then just multiply them together for 8 = (2sqrt(2))^2
#help-23
steel stag
warm prism
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plush wadi
Quick question for this function
safe radishBOT
Quick question for this function
plush wadi
To find if the function is odd, even, or neither you plug in a -x and and positive x right?
Since negative numbers aren't included in the domain does that mean that the function is neither odd or even?
steel stag
correct
white bison
it's correct
just expand it and youll figure it out on your own
that
-15 goes with 3 to make -45
-15 goes with 5 to make -75
-15 goes with -1 to make 15
you just got back the original equation
well i dont see how it matters, the same thing could be written as 15(-3x-5y+1)
worthy hemlock
If you're asking if both are correct, yes, both are correct ways to write this
You can pull out either positive or negative numbers
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rocky raft
Hi! How to do this efficiently?
rocky raft
<@&286206848099549185>
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rocky raft
Hi! How to do this efficiently?
<@&286206848099549185>
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steady dust
how do i do this trig identity
safe radishBOT
how do i do this trig identity
ember sinew
Expand tanx and cscx
ember sinew
sinx cancels
ember sinew
Gimme a sec
ember sinew
Isn't there an identity about 1+cosx?
steady dust
no
steady dust
are we doing right or left side
ember sinew
Left
steady dust
oh we're expanding?
ember sinew
hmm?
ember sinew
Gimme a sec I'll send it
steady dust
ok
steady dust
what does it say on the second line
steady dust
so that second term came from multiplying cosx with 1 and cosx?
ember sinew
1/cosx
ember sinew
what are the red arrows indicating
Distribution
steady dust
why does 1/cosx multiplied with cosx give cosx/cosx
ember sinew
cosx in denominator(1/cosx term), cosx in the numerator(cosx term)
ember sinew
Np
steady dust
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gilded delta
Let $M$ be a set with operations $\circ, \diamond: M \cross M \to M$ such that there exist identity elements $e_{\diamond}$ and $e_{\circ}$ in terms of the operations. Furthermore, it holds: $(a \diamond b) \circ(c \diamond d)=(a \circ c) \diamond(b \circ d)$ for all $a,b,c,d \in M$. I've already shown that $e_{\circ} = e_{\diamond}$. How do I show that $a \circ b = b \diamond a$ for all $a,b \in M$ now?
safe radishBOT
Let $M$ be a set with operations $\circ, \diamond: M \cross M \to M$ such that t...
flat frigateBOT
˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞
half cove
Maybe start with $a = a \diamond e{\diamond}$
flat frigateBOT
Kiretori
half cove
**Kiretori**
e subscript diamond
And same thing for b
And apply the relation that holds for all element of M
half cove
Let $M$ be a set with operations $\circ, \diamond: M \cross M \to M$ such that t...
And then use the fact that the identity elements are equal
And then I think you'll the solution, I don't have a pen paper with me so I'm not 100% sure but it seems that it's the right way
safe radishBOT
@gilded delta Has your question been resolved?
gilded delta
How do I use the relation then
gilded delta
And same thing for b
with this
flat frigateBOT
Kiretori
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gilded delta
.reopen
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✅
gilded delta
and what now?
half cove
$(a \diamond b) \circ(c \diamond d)$
flat frigateBOT
Kiretori
half cove
**Kiretori**
You'll have an equality of this type
gilded delta
that's what I have
half cove
With b and c both equal e diamond
half cove
We just applied the relation that holds true for all elements of M
gilded delta
maybe im getting confused with the variables
half cove
Which variables got you confused?
gilded delta
why is c b all of a sudden now
half cove
there is no variable c in what I wrote
gilded delta
but we are replacing c with b no?
and b with e diamond now
so wouldn't that mean that b is e diamond
half cove
The b in the first relation is not the same b I used here
b just means an element of M
we could've named it h or g or whatever
The relation can be applied not only with variables named a, b, c, d
It can be applied with EVERYTHING that is an element of M
They said prove the equation for all (a,b) of M
gilded delta
and using the relation, wouldnt it lead to $(a \circ b) \diamond [...]$
flat frigateBOT
˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞
half cove
I used alpha and beta instead of a and b
I considered alpha as the a from the relation that holds true for all elements of M
The first e diamond I considered it as the b
The beta as c
gilded delta
isnt the relation $(a \diamond b) \circ(c \diamond d)=(a \circ c) \diamond(b \circ d)$
flat frigateBOT
˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞
half cove
Yes this relation holds for everything that is an elemant of M
They don't have to be named a, b, c, d
I just used it with alpha, beta, e diamond, e diamond
Even though there is two e diamonds it doesn't matter
gilded delta
but then we would get $\alpha \circ \beta$ for the first term, no
flat frigateBOT
˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞
half cove
Even though there is two e diamonds it doesn't matter
Because the relation holds for all the elements it didn't say that they should be all different from each other
half cove
**˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞**
To prove alpha cirlce beta = alpha diamond beta
We can do it in two ways
We can start from alpha circle beta
And show that it's equal to alpha diamond beta
Or the other way around
I started from alpha circle beta
And then using the relations that are defined in M
gilded delta
but you used the relation incorrectly, no?
half cove
Why incorrectly?
gilded delta
look at the order
half cove
The order of what
gilded delta
$(a \diamond b) \circ(c \diamond d)=(a \circ c) \diamond(b \circ d)$
flat frigateBOT
˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞˞
gilded delta
so if we have alpha for a, e diamond for b, beta for c and e diamond for d
half cove
Oh I see, it wouldn't change much if circle and diamond are commutative
But since they didn't say it is
Let me rewrite it in the correct way
now it's clear I think
gilded delta
okay, yeah, it's clear to me now
half cove
It's just a question of commutativity, and I think for an operation to have an identity element it has to be commutative, no?
I don't really remember, I haven't studied this subject in quite some time
gilded delta
not necessarily
actually
it depends on how you define your identity element
it could left merely left identity
just a thing on how you define it
nonetheless, you can deduce that it is also right identity and therefore commutative
half cove
Yeah yeah, even if the operation isn't commutative, it has to be commutative for the identity element
gilded delta
yeah
half cove
I just didn't pay attention to the relation at first I thought it swaps the b and d not b and c
half cove
You're welcome buddy ^^
half cove
very much appreciated
Don't forget to close this channel
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lyric sable
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lyric sable
Hw problem, proofs with inductions. I'm having trouble of what to do next after applying the exponent law at the bottom
grizzled fossil
don't start with what you want to prove
steel stag
so you want to transform that into something like the inductive step
grizzled fossil
start with LHS and rewrite using equalities and inequalities until you get RHS
but use 3<5 and 4<5, then factor then apply inductive hypothesis
and you get 5^(k+1)
lyric sable
How would I go about transforming the left into the right
grizzled fossil
but use 3<5 and 4<5, then factor then apply inductive hypothesis
.
3^(k+1)+4^(k+1)=3 * 3^k+ 4 * 4^k<= 5 * 3^k + 5 * 4^k. can you finish now?
lyric sable
no im still not understanding very well
grizzled fossil
$$3^{k+1}+4^{k+1}=3 \cdot 3^k+ 4 \cdot 4^k \leq 5 \cdot 3^k + 5 \cdot 4^k = 5(3^k+4^k)\leq 5 \cdot 5^k= 5^{k+1}$$
flat frigateBOT
ScapeProf
lyric sable
so we're trying to get to 5^k+1 by using the lhs of our inductive step to prove. where does the (53k + 54k) part come from, did we just pick 5 to make the inequality true?
grizzled fossil
We had to do some factoring to get (3^k+4^k) so we could apply inductive hypothesis
That would give us the 5^k part so we were just missing a 5 multiplied by that
Hence i used 3<5 and 4<5
safe radishBOT
what do u do
raw wharf
what did you find f'(x) to be?
nova moth
how is this not -7/12
I got -5/12
stray palm
-1/3x^2 - 4/x^3
stray palm
I got -5/12
how
nova moth
Is it correct?
nova moth
Wait let me check again
nova moth
Oh yeah
raw wharf
it should be $1/(3x^2) - 4/(x^3)$
flat frigateBOT
reking
molten finch
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pseudo geode
could someone help me over at #multivariable-calculus
#multivariable-calculus message
would be greatly appreciated 🙏
gleaming shell
I NEEED HELP
thin bridge
what have you tried?
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@pseudo geode Has your question been resolved?
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@pseudo geode Has your question been resolved?
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@pseudo geode Has your question been resolved?
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humble jungle
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humble jungle
how do i do this?
i dont get it
if we take the product of the three numbers as y
then we will have 2 variables
hmm
yeah
i dont get it
spare pike
4x(x+3) = 55
proven mauve
Can someone help me with 6
safe radishBOT
Can someone help me with 6
white bison
Can someone help me with 6
there is no 6?
proven mauve
wait sry i meant level 2 1
polar shadow
proven mauve
OH OK
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@proven mauve Has your question been resolved?
quasi bison
did you just write 1/(a-b) as 1/a - 1/b??
the reciprocal of 1-x is not 1 - 1/x
replacing sqrt(3)/3 with x for simplicity, this is what you write as your first step:
$\frac{1+x}{1-x} = (1+x)\paren{\frac{1}{1} - \frac{1}{x}}$
flat frigateBOT
Ann
quasi bison
do i understand you correctly?
yes or no
i'm not saying anything.
no i'm not saying anything. i am asking you whether or not this is what you intended to write
and you are not answering...
yeah
that's wrong.
your first step is wrong.
$\frac{1}{a-b} \neq \frac{1}{a} - \frac{1}{b}$
flat frigateBOT
Ann
quasi bison
no this isn't correct!
you're still making that very same mistake, just further down the line!
...yes
you're making this more complicated than it needs to be
i don't want to see another attempt riddled with the same mistakes as before
ok fine
safe radishBOT
@lean otter Has your question been resolved?
covert forge
ghosted again 
quasi bison
,calc (1 + sqrt(3)/3)/(1 - sqrt(3)/3) - (2 + sqrt(3))
flat frigateBOT
Result:
-4.4408920985006e-16
okay, it appears you're correct now
even if your way of doing it was all kinds of roundabout
thin bridge
multiplying the numerator and denominator of the big fraction by 3
quasi bison
by sqrt(3) :p
thin bridge
whoops
quasi bison
you can multiply by 3 but multiplying by sqrt(3) is a bit faster
thin bridge
yeh
quasi bison
you get $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
flat frigateBOT
Ann
quasi bison
which you can then rationalize to get $\frac{(\sqrt{3}+1)^2}{2}$
flat frigateBOT
Ann
quasi bison
no it's not set in stone by any means
however, when dealing with nested fractions, it is usually a good idea to do something to immediately make the fractions not nested anymore
fractions within fractions
not really what i tried to allude to...
also, it's denominator
frozen sluice
Given is the set $A = {|x|+|y| \leq \sqrt{2}} $. The solution states that a parameterization is $(x,y)=\begin{pmatrix}\frac{1}{2}(u+v)\\frac{1}{2}(u-v)\end{pmatrix}$. As a hint in the task they said that one can use $(x,y)=\begin{pmatrix}u+v\u-v\end{pmatrix}$. How can i find the factor $\frac{1}{2}$ though?
safe radishBOT
Given is the set $A = \{|x|+|y| \leq \sqrt{2}\} $. The solution states that a pa...
flat frigateBOT
Kurama
frozen sluice
I know that the set describes a square thats turned in a way i guess
but im really stuck and dont know how to find the factor 1/2
oh also in the solution they said that $u,v\in[-\sqrt{2},\sqrt{2}]$
flat frigateBOT
Kurama
frozen sluice
i feel like this could be reduced to a simple system of equations, but somehow i cant get it right
safe radishBOT
@frozen sluice Has your question been resolved?
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@frozen sluice Has your question been resolved?
frozen sluice
<@&286206848099549185>
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rustic gust
I'm trying to study for inequalities and I keep getting stuck on this problem and am having a hard time trying to figure it out
safe radishBOT
I'm trying to study for inequalities and I keep getting stuck on this problem an...
quasi bison
@rustic gust which of the two conditions are you having trouble converting into an inequality? the weight one or the "4 times as much" one?
rustic gust
4 times as much
quasi bison
(number of cars) ≥ 4 * (number of games)
it can literally be translated word for word
the amount of remote controlled cars (y) must be at least (≥) four (4) times (*) the amount of board games (x)
rustic gust
Alright, That makes more sense now, Thank you very much!
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warm beacon
Hey guys, does the vertex formula (-b/2a) work with minimum turning points as well, or only maximum?
safe radishBOT
Hey guys, does the vertex formula (-b/2a) work with minimum turning points as we...
warm beacon
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lean otter
what is this asking me to do?
safe radishBOT
what is this asking me to do?
lean otter
like, the second and last line
what is y(8)|x²y" + 5xy' + 4y = 0 supposed to mean? i mean, i know " means double derivative and ' means derivative. but what about the y(8)| ?
humble venture
might be asking you to find y(8) given that y(x) satisfies that differential equation
lean otter
oh
third line doesn't seem to make sense, since each line is supposed to give a value..
humble venture
but idk about the set notation
humble venture
third line doesn't seem to make sense, since each line is supposed to give a val...
yea
pulsar dagger
Assume we roll 2 ten sided dice. What is P({first roll larger than second roll})? Answer in reduced fraction form - eg 1/5 instead of 2/10.
safe radishBOT
Assume we roll 2 ten sided dice. What is P({first roll larger than second roll})...
pulsar dagger
can you assume that the probability is half or do you have to think of every possible outcome?
we can have (1,10-2), ( 2, 10 - 3), (3, 10 -4) etc
how do I calculate that or do I have to think about the individual outcomes
.close
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warped roost
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misty lichen
can anyone give me the answer, im really struggling
safe radishBOT
can anyone give me the answer, im really struggling
lean otter
can anyone give me the answer, im really struggling
cosine rule
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@misty lichen Has your question been resolved?
lean otter
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lean otter
I just want to ask about the mark beside each mcq answer
When it comes to money what does the c with line over it mean
<@&286206848099549185>
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@lean otter Has your question been resolved?
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@lean otter Has your question been resolved?
lean otter
<@456226577798135808> cents
Wait cents have a sign oh...
Ok thank you
.Close
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surreal tartan
please help me with math question what is 1000002385+2833
safe radishBOT
please help me with math question what is 1000002385+2833
@surreal tartan Has your question been resolved?
median flint
1000005218 according to google calculator
runic crag
"How do you choose a value for k so that f(x)=x^3 + kx lacks any extreme points"
The answer is quite obvious when you take the derivative (3x^2 +k) just from intuition you know that the answer is k>0 (since you can't allow the derivative to touch the x-axis)
But is there some general method to prove it? Cause I wouldn't know how to solve it if a similar but more complicated question came on a test
runic crag
Sorry if translations are off
safe radishBOT
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wheat patrol
I suppose the general method would be to take the derivative and pick a value of k that makes the derivative not equal 0 or an undefined value
IF these questions are only asked with polynomials then the derivative would be another polynomial so just knowing when polynomials are going to equal 0 may help
runic crag
This is kinda what I already did
wheat patrol
like for ex if the above problem was $-x^3+kx$ Then the derivative would be $-3x^2+k$ and since it would be facing down you'd know that it works for $k<0$
flat frigateBOT
bubbles
runic crag
Yep, imagining it in your head works too
wheat patrol
I feel like this type question can really be asked with a cubic too, like if it was a quartic then there wouldn't be a value of k that works
runic crag
Yeah I already tried that, there's always an extreme point no matter what
I think it's for every odd it's k>0 and even has no solution
wheat patrol
yea i think the question would have to be something different at that point
runic crag
I think it's for every odd it's k>0 and even has no solution
assuming no other parameters change other than the exponent in the first term ofc
alright, I guess that answers my question then, unless you were about to type something?
wheat patrol
nah thats it
i was trying to think of an example question of something similar
but it was getting too long lol so nvm wouldn't be asked a test
lean otter
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inland ivy
Well theres three prices of stamps you can buy
3
10
And
15
So youre looking for the lowest combination of all of those that is still above 12
lean otter
No there are from 3 to 10, 4 5 6 7 and 8 are there too
inland ivy
Oh yeah
lean otter
How many stamp
inland ivy
So you have
3
4
5
6
7
8
9
10
And
15
Then
The fact that you can make exactly 12 cents makes this easier
Cause you can buy 4 3 cent stamps, 3 4 cent stamps
lean otter
5 and 7 or 9 and 3
lean otter
The fact that you can make exactly 12 cents makes this easier
It's not exactly 12 it's between 8 and 12
Meaning 9, 10 and 11 are allowed
12 is not allowed
lean otter
Still it has to be less than 12
inland ivy
Meaning youd have to have the highest possible price at least
When something says "a to b inclusive" its meant to include a and b
lean otter
No we need the least amount of price to but stamps not the highest
inland ivy
Well so if you buy 9 cents of stamps
And it ends up being 11 cents
What will you do
lean otter
When something says "a to b inclusive" its meant to include a and b
It said between 8 and 12 inclusive not from 8 to 12 can you please read the question
inland ivy
If you dont want help thats fine
inland ivy
Well I mean if it wasnt inclusive of 8 and 12
Then there would probably be an option in the multiple choice for something less than 12
Cause you can make prices down to 3 cents even
You can make 8, 9, 10 etc
Im fairly certain that its 12 cents
I mean maybe your teacher does stuff differently
But from what I know it should be from 8 to 12 including both 8 and 12
lean otter
Did you reach the answer?
inland ivy
Im still p sure its 12
lean otter
Im still p sure its 12
No that's wrong
inland ivy
You tried it?
lean otter
A is not the correct answer... no I have mark scheme
lean otter
Yeah I have the answer
lean otter
<@&286206848099549185>
inland ivy
Well if you have the answer itd be easier to explain the given answer than come up with one
inland ivy
What prevents you from just buying 4 3 cent stamps
lean otter
because you cant pay 8 cents with that
inland ivy
Oh I understand the question
lean otter
That's really great... can u explain it to me too
@lean otter
i am thinking of a logical solution
thin bridge
start by noting that there isn't much point in buying 6c+ stamps as you could just buy at least 2 lower value stamps for more combinations
lean otter
start by noting that there isn't much point in buying 6c+ stamps as you could ju...
.
@lean otter tell me how did you get it how did you think of the answer
inland ivy
I dont think theres any formula or easy way to get to the answer
You just think about it
lean otter
.
you are left with 3,4,5 coins so you could write out what you need
8=5+3=4+4
9=5+4=3+3+3
10=5+5=4+3+3
11=5+3+3=4+4+3
12=4+4+4=3+3+3+3=5+4+3
and then just brute force, cant really think of any solution
Ok but still why do you think she can cover any postage value from 8 to 12
inland ivy
Cause she can
lean otter
Ok but still why do you think she can cover any postage value from 8 to 12
because if i take 3,3,4,5
8=3+5
9=4+5
10=3+3+4
11=5+3+3
12=5+4+3
lean otter
Cause she can
I really appreciate your help but please stop
lean otter
because if i take 3,3,4,5
8=3+5
9=4+5
10=3+3+4
11=5+3+3
12=5+4+3
Yeah but that's the least amount she spend on stamps
Money on stamps should be separated from postage coverage
???
thin bridge
wdym
lean otter
Yeah OK I got it
lean otter
i found first derivative ,but both terms are constant
safe radishBOT
i found first derivative ,but both terms are constant
@lean otter Has your question been resolved?
lean otter
<@&286206848099549185>
safe radishBOT
@lean otter Has your question been resolved?
marsh walrus
i found first derivative ,but both terms are constant
they shouldnt be constant?
assuming you differentiate wrt v
does that make sense 👀
yes
marsh walrus
,w derivative with respect to v of (18)/(120-2v) + (15)/v
flat frigateBOT
marsh walrus
looks like a sign error?
marsh walrus
i should be setting the entire equation equal to 0?
yea, to find the critical point
then depending on your numeric reasoning skills, youll need to do some additional work to ensure you've actually found a minimum
the second derivative test works pretty well for this through concavity
you may also be able to think out the shape of this function, too
lean otter
im having trouble finding v
i got 0 and 60 as my cp's and i plugged them into the original equation,but you cant drive at 0 speed and 60 gives undefined 
lean otter
.close
safe radishBOT
lean otter
ty for your help,but im going to go back and reread the chapter
lean otter
i dont think so
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subtle python
Because for v=150 you get C(v)=0
safe radishBOT
Because for v=150 you get C(v)=0
subtle python
And that’s kinda nice to not pay
marsh walrus
ty for your help,but im going to go back and reread the chapter
0 and 60 are undefined
have you looked at a graph of the cost function?
that may help guide your answer
safe radishBOT
@subtle python Has your question been resolved?
lean otter
im getting an answer way off from the example problems answer
verbal wing
I'm having some trouble with a related rates problem!
There's a square with sides expanding at a rate of 6 cm/second. Find the rate at which the area of the square is expanding when the area of the square is 16 cm^2. The problem is, the book says the answer is 48 cm^2 per/second and I'm getting 8 cm^2/second. 😅
safe radishBOT
I'm having some trouble with a related rates problem!
There's a square with side...
verbal wing
So I know A = x^2, x' = 6 cm/s, and right now, A = 16 cm^2. So, I can set 16 = x^2, which means x is 4. When we differentiate A = x^2, we get A' = 2x, which means that A' = 2(4) which equals 8! Which is not what my books says. Where am I messing up?
safe radishBOT
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verbal wing
<@&286206848099549185>
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@verbal wing Has your question been resolved?
humble venture
So I know A = x^2, x' = 6 cm/s, and right now, A = 16 cm^2. So, I can set 16 = x...
you differentiated A with respect to x (side length) and not t (time)
dA/dt should be what you're looking for
use the chain rule to get it
safe radishBOT
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velvet raft
How do I determine if this infinite series converges or diverges?
velvet raft
Can I just use the ratio test?
pulsar condor
How do I determine if this infinite series converges or diverges?
Honestly I'd just attempt to find the limit since it'll have some amount of telescoping
find the Nth partial sum then limit
velvet raft
the limit of the series or partial sum
pulsar condor
limit of the partial sum
infinite series is the limit of the series of partial sums
$\sum_{i=0}^\infty =\lim_{N\to\infty}\sum_{i=0}^N$
flat frigateBOT
Mosh
velvet raft
oh I see
alright ill try it out
wait tho
@pulsar condor what partial sum formula do i use for this
pulsar condor
you have a telescoping series
velvet raft
yeah
oh
alright
im not so great with series
i gotta relearn it
teachr didnt teach it well at all
safe radishBOT
@velvet raft Has your question been resolved?
velvet raft
but is that the partial sum formula?
pulsar condor
yes if your sum goes up to n
velvet raft
@pulsar condor
pulsar condor
yes.
velvet raft
right
got it
thanks so much
wait
@pulsar condor
isnt the partial sum of the series the same thing as the sum
like
since we have the partial sum of n numbers
isnt that the same as just the sum
or am i confused
pulsar condor
Yeah, your notation was poor
hence why I made the upper bound of the partials not the index
worthy hemlock
You don't. That's notation for inverse function
You could, but as I stated, that -1 is notation for inverse function
So distribution is one way, or doing $g \circ f$ first then finding the inverse should result in the same thing
flat frigateBOT
dldh06
safe radishBOT
@tiny siren Has your question been resolved?
burnt obsidian
Question
burnt obsidian
For the complex number (-1-i) how isn't it's angle 45°??
This is how I calculated it
Arctan(-1/-1)
which is the same as
arctan(-1/-1)
Which is equal to 1
And according to this
That's 45°
But obvs I did something wrong so pls help
Ah nvm I figured it out
It's becuase -1-i is on the third quadrant
Which means I needed to subtract or add 180° to 45°
Thank you
.close
vague herald
I've tried putting this equation equal to each other on Maple, but it doesn't work
safe radishBOT
I've tried putting this equation equal to each other on Maple, but it doesn't wo...
vague herald
The right part is the equation for an circle, shown in yellow:
The left part is the function shown in blue
I can't seem to get the other x value for the other crossing point, which is kind of an issue
Since I have to find them both
blazing marten
,w solve 2cos(1.6x+3.85)+2.0=4-0.02sqrt(-2500x^2+7600x+18249)
flat frigateBOT
vague herald
Appreciate it 😄
vague herald
Yeah, seems really weird
blazing marten
But wolfram alpha comes to the rescue once again.
vague herald
Since all other solutions are complex numbers
vague herald
But wolfram alpha comes to the rescue once again.
Can you use the calculator for free?
Or is it a subscription?
blazing marten
Its free except for some functions i think
I just used the texit bot which is connected to wolfram alpha
vague herald
Ohh okay
Is there any possible way, you can insert both x values into the left equation?
Since I'm getting the y-coordinate to two different things on my calculator
blazing marten
There probably is
Maybe go to the website of wolfram alpha and use the equation solver
blazing marten
Let it solve your equation and i think it will give you the option to use your answer
blazing marten
Appreciate the help btw 😄
No problem
vague herald
.close
jaunty heath
Hello, I hope you're day is going well
I have no clue how to do this, can I get some help? Thanks!
safe radishBOT
Hello, I hope you're day is going well
I have no clue how to do this, can I get ...
@jaunty heath Has your question been resolved?
flat frigateBOT
tomzizek
so $\frac{4^n}{4^{2n+2}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text>
\par
<*> 244527830086451200.tex
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```
left igloo
$\frac{4^n}{4^{2n+2}}$
flat frigateBOT
tomzizek
left igloo
**tomzizek**
this
safe radishBOT
@jaunty heath Has your question been resolved?
kindred ivy
Hello, I have made this graph based on this data, How would I make a single line of trend that represents all of them?
safe radishBOT
Hello, I have made this graph based on this data, How would I make a single line...
kindred ivy
(this specific graph excludes the US and China since they are outliers and cause the data to be difficult to visualize)
(the values are military expenditure btw)
kindred ivy
like so?
lean otter
Is that the average or mean?
kindred ivy
yes
lean otter
Then yeah, that's what you want
kindred ivy
but more than this being a math question, I am trying to find out how I can put it into a graph as a straight line
lean otter
Oh as a straight line
lean otter
You could just connect the end and starting point
safe radishBOT
Available help channel!
silver axle
How do I solve this
safe radishBOT
How do I solve this
safe radishBOT
@silver axle Has your question been resolved?
vestal marsh
safe radishBOT
vestal marsh
Anyone get how theres 3 extrema?
I only see one at 0 and 15
because f'x changes sign
<@&286206848099549185>
safe radishBOT
@vestal marsh Has your question been resolved?
safe radishBOT
@vestal marsh Has your question been resolved?
ember bough
@vestal marsh there's one at... uh, 9 or something?
obsidian zephyr
safe radishBOT
acoustic pasture
8 is cube number
obsidian zephyr
thanks
obsidian zephyr
how do i grt my own channel
obsidian zephyr
huh
obsidian zephyr
tysmmmm
acoustic pasture
np
safe radishBOT
@obsidian zephyr Has your question been resolved?
ember bough
turn 2.1 into a fraction
add
then simplify
ember bough
for b and c, just do it
remember that 2 - signs make a +
safe radishBOT
@obsidian zephyr Has your question been resolved?
quasi bison
no
safe radishBOT
safe radishBOT
Available help channel!
opal belfry
I would need a way to get a y = f(x) function for y > 0 when :
safe radishBOT
I would need a way to get a y = f(x) function for y > 0 when :
opal belfry
$\frac{(e^{1+y})}{\left(x+e\right)}-\ln(y+1)-\ln(x+e)=0$
flat frigateBOT
Mr. Rotor
opal belfry
top side red curve
I'm looking for it but it seems quite doomed
$\frac{(e^{x})}{y}-\ln(x)-\ln(y)=0$
flat frigateBOT
Mr. Rotor
opal belfry
would work too
safe radishBOT
@opal belfry Has your question been resolved?
hot thistle
are you looking for a domain?
safe radishBOT
safe radishBOT
Available help channel!
lean otter
hello
lean otter
So regarding indices and doing sqrt and cbrt to them
if i have cbrt(x^5)
do i do 5-3
or 5/3
flint ledge
5/3
flint ledge
yes
lean otter
ty ty
flint ledge
it’s also the same as (2^6)^(1/2)
flint ledge
in that case you multiply the exponents
lean otter
have a great day
flint ledge
no problem
lean otter
.close
safe radishBOT
safe radishBOT
Available help channel!
lean otter
safe radishBOT
lean otter
Can someone help me with this?
flint ledge
what are you trying to do
worthy hemlock
,rotate
flat frigateBOT
flint ledge
what is the lcm of 2 and 5
blazing fiber
then first bring all x to one side and the rest on the other side
lean otter
yeah but i cant bring x to one side
flint ledge
you can
lean otter
dude i obv cant which is why im asking for help
flint ledge
it’s a linear equation, this won’t take long
blazing fiber
Ok, why do you think you can't bring them on the other side?
flint ledge
or preferably get rid of the denominators
flint ledge
multiply both sides by the lcm of 2 and 5
lean otter
then add that with x/2?
lean otter
yeah but if i multiply one side with 2 then i need to do the other
flint ledge
what do you mean by extend
worthy hemlock
so i extend the fractions first
What does this mean?
flint ledge
you multiply by the lcm of the two denominators
lean otter
multipy the numerator and denominator with 5
in the case of 2
then multiply the denominator 5 with 2
flint ledge
no that’s extra work
lean otter
idk what lcm is
flint ledge
least common multiple
lean otter
dk what that is either
flint ledge
the least common multiple is 10 here
between 2 and 5
basically call the numbers in the denominators a and b, multiply both sides by ab
which in this case is 2*5
blazing fiber
$\frac{1}{2} +\frac{4}{3} = ?$
lean otter
so lcm is denominators multipled by each other?
flat frigateBOT
Moriarty
lean otter
in this situation
i would find the number that i can multipy 2 and 3 with so they would be same
so it would be 6
worthy hemlock
so lcm is denominators multipled by each other?
Multiplying denominators is a method to finding a common denominator, for the fractions
lean otter
so i would mukltipoly 2 with 3
lean otter
and then 1 with 3 aswel
aha
oh i understand then
ok so
i multiply the fractions with 10?
lean otter
ok but what happens if i multipy with fraction by 10
doesint it just get bigger
x/2
will become
10x/20
blazing fiber
$\frac{a}{c} + \frac{b}{d} = \frac{a * d}{c * d} + \frac{b * c}{c * d} = \frac{ad}{cd} + \frac{bc}{cd} = \frac{ad + bc}{cd}$
lean otter
lean otter
is that what you mean?
flat frigateBOT
Moriarty
lean otter
cause i divided by 10
or shit
yeah
im tired
so what im supposed multiply both sides with 10?
blazing fiber
1st bring all x to one side and the rest on the other side,
2nd use the formular I write above
lean otter
you would get 5x-150 = 2x
dude what
how does it become 5x
if i just multipled with 10
flint ledge
x/2 * 10 = 10x/2 = 5x
lean otter
why does the 2 become 20
lean otter
1st bring all x to one side and the rest on the other side,
2nd use the formular...
bro i dont understand that formula
flint ledge
it doesn’t
lean otter
its confusing because its so many variables and stuff
i meant why does the 2 not become 20
flint ledge
there isn’t a 20 in what i just sent
lean otter
if i multipy with 10
flint ledge
yes
lean otter
i just understood that
wow
ok so ive probably hit a brickwall in my studying right?
should i sleep or just take a break
flint ledge
whatever you want considering you thought i meant 10/10 when i said 10 lol
flint ledge
ok
flint ledge
you’re the only one that can close it anyways
lean otter
ok
but im still stuck
what do i do after doing muliplying with 10
cause i still cant get rid of the frations
worthy hemlock
If you multiplied by 10, it gets rid of all the fractions
lean otter
why and how
worthy hemlock
Because of simplification
blazing fiber
10 / 2 = ?
10 / 5 = ?
blazing fiber
$\frac{10}{2} * x -150 = \frac{10}{5} * x$
flat frigateBOT
Moriarty
lean otter
Because of simplification
whatever im just gonna keep it as a law that if you muliply with the common multiple then the frations disappear, still dont understand why tho
blazing fiber
because of this:
$\frac{a}{c} + \frac{b}{d} = \frac{a * d}{c * d} + \frac{b * c}{c * d} = \frac{ad}{cd} + \frac{bc}{cd} = \frac{ad + bc}{cd}$
flat frigateBOT
Moriarty
lean otter
**Moriarty**
why did my x become a 10
worthy hemlock
dudei dont understand that one
Look up concepts about clearing fractions
That's what you're doing
worthy hemlock
whatever im just gonna keep it as a law that if you muliply with the common mult...
It's not some law, it's a concept of simplification