rocky tendon
or will it be deleted
#help-0
void solar
It'll always be here
lone heartBOT
https://i.gyazo.com/331e23e7560d3e52eecde76d595ddf93.png
alpine sable
- is the one i need help with
can I use algebra to solve it? how do you go about sketching it
lone heartBOT
@alpine sable Has your question been resolved?
rigid smelt
well you do need a bit of algebra to "sketch" the intequality
so the answer to that is yes
about sketching it, how familiar are you with sketching inequalities?
starting from inequalities involving linear functions to polynomials of higher degrees and other curves?
vital zenith
May i ask how do i factor this?
lone heartBOT
May i ask how do i factor this?
vital zenith
$9k^2 - 25 = 0$
ocean sealBOT
leewchaha
vital zenith
im using this method but it doesnt work out for me
slender marten
Write 9k^2 as (3k)^2 to obtain (3k)^2 - 5^2 and you might recognise this as the difference of two squares which can then be simplified.
vital zenith
im using this method but it doesnt work out for me
nono this is from a youtube video
vital zenith
Write 9k^2 as (3k)^2 to obtain (3k)^2 - 5^2 and you might recognise this as the ...
oh isee
$(3k - 5)^2$
ocean sealBOT
leewchaha
vital zenith
is this the final answer?
vital zenith
$(3k)^2 - 5^2$
ocean sealBOT
leewchaha
vital zenith
oh thanks
wanton pebble
or just simply take k on one side and other terms on other side of equality
slender marten
(a^2 - b^2) = (a - b)(a + b) so let a = 3k and b = 5 to obtain the desired result.
vital zenith
(a^2 - b^2) = (a - b)(a + b) so let a = 3k and b = 5 to obtain the desired resul...
can this (a^2 - b^2) = (a - b)(a + b) be applied for every (a^2 - b^2) ?
slender marten
Yes. To prove this just expand (a - b)(a + b).
vital zenith
oh thats awesome. are there other standardized solution like this and (a + b)^2 = a^2 + 2ab + b^2 ?
it would definitely be easier for me to udnerstand
slender marten
Well I'd have probably seen the simplification instantly due to experience of seeing it many times but there may be others I don't know of.
idle steppe
hi!
idle steppe
isnt it (y(t))^2 not t^2
ionic trail
u(v(t)) = (y(t))² with that definition of u and v
idle steppe
u(v(t)) = (y(t))² with that definition of u and v
so its: 2(y(t))(y'(t)) + 2(y'(t))(y''(t))
ionic trail
Yes!
ionic trail
We know that y'' = -y
(Sorry, -y, not -y')
ionic trail
oh wait u factor out a 2y'
Yeah that sounds good
ionic trail
Strictly speaking, y(t) is y evaluated at t. y is just the function itself
But yeah, it's y''(t) = -y(t) is more precise
Its true for all t
idle steppe
But yeah, it's y''(t) = -y(t) is more precise
so the answer is still 0
ionic trail
so the answer is still 0
Has to be. Then the original fct. is a constant
Derivative of a const is 0
ionic trail
Glad to help
idle steppe
Glad to help
what does this mean for an associated first order planar 2x2 system
ionic trail
what does this mean for an associated first order planar 2x2 system
Sorry, no idea what that is😅
keen plinth
i assume your system is (y, y')
so the phase plane is graphing y' against y
if y^2 + y'^2 = constant
then that is a circle on the plane
which means your solution is a periodic solution
which should make sense
since it is the equation of a harmonic oscillator
@idle steppe
idle steppe
since it is the equation of a harmonic oscillator
Does the reasoning above on why it’s a constant make sense?
keen plinth
it is fine
alpine sable
Since hashing functions will take in an arbitrary-sized stream of bytes and return a fixed sized byte stream, would it be correct to state that for a given output, there are infinite inputs that would produce it? Thanks.
keen plinth
this is probably not true in general
alpine sable
this is probably not true in general
Why'd that be?
keen plinth
suppose the fixed size output is of size 1
let the hashing function h send only 0 to 0
but any other input to 1
then for the given output of 0
there is only one input that will produce it
namely 0
alpine sable
But I reckon h wouldn't be considered a hashing function by definition. The output is meant to be completely different, no matter the difference in two given inputs. So both 1, 1.00000001 and 25987525 will have the exact same output.
keen plinth
then it will depend in your definition of what a hashing function is
i agree that a "good" hashing function should not do what i suggested
but that does not preclude it from being a hashing function
it would just be a very poor one
alpine sable
Yeah, I see. Let's pick a specific one, then. Say SHA512. Would it be right to make that statement about it?
keen plinth
then it is quite possibly true
but that is only a specific case
it is not a statement that can be said of every hashing function
keen plinth
np
alpine sable
.close
modest perch
Hello, how can I prove that the function $\frac{\left(e^{x}-1\right)}{x}$ is bounded on the interval ]0;1]?
ocean sealBOT
Keroro
modest perch
It's the derivative of the function $\left(\int_{1}^{x}\frac{\left(e^{t}\right)}{t}dt\ -\ln\ x\right)$ , i have to show that it's bounded and from that, prove that the function g is also bounded on the same interval
solemn juniper
modest perch
(g is the function with the integral)
ocean sealBOT
Keroro
slender marten
I think you just need to show that the limit as x goes to 0+ of (e^x - 1)/x is finite. L'hopital's rule will show it equals 1.
marsh rapids
depending on your level of knowledge of analysis, you can show it's continuous on [0, 1] (by setting f(0) = 1) then you know it's bounded because continuous on a closed set. From the derivative being bounded the MVT then gives that g is bounded
marsh rapids
no. Intervals are closed iff they include the bounds
marsh rapids
yes
marsh rapids
depending on your level of knowledge of analysis, you can show it's continuous o...
Normally these properties should be known btw, otherwise it would be good to know that we can't use these
lone heartBOT
lone heartBOT
Available help channel!
torpid shoal
lone heartBOT
torpid shoal
need help :>
ocean sealBOT
R00tKiT
torpid shoal
how did you get -1
I actually don't know how my friend got that answer. he is asking me for help but i also don't know how to do it can you teach me so I can help him?
gray isle
get them here?
torpid shoal
okok brb
gray isle
dunno why people are asking people that don't know
torpid shoal
he told me he needs to wait 10 mins before he can put a message
gray isle
mmk
torpid shoal
he told me he got it by replacing x with 1
gray isle
would like to see the work for it
because doing subbing in x=1 properly doesn't get what's in those boxes
stuck canyon
Hi
gray isle
are you the original owner of that question
stuck canyon
Yeah
gray isle
can you show all the work you did
gray isle
that would be the value of f(1),
which isn't what the question is asking for
lone heartBOT
@torpid shoal Has your question been resolved?
limpid spade
Lol
stuck canyon
How do i solve it?
gray isle
for
f(g(1)),
first find g(1)
alpine sable
Why is the intersection between 2 spans is the equality between them?
worn fox
What do you mean?
lone heartBOT
@alpine sable Has your question been resolved?
alpine sable
What do you mean?
basically
lets say span [A] intersection span[B]
my lecturer opened up the spans and made equality between them
aV1 + bV2 + cV3 = alphaU1 + betaU2 + gamaU3
my question is why "aV1 + bV2 + cV3 = alphaU1 + betaU2 + gamaU3" is the same as "lets say span [A] intersection span[B]"
worn fox
aV1 + bV2 + cV3 = alphaU1 + betaU2 + gamaU3
Guessing context here, but the left hand side is a general element in span(A) and the right hand side is a general element in span(B)
alpine sable
let me screenshot it 1 sec
worn fox
So if we want to find something in both spans, we need to set two general elements equal and see what conditions on the coefficients we get
worn fox
So if we want to find something in both spans, we need to set two general elemen...
Yes so this
small oasis
not sure what the answer is
lone heartBOT
not sure what the answer is
alpine sable
not sure what the answer is
is this assessed
lone heartBOT
@small oasis Has your question been resolved?
alpine sable
I have solved this question correctly.
I am wondering if there is an easier approach - specifically if there is a way to combine the first two combinations into a single combination.
Also is it correct to call those n choose k terms "a combination/combinations" or are they called something else?
Thanks
lone heartBOT
I have solved this question correctly.
I am wondering if there is an easier app...
@alpine sable Has your question been resolved?
prime spindle
how find in terms of alpha beta
lone heartBOT
how find in terms of alpha beta
glad osprey
🤡
alpine sable
how find in terms of alpha beta
find alpha and beta first
then solve the other equation like you would normally do
alpine sable
I found alpha+beta and alpha x beta
no you need each of thel
do you know how to calculate the discrimnant ?
alpine sable
Let the diophanite equation be ax² + by² = c
My thereom first equate ≥ any one term of x or y
Let y term, ⇒ by² ≥ 0
Adding ax² to both sides
ax² + by² ≥ ax²
Now there is a possible case
ax² ≤ c
⇒ - √c/a ≤ x ≤ √c/a
Now put integer values of x in this range so that output will be an integer let the process as function f(x)
Now to cross check, do same with x term or if you don't then no problem (note some rare cases may be)
Our solution set will be the inputs and outputs of the function f(x)
Please check one time
prime spindle
Get tf out of here
prime spindle
Get your own help channel
lone heartBOT
@prime spindle Has your question been resolved?
lone heartBOT
@prime spindle Has your question been resolved?
acoustic thunder
lone heartBOT
alpine sable
ok
acoustic thunder
hey how do i find b?
lone heartBOT
@acoustic thunder Has your question been resolved?
acoustic thunder
<@&286206848099549185>
lone heartBOT
@acoustic thunder Has your question been resolved?
acoustic thunder
what's your solution for a?
5,7,9
turbid violet
well the number of terms would be l-4 right?
do you see why that would be the case?
dark trail
i need help pls
lone heartBOT
i need help pls
hazy violet
Divison rule
dark trail
yeah i tried that
hazy violet
Ah
dark trail
lemme show u what ive doen so far
hazy violet
OK
dark trail
But I’m still confused
It’s sending btw @hazy violet
Potato Wi-Fi
It’s done that’s what I’ve tried to do but now I’m hella confused
@hazy violet how would u divide or do integration by parts
keen plinth
is this an integration by parts question?
dark trail
yes
hazy violet
<@867780015323611146> how would u divide or do integration by parts
The ILATe rule
keen plinth
it's not typically something you'd do by parts
dark trail
but i got to put it into partial fractions
it says
solve these derivatives
or find the integrals by first splitting the expression into partial fractions
Q11b
keen plinth
so it is not by parts
dark trail
The ILATe rule
Whats ILATe?
keen plinth
it's a rule you can use to decide how to integrate by parts
but it's not necessary here since you shouldn't even use by parts
dark trail
Is splitting into partial fractions and integrating by parts the same thing?
keen plinth
no
keen plinth
they are completely different methods that can be used for integration
dark trail
Ahhhh they sound so similar
keen plinth
by parts is the product rule of differentiation but backwards
partial fractions decomposes rational functions
dark trail
What’s rational functions?
hazy violet
but it's not necessary here since you shouldn't even use by parts
Oh..
keen plinth
p(x)/q(x)
dark trail
Ahhh
keen plinth
where both p and q are polynomial
keen plinth
not really partial fractions
dark trail
So 11d would be by parts?
keen plinth
since you only have 1 linear factor
nay
all of 11 are some form of partial fractions
dark trail
Bruh this is confusing . Maybe it’ll marinate later . But how do I do this then?
keen plinth
3ii is something you would do by parts
keen plinth
what you've done here is good
keen plinth
rewrite the integrand as something of the form ax + b + c/(x+2)
keen plinth
do you know how to do polynomial division
dark trail
No problem Ill research more
On polynomial division
The issue is I’ve never seen the answer turn out like he did
It’s like he took out the -x from the original quadratic on top
Divided it by x+2
keen plinth
yes that's how you do polynomial division without doing polynomial division
it's essentially magic
keen plinth
you only do that if you know what you're doing
but if you don't
then the systematic route is doing long division
you will get the same answer
with less brain and magic
dark trail
So did he do it like quadratic function/linear function= 0
Then take the x to the left since it was +
keen plinth
no this is just clever algebraic manipulation
dark trail
Oof
keen plinth
it has the same result as polynomial division
keen plinth
then the systematic route is doing long division
hence my suggestion
dark trail
Ya il do that the long division next time
dark trail
Oh th he said I missed the + sign or something
keen plinth
yeah
dark trail
Between the brackets
keen plinth
that part
dark trail
What level of algebra would what he did be? University or the year before university which is called a levels
I need to learn that method I hate long division
dark trail
Bruh
keen plinth
it is not something that is "taught" per se
it is something that you just have to build up intuition for
dark trail
Y’all are geniuses fr
keen plinth
you mistake experience for intelligence
dark trail
Now my brain is going blank I’m struggling to integrate x-2x
I’ve got x^2/2
But -2 should be -2x right
keen plinth
so the expression you're integrating should be something like
$x - 1 - \frac 2 {x + 2}$
i believe
ocean sealBOT
夢雪
keen plinth
But -2 should be -2x right
you would be right if we were integrating -2
dark trail
And should I split each by doing this
keen plinth
yes that will work
dark trail
Perfect lemme do that now
Final answer
11c would be polynomial division again without the polynomial division 😂
keen plinth
wonderful
dark trail
This plagued me for 4 days
I avoided revision because I couldn’t move forward on this question
Thanks so much bro
Could you quickly lead me in the right direction to do the magic method?
Like a video on elementary fractional division but on a PHd level
keen plinth
i'm not sure if such resources exist
dark trail
Damn bruh 😭
keen plinth
it's more just a thing you realise will work
dark trail
If only I had common sense and experience like that 💀
keen plinth
like say for the next one
dark trail
Ok so polynomial division for 11c
keen plinth
2x^2 + x - 7
x - 3
what you could do is see that the denominator is x - 3
and try to match it on the numerator
dark trail
Hmm
keen plinth
so you could write 2x^2 - 6x + 7x - 7
then the 2x^2 - 6x = 2x(x - 3)
which allows you to cancel the x - 3
then you're left with 2x + (7x - 7)/(x - 3)
then you just repeat
match the denominator by writing 7x - 21 + 14
and 7x - 21 = 7(x - 3)
which cancels
and you get 2x + 7 + 14/(x - 3)
and that's done
dark trail
Imma screenshot all this and try it continuously
keen plinth
like
it's polynomial division but using more brain power
there's not much point to it apart from stroking your own ego
dark trail
But how do u split those to be integrated ? Like all the plus signs act as walls where by what’s behind the wall gets integrated individually
keen plinth
the integral of a sum of things
dark trail
So 2x integrated then 7 then the fraction as we did before
keen plinth
is the sum of the integrals of those things
dark trail
Ahhh ok
So we split it like I did before integrate x then 1 then the 2(x+2)^-1
Kk lemme try understand your method and complete the question
keen plinth
yes
keen plinth
and you get 2x + 7 + 14/(x - 3)
i have no idea if that's correct
dark trail
Ooof 😭
dark trail
Lemme just find polynomial dividing calculator atp
keen plinth
i haven't checked it yet
keen plinth
it's not worth doing
dark trail
Fr? Ir seems quicker
keen plinth
yep
synthetic division will give you the answer without even needing you to use your brain
marble oxide
where should i go for help on business math
marble oxide
is there a specific one
dark trail
After finishing 11d I’m gonna eat a crack load of ramen to refill my expended brain power
marble oxide
im dying trying to figure these out
keen plinth
x-3
keen plinth
now you must partial fraction
keen plinth
the numerator has to have a degree less than the denominator
before you do partial fraction decomposition
dark trail
Uhhh
keen plinth
you'll never get the x^2 term on top otherwise
so essentially the deal is
do polynomial division to reduce the degree
then do partial fractions on the remainder
keen plinth
(x-1)^2
keen plinth
it'll be 1 + stuff
dark trail
so far ive split it so that it looks like xsquard over quadratic
+(-2x over quadratic)
+(-5 over quadratic)
keen plinth
that will not do what you want
keen plinth
so essentially it's like
x^2 + 2x + 5
over
x^2 - 2x + 1
if you rewrite the numerator as x^2 - 2x + 1 + 4x + 4
then you can separate the first 3 terms
keen plinth
because then you can cancel from the denominator
dark trail
is this that magic method agaun?
keen plinth
so then the first 3 terms cancel with the denominator
and the 4x + 4 is the remainder
keen plinth
so you get 1 + (4x + 4)/(x - 1)^2
dark trail
then i need to find
keen plinth
and that's the partial fraction you need to do
dark trail
some x-1)^2
keen plinth
the 4x+4 part
dark trail
or value of x-1 that can cancel 4x=4
4x+4
like 4x-4
but then i get 0
@keen plinth im confused sorry
dark trail
wrong channel
sour hill
where?
dark trail
its occupied bro
dark trail
np
dark trail
AHHHH
keen plinth
what your job is is to solve for a and b
dark trail
and i select a value tht cancels a or b
keen plinth
uh
i'm not sure about that
the typical way to solve for a and b is to multiply through by (x-1)^2
glossy current
integrate (x^2+x-4)/x+2 by parts
Are you still stuck with it
dark trail
Are you still stuck with it
kinda figuring it out
lemme shwo u
I multiplied what the a and I’m not sure if it’s accurate e @glossy current
keen plinth
so you get 4x + 4 = A(x - 1) + B right
dark trail
Ahhh
dark trail
So yes the B cancels that I know
keen plinth
we know B = 8
dark trail
Ok
keen plinth
so you get 4x + 4 = A(x - 1) + B right
so the only x comes from the A
which means A = 4
dark trail
Ahhh
So the relevance to 4 is that the x in 4x belongs to x=1
That’s why the coefficient is 4
keen plinth
its not that there's x=1
keen plinth
it's that if you write out the right
dark trail
Yes sir
keen plinth
you get Ax - A + B
that has to equal 4x + 4
the x coefficients must match
so A = 4
dark trail
Ahhh
Ok . Imma pretend I got all that💀💀 ahhh wait
Since coefficients of 4x+4 are equal
And Ax-A exists
The coefficients must be 4
Since it’s kind of of the same form
keen plinth
there's also B
we are equating 4x + 4 = Ax - A + B
this means A = 4 and -A + B = 4
dark trail
Ah ok ok
keen plinth
because the coefficients must be the same for the equation to hold over all values of x
keen plinth
so you get 4x + 4 = A(x - 1) + B right
plugging in x=1 here gives you 8 = B
so we know A = 4 and B = 8
dark trail
So u get this to be integrated
keen plinth
(4x + 4)/(x - 1)^2 = a/(x - 1) + b/(x - 1)^2
the denom of A only has 1 power not 2
otherwise it would be redundant
dark trail
Generally tho. With these types of questions. You’re trying to reduce and split the fractions to be integrated in parts
Ahh ok I’ll change that
keen plinth
Generally tho. With these types of questions. You’re trying to reduce and split ...
yes
dark trail
Will u always have an instance whereby ln|fx| will be integral
With these types of qs
keen plinth
it is common
dark trail
Ahh ok ok
keen plinth
partial fractions means you get stuff that looks like 1/(x - c)
so you'll almost always get logs
dark trail
Ahhh ok .
glossy current
If you've done, ping me. I'll show you another elegant method
glossy current
Yes
keen plinth
its not generally useful though
dark trail
If you've done, ping me. I'll show you another elegant method
Im almost done thank u I’ll also like to know the method
glossy current
Partion fraction is usually long
There's better ones
,tex
Another way to do it without original partion fraction is to do algebra manipulation.\\
$\begin{aligned}
I&=\int \frac{x^2+x-4}{x+2} dx\\
&=\int \frac{(x^2+4x+4)-(3x+8)}{x+2} dx\\
&=\int \frac{(x+2)^2}{x+2}-\frac{3x+6+2}{x+2} dx\\
&=\int (x+2) -3\frac{x+2}{x+2}-\frac{2}{x+2} dx\\
&= \frac{x^2}{2}-x+2\ln\abs{x+2}+C\\
\end{aligned}$
Darkness
keen plinth
no we did that one already
glossy current
o why don't you use this way then
keen plinth
because the integrand is (4x + 4)/(x - 1)^2
glossy current
Oh that
keen plinth
to do magic
keen plinth
in general partial fractions will be arbitrarily bad
glossy current
$\int \frac{y'}{y} dx= \ln|y|+C$
ocean sealBOT
Darkness
glossy current
We notice the denominator is a quadratic form
The numerator is linear one, which is derivative
Of the denominator
keen plinth
that's the same thing as doing magic
glossy current
Hm
dark trail
Final answer?
dark trail
Should be + right
keen plinth
i think so
glossy current
Hm it shouldn't be that i believe
keen plinth
like
we've been working through this question over quite a bit of length
so there's more to it than what i've just mentioned
glossy current
Nvm it's correct
I mislook at 4x-4 instead of 4x+4
keen plinth
if you want to help then i would recommend reading through the whole history
dark trail
This has been hell. I’ve changed it to x +4ln…
Im glad it’s all over thank you all I’ve learn so much
keen plinth
np
dark trail
Now I’ve gotta make myself some damn good ramen noodles
Air hug to u guys ❤️
.close
tropic sail
Is (x^2)/x equal to x?
lone heartBOT
Is (x^2)/x equal to x?
mortal trellis
they are not equal
however, you can continuously extend the function x^2/x to the function x at the point x=0
so we often don't care too much about the difference
tropic sail
they are not equal
If it isnt equal then what is the term for it?
mortal trellis
not equal?
simple ether
.help
lone heartBOT
Commands:
clopen: .close, .reopen
consensus: .poll
factoids: .tag
help: .help
Type .help <command name> for more info on a command.
tepid frigate
lone heartBOT
tepid frigate
hi i got a question, at what point if f discontinous? and what point is f not differentiable
how can i tell from.a graph if its discontinous?
alpine sable
hi i got a question, at what point if f discontinous? and what point is f not di...
The function is discontinous where the graph jumps and not differentiable where the graph has an angle
rose sigil
that's weird... it's not even the graph of a function
seems like a mistake and maybe the "vertical line" at x = 0 isn't supposed to be there
alpine sable
either somebody was using a stupid program or was trying to be too clever and not to show obvious discontinuity
urban pine
maybe it's just really steep
lone heartBOT
@tepid frigate Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
whole canopy
lone heartBOT
noble tusk
got no clue how that simplyfiys to 5 2squared
It simplifies to sqrt(1+49) = sqrt(50) = sqrt(25*2) = sqrt(25)*sqrt(2) = 5sqrt(2)
slow hound
can you send a screenshot instead
crimson carbon
Yeah, reuploading it
Hey, I was catching up with summer work and I'm stuck on this problem. I have the answer but I don't understand how to get the answer
The answer was 21%
slow hound
@crimson carbon
how many ways are there to choose 8 sketches out of 40 (with no restrictions?)
crimson carbon
Would it be 76,904,685?
lone heartBOT
Available help channel!
crimson carbon
Hello?
slow hound
did you do
ocean sealBOT
crimson carbon
I'm a little lost, what does this mean?
tacit arch
that color 👏🏻
slow hound
ocean sealBOT
crimson carbon
I thought I did do it
crimson carbon
Would it be 76,904,685?
this?
ocean sealBOT
crimson carbon
I don't know how to get the probability from there now
That was the first half of the problem
slow hound
so that's the total amount of ways
yeah
now we can tackle the second half
probability is # of successful outcomes / total outcomes
we can imagine the 40 sketches listed as follows
XXXXXAAAAAAAA....AAAA
where there are 5 X's (your sketches) and 35 others (other ones)
that total to 40
with me so far?
crimson carbon
Yes sorry, was rereading it a bit
slow hound
we want to pick exactly 2 of your sketches
how many ways are there to do that?
(choose 2 X's out of 5)
crimson carbon
10
slow hound
yes 5C2
now we need the other 6 sketches from the other ones
there are 35 of those
(choose 6 A's out of 35)
slow hound
,w (35 choose 6)
ocean sealBOT
slow hound
yep
multiply the two numbers we just got together and that will give you total number of ways to choose exactly 2
and thats your numerator
crimson carbon
Oh wow, I got it!
Thank you so much!
I'm new to this server, but is there like any way I could review you? I really appreciated how you walked me through the whole thing
slow hound
ah there's no need for that, your kind words are plenty :)
slow hound
you as well!
drowsy scroll
lone heartBOT
worn fox
What's 3/5 * 20?
drowsy scroll
Wait do you multiple the 3 or the 5
worn fox
$\frac35 \cdot 20$
ocean sealBOT
ΣAC
worn fox
You know how to multiply fractions
drowsy scroll
Is 60/5
worn fox
Which is?
worn fox
Uh huh it's the same for the rest, it's just also being timesed by x
lone heartBOT
worn fox
You're welcome
lone heartBOT
Available help channel!
dawn quail
lone heartBOT
worn fox
You can times it by $1 =\frac{-1}{-1}$
ocean sealBOT
ΣAC
dawn quail
Oh wow, OK
I would see that as an extra step involved
Not so easy to spot right away for me
mortal trellis
after you've seen it once it's pretty obvious
dawn quail
For me I would write -b + a for the numerator and then rearrange to put the positive a in front afterwards
I hope this will not be a problem for me in Calculus to take my time with writing the full steps rather than skipping steps
And if the instructor skips a step I hope I can spot it
mortal trellis
gotta practice more. then these steps are easier to see and you get faster
dawn quail
Perhaps writing full steps is better practice to get into for math? Takes longer but instructors will give full marks?
I dunno haha everyone is at a different level with understanding math. Someone else may see first to last right away someone else may only see it with middle steps involved
dawn quail
On my case I only see that with the middle step included lol
mortal trellis
never hurts to write more steps
lone heartBOT
@dawn quail Has your question been resolved?
dawn quail
Ty
mild saffron
lone heartBOT
mild saffron
Hello, this is a solids of revolution problem from Calc II
And the drawing pic (2nd pic) is my visual interpretation of the problem...
The problem says that the cross section is a semi-circle...
And I think the equation of the line is y=-x+14
How can I set up my integral to calculate the volume of the solid?
I guess the y-value (or the height) is going to be that length of the semi-circle
In order to get the radius of the semi-circle you would have to square it then divide it by 2, no?
last ether
Alright so the volume is going to the sum of the areas of each semicircle, which I think you get
mild saffron
I'm confused.
last ether
So what you're gonna do is treat f(x) is the diameter of each semicircle
Do you know your formula for the area of a semicircle?
@mild saffron
last ether
Btw what you have is right
last ether
What does it say
last ether
I don't see how
mild saffron
Yeah, neither do I.
last ether
Oh it shojld be 8, not 4
last ether
Gimmie time to type it out
mild saffron
Alright!
last ether
The area of a circle is $πr^2$, right? So if we sub in diameter:
$$\begin{align*}
A &= π\left(\frac{d}{2}\right)^2 \
&= \frac{πd^2}{4}
\end{align*}$$
If we wanna find half the circle area (which is the area of a semicircle), we divide both sides by 2:
$$0.5A = \frac{πd^2}{8}$$
ocean sealBOT
Umbraleviathan
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
mild saffron
Hmm, lemme read and interpret it on my own!
Ohhh okay
I finally understand it.
Thanks, @last ether.
last ether
Np
lone heartBOT
@mild saffron Has your question been resolved?
mild saffron
Yes
alpine sable
Two glass containers of different sizes are shaped like cubes. How many times bigger
volume does the larger container have than the smaller one?
(1) The larger container holds 113 dm^3 more than the smaller container.
(2) The smaller container has half the base area of the larger container.
Sufficient information for the solution is obtained
A in (1) but not in (2)
B in (2) but not in (1)
C in (1) together with (2)
D in (1) and (2) separately
E not by both statements
alpine sable
I'm thinking that there is a relationship between the larger cube and the smaller one, the smaller cube has a base area of a^2/2, doesn't that mean that the base area is half of a^2 for the smaller cube?
turbid violet
yeah if you assume the base of the larger cube is a^2
alpine sable
(2) Call the base surface of the larger cube x^2 (the side of the cube is thus x). Since the base area of the smaller cube is half as large, it is (x^2)/2. Each side of this smaller cube is therefore sqrt((x^2)/2)=x/sqrt(2)
The volume of the large cube is x^3
The volume of the smaller cube is (x/sqrt(2))^3=x^3/(2sqrt(2))
You can now derive a relationship between the volumes and it can be solved with B.
This is the solution provided by my friend, he is saying that this one is more correct. Do you know why?
turbid violet
well did you understand the question?
Let's say V_s is the volume of the smaller cube and V_l is the volume of the larger cube. You want to find an equation of the form V_l = kV_s (k is some constant)
so from just the information in statement 2 you're able to figure out how many times larger the bigger cube is (like how your friend showed)
what you'd want to check now is that can you do the same with only the information in statement 1?
try working it out
alpine sable
@turbid violet (2), so the side of the bigger cube is square root(2)×a?
turbid violet
yep
alpine sable
I'm thinking that there is a relationship between the larger cube and the smalle...
But is this wrong?
turbid violet
no, it's the same thing
but that was given in the question
you need to derive the relation between the volumes
alpine sable
So if we use my solution method, how would you calculate the ratio between the different volumes of the cubes?
I.e a^2 = a^2/2
2a^2 = a^2
Uh
turbid violet
that's wrong
turbid violet
call the side of the larger cube something else
it's different
b or something maybe
turbid violet
2a^2 = a^2
you just proved 2 = 1 there 😛
turbid violet
yep
turbid violet
yep
alpine sable
a^3
turbid violet
and for a cube with side b?
turbid violet
yep
livid bison
Hello! The instruction is to write the equation of a circle given the center and the radius. I am not sure if I am correct because the answer online says otherwise??
radiant halo
radius is √3/3
livid bison
how did it become that?
radiant halo
r²=1/3
livid bison
what happens with that 😅 sorry
radiant halo
I also don't have the full context of the problem
livid bison
Oh wait
"Write an equation of a circle given the following information" C: (3,0) R: 1/3
livid bison
ohh? so the answer is really 1/9?
radiant halo
R is a way of saying r²
livid bison
yes which is why i squared 1/3
radiant halo
you would square it if r was 1/3
I'm saying that R might be r²
In which r becomes √(1/3)
livid bison
and then what happens?
livid bison
my question is more of how did it turn into 3/3😭 what did u multiply to itt
livid bison
OHHHH
radiant halo
Then denominator rationalization
livid bison
OKOK I GET WHY IM WRONG NOW
the answer is really 1/9 but i
THANK U ANWAYS 😭
i got it
/close
close.
closed.
.close
wooden compass
lone heartBOT
nocturne dove
They can be skew lines
lone heartBOT
@wooden compass Has your question been resolved?
cunning trout
For part e) of this question
lone heartBOT
For part e) of this question
cunning trout
This is the graph
I don’t understand the answer here
Why are they finding the arm span of a student with foot size of 22cm, shouldn’t they be finding it for 31cm ?
kindred dagger
Yes just a mistake in the answer i think
cunning trout
Ya
Also
after we find the equation of line of best fit
Then can i put it the foot size (31cm) in the equation and find the arm span?
Also this is the equation of line of best fit
Or i guess we do that
kindred dagger
"Use the graph" so i think yes
cunning trout
Or i guess we do that
Wait i feel like something is wrong here
cunning trout
"Use the graph" so i think yes
Ya ok cool
kindred dagger
But you can use the equation too
cunning trout
Yeah i mean i find it easier ig so i would probably do that
cunning trout
Or i guess we do that
Also when do we stop the line here. So i made it go horizontally until it hit the red line i think
Yeah
So we stop at the red line ?
I mean when it hits the red line
kindred dagger
But if you want the arm span with the equation, it's better to solve the equation to have a S = smthing before doing calculations
cunning trout
True
I think its more better with the equation unless they explicitly ask but yeah
kindred dagger
In theory you can extand the line as far as you can, but in practice if it's not close to the black dots, we don't know if it's still linear or it does something else
It's why i think it's the question that got the wrong value, since 31 is quite far from other values
cunning trout
Hmm true
But also the more you extend the line the more it doesn’t make sense since its practically weird to have such a foot size then but yeah it makes sense now
Thanks!
cunning trout
In theory you can extand the line as far as you can, but in practice if it's not...
Ya true
.close
mellow crane
I drew a diagram but it's not really helping and I'm still confused as to what to do but I think the distance between c and b is 1800 miles
gray isle
show your diagram
gray isle
have you done any more work than this
gray isle
why are you multiplying 40 and 45
mellow crane
why are you multiplying 40 and 45
I think because speed*time is distance
mellow crane
time is in minutes, speed is in miles/**hour**
Oh, so then I divide by 60 and it's 30 miles?
gray isle
yes, that would be the distance between C and B
mellow crane
so to find the speed of the other truck it's (30+CA)/65?
gray isle
where's that coming from
mellow crane
because 20+45=65 minutes for the whole trip and then 30miles+the distance from AC is distance for the whole trip and then d/t=s so then that's the second truck's speed
gray isle
what's x
mellow crane
the time it takes to get from a to c for truck b
gray isle
no
also that's already given
and like before, you shouldn't be adding the times from different trucks like this
ideally in your diagram, you should separate the info for the two different trucks
gray isle
you mixed up your trucks
mellow crane
oh
gray isle
lone heartBOT
@mellow crane Has your question been resolved?
mellow crane
.close
hallow thicket
