dont confuse yourself, dont take anything as a "rule"
#help-28
muted kite
kinda
torn jolt
like
you understand the x axis are the points you choose, and the y axis are the points that get produced in a sense right?
muted kite
yes
torn jolt
ok great
so you understand here that for the first one we have (0,0), (1,1), (2,4) as points and for the second one (0,0), (1, 3), (2, 12) right?
muted kite
I do understand
torn jolt
ok
so compare (2,4) and (2,12)
on the x axis, they are on the same point x
but its obvious that the second one is much bigger than the first in terms of the y-values correct?
torn jolt
can u pls explain this to mee if u djnt mind
think about it this way
you have (-100000000)^3
what do u think that would be equal to? (dont actually calculate it)
torn jolt
ok so do you know the fact that like
[
(-x)^n = x^n \q \txs{if} n\tsx{is even} \
(-x)^n = -x^n \q \txs{if} n\tsx{is odd}
]
glossy valveBOT
torn jolt
Okay i think this will be a bit hard for me because i think you have a lot of background info missing. Would you like me to link you some Khan academy videos to study from instead?
muted kite
thank youuu for your time and effortt T-T!!
lean rock
Are Boole, event, and set algebra equivalent?
full forumBOT
Are Boole, event, and set algebra equivalent?
lean rock
Boole algebra has to do with binary values, true and false. Set algebra focuses on elements, while event algebra is concerned with probabilities of outcomes of events.
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desert musk
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desert musk
how to?
thick hedge
hint :- cos(60)-sqrt3/2
twilit leaf
hint :- cos(60)-sqrt3/2
What
desert musk
that is true right
desert musk
so i replace the sqrt(3)/2 with cos(30)
twilit leaf
use cos(pi/6) because radians are better
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tropic wedge
I need help proving this
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I need help proving this
tropic wedge
I got here
next sail
dang that looks annoying
idle kelp
Are you solving this question independently of the previous questions
idle kelp
You are making this kind of annoying by cropping things so that you remove all context of how these problems are being built up
tropic wedge
they are put in a word document very randomly
so you have to either span over in 4 places xD
or just at the cropped part
well these exercises were not related until now
the only knowledge base was Vandermonde's determinant
idle kelp
If you just want to show that identity is true, just use the vandermonde identity you apparently already have on the right side and then apply the rule of sarrus on the left side and apply polynomial long division on the vandermonde part. But I get the feeling with your arbitrary cropping that you are not getting the intended learning outcome from this.
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gritty rose
nono, I want to understand the matrix transformations needed
Understand what exactly
Needed for what
tropic wedge
needed to modify the left side's determinant
with column / row additions or subtractions
so I can get the same determinant as in the right side
tropic wedge
I need help proving this
this is what I got no idea how to modify
tropic wedge
I got here
I got here
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rugged narwhal
Why does it work this way?
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Why does it work this way?
rugged narwhal
Why does dy/(y-1) become ln, while dx becomes an antiderivative?
lime ether
because ur integrating both sides with respect to each variable
integrating 1/y-1
u get ln(y-1)
because the derivative of ln(y-1) is 1/y-1
rugged narwhal
integrating 1/y-1
But where did the dy go?
lime ether
the c u found using the initial condition of y(0)
rugged narwhal
the c u found using the initial condition of y(0)
I mean shouldnt there also be a c when you integrate dy/(y-1)
lime ether
because there are infinitely many functions that have that same derivative
lime ether
I mean shouldnt there also be a c when you integrate dy/(y-1)
but they combine to make one C
so u don’t have to put it on both sides
if u added C to both sides u still have a constant
so just add it to one
rugged narwhal
but they combine to make one C
Ok heres another question, why does e^x+c become Ce^x?
lime ether
use exponent rules
e^(x+c)=(e^x)(e^c)
but e^c is just a constant
so u can call it C
remember these constants aren’t all the same constant
it’s just a constant
rugged narwhal
so u can call it C
Ok i see
lime ether
u can differentiate between the constant by adding subscripts
rugged narwhal
Weird how it works though
rugged narwhal
Thanks @knief
lime ether
to show they’re not the same
rugged narwhal
.close
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lime ether
yea no problem
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harsh elk
are there values of a and b so that the equation sqrtroot (t+a) = b has more than one solution
help?
how do i even start this?
do i square both sides?
t+a=b^2?
harsh elk
aka, if you know a and b then you can calculate t
but its asking for more than one solution
for values of a and b
fast peak
well it asks whether there are such values a,b
gritty rose
What's t
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gritty rose
are there values of a and b so that the equation sqrtroot (t+a) = b has more tha...
Oh more than one solution in t
Easier way to do this is to just plot
harsh elk
Easier way to do this is to just plot
what do u mean?
gritty rose
Plot y=b and y=sqrt(t+a)
gritty rose
<@261933205387477002> u there?
What
gritty rose
which means t=b^2-a
Yes it's here.
harsh elk
Yes it's here.
i dont get that tho
gritty rose
Why not
gritty rose
Why
gritty rose
are there values of a and b so that the equation sqrtroot (t+a) = b has more tha...
You got the bottom equation
harsh elk
Why
maybe his wording?
if you know a and b then you can calculate t
i dont understand why this is important
gritty rose
He just subtracted a from both sides
harsh elk
You got the bottom equation
yea
harsh elk
He just subtracted a from both sides
why a?
gritty rose
Oh more than one solution in t
.
gritty rose
Oh more than one solution in t
Do you get this?
harsh elk
Do you get this?
no
gritty rose
Well you should always understand the problem and what it's asking for before doing any work
harsh elk
Well you should always understand the problem and what it's asking for before do...
i dont understand why ur including t into it
when its asking only for a and b
gritty rose
i dont understand why ur including t into it
That's what the question is asking for
More than 1 solution in t
harsh elk
That's what the question is asking for
its asking for a and b vlues
harsh elk
More than 1 solution in t
where?
are there values of a and b so that the equation sqrtroot (t+a) = b has more than one solution
gritty rose
Implied from the question
harsh elk
see a and b
gritty rose
gl = good luck
shell idol
Hi
shell idol
Need help with this
sacred sparrow
Need help with this
mind briefly explaining the context of Q* and P*?
sacred sparrow
<@1000186270078750740>
so can we assume P*=P and Q*=?
I'm still uncertain what the * stands for
shell idol
Same
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swift fulcrum
Hello
calm nacelle
∆
=
Probably "Q is defined to be equal to this integral"
Does that make sense with the context?
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calm nacelle
If not, you might want to post more context
swift fulcrum
Probably "Q is defined to be equal to this integral"
Do you have an idea what Q might mean?
calm nacelle
Yeah I think it just means "defined to be equal".
swift fulcrum
Ohh I see
calm nacelle
Q is specifically change in charge from t0 to t in this notation
swift fulcrum
Are you in engineering btw
calm nacelle
No I'm not
quasi bronze
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lime ether
add up the 4 unit exams
that is a total of 400 points
since the final exam is worth 200
the total 5 exams will be worth 600 points
for him to get a B he must be in the range of 80-89
the average of the exams
would be the total score/5
or wait so u need
no yea that works
lime ether
add the 4 exams
quasi bronze
add the 4 exams
Like this?
[
\text{Overall percentage} = \frac{\text{Total points from unit exams + Final exam score}}{\text{Total possible points}} \times 100
]
glossy valveBOT
Lia 👅
quasi bronze
[
\text{Overall percentage} = \frac{344 + 136}{600} \times 100
]
glossy valveBOT
Lia 👅
glossy valveBOT
Lia 👅
quasi bronze
[
\text{Overall percentage} = \frac{344 + 166}{600} \times 100
]
glossy valveBOT
Lia 👅
quasi bronze
[
\text{Overall percentage} = \frac{344 + 176}{600} \times 100
]
glossy valveBOT
Lia 👅
quasi bronze
[
\text{Overall percentage} = \frac{344 + 196}{600} \times 100
]
glossy valveBOT
Lia 👅
lime ether
yes
quasi bronze
- For F (136 points): 80%
- For G (156 points): 83.33%
- For H (166 points): 85%
- For J (176 points): 86.67%
- For K (196 points): 90%
lime ether
yes because 90% isn’t in the 80-89 range
196 would give a 90
which isn’t a B
it’s an A
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tribal stratus
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tribal stratus
Tried a) but I couldn't get the answer
I added two integrals but I couldn't solve the first one
wet patrol
do you need integration for this?
wet patrol
tbh this looks very solvable without it
tribal stratus
how would you do them both then
wet patrol
couldn't you work out the earnings by 60000(0.5)+45000 for the first 6 months then add 75000(0.5) for the second half of the year?
wet patrol
whats the answer?
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tribal stratus
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✅
tribal stratus
Still confused how to solve a and b
tribal stratus
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tribal stratus
🙏 please
austere cove
@tribal stratus hi
austere cove
So, an integral is the correct approach.
tribal stratus
Ok so phew yea
austere cove
The key is "continuously compounded"
The other option would be discretely compounded which would involve the use of a rather awkward formula.
tribal stratus
right
if it's continuously compounded
I have the formula
but I'm not sure how to integrate
I posted my two calculations above
but I couldn't figure out what to do with them
austere cove
So you'll want to do integration by parts
austere cove
Yup
tribal stratus
I have a class in a bit- but once I'm done I'll try it out
yea ofc
this channel might close, you mind if I dm you once I have it figured out?
austere cove
I don't mind, but I also cannot promise to be available
tribal stratus
I don't mind, but I also cannot promise to be available
that is fine
I'll try it right now
before I leave
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oak bluff
can someone please teach me how to divide using inspection
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can someone please teach me how to divide using inspection
austere cove
@oak bluff there's a theorem (rational root theorem) that says that if a polynomial has a rational root, then the factors of the polynomial: (ax - b), will be comprised of the prime factors (or 1) of the first and last coefficients of the polynomial with or without a negative sign. That is, any factor of your example polynomial up there will need a to be +/-3 or +/-1. And b will need to be +/- (1, 2, 4).
This gives you only a handful of possibilities to test.
oak bluff
i know how to get the (ax - b)
but after that i dont know how to solve to get the "3x^2 - x - 4) part
austere cove
That's the same thing, though you can just use the quadratic formula instead.
oak bluff
how so? don't i have to divide 3x^3 + 2x^2 -5x -4 by (x + 1) like in the example?
oak bluff
i heard theres a different way tho
austere cove
if you remember how to do long division for numbers, it's really quite similar.
oak bluff
inspection
oak bluff
if you remember how to do long division for numbers, it's really quite similar.
yeah, however that one is prone to sign errors
this "inspection" this is elusive
austere cove
I would argue that attempting to do polynomial long division by inspection is much more fraught with sign errors
at least in the general case.
oak bluff
but how would i execute it?
austere cove
Essentially, you do long division in your head
oak bluff
ah
austere cove
like, generally, if you can do a polynomial division by inspection, there's normally a trick
oak bluff
so for the long division, i have one question; do i divid x then the number that follows before moving onto the next number or do i do both
austere cove
like dividing (x^n - 1) by (x - 1)
oak bluff
if that makes sense
austere cove
I'll try to type up an example.
actually I'll write up an example.
and post the picture, one moment
oak bluff
that would be great thank you
austere cove
oak bluff
ahhh, yeah i think i remember what im talking about, not sure if waht im talking about is inspection
In q3c
Im not sure what that is
austere cove
,rotate
glossy valveBOT
austere cove
ah, that only specifically works for going from 3 -> 2. You set up an unknown coefficient in the middle, then figure out what it has to be.
You can do it from 4-> 3, but then you have two unknowns, and two equations, and so on.
oak bluff
?
austere cove
It works, never seen it done before, but it makes sense.
3 -> 2 meaning degree 3 to degree 2.
oak bluff
from 3 -> 2? i dont understand
austere cove
x^3 to x^2
oak bluff
what is a degree?
ahhh
[power
thank you very much for you help, have a nice day
.close
torn jolt
for proving left is a subset of right, why do you need to consider the case that x is in C?
agile zodiac
Problem 1) Let f(x) = P(x) if x ≤ 0 and Q(x) if x > 0
a continuously differentiable function whose function values and derivatives at the points x = −1 and x = 2 agree with the respective function values and derivatives of the function g(x) = x to the power of 3. Let P(x) and Q(x) be polynomials of degree at most 2.
Determine P(x) and Q(x).
Note: Here you can use your previous knowledge from school to derive quadratic functions. That’s my task and I can’t find anything helpful. Please help me
austere cove
!status
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
Between 2 and 3
austere cove
Ok can we see your work so far?
agile zodiac
austere cove
So you have set up f(x) as a cubic?
agile zodiac
Sorry my English is not the best but you mean with cubic x to the power of 2??
austere cove
x^3
agile zodiac
Yeah that was my problem in the digitally written, that’s my I made the other one because it seems to be wrong
austere cove
Ok, cool
So your second image seems like the right approach, but it's a big question, 6 variables and 6 equations
So let's take stock, you know all 6 equations are?
No need to be explicit, implicit is fine for now
agile zodiac
Actually there are only two unkown, p(x) and q(x) which belong to f(x) for the special cases like in the me provided task: Problem 1) Let f(x) = P(x) if x ≤ 0 and Q(x) if x > 0
a continuously differentiable function whose function values and derivatives at the points x = −1 and x = 2 agree with the respective function values and derivatives of the function g(x) = x to the power of 3. Let P(x) and Q(x) be polynomials of degree at most 2.
Determine P(x) and Q(x).
austere cove
Yes, but the 6 unknowns I'm referring to are (a,b,c,d,e,f)
agile zodiac
Because you have the form ax2 + box +c and this for the two parts from f(x)
austere cove
To be explicit.
P(x) = ax^2 + bx + c
Q(x) = dx^2 + ex + f
agile zodiac
Correct
agile zodiac
Yes, that’s what I meant
austere cove
Ok, just making sure
So we have 6 equations that we can use, do you know what they are?
agile zodiac
What do you specifically mean?
austere cove
So in order to solve a system of equations with k unknowns you need k equations
agile zodiac
True
austere cove
We have 6 unknowns, so we need to make 6 equations
And we can from the information in the problem
Do you know how?
agile zodiac
We’ve got the Points given in the hand written, what I’ve calculated with the g(x) and g’(x)
austere cove
It's not clear where these formulae are coming from
agile zodiac
Let’s say it there are not given in the task
austere cove
Using P, Q, and g
Can you make 6 equations?
From points provided in the problem statement?
agile zodiac
Maby 4 equations, because we get an x and y coordinate
agile zodiac
(-1 3) and (-1 -1) and (2 8) and (2 12)
austere cove
What I'm looking for is, for instance P(-1) = g(-1)
agile zodiac
And P(2) and g(2) ??
agile zodiac
Is that not also -1
austere cove
is what -1?
agile zodiac
P’(-1) g’(-1)
agile zodiac
Yeah you true
agile zodiac
g(2)=2^3 =8
austere cove
that doesn't involve P or Q
(We can always calculate explicit numbers a little bit later.)
(In fact, we'll be doing a hell of a lot of that, so it's better to make sure we have everything we need first)
So far we have:
P(-1) = g(-1), equality at -1
P'(-1) = g'(-1), equal slopes at -1
Q(2) = g(2), equality at 2
it should be easy to give me a 4th here.
agile zodiac
is g(2)=q(2) so it means q(2)=8
austere cove
yes
but for now, let's not do numbers.
let's finish setting up the problem before trying to knock it down
agile zodiac
Q'(2) =g'(2)=2
agile zodiac
its 12, sorry
austere cove
ok
now there's two more equations
and they come from this line of the problem:
Let f(x) = P(x) if x ≤ 0 and Q(x) if x > 0 be a continuously differentiable function
what does this mean?
agile zodiac
P has the cases equal to 0 and smaller than 0 and Q has no 0 as special case
austere cove
but what does continuously differentiable mean?
agile zodiac
simple if you can put a tangent to the function and the Differentialquotient is the derivation
austere cove
it means that the function is continuous, and the derivative of the function is continuous, which means that the limit of the function and derivative at each point exists and is equal to the actual value of the function at that point.
In particular, it means that the limit of f(x) as x->0 is f(0), and the limit of f'(x) as x->0 is f'(0)
which means that both the left and right limits are equal
which means that P(0) = Q(0) and P'(0) = Q'(0)
agile zodiac
so, that have i understood and now?
agile zodiac
whats the next step i have no plan
austere cove
edit cleared it up
ok
So we have 6 equations:
- g(-1) = P(-1)
- g'(-1) = P'(-1)
- g(2) = Q(2)
- g'(2) = Q'(2)
- P(0) = Q(0)
- P'(0) = Q'(0)
Now we need to figure out what P, Q, and g, and their derivatives are in terms of x.
P(x) = ax^2 + bx + c
P'(x) = 2ax + b
Q(x) = dx^2 + ex + f
Q'(x) = 2dx + e
g(x) = x^3
g'(x) = 3x^2
finally, replace x with the values in our equations to get explicit equations in a,b,c,d,e,f
then solve the system
I need to run some errands.
If you're still having trouble when I get back, I'll help out again, but it's going to be a few hours at least.
agile zodiac
okay
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agile zodiac
That’s my current solve process, if there is any thing incorrect please call it out, I now take a nap.
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storm dragon
How can I achieve this factorization?
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How can I achieve this factorization?
torn jolt
[
r^3 -2r^2 + r -2 = r^2(r-2)+r-2
]
glossy valveBOT
torn jolt
does this help you see it
storm dragon
does this help you see it
It does thanks
storm dragon
\[
r^3 -2r^2 + r -2 = r^2(r-2)+r-2
\]
But how did you find the right first step? Or does it just come with experience?
storm dragon
Factoring by grouping?
torn jolt
yeah
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twilit birch
I wanted to make the cantor set in desmos. I came up with this but it barely dosent work. m is any natural number >2
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I wanted to make the cantor set in desmos. I came up with this but it barely dos...
@twilit birch Has your question been resolved?
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@twilit birch Has your question been resolved?
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calm nacelle
I wanted to make the cantor set in desmos. I came up with this but it barely dos...
Can you explain your reasoning behind what you have so far?
It seems pretty random to me at first glance
twilit birch
Can you explain your reasoning behind what you have so far?
it works something like that b is a list in the picture [1,2,3,4,5] and this uses multiple "parts" which oscillate with a period in this case its the top part of the fraction. this picture is a simpler version which preforms iven worse
calm nacelle
Are you familiar with the fact that the Cantor set is just the numbers that don't have a "1" in their ternary expansion?
You could probably use that fact.
twilit birch
how?
calm nacelle
You can write an equation to detect if a number doesn't have a 1, like for example to check if it has a 1 in the first ternary place, you would multiply by 3, then floor, then check if it's 1.
To check the second ternary place, you could multiply by 9, mod by 3, then check if it's 1.
And so on.
twilit birch
You can write an equation to detect if a number doesn't have a 1, like for examp...
thanks
.close
velvet sandal
would there be a difference in the following problem if there weren't parenthesis around the -3?
warm harbor
no.
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wanton blade
I'm in a Python class and this is our first project. I don't need help with the coding aspect but I'm having diffculty with getting the actual problem solved in preparation for coding. The instructions say we can get outside help for the calculation.
wanton blade
We're also supposed to count the 3 coins each crew took for shore leave as part of their total share.
I don't understand this problem in general but I've tried to find the reverse percentage and just calculating the percentage value from the initial step but I didn't get the right answer.
chilly pine
I need help with my homework because i was sick on that day and nobody have told me how to do this(i‘german)
grizzled dagger
@chilly pine This help channel is occupied, maybe find a different one?
chilly pine
i‘sorry
wanton blade
So I added all of the values in the first image together by doing (39.62 x 18) + 127.15 + 159.60 and got 999.91, which is technically correct so I can probably just go backwards from there, but I don't feel like I understand it still.
Plus I'm supposed to go Yondu first, then Quill, then the Crew, and this way makes it go backwards.
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wanton blade
Okay so I just realized that the question has Yondu and Quill getting their own share plus the crew share. So for yondu's case it would be (13%/Y) + 39.62
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swift crane
What does it mean to multiply a real vector function to a normal function?
swift crane
I didn't think you can multiply a vector function to a real valued one
buoyant pewter
I didn't think you can multiply a vector function to a real valued one
$f\left( t \right)\cdot \overrightarrow{r\left( t \right)}=f\left( t \right)\cdot \left[ x\left( t \right),y\left( t \right),z\left( t \right) \right]=\\=\left[ f\left( t \right)\cdot x\left( t \right),f\left( t \right)\cdot y\left( t \right),f\left( t \right)\cdot z\left( t \right) \right]$
glossy valveBOT
Joanna Angel
buoyant pewter
yw
umbral dome
it's exactly the same concept as multiplying a scalar and a vector in any other context
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eager anchor
I pretty much need help getting an exponential function from the table to answer the questions below
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naive bison
best wqy is just to look up on youtube, how to convert table into expinential function
way*
eager anchor
i tried and couldn’t find anything so i was trying to get help here
i thinks it’s because this one has transformations
naive bison
how much time you got
eager anchor
I was hoping for an answer tonight but i have a test tmr morning and a question like this might be on it
shut torrent
I'm doing optimization in calculus and I'm really stuck on this problem
shut torrent
how would I write the length and width of the rectangle in respects to the triangle's length and width???
rough tundra
I'd suggest making a function that outputs the area $A$ at a certain time $t$ as the rectangle changes
glossy valveBOT
PajamaMamaLlama
rough tundra
so you'd maximize $A(t)$
glossy valveBOT
PajamaMamaLlama
shut torrent
so write the area of a rectangle and then take the derivative?
rough tundra
rough tundra
wait what variables would you recommend I write for the area, just l x w ?
use whatever variables you want but l and w are the accepted standard
shut torrent
and I would take the derivative in respects to time?
rough tundra
yes and time in this case how much it's moved along the base
rough tundra
can I suggest a different method?
shut torrent
yes please!
rough tundra
let's say A=(0,0)
shut torrent
yeah ok, got it
rough tundra
so can you get a line that passes through those two points?
rough tundra
no, just the line that passes through A and B :)
shut torrent
like y1 - y2/ x1 - x2?
rough tundra
shut torrent
lmao ok its 5/4
rough tundra
so at any time t, the point on the line will be 5t/3 right?
shut torrent
yeah that makes sense
rough tundra
okay so at any time t that will be the height of the rectangle, correct?
shut torrent
yeahhhh
rough tundra
can you figure out what the base would be at any time t?
shut torrent
um like somewhere between 0 and 8?
rough tundra
kidna budget
made it on desmos in like 1 minute
but
kinda shows what's happening
now that you have this video as aid, can you figure out b(t)?
shut torrent
oh wait its like b(t) is the x value of the equation we made?
and height is the y value?
rough tundra
the height is yes but the base is not quite
the base is 6-t, this is because the base of the triangle is 6 the base of the rectangle is the base of the triangle minus the time we've removed from the base
so 6-t
shut torrent
the base is `6-t`, this is because the base of the triangle is 6 the base of the...
wait but the base of the triangle is 8?
rough tundra
omg
wow I'm tired
same princple
so what's the base of the rectangle in your problem?
shut torrent
ohhhh ok wait I get it, its 8 - t?
rough tundra
yes!
so we have base and height in terms of t
so now we can get area in terms of t
shut torrent
ok so wait sorry, this is so tricky for me to understand
so our equation is: y = (5/4)(x)
and y is the height of the rectangle?
rough tundra
ok so wait sorry, this is so tricky for me to understand
no worries, I have no doubt it would be otherwise :) (i think I said that right, double negatives mess with my brain)
shut torrent
and (8-x) is our width of the rectangle?
glossy valveBOT
PajamaMamaLlama
rough tundra
now you can get $A(t)=(b\cdot h)(t)$
glossy valveBOT
PajamaMamaLlama
shut torrent
wait t is time?
shut torrent
**PajamaMamaLlama**
oh ok ok and is this equation for the triangle or rectangle?
rough tundra
rectangle
shut torrent
ok so area of a rectagle equals base times height, right? so why did you multiply by t as well?
sorry for asking so much, just rlly trying to understand wut is happening
rough tundra
no worries these types of problems involve lots of visualization and thinking
so where are you confused about multiplying by t?
shut torrent
so you said that this equation was for the rectangle and the area of a rectangle is just base times height so why did you add t as well?
glossy valveBOT
PajamaMamaLlama
rough tundra
shut torrent
ok so A(t) would equal to (8 - t) times (5/4)(t) ?
shut torrent
omg bro ur so smart fr, ok wait lemme take the derivative
ok so sorry, my work is kind of not organized but i got the derivative
now i just equate that to zero and find the original equation’s max and mins, rught?
rough tundra
rough tundra
(btw you don't need quotient rule!!!)
shut torrent
omg, how would u do it then?
rough tundra
$\frac{d}{dx}\left[10x-\frac{5x^2}{4}\right]=\frac{d}{dx}\left[10x\right]-\frac{d}{dx}\left[\frac{5x^2}{4}\right]=10-\frac{5}{4}\frac{d}{dx}[x^2]=10-\frac{5x}{2}=\frac{20-5x}{2}$
glossy valveBOT
PajamaMamaLlama
rough tundra
because $\frac{d}{dx}[cf(x)]=x\frac{d}{dx}[f(x)]$
glossy valveBOT
PajamaMamaLlama
fierce pier
quotient rule only for variable in denominator
rough tundra
you can pull constants out of the derivative if they're being multiplied :)
shut torrent
**PajamaMamaLlama**
ohhhh damn thats mad smart
rough tundra
but in the end your derivative is correct, honestly idek how but lmao it works! 
equate it to 0
plug that back into A(t) and you have your maximum area
shut torrent
ok i equated to 0 and got x = 4
fierce pier
but in the end your derivative is correct, honestly idek how but lmao it works! ...
doesnt quotient rule also work for a constant denominator but no one does it cause its so inefficient
rough tundra
doesnt quotient rule also work for a constant denominator but no one does it ca...
Oh maybe, my brain thought you take d/dx in the bottom for some reason
fierce pier
;-;
rough tundra
ok i equated to 0 and got x = 4
yep! now find A(4)
rough tundra
;-;
I just do $\frac{d}{dx}\left[f(x)\cdot\frac{1}{g(x)}\right]$ and avoid quotient rule altogether
glossy valveBOT
PajamaMamaLlama
fierce pier
interesting
shut torrent
ok so for A(4), I got 20 and thats the x value so the value of the base right?
rough tundra
not quite, t is the how long we've been travelling away on the x-axis so t=4 means we've travelled 4 units in the x-direction
and A(4)=20
means the area at that point (the maximum) is 20 :)
this is ineffect
everything you just did summed up in a Desmos graph
fierce pier
https://www.desmos.com/calculator/vcrmd4qfdw
people who use desmos >>>
shut torrent
everything you just did summed up in a Desmos graph
ohhhh bro thank u so much, i get it now
fierce pier
btw any of yall submitting for desmos art comp?
rough tundra
ohhhh bro thank u so much, i get it now
hope I helped a little! Again these types of problems take some visualization and thinking so don't get upset if you don't get it immediately and practice, is all I suggest for these problems!
rough tundra
btw any of yall submitting for desmos art comp?
I'd never win, some people are crazy 😭
shut torrent
btw any of yall submitting for desmos art comp?
oh nah, idts lmao
fierce pier
I'd never win, some people are crazy 😭
i wouldn't win either
this is my first time tryinf
shut torrent
I'd never win, some people are crazy 😭
FR im far from artistic if im being honest
analog nymph
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analog nymph
Can someone help me how to find the value of n
plush egret
the pieces of information are pretty evenly split
one of the quantitative pieces of info resolves k
the other resolves n
can you guess which is which?
plush egret
it is, yea
analog nymph
Yeah but i dont know how to solve for n
plush egret
well, whats the amplitude?
analog nymph
the heightbetween the middle and the top of the graph
i dont know how to explain properly
plush egret
i guess i mean in the context of this problem
i think youre identifying the correct graphical feature
analog nymph
ohhhh
plush egret
analog nymph
wait the amplitude is 4.5
plush egret
like this is kind of the picture
plush egret
wait the amplitude is 4.5
yea, meters
analog nymph
yup but how do i find n
plush egret
so the other piece of info gives n
analog nymph
the hours?
plush egret
12.2 hours, yea
oop
this image is for a sine, but i just didnt feel like typing the period formula lol
maybe its confusing
plush egret
period is just generally, how long until it gets back to where it was again
analog nymph
ohhhhhhh
plush egret
,w plot 4.5 cos( (2*pi)/(12.2) x )
glossy valveBOT
plush egret
you could check for the important graphical features here
but yea, thats all there is to it
make sense?
plush egret
hushed wasp
2 cos^2(x/2)sin^2x = x^2+1/x^2
Where 0<x<=π/2 has
No solution, one real solution, more than one real solution
open zinc
2 cos^2(x/2)sin^2x = x^2+1/x^2
Where 0<x<=π/2 has
No solution, one real soluti...
!status
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2 cos^2(x/2)sin^2x = x^2+1/x^2
Where 0<x<=π/2 has
No solution, one real soluti...
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
dusty musk
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vale sentinel
If G is a multiplicative abelian group of order 8 and f:G->G be a mapping such that f(x) = x³ for all x in G, then prove that f is an isomorphism
vale sentinel
I proved f is well defined and f is a homomorphism. But i am having trouble with proving f is injective
From f(a) = f(b) or a³ = b³ how can i show a = b
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vale sentinel
Ok
vale sentinel
.close
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lilac junco
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torn jolt
help
crimson ridge
help
excuse my bad handwritting, but if you foil the right hand side, you get 8x^3 -8x
and if you integrate that you get 2x^4 - 4x^2
and then you can simplify by dividing everything by 2
if you expand the answer they wrote ( x^2 -1)^2, you get x^4 - 2x^2 + 1
which is basically what we got, except they wrote it as x^4 - 2x^2 + A
idk why tho
crimson ridge
i was replying to the wrong guy ^
lilac junco
which is basically what we got, except they wrote it as x^4 - 2x^2 + A
yeah its strange cus A is meant to be the constant of integration not part of the expression right?
crimson ridge
yeah its strange cus A is meant to be the constant of integration not part of th...
yea tbh i didnt fully understand their explaination :')
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paper lagoon
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paper lagoon
Is D correct?
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paper lagoon
<@&286206848099549185>
wheat junco
There can't be any extremities
So Option D should correct
This function decreases as x decreases and increases as x increases
paper lagoon
are you 100%
wheat junco
are you 100%
yes
For x > 0 f'(x) > 0 so the funcn increases
For x < 0, f'(x) < 0 so the function decreases
swift crane
Why is it that the 'direction of motion' refers to the direction the velocity vector is pointing to rather than direction the posiiton vector of particle is pointing to?
gritty rose
Why is it that the 'direction of motion' refers to the direction the velocity ve...
If something isn't moving how does it have motion
swift crane
If something isn't moving how does it have motion
If we wanted to find the direction of motion at t=pi/6 for example, why would it not be r(pi/6)?
I don't understand what it means if it isn't moving or not, isn't r(t) the path of the particle?
gritty rose
I don't understand what it means if it isn't moving or not, isn't r(t) the path ...
Not moving implies velocity is zero
gritty rose
If we wanted to find the direction of motion at t=pi/6 for example, why would it...
You already said it. Motion is velocity
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limpid glen
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limpid glen
hello, for this part
im thinking of using systems of non linear equations
the problem is
we dont have something to equate to somehow
hot knoll
In order for a piecewise function to be continuous everywhere, it must be continuous at the breakpoints for the different pieces of the function
limpid glen
In order for a piecewise function to be continuous everywhere, it must be contin...
what are breakpoints again 😭
hot knoll
That is to say, f should be defined at the breakpoints and, more importantly for this type of problem, the limit as x approaches the breakpoints must equal the value of f
limpid glen
oh are those like endpoints?
hot knoll
Yea like the endpoints of the intervals where it changes
In this case there is only one such point to check, namely x = -1
limpid glen
That is to say, f should be defined at the breakpoints and, more importantly for...
so if a = 1,
one of subfunction should euate to 1 when subbed
ohhhh
hot knoll
Well what you'll find is that
limpid glen
That is to say, f should be defined at the breakpoints and, more importantly for...
f(-1) = 4
find a and b such that, when subbed equates to 4?
hot knoll
f(-1) = limit from the left and limit from the right when x approaches -1
Yea basically in this one you'll end up setting the top function and bottom function equal to 4
hot knoll
And eventually you'll be able to solve it all out
hot knoll
Well you do have x = -1
limpid glen
Well you do have x = -1
thats true
hot knoll
Yw
limpid glen
The limit is found by letting x approach, but not equal, - 1
ohhh
x = -1 is a breakpoint right
hot knoll
We find the limit by doing the left hand side limit (let x approach -1 from left) and right side limit
hot knoll
x = -1 is a breakpoint right
Yes for this function that is the point of interest
For the piecewise function to be continuous the pieces all have to connect to each other basically
So the top expression should be approaching 4 when x -> -1 from left
And the bottom expression should be approaching 4 when x-> -1 from right
limpid glen
$$lim_{x \rightarrow -1^{+}} x^2 + bx + (a + 4) = 4 $$
$$5 + a + b$$
limpid glen
That should equal 4
is it wise to use systems of non linear equations?
hot knoll
I think you have no choice
glossy valveBOT
Maladroit
hot knoll
But they made it easy enough to solve
Because you can get a or b directly from that first limit you took
Set
5 + a + b = 4
Then you can substitute into the other expressiom
limpid glen
$$lim_{x \rightarrow -1^{-}}ax^3+ 2bx + 1= 4 $$
$$-a -2b + 1 = 4$$
glossy valveBOT
Maladroit
hot knoll
Because we want the function to be continuous
We need
f(-1) = lim from left of f as x-> -1 = lim from right of f as x -> -1
limpid glen
Because we want the function to be continuous
and how does equating all subfunctions to 4 achieve that
im sorry 😭
wide wind
$$lim_{x \rightarrow -1^{-}}ax^3+ 2bx + 1= 4 $$
$$-a -2b + 1 = 4$$
You need \ before lim.
limpid glen
You need \ before lim.
ye hihi
hot knoll
and how does equating all subfunctions to 4 achieve that
In piecewise function, the region less than the breakpoint is the left side limit behavior
limpid glen
In piecewise function, the region less than the breakpoint is the left side limi...
yeah
hot knoll
Until you reach -1 itself, you act like that top function
limpid glen
Until you reach -1 itself, you act like that top function
x < -1?
hot knoll
Yea
So if i want the behavior as I approach -1 from the negative (left) side
I would use that top function's behavior
hot knoll
What's more likely is its two different curves that connect at the point (-1,4)
limpid glen
What's more likely is its two different curves that connect at the point (-1,4)
i created the graph, i dont know if based from here its continuous
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rotund wedge
Why can't I just use x = 2 here?
And how should I go about trying to solve this question? Do I differentiate? It looks messy to differentiate
rotund wedge
btw is this correct? $\frac{d}{dx} 3^x = 3^x ln(3)$
onyx glen
Why can't I just use x = 2 here?
And how should I go about trying to solve thi...
what happens when you try to put x=2?
glossy valveBOT
realtime
rotund wedge
what happens when you try to put x=2?
3^2 + 3 - 12 = 0
3^-1 - 3^-1 = 3^-1 - 3^-1
= 3- 3 = 0
onyx glen
1-2 = ?
rotund wedge
oh my god
im dumb
okay so its 0*0
so it shouldn't require much after that yeah or is it indefinite?
onyx glen
0/0 not 0*0
i think i would substitute t := 3^x here. unless this is one of those "SOLVE THIS IN 10 MILLISECONDS OR ELSE" questions.
rotund wedge
but its
0/(3^-1 - 3^-1) = 0 * (3-3) = 0
rotund wedge
i think i would substitute t := 3^x here. unless this is one of those "SOLVE THI...
lol haha
okay
so
onyx glen
but its
0/(3^-1 - 3^-1) = 0 * (3-3) = 0
??
1/(a-b) is not equal to 1/a - 1/b.
1/(1/a-1/b) isn't equal to a-b either.
rotund wedge
Why can't I just use x = 2 here?
And how should I go about trying to solve thi...
apologies for the bad drwaing its by mouse
wait would d/dx of 27/3^x not be 0/3^x?
and same for 3/3^x
fleet breach
and same for 3/3^x
same mistake with this^
fleet breach
cause it is an exponent
you need to use chain rule to evaluate this diverative
because you need to find diverative of function itself and the exponent
using l'hopital rule as the symbol is undefined when you put x=2 you can easily evaluate this limit as so:
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waxen flicker
Am I doing this right?
\begin{align*}
I_0 = \int \sqrt{x^2-1} , dx = x \sqrt{x^2-1}-\underbrace{\int x , d(\sqrt{x^2-1})}_{I_1}
\end{align*}
\begin{align*}
I_1 = \int x \frac{\cancel{2}x}{\cancel{2}\sqrt{x^2-1}}, dx &= \int \frac{x^2}{\sqrt{x^2-1}} ,dx \
&= \int \frac{x^2-1+1}{\sqrt{x^2-1}} , dx \
&= \underbrace{\int \sqrt{x^2-1} , dx}_{I_0} + \int \frac{1}{\sqrt{x^2-1}} , dx \
&= I_0 + \arcsin x
\end{align*}
\begin{align*}
I_0 &= x \sqrt{x^2-1} - I_0 + \arcsin x && \vert , + I_0 \text{ and then } \vert \cdot \frac{1}{2} \
I_ 0 &= \frac{1}{2} \left(x\sqrt{x^2-1} + \arcsin x\right)
\end{align*}
glossy valveBOT
Sweet Tea 🧋
waxen flicker
,w (1/2 (x sqrt(x^2-1))+arcsin(x))'
glossy valveBOT
waxen flicker
doesn't look like $\sqrt{x^2-1}$, altho I can't find a mistake
glossy valveBOT
Sweet Tea 🧋
pale jacinth
Sureley using a coshu substitution would work
waxen flicker
Sureley using a coshu substitution would work
we didn't learn this, so I wanna practice solving w/o it (at least for now)
pale jacinth
What about x = secu
waxen flicker
What about x = secu
i think it's a lot simpler (see my solution)
pale jacinth
Yea but i dont think that's right
waxen flicker
Yea but i dont think that's right
hm, why not? 
pale jacinth
well look at ur derivative
waxen flicker
well look at ur derivative
well, yea, but the steps seems to be correct lmao xD
maybe it's wolfram xD
pale jacinth
well, yea, but the steps seems to be correct lmao xD
maybe it's wolfram xD
nah i evaluated it on my calc it's not
I'm pretty sure x = secu or x = coshu is standard
buoyant pewter
well, yea, but the steps seems to be correct lmao xD
maybe it's wolfram xD
look:
$\int_{}^{}\frac{dx}{\sqrt{x^{2}-1}}=\int_{}^{}\frac{x+\sqrt{x^{2}-1}}{\sqrt{x^{2}-1}}\cdot \frac{dx}{x+\sqrt{x^{2}-1}}=\\=\int_{}^{}\frac{1+\frac{2x}{2\sqrt{x^{2}-1}}}{x+\sqrt{x^{2}-1}}dx=\ln\left( x+\sqrt{x^{2}-1} \right)$
glossy valveBOT
Joanna Angel
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@waxen flicker Has your question been resolved?
buoyant pewter
well, yea, but the steps seems to be correct lmao xD
maybe it's wolfram xD
$\text{Hence:}\\\int_{}^{}\sqrt{x^{2}-1}\text{ }dx=x\sqrt{x^{2}-1}-\int_{}^{}\frac{\left( x^{2}-1 \right)+1}{\sqrt{x^{2}-1}}dx=\\=x\sqrt{x^{2}-1}-\int_{}^{}\sqrt{x^{2}-1}\text{ }dx-\ln\left( x+\sqrt{x^{2}-1} \right)\\\Leftrightarrow \int_{}^{}\sqrt{x^{2}-1}\text{ }dx=\frac{x}{2}\sqrt{x^{2}-1}-\frac{1}{2}\ln\left( x+\sqrt{x^{2}-1} \right)+C$
glossy valveBOT
Joanna Angel
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waxen flicker
**Sweet Tea 🧋**
oh goooosh $$\int \frac{1}{\sqrt{x^2-1}} , dx \stackrel{?}{=} \arcsin x$$ 
glossy valveBOT
Sweet Tea 🧋
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