help-13

[manic escarp]

question?

[cedar kiln]

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[tawdry pagoda]

this is from arthur engel's book. The underlined part is not making much sense to me as for this to be true, doesn't the sequence formed by this formula make a line? but he has given formula x(n+1) and y(n+1)
basically, my question is how is x(n)*y(n) = x(n+1)*y(n+1)

[tawdry pagoda]

sorry i dunno latex

[zenith nacelle]

just multiply, $x_{n+1}y_{n+1}=\frac{x_n+y_n}2 \cdot \frac{2x_n y_n}{x_n+y_n}=x_n y_n$ ?

[tawdry pagoda]

when i put them in paren, i meant subscript

[wraith dagger]

moriaritie

[tawdry pagoda]

oh

sorry

i'll close it now

thanks

.close

[cedar kiln]

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[wet mist]

Let ABC be a triangle with AB < AC and let I be its incenter. The incircle is tangent to side BC at point D. Let E be the only point that satisfies that D is the midpoint of segment BE. The line perpendicular to BC that passes through E intersects CI at point P. Show that BP is perpendicular to AD.

[wet mist]

I am stuck mainly I think because I can’t find a way to use BD=DE

I’ve tried making an isosceles triangle with the incenter but it seems that it doesn’t work

I thought about proportional triangles as well in order to apply that equality but it doesn’t make any sense either

<@&286206848099549185>

[cedar kiln]

@wet mist Has your question been resolved?

[wet mist]

<@&286206848099549185>

[soft umbra]

What's the question?

Oh my bad It's there

Can't you say that for BE to be bisected, line AB must be perpendicular to BE because bisectors occur at 90° angles?

[wet mist]

I’m sorry, could you pls repeat it?

AB perpendicular to BE?

[soft umbra]

Oh wait no

I misread it

Sorry

Have you had a look at geogebra?

[soft umbra]

I'm not 100% sure on what to do on this question but this should help you to visualise it more clearly

[wet mist]

Ok thanks I hope that will be helpful

[cedar kiln]

@wet mist Has your question been resolved?

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[tribal radish]

Im trying to make a program that can predict when two cos and sin functions moving at differing speeds will start repeating the loop, the function I made works mostly, but sometimes it just gets it all wrong and I dont really understand how to make it work better, I will send some photos in a second

[tribal radish]

ok so as this values are the same, it creates a circle, and as its values are in radians (so we are changing the angles inside the cos and sin functions) it ends in a multiple of pi*2

https://gyazo.com/c85ecf83d692bc1ad6f87205f72e5532

Im using angular frequency but honestly? I dont really know

https://gyazo.com/b0b074217bfc5ed9edf6a278335154f9

here is a case where the end moment of the trajectory is correcly predicted (I think)

https://gyazo.com/ee223502e2a49b21257ff7b61e8f7d21

and here a moment where im just completely wrong

do some of them just never repeat??

so the formula im using is
omegaX = PI times 2 / X movespeed
omegaY = PI times 2/ Y movespeed
and the totalTime = PI times 2 times |omegaX-omegaY|

if both speed the same I just use Pi*2 / one of the speeds

[shy widget]

Do you have it set up in a form similar to $\sin(v_{y}t)$ and $\cos(v\_{x}t)$

[tribal radish]

yes

[wraith dagger]

The Great D

[tribal radish]

exactly like that on the trajectory part

inside a loop that ends when t > totalTime

[shy widget]

Yup. Trying to think with you.

I would think that there is always a period

[tribal radish]

yes same here

[shy widget]

with it being at most gcf(v_y, v_x)

[tribal radish]

I tried using the Least Common multiple but it was a mess with below 1 inputs

[shy widget]

sorry yes lcm

[shy widget]

yupp i see

[tribal radish]

what does gcf mean btw? english is not my first language

[shy widget]

it would be greatest common factor but no i meant least common multiple

[tribal radish]

oh I did see this but when I tried coding it in it didnt work, probably by a mistake on my part

[shy widget]

i think you should find the lcm of 1/a and 1/b

[cedar kiln]

@tribal radish Has your question been resolved?

[tribal radish]

hmm ok I will test it

its just crashing due to lack of memory damn 💀

so, when I get the lcm of like 1/0.05 and 1/0.1 its 20, when the correct number is closer to like, 125

sometimes its very freaking high

[cedar kiln]

@tribal radish Has your question been resolved?

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[formal glen]

Hello

[formal glen]

I need some help on this question

i need my work to be checked

for number 9

by simplifying it we get g(x)=f(x)-5

and then would that be a vertical shift?

<@&286206848099549185>

[dark flume]

Indeed the graph of g will be shifted vertically by 5 units to the bottom compared to that of f.

[formal glen]

vertically shift down by 5 right

also and this graph would be just a reflection across y axis right

[dark flume]

Right but I'm not so sure for the terminology we're using here because I come from a french background and the word we use in french for 'shift' is translation and I think it might be the same in english.

[formal glen]

yeah they just mean the same

[dark flume]

Just a second. From the looks of it, it seems so but I will further check it out.

[formal glen]

Ok

[dark flume]

h(x) = f(-x) = -3(-x) + 1 = 3x + 1 = 3x - 1 + 2 = -( -3x + 1) + 2 = -f(x) + 2. So no it's not a reflection across the y axis because a reflection across the y axis will yield to h(x) = - f(x) which is not the case here.

If what I wrote is right, it means that it should be a reflection across the x axis and then an upward shift by 2

Yess Indeed. h(x) = f(-x) = -3(-x) + 1 = 3x + 1 that means that h(x) = 3x + 1

[dark flume]

I was checking some stuff. I'm sorry, but actually for that graph you provided, the correct answer would indeed be it's a reflection along the y axis.

[dark flume]

This one.

In general. f(-x) is reflection along the y-axis of f(x) while -f(x) is a reflection of f(x) along the x-axis.

For the last one it's a horizontal stretch

A horizontal stretch by a factor of 4.

In general when you have r(x) = k f(x), if k > 1 then r(x) horizontally compresses the graph of f(x) by a factor of 1/k while if 0<k<1 it horizontally stretches the graph by a factor of 1/k.

In our case we have r(x) = (-1/4)x = (1/4)f(x). and we have 0<1/4<1 So the graph is horizontally stretched by a factor of 1/(1/4) = 4.

No, here you have again the format r(x) = k f(x), only this time k = 4 > 1 so this time the graph of r compresses that of f by a factor 1/4 = 0.25.

For it to be a vertical shift down by 6, we'd need to have r(x) = f(x) - 6

Which is not the case here

In order to get your answer, you'd need to try to write r(x) as a function of f(x), here we have been given the expression of r(x) as a function of f(x) namely r(x) = 4f(x) so you can figure out right away the transformation that happened without any further work.

[dark flume]

No as I already mentioned generally when we get r(x) = k f(x). It depends on the value of k. If 0<k<1 then it's a stretch by 1/k and if k>1 then it's a compression by 1/k. Here we have k=4 so it is a compression by 1/k = 1/4

[cedar kiln]

@formal glen Has your question been resolved?

[cedar kiln]

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[cerulean star]

!original

[cedar kiln]

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.

[cerulean star]

two circles have 4 tangents

oh, you did not say that.

ok, I guess you could find where they share a point then

ok

find the centers

the intersection point is the midpoint where r1 and r2 meet

I am doing that just by completing the squares in each equation.

[cedar kiln]

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[dire geode]

Why do you need to

[urban ermine]

what is $c_1$

[wraith dagger]

Joshii

[upper abyss]

Looks like an exact DE

[cerulean star]

that's pretty cool, I didn't remember that fact

[upper abyss]

Or wait, the signs aren't right for that, are they

[cerulean star]

I started by subtracting them but didn't think it was gonna go anywhere until I plotted it with Desmos

[dire geode]

That doesn't answer my question

That doesn't answer my question

!original

[cedar kiln]

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.

[dire geode]

Do your best to provide a translation.

No

That's different from what you said

If you're given multiple choices, you should show them

Here you can just implicitly differentiate each option to check

Show then

[cedar kiln]

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[gritty osprey]

i dont understand

[gritty osprey]

how does this and that simplify to that

[woeful helm]

This image might help you, (if you know how to interpret it), You could also break the equation down into latex and ask Bard to explain it (my go-to method)

[gritty osprey]

ok thanks

what does it mean

[woeful helm]

hmm

are you aware of this maths concept?

or is this new to you'

[gritty osprey]

new

ish

whats the concept?

[woeful helm]

if you're in the US this is Highschool Math Ciriculum

[gritty osprey]

oop

im not us

but ok

wait so i just expand the fraction?

[woeful helm]

Do you have access to bard?

bard.google.com

[gritty osprey]

oh ai?

[woeful helm]

yep

[gritty osprey]

i just opened it

[woeful helm]

it's actually way smarter than you think

It solves all my problems that take for ever, you can ask it to give a rundown on the subject or explain the equation

[gritty osprey]

ooo sounds good

[woeful helm]

do you know how to format this into latex?

[gritty osprey]

whats latex

[woeful helm]

oof

[gritty osprey]

i just pasted the pic

[woeful helm]

oh yeah

i forgot that exists

[gritty osprey]

ohh wait thanks

[woeful helm]

no prob, I do it all the time

[gritty osprey]

damn this is better than chatgpt

[woeful helm]

it uses google

you can even cite sources, and it* fact checks

(Which is my favorite part)

because I can use it in school,

[gritty osprey]

damn

thats cool

thanks

[woeful helm]

so yeah have fun with it! It'll give you that answer to absolutely anything

Except quantum physics :/

[cedar kiln]

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[fiery valley]

hi everyone, when you square toot something, will the answer be positive only, or 2 answers?

[fiery valley]

$$\sqrt{4} = \pm 2$$

[wraith dagger]

Maladroit

[fiery valley]

or

$$\sqrt{4} = 2$$ ?

[wraith dagger]

Maladroit

[tropic oxide]

positive only

[scenic brook]

√4=2
x²=4 means x=±2

[crimson sedge]

The principal square root (positive solution) is assumed by convention

[muted bear]

x²=4 two solutions
x=sqrt(4) one solution

The two equations are different

[crimson sedge]

why? because the square root function would not be a function snymore

[dull oxide]

Non negative only

[tropic oxide]

for 0 there is no ambiguity anyway

im going to bite you

[cedar kiln]

@fiery valley Has your question been resolved?

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[winter tulip]

3. Has to be an equilateral triangle right

[winter tulip]

4. I mean

[crimson sedge]

question 4?

equilateral triangle? no

its an isosceles

[tropic oxide]

for everybody else's benefit, the question ^

[winter tulip]

Let's assume (6,-3) is a and all of the choices are b and c, you just have to use the distance formula to get the 5 units rght

[tropic oxide]

yes...

that's it, don't overthink it

[winter tulip]

I was confused cuz I thought b and c will equal 5 too so I think it was equilateral 💀

.cancel

[tropic oxide]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[tropic oxide]

(but also "b and c equal 5" is bad wording)

[winter tulip]

oh yeah you type .close

ok thanks

.close

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[coral lily]

I need help with these two. I don't really know how you solve cubic equations so I can't get around the first one. In the second one I tried to bring a common denominator but then I don’t know how to equalize the problem.

[scenic brook]

Okay. You can use Horner's method or factoring to solve first one

[wanton grove]

usually for cubics you try to get the first root then reduce to a quadratic that you can solve

[scenic brook]

In the second you need to first find a domain of function and then put all of it to the left side then general denominator then solve two equations upper side and bottom side make equal to zero. You'll find roots, you check the domain of function then

[coral lily]

I'll try to do it now, thank you for the input.

.close

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[river creek]

If A is a skew-symmetric n×n matrix, verify that adjA
is symmetric or skew-symmetric according to whether n is odd or even.

I am trying to prove it for n = odd, using the formula A* adj(A) = det(A) * I_n

[crimson sedge]

,rccw

[river creek]

i also used an other approach

but i crossed it out

,rccw

[river creek]

I tried this approach but I am missing a -

.close

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[cedar kiln]

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[nova echo]

hey what formula should i use to get the answer to 4?

[marsh lake]

I think F=ma = mv^2/r

m = 1000 v = 20 r = 100

[crimson sedge]

m is not 9800

thats the weight

[marsh lake]

oh

[crimson sedge]

w = mg -> 9800/9.8 = m

so 10^3

[marsh lake]

ohh

my bad didnt learn that yet

[nova echo]

how does ma=mv^2/r?

[marsh lake]

did you learn centripetal acceleration = v^2/r?

[nova echo]

ohhh

[marsh lake]

replace the a in F=ma with v^2/r

[nova echo]

thanks appreciate it

[marsh lake]

yw

[nova echo]

there's this question to

could you inform me with what formula i should use

[marsh lake]

conservation of momentum?

m1u1+m2u2 = m1v1 + m2v2

[nova echo]

alright thanks

.close

[cedar kiln]

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[night karma]

9x+6≡10(mod5)

is x = 1?

[split pike]

yes

[lost adder]

if x belongs to integer

9x=4(mod5)

5x+4x=4(mod5)

4x=4(mod5)

x=1(mod5)

i can't find congruence symbol in keypad

[split pike]

it isnt there

[night karma]

ooh okay I calculated it this way:
9x+6 = 10 (mod5) I-6
9x = 4 (mod5) I * 4
36x = 16 (mod5)
x = 1

so when I understood it correctly, basically on both sides is a % 5?

so 9x%5 = 4%5 is 9x = 4(mod5)?

[lost adder]

yes

[night karma]

ok great

.close

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[steady elk]

what is 1^i where i is iota

[steady elk]

is this a quantity wether real or complex

[lost adder]

https://math.stackexchange.com/a/1859299

[steady elk]

u r from before

[crimson sedge]

huh

so like 1^i power

where i is the imaginary unit

[silver fable]

did u learn complex logarithm ?

[steady elk]

is ee thats very nice and ez thanks

[steady elk]

no

whats that

[crimson sedge]

Any number x^i = cos(ln x) + i sin(ln x)

[lost adder]

can't ln x have infinitely many values

[steady elk]

but how will u account for negative x values

[lost adder]

Using polar form

[steady elk]

can u again link example

plz

[steady elk]

for this

[lost adder]

Any complex number can be written as r*e^(iθ)

ln(r*e^iθ)=ln(r)+iθ

[steady elk]

ohhhhh i see so moduli and this positive base remain positive for all numbers

[steady elk]

btw is this similar natural logarithm ?

[silver fable]

https://en.wikipedia.org/wiki/Complex_logarithm

[crimson sedge]

Think this is how itd work

[steady elk]

how come i times pi in second line

[silver fable]

$-1=e^{i\pi} \ -x=e^{i\pi}\cdot x$

[wraith dagger]

Adam Chebil

[steady elk]

ohkm ohk

ig then this must be complex log?

[lost adder]

why can't we take -1=e^(-iπ)

[crimson sedge]

??

[lost adder]

ln(-1)=iπ but also ln(-1) can be iπ+2iπk

where k is integer

[steady elk]

general solution

[crimson sedge]

^^^

[steady elk]

thanks

.close

[cedar kiln]

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[kind sparrow]

I am stuck on the second one, even with the explaination given, can someone help me out a bit :))

[quasi wren]

Place three balls randomly into four boxes. Let 𝑋𝑖 denote the number of balls in the 𝑖-th box {𝑖 = 1, 2}. Determine the joint probability distribution and marginal probability distributions of 𝑋1 and 𝑋2.

[quasi wren]

How to solve this question😭

[kind sparrow]

sorry mate could you please ask again in an unoccupied channel??

[quasi wren]

ok, my fault

[lost adder]

you calculated the reiman sum

here 1/n² is constant hence will come out of summation

the formula for summation is given

you just have to multiply it by 3/n² then take limit as n goes to infinity

[cedar kiln]

@kind sparrow Has your question been resolved?

[kind sparrow]

ty

[cedar kiln]

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[sand cradle]

Five animals are in the wrong cages. The signs above the cages show who actually should be in that cage. The animals can move left and right to the cages, but no two animals are allowed to ever be in the same cage. The first, third and fifth cage have an opening to the exterior area that they can enter, but no two animals are allowed there either. What's the minimum amount of moves to get the animals back to their cages?

[sand cradle]

One could also view the animals as 5 4 2 1 3. We need to make them consecutive and can only send the numbers on odd places to the exterior area.

[crimson delta]

try experimenting

[sand cradle]

I did, at 14 moves, I had N L G B T

That's 2 1 4 3 5, still doesn't look very good
At 19 moves, it was still not near the end result. And apparently, the minimum is 16 moves.
Beginning again, I can make it become 5 4 3 2 1 in some moves, but even though that looks better, it's even worse for this problem

.close

[cedar kiln]

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[crimson sedge]

<enumitem>
Determine the base of the numbers in each case for the following operations to be
correct:
\env{enumerate}{[a)]
\ii $\ff{67}5 = 11$
\ii $15 \cd 3 = 51$
\ii $123 + 120 =303$
}

[crimson sedge]

whats the idea for this exactly?

for the first one

you get 13 with a remainder of 2

for the division

[scenic brook]

Maybe it's about 16 or 8 number system idk

[silver walrus]

well 6 and 7 exist in this base
so the base number is definitely >=8
however moving 5 to the right hand side will get 67=55, which is impossible for any (positive integer) base

[crimson sedge]

[opal basin]

In some way base -2 works

[opal basin]

(6b + 7)/5 = 1b + 1

[crimson sedge]

those are the solutions

i am unsure on how you are meant to deduce that

or maybe i am misinterperting something

[silver walrus]

67_8=55_10

[opal basin]

They just took 11 as base 10 apperently

weird

[silver walrus]

UGH this question is bad, taking the equation's two sides differently

[tropic oxide]

let the base be b and then write out what each base-b notation means lol

[opal basin]

that got me base -2

[silver walrus]

15_7=12_10
51_7=36_10

Hmm the (b) part is right

[tropic oxide]

wait which one are we doing

[crimson sedge]

well i guess lets do it one at a time

for a(

i am guessing you are meant to interpert the right hand as decimal 10

and the left as the mysterious base?

[opal basin]

Yes, I assume that's a mistake in the exercise

In the other exercises all numbers are in the mistery base

[crimson sedge]

Oh what

thats dumb 😅

oh god im not sure if i should trust this book anymore

there are so many inconsistencies

[opal basin]

To solve them, you can write numbers as x*b^2 + y*b + z

where b is the mystery base

[crimson sedge]

hm

so

[
(67){b} = (55){10}
]

[crimson sedge]

[
a_0 + a_1 b^1 + a_2 b^2 + \dots + a_k b^k = 55
]

[crimson sedge]

wait uhhh

how do i use 67 here

[opal basin]

67_10 = 6 * 10 + 7

that's base 10, base b is similar

[crimson sedge]

oh i see

it just clicked

[
55 = 6\cd b + 7
]

[opal basin]

Yes

[crimson sedge]

okay

for b) and c) how am i meant to interpert both sides of the equation

[opal basin]

all numbers are in the mistery base

[crimson sedge]

okay so hm

what am i supposed to do then because i cant use a decimal expansion anymore

maybe like

[
51 = 1 + b\cd 5
]

[crimson sedge]

no this doesnt make sense

[
15\cd 3 = 1 + b\cd 5
]

[crimson sedge]

nope still not on the right track

any hints please?

[opal basin]

this, but apply it to all numbers

[crimson sedge]

okay so we dont know the base 10 number

[opal basin]

15 * 3 = 55, you have to replace each number

[opal basin]

none of them are base 10

[crimson sedge]

exactly

[silver walrus]

15_b*3_b=51_b
which is, (1b+5)*3=5b+1

$15_b\cdot 3_b=51_b$

[wraith dagger]

CollinGao-Original

[crimson sedge]

i was hesitant to do that

like can we be sure that the multiplication operator will function the same here with a different base?

[silver walrus]

The numbers are the same, the only things which will change are their representations in different bases

[crimson sedge]

wait this is a bit weird

because like regardless

[crimson sedge]

how can we get [
(55)_b = (51)_b
]
ever?

[crimson sedge]

am i under a misapprehension

[opal basin]

sorry, I meant 15 * 3 = 51

the 55 should have been 51

[crimson sedge]

okay okay lets see

[sacred grail]

oh god you're here

[crimson sedge]

[
(51)_b = 5b + 1 \
(15)_b = 1b + 5 \
(3)_b = 3
]

[crimson sedge]

so like

[sacred grail]

you've become a quadratic equation solver

[crimson sedge]

[
3b + 15 = 5b + 1
]

[crimson sedge]

b = 7

[crimson sedge]

[sacred grail]

i guess not yet

soon

,w (7+5)*3 = 35+1

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tearfox

[crimson sedge]

but wait how does this make sense?

like

they are base b not 10 so how can we like

decimal expand them

oh wait

[sacred grail]

where did you decimal expand them

[crimson sedge]

nevermind im kinda like high sorr

okay speaking of quadratics

The solutions to the quadratic equation $x^2 - 13x + 22 = 0$ are $x = 7$ and $x = 2.$ What is the base of the numbers?

like do i have to use the quadratic formula

and just like

idk

expand everything

[crimson sedge]

so like

[sacred grail]

you're actually solving this?

[crimson sedge]

what

[sacred grail]

set up the system of equations

[crimson sedge]

okay so like what do i have going

uh

(x-7)(x-2) = x^2-13x+22

[
\bs{(x-7)(x-2)}_b = \bs{x^2 - 13x +22}_b
]

[sacred grail]

vee eh tah

[crimson sedge]

wait do i solve it in base 10 first and like

equate them

[sacred grail]

you cant really do that if you dont know the base of the coefficients

[crimson sedge]

bro

okay

[sacred grail]

do you know vietas formulae

[crimson sedge]

yeah

its just that

all this base crap makes it 10 times more confusing somehow

[sacred grail]

just like

blind yourself to the numbers

theyre just there and unimportant until you get into the weeds of the calculations

[crimson sedge]

[
p + q = -\f ba \
pq = \f ca
]

[crimson sedge]

so like

[
\p{7+2}_b = \p{13}_b \
\p{7\cd 2}_b = \p{22}_b
]

[sacred grail]

uh

[ 7_b + 2_b = 13_b, \quad 7_b \times 2_b = 22_b ]

[sacred grail]

putting theround the whole thing is a bit goofy

[crimson sedge]

wait was it fine to like simplify -(-13/1)

[sacred grail]

1 is always 1

[crimson sedge]

right

and the negative is always the negative

fair enough

okay

7 + 2 = b + 3

b = 6

7*2 = 2b + 2

b = 6

[sacred grail]

sus

[crimson sedge]

why

[sacred grail]

you cant have 7 in base 6

[crimson sedge]

so are they uh

just

wrong

[sacred grail]

weird

its the only solution though

sus

[crimson sedge]

the answer key agrees

[sacred grail]

yeah sus

[crimson sedge]

this textbook is really

interesting

[sacred grail]

[crimson sedge]

okay i think i get this

but basically

change + to 0

change - to 1

add and if overflow fuck it off

if negative subtrahend complement the negative and add

thats it right

okay i think this is fine

okay okay okayt

i will escape the binary shit now

thanks snow, jelle, and collin

.close

[cedar kiln]

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[serene pond]

what is the difference between the rank of A and the rank of the augmented Matrix AIB?

[serene pond]

I don't undrestand how the rank can differ

[lyric widget]

well if A is a zero matrix and B isn't then you'd expect the ranks to differ

[serene pond]

if A=0 and B=(1,2,3) the I'd have p(A)=0 and p(AIB)=1?

[lyric widget]

if p denotes rank then yes

[serene pond]

okay

thank you!

[lyric widget]

no problem ^^

[serene pond]

.close

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[winter tulip]

I think answer isn't in the choices at all

[opal basin]

It is, why?

[winter tulip]

How

[opal basin]

what did you get?

What's (-3) - 8?

[winter tulip]

?

Diana's x axis is -3

x2 - x1 I guess

[opal basin]

no, its value

[winter tulip]

huh

[opal basin]

it isn't -5 or 5, as you wrote

[winter tulip]

what is it

[scenic brook]

I think it's B but it's rounded

You can use Pythagorean theorem

To find hypotenuse

[winter tulip]

ohh

So you include Paula's house aswell?

[opal basin]

(-3) - 8 = -11, that's where your mistake is

[winter tulip]

I thought just Jolina and Diana

[opal basin]

the rest of this is fine

[scenic brook]

Am i stupid

[winter tulip]

Oh okay

[scenic brook]

Why it's not 11²+3²

[opal basin]

It is

at least sqrt(11^2 + 3^2)

[winter tulip]

Wait why did I write 5

😂

[scenic brook]

yes.

[scenic brook]

Yes, it's strange

[winter tulip]

Oh god

[opal basin]

probably because 8 - 3 = 5

[winter tulip]

Oh okay I got 11.4 now

thanks guys

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[cobalt gulch]

Please explain me Paschel's Triangle and it's uses in Logical Reasonings.

[frosty flax]

Can you be a little more specific?

[cobalt gulch]

Explain how to use Paschel's Triangle in Logical Reasonings questions

my friend told me there is a way to use in logical reasonings

[idle tusk]

what do you mean by "logical reasonings questions"?

any example?

[cobalt gulch]

leave it

[idle tusk]

pascal's triangle is useful when you are doing questions on binomial expansions

[fair geyser]

by logical reasonings you clearly mean math

[frosty flax]

I mean, one can argue that solving problems involving the pascal's triangle can improve your logical reasoning. Not because of the triangle per se, but because your are solving math problems

[raw gulch]

in my opinion, a person wants to know how to use Pacal's triangle if, for example, he wants to expand a+ b to the 9th power, etc.

[split pike]

I would use binomial theorem

[raw gulch]

of course, but for many people in school they do not know or do not understand this theorem, and that is why they ask about Pascal's triangle, because it seems easy for them

[cedar kiln]

@cobalt gulch Has your question been resolved?

[cedar kiln]

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[crimson sedge]

i need help with logarithm graphs like these ones

[mighty shuttle]

Hint:- The inverse of a function is its reflection about x=y

[idle tusk]

log is an increasing function that has a 0 at x=1

from just that you can see that the answer must be c

[daring fossil]

can you explain it to me in a call please

[split pike]

Okay..

[cedar kiln]

@crimson sedge Has your question been resolved?

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[prisma nymph]

Can someone help me understand how to get the answer to this?

[remote salmon]

show your work

[prisma nymph]

ok

[prisma nymph]

I tried

I picked 3/2 cuz idk it looked similar

[next hornet]

why are you doing f(-1) and g(-1)

[remote salmon]

exactly

[remote salmon]

you need to put x+1 in f(x)

[next hornet]

You should replace x with x+1 in the original function

[remote salmon]

do you know how to do that?

[prisma nymph]

lemme try i think i know

What did I do wrong

oh the sign

oh ok

I got it

[remote salmon]

nice

[prisma nymph]

thank you

.close

[cedar kiln]

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[remote salmon]

no probs

[cedar kiln]

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[cedar kiln]

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[dawn basin]

doing part a first:

[heady lotus]

c crosses x axis when y=0

[dawn basin]

first i multiplied everything by root(x) then I squared both sides and got $x^2+9x+4$

[wraith dagger]

sealpup321

[dawn basin]

but i believe this is wrong as when i completed the square I got horrible answers

[dawn basin]

ye

[heady lotus]

maybe you muktiplied wrong

3x^(1/2)-x-2=0

3x^(1/2)=x+2

[dawn basin]

$3x^(1/2)-x-2=0$

[wraith dagger]

sealpup321

[heady lotus]

9x=x^2+4x+4

x^2-5x+4=0

x=1,4

gn

[dawn basin]

huh

thanks

[heady lotus]

giod night

[dawn basin]

wow

[heady lotus]

good night

[dawn basin]

good night

sleep well

[heady lotus]

bye

you too

close

[dawn basin]

...

.close

[cedar kiln]

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[minor crystal]

what do you think?

[fringe tree]

.close

[cedar kiln]

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[fringe tree]

nm

nvm

[minor crystal]

okay lol

[fringe tree]

That was the wrong screenshot

lol

@minor crystal

this is the one i meant to do

The negative X-direction.

[minor crystal]

that's what you think is the answer?

[fringe tree]

hmm

I guess I don't really understand it to me

[minor crystal]

what does positive x direction mean to you?

or negative x direction

[fringe tree]

hmm

what i learned is that value

values right of origin are positive

and the inverse for the negative

actually

so its positive sorry

that was stupid

[minor crystal]

okay then

[fringe tree]

I'm confident

[minor crystal]

moving right means moving in positive x direction

moving left means moving in negative x direction

make sense?

[fringe tree]

ASlright

Yes

[minor crystal]

which force is bigger?

[fringe tree]

The force to the right

?

[minor crystal]

yeah so if you were the mass

your stronger friend is pulling you to the right

your weaker friend is pulling you to the left

[fringe tree]

I see

[minor crystal]

which way would you move?

left or right?

[fringe tree]

Right

[minor crystal]

yeah

[fringe tree]

Thx

that make sense

What if

The forces of an object are balanced

[minor crystal]

then think about the y direction

what's happening

for this object in the y direction?

[fringe tree]

its not accelerating?

[minor crystal]

start off first by

defining what is positive and negative y direction

which forces are positive y? and which are directed in the negative y direction?

[fringe tree]

hm idk

[minor crystal]

is positive y direction up or down?

[fringe tree]

Up would b epositive

[cedar kiln]

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[near aurora]

Need help with 7. 63 and 7. 70

[opaque root]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[near aurora]

I don’t know where to start

.close

[cedar kiln]

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[slow jewel]

The product of two things is 0(in 7.063)

[slow jewel]

Nvm

[cedar kiln]

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[dire turtle]

can i prove that AOB = 90?

i know that CB || AD

with 90 degrees

[vernal minnow]

Taht depends..what do we know?

[dire turtle]

we knot that DC is a diameter

and DA and CB and AB are tangents

[vast pike]

hint (?):

[dire turtle]

?

[vast pike]

blue+blue+orange+orange=180

[vernal minnow]

yeah but do we know them to be equal?

[dire turtle]

?

oh

thats nice

why is blue equal blue

and orange equal orange

[vast pike]

because the triangle ABC and AB'C are congruent

AB=AB'

AC=AC

and they are both right triangle

so you can use pythagora to get BC=B'C

[vernal minnow]

oh wait

I was confuse but

if 2(blue+red)=180° <=> blue+red=90° which solves the question

yeah you're right

[dire turtle]

yes but why is blue equal blue

why are they equal

[vast pike]

yeah ...but what's the sloppy spet ?

[dire turtle]

why is angle DOA = AOE?

[vast pike]

kssenia i answered with this..is it clear ?

[dire turtle]

oh

alright

because its a kite?

[vast pike]

in that pic BC and B'C are tangents

[dire turtle]

yeah alright

[vast pike]

i don t know what is a kite in euclidian geometry

[dire turtle]

in a kite the main diagnol is an angle bisector

[vast pike]

but yeah ..i have just proved that they are congruent with basic tools

[dire turtle]

what about how to prove that AB = AD + BC

well based on the last section i they all are parallel

that means that AE = EB

from there its eeasy i think

AB = 2x

AD + BC = x + x = 2x

hence AB = AD + BC

[vast pike]

i don t think that this is true

[dire turtle]

this is

do u agree that AE = EB?

[vast pike]

sorry yeah it is true

[dire turtle]

alright

thanks

.close

[cedar kiln]

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[dire turtle]

hello

[vast pike]

here we are again

[dire turtle]

heheh

yeah

so AC is a tangent to one circle

and AD is a tangent to another

i gotta prove that EFG is an isosceles triangle

[vernal minnow]

this time , the right triangle is trivial!

(it's ACD)

[dire turtle]

ooo good one

i didnt see it

lets see how it helps

[vernal minnow]

wait no... I don't think it is

[dire turtle]

it isnt yeah

because that isnt a radius

that is connected to the tangent

[vernal minnow]

where is G constructed from?

[dire turtle]

angle ABC = ABD

[dire turtle]

EC and FD meet in point G

[vernal minnow]

and E and F are points on the circles which lie on the perpendicular to AB through A?

[dire turtle]

yeah

A straight line passing through point a intersects the circle at point F and the other circle at point E

[vast pike]

but AB is not perpendicular to EF isn't it ?

[dire turtle]

its not given in the question

[vernal minnow]

then E and F are arbitrary?

[silent finch]

This is just alternate segment theorem

[vernal minnow]

because if so, I guarantee there exist such E,F that EFG is isosceles

but that's not a very insightful statement

[dire turtle]

anyone has any ideas?

<@&286206848099549185>

<@&286206848099549185>

[vast pike]

by kappa's comment

[dire turtle]

yeah

[vast pike]

orange=orange

[dire turtle]

this is right

[vast pike]

then red=red

now ECBA and ABFD are cyclic quadrilateral

so angles F=E iff CBA=ABD

[dire turtle]

why

oh

ok

ahts right

i get it

[vast pike]

gg

[dire turtle]

thanks everyone

[dire turtle]

indeed

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[cedar kiln]

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[peak spruce]

can someone help me on this? I couldn’t figure out how to transform it to y/x form

[slow jewel]

instead of y, substitute it with vx instead

And instead of dy/dx, sub it with v+x dv/dx

[peak spruce]

so it should be something like this?

[slow jewel]

There'll be no dx on the right side in the last step

Just v+x dv/dx = xsinx + v

[peak spruce]

how did we get rid of it?

[slow jewel]

That's coz dy/dx is v + x dv/dx
And not just dy

[cedar kiln]

@peak spruce Has your question been resolved?

[cedar kiln]

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[hot sigil]

How can I find the remainder modulo 7 of a number of n digits only containing 1s?

[hot sigil]

I am noticing that it’s periodic with n between 1 and 6 but I can’t prove it

[opaque root]

Do you mean 11 mod 7, 111 mod 7, etc?

[hot sigil]

Yes

[opaque root]

What have you tried?

[hot sigil]

I figured that if $u_n$ is the remainder, then it would be congruent to $3u_{n-1}+1$

[wraith dagger]

BeeReallyYum

[opaque root]

Try looking at the numbers in this way

For example, 11 would be = 10^1 + 10^0

111 = 10^2+10^1+10^0

this way you can prove it

you only need to know that 10^1 mod 7≡3; 10^2 mod 7≡2; 10^3 mod 7≡6

etc...

[hot sigil]

Oh and I can use euler’s theorem here

To prove that it’s periodic right

Because after 3^6 it would just repeat

[opaque root]

after 10^6 mod 7≡1

[hot sigil]

Yeah precisely

[opaque root]

10^7 would be 3 again

[hot sigil]

And 10^6 is congruent to 3^6 anyway so it makes calculations easier if done by hand

Could you just tell me how to write it please?

To prove that it’s periodic

I’m thinking I could write Sum of 10^k with k from 0 to n-1

And then how would I say it repeats

[opaque root]

a^(ϕ(n)) ≡1 mod n

if a and n are coprimes

so ϕ(7) = 6

that'd mean every 6 times

it restarts

[hot sigil]

Okay got it

Thank you so much!

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[cedar kiln]

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[dawn talon]

I've been given a task in my university and I have a hard time understanding Lagrange's theorem application and how it works. I need to find all non trivial cyclic subgroups of a multiplicative whole number group of order 16 <Z17/{0}; x>. How do i find and list all subgroups. Can I assume that the group itself is cyclic because 17 is a prime number, thus all of the subgroups are going to be cyclic?

[solid juniper]

whether you can assume the group is cyclic depends what results you’re allowed to use, but maybe

[dawn talon]

I believe I have to actually list all the cyclic subgroups as the task requires me to "find them"

[solid juniper]

yea

[dawn talon]

The issue is that I am shooting blanks here. What is the basic concept of finding subgroups in general?

[crimson delta]

look at what can generate them

[dawn talon]

So finding all the generators of the given group and take the generators raised by n where n can be either 2, 4 or 8 ?

All of the generators raised by either 2, 4, 8 are going to be all of the subgroups?

[crimson delta]

why those

[dawn talon]

Because they are divisors of 16, no ?

[crimson delta]

why does that matter

[dawn talon]

I don't know

Could you please explain what should I do and why?

I am trying to figure out all the subgroups of a Z5/{0} multiplicative group as an example, its generator is 2. Looking at the operation table, I can notice that it has 2 non trivial sub groups, which are - {1; 4} and {2; 3}. I can't notice a pattern how to get these 2 subgroups from generator itself, as raising generator, by 1 gives 2, by 2 gives 4, by 3 gives 3 and raising it by 4 gives 1. So the subgroups are generated by raising the generator by either 2 and 4 or 1 and 3.

How does that apply in much larger groups like the one i have?

<@&286206848099549185>

[random lintel]

In general, for a cyclic group of order n with generator g, the subgroups are generated by raising g to powers that are relatively prime to n, forming cyclic subgroups of various orders.

do get the point?

[dawn talon]

I get the idea. So if I take my original Z17 multiplicative group and it's generator 2, i need to raise it to powers 3, 5, 7, 9, 11, 13 ?

[dawn talon]

What then? Could you give a specific example?

[random lintel]

Yes, for the Z17 multiplicative group with generator 2, the subgroups are generated by raising 2 to the powers 3, 5, 7, 9, 11, and 13, as these powers are relatively prime to 17.

[random lintel]

For the Z17 multiplicative group with generator 2, one subgroup is generated by raising 2 to the power of 3, which gives 1, 2, 4, 8, 16, 15.

I hope my information helps you!

[dawn talon]

Shouldn't the subgroup only contain either 1, 2, 4 or 8 elements as per Lagrange's theorem, but you listed 6 elements, also how does raising 2 to the power of 3 yield 6 elements?

[random lintel]

ur correct I apologize

[random lintel]

Raising 2 to the power of 3 in Z17 gives 8, and the resulting subgroup is 1, 2, 4, 8, 16, 15, because these are the powers of 2 modulo 17, wrapping around to 15 instead of 9. The correct subgroup generated by 2^3 is 1, 2, 4, 8.

[cedar kiln]

@dawn talon Has your question been resolved?

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[silver fable]

what is the informal definition of f ∼ g (intuition)

[silver fable]

what does it mean exactly ?

.close

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[cedar kiln]

Please don't occupy multiple help channels.

[fringe tree]

.close

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[livid swan]

i think i found x but idk how to find h

[nimble veldt]

really? you know x but you do not know h?

if R is at the high position, where is P? if R is at the low position. where is P? whats the difference?

[cedar kiln]

@livid swan Has your question been resolved?

[livid swan]

sorry i was afk

[livid swan]

The thing is there is no low/high position technically

lets say the piston travels down

P will be on the other side of the circle in the same position

the difference is 2r in terms of distance

the displacement would be 2sin(1)

(in radians btw not degrees)

[nimble veldt]

there is no low/high?

[livid swan]

idk if u know what i mean or not

my bad i was thinking about the position of P

R cannot be in the high position the size of the piston is between R and the high position, the piston is not a 1d shape, it's a 2d rectangle

the piston is already in the high position

that is the max height R can go

im pretty darn sure

so R in that position, leaves P to be where it is in the diagram

if R is in the low position P will be on the same level/height but on the other side of the circle

(assuming the wheel moves in an anticlockwise direction)

regardless of rotational direction the distance between P given R is high and P given R is low is the same... 2r

I have substituted r for 1

so it resembles the unit circle

thus 2PQ=C(2/2pi)=C(~0.32)

[livid swan]

ig if r is in the high position P would be at the highest point of the circle

i tink

then if r is at the bottom P would be at the bottom of the circle

are you telling me that the piston is just a 1d line

[livid swan]

2r

or d

[nimble veldt]

so what is h?

[livid swan]

2r

[nimble veldt]

👍

[livid swan]

but hang on, I must know one thing

did you interpret the piston as a line or a rectangle

[nimble veldt]

in your sketch the piston is symbolized by the point R and the line RP symbolizes the connection rod.

[livid swan]

i know how a piston works and looks like

double edged sword though

predisposed to think the cylinder was realistic, and was a rectangle

alright thanks man

👍

Now i cant find x again

the answer is that x is 17% of h

im getting 23%