help-39

[vital estuary]

eh wouldnt work

[brittle tinsel]

well if we get that the sum is neither 0 nor 1 mod 4, it's done

otherwise yeah it's an issue

it's clearly not 0

[vital estuary]

you can use mod 4 to disprove odds and mos 3 to disprove evens

the only different case is n=1 becuase 2^n !≡ 0 mod 4

[sharp smelt]

so I find 4mod(2^n+7^n)?

I don't get any of this

[vital estuary]

no the other way around

[sharp smelt]

$2^n+7^nmod(4)$

[jolly parrot]

Why am. I here

[vital estuary]

yes

[sharp smelt]

which is $(2^n)mod(4)+ 7^n(mod4)$

[jolly parrot]

Why am. I here

[brittle tinsel]

👍

[vital estuary]

yes

[sharp smelt]

hmm, and $(2^n)mod4 =0$ for all n>1

[jolly parrot]

Why am. I here

[vital estuary]

yup

[brittle tinsel]

well this doesn't seem like it'll work

[vital estuary]

it will

trust the process

[sharp smelt]

now to find $7^nmod(4)$

[jolly parrot]

Why am. I here

[sharp smelt]

uh, 7mod(4)=3

$7^2mod(4)=1$

[jolly parrot]

Why am. I here

[vital estuary]

yes

[sharp smelt]

$7^3mod(4)=3$

[jolly parrot]

Why am. I here

[sharp smelt]

so I guess I prove 7^nmod(4) is never 0

[vital estuary]

thats not enough

[brittle tinsel]

ah I see what you can do

carry on

[vital estuary]

lets go over quadratic residues real quick

there are certain values which cannot be squares modulo a certain modulus

e.g. 2(mod4) is never square

another example is 3(mod4)

:O

if n is odd, what is 7^n(mod4)

[sharp smelt]

3?

[vital estuary]

yup

you can tell which are quadratic nonresidues by casework

in the modulo 3 world, what are the quadratic residues?

[sharp smelt]

modulo 3 is basically $nmod(3)$, right?

[jolly parrot]

Why am. I here

[vital estuary]

yes

mod and modulo are essentially the same

[sharp smelt]

hmm, 2mod(3)=2

4mod(3)=1

5mod(3)=2

7mod(3)=1

and so on

[sharp smelt]

when n is odd 1, when n is even 2

[vital estuary]

ok

thats good!

but thats not the qiadratic residues

[sharp smelt]

when n is 3a, it's 0

[vital estuary]

for quadratic residues, you only need to check 0^2, 1^2, and 2^2 mod 3

[sharp smelt]

wait, what?

[vital estuary]

quadratic as in squares

[sharp smelt]

yeah, I get that

[vital estuary]

what are the possible squares mod 3?

[sharp smelt]

0

1

4

[vital estuary]

4 mod 3 is also?

[sharp smelt]

1mod(3)?

[vital estuary]

yes

notice how it is never 2 mod 3

this is a quadratic __non__residue

in other words, cant be a square number

[sharp smelt]

hmm, ok a number is a perfect square if its quadratic residue is 0?

[vital estuary]

no

a number can be a perfect square if it is congruent to a square number modulo some number n

a number cannot be a perfect square (at all) if it is not congruent to a square number modulo some number n

[sharp smelt]

,w congruent in number theory

ok, what does that mean?

[vital estuary]

same remainder

eg 1, 4, 19999999 are all congruent mod 3

[sharp smelt]

huh,ok

[sharp smelt]

why is this?

[vital estuary]

thats kinda how congruence works idk what to say

[sharp smelt]

huh,ok

[sharp smelt]

so I now have to prove that $7^n+2^n$ isn't congruent to what?

[jolly parrot]

Why am. I here

[brittle tinsel]

depends on the modulus, but i would suggest trying modulo 4 first

[sharp smelt]

$(7^n+2^n)mod(4)$?

[jolly parrot]

Why am. I here

[vital estuary]

yes

[sharp smelt]

ok, so there are two cases, for n>1

when n is 2a, it's 2

when n is (2b+1) but not 3c it's 1

[vital estuary]

?

[brittle tinsel]

what are a b and c

just integers?

[sharp smelt]

yeah

[vital estuary]

i think youre mixing modulos and expressions and moduli

so your 2 cases are
n is odd
n is even

what happens to the expression mod 4 in each case?

[sharp smelt]

nmod(4) when n is even is either 2 or 0

nmod(4) when n is odd is either 1 or 3

[vital estuary]

when n is odd
what is 2^n + 7^n (mod 4)?

[sharp smelt]

if $n \neq 1$ then 3

[jolly parrot]

Why am. I here

[vital estuary]

good

is 3(mod 4) a quadratic residue?

[sharp smelt]

that means 9mod(4)

right?

[vital estuary]

no

are any square numbers congruent to 3 mod 4?

[sharp smelt]

yeah

9mod(4)

right?

[vital estuary]

what is 9 mod 4

[sharp smelt]

1

oh

3mod 4 is 3

[vital estuary]

yes

[sharp smelt]

I don't think so

[vital estuary]

no

you can check 0^2, 1^2, 2^2, 3^2

its either 0 or 1

from 4^2 on it just repeats

[sharp smelt]

oh,ok

[vital estuary]

so can n be odd if we want 2^n+7^n to be a square number?

[sharp smelt]

no, idts

[vital estuary]

no, because it is not a quadratic residue mod 4

[sharp smelt]

ah, I see, thanks

[vital estuary]

what about even n?

[sharp smelt]

I use the same logic, I think

[vital estuary]

what does the expression become?

[sharp smelt]

(7^n+2^n)mod(4)

[vital estuary]

ok

keep going

[sharp smelt]

so when n is even

2^nmod(4) is 0

7^nmod(4) is 1?

[vital estuary]

yes

since 1 is a quadratic residue, the test is inconclusive

lets try a different modulus

(modulus was 4 last time because it was the n in mod n)

what about 3

lets do odd n and even n again

what is 2^n+7^n mod 3 for even n?

[sharp smelt]

2^nmod(3) for even n is 1

[vital estuary]

good

[sharp smelt]

7^nmod(3) is 7?

atleast for n=2

[vital estuary]

7 mod 3 can be simplified

[sharp smelt]

what, how?

[vital estuary]

what is the remainder when 7 is divided by 3

[sharp smelt]

1

[vital estuary]

yes

so in total the expression is?

[sharp smelt]

2?

[vital estuary]

yes

[sharp smelt]

so that proves that the only sol is 1?

[vital estuary]

what are the quadratic residues mod 3?

[sharp smelt]

hmm, so I have to find (2^n+7^n) mod3?

[vital estuary]

what

[sharp smelt]

that's what this means, right?

[vital estuary]

no

what are the possible values of square numbers mod 3?

[sharp smelt]

I'm really confused

[vital estuary]

quadratic residues is a fancy way of saying possible square numbers

[sharp smelt]

so I square 2 here, right?

[vital estuary]

where did the two come from

[sharp smelt]

here

[vital estuary]

no

2^n + 7^n = 2 (mod 3)

now we look at all possible square numbers mod 3 and check if they are equal to 2

[sharp smelt]

hmm, ok

9^2mod(3)=0, so this is obviously false

or even 3^2 mod 3 for that matter

[vital estuary]

you have to check all square numbers up to 3

up to 3^2*

[sharp smelt]

4mod(3)=1

1mod(3)=1

[vital estuary]

ok

so are any square number congruent to 2 (mod 3)

[sharp smelt]

ok, yea

[vital estuary]

yea?

[sharp smelt]

understood

thanks

so this proves this, right?

[vital estuary]

what does it prove

[sharp smelt]

I'm going off this

[vital estuary]

ok

[sharp smelt]

so based on this, no even n satifies the given statement either, right?

[vital estuary]

because?

[sharp smelt]

a number cannot be a perfect square (at all) if it is not congruent to a square number modulo some number n

so for even n(s) this clearly isn't true

[vital estuary]

youre so close to finishing it off

[vital estuary]

even n with what modulus

[sharp smelt]

modulo 3?

[vital estuary]

yes

what about odd n?

[sharp smelt]

modulo 4, right?

[vital estuary]

yes

what is 1 an exception?

[sharp smelt]

why did you chose 3 and 4 though?

[vital estuary]

why*

[vital estuary]

3 and 4 are usually useful with quadratic residues

also theyre small numbers so theyre easy to work with

[sharp smelt]

In terms of modulos?

[vital estuary]

yes

[sharp smelt]

hmm, (2^n+7^n)modulo(3) is 0

as is (2^n+7^n)mod4

[vital estuary]

2^n+7^n when n=1 is congruent to what mod 4?

[sharp smelt]

the reminder is 1

[vital estuary]

why isnt it 3 like the rest of the odd ns?

[sharp smelt]

not sure

[vital estuary]

think about the two terms, which one might act different?

[sharp smelt]

2^n mod 4

2 mod 4

as 2<4

[vital estuary]

its more like 4 does not divide 2

[sharp smelt]

ah, OK. Got it

thanks

[vital estuary]

yeah thats why 1 is an exception

youre welcome!

this mightve been a lot to take in

[sharp smelt]

it is, i'll go though this again to revise it soon

thanks!

can I close this now?

[vital estuary]

sure

[sharp smelt]

tysm for the help again!

.close

[pearl pond]

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[vital estuary]

youre welcome

[pearl pond]

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[rigid trench]

the answer should be 400 but idk how to solve it

[brittle tinsel]

I would recommend making a table, then filling it in with what you know

here, N stands for national, P stands for private, and ~ means not

try filling in this table with what you know

[pearl pond]

@rigid trench Has your question been resolved?

[brittle tinsel]

btw this question actually has multiple solutions, 400 is just the smallest

bad question

[pearl pond]

@rigid trench Has your question been resolved?

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[severe magnet]

for this problem would i do -4 = a (-4+1)^2 (x-2)^3 (x-4)?

[midnight haven]

you can do [
-4 = a(0+1)^2(0-2)^3(0-4)
]
rather

[severe magnet]

ohh okay i see thank you i thought i would also plug in the y intercept to the x for some reason

.close

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[quick hinge]

I just want to corroborate if this is correct

[stoic path]

,w differentiate 1/(4cos^2 (x^2 +2x))

[jolly parrot]

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Requested by masked.sniper

[stoic path]

It's right, yeah

But there should be a +c in the final ans

[pearl pond]

@quick hinge Has your question been resolved?

[pearl pond]

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[reef quiver]

want to prove this lemma in my notes, just the first equality. the inequality is clear

[reef quiver]

here is what i have

idk what algebraic trick or reasoning to use

wait can we do this?

group the i terms and j terms together

[toxic fractal]

yes

[reef quiver]

great!

.close

[pearl pond]

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[autumn ravine]

How do u get from the last bit to ((b²-k)/a)<a

[pearl pond]

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[midnight haven]

Is this meant to be a typo by any chance?

kind of weird

[pearl pond]

@midnight haven Has your question been resolved?

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[daring wedge]

n men are to be seated around a circular table. Find the probability that two particular men sit together

[daring wedge]

@worldly glacier

still remember permutations?

or combinations wtv idk

[vestal tapir]

you don't need to calculate much

and you can't

since n is not given

[daring wedge]

so wat do i need to do?

[vestal tapir]

i don't know

i can only tell the answer

[daring wedge]

whats the ans?

[vestal tapir]

2/(n−2)

[daring wedge]

@midnight haven sister pls help

[daring wedge]

what

are you sure

thats the correct ans?

[vestal tapir]

2/(n−1)

[daring wedge]

oh wow bruh you smart

[vestal tapir]

there's n−1 seats for the second man

[daring wedge]

i just checked the answer

[vestal tapir]

2 of them are ok

[daring wedge]

its correct

but still i need to write the whole solution 😭

[vestal tapir]

unfortunate

[daring wedge]

dw ill figure smth out

ig no use writing the whole problem in here

ill have to do it myself

.close

[pearl pond]

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[cerulean linden]

How to do range reduction on 2^x in 0 <= x < 1?
I'd like to implement this in a program, and taylor series works very well on nearzero.
I came up with a normal range reduction with lookup table using some top bits, (because 2^x * 2^y == 2^(x+y), I can lookup the result of bigger x and calculate taylor series of smaller y then multiply together.) but I'm wondering if there are other ways.

[pearl pond]

@cerulean linden Has your question been resolved?

[pearl pond]

@cerulean linden Has your question been resolved?

[pearl pond]

@cerulean linden Has your question been resolved?

[cerulean linden]

<@&286206848099549185> hi!

[midnight haven]

i was asleep lol

[latent quail]

Now you’re awake

[midnight haven]

no im not

[latent quail]

Same

[midnight haven]

fr

[pearl pond]

@cerulean linden Has your question been resolved?

[pearl pond]

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[worthy hound]

im not sure how to find the general solution here

[midnight haven]

why cant you directly integrate it from here?

(y/y+1) y' = (1-1/y+1)y'

integrating this gives : y - ln(y+1) = x^2 + C

[worthy hound]

well i dont know how

[midnight haven]

I did it above. Any qs about that?

[worthy hound]

where did we get 1-1

[midnight haven]

y = (y+1 -1)

[worthy hound]

wat

[midnight haven]

just add and subtract 1 from y

y = y+1 - 1 = (y+1) - 1. Now dividing by (y+1) we get 1 - 1/(y+1)

[worthy hound]

yeah i dont follow, its hard reading text like this and trying to put it into an image

i got lost at the 4th line

[midnight haven]

Idk how to use latex on discord sorry.

[worthy hound]

where does the 1- blah blah come from

[midnight haven]

So look at the numerator. It's y right?

Whereas the denom. only differs by the numerator by 1. So just add and subtract the 1 in the numerator

So y = y+1 -1

Agree?

[worthy hound]

but why

is taht how u integrate?

[midnight haven]

I've not even gotten to integration yet

Do you agree that y/(y+1) = 1-1/(y+1)

[worthy hound]

no

i think thats where i got lost

[midnight haven]

Ok

So let's focus on that.

y = y+1-1

yes?

[worthy hound]

well how do we go from this to what you had

[midnight haven]

Ok let's do this. Call y+1 = z

Hence y/(y+1) = (z-1)/z = 1 - 1/z

agree?

[worthy hound]

y/(y+1) = (z-1)/z

agreed

but this part ??

1 - 1/z

[midnight haven]

(a+b)/c = a/c + b/c right?

[worthy hound]

yes

oh

i see now

so how shal we continue this

[midnight haven]

Ok so we're at (1-1/(y+1))y' = 2x dx right?

[worthy hound]

agreed

[midnight haven]

So now we're going to integrate this

$$\int dy - \frac{dy}{y+1} = \int 2x dx$$

[jolly parrot]

swimmingland

[midnight haven]

$$y - \log(y+1) = x^2 + C$$

[jolly parrot]

swimmingland

[worthy hound]

there are 2 dy's?

[worthy hound]

i agree with this

but natural log right

[midnight haven]

yes

[midnight haven]

you're just splitting the dy over the two terms in (1-1/y+1)

[worthy hound]

okok

how shal we isolate for the y now, there are 2 of them

[midnight haven]

You can't

That's the best you're getting

[worthy hound]

ah

so i am done

[midnight haven]

yup good job haha

[worthy hound]

thank you

.close

[pearl pond]

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[stark basin]

Simplify each expression.
13. sin θ sec θ

[pearl pond]

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[alpine elbow]

a+b (they are whole numbers) can be divided by 7
prothe that
a^7+b^7 can be divided by 49

[alpine elbow]

i proved that it can be divided by 7

[midnight haven]

use fermat's little theorem

[alpine elbow]

a+b = 0 (mod 7)
a = -b (mod 7)
a^7 = -b^7 (mod 7)
a^7 + b^7=0 (mod 7)

[alpine elbow]

huh

ooo

so
a^7 = a(mod 7)
and
b^7 = b(mod 7)
a^7 +b^7 = a+b (mod7) = 0 (mod 7)

but we need it to be divided by 49

@midnight haven

[pearl pond]

@alpine elbow Has your question been resolved?

[alpine elbow]

<@&286206848099549185>

<@&286206848099549185>

[fallen pilot]

i have trouble whit my times tables

[pearl pond]

@alpine elbow Has your question been resolved?

[pearl pond]

@alpine elbow Has your question been resolved?

[pearl pond]

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[sharp minnow]

was i wrong or was this question wrong

[plush bramble]

Show your work

[sharp minnow]

I did it again still got

[plush bramble]

Limits should be from 0 to pi

Trace it out and see

[sharp minnow]

you mean sketch the function?

[plush bramble]

Yes

Or maybe your answer is only for one of the petals

[sharp minnow]

Thats what my other friend told me

Cus only pi/2 and 3pi/2 = 0 for cos from 0 to 2pi

Idk how to sketch the function

Thx anyways

.close

[pearl pond]

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[feral owl]

New to scatter plots. Can someone help me interpret those scatter plots? Seems like there is no relationships?
X axis is electrical power
Y axis is gas consentration
Color are different temperatures

[regal herald]

would probably be of value to isolate the temperatures

there does seem to be a rather strong concentration on the 50ppm line

certainly for pink and gold

[feral owl]

So like no colors or like make them into ranges like 28-30 degrees

[regal herald]

yeah, split it into a number of graphs

[feral owl]

Okay, ty.
Yes, at 50 ppm there is most data points together but its quite linear

Idk can i like make some correlation or is it just coincidence. Between 6-8k power H2 in most rapidly generated?

Okay i will look more into it and ask again if i have further questions

Thank you

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[vivid pilot]

i was performing u-substitution on an integral and got this $\int^5_1{\frac{1}{\sqrt{u-1}}\sqrt{u}\mathrm{d}u}$

[jolly parrot]

talk_less

[vivid pilot]

and im honestly not too sure what to do with this lol

this isnt really much easier to integrate

[regal herald]

what was the original integrand

[vivid pilot]

$\int^4_0{\sqrt{1+(\frac{x}2)^2}{\mathrm{d}x$

[jolly parrot]

talk_less
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vivid pilot]

and i let $u=1+(\frac{x}2)^2$

[jolly parrot]

talk_less

[regal herald]

do you by any chance know hyperbolic functions

[vivid pilot]

no

[acoustic path]

would do this integral using trigonometric substitution tbh

[vivid pilot]

whats that?

i dont think i learned that yet

[acoustic path]

oh u don't know

[vivid pilot]

i only know u-sub

@regal herald is it even possible with usub

[regal herald]

would definitely be easier with a hyperbolic sub like x/2=sinh(u) or something

its a bit messy otherwise

[vivid pilot]

im not sure on what exactly to do though

[rare widget]

u=x/2, then use the substitution u=tan(v) and simplify with trig identities

using hyperbolic functions would be easier if uve learned them

trig sub works similarly to usub you write du in terms of v

[vivid pilot]

what is v

oh i see

thank you!

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[upbeat lily]

hello help me pls 🥲

[upbeat lily]

the pole is 50 feet but idk how to get the distance from the middle line to the top/bottom of the pole

if that makes sense?

this is what I have so far

wait hold on

there

[unkempt yacht]

tell me whats the distance from the pole's middle point to its top?

[upbeat lily]

no clue

it's not 25 which is what i tried

LOL

[upbeat lily]

oh hey u also like vocaloid

cool

[unkempt yacht]

[unkempt yacht]

but anyway

[unkempt yacht]

for easier reference, im going to label A as the person's POV, B is the pole top, C is the pole bottom and D is the middle point

can you tell me what is triangle ABD?

[upbeat lily]

like the angle?

72º

[unkempt yacht]

no, im asking what kind of triangle is it

[upbeat lily]

ohhh

right triangle

i think that's the english name

😭 idk proper terminology i take math in spanish

[unkempt yacht]

yes

[upbeat lily]

awesome

[unkempt yacht]

so based on that, how do you calculate BD?

[upbeat lily]

tangent?

tan(72)

something is missing though no?

[unkempt yacht]

thats right

but just do it

[upbeat lily]

3.078

[unkempt yacht]

no

dont use decimal

[upbeat lily]

ohh

[unkempt yacht]

keep using tan

so again, what is BD?

[upbeat lily]

so just tan(72) for now

tan(72)

[unkempt yacht]

no

[upbeat lily]

oh

[unkempt yacht]

multiplied by what

[upbeat lily]

ohhh

byyyyyyyy uhhhhhhhh

by D?

[unkempt yacht]

no lmao

[upbeat lily]

😭

[unkempt yacht]

recall how do you calculate tan of an angle

[upbeat lily]

it's gonna be so obvious and im gonna laugh at myself

tan=opposite/adjacent

[unkempt yacht]

right

[upbeat lily]

so tan(72)=

OH

OHIOH

OHHHHH

[unkempt yacht]

so im gonna ask again

whats BD?

how do you calculate it

[upbeat lily]

yeah ignore this i thought i got it but no

is it tan(72)(18)?

im so lost can u tell

💀

[unkempt yacht]

still no

how do you calculate tan72 in this case?

[upbeat lily]

but it does have to do w the angles

?

[unkempt yacht]

like you said, its opposite/adjacent

so for tan72, whats the opp, whats the adj

[upbeat lily]

i dont have any of the side measurements 😭

would it be

x/something

[unkempt yacht]

just do it

[upbeat lily]

x/whatever the blue line measures

[unkempt yacht]

the opp is x yes

and the adj is BD

[upbeat lily]

oh mhm

so

[unkempt yacht]

so tan72= x/BD therefore BD=...?

[upbeat lily]

tan(72)(BD)

[unkempt yacht]

no

try again

[upbeat lily]

okay!

uh

ghf3eoiwrho3eiwhfqio43re

[upbeat lily]

tan(72)(x)?

[unkempt yacht]

still no

dude

$\frac{x}{BD}=\tan72$

[jolly parrot]

FungusDesu

[unkempt yacht]

so BD=...?

[upbeat lily]

x/tan(72)?

[unkempt yacht]

thats right

[upbeat lily]

LETS GOOOOOOOOOO

[unkempt yacht]

for now, we do not care what x be

[unkempt yacht]

by the same process, can you find CD?

[upbeat lily]

x/tan(76)

[unkempt yacht]

thats right

now, what does BD+CD equal?

[upbeat lily]

2x/tan(76)tan(72)?

[unkempt yacht]

wrong

[upbeat lily]

2x/tan(76)+tan(72)?

[unkempt yacht]

put that aside, do you know what does BD+CD represent?

[upbeat lily]

wait

is it the 50?

[unkempt yacht]

yes

[upbeat lily]

oh awesome

[unkempt yacht]

its the height of the pole

so now

$BD=\frac{x}{\tan72}\newline CD=\frac{x}{\tan76}$

[jolly parrot]

FungusDesu

[unkempt yacht]

BD + CD=50

therefore...?

[upbeat lily]

x/tan(72) + x/tan(76) = 50

[unkempt yacht]

thats right

now, find x

[unkempt yacht]

once you found x, substitute it to this system of equations

you shall get the measurements of blue and green segments of the pole

[upbeat lily]

i got

308.598

for X

[unkempt yacht]

wrong

try again

[upbeat lily]

dang okay

how do u add tangents

omg

LMAO

[unkempt yacht]

well NOW i would recommend writing tan as decimal instead

[upbeat lily]

yeah i did that just now

what i did b4 was

just

move both tangents multiplying

which

clearly was wrong

hey r u sure it isnt 308.598 🧍‍♂️ i got that answer again

[unkempt yacht]

show your work

[upbeat lily]

x/tan(72) + x/tan(76) = 50
x/3.077 + x/4.011 = 50
4.011x/12.3439 + 3.077x/12.3439 = 50
7.088x/12.3439 = 50
7.088x = 50(12.3439)
7.088x = 617.1957
x = 617.1957/7.088
x = 87.0761

is that right

[unkempt yacht]

now youre correct

[upbeat lily]

YAY

[upbeat lily]

I'd forgotten to add the numbers multiplying the Xs

i just did 2x instead of 7.088x

💀

yay it was correct!!

thank u sm for ur help

i have a test on this . tomorrow 💀

[unkempt yacht]

good luck

!done

[pearl pond]

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[upbeat lily]

.close

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[fallen rose]

I need help on question 13

I havent been taught this since i missed a day

of school where they learned the component method

<@&286206848099549185>

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[green fulcrum]

calc 2

[green fulcrum]

Sketch the region enclosed by the given curves and find its area.

idk how to procceed with this

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[topaz urchin]

(game theory)

anyone know how and why the decided P_X = 1/3 and P_A = 2/3 ?

i understand everything else

[pearl pond]

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[pearl pond]

@topaz urchin Has your question been resolved?

[pearl pond]

@topaz urchin Has your question been resolved?

[pearl pond]

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[delicate dock]

is there an algorithm I can follow when simplifying boolean expressions like this:

ac + b’c’ + a’c’ + bc

I'm stuck and I don't know how I can proceed to simplify this

[cosmic charm]

k-maps maybe?

[delicate dock]

nope we're not there yet

[cosmic charm]

i see

[pearl pond]

@delicate dock Has your question been resolved?

[cosmic charm]

yeah i also dont know how to simplify this only using boolean algebra, unless you do some ingenious stuff with it

easiest way is to just use a kmap

but even after using a kmap it just reduces it down to 3 terms

and 2 of the terms are same as the original

[delicate dock]

hmm okay

thanks

[pearl pond]

@delicate dock Has your question been resolved?

[pearl pond]

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[edgy jetty]

hii the answer says its 4r^3/3s^2 but i keep getting it wrong

[midnight haven]

so 12/9 is 4/3 yes? now collect the powers of r and s

r^4/4 = r^3

s^6/s^{-8} = s^{6+8} = s^14

so it should be 4/3 r^3 s^14

[edgy jetty]

umm

[midnight haven]

if it's saying 1/s^2

[edgy jetty]

i got s^14 but it the book says otherwise 😭

[midnight haven]

then the denominator they've written is incorrect and it should be 9rs^8 instead of 9rs^{-8}

[edgy jetty]

hmmm

soo do i just put s^14?

[midnight haven]

if the q is correct then yes

if the q has a typo and the s^-8 is supposed to be s^8 then their answer would've been correct

[edgy jetty]

ahh okayy

also just realised its you again 😭 hii

thanks againnn

for the third time

[midnight haven]

haha hii

and you're very welcome again 🙂

[edgy jetty]

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[fossil lodge]

How would I go about showing that everything in $Z_{47}^*$ has a multiplicative inverse?

[jolly parrot]

WhoTao

[fossil lodge]

So $Z_{47}^* = {1,2, \dots, 46}$.

[jolly parrot]

WhoTao

[fossil lodge]

Since 47 is a prime number, then there is a result which says an integer $a$ has inverse modulo $n$ iff gcd(a,n) = 1

[jolly parrot]

WhoTao

[fossil lodge]

47 is prime. so gcd is going to be 1. But how can I show this result without using the thm

[plush bramble]

find an inverse for each number.

[fossil lodge]

sounds good.

ty

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[lunar scaffold]

Need help with exercise 2

[midnight haven]

you mean ex2 part 2?

[lunar scaffold]

Yesyes

[midnight haven]

its discontinous

[lunar scaffold]

I don't get it

Why

[midnight haven]

because the remaining values are all in the 220's range

whereas you have a sudden break

in the 3rd hour

because of value 0

i suggest plotting the 5 points

[lunar scaffold]

And then?

Wait, I'm sending a picture

It's my graph

[midnight haven]

sorry not sure, i'm just saying plot the 5 points, you'll see that the 3rd hour is a break

and hence a discont.

[grizzled dust]

kinda silly to talk about continuity when all that's given is 5 discrete data points and you know nothing about the behavior of the function in between anyway

[midnight haven]

yeah i know

but i figured that this is the intent

[grizzled dust]

yeah guess so

[lunar scaffold]

Sorry my internet sucks :(((

[midnight haven]

nw

[midnight haven]

eh i mean sure but like you should think of 220,221 and 223 as being very far from 0

[lunar scaffold]

How'd it look?😭

What's the behavior of the graph?

Made a mistake

[midnight haven]

yeah

so now you can see the discontinuity more clearly right?

[lunar scaffold]

Yeah

[midnight haven]

so i guess that's it?

[pearl pond]

@lunar scaffold Has your question been resolved?

[lunar scaffold]

Yes

[pearl pond]

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[trim seal]

My task is to design a wine glass algebraically using only functions but I made mine by changing the values repeatedly. How can I predict where the lines intersect using solveN on my graphics calculator.

[naive dirge]

it looks like a frog's lower body.

rather than wine glass

[trim seal]

pleaseee help meee

[naive dirge]

I can only offer some artistic advices.

[trim seal]

i dont need artistc advide i need mathematical advice🙏

<@&286206848099549185> please

<@&286206848099549185>

[spark parcel]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[spark parcel]

haiyaa

[trim seal]

crodie please hell\p

help

[pearl pond]

@trim seal Has your question been resolved?

[latent quail]

Solve the function, could you give me some examples?

[pearl pond]

@trim seal Has your question been resolved?

[latent quail]

In that case, I could better explain the idea

[trim seal]

how do i explain the math behind an intercept

[latent quail]

There is a mutual coordinate between two functions

That’s called intersection

[trim seal]

if i want 2 poraobals to interscet at (2,3) how can i do that. what do i put into my calculator?

[latent quail]

What is poraobal?

[trim seal]

parabola

[latent quail]

It depends on in which spot do you want the two functions intersect at
If you want them to intersect at the vertex, just set both of them has the same vertex and manipulate the rest value to get what you’d like it to be

But if you want them to intersect at the vertex of parabola A and the none vertex on parabola B, you have no choice but to alter the value consecutively to get what you want

Is it helpful?

@trim seal

Ping me when you still have any confusion

I’ll be on other channels

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[dusky ferry]

Is this correct?

[timid spindle]

You can verify yourself by checking that f(f^-1(x)) = x (and the other way around too to double check)

[pearl pond]

@dusky ferry Has your question been resolved?

[gloomy scroll]

Quick question, is there such thing as a function f that satisfies f(f^-1(x))=x and f^-1(f(x)) != x or the other way around? I feel like one implies the other

[timid spindle]

Think about squaring and square rooting

[gloomy scroll]

Ah, right. Thanks :)

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[atomic tundra]

I am doing a math project where i need to show the use of maths in real life
from grade 10 and 11

[atomic tundra]

but i cant find any good examples

[latent quail]

Go to the convenient store

And ask the clerk that you need cos^2 theta+ sin^2 theta of bottles of coke

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[celest flower]

Does this logic look correct? If not, how would I fix it?

[pearl pond]

@celest flower Has your question been resolved?

[celest flower]

<@&286206848099549185>

Not the end of the world if I don't hear back, but feedback would be greatly appreciated. :)

[pearl pond]

@celest flower Has your question been resolved?

[tropic saddle]

why is that set at all related to gcd(a,b) being a divisor of c or not

[pearl pond]

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[celest flower]

.reopen

[pearl pond]

✅

[celest flower]

Could you clarify what you're asking please? Right now it doesn't feel like you're trying to help with your comment

[tropic saddle]

you started with gcd(a,b) not being a divisor of c

then you wrote down a set which apparently now doesnt contain c

and then you are suddenly done

why

what do these steps have to do with each other

[celest flower]

The set S that I wrote was given during lecture as a property of the gcd, so that's where that came from

Since the gcd(a,b) divides all linear combinations of a and b, if the set containing all those linear combinations doesn't contain c, then ax+by=c cannot have a solution.

[tropic saddle]

so you already know that gcd(a,b) divides all linear combinations

[celest flower]

Ye

[tropic saddle]

then why are you even trying to prove that step

clearly then gcd(a,b) | ax+by=c

[celest flower]

Oh ur right, but what about the iff?

Would that still cover it with just the linear combination property?

[tropic saddle]

write down exactly what you know about the set S

gcd(a,b) | ax+by is the easy direction because gcd(a,b) divides a and it divides b

[celest flower]

Aight, so I wouldn't need the set as long as I specify that gcd(a,b)|ax+by?

Since if ax+by=c and we know that property, then there's no way that there'd be a solution if the gcd doesn't divide c.

[tropic saddle]

well now we have proved "there is a solution" => "gcd divides c"

but you still need the other one

thats harder

[celest flower]

That was given, that's why I did a proof by contradiction

[tropic saddle]

.

[celest flower]

S is the set of all linear combinations of a and b such that x,y is in the set of integers. The smallest positive integer in the set is the gcd.

At least that's what my prof told us

[tropic saddle]

good

ok so lets say gcd divides c

[celest flower]

Mhm

[tropic saddle]

for example c=gcd(a,b)*k

[celest flower]

Ye following

[tropic saddle]

and you know that there is a solution ax+by=gcd(a,b)

[celest flower]

Yes

Oh wait

Lemme see if I can finish

So the solution c would then be (ax+by)*k, right?

I.e:
k(ax+by)=c

For the x and y that give the gcd(a,b)

[tropic saddle]

yes

or equivalently, a(kx)+b(ky)=c

[celest flower]

Not necessarily for any x and y, just the specific gcd ones

[tropic saddle]

and now you can rename kx and ky as for example x' and y'

and then ax'+by'=c

so the equation has a solution

[celest flower]

So I wouldn't just do a proof by contradiction, it'd be a proof by cases right? One for disproving when gcd doesn't divide c and one for when it does?

[tropic saddle]

you do both directions by a direct proof

[celest flower]

Hm, if I had to do just one proof instead of two direct proofs, would a proof by cases work? My prof is very picky about us being as concise as possible.

[tropic saddle]

how is doing two cases better than two direct proofs

this is pretty much as concise as possible

[celest flower]

Idk, the guy grades more on how concise the work is than correctness, which feels odd to me for a formal proof. He took points off one time because I put too many words

[tropic saddle]

well proofs can be too long

but this one isnt

if you wrote down what we discussed it would fit into like 4 lines

[celest flower]

Meaning like, I explained "dividing by ___" instead of just doing it

I think I can see what you mean tho

So like:
gcd(a,b)|c:
ax+by = gcd
=> c = k(gcd)=k(ax+by)
=> c= a(kx) +b(ky)
QED
gcd [doesnt divide] c:
gcd=ax+by
Thus, ax+by=c cannot have a solution.
QED

Sumn like that?

Obv cleaned up just a bit

[tropic saddle]

you dont need to phrase the second part as a contradiction

just directly write gcd(a,b) | ax+by = c

[celest flower]

So like just that one line?

[tropic saddle]

well you can also write "because gcd(a,b) | a and gcd(a,b) | b"

I hope that you have such basic results as x|y and x|z implying x|y+z

otherwise you would also need to prove that

[celest flower]

Oh yeah, that's one of the lemmas we've covered

That's on the exam, but it's more of a "prove you know why" than a "I expect you to prove this every time"

How would that be more concise tho?

Also I have an appt soon, could we continue a bit more in DM later so I can open the channel for sumn else?

[tropic saddle]

well you introduce an unnecessary assumption that gcd does not divide c

[celest flower]

Totally fine if not tho

[tropic saddle]

then you show that gcd does divide c

and then you conclude a contradiction

just cut out the outer parts

[celest flower]

It's if and only if, so disproving is necessary

[tropic saddle]

no

thats not how iff works

your proof currently is:
need to show gcd(a,b) | c
assume gcd(a,b) does not divide c
show that gcd(a,b) | ax+by = c, so gcd(a,b) | c
conclude contradiction to assumption, therefore gcd(a,b) does divide c

the second and fourth line are completely pointless

[celest flower]

Really gotta go now, up to you

.close

[pearl pond]

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[tropic saddle]

I have to go aswell

[pearl pond]

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[stray stump]

Given a boolean function that's a product of sums, how do I implement it using NOR gates?
For example, does the following circuit represent this function?
change last 1 to 0
f = ( k' + l ) ( k + m + n' ) ( o )

assume gate can take as many inputs as needed and inputs are available in true and complement forms

[midnight haven]

are you given a product of sums function and u r asked to implement it with NORs is this what you are trying to find just to be clear

[stray stump]

Yep, the function looks almost like the one I've provided

And I'm asked to implement it using NOR gates only

[midnight haven]

well i guess just like use the fact that
xy = (x'+y')'
for each product

so with the function f you provided, ill shorten the triple product to be xyz. You can just use

xyz = (x'+y'+z')'

by negating the individual inputs, so for example , (k' +l)' = kl' now you need to apply the same process for kl' again

[stray stump]

so the way im doing it is correct?

because im getting the same result

[pearl pond]

@stray stump Has your question been resolved?

[pearl pond]

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[bleak bramble]

Need help with this review

[bleak bramble]

Ping me if you can help me

[sudden arch]

!status

[pearl pond]

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

[rigid gale]

Ever heard of sine law? sin(ß)/b=sin(y)/c...

[bleak bramble]

Yea I think so

I’ve heard of it idk what it means

[rigid gale]

It means that in every triangle the quotient of the sin of an angle and the length of the opposite side is the same for every pair of angle and side

[bleak bramble]

So how would I apply that to the problem

[rigid gale]

Well, you are given an angle and an opposite side. You are given another of the sides as well and you want to finde the opposite angle. Look at the lil equation i gave you. out of 4 variables, only one is unknown. Sub them in, and then just solve for sin(y). Apply arcsin and you got your sol.

[bleak bramble]

So the measurement for angle c would be 45 if I’m rounding to the nearest whole degree?

[rigid gale]

Mhhm, im getting 37°. Its: sin(55)/55=sin(y)/40
-> sin(y)=40/55*sin(55°) -> y≈37° (36.56..)

[bleak bramble]

Ohhh that makes sense thank you

@rigid gale So how would I solve for triangle abc would it be the same equation

[rigid gale]

Yeah, just multiple applications

[bleak bramble]

So how would that look like

@rigid gale

I’m not sure how to do multiple applications

[rigid gale]

You can solve for side c immediatly with sine law. For A, just use that A+B+C=180. Then use it again to solve for a.

[bleak bramble]

Oh okay

So would the sides of ABC A=13 B=17 and C=18 @rigid gale

[pearl pond]

@bleak bramble Has your question been resolved?

[bleak bramble]

<@&286206848099549185>

[dusky ferry]

alright ty

[pearl pond]

@bleak bramble Has your question been resolved?

[bleak bramble]

I need help with doing this review

[pearl pond]

@bleak bramble Has your question been resolved?

[pearl pond]

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[tepid panther]

I got 4pi

[tepid panther]

Should I just leave my answer as 4pi or should I round to the nearest whole number?

B = pir^2

B = 25pi

v = 1/3bh

100pi = 1/3 25pi h

ohh I see where I messed up on

.closw

.close

[pearl pond]

Channel closed

Closed by @tepid panther

Use .reopen if this was a mistake.

[tepid panther]

.reopen

[pearl pond]

✅

[tepid panther]

It marked it as wrong.

If the diameter is 10, the r is 5 right?

[pearl pond]

@tepid panther Has your question been resolved?

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[kindred kernel]

Pls help

[sharp karma]

Okay let's define what cotangent is first yeah?

in terms of our trigonometric functions, what is cotangent (using sine and cosine)

[kindred kernel]

its a ratio

[sharp karma]

yes

can we define it as a ratio between sine and cosine?

like how we know

[kindred kernel]

yes

im so lost

[sharp karma]

$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$

[jolly parrot]

Oreo

[sharp karma]

that's fine

we can figure this out

[kindred kernel]

Yes we can

[sharp karma]

so

if we can define it

do you know how to?

if not, that's okay

[kindred kernel]

no

[wild otter]

What is the y intercept of y = x

[kindred kernel]

i dont

[sharp karma]

okay

[wild otter]

I’ve been thinking it might be 0

[sharp karma]

please open another help ticket

[wild otter]

But my teacher says 1

[sharp karma]

but yes it 0

[wild otter]

I don’t get it

[wild otter]

Why

[kindred kernel]

..

[wild otter]

Oop

[sharp karma]

sorry, we can get back

[wild otter]

Sorry

[kindred kernel]

Yes

[wild otter]

Sorry I’m new

My bad

[sharp karma]

so cotangent is also

[wild otter]

14*

[sharp karma]

the reciprocal of tangent

[kindred kernel]

Oh

[sharp karma]

you're good

[kindred kernel]

so its 0?