on paper perhaps
#help-27
pseudo basin
spice valley
do i group together the e^x
aight imma give it a try
and come back if its wrong
THANKSS
.close
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faint totem
the question is
proof B is invertible if and only if A is invertible
now the answer uses a ton of stuff but I was wondering if it isnt just okay to say
B is a lower triangular matrix so its determinant will be det(A) * det(A)
so if we say det(A) = 0 then the det(B) wont equal 0 and if det(A) != 0 then det(B) != 0
stone stump
has your course covered those determinant properties?
faint totem
I translated it witch chatgpt but this is the answer the sheet gives
faint totem
has your course covered those determinant properties?
yeah it has
stone stump
also, determinant proofs manage to completely miss any possible intution you could gain, so maybe they dont want that
stone stump
I translated it witch chatgpt but this is the answer the sheet gives
(its not like this is a very involved proof either. its just two steps in both directions essentially)
faint totem
also, determinant proofs manage to completely miss any possible intution you cou...
well this is from a previous test so it'd be too late for that lol
stone stump
and it tells you more about whats going on
faint totem
and it tells you more about whats going on
yeah I get that but I felt like my answer was simpler and was wondering if I was missing something
stone stump
lots of problems like this can be solved in a lot of ways
if it can be solved with the det then thats often the shortest proof. but also tells you little else
faint totem
if it can be solved with the det then thats often the shortest proof. but also t...
I see, proofs are the hardest for me right now
like I know the theory and I know what properties I could use but actually forming the proof is hard
stone stump
related exercise: suppose A,B are invertible .prove that AB is invertible. do it either by det, by giving a formula for the inverse or by considering ABx=0
faint totem
related exercise: suppose A,B are invertible .prove that AB is invertible. do it...
mm, doing this by det is trivial. Ill try one of the other methods
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@faint totem Has your question been resolved?
faint totem
related exercise: suppose A,B are invertible .prove that AB is invertible. do it...
so the square matrices A and B will be invertible only if Null(A) and Null(B) = {0}. So Ax = 0 and Bx = 0 will only be true if x = 0.
So when were doing ABx = 0 It means either AB or x will be 0
now we can do B^−1(ABx) = B^-1 * 0 = (B^-1AB)x = 0 => A(Bx) = 0
so A(Bx) = 0 implies Bx = 0
and Bx = 0 implies that x = 0
so Nul(AB) = 0
something like that?
stone stump
not sure what that line with the B^-1 is about
the three lines after that are enough
the key insight being that you can "iterate" solving the system
which is for example relevant for LU factorizations later (you dont have to know what those are)
I dont wanna say that this or the det proof is "better"
just that sometimes proofs tell you more than just the theorem you proved
I have to go. spoiler for the formula for the inverse: ||(AB)^-1=B^-1A^-1. check that (B^-1A^-1)(AB)=id and (AB)(B^-1A^-1)=id||
faint totem
which is for example relevant for LU factorizations later (you dont have to know...
Weve already seen it
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analog lark
A right section of an oblique prism is a hexagon in which the sides are 3cm, 4cm, 4cm, 5cm, 6cm, and 7cm. The measure of one lateral edge is 7cm. Find the lateral area
can someone show the figure for this i dont quite get it
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A right section of an oblique prism is a hexagon in which the sides are 3cm, 4cm...
@analog lark Has your question been resolved?
midnight echo
!original "the measure of one lateral edge is 7cm the lateral area" doesn't make sense
devout snowBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
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analog smelt
Hello! This is really simple but can I get help on this
devout snowBOT
Hello! This is really simple but can I get help on this
analog smelt
I wrote it down
My answer was choice number 2 but my answer sheet says the correct answer is 1
How is it 1😭
olive snow
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vivid estuary
are s and t prime?
devout snowBOT
are s and t prime?
vivid estuary
Theorem 13.3
acoustic leaf
we're not assuming they're prime, just that they're factors of n
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vivid estuary
oh ok I think I see
vivid estuary
we're not assuming they're prime, just that they're factors of n
so they essentially showed the only number that can divide n is n here?
which shows n is prime
acoustic leaf
yes, they showed the only factorization of n is n = n * 1
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worldly tapir
I think the teacher is wrong for this because Im pretty sure its x-1 and not x+1
worldly tapir
this is the graph btw to make life easier
rain summit
I think the teacher is wrong for this because Im pretty sure its x-1 and not x+1
What's a?
rain summit
where does a appear?
That's what im asking
worldly tapir
That's what im asking
Wym by A
rain summit
Where is question a?
proud mural
I think the teacher is wrong for this because Im pretty sure its x-1 and not x+1
youre correct
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rancid wharf
I think the teacher is wrong for this because Im pretty sure its x-1 and not x+1
correct
oops it didnt update
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flat ember
Ok I’ll try to solve this here
devout snowBOT
Ok I’ll try to solve this here
flat ember
(x2 - 3x)(x2 - 3x) = x2
x4 - 9x2 -3x3 + 6x2 = x2
x4 - 6x2 + 6x2 = x2
x2 = 0
x = 0
atomic thunder
x4 - 9x2 -3x3 + 6x2 = x2
You've made a mistake on this step, what is 3x times 3x?
Actually you've made two, what is 3x times x^2?
flat ember
You've made a mistake on this step, what is 3x times 3x?
Oooh! I added this together
flat ember
Actually you've made two, what is 3x times x^2?
3x^3?
atomic thunder
Oooh! I added this together
I know you edited the original, but it is still wrong, because -3x times -3x is 9x^2, so it should have a plus
flat ember
But you didn’t mention the minuses
real helm
x4 - 9x2 -3x2 + 9x2 = x2
last term should be 9x^2
atomic thunder
But you didn’t mention the minuses
It shouldn't matter when you square a value, but you are right, my bad
flat ember
Also what is the scientific reasoning that - times - is +?
x4 - 9x2 -3x2 + 9x2 = x2
x4 -12x2 + 9x2 = x2
x4 - 3x2 = x2
-8x2 = x2
x = -8
OMG I’m so smart
atomic thunder
Also what is the scientific reasoning that - times - is +?
I could answer you, but I don't think it is really needed right now, probably it's better to solve the initial problem
real helm
x4 - 3x2 = x2
-12x^2 + 9x^2=?
flat ember
-12x^2 + 9x^2=?
-3x2?
flat ember
x4 - 3x2 = x2
atomic thunder
x4 - 9x2 -3x2 + 9x2 = x2
In this part there are still some mistakes, so let's look at the problem from the beginning
flat ember
OMGGGGGG
atomic thunder
So you have $(x^2-3x)^2=x^2$ this is equivalent to $x^4-3x^3-3x^3+9x^2=x^2$, do you agree with it or is there anything that is not clear?
woven radishBOT
KonoEmllikDa
flat ember
I’ll do this again
(x2 - 3x)(x2 - 3x) = x2
x^4 - 6x^2 - 3x^2 +9x^2 = x^2
Ok this should be correct now
x^4 -9x^2 + 9x^2 = x^2
atomic thunder
x^4 - 6x^2 - 3x^2 +9x^2 = x^2
The -6x^2 should be -3x^2 here
and again it is -3x^3
flat ember
x^4 - 3x^2 - 3x^2 +9x^2 = x^2
atomic thunder
still not completely right, the part with -3x^2-3x^2 should be -3x^3-3x^3
flat ember
still not completely right, the part with -3x^2-3x^2 should be -3x^3-3x^3
Huh why?
atomic thunder
because x^2 times -3x is equal to -3x^3
flat ember
But -3x isn’t squared
real helm
so basically
x^4-6x^3+9x^2=x^2
real helm
But -3x isn’t squared
but x^2 is
flat ember
Ugh
atomic thunder
But -3x isn’t squared
that's right, but -3x=-3x^1
real helm
powers with the same base get added
flat ember
that's right, but -3x=-3x^1
Oh
atomic thunder
and you multiply the coefficients and add the powers of x
flat ember
so basically
x^4-6x^3+9x^2=x^2
x^2 - 6x^3 + 9x^2 = x
real helm
x^2?
flat ember
X^4 - X^2 = X^2
real helm
X^4 - X^2 = X^2
that's not how exponents work
flat ember
Ohh it’s with division
real helm
lets just stay on track for now
real helm
so basically
x^4-6x^3+9x^2=x^2
simplify this
real helm
x^4 - 6x^3 + 9x^2 = x^2
x^4 + 3x^2 = x^2
Only the terms with the same power can be operated on
flat ember
Oh I didn’t see the 3
real helm
x^2 - 6x^3 + 9x^2 = 1
Powers cannot be subtracted like that
flat ember
Powers cannot be subtracted like that
I multiplied by both sides..
No I divided and then did the opposite
real helm
I multiplied by both sides..
if you did,you have to multiply it with the complete left hand side,not just one term
real helm
No I divided and then did the opposite
still the same
do it with all the terms on left hand side
I think you are getting confused
flat ember
x^4 - 6x^3 + 9x^2 = x^2
real helm
why dont you try it on a paper?
atomic thunder
But what do you do with the x^2?
You can subtract x^2 in both sides of the equation
real helm
yup
atomic thunder
That way you have something=0 which will be useful for later
atomic thunder
yes
flat ember
Ol then ik what to do next
atomic thunder
Can you factor something in the left hand side of the equation?
flat ember
x^4 + 4x^4 - 6x^3 = 0
atomic thunder
x^4 + 4x^4 - 6x^3 = 0
that's not how it works
real helm
Consider x^2 as x
atomic thunder
$8x^2\neq 4x^4$
real helm
it will simplfy things
woven radishBOT
KonoEmllikDa
real helm
Consider x^2 as x
i mean y
atomic thunder
Consider x^2 as x
I don't think it works here because he has a x^3
flat ember
It’s 2
real helm
I don't think it works here because he has a x^3
hmm
flat ember
When it doesn’t work
It’ll have to be 2x^6
Okay this is Ultimate result
x^4 - 6x^3 + 8x^2 = 0
Impossible to solve
real helm
x^4 - 6x^3 + 8x^2 = 0
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
its another quadratic but much simpler to solve
atomic thunder
It’ll have to be 2x^6
You are confused, to be sure you can substitute x with some different numbers, for example, if x=3, 8x^2=72 and 4x^4=324
flat ember
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
And it’s a single number in the brackets
flat ember
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
How did you turn this
flat ember
so x^2=0 or
x^2-6x+8=0
Into this
real helm
How did you turn this
when you multiply two numbers which give product 0,it implies that one of the two terms is 0
flat ember
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
But you have to multiply x^2 and x^2, you can’t just remove it…
real helm
so we can equate by assuming the two terms to be 0
real helm
But you have to multiply x^2 and x^2, you can’t just remove it…
i did the opposite of multiplication
i took the term which was common in all of them
which is x^2
is that understandable?
flat ember
No…
real helm
are you familiar with distributive property of multiplication?
flat ember
What?
real helm
like
a×(b+c)= a×b+a×c
flat ember
No one is teaching that in school
flat ember
You probably went to private school
real helm
yeah
flat ember
Yup exactly
real helm
anyways
look at it like this
you have 4 apples
you distributed it into 2 people giving two apples each
thus 4=2×2
where 2 is the number of the apples you distributed
basically
i broke the number down into simpler terms
real helm
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
thats what i did here too
flat ember
i broke the number down into simpler terms
Yeah ik how to do that
Ohh I understand this cause I thought x^2 * x^2 will be 2x^2 nvm
flat ember
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
Okay what is next
real helm
Leave x^2=0 as it is
for now
flat ember
How where’s one x^2??
x^2-6x+8=0
real helm
How where’s one x^2??
x^2-6x+8=0
I broke x^4 into two x^2s
one is in the quadratic above
flat ember
Also you cut the whole equation by half
real helm
Leave x^2=0 as it is
and one is here
real helm
Also you cut the whole equation by half
thats what simplifying does
you keep breaking the equation down
flat ember
**x^2**(x^2 -6x+8)=x^2?
What did you do with this?
Was it like that?
real helm
no
real helm
taking x^2 common from the equation will give us
x^2(x^2 -6x+8)=0
we already solved it and got this
real helm
Then what do you do to remove the x^2
We don't "remove" it
we treat them as two terms
then we assume that any one of the two terms is a zero which is why our product(the left hand side) is also zero
that is why we assume a condition where either the first term is zero or the second term is zero
flat ember
How’d you even know that x^2 is zero ?
x^2(x^2 -6x+8) is ZERO.
Who said one product must be always zero?
real helm
Who said one product must be always zero?
anything × 0= 0,right?
real helm
**x^2(x^2 -6x+8)** is ZERO.
The product of this equation is zero
that means either x^2 × 0=0
OR
0×x^2-6x+8=0
there are these two possibilities
flat ember
so x^2=0 or
x^2-6x+8=0
With this
flat ember
Do you have to use delta or whatever that is
real helm
Do you have to use delta or whatever that is
the discriminant method?
flat ember
Idk
flat ember
that means either x^2 × 0=0
**OR**
0×x^2-6x+8=0
there are these two possibiliti...
Well how to solve rjis?
real helm
do you know the factor theorem?
real helm
Idk about that
real helm
What method
There are plenty of method s to solve a quadratic like completing the square method,factor theorem,discriminant method
Well I can teach you the quickest method
flat ember
Yes
real helm
So firstly keep in mind the general form of ANY quadratic equation
which is
ax^2+bx+c=0
This is the general form
flat ember
lol never taught about this
real helm
In which grade are you?
flat ember
Like 11
flat ember
ax^2+bx+c=0
We don’t even have c or bx
real helm
We don’t even have c or bx
No that's just an interpretation
Our original equation
x^2-6x+8=0
Now compare this with the general form
You will get the values of a,b and c for this equation
flat ember
Our original equation
x^2-6x+8=0
But why can’t we just solve it?
real helm
But why can’t we just solve it?
Do you know how to?
flat ember
Like move -6x to the other side
real helm
It isn't making things better
Anyways
So the formula for x through discriminant method is
x=
-b+/-✓b^2-4ac whole divided by 2a
I would recommend you to write this on a paper
It might get confusing while typing
flat ember
Oh this is further beyond my knowledge
I don’t think I’m even on the correct level
Never done this
And it’s so soon for me now we spent an hour and half on this
real helm
Oh this is further beyond my knowledge
Grade 11 and you haven't even seen this?
Strange
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digital axle
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digital axle
help
digital axle
i dont know
why the formula has a different answer
than my own calculations
im looking for a mistake
because from my point of view ive used the formula correctly, but then i saw the answer written on the forum
and it was explained in the other way
i can give u a whole exercise
because its only a part of an bigger problem
The base of the pyramid ABCS is a right-angled triangle with a right angle at vertex C. The radius
of the circle circumscribed around this triangle is equal to 3. The side walls ACS and BCS are perpendicular to
the base. The area of wall ABS is equal to 12√2 and it is inclined to the base at an angle of 60
degrees. Calculate the volume and total surface area of this pyramid.
here u have the problem calculations, the writing is in my national languages however the calculations are in the math language so it should be ok
i think that the author of this answer also made a mistake as the first point is done incorrectly, however the second one isnt
im confused tbh
omg i also made a mistake
🙂
i will try to make it one more time, but how would you solve the second part of this problem?
restive river
i will try to make it one more time, but how would you solve the second part of ...
which part is the second part?
digital axle
total surface area of this pyramid.
restive river
total surface area of this pyramid.
does the original question have a diagram?
of the pyramid
And everything
digital axle
no
digital axle
The base of the pyramid ABCS is a right-angled triangle with a right angle at ve...
this is the question
restive river
total surface area of this pyramid.
are you familar with equations of for volumes of different pyramids?
digital axle
what:0
digital axle
oh okey
i know my mistake tho
i used the wrong formula\
but i will keep this tab open until i dont finish the exercise
restive river
alright good luck
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digital axle
bruhhhh i did it tho
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frail axle
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frail axle
Can my solution be evaluated??
restive river
first two limits are different?
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crimson knoll
is the answer not A
as f''(0) is not defined due to f'(x) being not continuious at x=0?
devout snowBOT
is the answer not A
as f''(0) is not defined due to f'(x) being not continuious ...
tribal ivy
is the answer not A
as f''(0) is not defined due to f'(x) being not continuious ...
f''(x) exists for x != 0
but it doesnt exists for x = 0 because the limit does not exist
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verbal knot
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verbal knot
0 clue
frail axle
first two limits are different?
why different?? could you elaborate more?
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@verbal knot Has your question been resolved?
lilac crescent
,, \frac{1}{2} \left(\frac{1}{x} + \frac{1}{y}\right) \leq \sqrt{x \cdot y}
woven radishBOT
lilac crescent
how would i prove it for $x,y \in \mathbb{R}^*_+$
woven radishBOT
glossy dew
woven radishBOT
mawzi
crimson knoll
soit and montrer means?
lilac crescent
soit = for
montrer = prove
the point of the exercice is to prove the inequality AM-GM
crimson knoll
yeah
hm <= gm <= am
it looks like a misprint
do you know about the jensen's inequality?
lilac crescent
for all x,y belong R
we would get xy but how would we deal with the square
,, x^2 + y^2 +2xy \ge 0 \implies xy \ge -\left(\frac{x^2 + y^2}{2}\right)
woven radishBOT
crimson knoll
oh hm wait we should prol start with the (x-y)^2
so that we can convert it into (x+y)^2 to get the inequality
lilac crescent
.close
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pure kelp
Question:
$\lim_{x\to 0} {\frac{e-(1+2x)^{\frac{1}{2x}}}{x}}$
woven radishBOT
T&C
devout snowBOT
Question:
$\lim_{x\to 0} {\frac{e-(1+2x)^{\frac{1}{2x}}}{x}}$
north roost
i wouldnt say it is the most ideal approach..
pure kelp
i wouldnt say it is the most ideal approach..
should i substitute 2x->t then use binomial expansion?
pure kelp
okay\
pseudo basin
prefer writing (1+2x)^(1/(2x)) as exp( log(1+2x)/(2x) ) and then taylor expanding that
coral anvil
hello everyone excuse me please anyone can you give me tips to calculate fast if it is possible ?
pure kelp
prefer writing (1+2x)^(1/(2x)) as exp( log(1+2x)/(2x) ) and then taylor expandin...
you mean ${(1+2x)^{\frac{1}{2x}}={\frac{e^{ln(1+2x)}{2x}}$
harsh sierra
hello everyone excuse me please anyone can you give me tips to calculate fast if...
!occupied
devout snowBOT
hello everyone excuse me please anyone can you give me tips to calculate fast if...
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
pure kelp
?
pure kelp
The 1/(2x) should be out of the ln
got it
woven radishBOT
T&C
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
atomic thunder
But in this limit, probably it is better to separate when x goes to 0+ and 0- and make a change of variable t=1/(2x), at least that's what I would do
pure kelp
But in this limit, probably it is better to separate when x goes to 0+ and 0- an...
okay.
.close
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harsh sierra
Its occupied
coral anvil
what is it?
harsh sierra
Open a help channel if you need help
coral anvil
were?
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pallid mist
part ii?
pallid mist
<@&286206848099549185>
midnight echo
pallid mist
honestly i dont even know where to start
pallid mist
i was thinking inclusion exclusion principle but idk if its even applicable
midnight echo
yeah, start by considering the naiive solution 10c2 * 12c2 * 8c2
That doesn't work cause double-counting, it's an overestimate
Instead you have to iterate. you have:
S = 6
SM = 4
M = 5
ML = 3
L = 5
You would iterate by finding the number of combinations with both short runers being from S, one from S one from SM, and two from SM. And then the same thing with the long runners.
chilly salmon
my guy is doing bos at 3am on the day of the exam
midnight echo
Like for the number of combinations where the short runners are only from S (not SM) and long runners are only from L (not ML), you would do:
6c2 * 4c0 * 5c2 * 3c0 * 12c2
Yeah kinda, like 9 of those^
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lyric tide
Can someone help me solve these vectors
devout snowBOT
Can someone help me solve these vectors
pseudo basin
image incoming?
lyric tide
Is it?
lyric tide
Sorry for the quality 💔
lyric tide
So it's x-2/3 = y-3/4 = z-5/2 ?
proud perch
Yes
Barring incorrect paranthesization
But I want you to notice something
Lets zoom in on the first expression
(x-2)/3
Where do 2 and 3 come from?
devout snowBOT
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merry ivy
devout snowBOT
polar chasm
the second term is twice the sum of all the terms in the series following that term
Can u put this to an equation?
merry ivy
> the second term is twice the sum of all the terms in the series following that...
Maybe?
polar chasm
try it then and send the equation u get
merry ivy
The problem is that i do not really understand it
polar chasm
what's the second term?
polar chasm
> the second term is twice the sum of all the terms in the series following that...
do u not understand the concept of infinite sum or do you not understand what this sentence says?
merry ivy
do u not understand the concept of infinite sum or do you not understand what th...
The later?
merry ivy
2(1+r)
polar chasm
that's twice the second partial sum
the second term is just r
first term is 1, second term is r, third term is r^2, ...
is this clear?
merry ivy
is this clear?
Yes
polar chasm
> the second term is twice the sum of all the terms in the series following that...
Okay so "the second term is twice the sum of all the terms in the series following that term" means
r = 2 * (the sum of all the terms in the series following r)
and the sum of all the terms in the series following r is simply r^2 + r^3 + r^4 + r^5 + ...
so we get r = 2 * (r^2 + r^3 + r^4 + r^5 + ...)
and this property, which holds for the second term, holds for each consecutive term as well. So we can get similar equation for r^2, r^3, etc
merry ivy
I see...
merry ivy
r ^2= 2 * (r^3 + r^4 + r^5 + r^5 + ...)
polar chasm
nice, now can u combine the 2 equations in some way to get rid of the infinite sum?
so that u can get an equation purely in r with no infinite sums
if you want a hint:
||Try expressing r^3 + r^4 + ... using the second equation and then substituting it in the first equation||
alternatively, you can use the formula for geometric series (if u know it) to get that equation
merry ivy
if you want a hint:
||Try expressing r^3 + r^4 + ... using the second equation a...
Got it
So
r ^2/2= (r^3 + r^4 + r^5 + r^5 + ...)
r = 2 * (r^2 +r^2/2)
polar chasm
now try solving it
merry ivy
r=1/3
merry ivy
now try solving it
Right?
polar chasm
r=1/3
or r=0
but the r = 0 is probably not what they wanted, although they shouldve mentioned it
anyway, to be absolutely sure that you got it right, u should now check that r = 1/3 indeed satisfies all the equations
r^2 = 2*(r^3 + r^4 + r^5 + ...)
r^3 = 2*(r^4 + r^5 + r^6 + ...)
...
merry ivy
but the r = 0 is probably not what they wanted, although they shouldve mentioned...
Thank you so much
polar chasm
np
btw also remember what the og question was, they asked for the value of 2nd and 3rd terms, so u cant just answer with r = 1/3
but i believe u can already find out what the 2nd and 3rd terms are by yourself
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peak patio
im doing SAT math but purely relying on desmos. i only want to solve this using the regression feature desmos provides
devout snowBOT
im doing SAT math but purely relying on desmos. i only want to solve this using ...
peak patio
i thought of something like this so far
but its definitely wrong
cuz it shows up as an error
how do i do this
okay so i know they have to be parallel
in order for there not to be a solution
i graphed both
i thought of something like this
like using the regression to figure out what k value would make the slope equal to the other equation's slope
that would make them parallel
and have no solutions
but i just dont know how to set it up
peak patio
this isnt really a regression problem
just help me do it this way please
lethal wraith
why do u need to solve this using regression
desmos allows you to use regression to solve stuff
peak patio
but its still like not right
bruh
the answer is supposed to be -.9333
for k
oh wait maybe
i can split it more
hold on
ahhh there we go
i can remove the n too
silk zenith
isnt this more effort than if you simply used algebra?
peak patio
isnt this more effort than if you simply used algebra?
if you get good at regressions its much easier
less rules to remember
silk zenith
if you get good at regressions its much easier
but its a 2 line answer
peak patio
but its a 2 line answer
wdym
-.9333 is the correct answer
it just asked me to find k
silk zenith
either or
ik theyre the same thing, i was taught to never write in decimals unless specifically asked
polar chasm
when u get to this, u can literally just divide it by 2x and u're done
peak patio
ik theyre the same thing, i was taught to never write in decimals unless specifi...
i see
peak patio
when u get to this, u can literally just divide it by 2x and u're done
wdym
it gave me the answer there
ohhh
you mean algebraically
yes
polar chasm
2kx = -28/15x
2k = -28/15
k = -28/30
peak patio
yeah but the regression does it for you
im just trying to figure out how to set it up
silk zenith
you mean algebraically
ye isnt it longer to put this in a calc?
peak patio
because it can solve pretty much most SAT problems
peak patio
ye isnt it longer to put this in a calc?
i can type faster
its a digital test
that allows you to use DESMOS
so im trying to utilize it
polar chasm
lmfao
silk zenith
its really unconventional
ya dont say
silk zenith
2kx = -28/15x
2k = -28/15
k = -28/30
interesting, i dont understand what regression is(havent used digital calcs)
peak patio
interesting, i dont understand what regression is(havent used digital calcs)
i like itt
okay thanks guys
.close
devout snowBOT
polar chasm
how about just drawing this and then tweaking k until there are no sols
peak patio
how about just drawing this and then tweaking k until there are no sols
yeah that too
polar chasm
thats a lot faster than regression
silk zenith
thats a lot faster than regression
isnt that still slower than algebra....
peak patio
you would have to set the count to .0001
to even get it
and then to identify it
with your eyes
polar chasm
isnt that still slower than algebra....
possibly, but it's at least a somewhat systematic way and for more difficult problems it's gonna be generalizable quite nicely
peak patio
i dont think my eyes would be able to differentiate between -.9333 and -.9332
silk zenith
possibly, but it's at least a somewhat systematic way and for more difficult pro...
hmm maybe
silk zenith
i dont think my eyes would be able to differentiate between -.9333 and -.9332
yall really overcomplicating a division by zero smh
peak patio
yall really overcomplicating a division by zero smh
lemme like NOT DO ALGEBRA
lmaoo
.reopen
devout snowBOT
✅ Original question: #help-27 message
peak patio
i cant set this up for the life of me
this is all i can think of
oh nvm
i literally just decided to use a slider to find when the bottom of the graph hits 0
.close
visual stone
A biochemist is conducting a study on a new treatment for a species of insect harmful to soybean crops. Before spraying the treatment, the number of insects ( N(t) ) (in hundreds of individuals) evolves according to the rule
N(t) = -21sqrt(t + 100) + 300
where ( t ) is the time elapsed (in days) since the start of the study. As soon as the number of insects reaches a critical level, the treatment is sprayed.
The graph shows the change in the number of insects from the moment the treatment was sprayed.
Determine the duration of this study, knowing that it will end when the insect population returns to the level it was at the beginning of the study.
devout snowBOT
A biochemist is conducting a study on a new treatment for a species of insect ha...
woven radishBOT
opal
visual stone
i dont understand why theres 2 equations for 1 function
also whats the insect population at thebeginning?
300 or 258
i found 100 but not sure if its correct
inland beacon
i dont understand why theres 2 equations for 1 function
There are actually two functions, one referring to the evolution before spraying the treatment, and the other referring to after spraying the treatment
N(0) seems to be 300 - 21 * 10 = 300 - 210 = 90, so there are 9000 individuals at the beginning
visual stone
There are actually two functions, one referring to the evolution before spraying...
oh ok
well i found 100 i can show u how
inland beacon
Go ahead
visual stone
inland beacon
21 * 10 is 210, not 200
inland beacon
Yep
But since it's in hundreds, the amount of individuals at the beginning would be 9000
visual stone
how do we know its in hundreds
magic wind
Are you sure, that N(t) is correct? The number would be decreasing with time without treatment.
inland beacon
Yeah I'm finding that odd too
visual stone
"the number of insects ( N(t) ) (in hundreds of individuals)"
oh ok
visual stone
Are you sure, that N(t) is correct? The number would be decreasing with time wit...
yes
inland beacon
The critical level is implied to be 25800 individuals by the graph, but since N(t) is decreasing that would be impossible
So I either interpreted the question wrong or there is an error
inland beacon
The (t - 100) is under the square root, correct?
visual stone
yeah
inland beacon
Ok
magic wind
Are you sure it is not N(t) = -21*sqrt(-t + 100) + 300, That would give us N(0) = 90 and N(96) = 258.
inland beacon
Are you sure it is not N(t) = -21*sqrt(-t + 100) + 300, That would give us N(0)...
Yeah this would make more sense
visual stone
Are you sure it is not N(t) = -21*sqrt(-t + 100) + 300, That would give us N(0)...
theres a - before the paranthese
i can show u
visual stone
my fault
inland beacon
So just equate the expression to 258
inland beacon
Are you sure it is not N(t) = -21*sqrt(-t + 100) + 300, That would give us N(0)...
I think the result is 96 days to reach the critical point
And then use the graph to check after how many days for it to return to 90
magic wind
That is not the total time of the study
inland beacon
That is not the total time of the study
Yeah ik
inland beacon
And then use the graph to check after how many days for it to return to 90
Alternatively you can equate the given expression to 90, but it's faster to check the graph
inland beacon
Because 258 is the critical value for N(x) according to the graph
inland beacon
its the summit
Not exactly, because sqrt(x) is boundless
It's just a cap for N(x)
It's 258 because it gives a fitting and integer solution for the problem
inland beacon
Sum it afterwards
Just sum the solutions for N(x) = 258 and P(x) = 90
P(x) is the function in the graph
visual stone
Just sum the solutions for N(x) = 258 and P(x) = 90
P(x) is the function in the ...
wait u found P?
inland beacon
P is given
visual stone
where
woven radishBOT
Mathlympian / Gab
visual stone
yes
inland beacon
in the graph
devout snowBOT
@visual stone Has your question been resolved?
visual stone
.close
waxen steeple
devout snowBOT
waxen steeple
[
\text{Let } S \text{ be a semiring of sets, and let}
]
[
L = \bigsqcup_{j=1}^{r} B_j,
\qquad
M = \bigsqcup_{n=1}^{m} C_n,
]
[
\text{where each } B_j, C_n \in S.
]
[
\text{Then}
]
[
L \setminus M
= \left( \bigsqcup_{j=1}^{r} B_j \right)
\setminus
\left( \bigsqcup_{n=1}^{m} C_n \right)
= \bigcap_{n=1}^{m}
\left(
\bigsqcup_{j=1}^{r} (B_j \setminus C_n)
\right).
]
[
\text{Since } S \text{ is a semiring, for each } j,n \text{ there exist finitely many disjoint sets }
D_{j,n,1}, \dots, D_{j,n,K_{j,n}} \in S
\text{ such that}
]
[
B_j \setminus C_n
= \bigsqcup_{k=1}^{K_{j,n}} D_{j,n,k}.
]
[
\text{Hence}
]
[
L \setminus M
= \bigcap_{n=1}^{m}
\left(
\bigcup_{j=1}^{r}
\bigsqcup_{k=1}^{K_{j,n}} D_{j,n,k}
\right).
]
woven radishBOT
ashyboi
waxen steeple
always get stuck here
idk how to show that this becomes a union of disjoint sets of S :/
devout snowBOT
@waxen steeple Has your question been resolved?
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@waxen steeple Has your question been resolved?
waxen steeple
it's a disjoint finite union of sets of S?
but then what about the intersection
but
wait if you're taking the intersection of disjoint finite unions of sets of S, then isnt this intersecting over elements ont in S?
(since union isnt closed in S)
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lost laurel
I was hoping someone could explain , why $V^* \not \simeq V$ when $V$ is infinite dimensional. And infact $\abs{V^*}> \abs{V}$
lost laurel
Intuitively I get it's due to infinite divergent series being possible
woven radishBOT
wai
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normal bolt
Here how to check the phase difference on graphs?
devout snowBOT
Here how to check the phase difference on graphs?
normal bolt
Wrt to x(t) plot
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sand girder
Two positive integers a and b are called a similar pair of numbers if they have the same set of prime factors. For example: 12 and 18 are considered similar because they have the same set of prime factors, that is {2, 3}. Given two positive integers l, r. (0 < l < r <= 10^6). Count the number of similar pairs {a, b} that satisfy l <= a < b <= r. (Time limit: 1s, Memory limit: 256 MB)
devout snowBOT
Two positive integers a and b are called a similar pair of numbers if they have ...
vagrant skiff
Two positive integers a and b are called a similar pair of numbers if they have ...
Have you tried anything? Also, python? C++?
sand girder
I have tried to code it, but it turned out to be over the memory limit
vagrant skiff
Can you send the code here?
sand girder
Have you tried anything? Also, python? C++?
C++
sand girder
my codeblocks isn't opening fsr
Also the tester said RTE before anything, so I don't know if my code is truly correct
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
ll l, r;
cin >> l >> r;
int n = r - l + 1;
vector<ll> a(n);
for (int i = 0; i < n; i++) a[i] = l + i;
vector<ll> mask(n, 0);
int lim = sqrt(r) + 1;
vector<int> p;
vector<bool> isPrime(lim + 1, true);
isPrime[0] = isPrime[1] = false;
for (int i = 2; i * i <= lim; i++)
if (isPrime[i])
for (int j = i * i; j <= lim; j += i)
isPrime[j] = false;
for (int i = 2; i <= lim; i++)
if (isPrime[i]) p.push_back(i);
for (int i = 0; i < (int)p.size(); i++) {
int P = p[i];
ll start = max(1LL * P * P, ((l + P - 1) / P) * 1LL * P);
for (ll j = start; j <= r; j += P) {
int id = j - l;
mask[id] |= (1LL << i);
while (a[id] % P == 0) a[id] /= P;
}
}
for (int i = 0; i < n; i++)
if (a[i] > 1)
mask[i] |= (1LL << p.size());
unordered_map<ll, int> dem;
dem.reserve(n * 2);
ll kq = 0;
for (auto m : mask) dem[m]++;
for (auto &it : dem) {
ll k = it.second;
kq += k * (k - 1) / 2;
}
cout << kq;
}
vagrant skiff
You use a 64-bit mask but therer are 168 primes <= 1000, so 1LL << i overflows when i >= 63.. That's UB and merges unrelated factor-sets. You also collapse every leftover prime > sqrt(r) into the same single bit, so numbers like 2 x 10007 and 2 x 10009 are indistinguishable in your approach.
In worse cases (e.g. where you have many primes in the range), you create up to n distinct keys. Node-based unoredered_map can and will blow up memory.
sand girder
Are there any ways to fix/improve those errors?
alpine geode
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() ...
by the way, you can format code in discord
#include <iostream>
...
I think so. I believe the best way to go about improving it is to compute the radical for each number in [l,r] with a segmented sieve, store it in an array, sort the radicals, then count equal runs.
alpine geode
#include <iostream>
...
Like this
sand girder
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
ll l, r;
cin >> l >> r;
int n = r - l + 1;
vector<ll> a(n);
for (int i = 0; i < n; i++) a[i] = l + i;
vector<ll> mask(n, 0);
int lim = sqrt(r) + 1;
vector<int> p;
vector<bool> isPrime(lim + 1, true);
isPrime[0] = isPrime[1] = false;
for (int i = 2; i * i <= lim; i++)
if (isPrime[i])
for (int j = i * i; j <= lim; j += i)
isPrime[j] = false;
for (int i = 2; i <= lim; i++)
if (isPrime[i]) p.push_back(i);
for (int i = 0; i < (int)p.size(); i++) {
int P = p[i];
ll start = max(1LL * P * P, ((l + P - 1) / P) * 1LL * P);
for (ll j = start; j <= r; j += P) {
int id = j - l;
mask[id] |= (1LL << i);
while (a[id] % P == 0) a[id] /= P;
}
}
for (int i = 0; i < n; i++)
if (a[i] > 1)
mask[i] |= (1LL << p.size());
unordered_map<ll, int> dem;
dem.reserve(n * 2);
ll kq = 0;
for (auto m : mask) dem[m]++;
for (auto &it : dem) {
ll k = it.second;
kq += k * (k - 1) / 2;
}
cout << kq;
}
like this?
vagrant skiff
```c++
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int ...
Is this just the same code as above?
sand girder
yeah, just formatted
vagrant skiff
I think so. I believe the best way to go about improving it is to compute the ra...
Okay. Have a think about this.
devout snowBOT
@sand girder Has your question been resolved?
sand girder
does this work?
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
freopen("SIMILAR.INP", "r", stdin);
freopen("SIMILAR.OUT", "w", stdout);
int l, r;
cin >> l >> r;
int n = r - l + 1;
vector<ll> rad(n, 1);
int sqr = sqrt(r) + 1;
vector<int> isPrime(sqr+1, 1);
vector<int> primes;
isPrime[0] = isPrime[1] = 0;
for (int i = 2; i <= sqr; i++){
if (isPrime[i]) {
primes.push_back(i);
for (int j = i * 2; j <= sqr; j += i) isPrime[j] = 0;
}
}
vector<ll> num(n);
for (int i = 0; i < n; i++) num[i] = l + i;
for (int p : primes){
ll start = ((l + p - 1) / p) * p;
for (ll j = start; j <= r; j += p){
int idx = j - l;
rad[idx] *= p;
while (num[idx] % p == 0) num[idx] /= p;
}
}
for (int i = 0; i < n; i++) {
if (num[i] > 1) rad[i] *= num[i];
}
sort(rad.begin(), rad.end());
ll kq = 0;
for (int i = 0; i < n;){
int j = i;
while (j < n && rad[j] == rad[i]) j++;
ll k = j - i;
kq += k * (k - 1) / 2;
i = j;
}
cout << kq;
return 0;
}
<@&286206848099549185>
dim mural
brisk topaz
Hmmm, good question...
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coral anvil
hi
soft umbra
hi
coral anvil
i want sumone help me improve my math calculation speed if it is possible
supple knot
coral anvil
?
supple knot
ask in ^
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swift knoll
I have perhaps a basic doubt. Consider a sequence of random variables $(X_i){i\in I}$ with $I=\mathbb{N}$, each entry with values in the measurable space $(E_i,\mathcal{E}i)$. Then $$\sigma((X_i){i\in I})=\sigma({X_i^{-1}(B_i):B_i\in\mathcal{E}i,i\in I}).$$Is it true that $$\mathcal{C}=\bigcup{i=1}^\infty \sigma(X_1,X_2,\ldots,X_i)$$is a generating set for $\sigma((X_i){i\in I})$? \
For a finite index set, e.g. ${1,2}$, I think we have that $\sigma(X_1,X_2)$ is generated by $\sigma(X_1)\cup\sigma(X_2)$. I don't see that $\mathcal{C}$ is a generalization of this finite case.
devout snowBOT
I have perhaps a basic doubt. Consider a sequence of random variables $(X_i)_{i\...
woven radishBOT
psie
swift knoll
Taking {1,2} is a perhaps dumb, but I think even if we have {1,2,3,4}, then o(X1,X2,X3,X4) is generated by o(X1) U o(X2) U o(X3) U o(X4).
.close
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lusty sapphire
I think you ended up making your original integral harder
Why not just u-sub $u=x^7$ at the very start?
severe nest
try setting u = x^7 I think that might help
woven radishBOT
SWR
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normal bolt
How to solve question 8
devout snowBOT
How to solve question 8
normal bolt
Also which diagram suits the above question
normal bolt
No I don't know the BF lenght but can be found using cosine law angle B in 𝜟ABF
correct?
untold ravine
nono, im jus tasking for BF right now
like in terms of the angle ABC and c
ok, tell me what cos ABC is
in triangle ABF
woven radishBOT
untold ravine
or you could just use that ABF is a right angled triangle to get
cos B = BF/c
so BF = c cos B
right?
normal bolt
Correct
normal bolt
CosB= a²+c²-b²/2ac
normal bolt
it should -b²
untold ravine
do you understand now, sorry
normal bolt
Yes cosB = BF/c
normal bolt
BM - BF
normal bolt
c²-b²/2a
normal bolt
but options
untold ravine
more like |(c^2-b^2)/2a|
because it can be BM - BF or BF - BM
depending on which one is larger
untold ravine
yes
normal bolt
Yo thanks
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solemn atlas
devout snowBOT
solemn atlas
I dont know how to best set this up
tame palm
I dont know how to best set this up
Set what up?
tame palm
You have a parabolic function. That means it either has an absolute maximum or an absolut minimum. Which do you think it has?
solemn atlas
Maximum?
tame palm
Yes. Do you know how to find the x-value for that absolute maximum?
solemn atlas
Ehh... no
tame palm
Are you learning about derivatives?
solemn atlas
No its application of Quadratic equations and models
tame palm
A useful formula for quadratic equations to determine the vertex is x = -b/2a.
If you have a quadratic equation in the form of ax^2 + bx + c = 0, you can use x = -b/2a to find the x-component of the vertex of the parabola.
tame palm
Hmm ... on further thinking, it may have been better to just solve for A(x) = 20000 and A(x) = 50000 using the Area formula you created.
Do you know how to Complete the Square?
solemn atlas
Hmm ... on further thinking, it may have been better to just solve for A(x) = 20...
Yes?
tame palm
Give that a try.
tame palm
By completing the square, that will allow you to solve for x.
solemn atlas
By completing the square, that will allow you to solve for **x**.
Ok so I have -2(x-380)² +50,000
I cant figure this out
...
@tame palm
I am not getting numbers close to the options
Ha...
Please hel0
<@&286206848099549185>
devout snowBOT
@solemn atlas Has your question been resolved?
solemn atlas
...
Im messing up
Seriously Im stuck
I have -2(x-190)²
<@&286206848099549185>
...
Hey uh... I understand if you are busy but a heads up would be nice
half comet
on d?
half comet
<@&286206848099549185>
replying so you get notification
solemn atlas
Yes on d
tame palm
I am not getting numbers close to the options
Can you show what you did to complete the square for A(x) = 20000?
solemn atlas
Can you show what you did to complete the square for A(x) = 20000?
I think?
Seriously I need some assistance
...
Do I need to show the full problem?
Hey everything alright?
solemn atlas
Can you show what you did to complete the square for A(x) = 20000?
You good?
Everything alright?
I really dont mean to be impatient
@half comet are you avaliable?
half comet
I'm afraid im not available, but from how i see it you should just be able to set 20000 and 50000 equal to 760x-2x^2 and give x in the range
from there solve by abc or square, up to you
currently playing mc soooooo yeah gl
alt go to other forums (redit etc) if none helps you
solemn atlas
Ah.
tame palm
Sorry. There was lightning and now I'm making dinner.
solemn atlas
No problemI understand
tame palm
Do you understand what you need to do?
solemn atlas
Ohhhh
half comet
do you get how you solve for x from there on?
tame palm
Yes. By completing the square, you will be able to isolate the x variable.
tame palm
tame palm
Yes, now complete the square.
tame palm
That's not how you complete the square.
tame palm
When completing the square, the first thing you want to do is make sure the coefficient of x^2 is equal to 1.
So that means dividing the entire equation by -2.
solemn atlas
Ah
tame palm
(x-190)^2 + 10000 - 190^2 = 0
solemn atlas
I
tame palm
Slight correction.
tame palm
b = -380 so the innards should be -190, but because it is squared we can change it to a positive 190.
crystal dawn
here to remind you to take your time doing each step 
tame palm
So the completed square becomes (x + b)^2 where b = -190.