grizzled orchid]

okay

[obsidian monolith]

hegel:

hegel:

[grizzled orchid]

so luckily this expression is very simple and very useful

its just $(a - \sqrt{b})$

[obsidian monolith]

hegel:

[upper flint]

so i multiply the denominator by itself?

[grizzled orchid]

no

close

[upper flint]

numerator

[grizzled orchid]

no

you multiply it by an expression where the second term is of the opposite sign

what i mean is

[upper flint]

ohhh i see

[grizzled orchid]

$(a + \sqrt{b})(a - \sqrt{b})$ produces an expression with no square roots

[obsidian monolith]

hegel:

[upper flint]

that makes sense

[obsidian monolith]

hegel:

[grizzled orchid]

you can FOIL this out yourself and see it in action pretty easily

[obsidian monolith]

hegel:

[grizzled orchid]

this all make sense, right?

i wanna make sure you get it and why it works very clearly because it is a very important concept

are you alive

[upper flint]

i am

so i would get 16-2?

over 5-sqrt2?

[grizzled orchid]

?

no

[wary plover]

I got a 94 on my second attempt of that test you guys helped me with 😄

[grizzled orchid]

ayy

@upper flint to simplify an expression with a radical in the denominator

by convention we say that we want to eliminate the radical in the denominator

so the only way to do this is to multiply the denominator by its conjugate

[upper flint]

i am so dumb

i did the numerator💀

[grizzled orchid]

lol just a small mix up, its fine

however

there is a problem in simply multiplying the denominator

[obsidian monolith]

hegel:

[upper flint]

yes but idk if ill word it right

its 1 over 5-sqrt2

so like i gotta multiply 4 + sqrt2 by the conjugate aswell?

[grizzled orchid]

yep!

[upper flint]

oh deadass

[obsidian monolith]

hegel:

[upper flint]

And thats tge same as the original problem

[grizzled orchid]

if we multiply $\frac{4 + \sqrt{2}}{5 - \sqrt{2}}$ by $\frac{5 - \sqrt{2}}{5 - \sqrt{2}}$, we're really just multiplying $\frac{4 + \sqrt{2}}{5 - \sqrt{2}}$ by 1 represented as a fraction

[obsidian monolith]

hegel:

[upper flint]

makes sense

[grizzled orchid]

yep

[upper flint]

so the answer is

22 + 9sqrt 2 /23

[grizzled orchid]

close

but remember

wait not close

yes

thats correct lol

[upper flint]

swear

[grizzled orchid]

oh fucc i just realized i've been writing 5 - sqrt(2) rather than 5 + sqrt(2) this whole time

oops

[upper flint]

lmao its cool

[grizzled orchid]

but yes i do swear haha

[upper flint]

what if in the conjugate

theres a radical divided by 2

is it like the same thing with division or do i simplify from the start

[grizzled orchid]

something like $\left(3 + \sqrt{\frac{2}{3}}\right)$

?

[obsidian monolith]

hegel:

[upper flint]

dude wtf

ur psychic or smtg

[grizzled orchid]

haha

thanks

in this case it's the same principle

$\left(3 + \sqrt{\frac{2}{3}}\right) \cdot \left(3 - \sqrt{\frac{2}{3}}\right) = 9 - \frac{2}{3}$

[obsidian monolith]

hegel:

[grizzled orchid]

no need to change the process at all

[upper flint]

the radical is only over the numerator

[grizzled orchid]

oh i see what you mean

[upper flint]

yeah

[grizzled orchid]

so something like $\frac{1}{\left(3 + \frac{\sqrt{2}}{3}\right)}$?

oops

[obsidian monolith]

hegel:

[grizzled orchid]

in this case i would do something like $3 = \frac{9}{3}$

[obsidian monolith]

hegel:

[grizzled orchid]

ah yeah i see

we can go through this example if you want

[upper flint]

its the only other thing i dont understand lmao

[grizzled orchid]

ok

so first off the basic idea behind this process is that $\frac{a}{\frac{b}{c}} = \frac{ca}{b}$

[obsidian monolith]

hegel:

[grizzled orchid]

in other words, if the denominator of fraction 1 contains fraction 2

the denominator of fraction 2 multiplies by the numerator of fraction 1

this make sense?

[proud sparrow]

Just multiply the top and bottom by the same thing

[upper flint]

i kinda get it

[grizzled orchid]

it makes intuitive sense when you think about it, right?

[upper flint]

ok yeah

[grizzled orchid]

so with $7 + \frac{\sqrt{3}}{2}$

[obsidian monolith]

hegel:

[upper flint]

so i multiply 2 with sqrt3?

[grizzled orchid]

uh

not quite

i think you'll see it if we go through it

we can rewrite 7 as 14/2 right?

[upper flint]

yes

do i do smtg with the reciprocal

[grizzled orchid]

so we can say $7 + \frac{\sqrt{3}}{2} = \frac{14}{2} + \frac{\sqrt{3}}{2} = \frac{14 + \sqrt{3}}{2}$

ye

[obsidian monolith]

hegel:

[upper flint]

ok im there

the same for the top now?

[grizzled orchid]

for the top it isn't necessary

[upper flint]

oh ok

[obsidian monolith]

hegel:

[upper flint]

ok

[grizzled orchid]

$\frac{3 - \frac{\sqrt{3}}{2}}{\frac{14 - \sqrt{3}}{2}} = \frac{2 \left(3 - \frac{\sqrt{3}}{2}\right)}{14 + \sqrt{3}}$

[obsidian monolith]

hegel:

[grizzled orchid]

sorry for the ridiculous delay, that latex sucked to type

this makes sense based on what we discussed before, right?

[upper flint]

Ohhh so this is when that rule

comes in

[grizzled orchid]

yep

so now just distribute the two

and apply the conjugate as normal

[upper flint]

u multiply the 2nd frac denominator with the top frac

[grizzled orchid]

yep

decreasing the size of the denominator and increasing the size of the numerator both increase the fraction's value overall, right?

[upper flint]

yes

[grizzled orchid]

so dividing the denominator (which decreases its size) by some constant a is the same as multiplying the numerator (increases size) by some constant a

anyway from here you just solve it like you would any other problem

[upper flint]

hmmm

[grizzled orchid]

[upper flint]

wait

on the numerator the 2s cancel out right

[grizzled orchid]

yep

you get 6 - sqrt(3)

even if they didn't cancel out the two underneath the sqrt it wouldn't matter

[upper flint]

hmm

the answer is

87-20sqrt3 / 193

[grizzled orchid]

uh

yeah i think so

[upper flint]

dope

[grizzled orchid]

[upper flint]

thank u so much dude

lmao like fr

[grizzled orchid]

you're welcome

[elder plume]

whos good at pre cal pls dm me

[long pond]

inb4 dmed by noone 😔

[full goblet]

Oof

[viscid thistle]

Oh wait

I think I know

[heady jewel]

you solved it?

[viscid thistle]

not yet, solving another problem first

I haven't been told about the g of g

or the f of f

[serene heath]

g o g =g(g(x))

[viscid thistle]

ohhhhhhhh ok

And wouldn't you know it...

I GOT THEM WRONG

I mean

Seriousy I'm trying

[uncut mulch]

f(g(2)) should return a value

[viscid thistle]

Fortunately i have more tries

I mean... what value?

It's really tough for me

[uncut mulch]

what is g(2)?
and what is f of that

[viscid thistle]

OHH

16x-18?

[uncut mulch]

there shouldn't be any 'x' in your answer

[viscid thistle]

but f (x) = 2x^2 + 1

[uncut mulch]

start with the value of g(2)

[viscid thistle]

Yeah it's 14x-18

[uncut mulch]

nope

[viscid thistle]

OH WAIT
The 2 also goes in the x of 7

So it's 14 - 9

5

[uncut mulch]

yeh

and then f( g(2)) = f(5)

[viscid thistle]

ohhhhhhhhhhh

but... what about this?

There's no value for x here

[uncut mulch]

substitute the 'x' terms in f(x) with g(x)

[viscid thistle]

I'm confused honestly

[uncut mulch]

eg
if f(x) = 5x
f( g(x)) = 5 * g(x)

[sour plinth]

what do you not understand about this?

[viscid thistle]

everything unfortunately... I have a hard time trying to think... it's just so complex

Also I don't think the 5 applies for problem (b)

It says find the following

[sour plinth]

he provided an example

[viscid thistle]

but what's the x here

I mean

[sour plinth]

ok lemme try and explain this as simply as i can

to find for example: f(2)

you replace all the values of x with 2

[viscid thistle]

I know he's trying to help... I really appreciate him for that. The problem is I'm confused a lot

He also said my answer shouldn't have an x value

[uncut mulch]

that was only for the 1st one

[viscid thistle]

Ah

[uncut mulch]

where they provided a value to evaluate

[sour plinth]

yeah, so if you have f(2), you replace all the x's with 2

so you shouldn't have an x left

[uncut mulch]

sry if that wasn't clear

[viscid thistle]

Hey it's okay man. I don't think clearly anyways...

It's really tough when i try

[sour plinth]

well do it slowly

what question are you up to

[viscid thistle]

f(g(x))

[sour plinth]

ok

so what does f(g(x)) mean

[viscid thistle]

2x^2 + 1 (7x - 9)

[sour plinth]

no

ok lemme give you an example

if f(2) means replace all x with 2

then what would f(g(x)) mean?

[viscid thistle]

There's no x value

I mean

There's no real number in that one

[sour plinth]

if f(2) means replace all x with 2, then following that pattern,
what would f(g(x)) mean?

[viscid thistle]

f(g(2))

[sour plinth]

no, assuming those two are separate questions

ok how about this

follow my pattern:
if f(1) means replace all x with 1
if f(2) means replace all x with 2
if f(a) means replace all x with a
if f(t) means replace all x with t
what does f(g(x)) mean?

[viscid thistle]

f(g(f))

[uncut mulch]

in words

[sour plinth]

no, follow the way i wrote it

[viscid thistle]

f(g) means replace x with g

[sour plinth]

but this isn't g

this is g(x)

[viscid thistle]

EXACTLY

[uncut mulch]

so you would replace all x with:

[viscid thistle]

g(x)

[sour plinth]

yes!

if f(1) means replace all x with 1
if f(2) means replace all x with 2
if f(a) means replace all x with a
if f(t) means replace all x with t
so f(g(x)) means replace all x with g(x)

do you understand how that works ^

[viscid thistle]

oh gee that's... understandable!

[sour plinth]

ok so now we have to replace all the x with g(x)

so we know that f(x) = 2x²+1 and g(x) = 7x-9

let's replace all the x's in f(x) with g(x)

[viscid thistle]

hmmmmm

2(7x-9)^2 +1

[sour plinth]

yes!

now we have to simplify that

by expanding it

[viscid thistle]

ok i think i got it

😛

[sour plinth]

great!

you can post ur answers here and im sure we can check if ur right or wrong

[uncut mulch]

dunno if the marking scheme requires you to actually expand it

[viscid thistle]

98x^2 - 252x + 163

[sour plinth]

👍

considering this is an online quiz, i'd say you do need to expand

they usually accept 1 answer only

[uncut mulch]

ok,
try c) yourself now

[viscid thistle]

ok

hmmmmm

14x^2 - 2

[sour plinth]

👍

ur getting the hang of it

[viscid thistle]

heh

so basically i saw that g o g =g(g(x))

so uh

maybe

I could try

7 (7x-9) -9

[sour plinth]

yep

[viscid thistle]

(f ∘ f)(−2)...

Wait

I think

I think i got it

2 (2 (-2)^2 +1) +1

[sour plinth]

yep

[uncut mulch]

missing a square

take it one step at a time
find f(-2) first

[sour plinth]

oops yeah

should be 2 (2 (-2)^2 +1)^2 +1

[uncut mulch]

less likely to make mistakes if you separate tedious processes

[viscid thistle]

163

oh yeah and uh supposedly this is wrong

OH WAIT

I forgot

sorry

[uncut mulch]

that was g(2) = 5
you needed to feed that into f

[viscid thistle]

sorry lol

But yeah i know

my stupid keyboard

forgot the 1

so yeah

it's 51 lol

@uncut mulch @sour plinth So I tried what you guys said

but

I mean

This is different

[sour plinth]

ok following the same principle

[viscid thistle]

but I tried the same procedure

[sour plinth]

f(g(x)) means replace x with g(x)

[viscid thistle]

Exactly

...wait

[sour plinth]

ok so if i put $\sqrt{x+2}$ into $x^2+1$ what do i get

[obsidian monolith]

187:

[viscid thistle]

$\sqrt{x+t}$^2

[obsidian monolith]

Le Super-Cewl Chicken Enby:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

I mean

Oh shoot

$\sqrt{x+2}$^2

[obsidian monolith]

Le Super-Cewl Chicken Enby:
Compile Error! Click the reaction for details. (You may edit your message)

[uncut mulch]

$ sign on the end, and what happened to your 1?

[viscid thistle]

Oh yeah + 1

I'm just testing

[uncut mulch]

and what do you get after simplifying

[viscid thistle]

So basically f(g(x)) is basically...
$\sqrt{$\sqrt{x+2}$
+2}$^2 + 1

[obsidian monolith]

Le Super-Cewl Chicken Enby:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

OH SHOOT

AGH

[uncut mulch]

if your tex is bad, just write it normally and use parentheses

[viscid thistle]

It's hard to use this bot

ok

[uncut mulch]

how did that happen...
why do you have 2 square roots
you are simply replacing the x term with sqrt(x+2)

[viscid thistle]

ohhhhhhh

x+4$\sqrt{x+2}$+7

[obsidian monolith]

Le Super-Cewl Chicken Enby:

[uncut mulch]

uh...

[viscid thistle]

it's tough

Here's how I interpreted it as

[uncut mulch]

why do you have an extra 2?

[viscid thistle]

OH SHOOT

right

gotcha

[uncut mulch]

what's your new answer

[viscid thistle]

x+3

And I did it with the other problem

[uncut mulch]

that's much better

[viscid thistle]

And got $\sqrt^2+3$

[obsidian monolith]

Le Super-Cewl Chicken Enby:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

AGH

I mean

$\sqrtx^2+3$

[obsidian monolith]

Le Super-Cewl Chicken Enby:
Compile Error! Click the reaction for details. (You may edit your message)

[viscid thistle]

AGHHH

[uncut mulch]

braces around terms ur rooting

[viscid thistle]

ah ok

also this

I'm putting $\sqrt[5-10x]$ into -4/x

AGH

[uncut mulch]

click trashcan to erase those screwups

[viscid thistle]

kk

but yeah here's what i did

huh.

[willow bear]

and do you have any reason to believe your answer is wrong?

[viscid thistle]

Well I just want to know

well i got the first part right

but

[willow bear]

are you sure 1/2 is the only number causing you trouble here?

[viscid thistle]

Oh wait i got it lol

[ashen furnace]

okay

can someone explain when to use the tangent equation and normal equation

[viscid thistle]

i don't think i'm on that part yet

I need help

[patent beacon]

Same as you were doing earlier this morning.
f(g(x))

Oh, you want the domain this time. Well, can you find what f(g(x)) is?

[viscid thistle]

Yeah it's basically f(g(f))

@patent beacon

[patent beacon]

I don't know what f(g(f)) means. Can you do the composition f(g(x))?

Remember, it's as simple as putting g(x) into f(x)

[viscid thistle]

If f = a, replace all x with a

I'm trying to follow along

@patent beacon

hello?

@willow bear Yeah I need help

[uncut mulch]

for f( g(x)), replace x with g(x)

[viscid thistle]

ah

[heady jewel]

and replace g(x) with the function in terms of x pls

dont leave it in g(x)

[viscid thistle]

ok

Also this thing

I'm confused about it honestly

[willow bear]

(f*g)(x) = f(x) * g(x)

likewise (f-g)(x) = f(x) - g(x)

[viscid thistle]

I mean

The graph

It doesn't say which is which

Which is f

And which is g

[uncut mulch]

There's a red f and blue g on the graph

[viscid thistle]

oh wait

aaaa

[willow bear]

the graph literally tells you

in big letters

[viscid thistle]

How stupid of me

[willow bear]

color coding yknow

[viscid thistle]

I thought they were just labeling the lines

But I mean in the QUESTIONS

Like (f * g) (2)

WAIT I GOT IT 😄

boy I'm stupid

how about this

it's confusing

[patent beacon]

What's tripping you up?

[viscid thistle]

It's just confusing

Like

How do I find f and g

[willow bear]

there are many possible answers here

but there's one that is sort of obvious

$f(x) = \sqrt{x}$ and $g(x) = 2x+6$

[obsidian monolith]

Ann:

[viscid thistle]

ah geez

Sorry

I'm stupid

[dense oar]

i'm starting precalc this year 😄

[elder plume]

hi

[keen root]

i get (3x^1/3+2x^3/2)/x

but then what?

pls help

[stuck lark]

Split the expression into two fractions... and recall exponent rules

[keen root]

i think the next step could be to multiply by x/x

to cancel out the x in the denominator

so by that logic the 1st part would be 3x^2/3

i think

so it might be C

but im still unsure

[viscid thistle]

someone pls help

[coarse solstice]

You substitute every T with 2t+1

And for the second question you got wrong, you just use the equation N(T) but in terms of t

But N(T) is equal to 1300

[uncut mulch]

for a) you wrote X instead of t

[coarse solstice]

Yeah

You put x instead of t

[shrewd urchin]

How do i read this

[patent beacon]

f is a function that sends real numbers to real numbers

[shrewd urchin]

Ok

[wraith idol]

What’s the latex of the arrow that’s like |->

For functions

[shrewd urchin]

does this mean excluding 0

[wraith idol]

Yes

All reals except 0

[valid flint]

$\mathbb{R}\rightarrow\mathbb{R}$

[obsidian monolith]

John Doe Smith:

[wraith idol]

Nope

[valid flint]

$\mathbb{R}\to\mathbb{R}$

[obsidian monolith]

John Doe Smith:

[valid flint]

hmm

[wraith idol]

$\mapsto$

[obsidian monolith]

Whoever:

[wraith idol]

That was what I was looking for xD

[valid flint]

oh i see

[wraith idol]

$\longmapsto$

[obsidian monolith]

Whoever:

[valid flint]

$f:\mathbb{R}\longmapsto\mathbb{R}$

[obsidian monolith]

John Doe Smith:

[wraith idol]

Nice

Also, you can use \bbR

[willow bear]

uh

no

that's not what \mapsto is meant to denote??

[obsidian monolith]

Whoever:

[wraith idol]

What is that meant to denote then???

Ohhh

I see

So it’s like $f:\bbR\to\bbR$, $x\mapsto x^2$

[obsidian monolith]

Whoever:

[willow bear]

yes

[wraith idol]

Something like that I think

Ok