It can’t be midway through a message
#help-28
wide sundial
dull stump
,w expand (x+2x² + 3x³ + 4x⁴ + 5x⁵ + 6x⁶)
glossy valveBOT
normal tree
So there are 10^3 combinations of 3 numbers you can pick
yeah but most of them turn out to be duds
dull stump
Hey frosst how do we find coefficient of a multinomial??
normal tree
like for the product to be one, you have 1 pair: (1, 1), for two, you have (1, 2) and (2, 1), for three, five, seven, it's the same. for four, it's (2, 2), (2, 2), and (4, 1) and (1, 4)
dull stump
i mean selecting from a set when nothing is specified then repetetition is disabled by default
dull stump
Hey frosst how do we find coefficient of a multinomial??
.?
wide sundial
In that case you can’t pick 1
dull stump
yes all combi with 1 is nope
dull stump
but answer is 1/40
wide sundial
,calc 6/(1098)
glossy valveBOT
Result:
0.0083333333333333
it might be wrong in the book lemme cross confirm from net
wide sundial
,calc 1/40
glossy valveBOT
Result:
0.025
Hmm
normal tree
,w 10 choose 3
dull stump
i think it is wrong
glossy valveBOT
dull stump
in answer
wide sundial
Oh
normal tree
2, 4, 8 is there too
2, 5, 10
so 3 groups of 3 numbers make it
there are 120 ways to choose 3 numbers out of 10
3/120 = 1/40
normal tree
it depends on how you construct your space of outcomes
dull stump
i think they are not considered different
normal tree
oh wait yeah
that's not exactly correct is it
,w 10 choose 3 * 3!
you have 720
but you can only permute the first two
unless they mean that any two numbers in the 3-tuple can multiply to the other one
as in (2, 4, 8), (4, 2, 8), and (8, 2, 4) are all valid
dull stump
why only first two?, oh yes 3rd is the product
normal tree
then the reasoning I gave before holds
dull stump
so in which case we reach 1/40
normal tree
yes
wide sundial
I mean with code these simple brute force things aren’t too hard
The question is pretty easy to code
normal tree
no that code computes something else
that computes it if a, b, c are independently uniformly chosen from 1 to 10
but they said that it's "different"
dull stump
so it is 1/40 our groups are (2,3,6),(2,4,8),(2,5,10)
3/120
i think they used wrong language in the question
normal tree
actually yeah I think the question is rather poorly written in the sense that it says that 3 different numbers are "randomly" selected from 1 through 10
clearly here, the actual distributions of these numbers matter and it's not immediately obvious how they're actually sampled
it turns out that it's probably some variant of picking so that every combination of 3 numbers is equally probable
but that should definitely be mentioned
dull stump
but isnt it the conditional probability P(A/B) A= product of first 2 = 3rd and B is selecting 3 from the set?
dunno
,w expand (x+x² + x³ + x⁴ + x⁵ + x⁶)^3
glossy valveBOT
dull stump
i cant find any article of getting coefficient of a term in multinomial
wide sundial
,w convert 0.006 to fraction
glossy valveBOT
wide sundial
Maybe I miscoded again
Ok I’m so dumb
Ok this one was hard
I had to debug it
The last few ones I just did the code in 1 go
glossy valveBOT
dull stump
umm what is 10!/7!
wide sundial
Ways of choosing 3 different numbers
glossy valveBOT
wide sundial
,w 10!/7!
glossy valveBOT
wide sundial
Hold up
Not the same
Isn’t it 10 numbers for the first slot
9 for the 2nd
8 for the 3rd
dull stump
technically ye
wide sundial
10C3 ignores repeats I think
dull stump
then wont it be just 1098 but then it will take like 1,2,3 and 3,2,1
but in 10C3 1,2,3 and 3,2,1 are same
wide sundial
Like 1, 3, 5 and 3, 1, 5 is the same in 10C3
wide sundial
Yeah exactly
dull stump
isnt that what we want
wide sundial
But it isn’t
dull stump
what is 10!/7!
wide sundial
10*9*8 = 10!/7!
dull stump
oh
ye
ye so we need combination not permutation
selecting those 3 numbers again is useless
selection not arrangement
wide sundial
It’s because the set we have where the 3rd is the product of the first 2 we have is as combinations
dull stump
that is for favourable cases we talkin bout denominator rn aint it?
wide sundial
2, 3, 6
2, 4, 8
2, 5 10
3, 2, 6
4, 2, 8
5, 2, 10
There’s 6 combinations of 3 numbers that lead to the 3rd number being a product of the first 2
Unless they mean without order
dull stump
but then again selecing 2,3,6 and 3,2,6 is selecting same group
we are selecting group, product is just a condition
dull stump
why arrange them they all independent cases
wide sundial
,calc 3*3!
glossy valveBOT
Result:
18
,calc 18/720
glossy valveBOT
Result:
0.025
That’s your 1/40
dull stump
:0 wth
wide sundial
So if you randomly pick 3 different numbers
And have to make it so any 2 of the numbers multiply to the 3rd
Then it’s 1/40
You see in my code it says near the bottom a*b == c
dull stump
ye
wide sundial
wide sundial
product of any two should be equal to third
That’s different to product of the first 2 equals the third
dull stump
ah yes
glossy valveBOT
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torn jolt
help a brother out
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gritty rose
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lunar cape
If the initial height of the rock is y1 = h = 5.6-m calculate the rock’s velocity when it has fallen to 2.4-m above the ground.
lunar cape
I don't understand what to start or do first.
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lunar cape
<@&286206848099549185>
void widget
Ok so initial height is 5.6 m therefore we could say that initial velocity is 0
Cause it's initial height the object starts moving from that place
v=?
u=0m/s
s=5.6-2.4=3.2m
And g=9.8 m/s
We can use the third equation of motion
±√(2gh)=v where u is 0m/s
void widget
wait <@518716603928543234> is h=height
Yeah ofc
lunar cape
√(2x9.8x5.6)
lunar cape
oh
void widget
Ok your approximated answer is right
void widget
That's it
lunar cape
so 7.9 m/s
void widget
Yea
void widget
Ok give me
lunar cape
If you throw a 148.83 g baseball straight up into the air at 70.8-mph how high will it get if air resistance is not present?
void widget
Ok so here g=-9.8m/s
And here you have to convert 70.8mph to m/s
u=0m/s
When you will convert 70.8 to m/s you will get v
Then you can use
V=u+gt ( time)
lunar cape
31.65
void widget
You will wille solve v=u+gt you will get the time
void widget
You will wille solve v=u+gt you will get the time
Then you could use h=ut+1/2(gt²) for height
void widget
converting 70.8 to m/s
Oh 70.8mph=31.65m/s
Ok
You can use directly
2gh=v²
For height
Remember g=-9.8 m/s here
Ok then do it yourself and there's no use of mass
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@lunar cape Has your question been resolved?
dense edge
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dense edge
Let’s say you are only given the first part, cos = 3/5
Could you determine the second part (quadrant) from only this?
onyx glen
no, of course not.
you could say that theta lies in either quadrant 1 or quadrant 4 (since those quadrants are where cosine is positive), but you could not say which.
dense edge
The second part would have to be provided with the question?
cosmic canopy
yes.
onyx glen
it depends on what the question wants of you.
cosmic canopy
to give a single solution yes.
onyx glen
it may well be that for your goal, the value of cos(θ) alone is enough.
cosmic canopy
^
dense edge
OK
cosmic canopy
it all depends on what you need
inner finch
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inner finch
Hi guys!
How would u guys see this c_n as?
Written in x_k?
I wrote it as this^^ But it's apparently wrong
c_n=9c_{n-2}
fast peak
$c_n = 0 c_{n-1} +9c_{n-2}$
glossy valveBOT
Denascite
fast peak
dont ignore that 0 coefficient
inner finch
**Denascite**
Hmm
Okay I guess i misunderstood that
I didn't think it HAD to have a cn-1
normal tree
I don't think you understand what the slide actually says
inner finch
No, probably not.. 😛
normal tree
it says that for a linear homogeneous recurrence relation of a_n = r_1 a_{n-1} + r_2 a_{n-2} + ..., you have this method of solving it
9 is not the coefficient of c_{n-1}
it's not about what it has or doesn't have to have
inner finch
it says that for a linear homogeneous recurrence relation of a_n = r_1 a\_{n-1} ...
No I get that part^^
inner finch
9 is not the coefficient of c_{n-1}
And I also didn't read that?
normal tree
that's what you wrote
inner finch
fast peak
I didn't think it HAD to have a cn-1
you always have cn-1. and you also always have eg cn-3 or cn-50. just most of these coefficients are 0
normal tree
the slide gives you a characteristic equation
inner finch
This is what I wrote?
normal tree
you didn't reproduce the characteristic equation correctly
normal tree
it explicitly says that the term in front of x^(2-1) should be the coefficient c_{n-1}
reread the slide
inner finch
inner finch
you always have cn-1. and you also always have eg cn-3 or cn-50. just most of th...
Ye okay
But I still don't exactly get what I did wrong, our k=2
inner finch
Yes, but you're kind of being a dick about it, don't you think? 😂
normal tree
I'm telling you that you don't understand as much as you think you understand
inner finch
Like I am not trying to offend you or anything, but I get that you don't think I understand the slide, but I am trying to understand it.
normal tree
whether you want to take that advice to heart or not is up to you
inner finch
Well, that's you putting words into my mouth lol.
normal tree
I'm telling you that you don't actually understand what they're saying if you're just shuttling the numbers from one thing to the other
it says a very specific thing
r_1 is defined to be the constant before a_{n-1}
r_1 is the thing that goes in front of x^{k-1}
inner finch
I see it as a way that if you have a linear homogenous recurrence relation you can rewrite it on the form x^k = ...
Yes, and I just see the 9 as the constant r?
normal tree
but the slide tells you that this is rewritten in a very specific manner
which you have not followed
you can't just "see" things the way you want to see them
normal tree
it tells you that r_1 is the constant corresponding to exactly the c_{k-1} term
nothing else
you can't reinterpret it
inner finch
Literally no reason to type all that. I haven't said any of that, I'm just trying to explain my thought process, but for some reason it's triggering the fuck out of you lmao
You don't need to help me mate, if it's making you so angry
I am not saying I understand the slides, or that I understand what I'm doing, or else I probably wouldn't be here anyways, right.
I was simply trying to tell you what my thought process was
So that perhaps, it was easier for you guys to see where it went wrong.
And then you start all this interpretation shit that I think that I know everything, like wtf is going on? 😂
Like I am not trying to hurt your ego, I know that you're smarter at math than me, don't worry bro
But so using this, 9 should correspond to the r_2
And r_1 is then 0
So you'd get x^k=0 x a_{n-1} + 9 x a_{n-2}
fast peak
do you know why we are looking at this specific equation?
inner finch
Lemme rewrite
Lemme write in Tex
that's easier to show
$x^k=0 \cdot x^{2-1}+9 \cdot x^{2-2}$
glossy valveBOT
SimonWin
fast peak
yes that is correct now
fast peak
yes
inner finch
Okay, awesome thanks, it makes sense now!! ❤️
static ore
how do i do this?
full forumBOT
how do i do this?
sharp flame
Do you understand how the notation for functions work?
static ore
Do you understand how the notation for functions work?
yeah
but I dont remeber the process
sharp flame
Go ahead then
Alright
If you have f(x) and you want to find f(4) then replace every x with a 4
So you need f(1) and f(0)
static ore
its 1 and 4
i dont know how to do the rest
the answer is supposed to be 96 somehow
but 4x1 is 4 and 6x4 is 20
omg
im an idiot
4*20 is 80
slender leaf
6*4 is ?
static ore
THIS NEVER HAPPEND
I NEVER ASKED THIS QUESITON
im an idiot
please forgive me
.close
livid gull
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gaunt orchid
I NEVER ASKED THIS QUESITON
did you get it though?
static ore
did you get it though?
4*6 is 24 not 20, 24x6= 96
gaunt orchid
yep thats right
static ore
do you know the pythagorean theorem
livid gull
do you know the pythagorean theorem
yes
livid gull
then use it
idk how
glossy valveBOT
adarw
glossy valveBOT
adarw
static ore
and then you can just plug it in @livid gull
livid gull
and then you can just plug it in <@766329608186232922>
Thanks I got it
one more thing
Do I do this the same way?
livid gull
i get the formula
yes
livid gull
a is the missing number right
yes
torn jolt
ok
torn jolt
yes
livid gull
2 and 3 are corresponding angels
2 and 3 are congruent
m3= 115 degree
1 and 2 are
vertical angles
and m1 = 115 degrees
1 and 3 are exterior alternate
they are congruent
torn jolt
i haven't done angles in a while
livid gull
dang
torn jolt
i can't remember anything
torn jolt
np
livid gull
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half remnant
I’m struggling with question 1
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chrome idol
Does this seem alright?
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Does this seem alright?
stiff musk
almost right, except you have n on both sides of the condition n > (n - N1)/2
oops wait, did flawed algebra while attempting to typeset, lemme fix
chrome idol
i see
stiff musk
notice that $$\frac{n - N_1}{n} = 1 - \frac{N_1}{n} \to 1$$ as $n \to \infty$, so you waon't be able to get it less than 1/2
glossy valveBOT
Bungo
stiff musk
that doesn't really matter though
you can get it less than say 2
and then your overall expression will be less than $\epsilon_1 / 2 + 2\epsilon_1$
glossy valveBOT
Bungo
stiff musk
which is good enough since $\epsilon_1$ can be chosen arbitrarily small
glossy valveBOT
Bungo
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static ore
y=-x^{2}-6x+7 how do i factor this?
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y=-x^{2}-6x+7 how do i factor this?
tame bobcat
$y = (-x)^{2} - 6x + 7$ ?
glossy valveBOT
barış
glossy valveBOT
barış
tame bobcat
thats whats what i did
but how would you get from the given to this one , I dont see what you did
static ore
but how would you get from the given to this one , I dont see what you did
you probably dont know this method
merry galleon
Can someone help me with this Descartes rule of signs question? I’m confused as to why it’s not a).
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Can someone help me with this Descartes rule of signs question? I’m confused as ...
@merry galleon Has your question been resolved?
merry galleon
<@&286206848099549185> pls help
sinful lintel
This feels weirdly familiar
plush egret
Descartes' Rule of Signs is a method for determining the number of positive and negative roots of a polynomial equation. It states that the number of positive roots of a polynomial equation is equal to the number of changes in sign of the coefficients of the equation, while the number of negative roots is equal to the number of times the leading coefficient changes sign. This rule only applies to polynomials with real coefficients and is not a proof but a heuristic method to count the number of positive roots of a polynomial equation.
merry galleon
never seen rule of signs before
you don't know pre calc?
plush egret
im non-traditional
i only know college level math 
the number of times the leading coefficient changes sign wtf
merry galleon
ik it's weird af but I gotta know it
plush egret
youre probably not getting help because this is some esoteric shit i would guess
but im trying to figure it out
okay so its + + + - -
rose rain
ok let's just cheat with wolfie
plush egret
the question confused the shit out of chat gpt
rose rain
,w plot 5x^4 + x^3 + 3x^2 - 3x - 1 x \in [-10, 10]
plush egret
ok let's just cheat with wolfie
were not trying to find roots
glossy valveBOT
Requested by DerpZ#9141
plush egret
trying to classify/enumerate them
rose rain
oh
plush egret
its sorta like rational root theorem
rose rain
I mean
plush egret
but for sign
rose rain
wow I have never seen descarte's rule of signs before
sinful lintel
I think you're just counting the number of roots there are?
plush egret
me either
glossy valveBOT
sinful lintel
I'm wondering if it's related to the logic (x+1)(x-1) and (x+1)^2 are completely different
I wish I could use English words to better describe what I'm thinking
plush egret
okay so
-
-
-
-
- means
-
-
-
one sign change
one positive root
ohh
wait
shit what a weird rule
not surprising i never seen it
rose rain
wolfie says that too yeah
rose rain
the proof?
plush egret
just a derivation of the statement of the theorem
rose rain
does it use galois theory 
plush egret
okay so its two statements i guess
the number of positive real roots is bounded by the number of changes in signs in coefficients
but also, the number of positive real roots, counted with multiplicity, is of the same parity as the number of changes in sign
plush egret
yea this shit makes no sense at all @merry galleon
are you able to provide us a statement of the rule as youre expected to apply it
so we dont have to search for it
merry galleon
nvm I found it
if there are 3 negative zeroes, there can be also less but they have to be odd since 3 is odd, so it was supposed to be 1 positive zero or 3 or 1 negative zeroes
plush egret
"We can determine also the number of true and false roots that any equation can have, as follows: An equation can have as many true roots as it contains changes of sign, from + to - or from - to +; and as many false roots as the number of times two + signs or two - signs are found in succession."
merry galleon
nvm I found it
if there are 3 negative zeroes, there can be also less but they h...
I actually found that from the link you sent me lol
rose rain
when I see a polynomial the first thing I do is open wolframalpha
plush egret
mission accomplished
rose rain
plush egret
if you see a polynomial on an exam or a quiz setting
sinful lintel
This rule is weird
rose rain
I guess this is probably something you're asked to memorize
merry galleon
mission accomplished
*cue gta san andreas music
rose rain
or plug it into your calculator
plush egret
*cue gta san andreas music
rose rain
most graphing calculators have a polynomial root finder
plush egret
i always think this
but im a boomer
the context for this image might not be like
as culturally present any more
merry galleon
or plug it into your calculator
wish I can but sometimes the tests have no calc :/
sinful lintel
W bush?
plush egret
ye
sinful lintel
or papa bush?
plush egret
dubya
rose rain
wish I can but sometimes the tests have no calc :/
ok then you probably need to memorize this 
sinful lintel
Damn the second to last real president we had
plush egret
he gave this speech for no reason since the war continued basically for another 2 decades
sinful lintel
ok then you probably need to memorize this <:sadcat:547124024585486359>
Agreed
merry galleon
oh shit this boutta turn into a political discussion
plush egret
yes
sinful lintel
Yeah sorry, it was supposed to be a one off joke
merry galleon
anyways thx for the help see ya
plush egret
here is george bush being greeted by the telletubies
still a non-political show for babies?
i think not.
.close
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obtuse light
How do I solve for x ?
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twilit leaf
you can use similar triangles
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dark escarp
Where did the pointed 2 move ?
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fast peak
did they plug a=22 in or something?
dark escarp
yess
fast peak
well then 2*a=2*22=44
vague kayak
You have an urn with 3 Red balls and 3 Blue balls. I pick 4 without replacement. What is the probability i pick at least 2 red balls. Need help finding what the complement is here
vague kayak
Is it probability of picking only one red ball?
frozen lark
you got picking up without replacement so you'll have to use Permutations
when you say at least 2 red balls
this means that you might have 2 red 2 blue
or might have 3 red 1 blue
that's an [or] for ya
@vague kayak ?
vague kayak
Just saw let me read
Ok let me write it out
Ok that makes sense but seems odd
Getting a really small number
frozen lark
which is ?
vague kayak
3/20
frozen lark
huh ... that's actually bigger than the correct solution
did find the number of all the cases omega ?
vague kayak
Im thinking like (3/6)(2/5)(1/4)(3/3)+ (3/6)(2/5)(3/4)(2/3) no?
For 3 red 1 blue plus 2 red 2 blue
vague kayak
Ok i see
glossy valveBOT
mino65
frozen lark
ok forget about that
we need to find the number of wanted cases which is 2R,2B and 3R,1B
and all the cases are 360
vague kayak
Yeah so im looking for numerator now
frozen lark
observe
it's 2 Red [AND} 2 blue
when dealing with probabilities AND means multiplication
vague kayak
Ok
frozen lark
so we'll pick up 2 reds out of three [and] 2 blues out of three
so it'll be like :
$3P2 \cdot 3P2$
glossy valveBOT
mino65
frozen lark
can you do the other case ?
vague kayak
Then its plus 3P1•3P3?
frozen lark
gotcha
vague kayak
Sweet
frozen lark
so the final answer is ?
vague kayak
But you taught me the intuition though
frozen lark
just like i messed the 'e' in something
vague kayak
So thats better than the way i was doing
frozen lark
look like i was useful in the end
vague kayak
For sure thank you again
frozen lark
no prob
vague kayak
.close
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tame bobcat
I have a question on unitary matrizes
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past atlas
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past atlas
could i have help with this question?
i have to create a polynomial function
that when i input an x value, it’ll spit out the corresponding y value
i figured out that it needs to be a cubic function
ayuda
rocky vale
Hi again
past atlas
hey there
rocky vale
Would you be comfortable solving a system of linear equations?
One with, say, four unknowns and four equations
rocky vale
There may be other ways to solve this but they involve some calculus
so I think this is the best
past atlas
i agree
rocky vale
You already figured out it's cubic, that's really helpful
we can say
y = ax^3 + bx^2 + cx + d
And then our goal is of course to find a, b, c, and d
past atlas
oh i see
rocky vale
See how you can get four linear equations?
rocky vale
okay
past atlas
i’m a little confused
so i need 4 different linear equations?
while inputting the same x value?
or can i make 4 different equations using different x values
rocky vale
or can i make 4 different equations using different x values
this
and their corresponding y values
past atlas
oh okok
rocky vale
you could actually make more than four equations
but you need four to find four unknowns
past atlas
rocky vale
Yep, that'll work
But for your own sanity, I suggest using (1,1) instead of (5,55)
it'll just be simpler
rocky vale
lol
past atlas
rocky vale
It's still gonna be kinda ugly
Yeah that's nicer
So, I'd rewrite that without all the brackets
like 30 = 64a+16b+4c+d
And then you have a system of equations
You can start doing some substitution or elimination
probably elimination
rocky vale
Not sure what you mean
rocky vale
Looks like you used the first and third equation together, then the second and fourth together
You can just choose another combination
like the third and fourth together
and eliminate a again
Then you'll have three equations and three variables
which is a step in the right direction
past atlas
so i can use any combination of equations?
rocky vale
yep
past atlas
could i use 1 and 2
then 1 and 3?
rocky vale
sure
You'll need to make three combinations, so you can eliminate one of the unknowns three times
Then you have a system of three equations/unknowns instead of four
past atlas
💀
rocky vale
it'll work out fine either way
past atlas
i’ll try that it’s okay
i’ve messed up
somehow
oh wait nvm
am i on the right track with this?
rocky vale
yeah but you don't need quite all those combinations
rocky vale
Go back to this, and try combining (7) and (6) on the right
past atlas
i’ll do that
rocky vale
leave what you have on the left
(9) is good, we just want to get something unique for (10)
rocky vale
I haven't checked for arithmetic mistakes but that looks good to me
rocky vale
but b = 31/3?
this is not what I got
rocky vale
would i have to repeat this for c and d
no, everytime you find one unknown, you can go back a step and plug it in
like you have equations involving just a and b
so once you find a, you can plug it in and find b
then you have equations involving a, b, and c
so once you know a and b, plug them in and find c
if that makes sense
Uh oh, I think I found a mistake a ways back though
rocky vale
your equation here that starts with a 9, shouldn't that be 5?
rocky vale
that's unfortunately gonna change a lot of the work 😶
past atlas
You'll need to make three combinations, so you can eliminate one of the unknowns...
i only need 3 combinations right?
rocky vale
right
rocky vale
No but I think I made a mistake, give me just a minute
past atlas
i think i should plug a and b into into an equation that contains a b and c
then find c
and then plug it into one of the original equations to find d
past atlas
yep
rocky vale
yes
past atlas
d=0
rocky vale
yep 👍
Here you can see those values for a, b, c, and d satisfy all four equations, and make a cubic that fits your points
past atlas
past atlas
Here you can see those values for a, b, c, and d satisfy all four equations, and...
awesome!
rocky vale
yep, just
past atlas
it all works out
rocky vale
1/3 x^3
past atlas
oh right right
rocky vale
that looks like (1/3)^3 x
past atlas
i’ll fix that
thank you so much for the help again
and thanks for your time, i’m really sorry that it took so long
rocky vale
No problem! I'm multitasking lol
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rocky vale
You too!
past atlas
:))
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Available help channel!
elfin lynx
Could someone help me out with this
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Could someone help me out with this
elfin lynx
twilit leaf
!status
full forumBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
I will send my working
twilit leaf
:)
elfin lynx
onyx glen
u' u + v v'
bad
that product rule is misstated
may be due in part to the striking similarity between the ways you handwrite the letters u and v, which makes you liable to mix them up
it's u' v + u v'
the product rule says that when you differentiate a product of two factors, the result looks like two copies of the original product, except each one has one of the factors differentiated.
elfin lynx
Oh ok
elfin lynx
may be due in part to the striking similarity between the ways you handwrite the...
Possibly
Okay is it would actually by: 5 * (x^2+3)^4 + 5x(x^2+3)^4
Right?
onyx glen
no
why'd you differentiate v if you are not going to put v' to use anywhere lol
$f'(x) = 5 \cdot (x^2+3)^4 + 5x \cdot 4(x^2+3)^3 \cdot 2x$
onyx glen
oh, i typoed, my bad.
glossy valveBOT
Ann
onyx glen
at this point, you simplify this by factoring out at least (x^2+3)^3.
elfin lynx
Can you clarify what you mean by factoring out?
onyx glen
i remember ramonov and others painstakingly explaining this to you not more than 12 hours ago.
depending on your timezone, that may be yesterday or today.
elfin lynx
Do you mind explaining it again?
onyx glen
when i say factor out, i mean factor out. as in apply the distributive property to transform an expression that looks like ab + ac into a(b+c).
you need to review factorization in general if you have not done so already.
elfin lynx
Yes
onyx glen
when we say "factor out <something>", the something is the common factor -- the a in ab + ac = a(b+c), whatever it may look like.
elfin lynx
Well I know that (x^2+3)^3 is the same thing as (x^2+3)^2 * (x^2+3)
I just get confused with terminology I guess
onyx glen
Well I know that (x^2+3)^3 is the same thing as (x^2+3)^2 * (x^2+3)
this specifically will not be of much use here.
if you want to shuffle exponents around, better to turn (x^2+3)^4 into (x^2+3)^3 * (x^2+3) in the first term.
though really this is not much different than, say, factoring $z^{11} + 8z^9$ as $z^9(z^2+8)$...
glossy valveBOT
Ann
elfin lynx
if you want to shuffle exponents around, better to turn (x^2+3)^4 into (x^2+3)^3...
Ok so in that case I can combine the two
Since they have the same base
onyx glen
... you'd be undoing what i just described, but sure.
onyx glen
$f'(x) = 5 \cdot (x^2+3)^4 + 5x \cdot 4(x^2+3)^3 \cdot 2x$
from this, factor out $(x^2+3)^3$ to get:
$f'(x) = (x^2+3)^3[5(x^2+3) + 5x\cdot 4\cdot 2x]$
elfin lynx
So my problem is I don’t understand how you arrived at the last line
onyx glen
oh, i typoed again, my bad.
glossy valveBOT
Ann
onyx glen
So my problem is I don’t understand how you arrived at the last line
i do not understand how to explain it in any more detail. if i try to write out explicitly both terms as products of (x^2+3)^3 with something else, then you'll keep wondering how THAT comes about, and we will keep walking in circles again.
it sounds as if you are reluctant to do a review of factorization, or you have not yet done it enough.
i've seen you run up against this EXACT SAME issue what feels like several weeks ago.
elfin lynx
I understand factoring but I just don’t understand what you mean by factor out, when I imagine factoring out I think of (x^2+3)^3 = x^6 + 9x^4 + 27x^2 + 27
elfin lynx
So factorising is finding common factors that would add to produce the original expression correct?
onyx glen
i would not describe it that way.
onyx glen
The common factor of x^2 and 3 is 1 correct?
sure but that is irrelevant
onyx glen
Then how would you describe it?
i don't know how to describe it without repeating myself
onyx glen
you wouldn't, (x^2+3)^3 by itself is already factored
what i am doing is pulling it out
treating it as a single unit
idk how else to say it
onyx glen
from the (x^2+3)^3 that gets pulled out of it, (x^2+3)^1 remains inside the brackets.
elfin lynx
Oh ok so (x^2+3)^4 - (x^2+3)^3?
elfin lynx
So expand the parentheses?
elfin lynx
Ok so 5x^2 +15 + 10x^2 * 4
onyx glen
not cleaned up in full yet
elfin lynx
(x^2 + 3) * 15x^2 +60
How is this
elfin lynx
@worthy tree
worthy tree
<@942843959783333890>
yes @elfin lynx ?
full forumBOT
@elfin lynx Has your question been resolved?
elfin lynx
yes <@655168192335052802> ?
could you help me out with this
worthy tree
the derivation of 5x(x² + 3)^4 ?Right ?
worthy tree
,align
\qty(5x(x² + 3)^4)'
&= (5x)' * (x² + 3)^4 + 5x * \qty((x² + 3)^4 )'\
&= 5 (x² + 3)^4 + 5x * \qty[4 (x² + 3)' (x² + 3)^{4 - 1}]\
&= 5 (x² + 3)^4 + 5x * 4 * 2x * (x² + 3)^{3}\
&={\color{red}5(x² + 3)^3} * (x² + 3) + {\color{red}5(x² + 3)^3} * \qty[x*4 *2x]
you reach here Right ?
worthy tree
but it is already factored?
nope
elfin lynx
how would I factor it?
elfin lynx
(x^2+3)^3 = (x² + 3)^2 * (x² + 3)
glossy valveBOT
Mehdi_Moulati
worthy tree
did you get it ?
worthy tree
**Mehdi\_Moulati**
5(x²+3)³ is common factor between them
so you can factorize by 5(x²+3)³
elfin lynx
Oh I understand
but where is the 5x?
I'm confused why there are two 5's
@worthy tree
glossy valveBOT
Mehdi_Moulati
worthy tree
**Mehdi\_Moulati**
this is what i did here ,
just follow it's step to understand
which step you didn't understand ?
elfin lynx
the last step
I understand everything else, I just don't understand the last step.
cause there was 5x on the third line but on the last line it is gone
worthy tree
,align
&= 5 (x² + 3)^4 + 5x * \qty[4 (x² + 3)' (x² + 3)^{4 - 1}]\
&= 5 (x² + 3)^4 + {\color{magenta}5}x * 4 * 2x * {\color{magenta}(x² + 3)^3 }\
&=5(x² + 3)^3 * (x² + 3) + {\color{magenta}5(x² + 3)^3 }* \qty[x*4 *2x]
what about now ?
elfin lynx
oh ok I see
glossy valveBOT
Mehdi_Moulati
worthy tree
5(x^2+3)^3 * (x^2+3) + 5(x^2+3)^3 * 8x^2
yes
factorize now by 5(x^2+3)^3
to get the answer
worthy tree
5 * 2 + 5 * 3
factorize by 5
what is the result ?
5 * 2 + 5 * 3 = 5 * (2 + 3)
${\color{red} 5(x^2+3)^3} * (x^2+3) +{\color{red} 5(x^2+3)^3} * 8x^2 ={\color{red} 5(x^2+3)^3} \qty[x^2 + 3 + 8x^2]$
do the same for this
glossy valveBOT
Mehdi_Moulati
${\color{red} 5(x^2+3)^3} \* (x^2+3) +{\color{red} 5(x^2+3)^3} \* 8x^2 ={\color{red} 5(x^2+3)^3} \qty[x^2 + 3 + 8x^2]$
worthy tree
@elfin lynx did you get it now ?
elfin lynx
yes I get it but I don't understand how to factor by 5(x^2+3)^3
worthy tree
**Mehdi\_Moulati**
```latex
${\color{red} 5(x^2+3)^3} \* (x^2+3) +{\color{red} 5...
i already did it bro
${\color{red} 5(x^2+3)^3} * \qty[x^2 + 3 + 8x^2]$
glossy valveBOT
Mehdi_Moulati
worthy tree
**Mehdi\_Moulati**
this is the factorize by 5(x² + 3)³
elfin lynx
oh ok, so to factor by 5(x^2+3)^3 I just multiply by 5(x^2+3)^3
worthy tree
oh ok, so to factor by 5(x^2+3)^3 I just multiply by 5(x^2+3)^3
yes
factorize by b for :
**a * b + b * c **
means
b * (a + c)
elfin lynx
ok I understand
but I have one more question
why does it become 5(x^2+3)^3 [x^2+3+8x^2]
do one of the 5(x^2+3)^3 cancel out?
simple totem
it was factored out
we can represent 5(x^2 + 3)^3 with the letter a
then the expression becomes a * (x^2 + 3) + a * (8x^2)
would you know how to factor an a from this?
elfin lynx
multiply by a
simple totem
a * (x^2 + 3) + a * (8x^2) = a * [(x^2 + 3) + (8x^2)]
elfin lynx
oh ok
simple totem
then just substitute back 5(x^2 + 3)^3 for a
elfin lynx
a * (x^2 + 3) + a * (8x^2) = a * [(x^2 + 3) + (8x^2)]
once i have this