3, 9, 27..
#help-23
twin prawn
fuck
wait
without recursion this is easy
a = 3; r = 3^n
this is a geometric sequence
but we're finding for recursion wait
i think i got it
(a[n - 1] * 3^(n)) + 2
frozen marlin
no it's
really close
again: look at the differences
11 - 5 = ?
29 - 11 = ?
83 - 29 = ?
twin prawn
6, 18, 54 respectively
frozen marlin
(a[n - 1] * 3^(n)) + 2
if you have actually there's just a minor mistake here
twin prawn
difference increases by factor of 3
twin prawn
(a[n - 1] * 3^(n)) + (2 * 3)
frozen marlin
(a[n - 1] * 3^(n)) + 2
again, minor mistake
twin prawn
(a[n - 1] + (3^(n) * 2)
11 = (5 + (3^(1) * 2)
11 = (5 + (6)
29 = (11 + (18))
i got it
thanks guys
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vague phoenix
Guys just 1 quick question how can c be explained?
safe radishBOT
Guys just 1 quick question how can c be explained?
delicate shore
continuous functions have the intermediate value property
vague phoenix
Hmm
This implies whether two distinct point a1 a2 such that f(a1) is differ from f(a2) must be a point between them?
dull sequoia
i'd say suppose f(x) ≠ f(y) for some x, y \in [0, 1], then you can argue that there's some point between f(x) and f(y) that doesn't lie in Q
so that contradicts what f is, so it can't be true that there eixsts some x, y in [0, 1] such that f(x) ≠ f(y) but then that means f(x) = f(y) everywhere so it's constant
vague phoenix
So if fx differs from fy, and since fx and fy are already rational so a value, say z, lies between them must be irrational?
vague phoenix
Then since continuous a point l must in domain interval x y such that fl=y?
dull sequoia
i mean if you just pick you might happen to pick a rational point
vague phoenix
Hmm
dull sequoia
continuous functions have the intermediate value property
the "you can argue..." uses this property
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raven heart
<@&268886789983436800>
magic junco
@modera
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austere goblet
three channels?
raven heart
three channels?
fastest channel switcher in the west
austere goblet
I thought the help channel limit is 2
guess the bot couldn't apply the role fast enough
frozen marlin
actually wild
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atomic parrot
Im stuck
atomic parrot
Because the centre is 0,0
smoky sandal
if you have two points (x1,y1) and (x2,y2) slope is defined as (y2-y1)/(x2-x1)
smoky sandal
Because the centre is 0,0
it is but they do not say that the line passes through the center
smoky sandal
How would i find change in y/change in x then
you need the y coordinate of Q
atomic parrot
How i get that
smoky sandal
do you know general form of equation of a circle?
smoky sandal
yeah that would be equation of a circle with center 0,0 and radius r
you can use the fact that point P and Q lie on the circle so their coordinates should satisfy equation of the circle
atomic parrot
So 5^2 +10^2 =r^2 ?
smoky sandal
yes find r
atomic parrot
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twin prawn
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wary cape
ye
twin prawn
In the gif above?
yup
neon locust
Its just simplifying the given term
twin prawn
Its just simplifying the given term
yeah, but we have a x^4 and skips x^3
midnight cypress
how is it x^3 + x^2 + x + constant?
a degree 4 polynomial divided degree 1 polynomial has to produce a quotient of degree 3 polynomial
neon locust
We have the divisionn
rich gazelle
Think about we are reducing the polynomial with the division.
neon locust
Nice
rich gazelle
i see... i think i get
😉
twin prawn
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pulsar river
fuck
open wedge
First question is a swear word👍
Fr
open wedge
First question is a swear word👍
Yep, this is called channel wasting
safe radishBOT
Asking the actual question right away is more likely to get responses.
Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.
If you are done with this channel, please mark your problem as solved by typing .close
hard ledge
Idk how to solve for n in these factorials
safe radishBOT
Idk how to solve for n in these factorials
manic patrol
hint: consider the definition of factorials, then consider what terms are in the larger factorial that are not in the smaller one.
manic patrol
start by considering the definition of factorials. what does n! mean?
manic patrol
then what is 7! ?
hard ledge
7654321=5,040
manic patrol
so you are multiplying by every single integer on the way down to 1. agreed?
manic patrol
then, it would make sense that n! = n x (n-1) x (n-2) x ... 3 x 2 x 1. agreed?
start from n, multiply everything on the way down until you reach 1.
hard ledge
Awwww yeh it looks more complicated when it’s like that but yeh that makes sence
manic patrol
cool, that's all we need to get started. let's do the bottom one first, as it's more obvious.
hard ledge
Ok
manic patrol
the left side of the bottom question has $\frac{(n+3)!}{n!}$.
so, the top factorial starts multiplying from (n+3) down to 1, and the bottom factorial multiplies from n down to 1. agreed?
flat frigateBOT
trilunar arithmetic (Columbina)
hard ledge
Ok, yeh following
manic patrol
so, if we write the factorials out as products...
$\frac{(n+3) \cdot (n+2) \cdot (n+1) \cdot \mathbf{n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1}}{\mathbf{n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1}}$
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
notice something about the bolded parts?
hard ledge
Yeh they’re the same, so we can cancel them out?
hard ledge
(N+3) (n+2) (n+1)
manic patrol
correct.
now, note that factorials are only defined for non-negative integers (that is, the number before the ! must be 0 or higher).
then, (n+1), (n+2), and (n+3) must definitely be positive integers, I'm sure you agree.
now, do you agree that (n+1) to (n+3) are consecutive positive integers?
hard ledge
Yes 👍🏻
manic patrol
so, can you find three consecutive positive integers that multiply to 24 (the right side)?
hint: think small.
hard ledge
4,3,2?
hard ledge
N=1??
manic patrol
correct!
if unsure, you can check against the original question.
indeed, 4!/1! = 24/1 = 24.
so yes, n = 1 indeed for the bottom question.
hard ledge
Ok ok
manic patrol
can you use this to do the top one?
hard ledge
I’ll try
manic patrol
alright. I'll be around.
hard ledge
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golden wren
Hi everyone, hope you guys are having a great night.
I am in grade 11 having trouble with the unit based on domain and range. Are there any tips someone could give me?
manic patrol
hi OP, what in the unit are you having trouble with?
golden wren
Hi, hope all is well 🙂
Mostly simple domain and range, combinations of transformations, reflections and properties of functions as well.
manic patrol
mm, could I trouble you to please be a little more specific? like, maybe what about domains and ranges do you not get?
for instance, do you:
- not know what domains and ranges are,
- know what domains and ranges are conceptually but have trouble applying it in questions,
- know what domains and ranges are conceptually but fail to see some rigor behind them,
- have a misunderstanding between ranges and codomains,
- have some other misunderstanding not stated?
golden wren
For the domain and range questions such as these i am struggling on.
golden wren
for instance, do you:
- not know what domains and ranges are,
- know what domain...
Im just not understanding the concept for some reason
manic patrol
then let's start from domains.
you have probably heard a thousand times that all functions need a domain for them to make sense.
what do you understand about what a domain is?
and bonus question that can help: what do you not understand about what a domain is?
golden wren
then let's start from domains.
you have probably heard a thousand times that all...
I understand that its a set of all possible numbers (all real numbers)
Sorry im not explaining it right
Im not understanding the basics of how to get the domain and range of a graph as well
manic patrol
you are right in that the domain is a set of numbers. not so right on how it relates to functions though.
the domain of a function is the set of numbers allowed to be used in the function, simply put.
for example, take a function f(x) = 2x + 3. this looks good, but you haven't technically specified what numbers you're allowed to use for x. this is important because it determined what numbers f(x) itself can and cannot produce.
for instance, if you allow x to only be integers (the same as saying that the domain of f(x) is all integers) then one look at the function tells you that you will only ever get integers out of it.
golden wren
the ending product has to include the number as the x in the f(x) = correct?
golden wren
for instance, if you allow x to only be integers (the same as saying that the do...
I see
manic patrol
I'm sorry, I don't fully understand what you meant by 'ending product'.
golden wren
For instance lets say f(6) and the equation is y = 2x - 1
To solve youd put the 6 in replacement of the 6 making the answer 11
So would it be f(6) = 11?
manic patrol
okay I see what you mean. yes, you can say f(6) = 11 or f(x) = 11 when x = 6. of course, the former is preferred.
then, most of the time, you want a function to have as large a domain as feasibly possible.
so what does feasibly possible mean in this case?
suppose you have f(x) = 1/x.
you know from third grade arithmetic that you cannot divide by 0.
well, if x = 0 here, then we're in a whole lot of trouble, aren't we?
so x = 0 is a forbidden value, also known as a domain restriction for f(x), because using that value causes headaches we really do not want to deal with in elementary algebra.
likewise, any values of x that cause the argument of a log to be negative or the portion under a square root to be negative (basically anything that would cause problems) are forbidden / restricted from the domain of the function.
once you have removed all such forbidden values, if you impose no further arbitrary restrictions, then you have created the largest domain feasibly possible for f(x).
(this is called its natural domain.)
manic patrol
Im not understanding the basics of how to get the domain and range of a graph as...
this connects back to this.
when you are told to find the domain of a graph, what you really want to find are values of x for which the graph is known to exist.
and that is essentially what function domains are.
I'm sorry if I lost you; please let me know if anything in there is still confusing.
golden wren
nono im here, thank you so much for doing this! I'm understanding a lot more so far
manic patrol
so are there any further questions about domains, or do you want to move on to another subtopic?
golden wren
Just 1 more question if u don't mind, when graphing an asymptote does it have to be exactly accurate or do u think i would be able to get away with full marks by just drawing the lines to approximate of what I think?
manic patrol
are you graphing on graph/grid paper or blank space?
golden wren
For the domain and range questions such as these i am struggling on.
The graph structure would be identical to this on a quiz/test
manic patrol
so you are graphing on a grid.
then less leeway is allowed but I imagine that if it was a couple of millimeters off your teacher wouldn't say much...?
I'm not your teacher though - you should ask them, they're the ones grading.
of course, if your asymptote is a negative value but your graph shows that it's clear on the other side of the axis then it wouldn't fly against any examiner.
golden wren
Got it
Sorry i've kept you a bother for quiet some time..
How much longer are you available for?
manic patrol
at least an hour more, but there are other helpers too.
also don't worry too much about it.
golden wren
Would it be okay if we quickly review one more topic? You've been superbly helpful and i feel guilty taking the time out of your day.
If that's okay please let me know :).
manic patrol
sure. you don't have to ask for permission!
golden wren
Perfect, thank you so much again!
For an example, i do have a quiz upcoming very soon and a big part of marks is writing the domain and range. I am super confused on how to put these into the actual written form if this makes sense.
I'm sure this is very easy but i'm not fully understanding...
Questions will most likely be more difficult to the ones on screen, but if I can figure out the basic structure i should be able to do it
manic patrol
then let's take two very quick examples that demonstrate the two different ways that you need to consider.
suppose $f(x) = \frac1x$.
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
now, the base domain we are working off of is always the set of real numbers unless specified otherwise.
so you can always start off by writing ${ x \in \mathbb{R} : ... }$.
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
the problem, then, is what goes in the ... space.
that is where our conditions will go to.
manic patrol
**trilunar arithmetic (Columbina)**
now, take a look at this function and tell me what value(s) x cannot take.
golden wren
It can't take 0 since it's indivisible by 1?
manic patrol
right answer, shaky reasoning.
the reason is simpler than you've phrased it - division by 0 is undefined.
golden wren
Got it
manic patrol
and by using x = 0, we are creating a division by 0 situation, so that's a no-no.
so we now need to restrict x = 0 from showing up.
if x cannot be equal to 0, what is the symbol you want to use to show this?
golden wren
The equal sign with the slash over it i believe
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
so then, x = 0 is restricted, and the condition then becomes $x \neq 0$.
thus, the domain of $f(x)$ is ${ x \in \mathbb{R} : x \neq 0 }$.
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
this is read as "all real x, except x = 0".
or, "all real x that is not x = 0", if you insist on a reading with the word not in it.
so that's how you notate a single-value restriction. some functions, however, are not so kind and will force you to restrict a range of values.
golden wren
I remember reading something about when writing u have to add "such that"..
What does that mean exactly?
manic patrol
the : here is taking the place of "such that" in this case.
"such that" here means "following these conditions".
for example, that domain above can be read as "all real x such that x is not 0".
if you have to use the phrase "such that", replace the colon with the phrase.
and by the way, if your function forbids several single values, you can add commas after the first excluded value, like so: ${ x \in \mathbb{R} : x \neq 0, 2, \pi, -\frac{1}{12}, ... }$
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
now, let's take a look at domains of functions that restrict a range of values. \
consider $g(x) = \sqrt{x}$. what value(s) can $x$ not take?
flat frigateBOT
trilunar arithmetic (Columbina)
golden wren
Would it not be able to take any odd numbers since u cant square root?
manic patrol
not really. sqrt(9) = 3.
golden wren
ah, brain fog lol
manic patrol
hint: we are working only in the reals.
manic patrol
correct! how would you notate negative numbers in one go?
golden wren
ahh sorry i'm a little stuck. Do you mean how would u write it in the brackets? Like =/0?
manic patrol
something like that, yes.
you won't be using an equal sign here, that much I can tell you.
safe radishBOT
@golden wren Has your question been resolved?
golden wren
Hi im so sorry my phone died..
safe radishBOT
golden wren
Oops
manic patrol
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safe radishBOT
✅ Original question: #help-23 message
golden wren
.reopen
manic patrol
no worries, you may continue.
golden wren
It would be the - and < right?
manic patrol
can you please write out what you mean?
golden wren
{XER | x - < 0, } or something?
manic patrol
do you mean to write $x \leq 0$?
flat frigateBOT
trilunar arithmetic (Columbina)
manic patrol
or do you actually mean $x -< 0$, which is bad notation?
flat frigateBOT
trilunar arithmetic (Columbina)
golden wren
Yes sorry
manic patrol
**trilunar arithmetic (Columbina)**
this is incorrect. yes, x = 0 works, but you are saying that the domain is all real x less than or equal to 0.
this is a bad domain - almost every single value here will cause the square root function to be undefined in the reals!
you're on the right track though!
manic patrol
still no, and even worse.
now, your only valid value is gone, and every single value of your new domain causes the function to be undefined.
a reminder that x cannot be negative.
manic patrol
yes! but remember to include 0.
golden wren
Got it!! Thank you so much and sorry i keep responding late my phone is acting up too much
manic patrol
to recap, the domain of $g(x)$ is ${x \in \mathbb{R} : x \geq 0 }$.
flat frigateBOT
trilunar arithmetic (Columbina)
golden wren
Perfect!!
Appreciated a lot, will be practicing. Have a great rest of your day/night!!
manic patrol
you too, OP. anything else?
safe radishBOT
@golden wren Has your question been resolved?
soft lava
I know I should use a conjugate here I just don’t remeber what my teacher said . She said either use the one with the square or without square sin
safe radishBOT
I know I should use a conjugate here I just don’t remeber what my teacher said ....
vague phoenix
you want to simplify this equation or?
you can just use (sec^2x -1 ) into (secx -1 )(secx +1)
vague phoenix
then you simplify secx-1
soft lava
well in theory if I was going to conjugate would I use the part with or without the square
vague phoenix
both numerator and denominator
vague phoenix
well in theory if I was going to conjugate would I use the part with or without ...
you could use whatever you comforts with
as long as it solve the question, but if the question explicitly ask for it, then follow its step
vague phoenix
thats correct
you can do by your way, not wrong
then it will be (secx-1)(secx+1)/secx-1
and cancel out secx-1
leaves secx+1
soft lava
How would I know if I would make it + or -
vague phoenix
well you apply a^2-b^2 method
soft lava
Wdym?
vague phoenix
How would I know if I would make it + or -
what do you mean by this first hand
soft lava
Oh wait I see one must be postive because two negatives would make a positive
safe radishBOT
@soft lava Has your question been resolved?
winged flare
which 1
soft lava
Wdym
winged flare
Where did this extra one come from
which is the extra one here
soft lava
I thought it was the highlighted one but I think was able to do it when I made it a box instead
soft lava
I don’t understand how we are keeping the highlighted thing if we are find the gcf
I really don’t understand gcf I tried googling but no videos show it with sin and cosine
How did they switch it from multiplication to addition
Ok I think I finally go the gcf for this one I got confused because of invisble 1
But now I don’t understand how they seperated these fractions
vague phoenix
this one splitting cos^2x into 1-sin^2x wad unnecessary
vague phoenix
But now I don’t understand how they seperated these fractions
This one correct
safe radishBOT
@soft lava Has your question been resolved?
lavish magnet
I'm probably just being paranoid but this is a valid way of solving for x right? This seems redundant to me for some reason.
safe radishBOT
I'm probably just being paranoid but this is a valid way of solving for x right?...
vague phoenix
thats correct tho
lavish magnet
ohh i thought it was weird because I had a sqr inside a bracket to the power of 3 and that whole thing getting cube rooted
open wedge
ohh i thought it was weird because I had a sqr inside a bracket to the power of ...
nw
are you allowed to use a calculator?
lavish magnet
are you allowed to use a calculator?
We are
I'd like to know how to do it myself tho
open wedge
i can give you a way that's kinda shorter
lavish magnet
Sure feel free
open wedge
so $x^{\frac{3}{2}} = 8$
flat frigateBOT
1 divided by 0 equals Infinity
flat frigateBOT
1 divided by 0 equals Infinity
lavish magnet
so the reciprocal of 3/2?
open wedge
so the reciprocal of 3/2?
yea you can say that
so RHS ig you can calculate immediately
using a calculator
lavish magnet
so multiply the RHS to the power of 2/3 in my calculator right?
Hmm thats quite neat
gets the answer much faster
Thanks for sharing @open wedge
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open wedge
nw
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lavish magnet
How is ACB an obtusce angle here? And in the future how am I supposed to know if an angle is obtusce or not just be looking at it? It wasn't explicitly stated in the question here, but the answer had ACB as an obtusce angle.
safe radishBOT
How is ACB an obtusce angle here? And in the future how am I supposed to know if...
lavish magnet
sinA/BC = sinC/AB
the answer comes out to 52.5 to 1dp
thats less than 90 so thats an acute angle
thats what i did when solving the question actually
lavish magnet
Maybe the image just for demonstration?
Well how else am I supposed to solve the question without using the img lol
vague phoenix
Not to scale...
lavish magnet
I don't understand
vague phoenix
Ahh i see why
The 52.5 degree here is angle ADB
You suppose to use sine theorem on triangle ADB
Then you find ADB
lavish magnet
I'm still a little lost
smoky sandal
,, sin\theta = sin(\pi- \theta)
flat frigateBOT
ghost
vague phoenix
But BC=BD then BCD isoceles
lavish magnet
You use sineA/BD = SineD/AB
So that Sin(35)/4.7=Sin(D)/6.5 right?
vague phoenix
Correct
lavish magnet
Remind me the law for isoceles triangles again if two sides have the same length they also have the same angle iirc?
lavish magnet
Okay I understand now, but in the future if i come across a question like this on my test how am I supposed to know if the angle I am looking for obtusce or not? Is there like a general rule I can follow or do I just have to pay attention
Since usually its stated in the question, but for this question it was not.
vague phoenix
Just use sine in biggest triangle
This question is really tricky, i fell for it too
Yea, just pay attention
lavish magnet
Gotcha, thanks for your help 
This discord server might be the only reason I'm not gonna fail on my exams i swear 🤞
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safe radishBOT
vague phoenix
Gotcha, thanks for your help <:thumb_rat:1282878538063544330>
This discord serv...
Me too it helped me
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Available help channel!
daring sonnet
How do I do part ii
fickle mantle
Question3 pls
safe radishBOT
Question3 pls
lone void
what did u do here?
thin bridge
why did the 4 disappear
fickle mantle
Bc it’s multiplying 0
lone void
🤔 could u elaborate? what is multiplied by 0
neon locust
🤔 could u elaborate? what is multiplied by 0
he means the 4
fickle mantle
4
neon locust
u dont do that tough
lone void
the question is 4sin(theta) -3 = 0 not 4sin(theta-3)=0
fickle mantle
My teacher said we leave the -3 by the theta
lone void
oh alright
thin bridge
the 10 is indicated to be part of the argument there
lone void
My teacher said we leave the -3 by the theta
are u sure about this?
fickle mantle
smoky sandal
,w arcsin 0.3 to degrees
flat frigateBOT
thin bridge
Not related to erasing the coefficient of the trig function
the first goal is to get
trig(angle) = value
with algebra
no different to solving
4t - 3 = 0
you can't just erase the 4 and say t = 3
@fickle mantle
neon locust
no different to solving
4t - 3 = 0
you can't just erase the 4 and say t = 3
so just 0.75
thin bridge
so just 0.75
Wasn't asking you
neon locust
sry for answering bro
fickle mantle
no different to solving
4t - 3 = 0
you can't just erase the 4 and say t = 3
Huh
Oh I have to do something rq
safe radishBOT
@fickle mantle Has your question been resolved?
summer coral
would someone be able to check my work on this?
safe radishBOT
would someone be able to check my work on this?
quasi goblet
would someone be able to check my work on this?
gimme 2 to check
frozen marlin
would someone be able to check my work on this?
seems to be fine
safe radishBOT
quasi goblet
uhm
quasi goblet
would someone be able to check my work on this?
(−3u^2,4uv,2u)
frozen marlin
no it's a double minus
the j component is negated anyway + there's one because it's the second term in the determinant
they cancel out
also, that wouldn't matter here, but yeah that bit's confusing
took me a minute lol
rich gazelle
they cancel out
Yeah, it don't take effects in the result.
frozen marlin
again: still had to check
and i appreciate fixer for that
but yeah that bit's annoying
quasi goblet
ok mb ik it doen't affect final result
summer coral
(−3u^2,4uv,2u)
oh right i forgot about i-j+k thanks but yeah its multiplied by 0 eventually anyways
safe radishBOT
Available help channel!
primal wasp
What am I doing wrong dude
safe radishBOT
What am I doing wrong dude
plucky elk
What am I doing wrong dude
what even is the problem
safe radishBOT
What am I doing wrong dude
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
primal wasp
what even is the problem
primal wasp
Please show the original problem, **exactly as it was stated to you**, with the ...
Oki
fathom jewel
he meant the original problem
like a photo from the textbook or where else you got it from
plucky elk
what are S_n and b_n supposed to be
plucky elk
mistake here. you didn't multiply by 3/4 correctly in the numerator
main mural
Here are the formulas
sakartvelo mentioned 🗣️🇬🇪🔥
primal wasp
sakartvelo mentioned 🗣️🇬🇪🔥
Let me complete it first
I'm still getting it wrong what the helly
The answer is supposed to be 1y
16
safe radishBOT
@primal wasp Has your question been resolved?
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lilac pewter
I dont want to look
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I dont want to look
lilac pewter
Want to know whys orthogonal projection minimize distance between point and line
Im thinking its probably a triangle inequality thingy
And like the degenerate triangle is formed by the perpendicular which is when triangle inequality is achieved
And your point is x and your line is ay where x and y are vectors
So trying to say min_a |x-ay| is given by |(I-yy’)x| i think
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@lilac pewter Has your question been resolved?
lilac pewter
Yeah probably
I need to write the idea down i think
Like |x+y|<=|x|+|y|
need to pick x and y to make sense in the analogy
So i guess I can treat it as x+(-ay)?
Nah this is wrong
crisp maple
confused about the solution of b)
I feel like I'm really close, but I have an extra sqrt(3)?
Its like the dots are there but I cant connect them rah
safe radishBOT
confused about the solution of b)
I feel like I'm really close, but I have an ex...
fathom jewel
what is the factor next to (2n+1) in the second line
oh wait nvm 3^n
though why did write arctan(x)
,w arctan(1/sqrt(3))
solar hazel
beautiful handwriting
safe radishBOT
@crisp maple Has your question been resolved?
fathom jewel
i dont see an error, you made me doubt myself like 10 times
you just wrote arctan(x) instead of arctan(1/sqrt(3)) but that's it
crisp maple
well thats the thing Im a bit confused about
fathom jewel
uhm why?
fathom jewel
it does, you should not divided by it both sides
crisp maple
Ik x should be 1/sqrt 3 but at the same time that feels wrong ig?
fathom jewel
arctan(1/sqrt(3)) = pi/6 but you have the factor sqrt(3)
so you end up the same with the solution
pi/6 * sqrt(3)
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fathom jewel
but ok one thing that confuses me is that you set it equal to arctan(x)
blud disappeared
crisp maple
well I guess x is pretty obvious
fathom jewel
blud appeared
crisp maple
but uhhh I wasn't sure so I left x there
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vagrant steppe
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clever tinsel
there are wrong signs on 10 and 12.
11 is correct.
vagrant steppe
I'm confused on my signs for 10 and 12
clever tinsel
you factored a -1 from the expression, which flipped the sign of the constant term from negative to positive.
however, you then factored the quadratic into two binomials with differing signs on the constant terms, ensuring that you will get a negative constant when you expand it back out.
runic torrent
Can anyone help me w this? Im really confused and im struggling sm
clever tinsel
this channel is occupied atm, unfortunately.
there's #help-44｜stanley-🌲-v2-dans which is open right now.
clever tinsel
you factored a -1 from the expression, which flipped the sign of the constant te...
anyway, unless that + on the 8 is a past mark from erasing, which might be possible considering you have -12x and I don't think you'd be factoring it that way if you saw the signs.
vagrant steppe
you factored a -1 from the expression, which flipped the sign of the constant te...
So both signs are wrong?
clever tinsel
if the +8 is a -8 then you are right, otherwise that +8 is wrong.
vagrant steppe
It's a +
clever tinsel
if the +8 is a -8 then you are right, otherwise that +8 is wrong.
then as mentioned here, that +8 is incorrect.
when expanding the quadratic back out you find that you would have -32 instead of +32 and +4x instead of -12x.
vagrant steppe
Shouldnt both be + the 8 and the 4 so then when you multiply them you get positive 32 and then -22
clever tinsel
then where will you get the -12x from?
vagrant steppe
-8 -4 = -12
clever tinsel
but you said both are +. if both are +, then I suppose you mean (x + 8)(x + 4)?
vagrant steppe
Yeah because when the - goes through they change
clever tinsel
what do you mean by, "when the - goes through they change"?
as I see it, you've already accounted for the factoring out of -1 by flipping the signs of the terms in the quadratic, which is correct.
so now you can just ignore that -1 out the front and focus on the inner quadratic, remembering to stick the -1 back in during the final answer.
vagrant steppe
clever tinsel
long story short, your current quadratic, after factoring out -1, is x^2 - 12x + 32.
vagrant steppe
I thought you like mentally did that and thats how you decide your signs, basically the opposite
clever tinsel
if you do it that way, then the original sign of the linear term is positive.
two +s will definitely give a positive linear term, but when you multiply the -1 back in you get a negative linear term, which is not what the question had.
also, your squares are missing from the x^2 terms after you factor the -1 out.
vagrant steppe
Oh I put them?
clever tinsel
I don't presume you have started the quadratic factoring at those steps, so yes, those squares should still be there.
(in fact I find this style of working a little confusing.)
clever tinsel
then you will see things like x - 12x and wonder why you hadn't simplified it already or where the quadratic even went to.
clever tinsel
okay, let's go through the whole thing with your proposed solution and see what happens.
vagrant steppe
I'm sorry I'm just confused
clever tinsel
you proposed:
$$-(x + 8)(x + 4)$$
as a factorization of
$$-x^2 + 12x - 32$$
expanding the binomials back out gives us $-[x^2 + 12x + 32]$.
then, multiplying the -1 back in gives $\color{green}{-x^2 } \color{red}{- 12x } \color{green}{- 32}$.
note the sign of the linear term.
flat frigateBOT
Yukari
vagrant steppe
Oh wait
I was thinking I was trying to get this
But it's all the way to the original?
clever tinsel
if you had indeed wanted to match coefficients with the quadratic with the -1 factored out, then you proposed: $$(x + 8)(x + 4)$$ as a factorization of $$x^2 - 12x + 32$$.
you can expand the factored form back out to confirm that you indeed have the sign of the linear term wrong again.
flat frigateBOT
Yukari
clever tinsel
I'll give you a tip for problems like this.
you decided to factor out a -1 to make $x^2$ positive, which is a good move. but once you have factored it out, you can just pretend it does not exist. why?
because suppose you have $-(x^2 + ax + b)$, and you factored it to $-(x + c)(x + d)$.
note that both the factored form and the original form have -1 outside already. since the factor of -1 is present in both forms, it suffices to check that the quadratics in both forms match (that is, expanding (x + c)(x + d) back out gives you $(x^2 + ax + b)$), because once they match, you guarantee that they are really the same quadratic.
flat frigateBOT
Yukari
vagrant steppe
So I shouldn't focus on that affecting my signs then?
clever tinsel
no, unless you somehow decided to multiply the - sign back in for your final answer, which is not a really good move.
a simpler visualization is this.
vagrant steppe
So ideallly I should be looking at my original equation not what I had circled here to try and match
clever tinsel
ideally you want to match these two if you're doing it this way.
vagrant steppe
So my signs should match that
clever tinsel
correct (without the -1).
vagrant steppe
So -(x-8) (x-4)?
clever tinsel
correct.
clever tinsel
yes.
clever tinsel
vagrant steppe
So I do look at that equation and not the original?
clever tinsel
yes, because you factored out a -1 and your final answer still has that -1 implicitly outside.
clever tinsel
please don't apologize. it's alright.
clever tinsel
correct.
clever tinsel
yes, but if you don't trust anyone else, you can multiply back out just to be sure.
clever tinsel
sure thing. I'll be around for another hour or so.
earnest nacelle
wtf miss Yukari helping 
vagrant steppe
sure thing. I'll be around for another hour or so.
I figured it out, it was a simple one
clever tinsel
well done! anything else I can help you with?
vagrant steppe
Thank you so so much for everything 💗 truly
clever tinsel
I'm glad to help!
vagrant steppe
Nope!! I appreciate it lots! Sorry for making it so confusing
clever tinsel
no worries! we've all been there at some point.
vagrant steppe
I appreciate it always
clever tinsel
if you have nothing else, then all the best up ahead, and see you around!
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exotic crystal
How can I do this? should I do it by cases? because today is not labor day, so the implication is true, but there are days when it's labor day but tomorrow is not tuesday
safe radishBOT
How can I do this? should I do it by cases? because today is not labor day, so t...
light shoal
"but there are days when it's labor day but tomorrow is not tuesday" - example of such a day?
exotic crystal
oh, I can't find any on the calendar lol
oh wait
2028
may 1st
at least in mexico
plucky elk
Why is the problem in English if you're in Mexico
static lava
Can anyone help me with this?
exotic crystal
Why is the problem in English if you're in Mexico
I'm using the book in english because it looks better, sorry for the lack of context
plucky elk
Is the book written for a Mexican audience?
exotic crystal
USA audience
my teacher just uses it and tries to adapt it to our context
but still this year's labor's day in the USA will be on Monday
light shoal
USA labor day is always on a monday, fwiw
light shoal
first monday in september, to be exact
exotic crystal
In Mexico is on May 1st always, that's why I was so confused
Ig the translators didn't account for that because the statement is exactly the same
light shoal
yea that's true in europe too, i think
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sleek hornet
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vague phoenix
from line 2 it supposed to be p and not(not(p or q)
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steep magnet
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mossy lotus
its S(T(x)) in that order
steep magnet
Also what do they mean by this line
mossy lotus
so if you use the definition of S and T, you get S(T(x)) = S(Ax) = BAx
mossy lotus
Also what do they mean by this line
column vector (1,0)^T becomes (0,1)^T when you apply the transform ST
and vice versa
steep magnet
Hmmmmmm
I think I get it
I will try some more examples and hopefully it will click in my head
Thank you!
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steep magnet
How do we know that w =
a
b
0
safe radishBOT
How do we know that w =
a
b
0
steep magnet
How do we know that w =
a
b
0
like i get that theres a 0 in the last spot of all the vectors in the span but how does that explain this ^^
mossy lotus
do you see why the two spans are the same?
steep magnet
do you see why the two spans are the same?
They just row reduced, right?
mossy lotus
well, yea. The linear combinations of the spanning vectors cover the same space. Thats W.
quasi goblet
the third coordinate is always 0 no matter what alpha and beta u pick so any w in W must have the form (a.b.0)
mossy lotus
So, like fixer said in his deleted message, w is an element of W, so it can be expressed as linear combination of those vectors
quasi goblet
yup
mossy lotus
since w is in span of e1 and e2, you say w is same as a*e1 + b*e2
quasi goblet
So, like fixer said in his deleted message, w is an element of W, so it can be e...
sorry botched it
mossy lotus
or of the form [a,b,0]^T
mossy lotus
sorry botched it
yea, nw
steep magnet
Ok so then what about z, how do we know this is the span
mossy lotus
coz z is in the complement of W
since W spans the first two rows, its complement must be 0 in those two rows
after all, the three spanning vectors together must span R^3
steep magnet
since W spans the first two rows, its complement must be 0 in those two rows
Ahhhhhhhhh I see
Dont think I would ever spot this in an exam 😅
But ok I get it now haha
Thank you guys!!
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open bridge
So, I got this inequality
x³ - 7x² + 5x + 1 > 0
I tried solving it 2 ways, both I failed
x²(x - 7) + 5(x + ⅕) > 0
Stuck at what next?
x(x² - 7x + 5) + 1 > 0
Δ = 29
sqrt(Δ) = sqrt(29)
``` positive, so should be possible to break it down further, but again I don't know how
So, what is the correct way to solve this?
broken yew
that's a cubic
often is the case, when you're given those, there's a trivial root enabling you to factorize
assuming that would help with the question
trivial meaning an integer between -5 and 5
generally when you're given polynomials of degree greater than 2, this is something you can quickly try
open bridge
often is the case, when you're given those, there's a trivial root enabling you ...
Could you please show an example of how it looks like?
I am having trouble with finding a way to apply it here
slate current
You evaluate the polynomial in 0,1,-1,2,-2 (and maybe a bit more), and if you find a root then you know you can factorize
broken yew
Could you please show an example of how it looks like?
I am having trouble with ...
have you been shown the root theorem
or factor theorem
open wedge
Could you please show an example of how it looks like?
I am having trouble with ...
let's try $x^3 - 6x^2 + 11x - 6$
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
we will only focus on factorizing this guy
so you can see what's going on here
is that alright?
@open bridge
open bridge
Sure
open wedge
So, I got this inequality
`x³ - 7x² + 5x + 1 > 0`
I tried solving it 2 ways, bot...
with a polynomial like in your inequality, you could do the same thing
open bridge
or factor theorem
No
Yes
open wedge
Sure
so normally, when we factor polynomials like these, we would want to find the roots of it
and as people mentioned, one way is to use the rational root theorem
consider each factor of -6
try x as each of the values in that and if it results in 0, there's one factor of it
open wedge
**1 divided by 0 equals Infinity**
so if i try $x = 1$ for example, i get $1^3 - 6 \cdot 1^2 + 11 \cdot 1 - 6$ which is 0, so $x - 1$ is one of the factors
flat frigateBOT
1 divided by 0 equals Infinity
broken yew
so a sensible order people would normally try is
0, 1, -1, 2, -2, ... yes?
just to be clear
open bridge
And then you can divide the plynominal by (x - y) y being the factor and go on from there
open bridge
And then you can divide the plynominal by (x - y) y being the factor and go on f...
Am I seeing this right?
broken yew
yes that's what you need to do
open wedge
And then you can divide the plynominal by (x - y) y being the factor and go on f...
that's basically the idea
well if you factored correctly, you get $(x - 1)(x - 2)(x - 3)$
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
which is the query i asked google to expand for me
open wedge
So, I got this inequality
`x³ - 7x² + 5x + 1 > 0`
I tried solving it 2 ways, bot...
now how about the polynomial in here?
open bridge
Trying rn
open wedge
in this case, try all factors of 1
open bridge
I found the (x - 1)
open bridge
(x - 1)(x² - 6x - 1) > 0
So
x² - 6x - 1 > 0
Δ = 40
sqrt(40) = 2 sqrt(10)
x1 = 3 - sqrt(10)
x2 = 3 + sqrt(10)
(x - 1)(x - 3 - sqrt(10))(x - 3 + sqrt(10)) > 0
open wedge
aight ic what you did there
it's technically not wrong there, but normally we work with integer values, and usually avoid roots/surds in factorizations like this
open wedge
How can it be done without the roots/surds?
so $x^2 - 6x - 1$
broken yew
well in this specific question you need surds
flat frigateBOT
1 divided by 0 equals Infinity
broken yew
since you are solving an inequality?
open wedge
you can complete the square
open bridge
since you are solving an inequality?
Yup
broken yew
am i being dumb
open wedge
**1 divided by 0 equals Infinity**
we can do the complete the square here
instead of surds
open bridge
So just (x - 1)(x² - 6x - 1) > 0?
broken yew
well i mean maybe you can, but if you prime factorize completely into linear factor you have the answer immediately no?
was my thoughts
open wedge
well i mean maybe you can, but if you prime factorize completely into linear fac...
yea ik
that's one way of factoring
broken yew
but go ahead - more techniques never hurts
open bridge
well i mean maybe you can, but if you prime factorize completely into linear fac...
Same thought here
open wedge
but go ahead - more techniques never hurts
damn
so let's try to consider $x^2 - 6x - 1$
flat frigateBOT
1 divided by 0 equals Infinity
open bridge
So just `(x - 1)(x² - 6x - 1) > 0`?
But then I need to solve the thing with sqrt(10) anyway to get the final answer no?
open wedge
**1 divided by 0 equals Infinity**
you can try completing the square
but still, you should still have your $\sqrt{10}$ anyway
flat frigateBOT
1 divided by 0 equals Infinity
open bridge
What do you reffer to as completing the square?
open wedge
What do you reffer to as completing the square?
you make it into a square add or subtract a number
broken yew
quadratic = (linear term) squared plus constant
open bridge
you make it into a square add or subtract a number
Ohh
broken yew
x^2 + 8x + 5 = (x + 4)^2 - 11
open wedge
**1 divided by 0 equals Infinity**
so how can you do it here?
broken yew
i assume you are just making sure the asker can solve a quadratic without directly plugging in the formula yes? 
broken yew
**1 divided by 0 equals Infinity**
since this
open bridge
x² - 6x - 1
x² - 6x + 9 - 10
(x - 3)² - 10
open wedge
x² - 6x - 1
x² - 6x + 9 - 10
(x - 3)² - 10
yes!
so if $x - 1 > 0$ then $(x - 3)^2 - 10 > 0$
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
can you see what you can do here?
open wedge
Where is the
if x - 1 > 0 then (x - 3)² - 10 > 0
From?
well, you got $(x - 1)((x - 3)^2 - 10) > 0$
flat frigateBOT
1 divided by 0 equals Infinity
open bridge
Yeah
open wedge
so if $x - 1 > 0$, what happens to $(x - 3)^2 - 10$?
flat frigateBOT
1 divided by 0 equals Infinity
broken yew
Where is the
if x - 1 > 0 then (x - 3)² - 10 > 0
From?
If you have ab > 0
if a > 0, then b > 0
open bridge
ohh I see
open wedge
so $x - 1 > 0$, you can solve that
flat frigateBOT
1 divided by 0 equals Infinity
open wedge
$(x - 3)^2 - 10 > 0$, how do you solve this?
flat frigateBOT
1 divided by 0 equals Infinity
open bridge
I see I can use the $a^2 - b^2 = (a-b)(a+b)$ by doing $(x - 3)^2 - 10 = (x - 3 - \sqrt{10})(x - 3 + \sqrt{10})$
broken yew
**1 divided by 0 equals Infinity**
if that inequality was an equality sign
how would you solve it.
same way.
flat frigateBOT
Noroi
open bridge
**1 divided by 0 equals Infinity**
kinda lost here
open bridge
that result in the first factorization method <:KEK:586240877358350341>
thought it was the goal for a sec
broken yew
you are trying to solve for x
so try rearranging the inequality
so you have only x on one side
same steps as you would do for an equation
open bridge
I mean
I sure can do
(x - 3)² - 10 > 0 into
(x - 3)² > 10
broken yew
yes.
exactly
if you do a mental sketch of what this graph looks like you can quickly figure out the answer to this inequality
and in fact regardless of what method you use to factorise the original inequality you had, a graph sketch would help a lot
open bridge
my teacher even asks for a sketch as a part of the answer, this one inequality is one I did incorrectly on a test a while back
open bridge
if you do a mental sketch of what this graph looks like you can quickly figure o...
I don't really know how to do that
open wedge
and in fact regardless of what method you use to factorise the original inequali...
i don't rlly need that much
open bridge
also I must say thank you all so so so much!!!
I don't know how I would figure it out on my own
open wedge
im not at calc yet 
open bridge
oh that's what you mean
open bridge
this helps you solve (x-3)^2 > 10
how?
broken yew
,,(x-3)^2 > (\sqrt{10})^2
flat frigateBOT
broken yew
so you are dealing with
x - 3 ? +-sqrt10
but you just don't know which way the inequality sign goes with the ?
right?
if this was an equation
you know x - 3 = +- sqrt10
i mean, i guess the sketch is not necessary - you can do it algebraically
i just personally prefer it to quickly make sure my inequality is going the correct direction
algebraically, you can notice:
,,(x-3)^2 > (\sqrt{10})^2 \geq 0
flat frigateBOT
broken yew
,,\abs{x-3}> \abs{\sqrt{10}}
flat frigateBOT
broken yew
which gives you this i think?
im rambling a bit
This should bring you to the correct answer sketch or no sketch
open bridge
I can see how this gives me the zero places
broken yew
but with the original cubic if you had just used quadratic formula
i would definitely use a sketch to tell me how it must be
(x - a)(x-b)(x-c) > 0 ?
broken yew
(x - a)(x-b)(x-c) > 0 ?
the graph must go this way
and that tells me the regions of x where the function is above 0
or if the coefficient of x^3 is negative, then the graph must go the other way
=
So does that make sense about the overall approaches you can take to solve this?
open bridge
I would definitely have to try the second way a few more times but I get the first one rather well
broken yew
or if only 1 root is real, you know the graph can only intersect the x axis once
open bridge
so you are dealing with
x - 3 `?` +-sqrt10
I still don't really see how to use this without going back to (x - 1)(x - 3 - sqrt(10))(x - 3 + sqrt(10))
open bridge
yeah
flat frigateBOT
broken yew
so now we have this, and you do the inequality algebra to figure out the answer
(no need for a sketch of the quadratic)
open bridge
does $\abs{\sqrt{x}} = \sqrt{x}$ ?
broken yew
i just didnt simplify
flat frigateBOT
Noroi
broken yew
yes it does, i just didnt write more steps
open bridge
fair enough
broken yew
,,\sqrt{10} < \abs{x-3}
open bridge
yeah I know how to solve it now
flat frigateBOT
strange flint
Wouldn't it have been easier to just factorize the the second parenthesis and check for positive intervals?
broken yew
ofc
open bridge
so
|x - 3|
x - 3 < sqrt(10)
or
x - 3 > sqrt(10)
x < sqrt(10) - 3
or
x > sqrt(10) - 3
x ∈ (-sqrt(10) + 3, sqrt(10) - 3)
``` ?I think I did something wrong
open bridge
yes it does, i just didnt write more steps
does this look any correct?
open bridge
okay
broken yew
and you've definitely got something wrong there
strange flint
Yeah
broken yew
,,\sqrt{10} < \abs{x-3}
flat frigateBOT
broken yew
,,\sqrt{10} < x-3
flat frigateBOT
flat frigateBOT
broken yew
you should be writing down these 2
open bridge
I see
hybrid spoke
Just square it and u will get 10 < (x-3)^2
broken yew
and that is going the opposite way to the solution?
hybrid spoke
No i am a idiit
open bridge
$\sqrt{10} < x-3$
$\sqrt{10} + 3 < x$
imagine extra newline here
$\sqrt{10} < -(x-3)$
$-\sqrt{10} > x - 3$
$-\sqrt{10} + 3> x $
broken yew
bro
hybrid spoke
Sowwy
broken yew
$\sqrt{10} < x-3$
$\sqrt{10} + 3 < x$
imagine extra newline here
$\sqrt{10} < ...
use double $ on each
ok
flat frigateBOT
Noroi
broken yew
yeah i think that looks correct
this is also on the original premise x - 1 > 0
for whatever reason
so you'd flip every single sign you see to check the opposite premise i suppose?
ive kinda lost tracked
(x - 1 > 0 and ... ) or (x - 1 < 0 and ...)
and figure out which one of these is impossible
(in the end this is obviously a long winded method)
strange flint
I'd summarize this in a table
open bridge
so now with this I get
x belongs to $(-\sqrt{10} + 3, \sqrt{10} + 3)$
broken yew
also, on the premise x > 1
broken yew
so now with this I get
x belongs to $(-\sqrt{10} + 3, \sqrt{10} + 3)$
no you don't
wrong way round
you've made some algebraic typos with your signs
and it isn't an 'and'
many things
- those 2 linear inequalities are an 'or'. either (or both) of them are true
- either of those inequalities are true assuming x > 1
- use this to figure which interval x can belong to
- then do the same thing with x < 1
broken yew
**1 divided by 0 equals Infinity**
see we have been assuming x - 1 > 0 all the way through
open bridge
yeah
open bridge
I am trying to solve only this for now assuming I can just add the x - 1 > 0 later
broken yew
,, (x - 1 > 0) ;\land; ((3 + \sqrt{10} < x);\lor;(3 - \sqrt{10} > x))
flat frigateBOT
flat frigateBOT
Noroi
broken yew
no
you are thinking wrongly
if i tell you x < -3 AND x > 3
then you know 3 < x < -3
so x in (3, -3) (which is empty)
broken yew
but if i tell you x < -3 OR x > 3 you cannot write something like that
open bridge
yeah makes sense
broken yew
i would just sketch a number line to make sense of what you have
open bridge
,, (x - 1 > 0) \;\land\; ((3 + \sqrt{10} < x)\;\lor\;(3 - \sqrt{10} > x))
how do i go further from this now?
broken yew
3 +- sqrt10, you should know the rough value of
most importantly how it compares with 1 on the number line
your sketch should look like this. then you need to apply the fact x - 1 > 0
strange flint
Sorry was too lazy to take a screenshot
But you could organize this in a table for cleaner results
Then your solution accounts for all positive intervals
This does contain the answer tho, so please try solving it urself first
open bridge
I mean I know the answer already
just trying to see how to arrive at it with this method
strange flint
I much prefer tables because it gives you a visual presentation of your work, without having to rely on interval intersection
So essentially, you just want to determine where your linear terms (the ones inside parentheses) are negative and positive
You then input that in a number line (the upper bar of the table)
Then plug + for pos and - for neg
All that's left is to multiply the signs -×- = + and -×+ = -
strange flint
Well done 👍