help-42

[sly kite]

I don’t think this will simplify to be independent of AE without additional work

This problem is much easier if you notice that BK and EK have the same length

This is since the arc BK and KE are the same, so should their chords be as well

[tall moon]

how do you know BK=EK?

[sly kite]

BAK and KAE are arcs subtended by the same angle

A/2, since that angle is bisected

If two arcs are identical, then the chords connecting their endpoints are the same length

[tall moon]

what is subtended?

[sly kite]

The arc BK is formed from the lines extending out of A

And the arc KE is formed from the lines extending out of A, AK and AE

[tall moon]

oh i see

[sly kite]

In a circle, an arc is defined by the angle of separation between the two lines

So if you have two separate arcs

But there angle is the same

Then we say the arcs are identical

[tall moon]

ok

[sly kite]

Which gives us both the length of the arcs ( the curved lengths) and the length of the chords (straight line between endpoints) are equal,

[tall moon]

but how can we use the fact that BK=KE?

[sly kite]

Consider the two triangles BKA and EKA

We know that BK and EK are equal

And we know that angle BAK = angle EAK

For ease let’s call this a

Then apply the cosine rule on one triangle

BK^2 = BA^2 + AK^2 - 2(BA)(AK)cos(a)

This is for triangle BAK

For triangle EAK, we’ll use the fact that EA is 1/2 BA

(This is because it’s both isosceles and midpoint of the side)

So EK^2 = (1/2 BA)^2 + AK^2 - 2(1/2 BA)(AK)cos(a)

Since BK=EK, so does BK^2=EK^2

So we set the two RHS equal

BA^2 + AK^2 - 2(BA)(AK)cos(a) = (1/2 BA)^2 + AK^2 - 2(1/2 BA)(AK)cos(a)

Cleaning up and simplifying gives AK = (3/4 BA)/cos(a)

The final step uses the extended sine rule, which gives 2r = BA/sin(c) where c is the angle BAC

By the fact that ABC is isosceles, we can calculate that c = 90-a

So 2r = BA/sin(90-a)

[tall moon]

what is r in this case?

[sly kite]

R is circumradius of ABC

[tall moon]

oh ok

[sly kite]

sin(90-a)=cos(a)

This is a standard trig identity, which can be proved by reflecting/translating the sin graph into a cos one, or by angle addition formulas etc

So re-arranging, cos(a)=BA/2r

AK = (3/4 BA)/cos(a) and cos(a) = BA/2r gives the result after simplification

Lmk if there’s anything that is unclear

[tall moon]

can i ask how you got 90-a?

a is angle BAK correct? and since the K lies within the line that subdivides the angle BAC, shouldnt it be 2a instead?

[sly kite]

This is the relationship between a and c

[tall moon]

oh, i was confused cause you said angle BAC

i see

ok ty

i have more questions, is it fine if i ask here?

[sly kite]

Sure

[tall moon]

if the shaded area is 777, find the unshaded area of the figure

[tall moon]

,rccw

[tall moon]

and it wants me to use $\pi=\frac{22}{7}$

[potent lotus]

Skill_Issue

[sly kite]

Consider this construction

We’ll analyse the small circle on the right, and the results for the left will follow by symmetry

The idea here is to use Pythagoras on triangle INJ to get a relationship between r1 and r2

I’ve given IN and IJ in the image, so only NJ, which is equal to KO is left

We use that O,J and L are collinear, since the little circle and semi circle share a tangent at L

So their radii pass through L and meet at 90 degrees, hence co linear

Since JL=r2 and OL= R, we establish that OJ = R-r2

The diameter of the big circle is the radius of the semi circle, so we can say that R = 2r1 and substitute that in

Apply Pythagoras on OJK to find KO, which is equal to NJ

KO^2 = (2r1-r2)^2-(r2)^2

Pythagoras on triangle INJ no gives NJ^2+NI^2=IJ^2, and substituting in KO = NJ, as well as the values in the image

(2r1-r2)^2-(r2)^2+(r1-r2)^2=(r1+r2)^2

Simplifying gives r2 in terms or r1, and you have r1 in terms of r, so all that remains is to calculate the shaded area in terms of r, and some algebra to change it to the unshaded area

[tall moon]

wow thats smart

is it 1295

[sly kite]

I’m haven’t calculated the value

I get that r1=2r2 tho

[tall moon]

ok

last question, this somehow feels like not enough info? it wants me to find the area thats marked with "?", and the numbers in the other triangles are its area

[sly kite]

If you look at the construction on the left, and split it into the two constructions on right

You’ll find that the ratio of [ABC] : [ACD] = [ECB] : [ECD] where [] means the area

This is because ABC and ACD share the same altitude H, so the ratio of their area is equal to the ratio of their bases

I.e.

[ABC]=H(BC) and [ACD]=H(CD)

So [ABC]/[ACD]=BC/CD

Since that H cancels

Apply the same process for the triangle on the bottom

Gives you [ABC]/[ACD]=[ECB]/[ECD]

[tall moon]

whats the diffrence between the triangle on the up right and down right? isnt it the same just the bottom one is just abit squashed

oh wait nvm

[sly kite]

Use this construction on the triangle

And form two equations in A1 and A2 to solve simultaneously

Then add A1 and A2 to get your answer

[tall moon]

12/10+A1+A2=4/A1
10+A1+A2=3A1
2A1-A2=-10
???

could you tell me where i went wrong, im gonna go to sleep, prematurely i want tot hank you for your time and inderstanding and i hope you have a great day/night, cya

[sly kite]

2A1-A2=10, so you got your sign wrong somewhere

Also you will need a second equation, by choosing a base of the big triangle

A2/10=(A1+A2+4)/18

And then solve both equations together

[weary oracle]

in this we derivate x^3 by power rule,but how we derivate for x. how does it become -1

[surreal cosmos]

The derivative of x becomes just the coefficient in front of it.

Sometimes it helps to think of it like a power rule: -x = -1x^1
Deriving this using power rule = -1*1x^0 = -1

[tall moon]

16A1-8A2=80
8A2=10A1+40
6A1=120
A1=20
A2=30
so is it 50?

[finite nexus]

hi there

[weary oracle]

ok got it , tqs

[green haven]

do you know the proof that if A and B are a line and they subtend an angle with 2 points C and D where the angles are the same, then all 4 points are on a circle

idk if this could help

[junior bane]

y=8

[tall moon]

.close

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[fierce cradle]

could someone help me with these problems?

[calm coral]

@fierce cradle Has your question been resolved?

[ancient helm]

what have you tried

[calm coral]

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[fierce cradle]

.reopen

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✅

[fierce cradle]

May I show you my work

[ancient helm]

yea

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[obtuse basin]

Could someone explain the logic behind why this diverges?

The n goes to 0? Shouldn't it converge? Or is it because it's a sum?

[ocean cypress]

the harmonic series diverges

here's a short proof

[obtuse basin]

No worries. I will look up the harmonic series. Thanks!

.close

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[old dew]

.reopen

hi i need help with my math

Like this is actually bad bro

Im so tied

Tired

Im js tryna get this done

<@&286206848099549185>

[green haven]

go to https://discord.com/channels/268882317391429632/488120190538743810

get a available help channel

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[calm coral]

@molten latch Has your question been resolved?

[calm coral]

@molten latch Has your question been resolved?

[pastel mango]

First：Draw the line of length AC

Second：A is center ,draw circle with the radius of AB

Third：C is center ,draw circle with the radius of BC

Fourth：the intersections of the circles in the second and the third step can be a position of point B(there are two intersections,choose arbitrary one)

Fifth：repeat the second to the fourth step to find the point D (Notice that as long as the position of B is chosen,there is only one corresponding position of D)

[calm coral]

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[sharp ember]

For the second part, do I just say that cauchy implies convergence?

[pallid halo]

is A a set of real numbers as well?

[sharp ember]

Yes

[pallid halo]

then sure

you don't even need uniform continuity

ordinary continuity is sufficient

[sharp ember]

is this sufficient for the first part tho?

[pallid halo]

yea, sees fine

[sharp ember]

Thanks 🙏🏼

.resolved

.resolve

.solved

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[molten trench]

.rotate

[molten trench]

.rotate

[calm coral]

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[molten trench]

oops wrong piicture

mb

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[median cloak]

I’m trying to solve this set of equations to see whether it has a solution when a=7.

By matrix operations

[half jolt]

Ok, so what's your question...?

Rightmost matrix looks correct

Though I don't know what the "1 -1 -1" means at the bottom

[median cloak]

That outside of the context , just do not mind it

look at the right end, how to I determine whether a=7 has a solution or not?

[half jolt]

Depends on b...

[median cloak]

wdy,

wdym

[half jolt]

Even if a = 7, this can still have a solution or not, depending on what b is

It's not only dependent on a.

[half jolt]

What did you do on the last step here?

[median cloak]

how do you know. There are manifold of relations between a and b implies from my result.

[half jolt]

You don't need to do the last step

Canonical form isn't needed if you just want to verify number of solutions

[median cloak]

why? I'm trying to solve the set of equation.

[median cloak]

why?

[half jolt]

Because all you care about is whether or not there's a contradiction

if there is, no solutions. If there isn't, at least one solution.

[median cloak]

so I should stop at the third step, for a-7(z)= b-9 is sufficient to prove that a=7 has many solutions.

[half jolt]

Stop here

That's enough

[half jolt]

Do you understand what I mean?

[median cloak]

the bottom line suggests that (a-7)z=(b-9)

[half jolt]

Yes

[median cloak]

does it means anything?

[half jolt]

If there is a row of the form 0 0 0 ... 0 c where c != 0, that's a contradiction.

Meaning there aren't any solutions.

Try to think why that's true

[median cloak]

why you mentioned c!=0

[half jolt]

Maybe I should say 0 0 0 ... 0 | c

Because if c = 0, the row 0 0 0 ... 0 | 0 is meaningless. You can ignore it.

That's just like writing 0x + 0y + 0z = 0

You can "ignore" it

But if you have 0x + 0y + 0z = 1, that's a contradiction. There are no solutions to this.

[median cloak]

(x,y,z)=(0,0,0) is a solution for that.

[half jolt]

No it's not

c != 0

[half jolt]

Here is an example

x = y = z = 0 isn't a solution. That gives you 0 = 1 which is obviously not true

Left hand side is always 0, while right hand side is always nonzero

[half jolt]

Meaning there are no solutions to this

@median cloak

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[whole dock]

Why does this rule work

[whole dock]

,rotate

[exotic falcon]

square both sides sir

[whole dock]

Oh i see

[exotic falcon]

$$\sqrt{6}=6^{\frac{1}{2}}=(2\cdot 3)^{\frac{1}{2}}=2^{\frac{1}{2}}\cdot 3^{\frac{1}{2}} = \sqrt{2}\sqrt{3}$$

another way to see it

[potent lotus]

Austin

[whole dock]

Nice thank you sir

[exotic falcon]

ABCD

[whole dock]

.close

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[safe moss]

Can someone help me with this idk what i did wrong but im sure the answer of mine is wrong

[safe moss]

The hint was given by the book btw

[minor knot]

,rotate

[minor knot]

the + sign vanished

it should be

$x^3(x^2+1)+8(x^2+1)$

[potent lotus]

Renz

[minor knot]

and you dont need to square x²+1, since you are simply factoring out x²+1 from the 2 terms

$(x^3+8)(x^2+1)$

[potent lotus]

Renz

[calm coral]

@safe moss Has your question been resolved?

[safe moss]

ohh okay

now i remember

thank you

.close

[calm coral]

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[safe moss]

.reopen

[calm coral]

✅

[safe moss]

wait

uhm then i got 3 roots?

[safe moss]

i get 3 roots here

[minor knot]

hmm, thats not quite right

[minor knot]

is the original expression equal to 0?

[safe moss]

not given

so i assume yes

[minor knot]

wait whats the question then

just to factor?

[safe moss]

lemme send it

[minor knot]

or find zeroes

[safe moss]

58

[safe moss]

Yes

[minor knot]

ah then its equated to 0

so now we have

$(x^3+8)(x^2+1)=0$

[potent lotus]

Renz

[minor knot]

wait ni

no

lets continue factoring x³+8

it can still be factored

[safe moss]

Oh right

x^3 + 2^3

[minor knot]

$x^3+y^3=(x+y)(x^2-xy+y^2)$

[potent lotus]

Renz

[safe moss]

Now we got 5

[minor knot]

correct!

[safe moss]

Thanks

[minor knot]

in total there should be 5 roots

[safe moss]

Btw i know that rule of roots etc

But 5 different roots?

right

we dont count the same ones like the multiplicity ones

or do we

[minor knot]

dunno if we count them as different roots

but the multipiclity should all add up to 5

for repeated factors

[safe moss]

Okay then we do

Probably

[safe moss]

thanks again im closing the chat ! have a great day

.close

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[hardy sorrel]

if g(x-1) = 5x + 2. find g(x) and g(-1)

[hardy sorrel]

idk where to start

ple h4lp

[unkempt drift]

Remember you can replace x with anything as long as you're consistent and replace it everywhere it appears

It's the same thing as $g \left( \Delta - 1 \right) = 5 \Delta + 2$

[potent lotus]

south

[unkempt drift]

with the triangle

So if we replace x with (x + 1), what do you get?

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[mental shale]

I'm not sure how to start this

can someone

[unkempt drift]

Can you come up with an equation that uses the fact (4, 1) is also on the graph?

Also the turning point has x-coordinate -3 gives you a

[mental shale]

uhm

an equation?

[unkempt drift]

Yes

[mental shale]

so

1 = (4-4)^2 + 1

?

[unkempt drift]

You've gone a step too far

[mental shale]

which is

[unkempt drift]

It's just 1 = (4 + a)^2 + b

And also a is not -4

[unkempt drift]

Replace x with 4 and replace y with 1

[mental shale]

oh ok

[mental shale]

yes

im following

[unkempt drift]

Ok so can you figure out a given the turning point has x-coordinate -3?

The turning point of y = (x + 0)^2 + b = x^2 + b is of course when x = 0

So we need to move the graph 3 units to the left

[unkempt drift]

So if we are moving a graph 5 units to the left, do we have x + 5 or do we have x - 5?

[mental shale]

wait

we move the graph?

[mental shale]

x-5

[unkempt drift]

Nope

If we have say y = x

And we now have y = x - 5

Then if y = 0, x = 0 for y = x

But if y = 0, x = 5 for y = x - 5

So we've translated it 5 units to the right

[unkempt drift]

Yes, well that's how you think of it graphically

[mental shale]

hmm

ok so

[mental shale]

okok

[unkempt drift]

Try graphing x, x - 5, and x + 5 yourself!

You learn more that way

[mental shale]

ok

so what now

[unkempt drift]

Yeah so this tells us a = 3

$1 = (4 + a)^2 + b$ so $1 = (4 + 3)^2 + b$

[potent lotus]

south

[mental shale]

yes

I understand

@unkempt drift

[unkempt drift]

Yeah cool

[mental shale]

so

whats next

is a constant?

[unkempt drift]

<@&268886789983436800>

[unkempt drift]

Yes, a = 3

[unkempt drift]

So literally you can find b from this

[mental shale]

oh ok

I thought b would change

for 4,1

so b would be

-15?

[mental shale]

why did u @ mod

[unkempt drift]

Spam link

[unkempt drift]

No, b is just 1 - (4 + 3)^2

= -48

[mental shale]

how did u get that

im sorry im confused

y = -48?

[unkempt drift]

No it's correct

[unkempt drift]

Yes

[unkempt drift]

Okay, so 1 = (4 + a)^2 + b right

But a = 3

Cause (x + 3)^2 + b has a turning point at x = -3 (if you sub x = -3 in, (x + 3)^2 becomes 0 so you know it is)

So 1 = (4 + 3)^2 + b

and then subtract both sides by (4 + 3)^2

[mental shale]

yep

[mental shale]

ohhh ok

thank youuu

[unkempt drift]

No worries

[mental shale]

.close

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[calm coral]

Please don't occupy multiple help channels.

[cursive linden]

you already have posted this question in another channel. stick to that channel

[idle fractal]

I don't know but I can't access that channel

[cursive linden]

https://discord.com/channels/268882317391429632/908076667224854568

[idle fractal]

Thanks

[cursive linden]

.close

[calm coral]

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[idle fractal]

Ahh

I closed other one 🤣

[cursive linden]

ah hell

[idle fractal]

.reopen

[calm coral]

✅

[idle fractal]

So any direction for this question?

[hidden tinsel]

what's A.P and G.P

[idle fractal]

Arithmetic progression

Geometric progression

[calm coral]

@idle fractal Has your question been resolved?

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@idle fractal Has your question been resolved?

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@idle fractal Has your question been resolved?

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[young stirrup]

calculate area

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[young stirrup]

.close

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[undone marlin]

Anyone need help?

[calm coral]

Please don't occupy multiple help channels.

[whole hinge]

.close

[calm coral]

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[late zenith]

My question is sum of: $6+66+666+ ... upto n terms$
I've gotten the answer $(10^n - 9)(n-10)/15$
The process I've done:
$6(1+11+111+...)
6(1 + 1 +10+1+10+100+...)
6[n10^0 + 10(n-1) + 10^2(n-2)+ ... + (10^(n-1))(n-(n-1)) + 10^n(n-n)]
6[(10^n + 10n + 10^2(n)+...10^(n-1) - (10^n-1)+(10+10^2+...+10^(n-1))
Using Geometric Progressions
6[n((10^(n-1) - 1)/10-1)-10((10^(n-1) - 1)/10-1)
Simplification
(10^n - 9)(n-10)/15 $

[potent lotus]

Stellar Ring

[late zenith]

welp that didnt work

I guess I'll just send a picture

[hasty fiber]

If your exponent has multiple characters you need to encase them in {}

$10^{(n-1)}$

[potent lotus]

Steakanator

[hasty fiber]

And line breaks are achieved by adding a double \

[late zenith]

Can i just send an image?

[hasty fiber]

Sure

But knowing proper tex is still good 😌

[late zenith]

Okay I think i know what is wrong with answer but i still want to know if it is possible to solve with my technique

$6+66+666+ ... upto n terms$

[potent lotus]

Stellar Ring

[late zenith]

Have to find some of n terms

my method was:

[potent lotus]

Stellar Ring

Stellar Ring

[late zenith]

Is it possible to expand this and use geometric proggressions to solve it?

[calm coral]

@late zenith Has your question been resolved?

[late zenith]

<@&286206848099549185>

[undone marlin]

?

[late zenith]

The question hasnt been answered yet

[undone marlin]

ok got it

[calm coral]

@late zenith Has your question been resolved?

[proud gull]

you could just quickly use the term cancelation thing
consideR:
S1 = 1+11+111+1111+.............+Tn
S1 = +1+11+111+.......................+Tn
=> S1-S1 = 1+10+100+1000 +.............. - Tn
=> Tn = " "
then just use summation on Tn

and then 6(S1) = your answer

[calm coral]

@late zenith Has your question been resolved?

[calm coral]

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[cedar finch]

Show that the set of all Turing-decidable languages is closed under union and intersection.

[cedar finch]

What is considered a sufficient and acceptable answer

I will show a couple i found online

<@&286206848099549185>

[calm coral]

@cedar finch Has your question been resolved?

[cedar finch]

<@&286206848099549185>

[calm coral]

@cedar finch Has your question been resolved?

[cedar finch]

<@&286206848099549185> lol

[calm coral]

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[calm coral]

@cedar finch Has your question been resolved?

[calm coral]

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[gusty cliff]

Help guys Im stuck

[ancient helm]

what have you tried

[gusty cliff]

Im confused

I tried putting them into a Sn= 1+1 +3 +1/3

Then idk💀 I got confused

[ancient helm]

do you notice any patterns

[gusty cliff]

Yes

And also one thing makes me confused

What are the first 2n terms? Is it the first 2 terms or the first 2 terms with indices n?

[ancient helm]

if i ask you for the first n terms would you understand what that means?

[gusty cliff]

first term of the sequence?

[ancient helm]

first n terms

like if i asked you

here's a sequence 1 , 1, 1, 1, ...

what is the sum of the first n terms?

[gusty cliff]

no idea💀

3

[ancient helm]

ok

what is the sum of the first 10 terms of that sequence

[gusty cliff]

10

[ancient helm]

it's 10 right

what about the first 100 terms

[gusty cliff]

100

[ancient helm]

what about the first 27 terms

what about the first n terms

[gusty cliff]

Mhm

[ancient helm]

the answer would be n

you add up 1 + 1 + 1+ ... + 1, n times

so similarly in your question

add up 2n things

and tell me what you get

does that make sense

[gusty cliff]

Wait a sec gotta process

Are they asking me to find S2n

Or like Sn + Sn

[ancient helm]

S_{2n}

[gusty cliff]

But why do they say the “first n terms”

Isn’t it just the sum of all n terms?

[ancient helm]

they don't, the say sum of the first 2n terms

and there's a reason for this particular problem

they say 2n

[gusty cliff]

Omagad

I’m fkn lost💀

[ancient helm]

ok

just

stop thinking about the indexing

and think about the actual question now

[gusty cliff]

Ok

[ancient helm]

what is the pattern that you notice

[gusty cliff]

Multiply by 3

[ancient helm]

is that it?

[gusty cliff]

And the 1/3 is multiplied by 1/3

The fractions are multiplied by 1/3

[ancient helm]

yes

we alternate

so with that in mind

i will rewrite this sequence as two separate sequences

$1, 3, 9, 27, 81, ... \n
1, 1/3, 1/9, 1/27, 1/81,...$

[gusty cliff]

Ohhh

It’s like 2 diff sequences put together

[ancient helm]

bad bot

but yes

so now

what is the nth term of the top and bottom sequences?

[gusty cliff]

Wait

1x3^n-1

Is it?

[ancient helm]

for the top

what is the bottom

[gusty cliff]

1x1/3^n-1

[ancient helm]

yes

so

[gusty cliff]

And do I just multiply these?

[ancient helm]

no

[gusty cliff]

Omfggg i just remembered what was an n term💀💀💀

[ancient helm]

huge

ok

i will let my top sum be R and the bottom sum be T

[gusty cliff]

Indeed

[ancient helm]

if i can find R_n and T_n, and add those two together

would you agree that the answer i get is equal to the sum of the first 2n terms in the original sequence

[gusty cliff]

Yes

[ancient helm]

i am taking n things from R, and i am taking n things from T

so in the end i have 2n things

ok perfect

i assume you know how to find the partial sums of R and T?

[gusty cliff]

Yes yes

I should be fine l

Lemme try real quick

Idk how to add them together lol

Im stuck here

Oh wait

[ancient helm]

factor out 1/2, combine like terms

which is -1 and 3

[gusty cliff]

i think

Yeah

got it got it

[ancient helm]

and you get the same thing the original picture

[gusty cliff]

ok thanks dude

U saved me

Got it now

.close

[calm coral]

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[calm coral]

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[high topaz]

Basically I have a bunch of individual points of data, is there anyway to find an equation for these data? They don't seem to be lying on a straight line

[high topaz]

These are the first 10 data, I got like 90 more of similar values

The data might be rounded to the nearest integer

[coral osprey]

My guy please use a table

Desmos has a table

I mean what you can do is like

Yeah no use a table please

You do not wanna be doing things like (A_9.y - A_8.y)/(A_9.x - A_8.x)

[high topaz]

You mean this?

[minor knot]

use that

[coral osprey]

That, yeah

There's also uh

I forgot hold on

[high topaz]

But I can't plug in values without defining f(x)

[coral osprey]

There's a way to track trajectory on desmos

No you have to like

Use all points

You could also do like

Eh

Manually place some points

[swift laurel]

desmos has the ability to fit regression lines/curves

[high topaz]

Oh

Thanks

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

[fleet raptor]

What are you asking

[idle fractal]

x+iy

x-iy

Angke will be same

So π/2 amgle

[ancient helm]

The angles aren't the same

and not that it matters but even if they were the same, that doesn't mean they would sum to pi/2

[fleet raptor]

Is this proper argument

If it is consider how Arg z can be neg

[idle fractal]

argzzconjugate

[fleet raptor]

Wdym branch

[ancient helm]

Wait nvm I'm dumb

[idle fractal]

argz z conjugate

Argz^2

Which real so 0

[ancient helm]

But branch is which interval [theta, theta + 2pi) you allow values of the argument to take

so you can choose the principal branch like you asked, which is [0, 2pi)

Or you can choose something like [-pi, pi)

[fleet raptor]

ok

[calm coral]

@idle fractal Has your question been resolved?

[calm coral]

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[neat talon]

Hello, I do not know what a unity and the units of a binary operation is, can anyone help me with this problem? (Abstract Algebra/Mathematical Linguistics)

[calm coral]

@neat talon Has your question been resolved?

[ancient helm]

a unit is a element that has multiplicative inverse

@neat talon

[neat talon]

ill get back to this once i get home

[calm coral]

@neat talon Has your question been resolved?

[neat talon]

i dont quite understand on how i want to find the unity in this operation

what must i do first?

nvm, i get it now

thanks

.close

[calm coral]

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[ornate summit]

Anyone know how you would solve for an unknown when you have one outside and inside sine or sine inverse.
I was setting up inequalities but it doesn't seem to work

[calm coral]

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.

[potent lotus]

²⁰Ne

[bronze adder]

It might be for specific values of the constants

[bronze adder]

So...

[ornate summit]

Oh

[ornate summit]

Sorry, this is the expression I got to

[bronze adder]

hmm

[crystal forge]

can you show the original original problem?

[bronze adder]

,w arccos(1/(sqrt(2)n)) = 1/n

[potent lotus]

Results provided by WolframAlpha

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Requested by neonperseus

[bronze adder]

crazy

[ornate summit]

Wants the minimum refractive index of glass which corresponds to the limiting case where the incidence angle at B is the critical angle

[ornate summit]

Thx

[potent lotus]

²⁰Ne

²⁰Ne

²⁰Ne

[bronze adder]

@ornate summit I think you might've messed up in the geometry

this is solvable pretty easily

[ornate summit]

It is?

[bronze adder]

yep

take one of the arcsins to the other side and

sin on both sides

you'll get a regular equation for mu

[ornate summit]

Sin( arccos(X) ) = ?

[bronze adder]

draw a triangle

[ornate summit]

I'm still confused

[bronze adder]

draw a triangle with cos A = x

then find sin A

[ornate summit]

$$ \sqrt{1-x^2} $$

[potent lotus]

Hemesfere

[bronze adder]

good

yeah

so you should be able to solve for 1/mu^2 now

its a linear equation

[ornate summit]

Alright, Ty very much

[calm coral]

@ornate summit Has your question been resolved?

[calm coral]

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[weary oracle]

Can you prove how theta = l/r

[remote mural]

This is by definition i believe

One radian is defined as the angle subtended at the center by an arc of length equal to the radius of a circle

so you just get theta = l/r

[calm coral]

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[granite flicker]

i need help finding sec

[idle marten]

what is the value in radians

use law of periodicity to simplify if you haven't done it already

[granite flicker]

its 3.25 pi.1

[granite flicker]

3.25pi/1 **

[idle marten]

ok simplify sec(3.25pi)

[granite flicker]

idk how thats whre im stuck

[minor knot]

what smaller angle is coterminal to 3.25pi

[inner delta]

,calc 585-360

[potent lotus]

Result:

225

[granite flicker]

on the unit circle tho i dont see a value of 3.25 pi

which is why im confused

[inner delta]

Note that 3.25pi>2pi

[granite flicker]

yes..

[abstract snow]

,w 225-180

[potent lotus]

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Requested by theprophetofthedamned

[minor knot]

thats why you should convert it to an coterminal angle between 0 and 2pi

[granite flicker]

so i convert it back to degrees?

[idle marten]

where do you find the vocab related to coterminal btw

[minor knot]

or use $sec(\theta)=sec(2pi-\theta)$

[potent lotus]

Renz

[granite flicker]

uh wdym, its in my lesson

[minor knot]

i was taught that as well

[granite flicker]

itsnt coterminal any angle under 360

and the reference angle is under 90

[minor knot]

its any angle where you add or subtract any multiples of 360degrees or 2pi

[granite flicker]

yeah that

[minor knot]

now

[granite flicker]

but whats the value for sec

[minor knot]

3.25pi is outside 0 and 2pi right

[granite flicker]

like on the unit circle

yes

[minor knot]

how do you make it between 0 and 2pi

by adding or subtracting 2pi

[granite flicker]

yeah i guess so

multiples right

ill be right back

[minor knot]

yea

[granite flicker]

in like 15 or so

im back for a bit

[minor knot]

so

[minor knot]

ye just do this

once you find an angle with the same terminal side inside 0 and 2pi, you can use the unit circle to solve for sec

[calm coral]

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[white mountain]

31c

[white mountain]

Don’t know how to start

[modern peak]

...?

shift the index

[white mountain]

How

[modern peak]

okay

we have k - 1

[white mountain]

There’s no way to change it in the formula

[modern peak]

and it starts from 1

so we get

[white mountain]

0

[modern peak]

0, 1, 2, so on
for
1, 2, 3, so on as k

you see what i'm getting at?

[white mountain]

Yes I know so I did (n-1)*k

But that’s wrong

[modern peak]

so

[potent lotus]

ren (what's lemonsaurus?)

[white mountain]

Yes

[modern peak]

you can simplify

[white mountain]

Well. =1

K=1

[modern peak]

yea that's what i meant

by shifting the index

FYI

[white mountain]

Okay wait so

[modern peak]

you don't have to shift the index for this question

[white mountain]

Can I do (n-1)(n-2) / 2

Using the sum formula

[modern peak]

you can split the thing into -1 and k, then substitute -n for the -1, and use 0.5k(k+1)

[modern peak]

that's not the sum formula

[white mountain]

n(n+1)

/2

Oh wait

So (n^2-n) /2

Is that right

[calm coral]

@white mountain Has your question been resolved?

[calm coral]

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[blazing coyote]

yes

what about it

it was for Trans visibility day

[graceful dust]

[unkempt drift]

That's not a maths question

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[unkempt drift]

lmaoooo

[strange lichen]

bruh

no way u just typed up all that

[quaint sphinx]

😂

[modern peak]

mqnic genuinely how much free time do you have

[quaint sphinx]

not enough unfortunately 😔

[calm coral]

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[weary oracle]

got it , tqs

[modern peak]

you've accidentally opened this channel
please do .close

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[waxen mica]

ive drawn a graph but idk what to do

i don't think hte graph is right but yeah

[worn rock]

Anyone good at binomial real option pricing model?
Should I apply the WACC or just the Rf when discounting values back to present if I apply it on a specific case. I'm to value the real option of expanding production capacity in Austria.

[waxen mica]

wrong place bro

[worn rock]

Ah I see. Sorry buddy

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[thick fog]

Am i correct ?

R = Berries ripe along the path
S = Walking is safe
A = Rabbits seen in the area

(R -> S) ^(and) (-A(not A))

[amber bolt]

no

i'm sure there are many sentences that are equivalent but this can't be one of them

[thick fog]

did i get the entire thing wrong ? or just part of it

its the last sentence that i am a bit confused on

[amber bolt]

there are many ways to say it, every individual part could be correct, just the entire thing is wrong

[calm coral]

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[remote mural]

need help with this silly question

[remote mural]

i found B and C

which are -2 and 3 respectively

but when i try to find A im stuck

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

you get Ax(X+1) + B(X+1) + C(X^2)

if im correct

the answer is A = 4 but i dont see how they get that

<@&286206848099549185>

[swift laurel]

if you multiply that out, you will get a quadratic with coefficients in terms of A, B, C

you need the coefficients to match on each side of the equation

[calm coral]

@remote mural Has your question been resolved?

[shut quail]

Im not a helper but stumbling across this, its partial fractions so you equate the coefficients

a(x^2)(x+1)+B(x+1)+C(x^2) = 7x^2+2x-2

and you just substitue for x=-1 and then x=0 to find values of a b and c or you just equate coefficients

[calm coral]

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[deft light]

Simple question: Does this mean E(X) = X?

cause expected value and random variables are not the same as far as I know

but how do they end up with the same expression

(M is a constant)

[carmine sorrel]

,rotate ccw

[ancient helm]

no

E(constant) = constant however, and E(X) itself is a constant

[remote mural]

Hello

[deft light]

yeah I mean since they end up with the same expression, it implies that both upper and lower equations are the same

so if we compare the left hand sides, E(M-X) = E(M-E(X))

and the only difference I see is X turned into E(X)

[ancient helm]

yes line 1 and line 2 are equivalent

no E(X) is not X

[strange lichen]

E[E[X]] = E[X]

[carmine sorrel]

that's not enough to imply X = E(X) though. Just that the expected value of those two things are the same.

[ancient helm]

you can't invert expected value like this

[deft light]

i think im confusing something lol

[remote mural]

I have no idea what yall are talking about, but i need help with a lot of stuff ( im a dumb 17 y/o college student who needs help with her maths because I NEED to pass. )

[carmine sorrel]

X is your variable, E(X) is a constant, the average of X
X and it's average are not the same thing.

[carmine sorrel]

See #❓how-to-get-help

[remote mural]

alr ty

[deft light]

hmmm I get what you are saying, the expected value and just the random variable are not the same thing

so you are saying both expressions are equivalent but not the same thing?

like they are not the same concept(?)

[carmine sorrel]

Correct. This, E(M-X) = E(M-E(X)), just says that M-X and M-E(X) have the same expected value (average), not that they are themselves equal.

[deft light]

ahh thank you for clarification

[carmine sorrel]

for example, 4 and 6 average to 5
3 and 7 also average to 5.
You can't say two things are equal because their expected values are the same.

[deft light]

OMG REAL

[ancient helm]

alternatively you can think about how Expected value isn't injective

just like how (-1)^2 = (1)^2 but -1 and 1 are unequal

[deft light]

so I was confusing those two concepts 😭

thank you so much I get it now!

.close

[calm coral]

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[calm coral]

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[normal wasp]

A guy needs 90 seconds to climb on a stopped escalator and when the escalator is moving but the guy is standing still he needs 60 seconds. How long would the guy need if both escalator and him are moving

[calm coral]

Please don't occupy multiple help channels.

[amber bolt]

you just write some kind of equation

[normal wasp]

yea ik

would it be a system or single one?

[clever ferry]

absolute speed = relative speed + translational speed

[amber bolt]

yes the speeds add up

either a system or a single one would work i expect

[clever ferry]

turn it into an equation of times

1/(t) = 1/(60) + 1/(90)

[amber bolt]

it's just hard to write a single one

[normal wasp]

i would prefer that

[clever ferry]

the answer is 36s

[normal wasp]

so whats the first step?

[clever ferry]

I literally just did

[amber bolt]

yeah

[normal wasp]

ok what is the equation?

[clever ferry]

start with this

they didnt mention anything about distance so you take an arbitrary one

v = d/t

cancel out the d

plug in the times

[amber bolt]

you assume the length is one, t is time we're looking for, then the combined speed is 1/t

[normal wasp]

what is the length?

of what

[clever ferry]

the escalator

like I said

it's the same for all 3 cases

so it cancels out

[normal wasp]

still dont get it lets start from the beggining

[amber bolt]

there's an escalator

it's 1 escalator long

[normal wasp]

ok

[amber bolt]

when it's moving and you walk on it at the same time these speeds add up to a single speed

which equals 1/t

1 escalator per t seconds

i think we can even solve it from just
90x = 60y

where x is walking speed and y is escalator speed

[clever ferry]

that's not correct

it also doesn't solve for anything

there are no unknowns in the equation

[amber bolt]

yeah I need 2 equations then
90x = 60y
60y = t(x+y)

solve first equation for x
x = (2/3)y

60y = t(5/3)y
t = 36

[calm coral]

@normal wasp Has your question been resolved?

[calm coral]

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[mild jetty]

How do you show $f(x) = x^2 + x + 1 \in \mathbb{Z}_2 [x]$ is irreducible?

[potent lotus]

WhoTao

[civic dirge]

If it is reducible, then it must factor into linear factors. Then, the linear factor would also be a root.

So go on, name a root.

[mild jetty]

it doesnt have a root

[civic dirge]

Bingo. Irreducible in this case

[mild jetty]

i see.

There is also this test I saw. Im not sure how to apply it

[civic dirge]

This works as high as cubics

[mild jetty]

in this case, what is p_i and n_i?

[civic dirge]

Quartics can be very difficult to determine irreducibility for though

I agree, I can't understand this

The "with" part looks like it's missing info

https://en.m.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields#:~:text=Rabin's test of irreducibility,-Like distinct-degree&text=Distinct-degree factorization algorithm tests,to distinct-degree factorization algorithm.

Does have "rabin's test"

Looks like p_i is a prime factor of n, and l is the number of prime factors
@mild jetty

[mild jetty]

i see

i dont know why i didnt think to send this

this example they included

[civic dirge]

There's not much there anyway lol

[mild jetty]

it checks out that l is the number of prime factors. For f(x) = x^2 + 1, there is only f(x) itself.

[civic dirge]

This seems like an easy way to get irreducibility for prime fields

[mild jetty]

yeah

ill verify tmrw with someone to be sure

thanks

.close

[calm coral]

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[mild jetty]

.reopen

[calm coral]

✅

[mild jetty]

oh nvm

.close

[calm coral]

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[calm coral]

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[honest kite]

why does this have a vertical translation

[calm coral]

@honest kite Has your question been resolved?

[dull wagon]

prod to sum law

[honest kite]

I know sum law, but prod?

[honest kite]

?

nvm i found it

[calm coral]

@honest kite Has your question been resolved?

[hollow perch]

you need to manipulate the function using log laws

specifically $\log_a(mn)=\log_am+\log_an$

[potent lotus]

chlamydia

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[calm coral]

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[eager cosmos]

Can someone explain the second question?

[calm coral]

Please don't occupy multiple help channels.

[eager cosmos]

.close

[calm coral]

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[minor sleet]

hey guys

[minor sleet]

whats the ideal way to do this for a)

im kind of lost with this gaussian integer stuff

like do i just guess and check random values lol

and make sure that |r| < |b|

[ancient helm]

it's the exact same procedure as doing traditional long division

[minor sleet]

in what way

[ancient helm]

how many multiples of 3 do you need to get to up to 52? then subtract and then how many multiples of -5i do you need?

[minor sleet]

17 and 1

[ancient helm]

just give it a try and see what you get

[minor sleet]

o

nvm

[ancient helm]

well it's not 1 multiple of -5i

[minor sleet]

yeah

[minor sleet]

this is what i did originally

i did 17 for q