help-33

[calm robin]

I have an egg yolk weight 31.5g

I've taken 3g and mixed it with 27ml 5% w/v buffer solution that was diluted to final a 1:50 dilution

I've then taken 50ul of this solution to figure out the protein concentration

I've found out it's 435.8ug/ml

I need to figure out how much protein is in the egg yolk

Do I need the volume of the egg yolk to figure this out or is there another way of doing it?

[marsh citrus]

@calm robin Has your question been resolved?

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[restive silo]

Hi

[restive silo]

I don't understand the ans

[rugged spindle]

there is only one choice for the ten-thousand digit (the choice is 0) because otherwise the integer is greater or equal to ten thousand. then, the thousands, hundreds, and tens place can be any of the five choices of number, but the ones place has to be 3, or else it will be divisible by 2.

the ones digit determines if it is odd, because any number with 5 digits can be described as $a_1+a_2\cdot 10 + a_3\cdot 100 + a_4\cdot 1000 +a_5\cdot 10000$, where $a_1$, is the number in the ones place, $a_2$ is the number in the tens place, etc. divide by 2, and you get $\frac{a_1}{2}+5a_2+50a_3+500a_4+5000a_5$

[elfin berry]

fish

[restive silo]

Why did u include the ten thousands tho question says it's less than 10k?

[rugged spindle]

just ignore it, doesn't matter. point is that the one's place shows that it is only odd when $a_1$ is odd

[elfin berry]

fish

[restive silo]

So

We need to check ones tens hundreds thousands possibilities

[rugged spindle]

yeah, pretty much

[restive silo]

Ones has to be 3 only

[rugged spindle]

yes

and ten thousand has to be 0

[restive silo]

Cuz it's out of our scope

[rugged spindle]

yes

[restive silo]

So 5x5x5 refers to tens hundreds thousands?

What abt that 1 at the end?

[rugged spindle]

yes, because you can choose any of the possible numbers

1, because the only choice for the ones place is 3

otherwise, it would be even

[restive silo]

But isn't the first 1 is about that?

I meant last 1 we have

[rugged spindle]

the first 1 is because you can only choose 0 for the ten thousand place

[restive silo]

Oh

[restive silo]

But isn't 0 is just nothing?

[restive silo]

@rugged spindle are you still there?

[rugged spindle]

yeah

well, you can also just think about it as $5\cdot 5\cdot 5\cdot 1$

[elfin berry]

fish

[restive silo]

Alr

That makes sense

Thanks alot

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[lofty shoal]

If the question states Two-Digits Number,
Can the number be 01?

[lofty shoal]

or does it have to start at 10

[spark otter]

usually 0 doesn't count as a first digit

otherwise they probably would have said "a number (integer) between 0 and 99"

[lofty shoal]

ight thx

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[celest wyvern]

i think i know the basic of circle and tangents and such

[celest wyvern]

i need hints to prove those two triangles are similar

i assumed the d(theta) line to be both straight and tangential

but nothing really works out

[static quarry]

rotate the small triangle 90 degrees

[celest wyvern]

i forgot to mention

the theta is just given for the larger triangle

i need to show that the smaller triangle too has an angle theta

[static quarry]

yea by rotating you can see that the rightmost angle of the small triangle equals the top angle of the big triangle

this is because the hypotenuses of the two triangles are at 90 degree angle relative to each other

and the same is true of the horizontal side of the small triangle and the vertical side of the large triangle

[celest wyvern]

i think i get the intuitive sense

how do i go about proving this

[static quarry]

proving which assertion?

the d(theta) line is at 90 degrees to the hypotenuse of the large triangle because as you said, it's tangential to the circle

and the horizontal side of the small triangle is at 90 degrees to the vertical side of the large triangle because they're horizontal and vertical

[celest wyvern]

i see that the two line would move by the same angle thus the angle created between them would in effect remain constant; how do i state this then. im looking for a "formal" statement

[static quarry]

if it helps, you can draw a third triangle by extending the vertical side of the small triangle downward until it reaches the hypotenuse of the large triangle

then use whatever theorems you have for showing various angles are equal

[celest wyvern]

ok so the black space between the two triangles would be angle "x" and x forms right angles with both the adjoining angles and thus they're equal

that seems so easy now 😭

thanks!

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[static quarry]

@celest wyvern in case you see this later, here's a picture

the two angles labeled phi are equal by a standard geometry theorem

the angle highlighted in red is 90 degrees because dtheta is tangent

so the angle at the top has to be theta

[marsh citrus]

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[solemn plank]

why for (a)
lamda is +- 2?

I only got positive 2

[solemn plank]

there has to be a mistake in my working but I cant spot it

[novel juniper]

for no solutions why does -x^2+3x-2 have to be 0?

[quaint elm]

last step of rref, bottom right entry

sign error

[solemn plank]

ahhh cuz for no solutions
1-lamda has to be the leading entry
meaning that everything before has to be 0
so we got two zeros, hence the quadratic hsould also be a 0

[solemn plank]

im sorry can you please circle the part i got wrong

the 1-lamda ?

i got everything right but idk where my course notes got -2 from

[quaint elm]

yeah

[solemn plank]

i got that from doing r3 + 2r2

so -1-lamda + 2

1-lamda

[quaint elm]

oh i misread

[solemn plank]

ye i just dont get how lamda can be -2
for no sols

i got lamda = 2 for no sols

i double checked everything, silly mistake seems highly unlikely
I think im missing a concept but i cant spot it

[quaint elm]

what does your reduced matrix look like if lambda = -2?

[solemn plank]

i plugged in lamda = -2
in a matrix calculator
and it said no sols

so -2 is also correct
but idk how to get it

@quaint elm

[quaint elm]

take your reduced matrix and plug -2 in

probbly what you want to do is complete the gaussian elimination / rref

so you can identify more 0 0 0 rows

[solemn plank]

ye then I get unique solution

ion get how this matrix calculator said -2 is no solution

maybe the calculator is wrong

[marsh citrus]

@solemn plank Has your question been resolved?

[cunning fiber]

You messed up then

Looking at the second and third rows, you can get the third row to be $$0 \text{ } (2+\lambda) \text{ } 0 \text{ } \text{(some nonzero thing)}$$

[elfin berry]

Civil Service Pigeon

[cunning fiber]

Setting lambda = -2 clearly yields 0 = some nonzero thing

Hence no solutions

[cunning fiber]

,w x+2y-2z=1, -x-2y-z=0, -2x-4y-2z=-1

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[cunning fiber]

In case you’re not convinced

[marsh citrus]

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[copper mist]

Im using GeoGebra and it has an annoying feature that forces a name for an equation in the field. This means that if I want to copy the equation with ctrl+A and paste it a new entry it with ctrl+V it takes this equation label with. But because it is the same it just ignores it. So now I manually have to edit the equation name to something else. And I have to do this every time I paste something into a new entry. Is there any way to disable this labeling function and just have each entry inherently be a new entry, without the need for labeling?

[copper mist]

I have set "Labelling" setting to "No New Objects", but it still does it.

[marsh citrus]

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[potent reef]

That's not a good question.
I mean how would you define a parallelogram? The properties will automatically apply and it will automatically be a rectangle, you got the use some of the properties of a parallelogram, and in doing so you will find that it must be a rectangle.

So you're only saying to not use some important properties of a parallelogram and do it some other way

It isn't a big task to 'prove that it is a rectangle first'. This shouldn't be an obstacle.

You could say for example that, since in a parallelogram opposite angles are equal, this must be only possible when the diagonal of the parallelogram passes through the centre (if it didn't then one angle had to be obtuse, and the other acute) so that both are equal.

Without 'calculating that each one of it is 90 deg'.

And 'saying' that it is a rectangle, if that was what you wanted.

[humble fog]

The definition of parallelogram is a shape that have 2 pairs of parallel side where the opposite side is in equal length

[potent reef]

No, you're not correct completely. We have to prove that opposite sides are equal in length from the fact that opposite sides are parallel.

[humble fog]

So is it proof the cyclic parallelogram is rectangle by using diagonal?

Actually what is ur question really meant I can’t really fully understand

[potent reef]

Ah

Yeah, your friend rather proved the converse of this theorem

So basically, opposite angles of a cyclic quadrilateral sum up ot 180 deg. Also, since it's given that the quadrilateral is a parallelogram, the opposite angles must be equal. Let each opposite angle be equal to x.
So x + x = 180 deg
x = 90 deg
Now, by the theorem in the image, angle BCD = 90 * 2 deg = 180 deg
So DCB must be a straight line, or DCB = 2 * radius, or DCB is the diameter, or the diagonal is the diameter, or the diagonal passes through the centre.

yeah.

[marsh citrus]

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[hazy dragon]

i need help finding the lim of

$[\frac{n-1}{2}] ^{-1 /2n}$

[elfin berry]

Shadow

[tulip idol]

$\left[\frac{n - 1}{2}\right]^{-\frac{1}{2n}}$ ??

[elfin berry]

Arya

[merry hill]

as n approaches what?

[tulip idol]

Limit of this? As n approaches?

[merry hill]

I assume as n approaches to 1

[marsh citrus]

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[modern sedge]

I get atan(-R) - atan(R) , but the integral calculator has something different

I get the same antiderivative, but atan(-R) is not the same as atan(1/R), is it?

[marsh citrus]

@modern sedge Has your question been resolved?

[marsh citrus]

@modern sedge Has your question been resolved?

[modern sedge]

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[lofty swan]

How do I claim lol

[lofty swan]

Alright

Circle on the left is r=30
circle on the right is r=20
O1O2 = 90cm
I need to find the ratio of o1e/o1c
I put O2E as X, and O1E as 90-x

The thing is, I'm not sure if I'm allowed to calculate the ratios

What I did next is O1A/O2B = O1E/O2E

30/20 = 90-x/x

Am I allowed to do that? And if so, why?

I need to type out the whole process (obviously) but I'm not certain if the process is right lmao

We know that AB is a tangent to both circles, if that helps us somehow

[calm vortex]

typing out the whole process probably means you need to show the triangles are similar

i.e. you need to show that angles in AO1E and BO2E are equal

[lofty swan]

Did you reach this conclusion simply via experience?

My math vision isn't exactly at its peak lmao

[calm vortex]

well, if that ratio is true it means that triangle AO1E and BO2E are similar

[lofty swan]

Ah I get what you mean lol

Yeah, I figured I wanna go towards that

The only issue is if I'm allowed to find o1e and o2e the way I did

[lofty swan]

Via this process

[calm vortex]

it's supposed to be O1A/O2B = O1E/O2E, not D

[lofty swan]

Yeah I corrected it as you were typing lmao

But yeah, am I allowed to do that?

[calm vortex]

then it's a mathimatically valid argument

[lofty swan]

I assume I can, since AB is a tangent to both circles

Mhm

[calm vortex]

if your teacher is picky you might need to do this though

[lofty swan]

Unfortunately, they are very picky lmao

Like, if that is a detail I leave out I likely fail the whole question

[calm vortex]

then show it

you already have one angle equal which is the right angle at the tangents

now you only need 1 more pair to be equal

[lofty swan]

Ah

Lmao so the whole process of finding O1E felt a bit useless in that case

Unless it does help me somehow here, but I don't really see how it does

[calm vortex]

it does help you

if you found O1E

then you can just plug it into O1E/O1C

to get what you need

[lofty swan]

Ah right right

[lofty swan]

As for this, my brain instantly goes towards Pythagorean to try and find AE, and then do AE/sinO1 = O1E/sin90

What would be the easier way tho, since this is obviously the longest way about it

[calm vortex]

you need to show that the angles are equal, not the sides

[lofty swan]

ah oops forgot to write the sines

[calm vortex]

basically
show angles are equal -> use the similar triangles to say that O1A/O2B = O1E/O2E -> solve for x and thus, O1E -> plug the value of O1E into O1E/O1C to get the desired ratio

[lofty swan]

Yeah I got that, my question is how do we show the angles are equal?

I might have just discovered a new way to solve this

Dude I love math so much, the fact that there's a billion ways to solve stuff is awesome

Well I think I got it from here, thank you! @calm vortex

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[upbeat hound]

$25x^4 -4(x^2 +2x -3)^2$

[elfin berry]

Simon James B

[upbeat hound]

i am not sure what to do to factor this out. I tried to say 25x^4 = (5x^2)^2 but from here?

[red nimbus]

have you tried the difference of squares formula?

[upbeat hound]

yea but i got stuck

$(5x^2)^2 -4(x^2 +2x -3)^2$

[elfin berry]

Simon James B

[red nimbus]

you can write the 4 as square as well and then use x²y² = (xy)²

[upbeat hound]

$(5x^2)^2 - 2^2(x^2 +2x -3)^2$

[elfin berry]

Simon James B

[upbeat hound]

that would be [(5x^2)(2^2)]^2 ?

[red nimbus]

no

2²(x²+2x-3)² = ( ? )²

[upbeat hound]

[2(x^2 +2x -3)]^2?

[red nimbus]

exactly

Now you got the form a²-b²

[upbeat hound]

$(5x^2)^2 - [2(x^2 +2x -3)]^2$

[elfin berry]

Simon James B

[upbeat hound]

$(5x^2)^2 - [2x^2 +4x -6]^2$

[elfin berry]

Simon James B

[upbeat hound]

$(5x^2 -2x^2 -4x +6)(5x^2 +2x +4x-6)$

[elfin berry]

Simon James B

[upbeat hound]

is that right

[red nimbus]

yes

[late geode]

missing ^2

[upbeat hound]

$(3x^2 -4x+6)(7x^2 +4x-6)$

[elfin berry]

Simon James B

[upbeat hound]

is that it

[red nimbus]

you could try to see if any of these quadratics can also be written in factors again

[upbeat hound]

in the first one with 3 and the second one with 2?

[red nimbus]

3x²-4x+6 and 7x²+4x-6 respectively

[upbeat hound]

yea but i can factor out in the first with 3 right?

[red nimbus]

well you'd have the fraction 4/3

[upbeat hound]

factor our x^2

[red nimbus]

no

[upbeat hound]

then idk

[red nimbus]

you can try to see if there exist any integer roots and if not you can leave it like that

[upbeat hound]

can't leave it like this because this is not the answer in my book

what interger roots what is this

[red nimbus]

ok then try to find the roots

roots are solutions of x where f(x) = 0

[upbeat hound]

i did not learn functions yet

[red nimbus]

but you know quadratics and they can have x-intercepts

[stoic slate]

and what is your book's answ cause ur answ is correct

[severe owl]

The first quadratic has complex roots !

[upbeat hound]

that has to be impossible we did not learn that in my grade

and the problem is form my grade textbook

[stoic slate]

as i said, u already factored it

[upbeat hound]

from*

[severe owl]

4²- 4×3×6 < 0

[upbeat hound]

(29-2x)(10x+11)

[stoic slate]

your book is wrong

that is obviously not equal to the original

[upbeat hound]

exactly what i said that is why i opened here

i will do other ones ig ty

[stoic slate]

probably refers to other problem

[severe owl]

The highest degree of x here is 2, and in question it's 4

[upbeat hound]

thanks anyways

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[clever abyss]

help with this please

[quaint elm]

just take it step by step

[clever abyss]

i can send what i did if you try to find any errors?

[vernal forge]

do that

[clever abyss]

if anyone can spot my mistake(s) i'd be very thankful

[main idol]

does this say 9-6i ?

if so that's not (3-i)^2

,calc (3-i)^2

[elfin berry]

Result:

8 - 6i

[quiet anvil]

9 - 6i - 1

Is how I read it

[main idol]

oh

,calc (8-6i)*(1+5i)

[elfin berry]

Result:

38 + 34i

[main idol]

i think mistake here. you didn't distribute the - sign

[clever abyss]

so than -3i/5 just becomes +3i/5 that's it?

[main idol]

fix that first and see

[clever abyss]

alr then result is then 36-34i seems normal.

thansk for help.

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[mint seal]

Hi guys I'm kind of stuck on the problem of the following: ABC is a triangle, S is its centroid, and P is on the ray of AS outside of the triangle (so not including segment AS), Q is on the ray of BS (same criteria as P), and R is on the ray of CS (also same criterias as P). How many PQR triangles are there if A is on side QR, B is on side PR and C is on side PQ?

[mint seal]

My guess was that there is only one, the one which is similar to triangle ABC

A rather chaotic graph of the progress so far: I picked a point on line BS and made a "triangle" which doesnt have its ends together (GHIJ)

you can see J's path highlighted there by the trace (HIJ are dependent on G) and as that only cut the ray of BS in 2 places, S and the Q in the guess, S cant be it due to the requirements, so I suppose there's only one possible answer which is the guess

However I can't prove that the trace of J is an ellipse

or rather I cant really decide that this is a correct way of proving the task at all

[marsh citrus]

@mint seal Has your question been resolved?

[mint seal]

<@&286206848099549185> could you guys look at it please

[cedar stone]

do you know phantom point?

[mint seal]

not really

whats that

[cedar stone]

Basically you show that a certain point works, and then you assume another point works, then show that the other point must be the first point

[mint seal]

kk thx ill work on it

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[civic blade]

Let $\mathbb{F}$ be a field and $f(x) \in \mathbb{F}[x]$ an irreducible polynomial. Suppose $\mathbb{E}$ is a field containing $\mathbb{F}$ such that $f(x)$ has a root $a\in \mathbb{E}$. Show that the map $\mathbb{F} \to \mathbb{E}$ defined by $g(x)\mapsto g(a)$ defines an isomorphism \[ \mathbb{F}[x]/<f(x)>\cong \mathbb{F}(a), \] where $\mathbb{F}(a)$ denotes the subfield of $\mathbb{E}$ generated by $\mathbb{F}$ and $a$.

[elfin berry]

Juke | ping me if no response

[civic blade]

what is \mathbb{F}[x]/<f(x)>\cong \mathbb{F}(a) ??

[harsh copper]

Bro

[civic blade]

get another channel

the $\mathbb{F}[x]/<f(x)>$ part doesn't mean anything to me, is it supposed to be
$$\mathbb{F}[x]/\langle f(x) \rangle \cong \mathbb{F}[a]$$

[elfin berry]

Juke | ping me if no response

[civic blade]

or am i trippin

[vernal forge]

it's definitely that

but also F(a), not F[a]

[civic blade]

yes, typo

but alright, thanks!

.close

[marsh citrus]

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[mellow crag]

i mean technically they're the same in this case lol

[vernal forge]

woops

[marsh citrus]

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[dim compass]

If we say m_A is the minimal polynomial of matrix A then what is the minimal polynomial of matrix A^4 + I_n

[dim compass]

??

[marsh citrus]

@dim compass Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[still temple]

,rccw

[still temple]

(a)

can we say
i th row --> c times i th row + b times j th row
is a single row operation ?

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[novel juniper]

which question

[novel juniper]

I don't think so

[still temple]

a

so

only the following will be considered as a single row operation
multiplying a row with a constant
interchanging two rows
ith row --> ith row + c times jth row

and a combination of any of the above 2 cant be an elementary matrix ?

[novel juniper]

That's two row operations

[still temple]

right

[novel juniper]

https://math.stackexchange.com/questions/200533/is-the-product-of-elementary-matrices-an-elementary-matrix

[still temple]

thanks

[novel juniper]

see this

[still temple]

i got few more questions

[novel juniper]

sure

[still temple]

the first and the last

d and g

[novel juniper]

What do you think

[still temple]

i can prove that if A is invertable then x can only have a trivial solution

but if A is not invertible

then

oh

lol

i forgot we can just use the equivalence theorem

what about g

i didnt quite understand what they are saying

[novel juniper]

Suppose A is invertible

Can you invert A using the product of elemenatry matrcies

[still temple]

yes

[novel juniper]

Is the decomposition unique

That's what my understanding of the question is

[still temple]

decomposition means product of the elementary matrices?

[novel juniper]

I think so

[still temple]

there can be many ways to convert A to an identity matrix through row operations right ?

so

[novel juniper]

yes

Let me confirm

yes

you're right

sorry, I just don't trust myself when it comes to matrices even if Iknow teh answer

[still temple]

so it would be false

[novel juniper]

Yes

[still temple]

there can me multiple ways

is it finite or infinite ?

[novel juniper]

I'm not sure

[still temple]

lite

thanks!

.close

[marsh citrus]

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[dense heath]

how to approach questions like these

[quartz lotus]

The top sum is a Newton binomial right?

Yes it totally is

So your 2^n is the n in the formula

[dense heath]

1 sec

[dense heath]

but theres extra factor of 2 in combinatrics part?

[quartz lotus]

Yes

Maybe extend the combinatorics formula, replace with what you have, simplify and convert back to combinatorics

I'm not sure if you can actually simplify

[dense heath]

i will try that

[quartz lotus]

Ok

[dense heath]

i dont think the (2^n+1 - 2i)! is helping in simplifying

[quartz lotus]

If you solve that, you may also simplify the other binomial which has the same form

[dense heath]

😢

[quartz lotus]

Okay

[marsh citrus]

@dense heath Has your question been resolved?

[quartz lotus]

@dense heath

[dense heath]

hii

[quartz lotus]

I have a way of dealing with this

[dense heath]

holy moly

[quartz lotus]

So first forget about 2^n and let's call that just n

Okay?

[dense heath]

ya

[quartz lotus]

So 2^(n+1) is clearly 2n

[dense heath]

yea

[quartz lotus]

Maybe it's not smart to use the same letter

Let's call it t

so 2^n is t and 2^(n+1) is 2t

[dense heath]

its fine i get what he is trying to say

[quartz lotus]

Bro don't mix things

We're doing something completely different

[quartz lotus]

I called it n for this formula

But now imagine the n on the formula is t

[dense heath]

😭

[quartz lotus]

So we have the sum from i to t of the combinatory of 2t over 2i times x^t-i

You see?

[dense heath]

yea

[quartz lotus]

Okay

Now magic will start

[dense heath]

yum

[quartz lotus]

So first imagine the polynomial (1+x)^m + (1-x)^m

If you do the expansion, many terms will cancel outy

[dense heath]

u sure?

[quartz lotus]

For example:

(1+x)6=1+6x+15x2+20x3+15x4+6x5+x6.

(1−x)6=1−6x+15x2−20x3+15x4−6x5+x6.

Now add vertically

you see we have 6x - 6x?

[dense heath]

ya

[quartz lotus]

and 20x^3 - 20x^3?

So what things are getting cancelled?

Every term of x^odd number

[dense heath]

ooo

i dont see any cancelling in this combi ;-;

[quartz lotus]

But we get everything multiplied by two here

[quartz lotus]

Think about this:

Is this what we have?

[dense heath]

yessir

i see it now

[quartz lotus]

Okay

Now we can replace

Do that while I go to lunch

[dense heath]

but there is x^n-k in this ..

and we have 2n-k

no we dont

my bad ..

i will do that

[dense heath]

i did it but its hard to write it

[quartz lotus]

I think the best path to follow is to make a change of variables in the limit

as 2^n = t, n= log_base2(t)

As n goes to inf, t also goes to inf

Then rewrite every 2^n term into t and every 2^(n+1) into 2t

[dense heath]

need to solve the pi notaion

[quartz lotus]

Okay first do this

every 2^n = t

[dense heath]

kk

let do u for 2^n and v for 2^k

[quartz lotus]

Okay

[quartz lotus]

Now I have a little doubt about this statement

If we expand the sum, won't we get all powers of x^n?

But the right hand side only uses even powers of x

I think inside the parenthesis it should be sqrt(x) instead of x

Does it make sense to you?

Okay I checked it, it should be sqrt(x) in the partenthesises

for n=3 for example, 1/2[(1+x)6+(1−x)6] = 1+15x^2+15x^4+x^6, while our sum is equal to 1 + 15^x +15x^2 +x^3

So every exponent is the double of what it should be, we can fix this by writting sqrt(x) instead of x

@dense heath

[dense heath]

oo

let me read allat

[dense heath]

got it

[quartz lotus]

Okay now it makes no sense to have v because we have the product from k=0

So i changed v to 2^k again

Now do you have the range of x in the problem?

[dense heath]

no range

it has to be some manipulation

[quartz lotus]

Okay so unless the product is tends to 0, the bottom part should be infinity

You see?

There is a t multiplying on the bottom

[dense heath]

yea

[quartz lotus]

So if we check that, and it converges to a non zero constant or diverges to infinity, the final result should be 0

[dense heath]

theres infinite in numerator too?

[quartz lotus]

I don't know

Let me think

If x in greater than 0 it should diverge right?

So we got two answers for these two cases

[dense heath]

yea

[quartz lotus]

All depends on what the product converges to

So let's analyze that first

[dense heath]

yeah right

should we use L hospital once

[quartz lotus]

We should check that

Now I don't know the product properties

But maybe there is a way to get rid of the 1/2

Or do something with it

[dense heath]

yes we can

it will become (log t base 2 )/2

[quartz lotus]

Okay I got this

[dense heath]

and we can take it out

[quartz lotus]

I think due to the second property, (1/2) turns into (1/2)^log_base2(t)

You see?

[quartz lotus]

Or is it this?

[dense heath]

1 sec

yeah your right

[quartz lotus]

okay

So let's simplify

we got 2^(-log_base2(t))

which is just (2^log_base2(t))^(-1)

Which is 1/t

You see?

[dense heath]

yep

[dense heath]

we can plug t = 2^n here

so its (1/2)^n

[quartz lotus]

Great

Now to the limit, both of these tend to 0

Let's keep that in mind

[dense heath]

yep

[quartz lotus]

Now if the product tends to any real number, and doesn't diverge, the whole result will be 0

If it diverges, as you said we may use L'Hôpital

[dense heath]

ye

[marsh citrus]

@dense heath Has your question been resolved?

[dense heath]

kinda , i ll try

[marsh citrus]

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[marble acorn]

hello im brainfarting badly rn
how do you solve 0.9 = sin(2x) - (4/20) * sin^2 (x)?

[marble acorn]

for x

is there some crazy trig identity or smth?

[main idol]

could try double angle formula then quadratic formula in sin(x)

[marble acorn]

i cant find a fitting formula :( maybe im not seeing it

[proud ice]

I have a strat for you. It might not be the most efficient, but it will get you to the solution

Write $\sin^2 x$ in terms of $\cos 2x$

[elfin berry]

SWR

[marble acorn]

oh snap does that mean theres no good general solution for these things?

well thanks ill try doing some approximation with that then

[proud ice]

you don't need to approximate

there's an exact solution

[marble acorn]

ahh sorry I misunderstood, now I get it hehe

thanks!

.close

[marsh citrus]

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[cloud marten]

I'm trying to make a card game, and I'm having some trouble calculating probabilities from combinations!

[cloud marten]

The game involves 2 decks of standard cards (values A-K, four suites), and ten cards are dealt to each player. The total number of possible hands are 104c10, or about 23.1 trillion.

I'm trying to calculate the number of hands that have at least a pair. From what I've read, the formula should be something like:

13c1 (choose a rank) * 8c2 (two cards of that rank) * 102c8 (the rest of the cards in the hand)

The problem with that is the resulting product is 3 times larger than the total number of possible hands, so obviously I did something wrong 😅

Could anyone help me figure out where I've gone wrong here?

(Also, I'm not sure if there's an easier way for writing combinations / choose in discord)

In latex, it would look like {104 \choose 10}

[frosty thicket]

you are not guaranteeing that the 8 remaining cards do not have more possible pairs

[cloud marten]

Yes, that's intentional.

[frosty thicket]

this will overcount

[cloud marten]

I'm trying to find the probability of finding all the possible hands that contain a pair or better.

[frosty thicket]

it would be easier find the hands that have a high card at best

poker hands must have either a pair or straight, so it would be easier to do a complement count

better than high card poker hands*

[cloud marten]

Yes, I know it's possible to find that: 104c10 - (13c10*(8c1^10))

But the problem is, I don't know how to extrapolate from that to find, say, 3-of-a-kind or better, 4-of-a-kind or better, etc

[frosty thicket]

if you avoid a pair, you guarantee avoiding trip, quad, and full house

the hard part about this problem is that you also have straights and flushes (i forgot about flushes)

[cloud marten]

Right, but complementary of 3-of-a-kind includes hands that have 5 pairs AND 4 pairs and two random cards, etc

I guess part of what I'm wondering is: why doesn't the intuitive solution actually work?

I would think that, for a given set n, your chances of being dealt a hand that contains that set or better would be:
13c1 * 8cn * (104-n)c(10-n)

But that's obviously not the case.

[frosty thicket]

this overcounts

[cloud marten]

What would be the more appropriate formula?

[frosty thicket]

a hand like AS AD 2S 2D 3S 3D 4S 4D 5S 5D will be counted for 5 times

a more appropriate formula is 104c10-13c10(8c1)^10

[cloud marten]

How could I adjust that formula for 3-of-a-kind?

[frosty thicket]

in other words, this is all hands minus the ones where all cards are different

you dont need to worry about 3 of a kind

if a hand has a 3 of a kind, it has a pair

[cloud marten]

Right, but I'm talking about for another set of probabilities. I've calculated the number of hands that have at least a pair, now how do I calculate the number of hands that have at least 3-of-a-kind?

[frosty thicket]

thats a lot harder, gimme a sec

also are you worried about other hands like straights?

[cloud marten]

Yes, but I figured I would work up to those

[frosty thicket]

if you want me to be completely honest, program it

oh wait trillions are way too mich cases

[cloud marten]

Yeah, I was wondering about that, writing a Rust program that just simulated drawing cards from a deck, with some functions for has_straight(), has_pair(), etc

These were the scenarios I was trying to identify for my card game:

• Royal Flush
• Straight Flush
• 8-of-a-kind
• 7-of-a-kind
• 6-of-a-kind
• 5-of-a-kind
• 4-of-a-kind
• 3-of-a-kind
• pair
• Twins (exact same card, Ex. ace of spades and ace of spades)
• Flush
• Mini-Flush (four cards same suit)
• Super-Flush (6 cards same suit)
• Straight
• Mini-straight (four cards in sequence)
• Super-Straight (6 cards in sequence)

[frosty thicket]

yeah that would be far easier to program, but i dont know how long it will take

[cloud marten]

Well, I guess I'll need to find out

Thanks anyway, I'll go work on that.

.close

[marsh citrus]

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[marsh citrus]

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[tall urchin]

Answer sheet says 24 for x but I keep getting negatives and decimals

[main idol]

show your work

[quaint valley]

Ok so what do you think you should do first

[tall urchin]

my first instinct is to subtract the +4 from the right

[quaint valley]

Well we need to try to get the x on the same side first

So to accomplish that what do you do

[tall urchin]

ye i think thats what im struggling with how do you get rid of 1 x without getting rid of the other

[quaint valley]

So they are both fractions like x/3 and x/6

Since x/3 is by itself you can multiply by 3 to get rid of it

So what do you end up with on the other side

[tall urchin]

x= x18+4?

but that doesnt seem right

[quaint valley]

Um close

Right idea but no

Ok so if you multiply 3(x/6 +4) what do you get

[tall urchin]

18x+12?

[quaint valley]

No

3 times x/6 is x/2

And then the 12 is right

[tall urchin]

see i dont understand that

[quaint valley]

Ok

[tall urchin]

3 times 6 is 18 why does it turn into x/2

[quaint valley]

Because the 6 is in the denominator

You’re dividing 3 by 6

If it was 3 times 6 it would be 18

Does that help?

[tall urchin]

im still confused as to why it divides when 3 is outside the brackets

[quaint valley]

No its 6

6 is the denominator

[tall urchin]

shouldnt u multiply everything in the bracket by 3

[quaint valley]

So basically it’s 3x/6

You do

3x/6 simplifies to x/2

We can go about this problem in a different way if this confuses you to much

[tall urchin]

how do you know when to treat it as a fraction vs treat it as division

[quaint valley]

A fraction is division

[tall urchin]

yeah i think im just getting more confused lol

[quaint valley]

You’re just simplifying it to make the math easier

[quaint valley]

Ok

We will start over then

[tall urchin]

ty

[quaint valley]

X/3 =x/6 +4

So what is the lcm

(Least common multiple)

[tall urchin]

i dont know how one would determine that

[quaint valley]

I’ll show you then

It might be greatest common factor 😭

Don’t quote me

Anyways
3 and 6 are both multiples of 3 so we can do 3,6,9,12 (multiples of 3)

Which is the greatest contained as a denominator in the equation?

[tall urchin]

6?

[quaint valley]

Yesss

[tall urchin]

keul

[quaint valley]

Uh oops

It’s different

Anyways
3x1, 3x1, 3x3,3x4,3x5,3x6,3x7

6x1, 6x2,6x3.6x4

Which give you the same number?

(Ignore what I said before)

[tall urchin]

3x6 and 6x3

[quaint valley]

Yess

Ok so for the one that is x/3 what would we multiply by to get 18

[tall urchin]

3x6

[quaint valley]

Yess

So do 6/6 on that side

Because 6/6 is equal to one

[tall urchin]

so 36x?

[quaint valley]

Um no

You do 6 times x on the numerator and then 3 times 6 on the denominator

[tall urchin]

why not 6x6 on the denominator?

[quaint valley]

Becuase the denominator is 3

Then we multiply by 6

$\frac{6}{6}x \frac{x}{3}=\frac{x}{6}+4$

[tall urchin]

forgive me i barely passed gr.3 im still confused

[elfin berry]

ℰ𝓁𝓁𝒶

[quaint valley]

Do you see how the 6 and 3 are both on the bottom

[tall urchin]

ye

[quaint valley]

That’s what we are multiplying

[tall urchin]

well shouldnt the 6 on top on the left still be x?

[quaint valley]

Yes

The top and the bottom get multiplied respectively

So 6 times x and 6 times 3

[tall urchin]

I'm still not getting it 😕

mb meant to write 3 for the answer

[quaint valley]

No so we only did 6/6 by the x/3 the whole equation on the other side still exists

It’s more like

$\frac{6}{6}x \frac{x}{3}=\frac{x}{6}+4$

[elfin berry]

ℰ𝓁𝓁𝒶

[tall urchin]

so where did the 6 over 6 come from?

and why 3 x's now

[quaint valley]

And then we would have $\frac{6x}{18}=\frac{x}{6}+4$ after we distribute

[elfin berry]

ℰ𝓁𝓁𝒶

[quaint valley]

We are using 6/6 to make a common denominator of 18 like I showed your earlier

6/6 can be used to manipulate the equation to make math easier since it is equal to one so mathematically we aren’t changing the equation at all

[tall urchin]

idk it still feels like you're just pulling numbers out of nowhere to make the equation work

[quaint valley]

😭 well you are you’re using them so you can subtract x/3 and x/6

You need a common denominator to do that

There are many ways to do this problem

[tall urchin]

is there one with minimal fraction usage lol?

[quaint valley]

The first one I went through with you is how I would personally do it

Uh no

The quickest one is the first example

This one is completely involved in fractions

I can send you examples of each of the ways with explanations for each step when I get home but the best I can do rn is walk you through the problem

[tall urchin]

okay so if i do the first way suggested and divide the denominator by 3 as i would in 3(x/6 +4) it gives me x0.5 + 12

but thats still a wack decimal

[quaint valley]

The decimal will go away soon

Yes that’s right

Ok so now we have x=.5x +12

Subtract x-.5x

Ok?

[tall urchin]

wouldnt that give me -.5x?

[quaint valley]

No

1 -.5 is?

[tall urchin]

.5

[quaint valley]

Yesss

So .5x =12

[tall urchin]

oh i c

ur subtracting the x not the .5

[quaint valley]

Yes

Kinda

But it works

[tall urchin]

okay so 12 divided by .5 gives me 24

[quaint valley]

Then for .5x =12 we need to get rid of .5 so what do we multiply the whole equation by

[tall urchin]

epic 😃

[quaint valley]

Yes that works but same

[tall urchin]

ty for the help 👍

[quaint valley]

You’re welcome

!done

[marsh citrus]

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[main idol]

.close

[marsh citrus]

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[tame cave]

Where does the x+1 come from in the sylindrical shell method? Is it because the graph until x = 1 gets rotated around the y axis?

[outer lodge]

Because it’s rotated about x = -1

[tame cave]

oh yeah you’re right thanks

!done

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[marsh citrus]

@tame cave Has your question been resolved?

[marsh citrus]

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[dapper field]

I got 5√3 and 10, but im pretty sure its wrong, why?

[quartz lotus]

So where is a=5?

If its the hypothenuse, you should use sine and cosine functions to obtain the other sides

[dapper field]

ohh ok

[proud ice]

!occupied

[marsh citrus]

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).

[proud ice]

(oops ignore me i thought you were asking a question)

[marsh citrus]

@dapper field Has your question been resolved?

[marsh citrus]

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[spark otter]

we discussed this earlier

those terms are useless at order 3

since they get eaten up by the o(x^3)

so remove them to get the correct approximation

I think the most correct definition of taylor approximation of a function of order n is:

$T_n$ is the taylor approximation of $f$ at $x=0$ of order $n$ if $T_n$ is a polynomial of order \textbf{at most $n$} and $f(x)-T_n(x) = o(x^n)$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

There's also a formula for that:

$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

yes

x^3 is not o(x^3)

so it would not get eaten

recall o is like "negligible compared to"

x^3 is not negligible compared to x^3

[spark otter]

I think this

[spark otter]

and this

can help you sometimes

the formula explicitely shows you that T_n can only have x^n at most in power

so order at most n

[marsh citrus]

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[marsh citrus]

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[olive fox]

this is a quick quiz from my calculus 2 class

[olive fox]

I evaluated it should equal 0.188706983

the instruction tell me to round up 2 decimal places

so it should be 0.19 but its wrong

did I solve it wrong somewhere

[spark otter]

ok so

$f(x,0) = \frac{\cos(e^x)}{-5(9x+7)}$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

so differentiate that at x = 2

$f_x(x,0) = -\frac 15 \frac{-(9x+7)e^x\sin(e^x)-9\cos(e^x)}{(9x+7)^2}$

did you get that?

[elfin berry]

rafilou is not not born in 2003

[spark otter]

oops small edit

[olive fox]

shouldn't I partial derivative before substitute y = 0 in ?

[spark otter]

if you recall the definition of partial derivative

sometimes it's easier finding $g(x) = f(x,0)$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

and then computing $f_x(x,0) = g'(x)$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

as is the case here, you get rid of a lot of complicated terms

$f_x(2,0) = -\frac 15 \frac{-25e^2\sin(e^2)-9\cos(e^2)}{25^2}$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

oops wait

$f_x(2,0) = \frac{25e^2\sin(e^2)+9\cos(e^2)}{5^5}$

[elfin berry]

rafilou is not not born in 2003

[spark otter]

,w \frac{25e^2\sin(e^2)+9\cos(e^2)}{5^5}

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by rafilou2003

[spark otter]

yeah it sounds like you messed up somewhere in your computations

can you show your work maybe?

[olive fox]

I used the quotient law step by step

[spark otter]

take a picture/screenshot if you're able

[olive fox]

gimme a sec Ill rewrite it in order

here it is

v * du/dx - u * dv/dx / v^2

wait

this is exactly equal with your result

I think I inputted something wrong in the calculator before

.close

[marsh citrus]

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[marsh citrus]

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[abstract cosmos]

If the vertex of y = x²-ax+a+1 is located at y = x line, then a = ....

[main idol]

You can plug in y=x to the quadratic

Or do complete the square and just get the vertex in terms of a

[abstract cosmos]

0 = x² - (a + 1)x + (a + 1)?

[abstract cosmos]

Wait I don't understand

[main idol]

,tex .cts

[elfin berry]

riemann

[main idol]

y=c(x-h)^2 + k is called vertex form

have you seen that before

[tulip idol]

No. Write given parabola in standard form $y - y_1 = a(x - x_1)^2$

[elfin berry]

Arya

[tulip idol]

(x_1, y_1) is then your vertex

[abstract cosmos]

Yes

[main idol]

?

[abstract cosmos]

Is this formula to change quadratic into vertex form?

[main idol]

That's one purpose yes

[abstract cosmos]

Yes, Ik this is the vertex. But I don't know the correlation to the y = x line

So like, the vertex of a parabola could be at some poin in this g line?

Im confused....

[main idol]

y = x means y1=x1

For the vertex

[tulip idol]

The correlation is, (x_1, y_1) lies on the line y = x. Do you see what that implies?

[tulip idol]

But first, you need to get this in standard form. I don't see why you're complaining about step 2 when we don't see your step 1 yet

[abstract cosmos]

AHHHH YESS

Waittt

Now I'm confused again

[main idol]

You followed this incorrectly

[tulip idol]

$y - (a + 1 - a^2/4 ) = (x - a/2)^2$

[elfin berry]

Arya

[abstract cosmos]

OHHHHH

I include the x.....

[abstract cosmos]

Alright I get the general idea

Thank you kind sir @main idol @tulip idol

.close

[marsh citrus]

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[marsh citrus]

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[hasty cave]

I got different answers than the answer key

is it -13/63

[regal coral]

seems right

[marsh citrus]

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[hot quartz]

a curve c has equation f(x) and P with x coordinate 3 lies on C, given f'(x)=4x^2+kx+3 and the normal to c at p has equation y = -1/24x + 5, show that k=-5 and find f(x)

[hot quartz]

this is the question, ive done the first part where i have to prove k

but i don't know how to go about finding f(x), i know how to find the integral but how do i find dx

stuck here

i guess i need to find y at x = 3 but how

<@&286206848099549185>

[pure jacinth]

You know that the normal to C at P goes through P

[hot quartz]

yeah

but how do i find the value of y at x(3)

[pure jacinth]

That's the point P. It has x value 3 and lies on the normal line

[hot quartz]

i still dont understand how to use that information to find y

[pure jacinth]

Ignore the original curve for a moment

What value does the normal line take when x is 3?

[pure jacinth]

When you wrote dx, you really meant +c, yes?

[hot quartz]

yeah

i need to find c but for that id have to find the value of y then sub it in with the x value 3 and then solve for c

[hot quartz]

yeah so i know the gradient and the x value

[pure jacinth]

Yes, but you know the value of y because of the normal line

[hot quartz]

how?

[pure jacinth]

Plug 3 into the normal line

[hot quartz]

wdym by the normal line

[pure jacinth]

The normal to C at P

[hot quartz]

oh,,,,

i'm so stupid i just realized

thank you!

.close

[marsh citrus]

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[gloomy merlin]

Consider the following transformations
\begin{align*}
x: \R &\to \mathbb C \[1ex]
T: (\R \to \mathbb C) &\to (\R \to \mathbb C) \[1ex]
\Theta : \R &\to \R
\end{align*}
Consider the specific mapping $\Theta: t \mapsto kt$ for $k\in\R$. Is there a special name for the following property: [
T(x\circ \Theta)(t) = T(x)(\Theta(t))
]

[elfin berry]

Aero

[marsh citrus]

@gloomy merlin Has your question been resolved?

[marsh citrus]

@gloomy merlin Has your question been resolved?

[quiet anvil]

λabcd.(a(bc))(d) = λabcd.(ab)(c(d))

So the first is the Bluebird combinator, applied to a value. The second is the Dove combinator.

I don't know of any more special terminology than that

@gloomy merlin

At least I think the second is the Dove combinator, I could be wrong

[gloomy merlin]

No relevent results come up when I search up the Dove combinator

Where did you get that name from

[quiet anvil]

https://github.com/loophp/combinator

[marsh citrus]

@gloomy merlin Has your question been resolved?

[marsh citrus]

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[mossy horizon]

Hi for question number 22

Is there a method to calculate it very fast?

I know the one where you have to get the lcd first but it takes quite a long time

[frosty thicket]

best i got is the combine the first and sixth, second and third, and fourth and fifth separately

then combine

because those pairs have large common factors

[mossy horizon]

That's slower then the current method I have lol but my method is still slow af

I js take lcd then divide the bottom by the lcd and then times the upper one

But the thing is for the exam I have 2 minutes for every question ●_●

[frosty thicket]

yeah this is quite a strange one

[merry hill]

I think the best one is factor out 3 from every numerator so itll become 3(1/24 + 1/40 + ... + 1/144) and I think you can go ahead and do the lcd method and I think this'll be faster since you dont have to multiply to 3 everytime but rather add them all then multiply to 3

[mossy horizon]

Yea I did this already 😭

Takes a long time so that's why I came here to ask if there's a faster method

I thought Cus of the 3 on top mb there's a method to quickly calculate them

[frosty thicket]

oh theres a pattern in the denominators

(x+3)(2x+4)

maybe a partial fraction decompisition will cause telescoping

[mossy horizon]

Wdym pattern?

[frosty thicket]

the xth fraction is of the form 3/((x+3)(2x+4))

[mossy horizon]

Uhh are u saying like this ●<●

[frosty thicket]

yeah but x=1,2,3

and so on

[mossy horizon]

Oh your right

[mossy horizon]

Doesn't do anything tho sadly

[frosty thicket]

aw darn

[mossy horizon]

It does do something but it makes everything harder to solve

[mossy horizon]

Yea 😭

[frosty thicket]

i guess its a matter of just knowing how to add fractions fast

there is probably a trick, but i dont know it, sorry

[marsh citrus]

@mossy horizon Has your question been resolved?

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[mossy horizon]

Np I'm js gonna give up on trying to win ●_● too difficult my goal now is to js get perfect score

[marsh citrus]

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[oak plume]

Is it possible to take the logarithm of a specific part of an equation?

[oak plume]

like lets say we have p2^x + log_2(y) = 2, can we just take the logarithm of p2^x, or would we need to take the entire logarithm of both sides?

[odd orchid]

to hold equality

whatever you do must be done to both sides

[oak plume]

Okay, understood. what about if you have 2^x + 3 = 4 for example. do you have to take a logarithm of the ENTIRE side, like log(2^x + 3) = log(4) or do you take logs of each individual component? the former right?

[odd orchid]

yea entire side

and log(a + b) does not equal log a + log b

[oak plume]

yeah, makes sense. Awesome, thank you.

Lets say we have this question here, what are you supposed to do with the logarithms and what not? We have logarithms in each equation, but I can't seem to find anything that you can do with the logarithms that is useful

i'm not quite sure what you can do with logarithms to simplify stuff like this

i thought about getting them into one equation and then getting the logs into just one logarithm, however once again I don't know how to move on from there

[oak plume]

please don't post your question in my question

What can you do when you're multiplying logarithms

(not using the product rule, like log(a * b) = log(a) + log(b))

but literally log(log(x))

or log(x)log(y)

[tulip idol]

First you need to solve the system for log y, and 2^x, then you can proceed ahead

So basically, assuming u = 2^x, v = log y, solve the equation

[oak plume]

I'm honestly not quite sure how to solve that equation

i keep rearranging and or substituting and I get nowhere with it

[tulip idol]

u + v = 1 => u = 1 - v

Now pu + v = 2 => p(1-v) + v = 2 => v(1-p) = 2-p

So, v = 1 + 1/(1-p) ✓ u = 1/(p-1)

Can you proceed from here? @oak plume

[marsh citrus]

@oak plume Has your question been resolved?

[marsh citrus]

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[fallen iris]

hello

[lofty gyro]

yes? math Q?