@lean otter Has your question been resolved?
#help-23
safe radishBOT
lean otter
<@&286206848099549185>
safe radishBOT
@lean otter Has your question been resolved?
lean otter
<@&286206848099549185>
safe radishBOT
@lean otter Has your question been resolved?
lean otter
<@&286206848099549185>
pastel verge
What part?
safe radishBOT
@lean otter Has your question been resolved?
lyric thunder
if something is inversely proportional, is it the same thing as indirect variation?
safe radishBOT
if something is inversely proportional, is it the same thing as indirect variati...
flat frigateBOT
Mosh
lyric thunder
?
pulsar condor
1st is inverse proportionality, 2nd is indirect variation.
lyric thunder
what are the formulas for them?
safe radishBOT
@lyric thunder Has your question been resolved?
lean otter
Could someone please explain to me how i can solve this question?
flint ledge
don’t immediately ping helpers
ancient escarp
have you drawn a picture
safe radishBOT
@lean otter Has your question been resolved?
mint flax
sohcahtoa
ancient escarp
i assume you know pythagorean's theorem
ancient escarp
that's all you need
lean otter
what is cos squared
ancient escarp
well what ratio can you get for cosine using your picture
lean otter
i honestly don't really know
flint ledge
$\cos^2(\theta) = (\cos(\theta))^2$
flat frigateBOT
quantum
mint flax
^
ancient escarp
you're supposed to draw the picture
lean otter
how does theta equal to 1
ancient escarp
it doesn't
mint flax
it doesn't
lean otter
i have drawn it myself
ancient escarp
then what does it look like
lean otter
kk
mint flax
resending for reference
lean otter
ancient escarp
label the sides
flint ledge
theta is an angle that isn’t the right angle
ancient escarp
that too
lean otter
so how am i suppose to answer the question with this knowledge
lean otter
lean otter
theta = a/c
flint ledge
no
mint flax
sin(theta)=a/c
lean otter
that's what i meant
lean otter
there wouldn't be cos theta
safe radishBOT
lean otter
.reopen
safe radishBOT
✅
flat frigateBOT
gmod
mint flax
$\sin^2\theta$ means $(\sin\theta)^2$
flat frigateBOT
gmod
lean otter
(a/c) squared
lean otter
would you add them
mint flax
so what would cos²(theta) be?
lean otter
algebraically
mint flax
yeah
mint flax
so it would be $(\frac{a}{c})^2+(\frac{b}{c})^2$
lean otter
b + a over c
flat frigateBOT
gmod
mint flax
sorry for the bad parentheses but this is what it would look like
flat frigateBOT
quantum
flint ledge
it doesn’t
flint ledge
not helper moment
flint ledge
@lean otter do .close
safe radishBOT
@lean otter Has your question been resolved?
brisk quest
436116003068178632599586304885865102436987283273192588087066946205825385420102245537617641171508064231363006853389201952225629937253532766998346708022315458543404066186915809029438078922052852453720811267502691138577239348756804955313693393091406303393978608038048294775593159201377364139187142305434603252709617151318958922901417601677040658907901408937709530384849660728157676437361951488873238454079633326062998564370475129597220298716911739158299214865235291416480348618758676919027696256066573122899301892912081375412160180698274455765997411725112824776559729034126511125728785207316456433746195093026529168420804699808556231371820824328024241019054519605469721555033733292097536/6197148516394364188900141323882322297210558701995146720297131484855101557855486731958221825570944947984004858553140186641155182016082966859736962550177180088600405321238825249746403835469727033342501800712896634341355449588496565230819852486507103432027694702173423912538079766299114717526424477880616639449302766419689877623466381205865605118390274355471509815278332251971759244543740352243076811277717664865760571719052087465244804777251295251995674217073974490490639368917793497383548456310727391232146480195472325811180553427557484142696840113044630459874629946903509481288793101524546602637214862483656004358674999005683914545225086663429140282193350054143291262884451455852063961426683890652699785527976797456599012104868536777488904438653913152719206119693088299602007397720839957389279772575408178106779022036761970796031083638898838300999211395119392678116082151174675257415694403682741632963982317824529718557167903433530955641359259898743752340961013280091956109688607349893209675381881203380524958393734988046719982595235111430847956368856989676534079417326438466595922976484817758530826916589148548056446877738990 possible combinations, and only 3133507531 work, what are the odds i guess a working combination
safe radishBOT
43611600306817863259958630488586510243698728327319258808706694620582538542010224...
ancient escarp
can you not
brisk quest
what?
brisk quest
can you not
did i do something wrong?
ancient escarp
yeah you're continually shitposting
flint ledge
too bad mods didn’t respond to me first time
long copper
43611600306817863259958630488586510243698728327319258808706694620582538542010224...
to answer your question, it's so small that it's 0
inland ivy
When the mods dont say b& 😔
ancient escarp
sad
flint ledge
rip
brisk quest
yeah you're continually shitposting
how tf is it a shitpost
its actual odds for a combination thing im guessing
im making a password guesser and want to know odds
inland ivy
That requires one channel
long copper
im making a password guesser and want to know odds
it doesn't seem like you have enough of a brain to make a password guesser if you can't work out the possibility of what you just asked. Either you're trolling or you're genuinely stupid
inland ivy
Just divide the two numbers
long copper
i just want to know odds
the chance of getting a correct combination is so small that no one can comprehend how small it is
brisk quest
Just divide the two numbers
i cant it gives me undefined
ancient escarp
the sun is gonna explode before your "password guesser" is correct
inland ivy
7.1850322138032947359625162212840189423241965162208148080546135972588521880931026266906874754026349729001904465802613146821853121E-675
In case you were wondering
flint ledge
what did you even use to get that answer
ancient escarp
oh that'll be fun
inland ivy
0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000071850322138032947359625162212840189423241965162208148080546135972588521880931026266906874754026349729001904465802613146821853121
About
ancient escarp
which is about 71850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 times the amount of atoms in the observable universe
flint ledge
lol
ancient escarp
have fun guessing that password 👍
inland ivy
Theres still a chance to get it first try
ancient escarp
indeed there is
there's also a chance to get it before the heat death of the universe
brisk quest
<@&268886789983436800>
flint ledge
you just pinged mods for no reason
brisk quest
😱
safe radishBOT
Available help channel!
fleet folio
hello so I'm going through my homework and i found that i did one question wrong in the solution and I'm not quite understanding what they are doing can someone pls explain
safe radishBOT
hello so I'm going through my homework and i found that i did one question wrong...
@fleet folio Has your question been resolved?
fleet folio
<@&286206848099549185>
royal kiln
<@&286206848099549185>
@fleet folio I got same answer. What is the correct answer?
royal kiln
second line under indirect should be
24C5 - (12C1)(12C4) - (12C0)(12C5)
the calculation is fine
fleet folio
but how did you get that?
royal kiln
is that the teacher's working ^^^ ?
nvrmnd
the logic is that you start by counting what all the possible hands are: 24C5
and remove the (comparably fewer) hands you don't care about
that's all the ways to pick hands with 1 black card and
all the ways to pick hands with 0 black cards
so in the 24 card deck described at the top of your teacher's explanation, half are black and half are red
that means 12 are black and 12 are red
so here we go, counting the hnads containing 1 black card:
we can choose 1 black card from any of the 12 black cards: 12C1
and we can choose 4 cards from any of the 12 red cards: 12C4
so that's (12C1)(12C4)
.
now we count the hands with 0 black cards.
that's choosing 0 black black cards from the group of 12 black cards: 12C0 = 1 (one way to do this; arguably a redundant calculation)
and we choose all 5 cards from the 12 red cards: 12C5
so that's (12C0)(12C5)
the complete calculation
24C5 - (12C1)(12C4)-(12C0)(12C5)
looks like
(24*23*22*21*20) - (12)*(12*11*10*9) - (1)(12*11*10*9*8)
np
fleet folio
.close
pure radish
quick question: can an obtuse triangle ever have 2 solutions?
safe radishBOT
quick question: can an obtuse triangle ever have 2 solutions?
steel stag
for side-side-angle sure
pure radish
cause there are only 2 rules for finding solutions for obtuse triangles
a<=b or a>b
which is why I was confused, since there isnt a rule abt comparing height
safe radishBOT
@pure radish Has your question been resolved?
pure radish
<@&286206848099549185>
safe radishBOT
@pure radish Has your question been resolved?
lean otter
$g(x)=2-\frac{1}{5x+2}$
safe radishBOT
$g(x)=2-\frac{1}{5x+2}$
flat frigateBOT
DoomKing
lean otter
$f(x)=\frac{1}{x}$
flat frigateBOT
DoomKing
lean otter
would the translations from fx to gx be
- left 2
- horiz shrink of 1/5
- reflect x axis
- up 2
inner tapir
oh sure yeah you can put the a up there
sorry didn't mean to compile onto yours haha
well anyway the +2 isn't really h here
do you see how a = 1/5 and h = -2/5?
lean otter
uhm
x is replaced by x+2
then you replaced the x in x+2 with 5x
if you did it your way
you get 5x first
then replace x with x+2
5(x+2)
inner tapir
also i see now what you mean by vertical, it doesn't really matter in this case (as long as you keep it straight), but you do need to factor either way
i'll do it another way maybe
flat frigateBOT
citrusmunch
$f(b(x-h)) + k = \frac{1}{b(x - h)} + k$
inner tapir
here b = 5, but still h = -2/5
lean otter
you shift left 2 first
then you are multiply x exclusively with 5
$f(bx-h)) + k = \frac{1}{(bx - h)} + k$
flat frigateBOT
DoomKing
inner tapir
do you see how g(x) is only shifting left 1? that's because h there is really 1 after you factor 2x + 2 = 2(x + 1)
lean otter
where are you getting 2x+2
inner tapir
sorry i just threw numbers in
you can change the coefficient there to 5
and you'll see f still stays shifted left 2, while g only shifts left 2/5
lean otter
yeah
cause VA:2x+5=0
x can't equal 2/5 cause it would give back 1/0 which is undefined
inner tapir
sure!
that's correct, but do you see that you need to factor to stay consistent with h in our equations above means the horizontal shift?
lean otter
yeah by the def of a horiz translation is to replace x with x-h
horiz scale is to replace x with (1/a)x
inner tapir
correct!
but there's a sort of order-of-operations with transformations of functions
lean otter
there is a rule of thumb
that rule of thumb isn't consistent
equations like these are the exceptions
inner tapir
so
take any function you want and graph f(bx - h)
your h will almost never be the horizontal shift
lean otter
cause imagine you have f(x)=x
you move that left to create g(x)=x+2
then take gx and horiz scale by 1/5
you get 5x+2
but reverse that order
f(x)=x
horiz scale makes the g(x)=5x
and now you are moving g(x) left 2
so 5(x+2)
or 5x+10
inner tapir
(just for the record the horiz scale here is 5 not 1/5)
but yes i agree? and i think that's what i'm saying?
maybe i'm not being clear haha
WAIT
lean otter
horiz scale is the reciprocal of a
inner tapir
no i disagree lmao
wait wait wait
in your first example the scale needs to apply to the whole function
sorry i missed that
lean otter
that would be vertical
inner tapir
horiz scale is the reciprocal of a
not for the simple linear case we are dealing with i'm so sorry i keep misreading what you're saying
lean otter
this isn't linear
inner tapir
yes you're correct
lean otter
it's a hyperbola
inner tapir
well the example you were using was
also correction above
for myself
let's start over haha
lean otter
horiz scale is always reciprocal
inner tapir
yes i read vertical i'm real sorry
let's start from the top
just these two steps you have written in this order:
- left 2
- horiz shrink of 1/5
i say f(x) => f(x+2) => f(5(x+2))
which would mean your denom must be 5x + 10, which you don't have
lean otter
but you are only replacing terms with x with 5x
inner tapir
you are replacing the argument of your function
which after you've been doing some transformations, is no longer simply x
lean otter
horiz scale is f(ax) where a is the reciprocal of what you are scaling it by
inner tapir
agreed
inner tapir
so that desmos link demonstrates that your shift is no longer left 2 if you say that
lean otter
it doesn't matter what my shifts say
my shifts change the graph in the order given
the result should give gx, doesn't matter if left 2 isn't physically there
inner tapir
it doesn't matter what my shifts say
it does matter, because when i follow your directions to construct a new function using your transformations, i don't get the function that we expect
lemme write this differently, it's really important that you are transforming the args of the function, not just x (which is usually the arg of functions we deal with)
f(u) take (u -> u + 2)
f(v) = f(u+2) take (v -> 5v)
f(5v) = f(5(u+2))
you take f(5v) and replace it with the terms that contain the variable ignoring parenthesis
inner tapir
you take f(5v) and replace it with the terms that contain the variable ignoring ...
absolutely not YOO i'm so dumb
inner tapir
you take f(5v) and replace it with the terms that contain the variable ignoring ...
lemme take this slower. can you clarify what you mean here?
are you saying the last line should be f(5u + 2)?
cause then i say absolutely not. that's inconsistent with how we are using transformations of functions
inner tapir
ok i've made a nice demonstration again (this time with sliders! :D)
play with h and tell me what you see
lean otter
https://www.youtube.com/watch?v=l3G_OIeKfh0 go to 9:10 and this is what i mean
lean otter
logically that doesn't make any sense
if i was too move every point, on the graph (yes, i know there is an infinite number of points), the graph's slope shouldn't be changing like the red is
it should be how the blue is changing by simply moving horizontally
inner tapir
ok so two things
- the vid is incomplete
- it's incomplete because while the
hthat you and the vid author are using will shift the graph horizontally, it's not byhunits! the effect is happening, but you are misreporting the value. unfortunately she only goes over a simple example where there is no horizontal stretch 😦
i can try to find a yt link if that's easier to learn from?
inner tapir
it should be how the blue is changing by simply moving horizontally
they are both moving horizontally, but notice f_0 is shifting the exact amount of h
while f_1 is shifting h/b
lean otter
either one of us is color blind or we aren't looking at the same thing
inner tapir
im so sorry i have a blue light thing i took a gamble lol
ill change to the names
i included the vertical line for h, but you can add the one for h/b so you can see that better
lean otter
ok, answer me this
set b=5 and h=-2
type the 2-(1/(5x+2))
and make both equations -f
why does mine give back the correct graph while yours gives a diff one
also add 2
for k
lean otter
i said b=5
lean otter
h is -2
inner tapir
gimme one more try haha
lean otter
going left is the same as --h
lean otter
you do x-(-h) which turns to x+h
lean otter
and now when you add 2 to the outside for k and do -f for the reflect
inner tapir
if this doesn't convince you i have to go and you need to sleep on it
hold up
just watch
i've been focused on this shift, but you should see which graph is yours and what the a, b, h, k values are
lean otter
...
this entire time
you are just reversing my steps
with the same outcome
-2/5=h is a very big difference between h=-2
lean otter
you are doing a horiz scale first then moving it l/r, im doing the reverse
it doesn't matter the order in which you do it, just cause it doesn't show perfectly on the graph doesn't mean it doesn't work
i just checked with one of the calc BC teachers, my method does work, your method is based on the making every transformation work perfectly for the graph while mine avoids fractions
inner tapir
the order matters
again your steps were
- left 2
- horiz shrink of 1/5
now take the asymptote for instance
it should end up where?
cerulean canyon
what's the original problem here
lean otter
2/5
cerulean canyon
the same 5x + 2?
inner tapir
what's the original problem here
one sec
lean otter
$g(x)=2-\frac{1}{5x+2}$
flat frigateBOT
DoomKing
lean otter
$f(x)=\frac{1}{x}$
cerulean canyon
and the question is?
lean otter
transformations from fx to gx
flat frigateBOT
DoomKing
lean otter
- left 2
- horiz shrink of 1/5
- reflect x axis
- up 2
inner tapir
g(x)=2 - 1/(5x+2)
would the translations from fx to gx be
- left 2
- horiz shrink of 1/5
- reflect x axis
- up 2
cerulean canyon
horizontal translations are only evident when they affect x directly
not its scaled form
in 5x + 2, +2 is a vertical shift
in 5(x + 2/5), 2/5 is a horizontal shift acting on x
you can see this through a change of variables
since you can just swap x + 2/5 = w and do all your computations in terms of this new variable w
flat frigateBOT
DoomKing
cerulean canyon
which effectively shifts the axis to the left
sure, but in that form, the vertical and horizontal offsets are not clear
you would rather write it in a form that makes it clear what is a vertical and what is a horizontal shift
cerulean canyon
that makes no sense, those are stretches and compressions
nothing to do with adding
lean otter
we are looking at just the denominator
cerulean canyon
just write it in a way that it is clear what is happening to x and what is happening to f
if you look only at the denom, to know what shifts and scaling operations are happening, you need to write it in a way that makes it evident
i.e. 5(x + 2/5), with a horiz shift of 2/5 and a compression of 5
either that, or a line 5x with a vertical shift of 2
lean otter
well 1, the 2nd step is flat out wrong
cerulean canyon
the problem is that you will be composing the function a few more times
lean otter
it's reciprocal of your a
cerulean canyon
what 2nd step?
lean otter
to get 5(x+2/5), assuming this is a deno, you have to first horiz shrink by 1/5 then move it left 2/5
cerulean canyon
the order of those 2 doesn't matter
lean otter
they do...
cerulean canyon
it doesn't
inner tapir
@lean otter hey i have to go here, but i'm really glad you stuck around and really tried to understand. honestly admirable, i work with a lot of students who would have quit a long time ago 😅
i'm not perfect, but i hope you can use what we worked with (especially that last desmos example). trust me when i say: if i could say the magic words that made you understand what i knew and where i was coming from i would say them !
goodnight both of you and gl with your math journies ofc ❤️
lean otter
do you agree that when moving something horizontally, you are replacing every instance of x with x-h or x+h depending on what direction?
cerulean canyon
surely, that's exactly what i meant when i told you to replace x + 2/5 with w
lean otter
do you also agree that when stretching/shrinking horizontally, you take the reciprocal of that value (let's define this new value as "a") and replace every instance of x with ax
cerulean canyon
if you want to do a change of variable again, yes, but this changes the axis
lean otter
then the order does matter
cerulean canyon
sure, but i'm not changing the axis
not in the second case
i didn't do 2 substitutions
only 1, for addition
you want to do 2, for whatever reason
and then in that case, it does matter
lean otter
im only doing 1 substitution for the addition of h
cerulean canyon
nope
when you took the reciprocal and replaced every x with ax
this is a change of variable
same as when you replaced x with x + h earlier
this is the reason why in your case it matters, but not in mine
i did the compression on the same original x axis, you transformed into a new ax axis
which i honestly find less intuitive
as soon as you say "replace"
you're changing variables and moving to a new axis
so you did 2 substitutions, if you read up what you wrote
so your method has to be different from the one i was suggesting
lean otter
you physically can't do 2 steps in 1 substitution
cerulean canyon
this has nothing to do with "physically"
and i didn't
i did one substitution and one multiplication
you did 2 substitutions
the order of substitutions matters
lean otter
how so?
cerulean canyon
multiplication is commutative
lean otter
i move left 2, then replace x with 5x
cerulean canyon
i said "let x + 2/5 = w", and then "compress it by multiplying by 5"
so i said "multiply 5 by w"
lean otter
that's a vertical transformation, and that would affect the numerator
cerulean canyon
you said "let x + 2/5 = w, then let u = w/5"
oh, sure it does, but you yourself said to focus on the denominator
lean otter
i never said that
lean otter
quote me
lean otter
numerator=denominator?
im still working in context of the problem, i'm just looking specifically at a smaller area
cerulean canyon
you need to understand, by the way, that there is no single sequence of transformations that will give you the same result
there is more than one way of doing it
you are sticking to a way that is very complicated, too
cerulean canyon
yeah cuz left 2 is wrong
the operation + 2 is acting on another function f(x) = 5x
so it's a vertical shift
flat frigateBOT
DoomKing
lean otter
$f(x)=\frac{1}{x+2}$
flat frigateBOT
DoomKing
lean otter
$f(x)=\frac{1}{5x+2}$
flat frigateBOT
DoomKing
lean otter
every instance of x gets replaced by 5x
cerulean canyon
ok, the issue is that this is an operation on x, and not on f(x)
if you want to transform f(x) into g(x), you need to do something to f(x)
but here you just swapped out the variable in a way that does not directly translate to an action on f
the overall action on f(x) in that step is (1/5)f(x+2/5)
cerulean canyon
well, this also depends on how you are seeing it in class
lean otter
reciprocal of 1/5=5. f(5x)
cerulean canyon
the definition i go by is this one
in my eyes, what you did is replace f(x) by f(w)
when you should have kept it as a function of x
so rather a(f(x+h))
lean otter
that's vertical
cerulean canyon
but if you saw it that way in class, that's also ok
cerulean canyon
that's entirely different
it isn't, you can reach the final transformation in more than one way. also i just used an arbitrary letter a, not the one you defined
cerulean canyon
so my comment would be "i prefer doing vertical compressions, you are doing horizontal ones"
what i would do is
start with 1/(x)
shift left by 2/5
stretch vertically as (1/5) (1/(x + 2/5))
make negative, so reflect along x axis
and then shift 2 up
cerulean canyon
and what you're doing is with horizontal compressions, which i honestly find a pain in the butt
but horizontal compressions are a change of axis, just like horizontal shifts
and in that case, order DOES matter
as you pointed out yourself
in my case it didn't
lean otter
yeah tahnks
cerulean canyon
as you see, there's more than one way
lean otter
thanks*
cerulean canyon
that aside, then, your way seems ok
lean otter
.close
safe radishBOT
safe radishBOT
Available help channel!
lean otter
I need help with 19 and 20
lean otter
<@&286206848099549185>
safe radishBOT
@lean otter Has your question been resolved?
lean otter
nvm i figured it out myself
safe radishBOT
@sudden oar Has your question been resolved?
tawny totem
in desmos.com:
if I were to have
f(x)=(x)^x
g(x)=(x+1)^x
is there anything I can use to plot two points and draw a line between them if one is f(x) and the other is always one more x and y value larger than f(x)?
tawny totem
to be more precise, there are two x^x functions
one starts at x=0, other at x=1
if the x=0 function has a point at x=1,y=1, how can I replicate that answer on to the second graph AND add one of both x and y to the answer
pulsar condor
just define f and g, then calculate the slope b/w 2 random points on them, then find the eqn of the line
hasty trout
find the max and minimum value in that set
safe radishBOT
find the max and minimum value in that set
hasty trout
i think i can either test some values and write down what happends
or maybe look into the first derivative?
median flint
First derivative looks like the way to go
hasty trout
$-\frac{e^{\frac{y+x}{2}}}{2}$ like that?
flat frigateBOT
LeonMarky
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lean otter
next gulch
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next gulch
3b
I'm not really sure how I should do this
I understand that the left denominator can be written as cos
But no idea of the right
ancient escarp
no it cannot
next gulch
Oh
ancient escarp
"use common denominator"
flint ledge
multiply by the conjugate of the denominator for both of them
flint ledge
@next gulch wait a second i have a slightly different approach
rewrite it as $\frac{1}{-1+\sin(x)} + \frac{1}{1+\sin(x)}$
flat frigateBOT
quantum
next gulch
Yeah let me try a few things
flint ledge
no problem
next gulch
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flint ledge
just realized i could have just multiplied the top and bottom by -1 instead 
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hexed cedar
@trim hollow you asked a question similar to this earlier if I’m not wrong. Apply that same principle here
snow pivot
Im trying to solve this, i tried making a taylor polynomial about x=0 of the numerator function but found it got a bit too complex/difficult that made me suspect there must be a better way
safe radishBOT
Im trying to solve this, i tried making a taylor polynomial about x=0 of the num...
lyric ember
Have you tried L'Hôpital?
snow pivot
ahh damn im havent learnt that yet but it is the next topic ill have a look at it
mental plover
lhopital isnt any better
lyric ember
You are right, it is horrendous lol
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flint ledge
you’ve learned taylor polynomials but not lhopitals rule?
snow pivot
im doing an online course thing and these are in the "challenging questions" part
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rough seal
you’ve learned taylor polynomials but not lhopitals rule?
This is a mad man running their class
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This is a mad man running their class
rough seal
Doing a 9th derivative using the definition of a limit 😭
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ember bough
im doing an online course thing and these are in the "challenging questions" par...
wait, where is that from?
flint ledge
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lean otter
Hello
lean otter
For the b) question the function will not be continuous at 5 and 4. But how to write that in mathematical form?
<@&286206848099549185>
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flat frigateBOT
HELLOBELLO
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merry igloo
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merry igloo
I know this is a question from text channel five
And I want to know how to solve the A and B and C part
And i am interested to know how to solve this one
night basin
thats easy
merry igloo
I thin I take 10000000/100 to find the vaccinated one and 1000000/900 to find not vaccinated one
Hold on B is 10000000/100
Hold on
ok
A is 10000000/25/100+10000000/75/900
<@&286206848099549185>
How to find A and B and C
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merry igloo
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brittle wave
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untold topaz
How to do 1?
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burnt skiff
Can someone help me with this please?
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Can someone help me with this please?
burnt skiff
<@&286206848099549185>
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neon dew
ionic hedge
.help
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mystic nacelle
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still carbon
thats the data i get
and thats the quastion
What is the Pearson correlation coefficient between the two variables? If so, was the decision to choose the t test for dependent samples really justified?
safe radishBOT
thats the data i get
and thats the quastion
What is the Pearson correlation coef...
grizzled fossil
thats the data i get
and thats the quastion
What is the Pearson correlation coef...
Read the picture you posted and you get the answer
still carbon
i know what the Pearson correlation but i dont know the second half of the question
grizzled fossil
So are the samples dependent?
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serene flicker
Does anyone know what the heck is going on here:
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Does anyone know what the heck is going on here:
serene flicker
6(38)^2 should be 8664 not 228
i feel like im taking crazy pills
NVM i figured it out, they messed up the annotation
it should not be (38)^2 its just 6(38)
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velvet raft
What kind of series is this?
safe radishBOT
What kind of series is this?
queen parcel
If you have to use a specific test, then the nth term test works best, given that the limit as n -> ∞ ≠ 0
velvet raft
If you have to use a specific test, then the nth term test works best, given tha...
there is no specific test i need to use
can you elaborate on the type of test
cosmic grove
there is no specific test i need to use
d'Alembert should be fine
velvet raft
ratio test?
pulsar condor
yeah ratio test should be sufficient
cosmic grove
what is ratio test ?
pulsar condor
sum converges iff ratio of terms is always less than 1
oh yeah abs value of the ratio
cosmic grove
well it's called "critère d'Alembert" where I am
pulsar condor
Im gonna guess D'Alembert proved it first then
balmy marlin
anyone know implicit derivatives
cosmic grove
no cap
balmy marlin
my bad i literally joined 2 mins ago lol
why did they multiply by y over y in the farmost right
ik everything else but that
pulsar condor
why did they multiply by y over y in the farmost right
#❓how-to-get-help wasn't a joke.
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cerulean dagger
shut steeple
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shut steeple
when will i get a answer to this
i am posting this for litterally 3rd time
and no answer yet
NO ONE KNEW
HOW TO SOLVE THIS
i solved this for fking 6hrs
i dont wanna solve anymore
mhm ok
but
later
ill just ppost here rn
for rn*
Wanna know a fact
This question is from russia 1991
Olympiad
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@shut steeple Has your question been resolved?
shut steeple
I dont know any names
But say the rule
Mhm
O m g
Ur god
Wait so
Can u give me a example of this
Rule
With a cubic
Equation
Ok
1 sign change
Mhm
Lets increse it a bit
X^3-x^2-3x+2
2 sign change
O
But qait
Wait
But
But
What will i do with number of positive roots
So basically
How to find number of negraive roots
True
So what can we do it
We just found out the total integral solutions
Which is 2
And we cant have 0 bcoz question already said integral roots lol
No i meant 0 positive integer and 0 negative integer
Will result in 0 roots
So others are imaginary
So we xant have 0 positive integer
Cant*
Lmao
Ok so
U said
Exactly
And we have 0 negative roots
B r o i saod that implies
If there are 0 pos integers and 0 negatice then others are imaginary
U said either zero or two
I am taking zero as the situation
Yup
Rhis is true
This*
shut steeple
If there are 0 pos integers and 0 negatice then others are imaginary
.
I said
Imaginary roots
In case of 0 pos integers
The question
Said
The qudratic
Has integral roots
yup
we just found that
fcn can have 2 possibilities
- 2 POSITIVE roots
- 0 positive roots
we also found we have 0 negative roots and 0 zero roots
SO in case 1) THERE ARE 0 IMAGINARY ROOTS
which satifies the question itself
- THERE ARE 2 IMAGINARY ROOTS
which does not satify the question since it already said its integral roots
shut steeple
yes sir
.............READ THE LAST LINE
it says it
-_-
yea the function
fcn is the function
yes thats what i have been saying how will it have complex roots it has to be integrals so the possible way is 2 positive integral roots
man
i am just saying
2nd case where the roots become imaginary is not possible
the first case is possible
of 2 positive integers
fine let me make it a bit better word
2nd case where the roots become imaginary is not satifying trhe given condition of question
the*
......................
.............................................
more dots
o m g
it already did tho
it alreayd said it has integral roots
its a condition
AAAAAAAAAAAAAA
either ur not understanding me or i am not understanding u
and this doesnt even lead to a answer
it just says if its positive or negative
or imaginary
and all cases
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brittle crane
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frigid osprey
Could someone explain me the relation between the sinus theorem and the circle
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silent forum
Hi Soy
silent forum
Could someone explain me the relation between the sinus theorem and the circle
Sure thing
Take triangle ABC inscribed in circle O
And now drop altitudes from the center Q to each of your sides
So far so good?
Let me know when you want me to continue
frigid osprey
Let me know when you want me to continue
Altitude u mean height?
silent forum
Yeah
So take for instance triangle QAB
As QA and QB are radii
They are congruent
So QAB is isosceles
And now angle AQB is twice angle ACB
All good up to this point? @frigid osprey
frigid osprey
Yep, @silent forum
silent forum
Ok
So then let D be the foot of the perpendicular from Q to AB
Then DQA is equal to DQB
As AQB is isosceles
So
Do u see where im going?
I think you should be able to finish from here
Wouldnt want to spoil the whole thing to you :))
frigid osprey
Okay let me represent
im so slow at math
q is the center of the cirle
or of the triangle?
@silent forum
silent forum
Center of the circle
frigid osprey
okay
silent forum
im so slow at math
This takes practice to improve
frigid osprey
alr
fiery flower
silent forum
Ok
frigid osprey
Is this alright?
frigid osprey
Okay
frigid osprey
I dont know
That their lenght is the same
and it is the radium of the circumference
frigid osprey
Alright
frigid osprey
QAB?
silent forum
The triangle
frigid osprey
its a isosceles trianglle
silent forum
Yes
frigid osprey
okay
silent forum
Keep in mind btw that were looking for the sin of the angles in our triangle
frigid osprey
Yep
Sin was the opposite of the angle/hipothenuse
or whatever is it said in english
silent forum
First, how are angles AQB and ACB related?
silent forum
Sin was the opposite of the angle/hipothenuse
Yup
frigid osprey
First, how are angles AQB and ACB related?
They both have the same sin?
silent forum
Diagram is fine
silent forum
Yeah they are congruent
Meaning that they have the same length
There's this property of circles
That an angle with the vertex on it is half an angle with a vertex in the center of the circle
Given that they intersect the circle in the same two point
To rephrase this more clearly
Angle ACB and angle AQB
Both rest on A and B
With A being on the circle and Q being the center
So AQB is twice ACB
We xan prove this too if you want
wild cape
ty
silent forum
Congruent means the same thing?
Yes congruent and equal are the same
frigid osprey
Yes congruent and equal are the same
okay
silent forum
Altho
Technically
Equal segment would coincide
But that's not usually considered
Now drop an altitude (or height) from Q to AB
In ur diagram
frigid osprey
So AQB is twice ACB
why?