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#help-40
peak ivy
remote sandal
frigid sundialBOT
||
remote sandal
have you made any progress so far?
Nooo I’m losing my mind over this :(
hushed wedge
as in you have no thoughts whatsoever on it? even that it is impossible?
as written, the problem says that the common ratio is an integer
this means the GP is either constant or increasing (in absolute value anyway)
yet 108 comes after 4343
so what gives?
remote sandal
Yea I spoke to my professor and he made an error while printing the question
Sorry for the trouble @hushed wedge
hushed wedge
ah so it was a typo
remote sandal
It’s 4/3 rather than 4343
hushed wedge
so 4/3 and 108 then
remote sandal
Which gives us the 4 root of 3
hushed wedge
what's the ratio of these?
remote sandal
And then we divide the q term and p term
hushed wedge
324/4 = ... 81?
remote sandal
Yes
hushed wedge
yes, ok, so the ratio is either 3 or 9
hushed wedge
yes, there's only finitely many combos that work
well actually hold on
can't R also be -3 or -9
remote sandal
It can be, which is why we get 4 possible soln.
And we can choose the smallest value
Sorry my internet is bad
Logically speaking you will have to take the smallest value for R
Which is why we take -9
R=-9
Q-P= 2
-9+2=-7
@hushed wedge this is how I solved it
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remote sandal
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hushed wedge
??
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frosty mica
hi, so for this question i cant figure out how they simplified the 4/3 power, i assume that they simplified it down to something like (3x2x2x2(1)^3/4)/4, or something like that, but im not sure how they did it, also, ignore my answer, i just put that in so i could see the solution
frigid sundialBOT
hi, so for this question i cant figure out how they simplified the 4/3 power, i ...
hybrid hollow
Wdym simplified?
frosty mica
so if you look at after at (8,-100)
they simply 3/4(8)^4/3 to 12
i was wondering how they did it
hybrid hollow
8^(4/3)=8^(3/3) * 8^(1/3)
frosty mica
ah
hybrid hollow
(Or just use a calculator)
But why not just use a calculator?
mighty agate
Hello 👋 I’m struggling to solve it. May be I’m doing something wrong
frigid sundialBOT
Hello 👋 I’m struggling to solve it. May be I’m doing something wrong
ember tulip
what part are you stuck at?
mighty agate
The answer is not correct. Therefore I can’t understand how to solve it
mighty agate
My answers are
f(0) = 4, f(4) = 0.00134, f(-2) = 218.392 but they’re incorrect.
Mb I need to use limit to find a function
Oh no I can’t use limit
😦 still don’t know why it’s incorrect
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plucky panther
hi, is (y+2)(y-9) and (y-2)(y+9) the same thing
frigid sundialBOT
hi, is (y+2)(y-9) and (y-2)(y+9) the same thing
ivory arch
yes, multiplication. ab=ba
wait
sorry didn't see the signs
it's not the same thing
(y+2)(y-9)=y^2-7y-18
(y-2)(y-9)=y^2+7y-18
ivory arch
What?
ivory arch
I don't understand what you've done.
ivory arch
ohhhhh
plucky panther
and we need to solve it
plucky panther
yea
plucky panther
im stuck
ivory arch
On solving it?
ivory arch
um
Can you explain what you have done in the pictures?
You multiply it like this, draw arrows (to make it easier) and multiply y with y then with -9, then 2 with y and -9
You get y^2-9y+2y-18
Then simplify, you get y^2-7y-18
Now you wanna solve it and find y1 and y2.
y^2-7y-18=0
You can factor it or use formulas. Do you know how to do that?
This is how it would look like for 1 and 2
Number 1 being (y+2)(y-9)
Number 2 being (y-2)(y+9)
You can solve it like this too by completing the square.
plucky panther
dang so complicated
ivory arch
y^2-7y=18
Take half of 7 raised to the 2nd power and add that to both sides.
y^2-7y+(7/2)^2=18+(7/2)^2
y^2-7y+(49/4)=18+(49/4)
Then factor it.
(y-(7/2))^2=121/4
ivory arch
dang so complicated
It's not so complicated, I think I might be making it harder. I'm trying my best to explain.
ivory arch
Watch these.
https://www.youtube.com/watch?v=qeByhTF8WEw
https://www.youtube.com/watch?v=nyk3UGwCAms
novel quiver
frigid sundialBOT
novel quiver
not exactly sure how to do this
my beautiful drawing
is the question asking if like matrix A exists in matrix B?
frail garnet
it's asking which of those product is defined
novel quiver
it's asking which of those product is defined
are they not all defined then?
oh
a 4by2 matrix can't multiply into a 2by4 matrix
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frigid sundialBOT
hybrid hollow
Find the value, x, such that P(X>x)<=0.01
.
H_0: p=0.3, H_A: p>0.3
What values will be critical then?
Small or large values?
Yes
Yes
hybrid hollow
Find the value, x, such that P(X>x)<=0.01
This
Say more than “I don’t understand”
1%
Says so in the question
Using a 1% sign level
It would be equal but our RV is discrete
Meaning you can’t actually hit equal
Random variable
30 is not a random variable
30 is a number
Says in the question
Is 0.3 “random”?
Seems like a number to me also
hybrid hollow
Says in the question
.
Read the question
Search for the words random variable
(Or just know very basic lecture 1 definition)
Try reading
?
It only says random variable one place in the question
Try reading what is right next to those words?
Like what?
hybrid hollow
(Or just know very basic lecture 1 definition)
Now we know like 1st definition given in your course
Just amazed you don’t know it
Seeing how that was one of the 1st things given to you
(Surely)
hybrid hollow
It would be equal but our RV is discrete
.
?
“You asked why less than and not just equal”
I answered
Then went on a detour because you didn’t know what a random variable was
hybrid hollow
Find the value, x, such that P(X>x)<=0.01
We are still here
Test values or use inverse distribution (implemented in any software)
I have a pc
What does CD compute?
So, P(?)
What does it compute
hybrid hollow
So, P(?)
Replace ? with what it calculates
What does it compute if you out x=1
P(?)
?
I don’t want a prob, I want the thing it calculates
So I know you understand it
hybrid hollow
I don’t want a prob, I want the thing it calculates
.
What does cumulative prob mean?
Grapping a book is the way then
Sounds like something you just googled
What does this compute then?
Instead of just typing words you don’t understand
I don’t want a probability
I want you to say what probability it computes
No
Your method is not learning a single definiton?
And just try random stuff
Without understanding anything
Try reading what you copy-pasted and understand it
It tells you what it computes
Try reading
In what you copy-posted or just grap your book
Finish reading
Probability of a RV is nonsense
What range?
Try reading sentence #2?
So what does what you typed in your calculator compute?
To what?
???
What did you plug in
What do you think it then calculates
I mean…
What is a binomial distribution?
frigid sundialBOT
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magic hinge
frigid sundialBOT
magic hinge
I already found AQR (22 degrees), but how do I find ABQ?
frigid sundialBOT
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magic hinge
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red crown
Hey im new here can anyone help me please
frigid sundialBOT
Hey im new here can anyone help me please
full gorge
How can I find the area of a regular polygon with only the radius?
frigid sundialBOT
How can I find the area of a regular polygon with only the radius?
barren kelp
full gorge
I get upto where he's using trig and then I get a negative
in my case it's a polygon with a radius of 25mm so I do 25sin(36)
Oh my calculator wasn't set to degree mode
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frigid sundialBOT
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undone crypt
I need help
frigid sundialBOT
I need help
candid olive
I need help
which two numbers (constants) multiply to get your x^2? then those two numbers = what? so then you take those (two numbers)*n=the number then you solve for n, sorry if this is confusing, I don’t wan to just give you the answer
frigid sundialBOT
@undone crypt Has your question been resolved?
undone crypt
@candid olive im not sure it’s a practice test
candid olive
if you solve it out what is multiplied by 2nx^2
candid olive
when ur foiling u multiply x and 2nx^2, x and 1, then what and 2nx^2?
@undone crypt
frigid sundialBOT
candid olive
rip
candid olive
type .reopen
undone crypt
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candid olive
well if i told you the answer that’d wouldn’t be helping you
undone crypt
The 2nd parenthesis @candid olive
I haven’t done the foil problem In forever 💀
Tho I think I remember some things
candid olive
do you at least know how to multiply out the two sides?
undone crypt
First outer inner last
I think so
But what is the value of x
If I’m multiplying that in
candid olive
(first + second)(third + fourth)
first * third + first * fourth + second * third + second * fourth
so multiply $x2nx^2+x1+262nx^2+261$
karmic cometBOT
FearMeNons
undone crypt
Lord have mercy
candid olive
what’s confusing
or just look at what will be in the x^2 position
first one will be x^3 position for example
undone crypt
Wait so 2nx1^2+x1+52nx^2+26
candid olive
because it’s 2nx^3
undone crypt
Wait so 2nx1^2+x1+52nx^2+26
Is that right?
candid olive
yes
well technically the first one is 2nx^2 not 2nx1^2 that’d be 2nx but the first one doesn’t matter
wait i mean the first one is 2nx^3
mb
so which one is in the x^2 position
undone crypt
Wym
candid olive
$2nx^3+x+52nx^2+26$
karmic cometBOT
FearMeNons
candid olive
which one is being multiplied by x^2
undone crypt
The 2nd
vivid grove
Can you simplify without n and compare with n and without?
candid olive
you set the 52nx^2=-520x^2
candid olive
wait should be -52nx^2 because there’s a negative on the outside of the ()
$-52nx^2=-520x^2$
karmic cometBOT
FearMeNons
candid olive
you can get rid of the x^2’s and solve for n
you could also do latex $6x^3-2nx^3=-14x^3$ but we’ll just stick with this
undone crypt
Ok
karmic cometBOT
FearMeNons
undone crypt
Then what?
candid olive
$-52n=-520$
karmic cometBOT
FearMeNons
candid olive
so just divide to get n alone
undone crypt
I get -572
candid olive
huh
undone crypt
The heck
candid olive
you divide -520/-52
undone crypt
Ok
candid olive
do you understand how we got here?
candid olive
I struggled at first but if you practice it can become very easy
undone crypt
Are you able to show the work step by step and explain it so I can get an idea
candid olive
sure
frigid sundialBOT
@undone crypt Has your question been resolved?
candid olive
I’ll do the -14x^3 one because now that I think about it it’s probably easier.
In order to find n we need to find something that corresponds between the product and the factors so need to have components with x^3 attached to them. we know on the left it's 6x^3, so need know what on the right side of the equation produces something with a x^3 attached to it. so if we foil out the right side we get 2nx^3+... and subtract it because there's a minus between the left and right side of the equations, so we need to know what is n so that we get a product of -14x^3. so we take latex $6x^3-2nx^3=-14x^3\\ -6x^3 -6x^3\\ -2nx^3=-20x^3\\ \frac{-2nx^3}{-2x^3}=\frac{-20x^3}{-2x^3}\\ n=10$
You should probably work on foiling so you can expand quadratics.
frigid sundialBOT
I’ll do the `-14x^3` one because now that I think about it it’s probably easier....
karmic cometBOT
FearMeNons
candid olive
Maybe your teacher can help you more
candid olive
I’m so tired 🥱
well I hope that helps and that you get good sleep, is there anything you want me to clarify?
undone crypt
No
simple ravine
Ok so I need help with this limit
frigid sundialBOT
Ok so I need help with this limit
proper arrow
what is the first symbol?
craggy maple
what is the first symbol?
Ig e
simple ravine
what is the first symbol?
e^x
hushed wedge
$\lim_{x \to -\infty} e^x(x^2 - 4x + 5)$
karmic cometBOT
Ann
craggy maple
Shouldn’t it be infinite?
simple ravine
if so then yes this is zero
Thank you
hushed wedge
Shouldn’t it be infinite?
x goes to negative infinity.
so e^x approaches 0 and does so faster than any polynomial
craggy maple
Ohh i see
proper arrow
technically indeterminate, but after some work, yes 0
craggy maple
x goes to _negative_ infinity.
I thought + infinity
jagged leaf
Do I have to figure out the value of a for me to answer the following ratios?
ember tulip
no
in the trig ratio you'll be dividing a side by another
like opposite/hypotonuse
so the a will cancel out
jagged leaf
If it'll be cancelled out what will happen to the adjacent of 60°?
Ah then what is this given about?
onyx flint
?
jagged leaf
I mean why are there two answer tan (30°) and rationalizing it
onyx flint
when your denominator is irrational, you have to rationalize it
well it's better to be rationalized
dusky pasture
How do I determine the number of permutations of order 4 in group of S5?
frigid sundialBOT
How do I determine the number of permutations of order 4 in group of S5?
@dusky pasture Has your question been resolved?
dusky pasture
<@&286206848099549185> 👉🏻👈🏻
dusky pasture
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frigid sundialBOT
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worn gull
i need some help on how to set up the intergrals for part c
i've looked through some of my lecture notes but im not too sure on how to get the bounds specifically
frigid sundialBOT
i need some help on how to set up the intergrals for part c
i've looked through ...
@worn gull Has your question been resolved?
worn gull
<@&286206848099549185>
haughty falcon
i need some help on how to set up the intergrals for part c
i've looked through ...
find the pdf of Y-X
worn gull
if you mean the f_X(x) and f_Y(y) then i have them already
tbh just some resources on how to solve this sort of problem would be helpful enough
i dont need to be walked through this example specifically
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simple ravine
so im not sure how to PROVE this is true
frigid sundialBOT
so im not sure how to PROVE this is true
simple ravine
Show that [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] → [(𝑝 → 𝑟) → 𝑠] is always true.
without having the implication of it being true
so like i know that is true unless [(𝑝 → 𝑟) → 𝑠] is false and [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] is true
but im not sure how im supposed to prove that its true without just attributing s=T or s=F and see what happens
frail garnet
assume otherwise
simple ravine
frail garnet
which means [(𝑝 → 𝑟) → 𝑠] is false and [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] is true
simple ravine
honestly i think im just too lazy to do this problem
because the truth table for it would be immense
frail garnet
if (𝑝 → 𝑟) → 𝑠 is false then (𝑝 → 𝑟) is false and 𝑠 is true
which implies p is false and r is true
simple ravine
i can assume 𝑠 ≡ F, and either 𝑝 ≡ F or 𝑞 ≡ F
and then that 𝑠 ≡ F, 𝑝 ≡ T, and 𝑞 ≡ T
oh
frail garnet
which implies p is false and r is true
substitute that in the other one and you get a contradiction
simple ravine
substitute that in the other one and you get a contradiction
allright
thanks
/close
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obtuse marten
ashen echo
lament lava
its 210 right
I got a different answer
obtuse marten
what
lament lava
Wait i missed a part of the question that said “three”
supple parcel
So boring
lament lava
So you might be right
obtuse marten
if it is could you help me with
ik its not the first and last options
obtuse marten
So you might be right
yo could you help with the other one
frigid sundialBOT
@obtuse marten Has your question been resolved?
stone coyote
frigid sundialBOT
stone coyote
Hey, I can't understand where I made my mistake
I feel like I solved it differently
however, it should be correct
The second one is the solution
and the first one is mine
sweet wren
stone coyote
this part?
proper arrow
yes
stone coyote
ah
so I wrote it twice
thank you so much!
I solved it finally!
Thank you so much
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stone coyote
would it be correct to write like this?
frigid sundialBOT
would it be correct to write like this?
stone coyote
the solution is here, but I solved it a bit differently
hearty flare
the solution is here, but I solved it a bit differently
integral of 1/u^2 du is not ln(u^2) lol
it’s power rule
integral 1/u du is a special case because it gives (x^0)/0 which is undefined
signal condor
Try differentiating $\ln(u^2)$ to convince yourself it's incorrect
karmic cometBOT
iCaird
frigid sundialBOT
@stone coyote Has your question been resolved?
Available help channel!
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bronze mica
What
junior radish
Just use the volume of a cylinder formula
grave hound
I feel like I'm forgetting stuff and I'm getting really really stressed
frigid sundialBOT
I feel like I'm forgetting stuff and I'm getting really really stressed
haughty falcon
,rotate
karmic cometBOT
dull lake
$$\frac{2x-15}{x} = \frac{x}{x+18}$$
karmic cometBOT
Denver Crossbourne
dull lake
I think you'd do this and then solve for x
The sides are equal because you're multiplying the terms of the geometric sequence by same thing, there is a common ratio
grave hound
?
dull lake
I could be wrong though, hopefully someone can confirm
It's been over 15 minutes, you can ping helpers
frigid sundialBOT
@grave hound Has your question been resolved?
grave hound
<@&286206848099549185>
grave hound
**Denver Crossbourne**
It's this but how does that let u get integers for x
frigid sundialBOT
@grave hound Has your question been resolved?
dull lake
Do you know what a geometric sequence is?
torn wind
whats a good technique for solving trig integrals? (yknow using the trig identities)
frigid sundialBOT
whats a good technique for solving trig integrals? (yknow using the trig identit...
@torn wind Has your question been resolved?
dull lake
Idk if there's a general technique, I'd say just do a lot of practice problems and look up techniques that are specific to the kind of integrals
Could take a look at this link, might be helpful
frigid sundialBOT
@torn wind Has your question been resolved?
unreal drift
frigid sundialBOT
unreal drift
calm geyser
@unreal drift you would need to find least greatest factor then you wi find the answer
unreal drift
ah thank you :))
novel quiver
frigid sundialBOT
novel quiver
Please someone help me to solve this question.
If the restrictions has not been imposed to put list in alphabetic order. The answer would be 11C6.
But here how to solve it? 🙂
amber carbon
Since we are discarding all combinations that are NOT in order, Combinations are the way to go
$\binom{7}{6}$
pft, knew I didn't remember the code. there we go
karmic cometBOT
Disorganized
amber carbon
the weird thing about all this is that order definitely does matter, but we can reduce the problem to just CHOOSING the letters involved...since they must always be ordered alphabetically (only one way to order what we chose). Understand?
frigid sundialBOT
novel quiver
.reopen
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amber carbon
just a comment
how polite! thank you
"You will never be bored in Combinatorics"
they all seem mysterious
and when they turn out to be simple, they're even better
that's it.
🙂
novel quiver
"You will never be bored in Combinatorics"
Love this statement.
Thank you for this insight 🙂
amber carbon
having said that
it feels...like something's wrong here
💩
ok, let's see if there's really only 7 ways
ohhh
that was WITHOUT repetition.
ok, so we need to deal with that
btw (stalling)
"7C6" is the number of ways to choose 6 elements from the set of 7 to be included,
AND the number of ways to pick the ONE element that is discarded.
...but you probably remember that
oh. I know
it's the "buckets"
or the dividers. Buckets and dividers.
let me draw a picture
_!_!_!_!_!_
Sure 🙂 Thank you
amber carbon
we have n=6 spaces we need to fill, and at most n-1 = 5 dividers
we can move the dividers according to the number of distinguished buckets we have
for example, say we chose only 4 distinct letters.
then there should only be 4 buckets
ugh...lol
amber carbon
was this help? lol
I'm not sure how to proceed here
I am not sure if the previous logic is compatible with this visualization now
oh wait
(maybe)
if we did this FIRST
we can choose the letters last...
...meaning we can choose them first...by commutative property 🤔
so I guess that still holds
novel quiver
Yes, Bars and stars 🙂
novel quiver
Thank you for sharing this book. It looks beautiful.
Do you any book like this for Probability?
I really like HTML versions 😀
amber carbon
I actually did order a book on combinatorics but...wait a min.
I think it's in the garage 😳
it's buried though 😒
lemme just peak if it's on top 👀
in the bin
novel quiver
Ohh. I am sorry.
If you have ordered online then you might be able to find book name in your mailbox.
amber carbon
ok I got it
like, completely
A A A A A A
B B B B B B
C C C C C C
D D D D D D
E E E E E E
F F F F F F
G G G G G G
we can pick any 6 of these characters
and would need to divide out the repetitious characters...on a case-by-case basis
...and now I see that's where we're already stuck 😒
but it's stars-and-bars after that
we need helpers
<@&286206848099549185>
novel quiver
...and now I see that's where we're already stuck 😒
Haha
novel quiver
<@&286206848099549185>
Thank you for pining them 🙂
amber carbon
Thank you for pining them 🙂
yes, I am very talented
novel quiver
Okay 😀
amber carbon
(working on it)
novel quiver
Thank you :)
frigid sundialBOT
@novel quiver Has your question been resolved?
amber carbon
multiset
The number of 6-symbol multisets chosen from 7 symbols (with repetition), where order is unimportant, is
$\binom{k+n-1}{k}$
karmic cometBOT
Disorganized
amber carbon
$\binom{6+7-1}{6}=\binom{12}{6} = 924$
karmic cometBOT
Disorganized
amber carbon
I think this is correct. It doesn't count alternative orderings of multisets. This means we assume symbols are in alphabetical order
btw, this is the out-of-the-box multiset formula. I just wanted to be more sure before I suggested it
was the solution provided to you in your textbook?
novel quiver
was the solution provided to you in your textbook?
Yes, the answer is correct.
Thank you for putting so much effort for me.
Please convey my gratitude to your friend also.
If possible then would you please elaborate this line?
I think this is correct. It doesn't count alternative orderings of multisets. This means we assume symbols are in alphabetical order.
he didn't do shit
amber carbon
he layed on his bed and ate "wine cheese"
he showed me the book link
he's on audio with me
we are laughing about this
novel quiver
Hahaha
amber carbon
he was a little helpful
amber carbon
```
I think this is correct. It doesn't count alternative orderings of multisets...
here is what I mean about this
[a,b]
[b,a]
we can arrange a set of two symbols 2 ways
if we don't care about the order, we would only need to count one of these cases
here's a better example. Say we have 3 symbols {a,b,c} and want to count the number ways to make a set of 2 symbols from this.
if order matters, there are 3P2 ways to do this
[a,b], [b,a],
[a,c], [c,a],
[b,c], [c,b].
Yes, order doesn't matter.
So, it will also solve an issue of alphabetic order.
Gotcha.
Thank you for putting so much effort for me.
I am very grateful to you ☺️
amber carbon
I was gonna try and show that this isn't as simple as cutting the number of counts in half for combinations
but it's tedious and we have done this a while, huh?
lol
sorry this took me so long
I am going primarily by the definition of what a multiset is/does
to justify it's straight-up use
in this situation
I am glad I could help
good night
frigid sundialBOT
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fathom hollow
frigid sundialBOT
fathom hollow
I'm confused on this question, how do you find the table without converting it to exponential form?
I was in class (these are my teacher's notes) and she was doing it without making it x/3 = 2^y first
Talking about #1 here
buoyant spruce
I'm confused on this question, how do you find the table without converting it t...
i think this is a better way to write that table
buoyant spruce
I'm confused on this question, how do you find the table without converting it t...
ion think there is any other way to solve the table apart from converting to exponential form or using calci.
buoyant spruce
this is just long and confusing way
buoyant spruce
I was in class (these are my teacher's notes) and she was doing it without makin...
just do it this way by converting in to exponential form and substituting -1,0,1,2 in place of y
fathom hollow
ion think there is any other way to solve the table apart from converting to exp...
Oh
Calci?
buoyant spruce
no i meant the way u said
fathom hollow
Calculator or calculus?
buoyant spruce
calculator, yeah.
buoyant spruce
yes prolly that's just a stupid way
fathom hollow
just do it this way by converting in to exponential form and substituting -1,0,1...
Alright thanks
Got it
buoyant spruce
just makin it long unnecessarily when u could solve it easily in the way u mentioned
buoyant spruce
alright bud
fathom hollow
.close
fast nimbus
I don't get it, I have to Simplify this in the simplest form but idk what to do
white bridge
,rotate 270
karmic cometBOT
white bridge
$$\frac{x}{3-5\sqrt5}-\frac{2+3\sqrt{2}}{1}$$
karmic cometBOT
Chunkin
fast nimbus
Don't we have to multiply the denominators so we can minus them?
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frigid sundialBOT
@fast nimbus Has your question been resolved?
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gloomy pewter
in probability, i often get confused where to use the bayes theorem. any tips? this is not exactly a question, its more of a ask for guidance on approach of the question
frigid sundialBOT
in probability, i often get confused where to use the bayes theorem. any tips? t...
elfin spire
You almost always use it when there is more the 1 variable involved in a probability and the variables are independent
frigid sundialBOT
@gloomy pewter Has your question been resolved?
frigid sundialBOT