precalculus

[viscid thistle]

I dont know how to explain

domain could be 0<x<2pi but solutions could be everywhere in the domain

[wide ocean]

Ah okay, ty

[muted granite]

how do you input the x>=0 condition on a ti84 via the y= button? can't find a video on it. thanks.

[pale bison]

$x\geq0$ and $y\geq0$ just means that you're focusing on the first quadrant only...

[obsidian monolith]

Publius:

[muted granite]

I know what it means. Just got this calculator today. And am trying to learn how to use it.

[remote veldt]

@muted granite I think what you're looking to do is change the window; there should be a button that says "window"

[muted granite]

is there away i can input the inequalities as is, or do i have to always put it in y=whatever by hand first each time?

[remote veldt]

https://www.dummies.com/education/graphing-calculators/how-to-enter-inequalities-in-the-y-editor/

there is a way to change it apparently

[pale bison]

@muted granite https://www.youtube.com/watch?v=pbaW0e1VfkE

it looks like stuff are mapped to 0 outside of the domain at least when i tried it

[muted granite]

@pale bison Thanks I will take a look.

[visual finch]

Dude I am taking precal right now and it is kicking my butt

My friend was saying that calculus is easier than precal

Is that true?

[viscid thistle]

generally no

but maybe it would depend on how your school teaches them

[willow bear]

the hardest part of calculus is algebra

[wide ocean]

@reef jasper What was the answer?

[harsh cipher]

this question is very weird

oh maybe not nm

[wide ocean]

@harsh cipher nice! is this homework?

[rare zephyr]

just a question i learned that sin x = cos (90 - x) and is there a visualization for this

cant wrap my head around it

[willow bear]

sin(x) = CB/AB

cos(90° - x) = CB/AB

does this answer your question? @rare zephyr

[rare zephyr]

yeah

i didnt expect it would be a simple visualization

[elfin crescent]

for a function if x doesnt equal 2, the correct function notation for the domain of that function would be [-∞,-2)u(-2,∞] right?

[willow bear]

replace the [] with ()

you don't use [] with ±∞

[elfin crescent]

@willow bear i thought the brackets meant that value is included, and the parenthesis means that value is excluded?

[willow bear]

exactly

but you don't want to include +∞ and -∞ themselves

[elfin crescent]

y

[willow bear]

these aren't real numbers lol

[elfin crescent]

o

if i was a real number i would put brackets though right?

[willow bear]

i mean.... h

[elfin crescent]

huh?

[willow bear]

i really am not sure what you mean

[elfin crescent]

like say if we had real numbers in replace with ∞

we would put the brackets around the real numbers since they're are real right?

[viscid thistle]

You put [] for real numbers if x equals those numbers and you put () if x doesn't equal those numbers. You always out () around infinity and negative infinity

[elfin crescent]

alright ty

[viscid thistle]

I'm not to sure what I'm doing wrong, but insight would be greatly appreciated.

how did you get the first one

@viscid thistle

oh

so I know that |z+w|^2 is |4+2|^2

that would result in me getting

36

hold one, |z| is not z

i don't quite follow

|z| is not equal to z

its a real number

so it should right?

nope

hmm

then what would I have to chnage?

the absolute value of a complex number is always a real number

right

but like the magnitude of |1+3| is 4

right?

|a+b|!=|a|+|b|

how?

https://en.wikipedia.org/wiki/Triangle_inequality

isn't it the same properties

not at all

oh 😦

that's

hmm

interesting

so then how would i add |a + b|

you cannot

if you only know |a| and |b|

may you elaborate?

or explain how id go about answering that

$\abs{a+b}\leq \abs{a} + \abs{b}$

|a+b|

[obsidian monolith]

HoboSas:

[viscid thistle]

you can answer "?"

it is allowed in the exercise

oooh

wait

so then according to that logic

|z-w| is also ?

yes

but

same reasoning

|zw| works

yes

and so does

it's a property of complex numbers

\frac{z}{w}

i made an error

$\abs{a\cdot b}=\abs{a}\cdot \abs{b}$

[obsidian monolith]

HoboSas:

[viscid thistle]

but i think you get what im trying to show

ooooo

so |4(2)|^2 =64

$\abs{\frac{a}{b}}=\frac{\abs{a}}{\abs{b}}$

[obsidian monolith]

HoboSas:

[viscid thistle]

right

okay that helps

a lot

so I |4/2|^2 also equals 4

that right corret?

correct*

$\abs{a+b}=\abs{a}+\abs{b}$ only if $\Im(a)=0$ and $\Im(b)=0$

[obsidian monolith]

HoboSas:

[viscid thistle]

so I |4/2|^2 also equals 4
yes

ooooo

okay thank you so much

Im going to try it

which is the case of normal real numbers

right

nice

thank you so much

btw the properties I have shown you are pretty easy to prove, I suggest you trying it for exercise

oh yea

so this would be correct @viscid thistle

this is more interesting

we have one more equation

right

but I don't think it should affect it

have you read the explanation?

exactly what we said (but better explained)

slightky

now I have to go

I'll take a look later

try submitting it though

okay

ty'

[viscid thistle]

I think you can actually solve for |z+w| and the other one now

If you let w=a+ib and z=c+id you got 4 equations with 4 solutions

A lot of calculations though...

i dont follow

and rn im so lost

@viscid thistle

First equation is $\abs{z}=\sqrt{c^2+d^2}=5$

[obsidian monolith]

HoboSas:

[viscid thistle]

right

Second is the same for w

$\abs{w}=\sqrt{a^2+b^2}2$

rightt

but what do i do from there

im so lost

Then you have $z\bar w =(c+id)(a-ib) = 6 + 8i$

[obsidian monolith]

HoboSas:

[viscid thistle]

Which gives you 2 equations

right but

$ac+bd=6$ and $ad-cb=8$

[obsidian monolith]

HoboSas:

[viscid thistle]

there aren't any imaginary numbers in 5 and 2

What?

lol nvm

i misinterpreted what you said

Now you have 4 equations

And 4 variables

yes, i understand that

but i was looking at it and i was thinking that

|z + w|^2

is equal to $(a+c)^2+(b+d)^2$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

and I got the result of 29

You expanded that

which was subsequently incorrect

yea

its just expanding it

but im not sure the following steps

Should be 25 + 4 + 2*6

You forgot 2ac + 2bd

It's a clever idea, nice one

41? 🙂

Same thing for |z-w|

41

and z-w

Try expanding that too

is 9

Try

okay

so $|z-w|^2 = (a+c)^2 - (b+d)^2$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

Not really

hm?

Try writing it

z-w

lol i realized my mistakr

so $|z-w|^2 = (-w+z) (-w+z)$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

?

In terms of a, b, c and d

yea so the result is 9

Aight

that right?

no its not

$ac-bd=6$ and $ad+cb=8$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

Hold on

ok

25+4-2*6

so 17

18?

$a^2 +b^2+c^2+d^2-2(ac+db)$

[obsidian monolith]

HoboSas:

[viscid thistle]

it was 25 my bad

lmao at this point

u gppg

?

not 25

my bad

ur goodo

is it correct?

yea lol

ty so much

[copper dirge]

Hia

Could someone help me with this

x+2b=3

[viscid thistle]

what's the goal?

[copper dirge]

Solving it seems fairly simple

x=3-2b

just wondering what the restrictions on x would be

[viscid thistle]

it depends on what b is

[copper dirge]

no restrictions or x has to be greater than or equal to zero

hmm

ok

[viscid thistle]

what is the problem

maybe more context would help

[copper dirge]

it just says Solve for "x" and state any restrictions

i'm thinking x = 3 - 2b with no restrictions

[viscid thistle]

seems good

[copper dirge]

yolo

[copper dirge]

Heyo me again

Just checking my answer

$5x-1/4(x-2)/5/x-2$

[obsidian monolith]

Tuki:

[copper dirge]

The non permissible value would be 2 right?

[viscid thistle]

What is this

[uncut mulch]

better use of parentheses and/or bot please

[viscid thistle]

Btw this is #prealg-and-algebra

[copper dirge]

Oh ok

Sorry just joined

[uncut mulch]

there isn't much of a distinction between pre-alg and pre-calc

[harsh cipher]

@wide ocean Yea

[wide ocean]

can't figure the right steps to isolate x

10^2/3 = third root of (x^2+48x)

[viscid thistle]

what is the inverse process of roots?

[wide ocean]

not sure

[viscid thistle]

consider using exponentiation in some way

before or after converting the log to exponent form, whichever you please

[wide ocean]

hmm still can't wrap my head around it

thanks for your help though

[past meadow]

Whats $(\sqrt[3]{a})^3$?

[obsidian monolith]

Sneaky:

[willow bear]

what do you think it is

or was it a socratic question

[past meadow]

It was in response to the above

They werent sure how to get rid of the cuberoot in an equation

It was meant to prompt cordial to realise that cubing is the inverse of cuberoot but they seem to have left

[willow bear]

theyre still on the server theyre just ghosting you

[wide ocean]

oh so i should cube root everything

to get rid of the root of 3

[willow bear]

why would you cuberoot both sides if you already have a cube root you wanna get rid of lol

[wide ocean]

just one side?

[willow bear]

that'll fuck up the equation lol

you're just gonna get $\sqrt[3]{10^{2/3}} = \sqrt[9]{x^2 + 48x}$

[obsidian monolith]

Ann:

[willow bear]

which is unhelpful to say the least

but that is what you will get upon cube-rooting both sides of $$10^{2/3} = \sqrt[3]{x^2 + 48x}$$

[obsidian monolith]

Ann:

[wide ocean]

i see

okay ty

so just 1 side?

[willow bear]

what do you mean, "just one side"?

do you even algebra?

you can't just cuberoot only one side! what you'll get has nothing to do with the equation you started with!

[wide ocean]

like ^3

[willow bear]

you wouldn't "add 5 to just the LHS" in z - 5 = 11 would you?

[wide ocean]

ah okay

so both sides

[willow bear]

OF COURSE!!!!

yes, CUBING both sides will help you.

[wide ocean]

thanks 👍

cubing

not cube root

[viscid thistle]

You could also have pulled a 1/3 out of the logarithm to remove the cube root at the beginning instead

[pallid cargo]

a1= x; r=x-3; x E r What is the corresponding equation for s10= (9x+27).5 9x-135 Or 55x-135

[alpine basin]

How do I find the inverse of this function?

[viscid thistle]

https://youtu.be/2zeYEx4eTdc

@alpine basin

[alpine basin]

Sorry I do know how to find them usually

but this one Im stuck

[viscid thistle]

A video will do more than an text explanation here

Really?

Well

[alpine basin]

You're supposed to replace the x's with y's and vice versa right?

[viscid thistle]

What have you tried

[alpine basin]

I got x=y(y+2)

Then I figured I multiply y+2 to both sides

And then thats where im stuck, y = x(y+2)

[viscid thistle]

I got x=y(y+2)
@alpine basin how? The q is x=y/(y+2)

You meant division there prolly right?

[alpine basin]

Oh

I was explaining how I got to y = x(y+2)

It was originally x=y/(y+2), then I multiplied y+2 to both sides

[viscid thistle]

Oh mb

Right

[alpine basin]

Im not sure how to get y by itself after that part

[viscid thistle]

Hold on

Right so

[alpine basin]

ye

[viscid thistle]

So from $y=x(y+2)$ expand it

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@alpine basin

[alpine basin]

yea

[viscid thistle]

$y=xy+2x$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Now bring the xy to the LHS

[alpine basin]

whats the lhs, not sure what that abbreviation is lol

[viscid thistle]

Left hand side lol

[alpine basin]

oh mb

[viscid thistle]

Nw

[alpine basin]

so y - xy = 2x

[viscid thistle]

Yes

Factor out y

[alpine basin]

oh..

[viscid thistle]

Yeah it was a lil bit tricky ngl

[alpine basin]

yeah i think i got it now, that shoulda been obvious lol

[viscid thistle]

Lol good

[harsh cipher]

Hi

1b

d) the exact value of the sum for each of the following series

I solved a= 1 which is the first term

number of terms = 8

ratio = -1/2

plugging this into the equation

a(1-r^n)/(1-r)

after plugging in the values. How is the sum 85/128 ?

nm got it

😛

[harsh cipher]

can someone help me with this question

[viscid thistle]

try writing all the terms

@harsh cipher are you there?

[willow bear]

sum literally contains three terms

how else could one possibly simplify it

[pale bison]

shh

[willow bear]

literally three goddamn terms just write it out lmfao

[pale bison]

lmao

[viscid thistle]

I guess we wait

[willow bear]

g h o s t

[pale bison]

[desert pendant]

@harsh cipher there is a logarithm property that log(a^b)=b log(a), then log(2) will be a constant number, which you can take out of the sum, then you have the sum from 3 to 5 of "x", which is 3+4+5=12, so the answer is 12 log(2) or log(4096)

[viscid thistle]

So the composition of functions is almost never commutative. Are there cases where it is, excluding when f(x) and g(x) are defined by the same expression, or when they are each monomials where the exponents have the same product?

[patent beacon]

What does your class consider "invertibility"?

[wide lynx]

I'm not sure?

It's the first day

[patent beacon]

Odds are, they only care about a horizontal line test

But that isn't always the case haha. That's why I ask

[wide lynx]

yea ur right

i just checked the key

[tawny nacelle]

they dont even give u the domain and the codomain ?

you might as well blindly answer that

[wide lynx]

the key said it is not

[patent beacon]

It's entry stuff. Domain is "whatever makes sense" and codomain is R

[tawny nacelle]

so did they talk about injectivity ?

or like one-one ?

[wide lynx]

[tawny nacelle]

plz

[patent beacon]

We call a function one-to-one if it passes a horizontal line test

Or "injective"

[wide lynx]

so functions can only be invertible

if it passes horizontal line test

[tawny nacelle]

see idk how much im supposed to tell you

[wide lynx]

.....

i see

usually is that the case?

[tawny nacelle]

but no just passing the horizontal line test means its one to one

you also need to check if the function is onto

[thorn mountain]

A function doesn’t have to be onto in order to be invertible

[obsidian monolith]

Malix:

Malix:

[tawny nacelle]

no wth

lets say u have the inverse and its ln(x)

[thorn mountain]

Invertible means having an inverse which is a function, yes?

[tawny nacelle]

whats ln gonna map -1 to ?

invertible means being injective and surjective

[thorn mountain]

That is not a definition of invertible I have encountered

Nor is it what comes up when searching for a definition of invertible function

A bijective function is certainly invertible

But invertible does not imply bijective

[jagged glade]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:composite/x9e81a4f98389efdf:invertible/a/intro-to-invertible-functions

[tawny nacelle]

theres a better definition of invertible with like if $f: A \to B$ and there exists $f^{-1}:B \to A$ such that $f \circ f^{-1}= i_{B}$ and $f ^{-1}\circ f= i_{A}$ then $f$ is invertible

[obsidian monolith]

soαρ:

[tawny nacelle]

but even according to this definition, invertibility is equivalent to injectivity and surjectivity

[patent beacon]

We are getting bogged down in different definitions of the word "function"

[thorn mountain]

No

[patent beacon]

In an early math class, all functions are assumed to be onto

[thorn mountain]

Also, no

$e^x$ is injective but not surjective

[tawny nacelle]

you gotta say what the codomain is man

[obsidian monolith]

Malix:

[patent beacon]

For example, sqrt(x) is invertible. That's because the codomain is automatically set to the range.

[tawny nacelle]

it doesnt make sense to say "not surjective" without even knowing the codomain

[thorn mountain]

Then it is likewise useless to tell a precalculus student that they have to check for surjectivity

[tawny nacelle]

i dont think u were talking about what precalc people need to check

[thorn mountain]

Is this channel for helping precalculus students?

[tawny nacelle]

u literally chopped off surjectivity from the requirement

yes indeed

[patent beacon]

I'm also not certain this stuff is taught the same everywhere

[thorn mountain]

It could be argued that if the function is not surjective that the inverse function does not map some portions of the codomain

But that doesn’t change whether it is a function or not

And it doesn’t change whether the function is invertible or not

[tawny nacelle]

not sure if its even a function if it doesnt use up all of its domain

its like having 1/x defined all over R

also if u chop off surjectivity, surely you're gonna lose a lot of the nice properties

[frozen needle]

there are two different concepts for "functions" but unfortunately they appear to have the same name in English

[tawny nacelle]

[willow bear]

[serene heath]

2 different definitions

[patent beacon]

At least where I'm from, no precalc class considers surjectivity. Every function automatically has the codomain set to the range and surjectivity is assumed

[serene heath]

the right one and the wrong one

[patent beacon]

Is this the best way to do it? No.

[thorn mountain]

Can you think of a situation where that automatic reduction of the codomain to the range of a 1-1 function would cause problems for answering the question “is this function invertible?”

[patent beacon]

Note that when you use this convention, you don't have to define a codomain. There are problems with this, where not knowing where a function maps to is awkward

It's never a problem when you're just mapping into R^n all the time

Non-surjectivity is also a beautiful thing to recognize, linear algebra makes a lot of use of it

[viscid thistle]

you're beautiful

[patent beacon]

Yes

[thorn mountain]

Thanks Kaynex. It’s still beyond the scope of this channel, but I concede it is important.

[fluid sable]

how would you know how many voted for lit?

[thorn mountain]

If the probability that math is favorite is 1/5 how many votes for math are there total?

[fluid sable]

100

[thorn mountain]

Ok, so x+y=100

[fluid sable]

Yeah

is it 50

for lit

cause theres 500

[thorn mountain]

What’s the sum of English, math, lit, and sci?

[fluid sable]

450

[thorn mountain]

Umm, no

[fluid sable]

150 + 100 + 200 = 450

[thorn mountain]

Adding all 4 classes together gets you all the votes

[fluid sable]

yes

150 + 100 + 200 + (50 for lit) = 500

[thorn mountain]

350+2x+y=500

[fluid sable]

?

[thorn mountain]

So 2 equations, 2 unknowns

2x+y=150 and x+y=100

[fluid sable]

Ok

[thorn mountain]

This is not to say your earlier deduction of x=50 was wrong btw

[fluid sable]

Oh

but if i were to do it mathematically

it'd be more like this

[thorn mountain]

Well, when you added all you knew together to get 450+x=500 that’s mathematical too

[fluid sable]

i guess yh

Say you had something like this

would you do

[thorn mountain]

But where I was going is probably what your class is set up to encourage because you are probably in a section on solving systems of equations or having a review of that

[fluid sable]

64G = 376

[thorn mountain]

No

[fluid sable]

lmao im bad

w these questons

[thorn mountain]

It’s ok

You have 2 different sizes of groups

[fluid sable]

5 and 7

yeah

[thorn mountain]

Say x groups have 5 people and y groups have 7

[fluid sable]

Ah ok

x + y = 64 ?

[thorn mountain]

Then x+y is the total number of groups (64)

[fluid sable]

alrighty

[thorn mountain]

And then if we add 5x +7y we should have the total number of people

[fluid sable]

how

[thorn mountain]

Each group of type x has 5 people in it

So the number of people in the x sized groups is 5x

Similarly for the number of people in the y groups

[fluid sable]

ah i see

5x + 7y = 376

[thorn mountain]

Yes

[fluid sable]

:))

Now

7y = 376 - 5x

[thorn mountain]

You could

I recommend y=64-x

[fluid sable]

y=64 - x

Okay

[thorn mountain]

Then 5x+7(64-x)=376

[fluid sable]

36 ?

[thorn mountain]

Yes, that is x

[fluid sable]

Alright,

How would you find the measure of QPC

[jagged glade]

Should this be in #geometry-and-trigonometry ?

[fluid sable]

I solved it.

[jagged glade]

Okay.

[fluid sable]

How would you go about solving thos

[wide lynx]

write an equation for f(x) & the inverse

i already got the inverse but cant figure out the normal

[fluid sable]

What's the inverse ?

[wide lynx]

whats the notation in typing for it

is it f^-1(x)

[fluid sable]

?

[wide lynx]

i got f^1(x)=3^x

for the inverse of that

[fluid sable]

Well first

establish the rate

In our case, you are adding one each time ?

and multiplying by 3 in the x section ?

[wide lynx]

x is multiplying by 3 each time

and f(x) is adding 1

yes

[fluid sable]

alright well

as you know

y=mx + b

[wide lynx]

yes but

would that just be 1x/3

[fluid sable]

first calculate

the slope

y2 - y1/x2 - x1

choose two points

(1,0)
(3,1)

[jade heron]

its not an arithmetic progression

[fluid sable]

hm ?

[jade heron]

@wide lynx 3^f(x) = x right?

f(x) isn't of the form Ax+B @fluid sable

[wide lynx]

well for the inverse of it I got f^-1(x)= 3^x

[jade heron]

https://cdn.discordapp.com/attachments/363224154469826562/724672968579612903/b7b37e5864cb343db31b17f429ca61a7.png

[wide lynx]

im just trying to find the normal

[jade heron]

Yh, f inverse is 3^x

so

if you replace x with f(x)

you should get

f^-1(f(x)) = 3^f(x)

x = 3^f(x)

which makes sense if u look at ur table too

3^0 =1
3^1 = 3
3^2 = 9 etc...

[wide lynx]

i see

so I just replace x with f(x)

ok thank you

[jade heron]

so you should get 3^f(x) =x so, f(x) should be the logarithm with base 3

[wide lynx]

not f(x)=3^f(x)

[jade heron]

It should be like $x = 3^{f(x)}$

[obsidian monolith]

L'Âne 🍐:

[jade heron]

not f(x)=3^f(x)

[jade heron]

@elfin crescent f(x)/g(x) can not be negative ever can it?

sqrt(1-x) is strictly positive, and (x^2+5) is strictly positive too

[elfin crescent]

yeah i just realized

ty

[wide lynx]

ln(x^2 -3)?

[viscid thistle]

...

Show more context @wide lynx

[wide lynx]

sorry

[viscid thistle]

You are doing d. i suppose

[wide lynx]

yes

[viscid thistle]

No c

[wide lynx]

yes but also d

i just didnt want to spam

[viscid thistle]

Right so just leave it like that

bruh to find the original function, take the ln of both sides, then swap x and y

[wide lynx]

theres nothing else i can do?

[viscid thistle]

c) g(x²-3)=ln(x²-3) i can't think of anymore

bruh to find the original function, take the ln of both sides, then swap x and y
@viscid thistle wut

[wide lynx]

^

so theres no solving that or anything?

[viscid thistle]

so u have f^-1(x) =3^x correct?

What the hell are you taking about lol @viscid thistle

R u memeing or what

look at the earlier chat

so theres no solving that or anything?
Not really

There is no more simplification

@viscid thistle he needed help with finding the normal function of f^-1(x) = 3^x, so i responded

Oh ok

thats what i was referring to when i said what i said

i'll leave now. goodbye

[wide lynx]

oh lol ty

yes i got that

x^2 is correct for f right?

[viscid thistle]

Who you are asking

[wide lynx]

u

[viscid thistle]

g(h(x)) is not x²

Where did the ln go

5 4 3 2 where the ln go

@wide lynx

[wide lynx]

i thought ln and e cancel

[viscid thistle]

Jeez im blind hold on

Yes you are right

@wide lynx i was looking at g(f(x)), yes g(h(x)) is x²

[wide lynx]

pog

[viscid thistle]

I wouldn't call it cancel but yeah ln(e)=1 so the ans is x² as you said

Good job

[wide lynx]

can an answer come out to be just ln

[viscid thistle]

ln itself is not a thing. Ln is a function, ln(x) can come out but not "ln" lonely

[azure dove]

Sequence and Series, try out Q8 from the CIE maths 9709 paper 12 winter 2019, if you get stuck check out the video solution
https://youtu.be/J4LsgjZSZQE

[wide ocean]

can someone explain why cosine graph becomes on angle like this?

is it because of a translation by x to the right? i'm confused

[viscid thistle]

Think of the - in -x as a slope

[copper vigil]

@wide ocean try actually plugging in values

you know that f(x)-g(x)= cosx-x

[wide ocean]

thanks @copper vigil

logarithms exponential growth help?

[lilac pier]

@wide ocean Use $P(t) = P_o e^{kt}$

[obsidian monolith]

Sup?:

[alpine basin]

How do I simplify i^-3 + 5i^7?

[viscid thistle]

$ i^{-3} + 5i^7$

[obsidian monolith]

HoboSas:

[viscid thistle]

this

?

@alpine basin

[alpine basin]

yea

[viscid thistle]

what is i^3

[alpine basin]

-i

[viscid thistle]

what is 1/(-i)?

[alpine basin]

-i

wait wut

[viscid thistle]

not really

[alpine basin]

1/-i is 1/-i

[viscid thistle]

yes but denominators are bad

try multiplying by -i/-i

[alpine basin]

i

[viscid thistle]

yes

[alpine basin]

hmm ok

yeah that was the only part i was havin trouble with

[viscid thistle]

alright

[alpine basin]

then it'd be i-5i whcih is -4i right?

[viscid thistle]

yes

[alpine basin]

okay thanks

[viscid thistle]

np

[copper vigil]

provided that you have 2 vectors in R2 that are not colinear, can you form every single vector in R2

through a linear combination of those vectors

[stuck lark]

what do you think

[copper vigil]

@stuck lark i think it's true but i don't know how to formally prove it

[finite wraith]

informally prove it then. /s

[viscid thistle]

^

Informally explain it here and we'll help you write it formally @copper vigil

[stuck lark]

i suppose it helps to break out definitions of things, eg collinear, linear combination, etc, then spitball as much as you can toward a cohesive proof

[copper vigil]

provided that A is not 0, the equation Ax=B always has one x value for which it is true

that's where my intuition is getting me

[wide lynx]

would the function be discontinuous at points (-3,-1),(-1,1),(1,0),(2,1),(4,-1)

[viscid thistle]

JEEZ

[wide lynx]

..................

[viscid thistle]

What i meant is: isn't it easier to write where it IS continuous?

And not where its not

[wide lynx]

the question is the x values where it is not continuous

so -3,-1,1,2,4

[viscid thistle]

I mean

[wide ocean]

what went wrong?

[viscid thistle]

Yeah okay that's correct but could be written better (prev q)

Actual answer=5.83h?

[wide ocean]

yes @viscid thistle

oh should i have expanded 3log200(1-r) ?

[copper vigil]

@viscid thistle is it sufficient to prove that 2 colinear vectors can only form another vector that is also colinear

is that equivalent to stating that any 2 vectors that are not colinear can form any other vector

[viscid thistle]

Hmm, that's uh.. too short imo

i suppose it helps to break out definitions of things, eg collinear, linear combination, etc, then spitball as much as you can toward a cohesive proof
@copper vigil

[copper vigil]

c1(x1,y1)+c2(x2,y2)=(x3,y3). so c1x1+c2x2=x3. if (x1,y1) is colinear with (x2,y2), it is the same as stating that A(x1,y1)=(x3,y3), implying that the linear combination of colinear vectors must also be colinear with those vectors

[patent beacon]

You haven't touched what x3 and y3 can/can't be, which is the crux of your proof

Lets talk impossibilities. Assume there's some value (x3, y3) cannot be. What does that say about (x1,y1) and (x2,y2)?

[copper vigil]

they must be colinear then

[patent beacon]

That's true! Can you prove it?

[copper vigil]

didn't i just do that? by proving that the linear combination of colinear vectors must also be colinear with those vectors. so to represent any vector in R2, you need a combination of vectors that are not colinear

the gap in my logic is proving that ANY vectors which are not colinear can form any vector in R2

which is where i'm stuck right now

[patent beacon]

Yeah yeah okay I see where you are with this. Let's say (a,b) is one of these colinear vectors you can make. That is:
a = kx1
b = ky1

Clearly, (a, b+1) is also a member of R². But it's not colinear to (x1,y1). That's because k = a/x1 doesn't work for the other equation. So, you can't express (a, b+1) and so you can't use two colinear vectors to express any vector in R²

[copper vigil]

huh i looked up the proof

i just don't know enough lin alg yet to prove it

[fleet yew]

Lmao from what it sounds like you literally just got into lin alg

Learn a bit more about matrices

Then you'll be able to do this proof

[fleet yew]

Specifically it has to do with taking the determinant of the matrix and realizing the conditions

[viscid thistle]

Is anyone available right now?

I need help with this problem

[finite wraith]

@viscid thistle is that x-> infinity?

[viscid thistle]

Yeah

I just figured it out, thank you though

[finite wraith]

you just plug infinity in

so yeah

[viscid thistle]

How would I explain this?

[finite wraith]

🙄 you forgot to put the limit

lim as x goes to what?

[viscid thistle]

Infinity

My bad

[finite wraith]

the answer is

-infinity to +infinity

idk what else to say tbh

its not like the limit really exists

sinx alternates between -1 and 1

we do not know if x is being multiplied by -1 or 1 at infinity

[viscid thistle]

That’s a little bit more understandable

[jagged glade]

quick answer, find y' 😄

parallel-slope relationship?

note: y'(x) is the slope of the equation at x.

https://www.mathsisfun.com/algebra/line-parallel-perpendicular.html

cool

well, for (a) 12(4x-7)^2 = ??

12(4x-7)^2 is the slope of tangent of the equation at x, and it is parallel to y=108+7

so...?

ah, for your convention, your gradient means slope i guess

sub 108 into the slope of tangent equation
very close, you just have to equate them 😄

yep

12(4x-7)^2=108

yea, gradient = gradient XD

[craggy dune]

well the root of (2x - 4) should be simple enough

and x^2 = 4x should be manageable as well

whats 0^3?

[harsh cipher]

hi

I have a question about point of discontinuity

y= (x+2)(x-6)/(x+2)(x-2)

why would the horizontal asymptote be on y= -1?

[jagged glade]

the horizontal asymptote should be on y=1

[harsh cipher]

well on the graph its on y= -1

I can show you the picture

[jagged glade]

,w graph (x+2)(x-6)/((x+2)(x-2))

[obsidian monolith]

Display results online and refine query

Requested by Biscuit#5208

[harsh cipher]

question b

[jagged glade]

I think you've missed out a negative sign

[harsh cipher]

well damn

answer key doesn't include a negative sign

so it's a mistake.

[jagged glade]

yep

https://www.wolframalpha.com/input/?i=graph+-(x%2B2)(x-6)%2F((x%2B2)(x-2))

[patent beacon]

Let's say the function is continuous. What should f(18) be?

[wide lynx]

for the top one it would be k*18 x<=18

[pale bison]

so f(18) = 18k

[wide lynx]

but its find k?

is that my answer or

[pale bison]

do you know limits?

[wide lynx]

kinda

just learning them

[pale bison]

ok, so $f(x)$ is continuous at $x=18$ if

$\lim_{x\to18}=f(18)$

[obsidian monolith]

Publius:

[patent beacon]

I mean you also want the function to connect to the other part

[wide lynx]

yeah

[patent beacon]

So yes, f(18) = 18k
But also you want f(18) = 17

You can connect this to a limit yes. Find the left-sided and right-sided limit x → 18. Since the function is continuous, these should match

[wide lynx]

is this finding k

it just doesnt really seem like 18k = k

[patent beacon]

No lol

Give the above one more look-over. You'll be suprised by how simple it can be

[wide lynx]

fr

[patent beacon]

Or at least, don't overthink it

[wide lynx]

im probably overcomplicating it

yea

[shrewd storm]

Do you know what a limit is?

[wide lynx]

yeah

[shrewd storm]

Ok so given that we know that as x approaches the limit your piece function must be continuous so the expressions must be equal

[patent beacon]

What's the left-sided limit, x → 18?

[wide lynx]

oh

wait

i see

because left side limit = right side limit

18k=17?

or no

[patent beacon]

Bang you got it fam

[wide lynx]

i see

[shrewd storm]

Yep because think about it if the function could approach to different values from two different sides it wouldnt be continuous

Even for an English definition cause the function would jump

[wide lynx]

would i use the same concept for this

[patent beacon]

Yup lol

But you're looking at x = 2

[wide lynx]

yea

i got it sorry fellas

[patent beacon]

Both of those pieces had better be equal if the limit should exist

Np! Glad it makes sense

[shrewd storm]

What there is nothing to apologize for I’m about to go get some help too

[willow bear]

@pale bison that notation merits a copswing

[pale bison]

it's not a notation, it's a typo and kaynex didn't correct me

no copswing pls

[wide lynx]

[willow bear]

what's giving you trouble

[wide lynx]

oh sick

[pale bison]

Nah Ann. They just need a confirmation and move on, no explanation or more thinking is required

[willow bear]

they could have asked "hey is this the correct answer"

[frigid sapphire]

I thought of ) as infinity sorry lol

Can't infinity simply be an undefined number, like 1 million?

[fluid shore]

1. Infinity is an undefined number
2. Infinity is 1 million, which is perfectly defined.

Do you see how that doesn't make sense?

[willow bear]

i mean clearly infinity is 1.7977*10^308 /j

[fluid shore]

10*infty = infty => 10 = 1

[obsidian monolith]

HoboSas:

[stuck lark]

$^\infty2$, ftfy

[obsidian monolith]

RokettoJanpu:

[fluid shore]

$\infty^{\int \infty d\infty}$

[obsidian monolith]

Abhijeet Vats:

[viscid thistle]

wrong channel

[desert pendant]

infnity=lim x->1 erf^-1(x)

[timber wraith]

Shibe

[maiden pebble]

https://i.imgur.com/aiebR7j.png

is the lim(x->1) = 1?

if so, is that because the left side of the graph doesnt have a limit, where as the right side does?

[viscid thistle]

it DNE

[maiden pebble]

https://i.imgur.com/nC5D9xc.png

So would the answer to this be DNE?

[viscid thistle]

p sure it is

[maiden pebble]

Thats what I said too

but I got the question wrong

https://i.imgur.com/ZHs6wME.png

[viscid thistle]

4?

might be 4

@maiden pebble

[maiden pebble]

How did you get that @viscid thistle @viscid thistle

(that was correct)

[viscid thistle]

the average of each limit in each graph is 2

and there are 2 graphs

so 2x2=4

makes sense

[maiden pebble]

So what are "each limit"

https://i.imgur.com/UAkLOnA.png
Are you saying that the lim(x->1) of red = 3 and of green = 1

[viscid thistle]

yah

so 3+1 = 4, 4/2 = 2

[maiden pebble]

yes, i udnerstand the average part

I just didnt know where the numbers 2+2 were comming from

[viscid thistle]

oh

ok

[uncut mulch]

it has nothing to do with averages.

consider the limits from the left and right

[maiden pebble]

Yes I was just about to ask about that

because my next question is very similar but the answer is DNE

[uncut mulch]

apply it here, show work and tell me what you get

[grizzled orchid]

this clearly looks like a test lol

[hallow cave]

its not

[grizzled orchid]

especially if it's "urgent"

[hallow cave]

its homework points for a test

[grizzled orchid]

[hallow cave]

yes bruh

[viscid thistle]

welp brööther, we shalln’t give you the answers but we might help you through it.

[alpine basin]

How would I write the polynomial for this graph?

I got -(x+1)(x-2), but it has to be a degree of 3 but Im not sure which x Im supposed to square

[viscid thistle]

try using stationary points

[alpine basin]

wut

[viscid thistle]

derivative

[alpine basin]

Im not sure how id do that

[viscid thistle]

alright then try plugging some x's in

[alpine basin]

I mean the answer is -(x+1)^2(x-2)

But I was asking how am I supposed to know it should be x+1 thats ssquared and not x-2

[viscid thistle]

plug in numbers

and see what's the y

[alpine basin]

@viscid thistle Is there a quicker way to do that or is that the only way

[viscid thistle]

I'd differentiate

but I think you have not studied that method yet

[alpine basin]

mk

[viscid thistle]

@alpine basin (x+1)is squared because the graph switched from decreasing to increasing when it hits the point

[stray dagger]

Im trying to figure out how to complete this square to find the center and radius

can anyone help?

[viscid thistle]

ideas?

[stray dagger]

i know its b/2a

[uncut mulch]

the x and z components are already in square form
so you only need to focus on completing the square for
y^2 - 2y

what's it?

[stray dagger]

is it 2

[uncut mulch]

what's it?

[stray dagger]

its 1

2/2(1)

[uncut mulch]

what does that represent?

[stray dagger]

the square???

[uncut mulch]

can you please be less vague?

[stray dagger]

Ok i dont know what it represents thats why

[uncut mulch]

$\cts$

[obsidian monolith]

ramonov:

[stray dagger]

ok so it represents the missing square

in y^2-2y

[uncut mulch]

alternatively if you have an equation, you can add that value to both sides

[stray dagger]

so is it y^2+2y+1

[viscid thistle]

I suggest ignoring the formula and thinking about it

[uncut mulch]

please try to stop using "it"

as it is unclear what it refers to

[viscid thistle]

what if I wanted it to have multiple meanings

im joking alright 👍

thanks for that

[uncut mulch]

so is it y^2+2y+1
also no. that isn't anything you should have in any context here

[stray dagger]

ok so im confused what im suppose to be doing then

[uncut mulch]

read that summary above
gives a decent intro to completing the square

and/or i'll guide you through it if you prefer

[stray dagger]

i understand the b/2a and then adding subtracting it but i need a guide for it

[uncut mulch]

just stating b/2a is also meaningless without full context of how you're using it

did you read the guide i made above?

[stray dagger]

yes

[uncut mulch]

so currently you need to complete the square for
y^2 - 2y

what would be the b value?

[stray dagger]

2

[uncut mulch]

not quite

[stray dagger]

-2?

[uncut mulch]

yes

[stray dagger]

oh so a is 1

then the square is -1

[uncut mulch]

just stating b/2a is also meaningless without full context of how you're using it
completing the square is slightly more complex if the coefficient of the ^2 term isn't 1

then the square is -1
NO

there isn't any need to involve a here

don't jump ahead

adding (b/2)^2 = (-1)^2 = 1 would complete the square for y^2 - 2y

based on the binomial expansion presented above

however you would need to +1 - 1 or add 1 to both sides of the equation

[stray dagger]

ok so wouldent that just make it x + 1)2 + y2 – 2y + z2 + 1 = 0

by adding 1 and subtracting on the other side

[uncut mulch]

$(x+1)^2 + (y^2 - 2y {\color{red}{\ + 1) - 1}} + z^2 = 1 \
OR \
(x+1)^2 + y^2 - 2y {\color{red}{\ + 1}} + z^2 = 1 {\color{red}{\ + 1}}$

[obsidian monolith]

ramonov:

[stray dagger]

ok so once i have that i need to find the center and rsdius

[uncut mulch]

that then allows you to factorise y^2 - 2y + 1 to (y-1)^2

and then determine the properties with the equation in an appropriate form

[stray dagger]

so the center is at -1,1 and the radius is 2?

[uncut mulch]

no

this is 3-dimensional

you should use () when representing points

and r isn't 2

[stray dagger]

well isnt it 2 because u add +1 to the 1 on the other side of the =

ok so then the center should be at (-1,1,1)??

[uncut mulch]

just because the isolated integer on the right is 2, doesn't mean the radius is 2

check the z-coodinate

[stray dagger]

ok so (-1,1,-1)

[uncut mulch]

still no

note: $z^2 = (z-0)^2$

[obsidian monolith]

ramonov:

[stray dagger]

so the coefficent is 0?

[uncut mulch]

not "coefficient"

the z-coordinate of the centre (of the sphere) is 0

[stray dagger]

ok so the quardanant of the sphere is (-1,1,0)

[uncut mulch]

quardanant ?

[stray dagger]

center i meanmt sorru

sorry

cordinates ofd the center are

[uncut mulch]

yes, that will be the center

still need to fix the radius

[stray dagger]

ok so then the radius is =2^2 right?

[uncut mulch]

no

a sphere with the equation : $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ will have the center: $(a,b,c)$ and radius: $r$

[obsidian monolith]

ramonov:

[stray dagger]

ok so the radius is 2 then

[uncut mulch]

no

2 is the isolated integer on the right side

$r^2 = 2 \
r = , ?$

[obsidian monolith]

ramonov:

[stray dagger]

idk it cant be 1

[uncut mulch]

this is very basic algebra at this point

solve the thing above for r>0

[stray dagger]

so r = ^2

1.4

so the radius is 1.4

[uncut mulch]

no

$r = {}^2$

[obsidian monolith]

ramonov:

[uncut mulch]

doesn't make sense

if you wanted to indicate the square root of 2

[stray dagger]

sqrt2

is what i mean

[uncut mulch]

you could write sqrt(2) in plain text

but that definitely isn't 1.4

and you should just leave r = sqrt(2) for your radius

[stray dagger]

ok

Thanks for your help

[small cedar]

$r=\sqrt{2}$

[obsidian monolith]

Orange905:

[wide ocean]

" If P(x)= -4x3 + 8x2 - 8x -3 is divided by (x-4), what is the remainder? "
A) P(x - 4)
B) P(4)
C) P(x + 4)
D) P(-4)
I did synthetic division and the remainder I got the first factoring was -163. Not sure why the answer is "B". Isn't remainder the number after synthetic division? Why is it the zero of the divisor? or is this coincidence?
thank you, I am really stumped

[jade heron]

remainder theorem

@wide ocean

If $(x-a)$ is a factor of $f(x)$, then $a$ is a root of $f$. Otherwise, the remainder given when $f$ is divided by $x-a$ is $f(a)$.

[obsidian monolith]

L'Âne 🍐:

[wide lynx]

what does it mean to illustrate

[pale bison]

by plotting the point maybe

[wide lynx]

ok

is there any way to do this without a derivative